Fluid Mechanics, 6th Ed.

Kundu, Cohen, and Dowling

Exercise 1.1. Many centuries ago, a mariner poured 100 cm3 of water into the ocean. As time
passed, the action of currents, tides, and weather mixed the liquid uniformly throughout the
earth’s oceans, lakes, and rivers. Ignoring salinity, estimate the probability that the next sip (5
ml) of water you drink will contain at least one water molecule that was dumped by the mariner.
Assess your chances of ever drinking truly pristine water. (Consider the following facts: Mw for
water is 18.0 kg per kg-mole, the radius of the earth is 6370 km, the mean depth of the oceans is
approximately 3.8 km and they cover 71% of the surface of the earth.)

Solution 1.1. To get started, first list or determine the volumes involved:
υd = volume of water dumped = 100 cm3, υc = volume of a sip = 5 cm3, and
V = volume of water in the oceans = 4 πR 2 Dγ ,
where, R is the radius of the earth, D is the mean depth of the oceans, and γ is the oceans'
coverage fraction. Here we've ignored the ocean volume occupied by salt and have assumed that
the oceans' depth is small compared to the earth's diameter. Putting in the numbers produces:
€
V = 4 π (6.37 ×10 6 m) 2 (3.8 ×10 3 m)(0.71) = 1.376 ×1018 m 3 .
For well-mixed oceans, the probability Po that any water molecule in the ocean came from the
dumped water is:
(100 cm3 of water) υ d 1.0 ×10−4 m 3
€ Po = = = 18 3
= 7.27 ×10−23 ,
(oceans' volume) V 1.376 ×10 m
Denote the probability that at least one molecule from the dumped water is part of your next sip
as P1 (this is the answer to the question). Without a lot of combinatorial analysis, P1 is not easy
to calculate directly. It is easier to proceed by determining the probability P2 that all the
molecules€ in your cup are not from the dumped water. With these definitions, P can be
1
determined from: P1 = 1 – P2. Here, we can calculate P2 from:
P2 = (the probability that a molecule was not in the dumped water)[number of molecules in a sip].
The number of molecules, Nc, in one sip of water is (approximately)
1.00g gmole molecules
N c = 5cm 3 × 3
× × 6.023×10 23 = 1.673×10 23 molecules
cm 18.0g gmole
23
Thus, P2 = (1− Po ) Nc = (1− 7.27 ×10 −23 )1.673×10 . Unfortunately, electronic calculators and modern
computer math programs cannot evaluate this expression, so analytical techniques are required.
First, take the natural log of both sides, i.e.
ln(P2 ) = N c ln(1− Po ) = 1.673×10 23 ln(1− 7.27 ×10 −23 )
then expand the natural logarithm using ln(1–ε) ≈ –ε (the first term of a standard Taylor series
for ε → 0 )
ln(P2 ) ≅ −N c ⋅ Po = −1.673×10 23 ⋅ 7.27 ×10 −23 = −12.16 ,
and exponentiate to find:
€ P2 ≅ e−12.16 ≅ 5 ×10 −6 ... (!)
Therefore, P1 = 1 – P2 is very close to unity, so there is a virtual certainty that the next sip of
water you drink will have at least one molecule in it from the 100 cm3 of water dumped many
years ago. So, if one considers the rate at which they themselves and everyone else on the planet
uses water it is essentially impossible to enjoy a truly fresh sip.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.2. An adult human expels approximately 500 ml of air with each breath during
ordinary breathing. Imagining that two people exchanged greetings (one breath each) many
centuries ago, and that their breath subsequently has been mixed uniformly throughout the
atmosphere, estimate the probability that the next breath you take will contain at least one air
molecule from that age-old verbal exchange. Assess your chances of ever getting a truly fresh
breath of air. For this problem, assume that air is composed of identical molecules having Mw =
29.0 kg per kg-mole and that the average atmospheric pressure on the surface of the earth is 100
kPa. Use 6370 km for the radius of the earth and 1.20 kg/m3 for the density of air at room
temperature and pressure.

Solution 1.2. To get started, first determine the masses involved.

m = mass of air in one breath = density x volume = (1.20kg /m 3 )(0.5 ×10−3 m 3 ) = 0.60 ×10−3 kg
∞
2
M = mass of air in the atmosphere = 4 πR ∫ ρ(z)dz
z= 0
Here, R is the radius of the earth, z is the elevation
€ above the surface of€the earth, and ρ(z) is the
air density as function of elevation. From the law for static pressure in a gravitational field,
z= +∞
€
dP dz = −ρg , the surface pressure, Ps, on the earth is determined from Ps − P∞ = ∫ ρ(z)gdz so
z= 0
Ps − P∞ (10 5 Pa)
that: M = 4π R 2 = 4π (6.37 ×10 6 m)2 −2
= 5.2 ×1018 kg .
g 9.81ms
€
where the pressure (vacuum) in outer space = P∞ = 0, and g is€assumed constant throughout the
atmosphere. For a well-mixed atmosphere, the probability Po that any molecule in the
atmosphere came from the age-old verbal exchange is
2 × (mass of one breath) 2m 1.2 ×10−3 kg
Po = = = 18
= 2.31×10−22 ,
(mass of the whole atmosphere) M 5.2 ×10 kg
where the factor of two comes from one breath for each person. Denote the probability that at
least one molecule from the age-old verbal exchange is part of your next breath as P1 (this is the
answer to the question). Without a lot of combinatorial analysis, P1 is not easy to calculate
€
directly. It is easier to proceed by determining the probability P2 that all the molecules in your
next breath are not from the age-old verbal exchange. With these definitions, P1 can be
determined from: P1 = 1 – P2. Here, we can calculate P2 from:
P2 = (the probability that a molecule was not in the verbal exchange)[number of molecules in one breath].
The number of molecules, Nb, involved in one breath is
0.6 ×10−3 kg 10 3 g molecules
Nb = × × 6.023 ×10 23 = 1.25 ×10 22 molecules
29.0g /gmole kg gmole
Nb −22 1.25×10 22
Thus, P2 = (1− Po ) = (1− 2.31×10 ) . Unfortunately, electronic calculators and modern
computer math programs cannot evaluate this expression, so analytical techniques are required.
First,
€ take the natural log of both sides, i.e.
ln(P2 ) = N b ln(1− Po ) = 1.25 ×10 22 ln(1− 2.31×10−22 )
€
then expand the natural logarithm using ln(1–ε) ≈ –ε (the first term of a standard Taylor series
for ε → 0 )
ln(P2 ) ≅ −N b ⋅ Po = −1.25 ×10 22 ⋅ 2.31×10−22 = −2.89 ,
€
and exponentiate to find:
€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

P2 ≅ e−2.89 = 0.056 .
Therefore, P1 = 1 – P2 = 0.944 so there is a better than 94% chance that the next breath you take
will have at least one molecule in it from the age-old verbal exchange. So, if one considers how
often they themselves and everyone else breathes, it is essentially impossible to get a breath of
truly fresh air. €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.3. The Maxwell probability distribution, f(v) = f(v1,v2,v3), of molecular velocities in a
gas flow at a point in space with average velocity u is given by (1.1).
a) Verify that u is the average molecular velocity, and determine the standard deviations (σ1,
12
#1 2 3
&
σ2, σ3) of each component of u using σ i = % ∫∫∫ (vi − ui ) f (v)d v( for i = 1, 2, and 3.
$ n all v '
b) Using (1.27) or (1.28), determine n = N/V at room temperature T = 295 K and atmospheric
pressure p = 101.3 kPa.
c) Determine N = nV = number of molecules in volumes V = (10 µm)3, 1 µm3, and (0.1 µm)3.
d) For the ith velocity component, the standard deviation of the average, σa,i, over N molecules
is σa,i = σ i N when N >> 1. For an airflow at u = (1.0 ms–1, 0, 0), compute the relative
uncertainty, 2σ a,1 u1 , at the 95% confidence level for the average velocity for the three volumes
listed in part c).
e) For the conditions specified in parts b) and d), what is the smallest volume of gas that ensures
a relative uncertainty in U of one percent or less?

Solution 1.3. a) Use the given distribution, and the definition of an average:
32
1 3
" m % +∞ +∞ +∞ * m 2-
(v)ave = ∫∫∫ v f (v)d v = $ ' ∫ ∫ ∫ v exp +− v − u . d 3v .
n all u # 2π kBT & −∞ −∞ −∞ , 2kBT /
Consider the first component of v, and separate out the integrations in the "2" and "3" directions.
32
! m $ +∞ +∞ +∞ . m 1
(v1 )ave = # & ∫ ∫ ∫ v1 exp /− *+(v1 − u1 )2 + (v2 − u2 )2 + (v3 − u3 )2 ,-2 dv1dv2 dv3
" 2π kBT % −∞ −∞ −∞ 0 2kBT 3
3 2 +∞
! m $ * m(v1 − u1 )2 - +∞ * m(v2 − u2 )2 - +∞ * m(v3 − u3 )2 -
=# &
" 2 π k BT %
∫ 1 , 2k T / 1 ∫ , 2k T / 2 ∫ +,− 2k T ./dv3
v exp + − . dv exp + − . dv exp
−∞ B −∞ B −∞ B
12
The integrations in the "2" and "3" directions are equal to: (2πk B T m) , so
1 2 +∞
! m $ * m(v1 − u1 )2 -
(v1 )ave = # &
" 2 π k BT %
∫ v exp +,−
1
2kBT /
. dv1
−∞
€ 12
The change of integration variable to β = (v1 − u1 ) ( m 2kBT ) changes this integral to:
1 ! ! 2k T $1 2
+∞ $ 1
(v1 )ave = ∫ ## β #" mB &% + u1 && exp {−β 2 } d β = 0 + π u1 π = u1 ,
π
−∞ " %
where the first term of the integrand is an odd function integrated on an even interval so its
contribution is zero. This procedure is readily repeated for the other directions to find (v2)ave = u2,
and (v3)ave = u3. Thus, u = (u1, u2, u3) is the average molecular velocity.
Using the same simplifications and change of integration variables produces:
32
! m $ +∞ +∞ +∞ . m 1
2
σ1 = # 2
& ∫ ∫ ∫ (v1 − u1 ) exp /− *+(v1 − u1 )2 + (v2 − u2 )2 + (v3 − u3 )2 ,-2 dv1dv2 dv3
" 2π kBT % −∞ −∞ −∞ 0 2kBT 3
1 2 +∞
! m $ * m(v1 − u1 )2 - 1 ! 2kBT $ +∞ 2
=# &
" 2 π k BT %
∫ (v1 − u1 ) exp +,− 2k T ./ dv1 = π #" m &% ∫ β exp {−β 2 } d β .
2

±∞ B ±∞
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

The final integral over β is: π 2 , so the standard deviations of molecular speed are
12
σ1 = ( k B T m) = σ 2 = σ 3 ,
where the second two equalities follow from repeating this calculation for the second and third
directions. €
b) From (1.27), n V = p kBT = (101.3kPa) [1.381×10 −23 J / K ⋅ 295K ] = 2.487 ×10 25 m −3
€
c) From n/V from part b): n = 2.487 ×1010 for V = 103 µm3 = 10–15 m3
n = 2.487 ×10 7 for V = 1.0 µm3 = 10–18 m3
n = 2.487 ×10 4 for V = 0.001 µm3 = 10–21 m3
d) From (1.29), the gas constant is R = (kB/m), and R = 287 m2/s2K for air. Compute:
€ 12 12 12
2σ a,1 u1 = 2 ( kBT m€ n ) [1m / s ] = 2 ( RT n ) 1m / s = 2 ( 287 ⋅ 295 n ) = 582 n . Thus,
for V = 10–15 m3 € : 2σ a,1 u1 = 0.00369,
–18 3
V = 10 m : 2σ a,1 u1 = 0.117, and
V = 10–21 m3 : 2σ a,1 u1 = 3.69.
e) To achieve a relative uncertainty of 1% we need n ≈ (582/0.01)2 = 3.39 × 109, and this
corresponds to a volume of 1.36 × 10-16 m3 which is a cube with side dimension ≈ 5 µm.

€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.4. Using the Maxwell molecular speed distribution given by (1.4),
a) determine the most probable molecular speed,
b) show that the average molecular speed is as given in (1.5),
12
#1 ∞ &
c) determine the root-mean square molecular speed = vrms = % ∫ 0 v 2 f (v)dv( ,
$n '
d) and compare the results from parts a), b) and c) with c = speed of sound in a perfect gas
under the same conditions.

Solution 1.4. a) The most probable speed, vmp, occurs where f(v) is maximum. Thus, differentiate
(1.4) with respect v, set this derivative equal to zero, and solve for vmp. Start from:
32
! m $ 2 ( mv 2 +
f (v) = 4π n # & v exp )− , , and differentiate
" 2 π k BT % * 2kBT -
32
df ! m $ / (* mvmp2 ,
* mvmp 3 (* mv 2 ,*2
= 4π n # & 12vmp exp )− -− exp )− mp -4 = 0
dv " 2π kBT % 10 +* 2kBT .* kBT +* 2kBT .*43
Divide out common factors to find:
mv 2 2kBT
2 − mp = 0 or vmp = .
k BT m
b) From (1.5), the average molecular speed v is given by:
32
1∞ # m & ∞ 3 * mv 2 -
v = ∫ v f (v)dv = 4π % ( ∫ v exp +− . dv .
n 0 $ 2 π k BT ' 0 , 2kBT /
Change the integration variable to β = mv 2 2kBT to simplify the integral:
12 12 12
! m $ k BT ∞ ! 8kBT $ −β ∞ ! 8kBT $
v = 4# &
" 2 π k BT % m 0
∫ β exp {−β } d β = #" π m &% (−β e − e )0 = #" π m &% ,
−β

and this matches the result provided in (1.5).

c) The root-mean-square molecular speed vrms is given by:
32
2 1∞ 2 # m & ∞ 4 * mv 2 -
vrms = ∫ v f (v)dv = 4π % ( ∫ v exp +− . dv .
n 0 $ 2 π k BT ' 0 , 2kBT /
12
Change the integration variable to β = v ( m 2kBT ) to simplify the integral:
12 ∞
2 4 ! 2kBT $ 4 2 4 ! 2kBT $ 3 π 3kBT
vrms = # & ∫ β exp {−β } d β = # & = .
π" m % 0 π" m % 8 m
Thus, vrms = (3kBT/m)1/2.
d) From (1.28), R = (kB/m) so vmp = 2RT , v = (8 / π )RT , and vrms = 3RT . All three speeds
have the same temperature dependence the speed of sound in a perfect gas: c = γRT , but are
factors of 2γ, 8 πγ and 3 γ , respectively, larger than c.

€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.5. By considering the volume swept out by a moving molecule, estimate how the
mean-free path, l, depends on the average molecular cross section dimension d and the
molecular number density n for nominally spherical molecules. Find a formula for ln1 3 (the ratio
of the mean-free path to the mean intermolecular spacing) in terms of the nominal molecular
volume ( d 3 ) and the available volume per molecule (1/n). Is this ratio typically bigger or smaller
than one?

Solution 1.5. The combined collision cross section for two spherical molecules having diameter
d is πd 2 . The mean free path l is the average distance traveled by a molecule between
collisions. Thus, the average molecule should experience one
collision when sweeping a volume equal to πd 2 l . If the molecular
number density is n, then the volume per molecule is n–1, and the
€ € mean intermolecular spacing is n–1/3. Assuming that the swept volume d
necessary to produce one collision is proportional to the volume per
molecule produces: €
π d l = C n or l = C ( nπ d 2 ) ,
2
€
where C is a dimensionless constant presumed to be of order unity. The dimensionless version of
this equation is:
mean free path l
= −1 3 = l n1 3
mean intermolecular spacing n
23 23
C C " n −1 % " volume per molecule %
= 23 2 =
3 23
= C $ 3' = C$ ' ,
n πd ( )
nd #d & # molecular volume &

where all numerical constants like π have been combined into C. Under ordinary conditions in
gases, the molecules are not tightly packed so l >> n −1 3 . In liquids, the molecules are tightly
packed so l ~ n −1 3 .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.6. Compute the average relative speed, vr , between molecules in a gas using the
Maxwell speed distribution f given by (1.4) via the following steps.
a) If u and v are the velocities of two molecules then their relative velocity is: vr = u – v. If the
angle between u and v is θ, show that the relative speed is: vr = |vr| = u 2 + v 2 − 2uv cosθ where
u = |u|, and v = |v|.
b) The averaging of vr necessary to determine vr must include all possible values of the two
speeds (u and v) and all possible angles θ. Therefore, start from:
1
vr = 2 ∫ vr f (u) f (v)sin θ dθ dvdu ,
2n all u,v,θ
and note that vr is unchanged by exchange of u and v, to reach:
1 ∞ ∞ π
∫ ∫ ∫ u2 + v2 − 2uv cosθ sinθ f (u) f (v)dθ dvdu
vr =
n 2 u=0 v=u θ =0
c) Note that vr must always be positive and perform the integrations, starting with the angular
one, to find:
12
1 ∞ ∞ 2u3 + 6uv 2 # 16k T &
vr = 2 ∫ ∫ f (u) f (v)dvdu = % B ( = 2 v .
3n u=0 v=u uv $ π '

Solution 1.6. a) Compute the dot produce of vr with itself:

2
v r = v r ⋅ v r = (u − v)⋅ (u − v) = u ⋅ u − 2u ⋅ v + v ⋅ v = u 2 − 2uv cosθ + v 2 .
Take the square root to find: |vr| = u 2 + v 2 − 2uv cosθ .
b) The average relative speed must account for all possible molecular speeds and all possible
angles between the two molecules. [The coefficient 1/2 appears in the first equality below
because the probability density function of for the angle θ in the interval 0 ≤ θ ≤ π is (1/2)sinθ.]
1
vr = 2 ∫ vr f (u) f (v)sin θ dθ dvdu
2n all u,v,θ
1
= ∫ u 2 + v 2 − 2uv cosθ f (u) f (v)sin θ dθ dvdu
2n 2 all u,v,θ

1 ∞ ∞ π
= 2 ∫ ∫ ∫ u 2 + v 2 − 2uv cosθ sin θ f (u) f (v)dθ dvdu.
2n u=0 v=0 θ =0
In u-v coordinates, the integration domain covers the first
quadrant, and the integrand is unchanged when u and v are
v! dv!
v!

swapped. Thus, the u-v integration can be completed above the

=

line u = v if the final result is doubled. Thus,

u

1 ∞ ∞ π
vr = 2 ∫ ∫ ∫ u 2 + v 2 − 2uv cosθ sin θ f (u) f (v)dθ dvdu .
n u=0 v=u θ =0
Now tackle the angular integration, by setting
β = u 2 + v 2 − 2uv cosθ so that d β = +2uvsin θ dθ . This leads to du!
2
1 ∞ ∞ (v+u) dβ u!
vr = 2 ∫ ∫ ∫ β 1 2 f (u) f (v)dvdu ,
n u=0 v=u β =(v−u)2 2uv
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

and the β-integration can be performed:

1 ∞ ∞ 2 3 2 (v+u)2 f (u) f (v) 1 ∞ ∞ f (u) f (v)
vr = 2 ∫ ∫ β
2n u=0 v=u 3
( )
(v−u) 2
u v 3n u=0 v=u
(
dvdu = 2 ∫ ∫ (v + u)3 − (v − u)3
u v
)
dvdu .

Expand the cubic terms, simplify the integrand, and prepare to evaluate the v-integration:
1 ∞ ∞ f (v) f (u)
3n u=0 v=u
(
vr = 2 ∫ ∫ 2u3 + 6uv 2
v
dv
u
)du
32
1 ∞ ∞ 2 4π
# m & 2 * mv 2 - f (u)
= ∫ ∫ (2u + 6uv ) v %$ 2π k T (' v exp +,− 2k T ./ dv u du.
3n u=0 v=u
3

B B

Use the variable substitution: α = mv2/2kBT so that dα = mvdv/kBT, which reduces the v-
integration to:
32
1 ∞ ∞ ! 3 k BT $ ! m $ kT f (u)
vr = ∫ ∫ # 2u + 6u
3n u=0 mu2 kBT "
α &4 π # & exp {−α } B dα
m % " 2 π k BT % m u
du

12 ∞
2! m $ ∞
! 3 k T $ −α f (u)
= # &
3n " 2π kBT %
∫ ∫ # 2u + 6u B α &e dα
" m % u
du
u=0 mu2 kBT
12 ∞ ∞
2! m $ ! kT $ f (u)
= # &
3n " 2π kBT %
∫ #"−2u3e−α + 6u mB (−α e−α − e−α )&% 2 u du
u=0 mu kBT
12
2! m $ ∞! 3 k BT $ ! mu 2 $ f (u)
= # & ∫# 8u +12u & exp #− & du.
3n " 2π kBT % u=0 " m % " 2kBT % u
The final u-integration may be completed by substituting in for f(u) and using the variable
12
substitution γ = u ( m kBT ) .
12 ∞ 32
1 ! 2m $ ! ! k T $3 2 ! k BT $ $ ! m $ ! k BT $
32

vr = # &
3 " π k BT %
∫ ##8#" m &% γ +12 #" m &% γ &&4π #" 2π k T &% #" m &%γ exp (−γ 2 ) dγ
B 3

0 " % B
12 ∞ 12
2 !k T $ 2 !k T $ ! 3 1 $
= # B &
3π " m %
∫ (8γ +12γ ) exp (−γ ) dγ = 3π #" mB &% #"8 8 π +12 4 π &%
4 2 2

0
12 12
4 ! k BT $ ! 16k T $
= # & = # B & = 2v
π" m % " πm %
Here, v is the mean molecular speed from (1.5).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.7. In a gas, the molecular momentum flux (MFij) in the j-coordinate direction that
crosses a flat surface of unit area with coordinate normal direction i is:
1
MFij = ∫∫∫ mvi v j f (v)d 3v where f(v) is the Maxwell velocity distribution (1.1). For a perfect
V all v
gas that is not moving on average (i.e., u = 0), show that MFij = p (the pressure), when i = j, and
that MFij = 0, when i ≠ j.

Solution 1.7. Start from the given equation using the Maxwell distribution:
32
1 nm " m % +∞ +∞ +∞ * m -
MFij = ∫∫∫ mvi v j f (v)d v =
V all u
3
$ ' ∫ ∫ ∫ vi v j exp +−
V # 2π kBT & −∞ −∞ −∞ , 2kBT
( v12 + v22 + v32 ). dv1dv2 dv3
/
and first consider i = j = 1, and recognize ρ = nm/V as the gas density, as in (1.28).
32
! m $ +∞ +∞ +∞ 2 * m -
MF11 = ρ # & ∫ ∫ ∫ u1 exp +−
" 2π kBT % −∞ −∞ −∞ , 2kBT
( v12 + v22 + v32 ). dv1dv2 dv3
/
3 2 +∞
! m $ 2
( mv12 + +∞ ( mv22 + +∞ ( mv32 +
= ρ# &
" 2 π k BT %
∫v 1 exp )− , dv1 ∫ exp )−
* 2kBT - −∞
, dv2 ∫ exp )−
* 2kBT -
,dv3
* 2kBT -
−∞ −∞
32
The first integral is equal to ( 2kBT m ) ( )
π 2 while the second two integrals are each equal to
12
( 2 π k BT m ) . Thus:
32
! m $ ! 2kBT $3 2 π ! 2π kBT $1 2 ! 2π kBT $1 2 kT
MF11 = ρ # & # & # & # & = ρ B = ρ RT = p
" 2 π k BT % " m % 2 " m % " m % m
where kB/m = R from (1.28). This analysis may be repeated with i = j = 2, and i = j = 3 to find:
MF22 = MF33 = p, as well.
Now consider the case i ≠ j. First note that MFij = MFji because the velocity product under
the triple integral may be written in either order vivj = vjvi, so there are only three cases of
interest. Start with i = 1, and j = 2 to find:
32
! m $ +∞ +∞ +∞ * m -
MF12 = ρ # & ∫ ∫ ∫ v1v2 exp +−
" 2π kBT % −∞ −∞ −∞ , 2kBT
( v12 + v22 + v32 ). dv1dv2 dv3
/
32
! m $ +∞ ( mv12 + +∞ ( mv22 + +∞ ( mv32 +
= ρ# & ∫ v1 exp )− , dv1 ∫ v2 exp )− , dv2 ∫ exp )− ,dv3
" 2π kBT % −∞ * 2kBT - −∞ * 2kBT - −∞ * 2kBT -
Here we need only consider the first integral. The integrand of this integral is an odd function
because it is product of an odd function, v1, and an even function, exp {−mv12 2kBT } . The
integral of an odd function on an even interval [–∞,+∞] is zero, so MF12 = 0. And, this analysis
may be repeated for i = 1 and j = 3, and i = 2 and j = 3 to find MF13 = MF23 = 0.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.8. Consider the viscous flow in a channel of width 2b. The channel is aligned in the
x-direction, and the velocity u in the x-direction at a distance y from the channel centerline is
[ 2
]
given by the parabolic distribution u(y) = U 0 1− ( y b) . Calculate the shear stress τ as a
function y, µ, b, and Uo. What is the shear stress at y = 0?

du d * $ y '2- y
€
Solution 1.8. Start from (1.3): τ = µ = µ U o ,1− & ) / = –2µU o 2 . At y = 0 (the location of
dy dy + % b ( . b
maximum velocity) τ = 0. At At y = ±b (the locations of zero velocity), τ = 2µ U o b .

€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.9. Hydroplaning occurs on wet roadways when sudden braking causes a moving
vehicle’s tires to stop turning when the tires are separated from the road surface by a thin film of
water. When hydroplaning occurs the vehicle may slide a significant distance before the film
breaks down and the tires again contact the road. For simplicity, consider a hypothetical version
of this scenario where the water film is somehow maintained until the vehicle comes to rest.
a) Develop a formula for the friction force delivered to a vehicle of mass M and tire-contact area
A that is moving at speed u on a water film with constant thickness h and viscosity µ.
b) Using Newton’s second law, derive a formula for the hypothetical sliding distance D traveled
by a vehicle that started hydroplaning at speed Uo
c) Evaluate this hypothetical distance for M = 1200 kg, A = 0.1 m2, Uo = 20 m/s, h = 0.1 mm, and
µ = 0.001 kgm–1s–1. Compare this to the dry-pavement stopping distance assuming a tire-road
coefficient of kinetic friction of 0.8.

Solution 1.9. a) Assume that viscous friction from the water layer transmitted to the tires is the
only force on the sliding vehicle. Here viscous shear stress at any time will be µu(t)/h, where u(t)
is the vehicle's speed. Thus, the friction force will be Aµu(t)/h.
du u
b) The friction force will oppose the motion so Newton’s second law implies: M = −Aµ .
dt h
This equation is readily integrated to find an exponential solution: u(t) = Uo exp (−Aµt Mh ) ,
where the initial condition, u(0) = Uo, has been used to evaluate the constant of integration. The
distance traveled at time t can be found from integrating the velocity:
t t
x(t) = ∫ u(t!)dt! = Uo ∫ exp (−Aµt! Mh ) dt! = (Uo Mh Aµ )$%1− exp (−Aµt Mh )&' .
o o
The total sliding distance occurs for large times where the exponential term will be negligible so:
D = Uo Mh Aµ
c) For M = 1200 kg, A = 0.1 m , Uo = 20 m/s, h = 0.1 mm, and µ = 0.001 kgm–1s–1, the stopping
2

distance is: D = (20)(1200)(10–4)/(0.1)(0.001) = 24 km! This is an impressively long distance and

highlights the dangers of driving quickly on water covered roads.
For comparison, the friction force on dry pavement will be –0.8Mg, which leads to a
vehicle velocity of: u(t) = Uo − 0.8gt , and a distance traveled of x(t) = Uot − 0.4gt 2 . The vehicle
stops when u = 0, and this occurs at t = Uo/(0.8g), so the stopping distance is
2
! Uo $ ! Uo $ Uo2
D = Uo # & − 0.4g # & = ,
" 0.8g % " 0.8g % 1.6g
which is equal to 25.5 m for the conditions given. (This is nearly three orders of magnitude less
than the estimated stopping distance for hydroplaning.)
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.10. Estimate the height to which water at 20°C will rise in a capillary glass tube 3
mm in diameter that is exposed to the atmosphere. For water in contact with glass the contact
angle is nearly 0°. At 20°C, the surface tension of a water-air interface is σ = 0.073 N/m.

Solution 1.10. Start from the result of Example 1.4.

2σ cos α 2(0.073N / m)cos(0°)
h= = = 9.92mm
ρ gR (10 kg / m 3 )(9.81m / s 2 )(1.5 ×10 −3 m)
3
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.11. A manometer is a U-shaped tube containing mercury of density ρm. Manometers
are used as pressure-measuring devices. If the fluid in tank A has a pressure p and density ρ, then
show that the gauge pressure in the tank is: p − patm = ρmgh − ρga. Note that the last term on the
right side is negligible if ρ « ρm. (Hint: Equate the pressures at X and Y.)

Solution 1.9. Start by equating the pressures at X and Y.

pX = p + ρga = patm + ρmgh = pY.
Rearrange to find:
p – patm = ρmgh – ρga.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.12. Prove that if e(T, υ) = e(T) only and if h(T, p) = h(T) only, then the (thermal)
equation of state is (1.28) or pυ = kT, where k is constant.

Solution 1.12. Start with the first equation of (1.24): de = Tds – pdυ, and rearrange it:
1 p $ ∂s ' $ ∂s '
ds = de + dυ = & ) de + & ) dυ ,
T T % ∂e (υ % ∂υ ( e
where the second equality holds assuming the entropy depends on e and υ. Here we see that:
1 # ∂s & p $ ∂s '
= % ( , and = & ) .
€ T $ ∂e 'υ T % ∂υ ( e
$ ∂ $ ∂s ' ' $ ∂ $ ∂ s ' '
Equality of the crossed second derivatives of s, & & ) ) = & & ) ) , implies:
% ∂υ % ∂e (υ ( e % ∂e % ∂υ ( e (υ
€ $ ∂ (1€T ) ' $ ∂ ( p T ) '
& ) =& ) .
% ∂υ ( e % ∂e (υ
€ $ ∂ (1 T ) ' $ ∂ (1 T ) '
However, if e depends only on T, then (∂/∂υ)e = (∂/∂υ)T, thus & ) =& ) = 0 , so
% ∂υ ( e % ∂υ (T
# ∂( p T) & €
% ( = 0 , which can be integrated to find: p/T = f1(υ), where f1 is an undetermined function.
$ ∂e 'υ
Now repeat this procedure using the second equation € of (1.24), dh = Tds + υdp.
1 υ % ∂s ( % ∂s (
ds = dh − dp = ' * dh + ' * dp .
€ T T & ∂h ) p & ∂p ) h
1 # ∂s & υ % ∂s (
Here equality of the coefficients of the differentials implies: = % ( , and − = ' * .
T $ ∂h ' p T & ∂p ) h
€ # ∂ (1 T ) & # ∂ (υ T ) &
So, equality of the crossed second derivatives implies: % ( = −% ( . Yet, if h depends
$ ∂p ' h $ ∂h ' p
# ∂ (1 T ) & €# ∂ (1 T ) & € % ∂ (υ T ) (
only on T, then (∂/∂p)h = (∂/∂p)T, thus % ( =% ( = 0 , so −' * = 0 , which can
$ ∂p ' h $ ∂p 'T & ∂h ) p
be integrated to find: υ/T = f2(p), where f2 is an € undetermined function.
Collecting the two results involving f1 and f2, and solving for T produces:
p υ
€ =T = or pf 2 ( p) =€υf1 (υ ) = k ,
f1 (υ ) f 2 ( p)
where k must be is a constant since p and υ are independent thermodynamic variables.
Eliminating f1 or f2 from either equation on the left, produces pυ = kT.
And finally, using both versions € of (1.24) we can write: dh – de = υdp + pdυ = d(pυ).
€
When e and h only depend on T, then dh = cpdT and de = cvdT, so
dh – de = (cp – cv)dT = d(pυ) = kdT , thus k = cp – cv = R,
where R is the gas constant. Thus, the final result is the perfect gas law: p = kT/υ = ρRT.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.13. Starting from the property relationships (1.24) prove (1.31) and (1.32) for a
reversible adiabatic process involving a perfect gas when the specific heats cp and cv are constant.

Solution 1.13. For an isentropic process: de = Tds – pdυ = –pdυ, and dh = Tds + υdp = +υdp.
Equations (1.31) and (1.32) apply to a perfect gas so the definition of the specific heat capacities
(1.20), and (1.21) for a perfect gas, dh = cpdT, and de = cvdT , can be used to form the ratio
dh/de:
dh cp dT cp υ dp dυ dρ dp
= = =γ =− or −γ =γ = .
de cv dT cv pdυ υ ρ p
The final equality integrates to: ln(p) = γln(ρ) + const which can be exponentiated to find:
p = const.ργ,
which is (1.31). The constant may be evaluated € at a reference condition po and ρo to find:
γ
p po = ( ρ ρ o ) and this may be inverted to put the density ratio on the left
1γ
ρ ρ o = ( p po ) ,
which is the second equation of (1.32). The remaining relationship involving the temperature is
€ found by using the perfect gas law, p = ρRT, to eliminate ρ = p/RT:
1γ
ρ p RT € pTo # p & T p" p%
−1 γ
" p %(γ −1) γ
= = = % ( or = $ ' =$ ' ,
ρ o po RTo poT $ po ' To po # po & # po &
which is the first equation of (1.32).

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.14. A cylinder contains 2 kg of air at 50°C and a pressure of 3 bars. The air is
compressed until its pressure rises to 8 bars. What is the initial volume? Find the final volume for
both isothermal compression and isentropic compression.

Solution 1.14. Use the perfect gas law but explicitly separate the mass M of the air and the
volume V it occupies via the substitution ρ = M/V:
p = ρRT = (M/V)RT.
Solve for V at the initial time:
Vi = initial volume = MRT/pi = (2 kg)(287 m2/s2K)(273 + 50°)/(300 kPa) = 0.618 m3.
For an isothermal process:
Vf = final volume = MRT/pf = (2 kg)(287 m2/s2K)(273 + 50°)/(800 kPa) = 0.232 m3.
For an isentropic process:
1γ 1 1.4
V f = Vi ( pi p f ) = 0.618m 3 (300kPa 800kPa ) = 0.307m 3 .
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.15. Derive (1.35) starting from Figure 1.9 and the discussion at the beginning of
Section 1.10.

Solution 1.15. Take the z axis vertical, and consider a small fluid element δm of fluid having
volume δV that starts at height z0 in a stratified fluid medium having a vertical density profile =
ρ(z), and a vertical pressure profile p(z). Without any vertical displacement, the small mass and
its volume are related by δm = ρ(z0)δV. If the small mass is displaced vertically a small distance
ζ via an isentropic process, its density will change isentropically according to:
ρ a (z0 + ζ ) = ρ (z0 ) + ( dρ a dz)ζ + ...
where dρa/dz is the isentropic density gradient at z0. For a constant δm, the volume of the fluid
element will be:
δm δm δm & 1 dρ a )
δV =€ = = (1− ζ + ...+
ρ a ρ (z0 ) + ( dρ a dz)ζ + ... ρ(z0 ) ' ρ(z0 ) dz *
The background density at z0 + ζ is:
ρ(z0 + ζ ) = ρ (z0 ) + ( dρ dz)ζ + ...
If g is the acceleration of gravity, the (upward) buoyant force on the element at the vertically
€
displaced location will be gρ(z0 + ζ)δV, while the (downward) weight of the fluid element at any
vertical location is gδm. Thus, a vertical application Newton's second law implies:
€
d 2ζ δm & 1 dρ a )
δm 2 = +gρ (z0 + ζ )δV − gδm = g( ρ(z0 ) + ( dρ dz)ζ + ...) (1− ζ + ...+ − gδm ,
dt ρ(z0 ) ' ρ(z0 ) dz *
where the second equality follows from substituting for ρ(z0 + ζ) and δV from the above
equations. Multiplying out the terms in (,)-parentheses and dropping second order terms
produces:
€
d 2ζ gδm dρ gδm dρ a gδm ' dρ dρ a *
δm 2 = gδm + ζ− ζ + ...− gδm ≅ ) − ,ζ
dt ρ (z0 ) dz ρ(z0 ) dz ρ (z0 ) ( dz dz +
Dividing by δm and moving all the terms to the right side of the equation produces:
d 2ζ g % dρ dρ a (
2
− ' − *ζ = 0
€ dt ρ (z 0 ) & dz dz )
Thus, for oscillatory motion at frequency N, we must have
g $ dρ dρ a '
N2 = − & − ),
ρ(z0 ) % dz dz (
€
which is (1.35).

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.16. Starting with the hydrostatic pressure law (1.14), prove (1.36) without using
perfect gas relationships.

Solution 1.16. The adiabatic temperature gradient dTa/dz, can be written terms of the pressure
gradient:
dTa # ∂T & dp # ∂T &
=% ( = −gρ% (
dz $ ∂p ' s dz $ ∂p ' s
where the hydrostatic law dp/dz = –ρg has been used to reach the second equality. Here, the final
partial derivative can be exchanged for one involving υ = 1/ρ and s, by considering:
# ∂h & # ∂h &
€ dh = % ( ds + % ( dp = Tds + υdp .
$ ∂s ' p $ ∂p ' s
# ∂ # ∂h & & # ∂ # ∂ h & &
Equality of the crossed second derivatives of h, % % ( ( = % % ( ( , implies:
$ ∂p $ ∂s ' p ' s $ ∂s $ ∂ p ' s ' p
€ # ∂T & # ∂υ & # ∂υ & # ∂T & # ∂υ & # ∂s &
% ( =% ( =% ( % ( =% ( % ( ,
$ ∂p ' s $ ∂ s ' p $ ∂T ' p $ ∂ s ' p $ ∂T ' p $ ∂T ' p
where the second two equalities are mathematical
€ manipulations that allow the introduction of
1 & ∂ρ ) & ∂υ ) ! ∂h $ ! ∂s $
α = − ( + = ρ( + , and cp = # & = T # & .
€ ρ ' ∂T * p ' ∂T * p " ∂T % p " ∂T % p
Thus,
dTa " ∂T % " ∂υ % " ∂ s % "c % gαT
= −gρ $ ' = −gρ $ ' $ ' = −gα $ p ' = − .
€ dz # ∂ p &s # ∂T & p # ∂T & p #T & cp
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.17. Assume that the temperature of the atmosphere varies with height z as T = T0 +
g KR
! T0 $
Kz where K is a constant. Show that the pressure varies with height as p = p0 # & where
" T0 + Kz %
g is the acceleration of gravity and R is the gas constant for the atmospheric gas.

Solution 1.17. Start with the hydrostatic and perfect gas laws, dp/dz = –ρg, and p = ρRT,
eliminate the density, and substitute in the given temperature profile to find:
dp p p dp g dz
= −ρg = − g=− g or =− .
dz RT R(T0 + Kz) p R (T0 + Kz)
The final form may be integrated to find:
g
ln p = − ln(T0 + Kz) + const.
RK
At z = 0, €
the pressure must be p0, therefore: €
g
ln p0 = − ln(T0 ) + const.
RK
€ equation above and invoking the properties of logarithms produces:
Subtracting this from the
" p% g " T0 + Kz %
ln$ ' = − ln$ '
€ # p0 & RK # T0 &
Exponentiating produces:
−g/KR g/KR
p "T0 + Kz % " T0 %
=$ ' , which is the same as: p = p0 $ ' .
p0 # €T0 & # T0 + Kz &

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.18. Suppose the atmospheric temperature varies according to: T = 15 − 0.001z, where
T is in degrees Celsius and height z is in meters. Is this atmosphere stable?

Solution 1.18. Compute the temperature gradient:

dT d °C °C
= (15 − 0.001z) = −0.001 = −1.0 .
dz dz m km
For air in the earth's gravitational field, the adiabatic temperature gradient is:
dTa gαT (9.81m / s 2 )(1 / T )T °C
=− = 2 2
= −9.8 .
€ dz cp 1004m / s °C km
Thus, the given temperature profile is stable because the magnitude of its gradient is less than
the magnitude of the adiabatic temperature gradient.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.19. A hemispherical bowl with inner radius r containing a liquid with density ρ is
inverted on a smooth flat surface. Gravity with acceleration g acts downward. Determine the
weight W of the bowl necessary to prevent the liquid from escaping. Consider two cases:
a) the pressure around the rim of the bowl where it meets the plate is atmospheric, and
b) the pressure at the highest point of the bowl’s interior is atmospheric.
c) Investigate which case applies via a simple experiment. Completely fill an ordinary soup bowl
with water and concentrically cover it with an ordinary dinner plate. While holding the bowl and
plate together, quickly invert the water-bowl-plate combination, set it on the level surface at the
bottom of a kitchen sink, and let go. Does the water escape? If no water escapes after release,
hold onto the bowl only and try to lift the water-bowl-plate combination a few centimeters off
the bottom of the sink. Does the plate remain in contact with the bowl? Do your answers to the
first two parts of this problem help explain your observations?

Solution 1.19. a) When the pressure around the inverted bowl's rim is atmospheric, the surface
does not support the weight of the liquid and any increase in elevation above the rim must
produce a lower pressure in the liquid. In this case the weight of the liquid is supported by a
reduced pressure on the interior of the bowl. Thus the vertical force Fz that the bowl applies to
the surface is negative and includes its own weight, W, and the weight of the liquid ( 2 3) π r 3ρ g :
Fz = −W − ( 2 3) π r 3ρ g .
So, for Fz to reach zero, the point of bowl lift-off, W must be negative! This implies that the bowl
will adhere to the surface and will not separate unless it's pulled upward.
This finding concerning the pressure force on the bowl can be reached via a more
traditional hydrostatic calculation using spherical coordinates (see Appendix B.5). The net
vertical pressure force on the inverted bowl will be:
π 2 π 2
Fz, pressure = 2π ∫ ( p − po ) ( e r ⋅ e z ) r 2 sin θ dθ = 2π r 2 ∫ ( p − p )cosθ sinθ dθ ,
o
θ =0 θ =0
where p is the pressure inside the bowl and po is the pressure outside the bowl (atmospheric
pressure). Here θ is the polar angle (θ = 0 defines the positive z-axis). In this coordinate system z
= rcosθ, and the pressure inside and outside the bowl must match when z = 0, so p – po = –ρgz =
–ρgrcosθ, so
π 2
2π 3
Fz, pressure = −ρ g2π r ∫ cos2 θ sin θ dθ = −
3
r ρg .
θ =0 3
So, the bowl's "weight" must be less than − ( 2 3) π r 3ρ g to keep the liquid trapped.
b) In this case the pressure force on the bowl is different because atmospheric pressure is reached
at the inside top of the bowl. In this case, p – po = –ρg(z – r), and the vertical pressure force
calculation looks like:
π 2 π 2
2 3
Fz, pressure = 2π r ∫ ( p − p )cosθ sinθ dθ = −ρ g2π r ∫ (cosθ −1)cosθ sinθ dθ
o
θ =0 θ =0

1
= ρ gπ r 3.
3
So, the static force balance for the bowl is: Fz = −W + (1 3) π r 3ρ g . Thus, the bowl's weight must
be more than (1 3) π r 3ρ g to keep the liquid trapped.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

c) Interestingly, even though a little bit of water may escape from underneath the inverted bowl,
nearly all the water remains trapped under the bowl when the bowl-plate combination is released,
and the inverted bowl does adhere to the plate when it is lifted. These observations closely match
the part a) answer, so it is the physically meaningful one.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.20. Consider the case of a pure gas planet where the hydrostatic law is:
z
dp dz = −ρ (z)Gm(z) z 2 , where G is the gravitational constant and m(z) = 4π ∫ ρ (ζ )ζ 2 dζ is the
o
planetary mass up to distance z from the center of the planet. If the planetary gas is perfect with
gas constant R, determine ρ(z) and p(z) if this atmosphere is isothermal at temperature T. Are
these vertical profiles of ρ and p valid as z increases without bound?

Solution 1.20. Start with the given relationship for m(z), differentiate it with respect to z, and use
the perfect gas law, p = ρRT to replace the ρ with p.
dm d & z ) p(z)
= ( 4 π ∫ ρ (ζ )ζ 2 dζ + = 4 πz 2 ρ(z) = 4 πz 2 .
dz dz ' 0 * RT
Now use this and the hydrostatic law to obtain a differential equation for m(z),
dp Gm(z) d # RT dm & # 1 dm & Gm(z)
= −ρ (z) 2 → % ( = −% ( .
dz z dz $ 4π z 2 dz ' $ 4π z 2 dz ' z 2
€
After recognizing T as a constant, the nonlinear second-order differential equation for m(z)
simplifies to:
RT d " 1 dm % 1 dm
$ 2 '=− 4 m .
G dz # z dz & z dz
This equation can be solved by assuming a power law: m(z) = Azn. When substituted in, this trial
solution produces:
RT d −2 RT
€
G dz
( z Anz n−1 ) =
G
(n − 3) Anz n−4 = −z−4 A 2 nz 2n−1.
Matching exponents of z across the last equality produces: n – 4 = 2n – 5, and this requires n = 1.
For this value of n, the remainder of the equation is:
RT RT
€ (−2) Az−3 = −z−4 A 2 z1 , which reduces to: A = 2 .
G G
Thus, we have m(z) = 2RTz/G, and this leads to:
2RT 1 2R 2T 2 1
ρ(z) = , and p(z) = .
€ G 4 πz 2 € G 4 πz
2

Unfortunately, these profiles are not valid as z increases without bound, because this leads to an
unbounded planetary mass.
€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.21. Consider a gas atmosphere with pressure distribution

p(z) = p0 (1− (2 / π )tan −1 (z / H ))
where z is the vertical coordinate and H is a constant length scale.
a) Determine the vertical profile of ρ from (1.14)
b) Determine N2 from (1.35) as function of vertical distance, z.
c) Near the ground where z << H, this atmosphere is unstable, but it is stable at greater heights
where z >> H. Specify the value of z/H above which this atmosphere is stable.

Solution 1.21. a) Start from (1.14):

dp 2 1 1 2 po 1
= −ρ g = − po 2
so ρ (z) = .
dz π 1+ (z / H ) H π gH 1+ (z / H )2
b) First determine dρ/dz:
dρ 2 po 1 2z " 2z H 2 %
=− = − ρ $ 2'
.
dz π gH (1+ (z / H )2 )2 H 2 # 1+ (z / H ) &
Determine the local adiabatic density gradient by expanding (1.31) for small vertical
displacement ζ that leads to an adiabatic density change:
γ γ
p(z) + ζ ( dp dz ) +... ! ρ (z) + ζ ( d ρ a dz ) +... $ ζ dp ! ζ d ρa $
=# & or 1+ +... = #1+ +... & .
p(z) " ρ (z) % p dz " ρ dz %
For small ζ, expand the final term and cancel the leading 1's, to find:
1 dp γ d ρ a ρg d ρa ρ 2g
= =− , thus: =− .
p dz ρ dz p dz γp
where the second equality follows from (1.14). Now use (1.35)
g " d ρ d ρa % g " " 2z H 2 % ρ 2 g % " 2z H 2 ρg %
N2 = − $ − ' = − $ −ρ $ ' + ' = g $ − '
ρ # dz dz & ρ # # 1+ (z / H )2 & γ p & # 1+ (z / H )2 γ p &
" %
2z H 2 2 1
= g$ − ' .
$ 1+ (z / H )2 γπ H (1− (2 / π )tan −1 (z / H )) 1+ (z / H )2 '
# &
" %
2g H $ z 1 '
= −
1+ (z / H )2 $# H γπ (1− (2 / π )tan −1 (z / H )) '&
c) In the limit as z H → 0 , the terms inside the big parentheses go to –1/γπ, which is negative,
so N2 is negative, and this indicates an unstable atmosphere. In the limit as z H → ∞ , both terms
inside the big parentheses become large, but the first one is larger so N2 is positive, and this
indicates a stable atmosphere. The boundary between the two regimes occurs when the terms
inside the big parentheses sum to zero. This condition leads to an implicit equation for z/H:
z 1 z" −1 " z %
%
− = 0 , or γ $ π − 2 tan $ ' ' =1,
H γπ (1− (2 / π )tan −1 (z / H )) H# # H &&
which has numerical solution z/H = 0.274 when γ = 1.4.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.22. Consider a heat-insulated enclosure that is separated into two compartments of
volumes V1 and V2, containing perfect gases with pressures and temperatures of p1 and p2, and T1
and T2, respectively. The compartments are separated by an impermeable membrane that
conducts heat (but not mass). Calculate the final steady-state temperature assuming each gas has
constant specific heats.

Solution 1.22. Since no work is done and no heat is transferred out of the enclosure, the final
energy Ef is the sum of the energies, E1 and E2, in the two compartments.
E1 + E2 = Ef implies ρ1V1cv1T1 + ρ2V2cv2T2 = (ρ1V1cv1 + ρ2V2cv2)Tf,
where the cv's are the specific heats at constant volume for the two gases. The perfect gas law can
be used to find the densities: ρ1 = p1/R1T1 and ρ2 = p2/R2T2, so
p1V1cv1/R1 + p2V2cv2/R2 = (p1V1cv1/R1T1 + p2V2cv2/R2T2)Tf.
A little more simplification is possible, cv1/R1 = 1/(γ1 – 1) and cv2/R1 = 1/(γ2 – 1). Thus, the final
temperature is:
p1V1 (γ1 −1) + p2V2 (γ 2 −1)
Tf = .
p1V1 [(γ1 −1)T1 ] + p2V2 [(γ 2 −1)T2 ]

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.23. Consider the initial state of an enclosure with two compartments as described in
Exercise 1.22. At t = 0, the membrane is broken and the gases are mixed. Calculate the final
temperature.

Solution 1.23. No heat is transferred out of the enclosure and the work done by either gas is
delivered to the other so the total energy is unchanged. First consider the energy of either gas at
temperature T, and pressure P in a container of volume V. The energy E of this gas will be:
E = ρVcvT = (p/RT)VcvT = pV(cv/R) = pV/(γ – 1).
where γ is the ratio of specific heats. For the problem at hand the final energy Ef will be the sum
of the gas energies, E1 and E2, in the two compartments. Using the above formula:
E f = p1V1 (γ1 −1) + p2V2 (γ 2 −1) .
Now consider the mixture. The final volume and temperature for both gases is V1+V2, and Tf.
However, from Dalton's law of partial pressures, the final pressure of the mixture pf can be
considered a sum of the final partial pressures of gases "1" and "2", p1f and p2f:
€ pf = p1f + p2f.
Thus, the final energy of the mixture is a sum involving each gases' partial pressure and the total
volume:
E f = p1 f (V1 + V2 ) (γ1 −1) + p2 f (V1 + V2 ) (γ 2 −1) .
However, the perfect gas law implies: p1f(V1+V2) = n1RuTf, and p2f(V1+V2) = n2RuTf where n1 and
n2 are the mole numbers of gases "1" and "2", and Ru is the universal gas constant. The mole
numbers are obtained from:
€ n1 = p1V1/RuT1, and n2 = p2V2/RuT2,
Thus, final energy determined from the mixture is:
nRT n R T $ pV ' R T $ p V ' R T $ pV ' T $pV ' T
Ef = 1 u f + 2 u f =& 1 1) u f +& 2 2 ) u f =& 1 1) f +& 2 2 ) f .
γ1 −1 γ1 −1 % RuT1 ( γ1 −1 % RuT2 ( γ 2 −1 % T1 ( γ1 −1 % T2 ( γ 2 −1
Equating this and the first relationship for Ef above then produces:
p1V1 (γ1 −1) + p2V2 (γ 2 −1)
Tf = .
p1V1 [(γ1 −1)T1 ] + p2V2 [(γ 2 −1)T2 ]
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.24. A heavy piston of weight W is dropped onto a thermally insulated cylinder of
cross-sectional area A containing a perfect gas of constant specific heats, and initially having the
external pressure p1, temperature T1, and volume V1. After some oscillations, the piston reaches
an equilibrium position L meters below the equilibrium position of a weightless piston. Find L. Is
there an entropy increase?

Solution 1.24. From the first law of thermodynamics, with Q = 0, ΔE = Work = (W + p1A)L. For
a total mass m of a perfect gas with constant specific heats, E = mcvT so ΔE = E2 – E1 = mcv(T2 –
T1) = (W + p1A)L. Then T2 = T1 + (W + p1A)L/mcv. Also, for a perfect gas, PV/T = constant so
p1V1/T1 = p2V2/T2. For the cylinder, V2 = V1 – AL, and p2 = p1 + W/A. Therefore:
p1V1 ( p1 + W A)(V1 − AL)
= .
T1 T1 + (W + p1 A)L mcv
Here, m = p1V1/RT1 so mcv = (p1V1/T1)(cv/R) = (p1V1/T1)(γ – 1)–1, and this leaves:
p1V1 ( p1 + W A)(V1 − AL) ( p1 + W A)(V1 − AL)
= , or p1V1 = ,
T1 T1 + (γ −1)(W + p1 A)LT1 p1V1 1+ (γ −1)(W + p1 A)L p1V1
after multiplication on both sides by T1. Solve for L.
p1V1 + (γ −1)(W + p1 A)L = p1V1 + WV1 A − p1 AL − WL ,
W WV1 A
L [(γ −1)(W + p1 A)L + p1 A + W ] = V1 , or L = .
A γ (W + p1 A)
The initial length Lo of the column of gas is V1/A, so this final answer can be written:
L W p1 A
= .
Lo γ (1+ W p1 A)
For an isentropic process, pVγ = constant, so for a constant cross section cylinder:
1
L " 1 %γ
p1Lγo = ( p1 + W A) (Lo − L)γ , or = 1− $ ' .
Lo # 1+ W p1 A &
This isentropic compression distance is always greater than the "sudden compression" distance
determined above. Therefore, the sudden compression does lead to an increase in entropy.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.25. Starting from 295 K and atmospheric pressure, what is the final pressure of an
isentropic compression of air that raises the temperature 1, 10, and 100 K.

Solution 1.25. Start from the first equation of (1.32), and solve for the pressure:
γ −1 γ 7
T ! p$γ ! T $γ −1 ! 295 + ΔT $ 2
= # & , or p = po # & = (101.3kPa) # & ,
To " po % " To % " 295 %
where the final equality applies for the values specified in the problem statement when γ = 1.4.
For the three given values of ΔT, the final pressures are: 102.5 kPa, 113.8 kPa, and 281.4 kPa.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.26. Compute the speed of sound in air at –40°C (very cold winter temperature), at
+45°C (very hot summer temperature), at 400°C (automobile exhaust temperature), and 2000°C
(nominal hydrocarbon adiabatic flame temperature)

Solution 1.26. The speed of sound c of a perfect gas is given by (1.33):

c = γ RT , or c = γ R[273+ (T in °C)] .
Presuming that the R = 287 m2s–2K–1 applies at the four temperatures given, the four sound
speeds are: 306 ms–1, 357 ms–1, 520 ms–1, and 956 ms–1, respectively.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.27. The oscillation frequency Ω of a simple pendulum depends on the acceleration of
gravity g, and the length L of the pendulum.
a) Using dimensional analysis, determine single dimensionless group involving Ω, g and L.
b) Perform an experiment to see if the dimensionless group is constant. Using a piece of string
slightly longer than 2 m and any small heavy object, attach the object to one end of the piece of
string with tape or a knot. Mark distances of 0.25, 0.5, 1.0 and 2.0 m on the string from the
center of gravity of the object. Hold the string at the marked locations, stand in front of a clock
with a second hand or second readout, and count the number (N) of pendulum oscillations in 20
seconds to determine Ω = N/(20 s) in Hz. Evaluate the dimensionless group for these four
lengths.
c) Based only on the results of parts a) and b), what pendulum frequency do you predict when L
= 1.0 m but g is 16.6 m/s? How confident should you be of this prediction?

Solution 1.27. a) Construct the parameter & units matrix using Ω, g, and L.

Ω g L
M 0 0 0
L 0 1 1
T -1 -2 0

This rank of this matrix is two, so there is one dimensionless group that is readily found by
inspection to be Ω(L/g)1/2.
b) The following table lists the results of the simple pendulum experiments:

L (m) N Ω (s–1) Ω(L/g)1/2

0.25 20 1.00 0.160
0.50 14.5 0.725 0.164
1.00 10 0.500 0.160
2.00 7.0 0.350 0.158

Here g = 9.81 ms–2 has been used to evaluate the final column. These results do suggest that the
dimensionless group is constant within the uncertainty of these simple experiments, and that the
constant is ~0.1605. (For small angular oscillations in the absence of air resistance, this constant
should be 1/2π ≈ 0.159)
c) Using the results of part b), Ω(L/g)1/2 = 0.1605 , so Ω = 0.1605 (g/L)1/2, which implies: Ω =
0.1605(16.6/1.0)1/2 = 0.654 s–1 for the specified conditions. Given the results of part b), the
confidence in this prediction should be high; it might have ± 1 or 2% error at most.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.28. The spectrum of wind waves, S(ω), on the surface of the deep sea may depend on
the wave frequency ω, gravity g, the wind speed U, and the fetch distance F (the distance from
the upwind shore over which the wind blows with constant velocity).
a) Using dimensional analysis, determine how S(ω) must depend on the other parameters.
∞
b) It is observed that the mean-square wave amplitude, η 2 = ∫ S(ω )dω , is proportional to F.
0

Use this fact to revise the result of part a).

c) How must η2 depend on U and g?

Solution 1.28. a) Construct the parameter & units matrix using S, ω, g, U, and F. The units of S
are (length2)(time).

S ω g U F
M 0 0 0 0 0
L 2 0 1 1 1
T 1 -1 -2 -1 0

This rank of this matrix is two, so there are 5 – 2 = 3 dimensionless groups. Three suitable
groups are readily found by inspection to be Π1 = SωF–2, Π2 = ωFU–1, and Π3 = U2F–1g–1. Thus, S
F 2 "ωF U 2 %
must depend on the other parameters as follows: S = Ψ$ , '.
ω # U gF &
b) Use the given equation and the result of part a):
∞ ∞
F 2 $ωF U 2 ' ∞
F 2 $ωF U 2 ' ∞ $ U 2 ' dγ
η 2 = ∫ Sdω = ∫ Ψ& , )dω = ∫ ω &% U gF )(
Ψ , dω = F 2
∫ &%γ, gF )( γ ,
Ψ
0 0 ω % U gF ( 0 0

where the final equality follows from changing the integration variable to γ = ωF/U. From this
equation, the only way that η2 can be proportional to F is for the undetermined function be
proportional to its second argument: Ψ (γ ,U 2 gF ) = (U 2 gF ) Φ(γ ) . Combine this result with the
F 2 "ωF U 2 % F 2 U 2 "ωF % U 2F "ωF %
result of part a): S = Ψ$ , '= Φ$ '= Φ$ '.
ω # U gF & ω gF # U & gω # U &
c) Substitute the final answer from part b) back into the integral relationships to find:
2
∞
U 2F ∞ 1 $ωF ' U 2F ∞ dγ U 2 F
η = ∫ Sdω = ∫ Φ& )dω = ∫ Φ (γ ) = ⋅ const.
0 g 0ω %U ( g 0 γ g
Thus, η2 must be proportional to U2 and inversely proportional to g.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.29. One military technology for clearing a path through a minefield is to deploy a
powerfully exploding cable across the minefield that, when detonated, creates a large trench
through which soldiers and vehicles may safely travel. If the expanding cylindrical blast wave
from such a line-explosive has radius R at time t after detonation, use dimensional analysis to
determine how R and the blast wave speed dR/dt must depend on t, r = air density, and E´ =
energy released per unit length of exploding cable.

Solution 1.29. Follow example 1.10 and construct the parameter & units matrix using R, t, r, and
E´. The units of E´ are energy/length. Construct the parameter & units matrix.

R t ρ E´
M 0 0 1 1
L 1 0 -3 1
T 0 1 0 -2

This rank of this matrix is three, so there is 4 – 3 = 1 dimensionless group, and it may be found
by inspection: Π1 = E´t2/ρR4. Since there is only one dimensionless group, it must be constant so:
R(t) = const.[E´/ρ]1/4t1/2. To find dR/dt, differentiate this with respect to time to find: dR/dt =
(const./2)[E´/ρ]1/4t–1/2 = R/(2t).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.30. One of the triumphs of classical thermodynamics for a simple compressible
substance was the identification of entropy s as a state variable along with pressure p, density ρ,
and temperature T. Interestingly, this identification foreshadowed the existence of quantum
physics because of the requirement that it must be possible to state all physically meaningful
laws in dimensionless form. To see this foreshadowing, consider an entropic equation of state for
a system of N elements each having mass m.
a) Determine which thermodynamic variables amongst s, p, ρ, and T can be made dimensionless
using N, m, and the non-quantum mechanical physical constants kB = Boltzmann’s constant and c
= speed of light. What do these results imply about an entropic equation of state in any of the
following forms: s = s(p,ρ), s = s(ρ,T), or s = s(T,p)?
b) Repeat part a) including  = Planck’s constant (the fundamental constant of quantum
physics). Can an entropic equation of state be stated in dimensionless form without  ?

Solution 1.30. a) Select each thermodynamic variable in turn, and see if it can be made
dimensionless using m, kB, and c. Here N is dimensionless and it is the only extensive variable so
it need not be considered when seeking the dimensionless forms of the intensive (per unit mass)
thermodynamic variables. The dimensions of the remaining parameters and constants are:

s p ρ T m kB c
M 0 1 1 0 1 1 0
L 2 -1 -3 0 0 2 1
T -2 -2 0 0 0 -2 -1
θ -1 0 0 1 0 -1 0

Exponent algebra can be used to reach the following results:

• entropy s can be made dimensionless via s/mkB;
• pressure p cannot be made dimensionless using m, kB, and c;
• density ρ cannot be made dimensionless using m, kB, and c; and
• temperature can be dimensionless via kBT/mc2.
These results imply that an entropic equation of state in any of the forms s = s(p,ρ), s =
s(ρ,T), or s = s(T,p) cannot be made dimensionless using non-quantum mechanical physical
constants.
b) The units of  are (energy)(time) = ML2T–1. With the addition of this parameter, exponent
algebra can be used to reach the following results:
• entropy s can be made dimensionless via s/mkB;
• pressure p can be made dimensionless via p 3 m 4 c 5 ;
• density ρ can be made dimensionless via ρ 3 m 4 c 3 ; and
• temperature can be dimensionless via kBT/mc2.
These results imply that an entropic equation of state in any of the forms s = s(p,ρ), s =
s(ρ,T), or s = s(T,p) can be made dimensionless with  .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.31. The natural variables of the system enthalpy H are the system entropy S and the
pressure p, which leads to an equation of state in the form: H = H(S, p, N), where N is the
number of system elements.
a) After creating ratios of extensive variables, use exponent algebra to independently render H/N,
S/N, and p dimensionless using m = the mass of a system element, and the fundamental constants
kB = Boltzmann’s constant,  = Planck’s constant, and c = speed of light.
b) Simplify the result of part a) for non-relativistic elements by eliminating c.
c) Based on the property relationship (1.24), determine the specific volume = υ = 1 ρ = (∂h ∂p)s
from the result of part b).
d) Use the result of part c) and (1.25) to show that the sound speed in this case is 5p 3ρ , and
compare this result to that for a monotonic perfect gas.

Solution 1.31. a) Here H and S are extensive variables so they must be proportional to N. So
parameter & units matrix can be constructed as:

H/N S/N p m kB  c
M 1 1 1 0 1 1 0
L 2 2 -3 0 2 2 1
T -2 -2 0 0 -2 -1 -1
θ 0 -1 0 1 -1 0 0

This rank of this matrix is four, so there are 7 – 4 = 3 dimensionless groups. Three suitable
groups that isolate H/N, p and S/N are readily found by inspection or exponent algebra to be: Π1
= HN–1m–1c–2, Π2 = p3m −4 c −5 , and Π3 = SN −1kB−1 . The following scaling law then applies to H/N:
H " p3 S %
= Θ $ 4 5, '
Nmc 2 # m c NkB &
where Θ is an undetermined function.
b) To eliminate c, extract the second dimensionless group from the argument of Θ, raise it to the
–2/5 power, and multiply on the left side with the altered group involving H:
−2 5 25
H ! p3 $ ! S $ H ! m 4c 5 $ hm8 5 ! sm $
2 # 4 5&
= Θ # & or 2 # 3 &
= = Θ# & ,
Nmc " m c % " NkB % Nmc " p % p 
25 65
" kB %
where H/Nm = h, and s = S/Nm has been used for the second-to-last equality.
c) Use the final equality of part b), and perform the indicated differentiation:
1 " ∂h % " ∂ ) p 2 5 6 5 " sm %,% 2  6 5 " s %
υ = = $ ' = $$ + 8 5 Θ $ '.'' = 85 35
Θ $ '.
ρ # ∂p &s # ∂p * m # kB &-&s 5 m p # mkB &
53
 2 ( 2 " s %+ 53
d) Use the final result of part c) get an equation for the pressure, p = 8 3 * Θ $ '- ρ , and
m ) 5 # mkB &,
53
" ∂p % 5  2 ) 2 " s %, −2 3 5p
differentiate as in (1.25) to find: c = $ ' = 83
+ Θ$ '. ρ = , which is the
# ∂ρ &s 3 m * 5 # mkB &- 3ρ
correct relationship for the speed of sound in a monotonic perfect gas.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.32. A gas of noninteracting particles of mass m at temperature T has density ρ, and
internal energy per unit volume ε.
a) Using dimensional analysis, determine how ε must depend on ρ, T, and m. In your formulation
use kB = Boltzmann’s constant,  = Plank’s constant, and c = speed of light to include possible
quantum and relativistic effects.
b) Consider the limit of slow-moving particles without quantum effects by requiring c and  to
drop out of your dimensionless formulation. How does ε depend on ρ and T? What type of gas
follows this thermodynamic law?
c) Consider the limit of massless particles (i.e., photons) by requiring m and ρ to drop out of
your dimensionless formulation of part a). How does ε depend on T in this case? What is the
name of this radiation law?

Solution 1.32. a) Construct the parameter & units matrix noting that kB and T must go together
since they are the only parameters that involve temperature units.

ε ρ kBT m  c
M 1 1 1 1 1 0
L -1 -3 2 0 2 1
T -2 0 -2 0 -1 -1

This rank of this matrix is three. There are 6 parameters and 3 independent units, so there will be
3 dimensionless groups. Two of the dimensionless groups are energy ratios that are easy spot:
Π1 = ε ρc 2 and Π 2 = kB T mc 2 . There is one dimensionless group left that must contain  . A
ρ 3 ε ! k BT ρ  3 $
bit of work produces: Π3 = 4 3 , so = ϕ1 # 2 , 4 3 & .
mc ρc 2 " mc m c %
€ b) Dropping
€  means dropping Π3. Eliminating c means combining Π1 and Π2 to create a new
Π1 ε ρc 2 εm
dimensionless group that lacks c: = 2
= . However, now there is only one
Π 2 kB T mc ρk B T
% ρk T (
dimensionless group so it must be a constant. This implies: ε = const ⋅ ' B * which is the
& m )
caloric equation of state for a perfect gas.
€
c) Eliminating ρ means combining Π1 and Π3 to create a new dimensionless group that lacks ρ:
ε ρ 3 ε 3
Π1 ⋅ Π3 = 2 ⋅ 4 3 = 4 5 . Now combine this new € dimensionless group with Π2 to eliminate
ρc m c m c
4
ε 3 1 ε 3 # mc 2 & ε  3c 3
m: 4 5 ⋅ 4 = 4 5 ⋅ % ( = . Again there is only a single dimensionless group so it
m c Π 2 m c $ k BT ' ( k BT ) 4
const 4
must equal a constant; therefore ε = ⋅ ( kBT ) . This is the Stephan-Boltzmann radiation law.
c
3 3
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.33. A compression wave in a long gas-filled constant-area duct propagates to the left
at speed U. To the left of the wave, the gas is quiescent with uniform density ρ1 and uniform
pressure p1. To the right of the wave, the gas has uniform density ρ2 (> ρ1) and uniform pressure
is p2 (> p1). Ignore the effects of viscosity in this problem. Formulate a dimensionless scaling law
for U in terms of the pressures and densities.

U!

p1, ρ1! p2, ρ2!

u1 = 0!

Solution 1.33. a) Construct the parameter & units matrix:

U ρ1 ρ2 p1 p2
M 0 1 1 1 1
L 1 -3 -3 -1 -1
T -1 0 0 -2 -2

This rank of this matrix is just two. There are 5 parameters but just 2 independent units, so there
will be 3 dimensionless groups. These are readily found by inspection: Π1 = U ρ1 p1 ,
Π 2 = ρ1 ρ2 , and Π3 = p1 p2 . Thus, the scaling law is:
U ρ1 p1 = Φ ( ρ1 ρ2 , p1 p2 ) .
where Φ is an undetermined function.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.34. Many flying and swimming animals – as well as human-engineered vehicles –
rely on some type of repetitive motion for propulsion through air or water. For this problem,
assume the average travel speed U, depends on the repetition frequency f, the characteristic
length scale of the animal or vehicle L, the acceleration of gravity g, the density of the animal or
vehicle ρo, the density of the fluid ρ, and the viscosity of the fluid µ.
a) Formulate a dimensionless scaling law for U involving all the other parameters.
b) Simplify your answer for a) for turbulent flow where µ is no longer a parameter.
c) Fish and animals that swim at or near a water surface generate waves that move and propagate
because of gravity, so g clearly plays a role in determining U. However, if fluctuations in the
propulsive thrust are small, then f may not be important. Thus, eliminate f from your answer for
b) while retaining L, and determine how U depends on L. Are successful competitive human
swimmers likely to be shorter or taller than the average person?
d) When the propulsive fluctuations of a surface swimmer are large, the characteristic length
scale may be U/f instead of L. Therefore, drop L from your answer for b). In this case, will higher
speeds be achieved at lower or higher frequencies?
e) While traveling submerged, fish, marine mammals, and submarines are usually neutrally
buoyant (ρo ≈ ρ) or very nearly so. Thus, simplify your answer for b) so that g drops out. For this
situation, how does the speed U depend on the repetition frequency f?
f) Although fully submerged, aircraft and birds are far from neutrally buoyant in air, so their
travel speed is predominately set by balancing lift and weight. Ignoring frequency and viscosity,
use the remaining parameters to construct dimensionally accurate surrogates for lift and weight
to determine how U depends on ρo/ρ, L, and g.

Solution 1.34. a) Construct the parameter & units matrix

U f L g ρo ρ µ
M 0 0 0 0 1 1 1
L 1 0 1 1 -3 -3 -1
T -1 -1 0 -2 0 0 -1

The rank of this matrix is three. There are 7 parameters and 3 independent units, so there will be
4 dimensionless groups. First try to assemble traditional dimensionless groups, but its best to use
the solution parameter U only once. Here U is used in the Froude number, so its dimensional
counter part, gL , is used in place of U in the Reynolds number.
U ρ gL3
Π1 = = Froude number, Π 2 = = a Reynolds number
gL µ
The€next two groups can be found by inspection:
ρ f
Π 3 = o = a density ratio , and the final group must include f: Π 4 = , and is a frequency
ρ € g L
€
ratio between f and that of simple pendulum with length L. Putting these together produces:
U $ ρ gL3 ρ f '
& o ) where, throughout this problem solution, ψi , i = 1, 2, 3, … are
= ψ1& , , )
€ gL % µ ρ g L ( €
unknown functions.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

U $ρ f '
b) When µ is no longer a parameter, the Reynolds number drops out: = ψ 2 && o , )) .
gL % ρ g L(
c) When f is no longer a parameter, then U = gL ⋅ ψ 3 ( ρ o ρ ) , so that U is proportional to L .
This scaling suggests that taller swimmers have an advantage over shorter ones. [Human
swimmers best approach the necessary conditions for this part € of this problem while doing
freestyle (crawl) or backstroke where the arms (and legs) are used for propulsion in an
€
alternating (instead of simultaneous) fashion. Interestingly, this length advantage € also applies to
ships and sailboats. Aircraft carriers are the longest and fastest (non-planing) ships in any Navy,
and historically the longer boat typically won the America’s Cup races under the 12-meter rule.
Thus, if you bet on a swimming or sailing race where the competitors aren’t known to you but
appear to be evenly matched, choose the taller swimmer or the longer boat.]
d) Dropping L from the answer for b) requires the creation of a new dimensionless group from f,
g, and U to replace Π1 and Π4. The new group can be obtained via a product of original
U f Uf Uf $ρ ' g $ρ '
dimensionless groups: Π1Π 4 = = . Thus, = ψ 4 & o ) , or U = ψ 4 & o ) . Here,
gL g L g g %ρ( f %ρ(
U is inversely proportional to f which suggests that higher speeds should be obtained at lower
frequencies. [Human swimmers of butterfly (and breaststroke to a lesser degree) approach the
conditions required for this part of this problem. Fewer longer strokes are typically preferred
€ ones. Of course, the trick for €
over many short reaching top speed € is to properly lengthen each
stroke without losing propulsive force].
e) When g is no longer a parameter, a new dimensionless group that lacks g must be made to
Π U gL U
replace Π1 and Π5. This new dimensionless group is 1 = = , so the overall scaling
Π 5 f g L fL
%ρ (
law must be: U = fL ⋅ ψ 5 ' o * . Thus, U will be directly proportional to f. Simple observations of
&ρ)
swimming fish, dolphins, whales, etc. verify€that their tail oscillation frequency increases at
higher swimming speeds, as does the rotation speed of a submarine or torpedo’s propeller.
f) Dimensionally-accurate surrogates for weight and lift are: ρ o L3 g and ρU 2 L2 , respectively. Set
€
these proportional to each other, ρ o L3 g ∝ ρU 2 L2 , to find U ∝ ρ o gL ρ , which implies that
larger denser flying objects must fly faster. This result is certainly reasonable when comparing
similarly shaped aircraft (or birds) of different sizes. € €
€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.35. The acoustic power W generated by a large industrial blower depends on its
volume flow rate Q, the pressure rise ΔP it works against, the air density ρ, and the speed of
sound c. If hired as an acoustic consultant to quiet this blower by changing its operating
conditions, what is your first suggestion?

Solution 1.35. The boundary condition and material parameters are: Q, ρ, ΔP, and c. The
solution parameter is W. Create the parameter matrix:

W Q ΔP ρ c
––––––––––––––––––––––––––––
Mass: 1 0 1 1 0
Length: 2 3 -1 -3 1
Time: -3 -1 -2 0 -1

This rank of this matrix is three. Next, determine the number of dimensionless groups: 5
parameters - 3 dimensions = 2 groups. Construct the dimensionless groups: ∏1 = W/QΔP, ∏2 =
ΔP/ρc2. Now write the dimensionless law: W = QΔPΦ(ΔP/ρc2), where Φ is an unknown
function. Since the sound power W must be proportional to volume flow rate Q, you can
immediately suggest a decrease in Q as means of lowering W. At this point you do not know if
Q must be maintained at high level, so this solution may be viable even though it may oppose
many of the usual reasons for using a blower. Note that since Φ is unknown the dependence of
W on ΔP cannot be determined from dimensional analysis alone.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.36. The horizontal displacement Δ of the trajectory of a spinning ball depends on the
mass m and diameter d of the ball, the air density ρ and viscosity µ, the ball's rotation rate ω, the
ball’s speed U, and the distance L traveled.
a) Use dimensional analysis to predict how Δ can depend on the other parameters.
b) Simplify your result from part a) for negligible viscous forces.
c) It is experimentally observed that Δ for a spinning sphere becomes essentially independent of
the rotation rate once the surface rotation speed, ωd/2, exceeds twice U. Simplify your result
from part b) for this high-spin regime.
d) Based on the result in part c), how does Δ depend on U?

Top view! spinning ball trajectory !

Δ"
U! L!

Solution 1.36. a) Create the parameter matrix using the solution parameter is Δ, and the
boundary condition and material parameters are: Q, ρ, ΔP, and c.

Δ m d ρ µ ω U L
Mass: 0 1 0 1 1 0 0 0
Length: 1 0 1 -3 -1 0 1 1
Time: 0 0 0 0 -1 -1 -1 0

This rank of this matrix is three. Next, determine the number of dimensionless groups: 8
parameters - 3 dimensions = 5 groups. Construct the dimensionless groups: Π1 = Δ/d, Π2 =
m/ρd3, Π3 = ρUd/µ, Π4 = ωd/U, and Π5 = L/d. Thus, the dimensionless law for Δ is:
Δ # m ρUd ω d L &
= Φ% 3 , , , (,
d $ ρd µ U d'
where Φ is an undetermined function.
b) When the viscosity is no longer a parameter, then the third dimensionless group (the Reynolds
number) must drop out, so the part a) result simplifies to:
Δ # m ωd L &
= Ψ% 3 , , (,
d $ ρd U d '
where Ψ is another undetermined function.
c) When the rotation rate is no longer a parameter, the fourth dimensionless group from part a)
(the Strouhal number) must drop out, so the part b) result simplifies to:
Δ # m L&
= Θ% 3 , ( ,
d $ ρd d '
where Θ is another undetermined function.
d) Interestingly, the part c) result suggests that Δ does not depend on U at all!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.37. A machine that fills peanut-butter jars must be reset to accommodate larger jars.
The new jars are twice as large as the old ones but they must be filled in the same amount of time
by the same machine. Fortunately, the viscosity of peanut butter decreases with increasing
temperature, and this property of peanut butter can be exploited to achieve the desired results
since the existing machine allows for temperature control.
a) Write a dimensionless law for the jar-filling time tf based on: the density of peanut butter ρ,
the jar volume V, the viscosity of peanut butter µ, the driving pressure that forces peanut butter
out of the machine P, and the diameter of the peanut butter-delivery tube d.
b) Assuming that the peanut butter flow is dominated by viscous forces, modify the relationship
you have written for part a) to eliminate the effects of fluid inertia.
c) Make a reasonable assumption concerning the relationship between tf and V when the other
variables are fixed so that you can determine the viscosity ratio µnew/µold necessary for proper
operation of the old machine with the new jars.
d) Unfortunately, the auger mechanism that pumps the liquid peanut butter develops driving
pressure through viscous forces so that P is proportional to µ. Therefore, to meet the new jar-
filling requirement, what part of the machine should be changed and how much larger should it
be?

Solution 1.37. a) First create the parameter matrix. The solution parameter is tf. The boundary
condition and material parameters are: V, ρ, P, µ, and d.

tf V P ρ d µ
Mass: 0 0 1 1 0 1
Length: 0 3 -1 -3 1 -1
Time: 1 0 -2 0 0 -1

This rank of this matrix is three. Next, determine the number of dimensionless groups: 6
parameters - 3 dimensions = 3 groups. Construct the dimensionless groups: Π1 = Ptf/µ, Π2 =
µ2/ρd2P, Π3 = V/d3, and write a dimensionless law: tf = (µ/P)Φ(µ2/ρd2P,V/d3), where Φ is an
unknown function.
b) When fluid inertia is not important the fluid's density is not a parameter. Therefore, drop ∏2
from the dimensional analysis formula: tf = (µ/P)Ψ(V/d3), where Ψ is yet another unknown
function.
c) One might reasonably expect that tf ∝ V (these are the two extensive variables). Therefore, we
end up with tf = const⋅µV/Pd3. Now form a ratio between the old and new conditions and cancel
common terms:
(t f ) new (µV /Pd 3 ) new (µV ) new V µnew 1
=1= 3 = , so new = 2 → =
(t f ) old (µV /Pd ) old (µV ) old Vold µold 2
d) If P is proportional to µ, then to achieve the same filling time for twice the volume using the
part c) result for tf implies,
Vold 2V €
€ € € 3
=€ old 3 €
(d old ) (d new )
Thus, the machine’s nozzle diameter must be increased so that dnew = 3 2 dold .
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.38. As an idealization of fuel injection in a Diesel engine, consider a stream of high-
speed fluid (called a jet) that emerges into a quiescent air reservoir at t = 0 from a small hole in
an infinite plate to form a plume where the fuel and air mix.
a) Develop a scaling law via dimensional analysis for the penetration distance D of the plume as
a function of: Δp the pressure difference across the orifice that drives the jet, do the diameter of
the jet orifice, ρo the density of the fuel, µ∞ and ρ∞ the viscosity and density of the air, and t the
time since the jet was turned on.
b) Simplify this scaling law for turbulent flow where air viscosity is no longer a parameter.
c) For turbulent flow and D << do, do and ρ∞ are not parameters. Recreate the dimensionless law
for D.
d) For turbulent flow and D >> do, only the momentum flux of the jet matters, so Δp and do are
replaced by the single parameter Jo = jet momentum flux (Jo has the units of force and is
approximately equal to Δpdo2 ). Recreate the dimensionless law for D using the new parameter Jo.

Solution 1.38. a) The parameters are: D, t, Δp, ρo, ρ∞, µ∞, and do. With D as the solution
parameter, create the parameter matrix:
€
D t Δp ρo ρ∞ µ∞ do
––––––––––––––––––––––––––––––––––––––––
Mass: 0 0 1 1 1 1 0
Length: 1 0 -1 -3 -3 -1 1
Time: 0 1 -2 0 0 -1 0

Next, determine the number of dimensionless groups. This rank of this matrix is three so 7
parameters - 3 dimensions = 4 groups, and construct the groups: Π1 = D do , Π 2 = ρ o ρ∞ ,
Π 3 = Δpt 2 ρ∞ do2 , and Π 4 = ρ∞Δpdo2 µ∞2 . Thus, the dimensionless law is:
D % ρ o Δpt 2 ρ∞Δpdo2 (
= f' , 2
, * , where f is an unknown function.
do & ρ∞ ρ∞ do µ∞2 ) € €
€ b) For high Reynolds
€ number turbulent flow when the reservoir viscosity is no longer a
parameter, the above result becomes:
D % ρ Δpt 2 (
€ = g' o , 2*
,
do & ρ∞ ρ∞ do )
where g is an unknown function.
c) When do and ρ∞ are not parameters, there is only one dimensionless group: Δpt 2 ρ∞ D2 , so
the dimensionless law becomes: € D = const ⋅ t Δp ρ o .
d) When Δp and do are replaced by the single parameter Jo = jet momentum flux, there are two
dimensionless parameters: J o t 2 ρ∞ D4 , and ρ o ρ∞ , so the dimensionless
€ law becomes:
14
€ D = ( J o t 2 ρ∞ ) F ( ρ o ρ∞ ) ,
where F is an unknown function.
[The results presented
€ here are the€fuel-plume penetration scaling laws for fuel injection in
Diesel engines where more than half of the world's petroleum ends up being burned.]
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.39. One of the simplest types of gasoline carburetors is a tube with small port for
transverse injection of fuel. It is desirable to have the fuel uniformly mixed in the passing air
stream as quickly as possible. A prediction of the mixing length L is sought. The parameters of
this problem are: ρ = density of the flowing air, d = diameter of the tube, µ = viscosity of the
flowing air, U = mean axial velocity of the flowing air, and J = momentum flux of the fuel
stream.
a) Write a dimensionless law for L.
b) Simplify your result from part a) for turbulent flow where µ must drop out of your
dimensional analysis.
c) When this flow is turbulent, it is observed that mixing is essentially complete after one
rotation of the counter rotating vortices driven by the injected-fuel momentum (see downstream-
view of the drawing for this problem), and that the vortex rotation rate is directly proportional to
J. Based on this information, assume that L ∝ (rotation time)(U) to eliminate the arbitrary
function in the result of part b). The final formula for L should contain an undetermined
dimensionless constant.

Solution 1.39. a) The parameters are: L, J, d, µ, ρ, and U. Use these to create the parameter
matrix with L as the solution parameter:
L J d µ ρ U
–––––––––––––––––––––––––––––––––––
Mass: 0 1 0 1 1 0
Length: 1 1 1 -1 -3 1
Time: 0 -2 0 -1 0 -1

Next, determine the number of dimensionless groups. This rank of this matrix is three so 6
parameters - 3 dimensions = 3 groups, and construct them: Π1 = L/d, Π2 = ρUd/µ, Π3 = ρU2d2/J.
And, finally write a dimensionless law: L = d Φ(ρUd/µ, ρU2d2/J), where Φ is an unknown
function.
b) At high Reynolds numbers, µ must not be a parameter. Therefore: L = dΨ(ρU2d2/J) where Ψ
is an unknown function.
c) Let Ω = vortex rotation rate. The units of Ω are 1/time and Ω must be proportional to J.
Putting this statement in dimensionless terms based on the boundary condition and material
J
parameters of this problem means: Ω = const = (rotation time)-1
ρUd 3

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

ρU 2 d 3 L ρU 2 d 2
Therefore: L = const (Ω-1)U = const , or = const . Thus, for transverse
J d J
injection, more rapid mixing occurs (L decreases) when the injection momentum increases.

€ € €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.40. Consider dune formation in a large horizontal desert of deep sand.
a) Develop a scaling relationship that describes how the height h of the dunes depends on the
average wind speed U, the length of time the wind has been blowing Δt, the average weight and
diameter of a sand grain w and d, and the air’s density ρ and kinematic viscosity ν.
b) Simplify the result of part a) when the sand-air interface is fully rough and ν is no longer a
parameter.
c) If the sand dune height is determined to be proportional to the density of the air, how do you
expect it to depend on the weight of a sand grain?

Solution 1.40. a) The solution parameter is h. The boundary condition and material parameters
are: U, Δt, w, d, ρ, and ν. First create the parameter matrix:
h U Δt w d ρ ν
–––––––––––––––––––––––––––––––––––––––
Mass: 0 0 0 1 0 1 0
Length: 1 1 0 1 1 -3 2
Time: 0 -1 1 -2 0 0 -1

Next determine the number of dimensionless groups. This rank of this matrix is three so 7
parameters - 3 dimensions = 4 groups. Construct the dimensionless groups: Π1 = h/d, Π2 = Ud/ν,
Π3 = w/ρU2d2, and Π4 = UΔt/d. Thus, the dimensionless law is
h & Ud w UΔt )
= Φ( , 2 2 , +,
d ' ν ρU d d *
where Φ is an unknown function.
b) When ν is no longer a parameter, Π2 drops out:
h % w UΔt (
€ = Ψ' 2 2 , *,
d & ρU d d )
where Ψ is another unknown function.
c) When h is proportional to ρ, then
h ρU 2 d 2 % UΔt (
€ = Θ' *,
d w & d )
where Θ is another unknown function. Under this condition, dune height will be inversely
proportional to w the sand grain weight.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.41. The rim-to-rim diameter D of the impact crater produced by a vertically-falling
object depends on d = average diameter of the object, E = kinetic energy of the object lost on
impact, ρ = density of the ground at the impact site, Σ = yield stress of the ground at the impact
site, and g = acceleration of gravity.
a) Using dimensional analysis, determine a scaling law for D.
b) Simplify the result of part a) when D >> d, and d is no longer a parameter.
c) Further simplify the result of part b) when the ground plastically deforms to absorb the impact
energy and ρ is irrelevant. In this case, does gravity influence D? And, if E is doubled how much
bigger is D?
d) Alternatively, further simplify the result of part b) when the ground at the impact site is an
unconsolidated material like sand where Σ is irrelevant. In this case, does gravity influence D?
And, if E is doubled how much bigger is D?
e) Assume the relevant constant is unity and invert the algebraic relationship found in part d) to
estimate the impact energy that formed the 1.2-km-diameter Barringer Meteor Crater in Arizona
using the density of Coconino sandstone, 2.3 g/cm3, at the impact site. The impact energy that
formed this crater is likely between 1016 and 1017 J. How close to this range is your dimensional
analysis estimate?
g! D!
Ε#
ρ, Σ"
d!

Solution 1.41. The solution parameter is D. The boundary condition and material parameters are:
d, E, θ, ρ, Σ, and g. First create the parameter matrix:

D d E ρ Σ g
M 0 0 1 1 1 0
L 1 1 2 –3 –1 1
T 0 0 –2 0 –2 –2

The rank of this matrix is 3, so there are 6 – 3 = 4 dimensionless groups. These groups are:
Π1 = D d , Π 2 = E ρ gd 4 , and Π3 = Σ ρ gd . Thus, the scaling law is:
D / d = fn(E / ρ gd 4 , Σ / ρ gd) ,
where fn is an undetermined function.
b) Use the second dimensionless group to remove d from the other two:
−1 4 ! Σ ! E $−1 4 $ # &
D! E $ # & D % Σ (.
# & = Φ# # 4&
, or =Φ
d " ρ gd 4 % ρ gd ρ gd & 14 % 14(
" " % % ( E ρ g) $ (ρ g E ) '
3 3

where Φ is an undetermined function.

c) In this case, the two dimensionless groups in the part b) must be combined to eliminate ρ,
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

D Σ1 3 D
14 1 12
= 13
= const. ,
( E ρ g) (ρ g E )
3 3
( E Σ)
and this lone dimensionless group must be constant. In this case, gravity does not influence the
crater diameter, and a doubling of the energy E increases D by a factor of 21 3 ≅ 1.26 .
d) When S is irrelevant, then the part b) result reduces to:
D
14
= const.
( E ρ g)
In this case, gravity does influence the crater diameter, and a doubling of the energy E increases
D a factor of 21 4 ≅ 1.19 .
e) If the part d) constant is unity, then that result implies: E = ρ gD 4 . For the Barringer crater,
this energy is:
E = (2300 kgm–3)(9.81 ms–2)(1200m)4 ≈ 4.7 x 1016 J,
an estimate that falls in the correct range, a remarkable result given the simplicity of the analysis.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.42. An isolated nominally spherical bubble with radius R undergoes shape
oscillations at frequency f. It is filled with air having density ρa and resides in water with density
ρw and surface tension σ. What frequency ratio should be expected between two isolated bubbles
with 2 cm and 4 cm diameters undergoing geometrically similar shape oscillations? If a soluble
surfactant is added to the water that lowers σ by a factor of two, by what factor should air bubble
oscillation frequencies increase or decrease?

Solution 1.42. The boundary condition and material parameters are: R, ρa, ρw, and σ. The
solution parameter is f. First create the parameter matrix:
f R ρa ρw σ
––––––––––––––––––––––––––––
Mass: 0 0 1 1 1
Length: 0 1 -3 -3 0
Time: -1 0 0 0 -2

Next determine the number of dimensionless groups. This rank of this matrix is three, so 5
parameters - 3 dimensions = 2 groups. Construct the dimensionless groups: Π1 = f ρ w R 3 σ ,
and Π2 = ρw/ρa. Thus, the dimensionless law is
σ %ρ (
f = Φ w *,
3 '
ρw R & ρa ) €
where Φ is an unknown function. For a fixed density ratio, Φ(ρw/ρa) will be constant so f is
proportional to R–3/2 and to σ1/2. Thus, the required frequency ratio between different sizes
bubbles is:
€ −3 2
( f ) 2cm " 2cm %
=$ ' = 2 2 ≅ 2.83 .
( f ) 4 cm # 4cm &
Similarly, if the surface tension is decreased by a factor of two, then
( f )σ 2 #1/2 &−1 2 1
=% ( = ≅ 0.707 .
€ ( f )σ $ 1 ' 2

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.43. In general, boundary layer skin friction, τw, depends on the fluid velocity U above
the boundary layer, the fluid density ρ, the fluid viscosity µ, the nominal boundary layer
thickness δ, and the surface roughness length scale ε.
a) Generate a dimensionless scaling law for boundary layer skin friction.
b) For laminar boundary layers, the skin friction is proportional to µ. When this is true, how must
τw depend on U and ρ?
c) For turbulent boundary layers, the dominant mechanisms for momentum exchange within the
flow do not directly involve the viscosity µ. Reformulate your dimensional analysis without it.
How must τw depend on U and ρ when µ is not a parameter?
d) For turbulent boundary layers on smooth surfaces, the skin friction on a solid wall occurs in a
viscous sublayer that is very thin compared to δ. In fact, because the boundary layer provides a
buffer between the outer flow and this viscous sub-layer, the viscous sublayer thickness lv does
not depend directly on U or δ. Determine how lv depends on the remaining parameters.
e) Now consider nontrivial roughness. When ε is larger than lv a surface can no longer be
considered fluid-dynamically smooth. Thus, based on the results from parts a) through d) and
anything you may know about the relative friction levels in laminar and turbulent boundary
layers, are high- or low-speed boundary layer flows more likely to be influenced by surface
roughness?

Solution 1.43. a) Construct the parameter & units matrix and recognizing that τw is a stress and
has units of pressure.
τw U ρ µ δ ε
–––––––––––––––––––––––––––––––
M 1 0 1 1 0 0
L -1 1 -3 -1 1 1
T -2 -1 0 1 0 0

The rank of this matrix is three. There are 6 parameters and 3 independent units, thus there will
be 6 – 3 = 3 dimensionless groups. By inspection these groups are: a skin-friction coefficient =
τ ρUδ ε
Π1 = w 2 , a Reynolds number = Π 2 = , and the relative roughness = Π 3 = . Thus the
ρU µ δ
τ & ρUδ ε )
dimensionless law is: w 2 = f ( , + where f is an undetermined function.
ρU ' µ δ*
€ b) Use the result of part €
a) and set τ w ∝ µ . This involves requiring€Π1 to be proportional to 1/ Π2
τ µ &ε)
so the revised form of the dimensionless law in part a) is: w 2 = g( + , where g is an
€ ρU ρUδ ' δ *
µU % ε (
undetermined function. € Simplify this relationship to find: τ w = g' * . Thus, in laminar
δ &δ )
boundary layers, τw is proportional to U and independent
€ of ρ.
c) When µ is not a parameter the second dimensionless group from part a) must be dropped.
τ &ε )
Thus, the dimensionless law becomes: w 2 = h(€ + where h is an undetermined function. Here
ρU 'δ *
2
we see that τ w ∝ ρU . Thus, in turbulent boundary layers, τw is linearly proportional to ρ and

€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

quadratically proportional to U. In reality, completely dropping µ from the dimensional analysis

is not quite right, and the skin-friction coefficient (Π1 in the this problem) maintains a weak
dependence on the Reynolds number when ε/δ << 1.
d) For this part of this problem, it is necessary to redo the dimensional analysis with the new
length scale lv and the three remaining parameters: τw, ρ, and µ. Here there are four parameters
l ρτ w
and three units, so there is only one dimensionless group: Π = ν . This means that:
µ
lν ∝ µ ρτ w = ν τ w ρ = ν u* .
In the study of wall bounded turbulent flows, the length scale lv is commonly known as the
viscous wall unit and u* is known as the friction or shear velocity.
€
e) The results of part b) and part c) both suggest that τw will be larger at high flow speeds than at
lower flow speeds. This € means that l will be smaller at high flow speeds for both laminar and
v
turbulent boundary layers. Thus, boundary layers in high-speed flows are more likely to be
influenced by constant-size surface roughness.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.44. Turbulent boundary layer skin friction is one of the fluid phenomena that limit
the travel speed of aircraft and ships. One means for reducing the skin friction of liquid boundary
layers is to inject a gas (typically air) from the surface on which the boundary layer forms. The
shear stress, τw, that is felt a distance L downstream of such an air injector depends on: the
volumetric gas flux per unit span q (in m2/s), the free stream flow speed U, the liquid density ρ,
the liquid viscosity µ, the surface tension σ, and gravitational acceleration g.
a) Formulate a dimensionless law for τw in terms of the other parameters.
b) Experimental studies of air injection into liquid turbulent boundary layers on flat plates has
found that the bubbles may coalesce to form an air film that provides near perfect lubrication,
τ w → 0 for L > 0, when q is high enough and gravity tends to push the injected gas toward the
plate surface. Reformulate your answer to part a) by dropping τw and L to determine a
dimensionless law for the minimum air injection rate, qc, necessary to form an air layer.
c) Simplify the result of part c) when surface tension can be neglected.
d) Experimental studies (Elbing et al. 2008) find that qc is proportional to U2. Using this
information, determine a scaling law for qc involving the other parameters. Would an increase in
g cause qc to increase or decrease?

Solution 1.44. a) Construct the parameter & units matrix and recognizing that τw is a stress and
has units of pressure.
τw L q U ρ µ σ g
–––––––––––––––––––––––––––––––––––––––––––
M 1 0 0 0 1 1 1 0
L -1 1 2 1 -3 -1 0 1
T -2 0 -1 -1 0 -1 -2 -2
This rank of this matrix is three. There are 8 parameters and 3 independent units, thus there will
be 8 – 3 = 5 dimensionless groups. By inspection these groups are: a skin-friction coefficient =
τ ρUL U
Π1 = w 2 , a Reynolds number = Π 2 = , a Froude number = Π 3 = , a capillary number
ρU µ gL
µU ρq
= Π4 = , and flux ratio = Π 5 = . Thus the dimensionless law is:
σ µ
€ τw % ρUL U µU € ρq (
= f' , , € function.
, * where f is an undetermined
2
ρU & µ gL σ µ )
€ b) Dropping τw means € dropping Π1. Dropping L means combining Π2 and Π3 to form a new
ρUL U 3 ρU 3
dimensionless group: Π 2Π 23 = = . Thus, with Π5 as the solution parameter, the
€ µ gL µg

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

scaling law for the minimum air injection rate, qc, necessary to form an air layer is:
ρqc µ = φ ( ρU 3 µg, µU σ ) where φ is an undetermined function.
c) When σ is not a parameter, Π4 can be dropped leaving: ρqc µ = ϕ ( ρU 3 µg) where ϕ is an
undetermined function.
€ d) When qc is proportional to U2, then dimensional analysis requires:
23 13
qc = (µ ρ)const.( ρU 3 µg€) = const.U 2 (µ ρg 2 ) .
So, an increase in g would cause qc to decrease.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.45. An industrial cooling system is in the design stage. The pumping requirements
are known and the drive motors have been selected. For maximum efficiency the pumps will be
directly driven (no gear boxes). The number Np and type of water pumps are to be determined
based on pump efficiency η (dimensionless), the total required volume flow rate Q, the required
pressure rise ΔP, the motor rotation rate Ω, and the power delivered by one motor W. Use
dimensional analysis and simple physical reasoning for the following items.
a) Determine a formula for the number of pumps.
b) Using Q, Np, ΔP, Ω, and the density (ρ) and viscosity (µ) of water, create the appropriate
number of dimensionless groups using ΔP as the dependent parameter.
c) Simplify the result of part b) by requiring the two extensive variables to appear as a ratio.
d) Simplify the result of part c) for high Reynolds number pumping where µ is no longer a
parameter.
e) Manipulate the remaining dimensionless group until Ω appears to the first power in the
numerator. This dimensionless group is known as the specific speed, and its value allows the
most efficient type of pump to be chosen (see Sabersky et al. 1999).

Solution 1.45. a) The total power that must be delivered to the fluid is QΔP. The power that one
pump delivers to the fluid will be ηW. Thus, Np will be the next integer larger than QΔP/ηW.
b) Construct the parameter & units matrix using ΔP as the solution parameter

ΔP Q Np Ω ρ µ
M 1 0 0 0 1 1
L -1 3 0 0 -3 -1
T -2 -1 0 -1 -0 -1

The rank of this matrix is three. There are 6 parameters and 3 independent units, thus there will
be 6 – 3 = 2 dimensionless groups. By inspection these groups are: a pressure coefficient =
ΔP ρQ 2 3Ω1 3
Π1 = 23
, the number of pumps = Π 2 = Np , and a Reynolds number = Π 3 = .
ρ (Ω2Q ) µ
c) The two extensive parameters (Q & Np) must for a ratio, so defining q = Q/Np, the two
ΔP ρ q 2 3Ω1 3
dimensionless groups are: 23
, and .
ρ (Ω2 q ) µ
ΔP
d) At high Reynolds number, the second dimensionless group will not matter. Thus, 23
ρ (Ω2 q )
alone will characterize the pump, and this group will be a constant.
Ωq1 2
e) The specific speed = 34
. Low values of the specific speed (below ~1/2 or so)
( ΔP ρ )
correspond centrifugal pumps that move relatively small amounts of liquid against relatively-
high pressure differences (a water pump for circulating a coolant through narrow passageways).
High values of the specific speed (above ~2 or so) correspond to propeller pumps that move
relatively high volumes of fluid against relatively-low pressure differences (a ventilation fan).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.46. Nearly all types of fluid filtration involve pressure driven flow through a porous
material.
a) For a given volume flow rate per unit area = Q/A, predict how the pressure difference across
the porous material = Δp, depends on the thickness of the filter material = L, the surface area per
unit volume of the filter material = Ψ, and other relevant parameters using dimensional analysis.
b) Often the Reynolds number of the flow in the filter pores is very much less than unity so fluid
inertia becomes unimportant. Redo the dimensional analysis for this situation.
c) To minimize pressure losses in heating, ventilating, and air-conditioning (HVAC) ductwork,
should hot or cold air be filtered?
d) If the filter material is changed and Ψ is lowered to one half its previous value, estimate the
change in Δp if all other parameters are constant. (Hint: make a reasonable assumption about the
dependence of Δp on L; they are both extensive variables in this situation).
Porous Material!

Q/A! Q/A!

Gage pressure ! Gage pressure !

= Δp! = 0!

L!

Solution 1.46. This question can be answered with dimensional analysis. The parameters are
drawn from the problem statement and the two fluid properties ρ = density and µ = viscosity.
The solution parameter is Δp, and the unit matrix is:

Δp Q/A L Ψ ρ µ
––––––––––––––––––––––––––––––––––––––
M 1 0 0 0 1 1
L –1 1 1 –1 –3 –1
T –2 –1 0 0 0 –1

There will be: 6 – 3 = 3 dimensionless groups. These groups are: a pressure coefficient =
Π1 = Δp ρ (Q / A)2 , a dimensionless thickness = Π 2 = LΨ , and a thickness-based Reynolds
number Π3 = ρ (Q / A)L µ . The dimensionless relationship must take the form:
Δp # ρ (Q / A)L &
2
= fn % LΨ, (.
ρ (Q / A) $ µ '
b) Dropping the density reduces the number of dimensionless groups. The product of the first
and third group is independent of the density, thus the revised dimensional analysis result is:
Δp ρ (Q / A)L ΔpL
2
⋅ = = fn ( LΨ ) .
ρ (Q / A) µ (Q / A)µ
c) The viscosity of gases increases with increasing temperature, thus to keep Δp low for a given
filter element (LΨ) the viscosity should be low. So, filter cold air.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

d) A reasonable assumption is that Δp will be proportional to the thickness of the porous

material, and this implies:
(Q / A)µ 2
Δp = ⋅ co nst. ( LΨ ) = const.(Q / A)µ LΨ 2 .
L
So, if Ψ is lowered by a factor of ½, then Δp will be lowered to ¼ of is previous value.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.47. A new industrial process requires a volume V of hot air with initial density ρ to be
moved quickly from a spherical reaction chamber to a larger evacuated chamber using a single
pipe of length L and interior diameter of d. The vacuum chamber is also spherical and has a
volume of Vf. If the hot air cannot be transferred fast enough, the process fails. Thus, a prediction
of the transfer time t is needed based on these parameters, the air’s ratio of specific heats γ, and
initial values of the air’s speed of sound c and viscosity µ.
a) Formulate a dimensionless scaling law for t, involving six dimensionless groups.
b) Inexpensive small-scale tests of the air-transfer process are untaken before construction of the
commercial-scale reaction facility. Can all these dimensionless groups be matched if the target
size for the pipe diameter in the small-scale tests is d´ = d/10? Would lowering or raising the
initial air temperature in the small-scale experiments help match the dimensionless numbers?

Solution 1.47. a) This question can be answered with dimensional analysis. The parameters are
drawn from the problem statement. The solution parameter is t, and the unit matrix is:

t V Vf L d γ c ρ µ
–––––––––––––––––––––––––––––––––––––––––––––––––
M 0 0 0 0 0 0 0 1 1
L 0 3 3 1 1 0 1 -3 -1
T 1 0 0 0 0 0 -1 0 -1

The rank of this matrix is 3, so there will be: 9 – 3 = 6 dimensionless groups. These groups are:
a dimensionless time: Π1 = ct/d; a volume ratio: Π2 = V/Vf; another volume ratio: Π2 = d3/Vf; an
aspect ratio: Π4 = L/d; the ratio of specific heats: Π5 = γ; and a sonic Reynolds number: Π6 =
ρcd/µ. Thus, the scaling law is:
ct ! V d 3 L ρ cd $
= φ ## , , , γ , && .
d V V
" f f d µ %
b) The first dimensionless group will be matched if the other five are matched. So, let primes
denote the small-scale test parameter. Matching the five independent dimensionless groups
means:
V V ! d 3 d !3 L L! ρ cd ρ !c!d !
= , = , = , γ = γ´, and = . (1, 2, 3, 4, 5)
V f V f! V f V f! d d ! µ µ!
Starting from the target pipe size ratio, d´ = d/10, (2) implies:
d 3 Vf
3
= = 10 3 or V f! = V f 10 3 , and V ! = V 10 3 .
d ! V f!
where the finding for V´ follows from (1). Similarly, from (3), the results for the length of the
pipe are:
d L
= = 10 , so L! = L 10 .
d ! L!
The next independent dimensionless group, γ, can be matched by using air in the scale model
tests since it will be used in the full-scale device. The final independent dimensionless group is
likely to be the most difficult to match. Using (5) and d´ = d/10, implies:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

ρ cd ρ !c!d ρ !c! ρc
= or = 10 ,
µ 10µ ! µ! µ
and this relationship only involves material properties. The initial temperature (and pressure P)
of the small scale experiments can be varied, ρ is proportional to T–1 (and P+1), while both c and
µ are proportional to T1/2 (and nearly independent of P). Thus, small scale testing at reduced
temperature (and elevated pressure) might make it possible to match values of this dimensionless
group between the large- and small-scale experiments.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 1.48. Create a small passive helicopter from ordinary photocopy-machine paper (as
shown) and drop it from a height of 2 m or so. Note the helicopter’s rotation and decent rates
once it’s rotating steadily. Repeat this simple experiment with different sizes of paper clips to
change the helicopter’s weight, and observe changes in the rotation and decent rates.
a) Using the helicopter’s weight W, blade length l, and blade width (chord) c, and the air’s
density ρ and viscosity µ as independent parameters, formulate two independent dimensionless
scaling laws for the helicopter’s rotation rate Ω, and decent rate dz/dt.
b) Simplify both scaling laws for the situation where µ is no longer a parameter.
c) Do the dimensionless scaling laws correctly predict the experimental trends?
d) If a new paper helicopter is made with all dimensions smaller by a factor of two. Use the
scaling laws found in part b) to predict changes in the rotation and decent rates. Make the new
smaller paper helicopter and see if the predictions are correct.

7 cm!

7 cm!

paper clip!
2 cm!
4 cm!

Solution 1.48. The experiments clearly show that Ω and dz/dt increase with increasing W.
a) This question can be answered with dimensional analysis. The parameters are drawn from the
problem statement. The first solution parameter is Ω (the rotation rate) and the units matrix is:

Ω W l c ρ µ
––––––––––––––––––––––––––––––––
M 0 1 0 0 1 1
L 0 1 1 1 -3 -1
T -1 -2 0 0 0 -1

The rank of this matrix is 3, so there will be: 6 – 3 = 3 dimensionless groups. These groups are:
Π1 = Ωρ1/2l2/W1/2, Π2 = l/c, and Π3 = ρW/µ2. Thus, the scaling law is:
ρl 4 " l ρW %
Ω = φ$ , 2 ',
W #c µ &
where φ is an undetermined function. The second solution parameter is dz/dt (the descent rate),
and it has the same units as Ωl, so it's scaling law is:
! dz $ ρl 2 ! l ρW $
# & =ψ# , 2 &.
" dt % W "c µ %
where ψ is an undetermined function.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

b) When µ is no longer a parameter, both scaling laws simplify

ρl 4 #l& ! dz $ ρl 2 !l$
Ω = Φ % ( and # & = Ψ# &,
W $c' " dt % W "c%
where Φ and Ψ are different undetermined functions. These laws imply
1 W #l& dz 1 W " l %
Ω= 2 Φ % ( and = Ψ$ ' ,
l ρ $c' dt l ρ # c &
c) These scaling laws are consistent with experimental results; the rotation and decent rates both
increase when W increases.
d) If the aspect ratio, l/c, is fixed, then decreasing l by a factor of two should increase Ω by a
factor of 4, and dz/dt by a factor of two. When the smaller helicopter is made, the experiments do
appear to confirm these predictions; both Ω and dz/dt do increase, and the increase for Ω is
larger.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.1. For three spatial dimensions, rewrite the following expressions in index notation
and evaluate or simplify them using the values or parameters given, and the definitions of δij and
εijk wherever possible. In b) through e), x is the position vector, with components xi.
a) b ⋅ c where b = (1, 4, 17) and c = (–4, –3, 1)
b) (u ⋅ ∇ ) x where u a vector with components ui.
c) ∇φ , where φ = h ⋅ x and h is a constant vector with components hi.
€ d) ∇ × u, where u = Ω × x and Ω is a constant vector with components Ωi.
"1 2 3&
€ $ $
€ e) C€ ⋅ x , where C = #0 1 2'
€ $0 0 1$
% (

€ Solution 2.1. a) b ⋅ c = bic i = 1(−4) + 4(−3) + 17(1) = −4 −12 + 17 = +1

+x .
€ ∂ + % ∂ ( % ∂ ( % ∂ (.- 1 0
b) (u ⋅ ∇ ) x = u j x i = -u1' * + u2 ' * + u3 ' *0 x 2
∂x j , & ∂x1 ) & ∂x 2 ) & ∂x 3 )/- 0
€ -,x 3 0/
) # ∂x & # ∂x & # ∂x & ,
+ u1% 1 ( + u2 % 1 ( + u3 % 1 (.
+ $ ∂x1 ' $ ∂x 2 ' $ ∂x 3 '. )u ⋅1+ u ⋅ 0 + u ⋅ 0, ) u1 ,
€ + # ∂x 2 & # ∂x 2 & # ∂x 2 &. + 1 2 3
. + .
= +u1% ( + u2 % ( + u3 % (. = +u1 ⋅ 0 + u2 ⋅1+ u3 ⋅ 0. = u jδij = +u2 . = ui
+ $ ∂x1 ' $ ∂x 2 ' $ ∂x 3 '.
+ . +*u3 .-
#
+ ∂x 3 & # ∂x 3 & # ∂x 3 &. *u1 ⋅ 0 + u2 ⋅ 0 + u3 ⋅1-
+u1%$ ∂x (' + u2 %$ ∂x (' + u3 %$ ∂x ('.
* 1 2 3 -
∂φ ∂ ∂x i
c) ∇φ = = ( hi x i ) = hi = hiδij = h j = h
∂x j ∂x j ∂x j
d) ∇€ × u = ∇ × (Ω × x ) = ε ∂ (ε Ω x ) = ε ε Ω δ = (δ δ − δ δ )Ω δ = (δ δ − δ δ )Ω
ijk klm l m ijk klm l jm il jm im jl l jm il jj ij jl l
∂x j
€ = ( 3δil − δil )Ωl = 2δil Ωl = 2Ωl = 2Ω
Here, the following identities have been used: εijkεklm = δilδ jm − δimδ jl , δijδ jk = δik , δ jj = 3 , and
€ δij Ω j = Ωi
€ #1 2 3'# x1 ' # x1 + 2x 2 + 3x 3 '
% %% % % %
e) C ⋅ x = Cij x j = $0 1 2($ x 2 ( = $€ x 2 + 2x 3 ( € €
%0 0 1%% x % % x3 %
€ & )& 3 ) & )

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.2. Starting from (2.1) and (2.3), prove (2.7).

Solution 2.2. The two representations for the position vector are:
x = x1e1 + x 2e 2 + x 3e 3 , or x = x1"e1" + x 2" e"2 + x 3" e"3 .
Develop the dot product of x with e1 from each representation,
e1 ⋅ x = e1 ⋅ ( x1e1 + x 2e 2 + x 3e 3 ) = x1e1 ⋅ e1 + x 2e1 ⋅ e 2 + x 3e1 ⋅ e 3 = x1 ⋅1+ x 2 ⋅ 0 + x 3 ⋅ 0 = x1 ,
and e1 ⋅ x = e1 ⋅ ( x1#e1# + x #2e#2 +€x #3e#3 ) = x1#e1 ⋅ e1# + x #2e1 ⋅ e#2 + x #3e1 ⋅ e#3 = x #iC1i ,
€
set these equal to find:
x1 = x "iC1i ,
€
where Cij = e i ⋅ e#j is a 3 × 3 matrix of direction cosines. In an entirely parallel fashion, forming
€
the dot product of x with e2, and x with e2 produces:
€ x 2 = x "iC2i and x 3 = x "iC3i .
€
Thus, for any component xj, where j = 1, 2, or 3, we have:
x j = x "iC ji ,
which is (2.7).
€ €

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.3. For two three-dimensional vectors with Cartesian components ai and bi, prove the
Cauchy-Schwartz inequality: (aibi)2 ≤ (ai)2(bi)2.

Solution 2.3. Expand the left side term,

(ai bi )2 = (a1b1 + a2 b2 + a3b3 )2 = a12 b12 + a22 b22 + a32 b32 + 2a1b1a2 b2 + 2a1b1a3b3 + 2a2 b2 a3b3 ,
then expand the right side term,
(ai )2 (bi )2 = (a12 + a22 + a32 )(b12 + b22 + b32 )
= a12 b12 + a22 b22 + a32 b32 + (a12 b22 + a22 b12 ) + (a12 b32 + a32 b12 ) + (a32 b22 + a22 b32 ).
Subtract the left side term from the right side term to find:
(ai )2 (bi )2 − (ai bi )2
= (a12 b22 − 2a1b1a2 b2 + a22 b12 ) + (a12 b32 − 2a1b1a3b3 + a32 b12 ) + (a32 b22 − 2a2 b2 a3b3 + a22 b32 )
2
= (a1b2 − a2 b1 )2 + (a1b3 − a3b1 )2 + (a3b2 − a2 b3 )2 = a × b .
Thus, the difference (ai )2 (bi )2 − (ai bi )2 is greater than zero unless a = (const.)b then the
difference is zero.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.4. For two three-dimensional vectors with Cartesian components ai and bi, prove the
triangle inequality: a + b ≥ a + b .

Solution 2.4. To avoid square roots, square both side of the equation; this operation does not
change the equation's meaning. The left side becomes:
2 2 2
(a + b) = a +2 a b + b ,
and the right side becomes:
2 2 2
a + b = (a + b)⋅ (a + b) = a ⋅ a + 2a ⋅ b + b ⋅ b = a + 2a ⋅ b + b .
So,
2 2
(a + b) − a + b = 2 a b − 2a ⋅ b .
Thus, to prove the triangle equality, the right side of this last equation must be greater than or
equal to zero. This requires:
a b ≥ a ⋅ b or using index notation: ai2 bi2 ≥ ai bi ,
which can be squared to find:
ai2 bi2 ≥ (ai bi )2 ,
and this is the Cauchy-Schwartz inequality proved in Exercise 2.3. Thus, the triangle equality is
proved.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.5. Using Cartesian coordinates where the position vector is x = (x1, x2, x3) and the
fluid velocity is u = (u1, u2, u3), write out the three components of the vector:
(
(u ⋅ ∇)u = ui ∂u j ∂x i . )
Solution 2.5.
+ % ∂u ( % ∂u ( % ∂u ( /
€ - u1' 1 * + u2 ' 1 * + u3 ' 1 * -
- & ∂x1 ) & ∂x 2 ) & ∂x 3 ) -
% ∂u ( % ∂u ( % ∂u ( % ∂ u ( - % ∂u ( % ∂u ( % ∂u (-
a) (u ⋅ ∇ )u = ui ' j * = u1' j * + u2 ' j * + u3 ' j * = , u1' 2 * + u2 ' 2 * + u3 ' 2 *0
& ∂x i ) & ∂x1 ) & ∂x 2 ) & ∂x 3 ) - & ∂x1 ) & ∂x 2 ) & ∂x 3 )-
%
- ∂u3 ( % ∂u3 ( % ∂u3 (-
u
- 1 ' * + u 2 ' * + u 3' *-
. & ∂x1 ) & ∂x 2 ) & ∂x 3 )1
) # ∂u & # ∂u & # ∂u & -
+ u% ( + v% ( + w% ( +
+ $ ∂x ' $ ∂y ' $ ∂z ' +
€ + # ∂v & # ∂v & # ∂v & +
= * u% ( + v% ( + w% ( .
+ $ ∂x ' $ ∂y ' $ ∂z ' +
+ # ∂w & # ∂w & # ∂w &+
+ u%$ (' + v% ( + w%$ ('+
, ∂x $ ∂y ' ∂z /
( )
The vector in this exercise, (u ⋅ ∇ )u = ui ∂u j ∂x i , is an important one in fluid mechanics. As
described in Ch. 3, it is the nonlinear advective acceleration.
€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.6. Convert ∇ × ∇ρ to indicial notation and show that it is zero in Cartesian
coordinates for any twice-differentiable scalar function ρ.

Solution 2.6. Start with the definitions of the cross product and the gradient.
∂ ∂ 2ρ
∇ × ( ∇ρ ) = εijk (∇ρ )k = εijk
∂xj ∂ x j∂ xk
Write out the vector component by component recalling that εijk = 0 if any two indices are equal.
Here the "i" index is the free index.
! 2 2 % ! 2 2 %
# ε123 ∂ ρ + ε132 ∂ ρ # # ∂ ρ – ∂ ρ #
# ∂ x2∂ x3 ∂ x3∂ x2 # # ∂ x2∂ x3 ∂ x3∂ x2 #
2 # 2 # # #
∂ ρ # ∂ ρ ∂ 2ρ # # ∂ 2ρ ∂ 2ρ #
εijk = " ε 213 + ε 231 =
& " – + &=0,
∂ x j∂ xk # ∂ x1∂ x3 ∂ x3∂ x1 # # ∂ x1∂ x3 ∂ x3∂ x1 #
# 2
∂ ρ ∂ 2ρ # # ∂ 2ρ ∂ 2ρ #
# ε312 + ε321 # # − #
#
$ ∂ x1∂ x2 ∂ x2∂ x1 #
' # $ ∂ x1∂ x2 ∂ x2∂ x1 #
'
where the middle equality follows from the definition of εijk (2.18), and the final equality follows
∂ 2ρ ∂ 2ρ
when ρ is twice differentiable so that = .
∂x j∂x k ∂x k∂x j

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.7. Using indicial notation, show that a × (b × c) = (a ⋅ c)b − (a ⋅ b)c. [Hint: Call d ≡ b
× c. Then (a × d)m = εpqmapdq = εpqmapεijqbicj. Using (2.19), show that (a × d)m = (a ⋅ c)bm − (a ⋅
b)cm.]

Solution 2.7. Using the hint and the definition of εijk produces:
(a × d)m = εpqmapdq = εpqmapεijqbicj = εpqmεijq bicjap = –εijqεqpm bicjap.
Now use the identity (2.19) for the product of epsilons:
(a × d)m = – (δipδjm – δimδpj) bicjap = – bpcmap + bmcpap.
Each term in the final expression involves a sum over "p", and this is a dot product; therefore
(a × d)m = – (a ⋅ b)cm + bm(a ⋅ c).
Thus, for any component m = 1, 2, or 3,
a × (b × c) = − (a ⋅ b)c + (a ⋅ c)b = (a ⋅ c)b − (a ⋅ b)c.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.8. Show that the condition for the vectors a, b, and c to be coplanar is εijkaibjck = 0.

Solution 2.8. The vector b × c is perpendicular to b and c. Thus, a will be coplanar with b and c
if it too is perpendicular to b × c. The condition for a to be perpendicular with b × c is:
a ⋅ (b × c) = 0.
In index notation, this is aiεijkbjck = 0 = εijkaibjck.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.9. Prove the following relationships: δijδij = 3, εpqrεpqr = 6, and εpqiεpqj = 2δij.

Solution 2.9. (i) δijδij = δii = δ11 + δ22 + δ33 = 1 + 1 + 1 = 3.

For the second two, the identity (2.19) is useful.
(ii) εpqrεpqr = εpqrεrpq = δppδqq – δpqδpq = 3(3) – δpp = 9 – 3 = 6.
(iii) εpqiεpqj = εipqεpqj = – εipqεqpj = – (δipδpj – δijδpp) = – δij + 3δij = 2δij.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.10. Show that C⋅CT = CT⋅C = δ, where C is the direction cosine matrix and δ is the
matrix of the Kronecker delta. Any matrix obeying such a relationship is called an orthogonal
matrix because it represents transformation of one set of orthogonal axes into another.

Solution 2.10. To show that C⋅CT = CT⋅C = δ, where C is the direction cosine matrix and δ is
the matrix of the Kronecker delta. Start from (2.5) and (2.7), which are
x "j = x iCij and x j = x "iC ji ,
respectively, and change the index "i" into "m" on (2.5): x "j = x m Cmj . Substitute this into (2.7) to
find:
x j = x "iC ji = ( x m Cmi )C ji = CmiC ji x m .
€ €
However, we also have xj = δjmxm, so €
δ jm x m = CmiC ji x m → δ jm = CmiC ji ,
which can be written:€
δ jm = CmiCijT = C⋅CT,
and taking the transpose
€ of the thisT produces: T
(δ jm ) = δmj = (CmiCijT ) = CmiT Cij = CT⋅C.
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.11. Show that for a second-order tensor A, the following quantities are invariant
under the rotation of axes:
A11 A12 A22 A23 A11 A13
I1 = Aii , I 2 = + + , and I3 = det(Aij).
A21 A22 A32 A33 A31 A33
[Hint: Use the result of Exercise 2.8 and the transformation rule (2.12) to show that Iʹ′1 = Aʹ′ii =
Aii.= I1. Then show that AijAji and AijAjkAki are also invariants. In fact, all contracted scalars of the
form AijAjk ⋅⋅⋅ Ami are invariants. Finally, verify that I 2 = 12 "# I12 − Aij A ji $% ,
I 3 = 13 "# Aij A jk Aki − I1 Aij A ji + I 2 Aii $% . Because the right-hand sides are invariant, so are I2 and I3.]

Solution 2.11. First prove I1 is invariant by using the second order tensor transformation rule
(2.12):
" = Cim C jn Aij .
Amn
Replace Cjn by CnjT and set n = m,
" = CimCnjT Aij → Amm
Amn " = Cim CmjT Aij .
Use the result of Exercise 2.8, € T
δij = Cim Cmj = , to find:
€ " = δij Aij = Aii .
I1 = Amm
Thus, the first invariant€ is does not depend on a rotation of the coordinate axes.
Now consider€whether or not AmnAnm is invariant under a rotation of the coordinate axes.
Start with a double application of (2.12):
€
Amn ( )(
" = Cim C jn Aij C pn Cqm A pq = C jn CnpT Cim Cmq
" Anm ) ( T
)(
Aij A pq . )
From the result of Exercise 2.8, the factors in parentheses in the last equality are Kronecker delta
functions, so
Amn " = δ jpδiq Aij A pq = Aij A ji .
" Anm
€
Thus, the matrix contraction AmnAnm does not depend on a rotation of the coordinate axes.
The manipulations for AmnAnpApm are a straightforward extension of the prior efforts for
Aii and AijAji.
€
Amn
" Anp ( )( )( ) ( )(
" A"pm = Cim C jn Aij Cqn Crp Aqr Csp Ctm Ast = C jn CnqT Crp C psT CimCmtT Aij Aqr Ast . )( )
Again, the factors in parentheses are Kronecker delta functions, so
Amn " A"pm = δ jqδrsδit Aij Aqr Ast = Aiq Aqs Asi ,
" Anp
which implies that the matrix contraction AijAjkAki does not depend on a rotation of the coordinate
€ axes.
Now, for the second invariant, verify the given identity, starting from the given definition
for I2. €
A11 A12 A22 A23 A11 A13
I2 = + +
A21 A22 A32 A33 A31 A33
= A11 A22 − A12 A21 + A22 A33 − A23 A32 + A11 A33 − A13 A31
= A11 A22 + A22 A33 + A11 A33 − ( A12 A21 + A23 A32 + A13 A31 )
€ (
= 12 A112 + 12 A222 + 12 A332 + A11 A22 + A22 A33 + A11 A33 − A12 A21 + A23 A32 + A13 A31 + 12 A112 + 12 A222 + 12 A332 )
2
€ = 1
[ A11 + A22 + A33 ] − 1
(2A 2
A21 + 2A23 A32 + 2A13 A31 + A + A + A 2 2
)
2 2 12 11 22 33
€
€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

= 12 I12 − 12 ( A11 A11 + A12 A21 + A13 A31 + A12 A21 + A22 A22 + A23 A32 + A13 A31 + A23 A32 + A33 A33 )
= 12 I12 − 12 ( Aij A ji ) = 12 (I12 − Aij A ji )
Thus, since I2 only depends on I1 and AijAji, it is invariant under a rotation of the coordinate axes
€ because I1 and AijAji are invariant under a rotation of the coordinate axes.
The manipulations for the third invariant are a tedious but not remarkable. Start from the
€ given definition for I3, and group like terms.
I3 = det ( Aij ) = A11 (A22 A33 − A23 A32 ) − A12 (A21 A33 − A23 A31) + A13 (A21 A32 − A22 A31 )
= A11 A22 A33 + A12 A23 A31 + A13 A32 A21 − ( A11 A23 A32 + A22 A13 A31 + A33 A12 A21 ) (a)
Now work from the given identity. The triple matrix product AijAjkAki has twenty-seven terms:
A113 + A11 A12 A21 + A11 A13 A31 + A12 A21 A11 + A12 A22 A21 + A12 A23 A31 + A13 A31 A11 + A13 A32 A21 + A13 A33 A31 +
€
3
€ A21 A11 A12 + A21 A12 A22 + A21 A13 A32 + A22 A21 A12 + A22 + A22 A23 A32 + A23 A31 A12 + A23 A32 A22 + A23 A33 A32 +
A31 A11 A13 + A31 A12 A23 + A31 A13 A33 + A32 A21 A13 + A32 A22 A23 + A32 A23 A33 + A33 A31 A13 + A33 A32 A23 + A333
These can be grouped as follows:
Aij A jk Aki = 3(A12 A23 A31 + A13 A32 A21 ) + A11(A112 + 3A12 A21 + 3A13 A31) +
2 2
A22 (3A21 A12 + A22 + 3A23 A32 ) + A33 (3A31 A13 + 3A32 A23 + A33 ) (b)
The remaining terms of the given identity are:
−I1 Aij A ji + I2 Aii = I1(I2 – Aij A ji ) = I1 (I2 + 2I2 − I12 ) = 3I1I2 – I13 ,
€
where€the result for I2 has been used. Evaluating the first of these two terms leads to:
3I1I2 = 3(A11 + A22 + A33 )(A11 A22 − A12 A21 + A22 A33 − A23 A32 + A11 A33 − A13 A31 )
= 3(A11 + A22 + A33 )(A11 A22 + A22 A33 + A11 A33 ) − 3(A11 + A22 + A33 )(A12 A21 + A23 A32 + A13 A31 ) .
Adding this € to (b) produces:
Aij A jk Aki + 3I1I2 = 3(A12 A23 A31 + A13 A32 A21 ) + 3(A11 + A22 + A33 )(A11 A22 + A22 A33 + A11 A33 ) +
€
€ A11 (A112 − 3A23 A32 ) + A22 (A22 2
− 3A13 A31 ) + A33 (A33 2
− 3A12 A21 )
= 3(A12 A23 A31 + A13 A32 A21 − A11 A23 A32 − A22 A13 A31 − A33 A12 A21 ) +
3(A11 + A22 + A33 )(A11 A22 + A22 A33 + A11 A33 ) + A113 + A22 3 3
+ A33 (c)
€
The last term of the given identity is:
€3
I13 = A€ 3 3 2 2 2 2 2 2
11 + A22 + A33 + 3(A11 A22 + A11 A33 + A22 A11 + A22 A33 + A33 A11 + A33 A22 ) + 6A11 A22 A33
= A113 + A223 3
+ A33 + 3(A11 + A22 + A33 )(A11 A22 + A11 A33 + A22 A33 ) – 3A11 A22 A33
€
Subtracting this from (c) produces:
€ Aij A jk Aki + 3I1I2 − I13 = 3(A12 A23 A31 + A13 A32 A21 − A11 A23 A32 − A22 A13 A31 − A33 A12 A21 + A11 A22 A33 )
€ = 3I3 .
This verifies that the given identity for I3 is correct. Thus, since I3 only depends on I1, I2, and
AijAjkAki, it is invariant under a rotation of the coordinate axes because these quantities are
€ €
invariant under a rotation of the coordinate axes as shown above.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.12. If u and v are vectors, show that the products uiυj obey the transformation rule
(2.12), and therefore represent a second-order tensor.

Solution 2.12. Start by applying the vector transformation rule (2.5 or 2.6) to the components of
u and v separately,
u"m = Cim ui , and v "n = C jn v j .
The product of these two equations produces:
u"m v "n = Cim C jn uiv j ,
which is the same as (2.12)
€ for second order tensors.
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.13. Show that δij is an isotropic tensor. That is, show that δʹ′ij = δij under rotation of
the coordinate system. [Hint: Use the transformation rule (2.12) and the results of Exercise 2.10.]

Solution 2.13. Apply (2.12) to δij,

" = CimC jnδij = CimCin = CmiT Cin = δmn .
δmn
where the final equality follows from the result of Exercise 2.10. Thus, the Kronecker delta is
invariant under coordinate rotations.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.14. If u and v are arbitrary vectors resolved in three-dimensional Cartesian

2
coordinates, use the definition of vector magnitude, a = a ⋅ a , and the Pythagorean theorem to
show that u⋅v = 0 when u and v are perpendicular.

Solution 2.14. Consider the magnitude of the sum u + v,

2
u + v = (u1 + v1 ) 2 + (u2 + v 2 ) 2 + (u3 + v 3 ) 2
= u12 + u22 + u32 + v12 + v 22 + v 32 + 2u1v1 + 2u2v 2 + 2u3v 3
2 2
= u + v + 2u ⋅ v ,
€ which can be rewritten:
2 2 2
€ u + v − u − v = 2u ⋅ v .
€When u and v are perpendicular, the Pythagorean theorem requires the left side to be zero. Thus,
u ⋅ v = 0.
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.15. If u and v are vectors with magnitudes u and υ, use the finding of Exercise 2.14
to show that u⋅v = uυcosθ where θ is the angle between u and v.

Solution 2.15. Start with two arbitrary vectors (u and v), and view them so that the plane they
define is coincident with the page and v is horizontal. Consider two additional vectors, βv and w,
that are perpendicular (v⋅w = 0) and can be summed together to produce u: w + βv = u.
u

w
θ v

βv
Compute the dot-product of u and v:
u⋅v = (w + βv) ⋅v = w⋅v + βv⋅v = βυ2.
where the final equality holds because v⋅w = 0. From the geometry of the figure:
βv βυ u
cosθ ≡ = , or β = cosθ .
u u υ
Insert this into the final equality for u⋅v to find:
%u (
u ⋅ v = ' cos θ *υ 2 = uυ cosθ .
€ &υ € )

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.16. Determine the components of the vector w in three-dimensional Cartesian

coordinates when w is defined by: u⋅w = 0, v⋅w = 0, and w⋅w = u2υ2sin2θ, where u and v are
known vectors with components ui and υi and magnitudes u and υ, respectively, and θ is the
angle between u and v. Choose the sign(s) of the components of w so that w = e3 when u = e1
and v = e2.

Solution 2.16. The effort here is primarily algebraic. Write the three constraints in component
form:
u⋅w = 0, or u1w1 + u2 w 2 + u3 w 3 = 0 , (1)
v⋅w = 0, or υ1w1 + υ 2 w 2 + υ 3 w 3 = 0 , and (2)
The third one requires a little more effort since the angle needs to be eliminated via a dot
product:
€
w⋅w = u2υ2sin2θ = u2υ2(1 – cos2θ) = u2υ2 – (u⋅w)2 or
€
w12 + w 22 + w 32 = (u12 + u22 + u32 )(υ12 + υ 22 + υ 32 ) − (u1υ1 + u2υ 2 + u3υ 3 ) 2 , which leads to
w12 + w 22 + w 32 = (u1υ 2 − u2υ1 ) 2 + (u1υ 3 − u3υ1 ) 2 + (u2υ 3 − u3υ 2 ) 2 . (3)
Equation (1) implies:
w1 = −(w 2 u2 + w 3 u3 ) u1 (4)
€
Combine (2) and (4) to eliminate w1, and solve the resulting equation for w2:
€ $ υ ' $ υ '
−υ1 (w 2 u2 + w 3 u3 ) u1 + υ 2 w 2 + υ 3 w 3 = 0 , or &− 1 u2 + υ 2 )w 2 + & − 1 u3 + υ 3 ) w 3 = 0 .
€ % u1 ( % u1 (
Thus:
$υ ' $ υ ' $u υ − uυ '
€ w 2 = +w 3 & 1 u3 − υ 3 ) &− 1 u2 + υ 2 ) = w 3 & 3 1 1 3 ) . (5)
% u1 ( € % u1 ( % u1υ 2 − u2υ1 (
Combine (4) and (5) to find:
w $$ υ u − υ 3 u1 ' ' w 3 $ υ1u3 u2 − υ 3 u1u2 + υ 2 u1u3 − υ1u2 u3 '
w1 = − 3 && 1 3 ) u2 + u3 ) = − & +)
€ u1 %% υ 2 u1 − υ1u2 ( ( u1 % υ 2 u1 − υ1u2 (
w $ −υ u u + υ 2 u1u3 ' $ u2υ 3 − u3υ 2 '
=− 3& 3 1 2 ) = w3& ). (6)
u1 % υ 2 u1 − υ1u2 ( % u1υ 2 − u2υ1 (
Put
€ (5) and (6) into (3) and factor out w3 on the left side, then divide out the extensive common
factor that (luckily) appears on the right and as the numerator inside the big parentheses.

2
$€(u2υ 3 − u3υ 2 ) 2 + (u3υ1 − u1υ 3 ) 2 + (u1υ 2 − u2υ1 ) 2 ' 2 2 2
w &
3 2 ) = (u1υ 2 − u2υ1 ) + (u1υ 3 − u3υ1 ) + (u2υ 3 − u3υ 2 )
% (u1υ 2 − 2u$2υ1 ) 1 ' (
w3 & 2 ) = 1, so w 3 = ±(u1υ 2 − u2υ1 ) .
% (u1υ 2 − u2υ1 ) (
If u = (1,0,0), and v = (0,1,0), then using the plus sign produces w3 = +1, so w 3 = +(u1υ 2 − u2υ1 ) .
€ Cyclic permutation of the indices allows the other components of w to be determined:
w1€= u2υ 3 − u3υ 2 ,
€
w 2 = u3υ1 − u1υ 3 ,
€
w 3 = u1υ 2 − u2υ1 .
€
€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.17. If a is a positive constant and b is a constant vector, determine the divergence and
the curl of u = ax/x3 and u = b×(x/x2) where x = x12 + x 22 + x 32 ≡ x i x i is the length of x.

Solution 2.17. Start with the divergence calculations, and use x = x12 + x 22 + x 32 to save writing.
$ ' $
$ ax ' $ ∂ ∂ ∂ ' & x1€, x 2 , x 3 '
) = a& ∂ , ∂ , ∂ ) ⋅ $& x1, x 2 , x 3 ')
∇ ⋅ & 3 ) = a& , , )⋅
% x ( % ∂x1 ∂x 2 ∂x 3 ( & [ x 2 + x 2 + x 2 ] 3 2 ) % ∂x1 ∂x 2 ∂x 3 ( % x 3 (
% 1 2 3 ( €
# ∂ # x1 & ∂ # x 2 & ∂ # x 3 && # 1 3 x1 1 3 x2 1 3 x3 &
5 (
= a% % 3 ( + % 3(+ % 3 (( = a% 3 − 2x1 ) + 3 − 5 (
2x 2 ) + 3 − 5 (
2x 3 )(
$ ∂x1 $ x ' ∂x 2 $ x ' ∂x 3 $ x '' $ x 2x x 2x x 2x '
€ # 3 3( x 2 + x 2 + x 2 ) & # 3 3&
1 2 3
= a%% 3 − 5
( = a% 3 − 3 ( = 0 .
(
$x x ' $x x '
€ Thus, the vector field ax/x3 is divergence free even though it points away from the origin
everywhere.
% b × x ( % ∂ ∂ ∂ ( % b2 x 3 − b3 x 2 ,b3 x1 − b1 x 3 ,b1 x 3 − b2 x1 (
€ ∇⋅' 2 * =' , , *⋅' *
& x ) & ∂x1 ∂x 2 ∂x 3 ) & x12 + x 22 + x 32 )
$ ∂ $ b x − b x ' ∂ $ b3 x1 − b1 x 3 ' ∂ $ b1 x 2 − b2 x1 ''
=& & 2 3 2 3 2)+ & )+ & ))
% ∂x1 % x ( ∂x 2 % x2 ( ∂x 3 % x2 ((
€ # 2 & # 2 & # 2 &
= (b2 x 3 − b3 x 2 )%− 4 (2x1 )( + (b3 x1 − b1 x 3 )% − 4 (2x 2 )( + (b1 x 2 − b2 x1 )%− 4 (2x 3 )(
$ x ' $ x ' $ x '
€ 4
= − 4 (b2 x 3 x1 − b3 x 2 x1 + b3 x1 x 2 − b1 x 3 x 2 + b1 x 2 x 3 − b2 x1 x 3 ) = 0 .
x
This field is divergence free, too. The curl calculations produce:
€
$ ax ' $ ∂ ∂ ∂ ' $ x1, x 2 , x 3 ' $ ∂x −3 ∂x −3 ∂x −3 ∂x −3 ∂x −3 ∂x −3 '
∇ × & 3 ) = a& , , ×
) & =
) & a x − x , x − x , x − x )
€ % x ( % ∂x1 ∂x 2 ∂x 3 ( % x 3 ( % 3 ∂x 2 2
∂x 3
1
∂x 3
3
∂x1
2
∂x1
1
∂x 2 (
# 3 x3 3 x2 3x 3 x3 3 x2 3 x1 &
= a% − 5 (
2x 2 ) + 5 (
2x 3 ),− 15 (2x 3 ) + 5 (
2x1 ),− 5 (
2x1 ) + 5 (
2x 2 )( = (0,0,0)
$ 2x 2x 2x 2x 2x 2x '
3
€ Thus, thus the vector field ax/x is also irrotational.
$ b × x ' $ ∂ ∂ ∂ ' $ b2 x 3 − b3 x 2 ,b3 x1 − b1 x 3 ,b1 x 2 − b2 x1 '
∇ ×& 2 ) = & , , ) ×& ).
€ % x ( % ∂x1 ∂x 2 ∂x 3 ( % x12 + x 22 + x 32 (
There are no obvious simplifications here. Therefore, compute the first component and obtain the
others by cyclic permutation of the indices.
$b × x' ∂ $ b1 x 2 − b2 x1 ' ∂ $ b3 x1 − b1 x 3 '
€ ∇ ×& 2 ) = & )− & )
% x (1 ∂x 2 % x2 ( ∂x 3 % x2 (
b # −2 & b # −2 &
= 12 + (b1 x 2 − b2 x1 )% 4 (2x 2 + 12 − (b3 x1 − b1 x 3 )% 4 (2x 3
x $x ' x $x '
2 2 2
€ 2b x − 4b1 x 2 + 4b2 x1 x 2 + 4b3 x1 x 3 − 4b1 x 3 2b 4 x
= 1 4
= − 21 + 41 (b1 x1 + b2 x 2 + b3 x 3 )
x x x
This field is rotational. The other two components of its curl are:
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

$b × x' 2b 4 x $b × x' 2b 4 x
∇ × & 2 ) = − 22 + 42 (b1 x1 + b2 x 2 + b3 x 3 ) , ∇ × & 2 ) = − 23 + 43 (b1 x1 + b2 x 2 + b3 x 3 ) .
% x (2 x x % x (3 x x

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.18. Obtain the recipe for the gradient of a scalar function in cylindrical polar
coordinates from the integral definition (2.32).

Solution 2.18. Start from the appropriate form of (2.32),

1 ez
∇Ψ = lim ∫∫ ΨndA , where Ψ is a scalar function of "z z
V →0 V e!
A
position x. Here we choose a nearly rectangular volume "!
"R eR
V = (RΔϕ)(ΔR)(Δz) centered on the point x = (R, ϕ, z)
with sides aligned perpendicular to the coordinate
€ y
directions. Here the e unit vector depends on ϕ so its
ϕ

direction is slightly different at ϕ ± Δϕ/2. For small Δϕ, ! R

x
this can be handled by keeping the linear term of a
simple Taylor series: [eϕ ] ≅ eϕ ± (Δϕ 2)(∂eϕ ∂ϕ ) = eϕ  (Δϕ 2)e R . Considering the drawing
ϕ ±Δϕ 2
and noting that n is an outward normal, there are six contributions to ndA:
# ΔR & $ ΔR '
outside = % R + (ΔϕΔze R , inside = −& R − )ΔϕΔze R ,
$ 2 ' % 2 (
€
% Δϕ ( % Δϕ (
close vertical side = ΔRΔz'−eϕ − e R *, more distant vertical side = ΔRΔz'eϕ − eR * ,
& 2 ) & 2 )
top = RΔϕΔRe z ,
€ €and bottom = −RΔϕΔRe z .
Here all the unit vectors are evaluated at the center of the volume. Using a two term Taylor series
approximation for Ψ on each of the six surfaces, and taking the six contributions in the same
€ €
order, the integral definition becomes a sum of six terms representing ΨndA.
€ 5.( € 9
ΔR ∂Ψ +( ΔR + 1 .( ΔR ∂Ψ +( ΔR + 1
70* Ψ + -* R + -e R ΔϕΔz3 − 0* Ψ − -* R − -e R ΔϕΔz3 + 7
7/) 2 ∂R ,) 2 , 2 /) 2 ∂R ,) 2 , 2 7
1 7
7(. Δϕ ∂Ψ +( Δϕ + 1 .( Δϕ ∂Ψ +( Δϕ + 1 77
∇Ψ = lim 60* Ψ − *
- ϕ−e − e R -ΔRΔz 3 + 0* Ψ + *
- ϕe − e R -ΔRΔz 3 + :
ΔR →0 RΔϕΔRΔz ) 2 ∂ϕ ,) 2 , ) 2 ∂ϕ ,) 2 ,
Δϕ →0 7 / 2 / 2 7
Δz →0 7.( + 1 .( + 1 7
70* Ψ + Δz ∂Ψ -e z RΔϕΔR3 − 0* Ψ − Δz ∂Ψ -e z RΔϕΔR3 + ... 7
78/) 2 ∂z , 2 /) 2 ∂z , 2 7;
Here the mean value theorem has been used and all listings of Ψ and its derivatives above are
evaluated at the center of the volume. The largest terms inside the big {,}-brackets are
proportional to ΔϕΔRΔz. The remaining higher order terms vanish when the limit is taken.
€
/( Ψ R ∂Ψ + ( Ψ R ∂Ψ + 3
1* e R + e R -ΔϕΔRΔz − *− e R − e R -ΔϕΔRΔz +1
1) 2 2 ∂R , ) 2 2 ∂R , 1
1 1(eϕ ∂Ψ e R + ( eϕ ∂Ψ e R + 1
∇Ψ = lim 0* − Ψ-ΔϕΔRΔz + * − Ψ-ΔϕΔRΔz + 4
ΔR →0 RΔϕΔRΔz ) 2 ∂ϕ 2 , ) 2 ∂ϕ 2 ,
Δϕ →0 1 1
Δz →0 1( R ∂Ψ + ( R ∂Ψ + 1
1* e z -ΔϕΔRΔz − *− e z -ΔϕΔRΔz + ... 1
2) 2 ∂z , ) 2 ∂z , 5
'Ψ ∂Ψ 1 ∂Ψ Ψ ∂Ψ * ∂Ψ 1 ∂Ψ ∂Ψ
∇Ψ = ( e R + eR + eϕ − e R + ez + = eR + eϕ + ez
)R ∂R R ∂ϕ R ∂z , ∂R R ∂ϕ ∂z
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.19. Obtain the recipe for the divergence of a vector function in cylindrical polar
coordinates from the integral definition (2.32).

Solution 2.19. Start from the appropriate form of (2.32),

1 ez
∇ ⋅ Q = lim ∫∫ n ⋅ QdA , where Q = (QR, Q , Qz) is a vector "z z
V →0 V
ϕ
e!
A
"!
function of position x. Here we choose a nearly rectangular
"R eR
volume V = (RΔϕ)(ΔR)(Δz) centered on the point x = (R, ϕ,
z) with sides aligned perpendicular to the coordinate y
€
directions. Here the e unit vector depends on ϕ so its
ϕ

direction is slightly different at ϕ ± Δϕ/2. Considering the x ! R

drawing and noting that n is an outward normal, there are
six contributions to ndA:
# ΔR & $ ΔR '
outside = % R + (ΔϕΔze R , inside = −& R − )ΔϕΔze R , close vertical side = −ΔRΔz[eϕ ]ϕ −Δϕ 2 ,
$ 2 ' % 2 (

more distant vertical side = ΔRΔz[eϕ ] , top = RΔϕΔRe z , and bottom = −RΔϕΔRe z .
ϕ +Δϕ 2
Here the unit vectors are evaluated
€ € at the center of the volume unless € otherwise specified. Using
a two-term Taylor series approximation for the components of Q on each of the six surfaces, and
taking the six contributions to n ⋅ QdA in€the same order, the integral € definition becomes:
€ 5.( 1 . 1 9
ΔR ∂QR +( ΔR + ( ΔR ∂QR +( ΔR +
70* QR + -* R + -ΔϕΔz3 − 0* QR − -* R − -ΔϕΔz3 +7
7/) 2 ∂R ,) 2 , 2 /) 2 ∂R ,) 2 , 2 7
1 € 77.( Δϕ ∂Qϕ + 1 .( Δϕ ∂Qϕ + 1 77
∇ ⋅ Q = lim
ΔR →0 RΔϕΔRΔz
60* Qϕ − -( −ΔRΔz )3 0* ϕ
+ Q + -ΔRΔz 3 + :
Δϕ →0 7 /) 2 ∂ϕ , 2 /) 2 ∂ϕ , 2 7
Δz →0 7.( + 1 .( + 1 7
70* Qz + Δz ∂Qz -RΔϕΔR3 − 0*Qz − Δz ∂Qz - RΔϕΔR3 + ... 7
78/) 2 ∂z , 2 /) 2 ∂z , 2 7;
Here the mean value theorem has been used and all listings of the components of Q and their
derivatives are evaluated at the center of the volume. The largest terms inside the big {,}-
brackets are proportional to ΔϕΔRΔz. The remaining higher order terms vanish when the limit is
€ taken.
0(QR R ∂QR + ( QR R ∂QR + 4
2* + Δϕ ΔRΔz − − − Δϕ ΔRΔz + 2
2) 2 2 ∂R -, *) 2 2 ∂R -,
2
1 2 ( 1 ∂Qϕ + ( 1 ∂Qϕ + 2
∇ ⋅ Q = lim 1−*− -ΔϕΔRΔz + * -ΔϕΔRΔz + 5
ΔR →0 RΔϕΔRΔz
Δϕ →0 2 ) 2 ∂ϕ , )2 ∂ϕ , 2
Δz →0 2( R ∂Ψ + ( R ∂Ψ + 2
2* -Δϕ ΔRΔz − −
*) 2 ∂z -, Δϕ ΔRΔz + ... 2
3) 2 ∂z , 6
&Q ∂Q 1 ∂Qϕ ∂Qz ) 1 ∂ 1 ∂Qϕ ∂Qz
∇⋅Q =' R + R + + *= (RQR ) + +
(R ∂R R ∂ϕ ∂z + R ∂ R R ∂ϕ ∂z
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.20. Obtain the recipe for the divergence of a vector function in spherical polar
coordinates from the integral definition (2.32).

Solution 2.20. Start from the appropriate z

1
form of (2.32), ∇ ⋅ Q = lim ∫∫ n ⋅ QdA ,
V →0 V
A
where Q = (Qr, Q , Q ) is a vector function
θ ϕ
#r
of position x. Here we choose a nearly er
rectangular volume V = (rΔ θ )(rsin θ Δϕ )(Δr) #"
€
centered on the point x = (r, θ, ϕ) with sides "
aligned perpendicular to the coordinate e"
directions. Here the unit vectors depend on y
θ and ϕ so directions are slightly different at e!
! rsin"
θ ± Δθ/2, and ϕ ± Δϕ/2. Considering the x #!
drawing and noting that n is an outward
normal, there are six contributions to ndA:
# Δr & # Δr & $ Δr ' $ Δr '
outside = % r + (Δθ% r + ( sin θΔϕ (e r ) , inside = & r − )Δθ& r − ) sin θΔϕ (−e r ) ,
$ 2' $ 2' % 2( % 2(
bottom = [ r sin(θ + Δθ 2)ΔϕΔr](eθ )θ +Δθ 2 , top = [ r sin(θ − Δθ 2)ΔϕΔr](−eθ )θ −Δθ 2 ,
close vertical side = rΔθΔr(−eϕ ) , and more distant vertical side = rΔθΔr(+eϕ ) .
ϕ −Δϕ 2 ϕ +Δϕ 2
€Here the unit vectors are evaluated at the center
€ of the volume unless otherwise specified. Using
a two-term Taylor series approximation €
€ for the corresponding components of Q on each of the
six surfaces produces:
$ € Δr ∂Qr ' % Δr ∂Qr ( % € Δθ ∂Qθ (
outside: &Qr + )e r , inside: 'Qr − *e r , bottom: 'Qθ + *(eθ ) ,
% 2 ∂r ( & 2 ∂r ) & 2 ∂θ ) θ +Δθ 2
& Δθ ∂Qθ ) & Δϕ ∂Qϕ )
top: (Qθ − +(eθ )θ −Δθ 2 , close vertical side : (Qϕ − +(−eϕ )ϕ −Δϕ 2 , and
' 2 ∂θ * ' 2 ∂ϕ *
€ € % Δϕ ∂Qϕ ( €
more distant vertical side : 'Qϕ + *(eϕ )ϕ +Δϕ 2 .
& 2 ∂ϕ )
€ Collecting and summing the six contributions € to n ⋅ QdA , the integral definition becomes:
1
∇ ⋅ Q = lim ×
Δr →0 (rΔθ )(r sin θΔϕ )Δr
€ ΔΔθϕ →0
→0
€
70* Δr ∂Qr - * Δr -
2 3 0* Δr ∂Qr - * Δr -
2 3;
92,Qr + /Δθ, r + / sin θΔϕ5 − 2,Qr − /Δθ, r − / sin θΔϕ5 9
91+ 2 ∂r . + 2. 4 1+ 2 ∂r . + 2. 49
9 9
9 0* Δθ ∂Qθ - * Δθ - 3 0* Δθ ∂Qθ - * Δθ - 39
8+2, Qθ + /r sin,θ + /ΔϕΔr5 − 2, Qθ − / r sin,θ − /ΔϕΔr5<
9 1+ 2 ∂θ . + 2 . 4 1+ 2 ∂θ . + 2 . 49
9 0* ∂Q - 3 0* ∂Q - 3 9
9−2, Qϕ − Δϕ ϕ /rΔθΔr5 + 2,Qϕ + Δϕ ϕ / rΔθΔr5 + ... 9
9: 1+ 2 ∂ϕ . 4 1+ 2 ∂ϕ . 4 9=

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

The largest terms inside the big {,}-brackets are proportional to ΔθΔϕΔr. The remaining higher
order terms vanish when the limit is taken.
0* ∂Q - 4
2,r 2 r + 2rQr /Δθ sin θΔϕΔr 2
2+ ∂r . 2
1 2 * ∂Q - 2
∇ ⋅ Q = lim × 1+,sin θ θ + cos θQθ /rΔθΔϕΔr5
Δr →0 (rΔθ )(r sin θΔϕ )Δr
Δθ →0 2 + ∂θ . 2
Δϕ →0 2 *∂Qϕ - 2
2+, /rΔθΔϕΔr + ... 2
3 + ∂ϕ . 6
Cancel the common factors and take the limit, to find:
1 .' ∂Q * ' ∂Q * '∂Q * 1
∇⋅Q = × /)r 2 r + 2rQr , sin θ + )sin θ θ + cosθQθ ,r + ) ϕ ,r2
€ (r)(r sin θ ) 0( ∂r + ( ∂θ + ( ∂ϕ + 3
1 &∂ ∂ ∂Q )
= 2 × ' ( r 2Qr ) sin θ + r (sin θQθ ) + r ϕ *
r sin θ ( ∂r ∂θ ∂ϕ +
€ 1 ∂ 1 ∂ 1 ∂Qϕ
= 2 ( r 2Qr ) + (sin θQθ ) +
r ∂r r sin θ ∂θ r sin θ ∂ϕ
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.21. Use the vector integral theorems to prove that ∇ ⋅ (∇ × u) = 0 for any twice-
differentiable vector function u regardless of the coordinate system.

Solution 2.21. Start with the divergence theorem for a vector n t1

function Q that depends on the spatial coordinates, € V
nc1
∫∫∫ ∇ ⋅ QdV = ∫∫ n ⋅ QdA nc2
V A
where the arbitrary closed volume V has surface A, and the t2
A1
outward normal is n. For this exercise, let Q = ∇ × u so that C A2
€
∫∫∫ ∇ ⋅ (∇ × u)dV = ∫∫ n ⋅ (∇ × u) dA .
V A
Now split V into two sub-volumes V1 and V2, where the surface of V1 is A1 and the surface of V2
is A2. Here A1 and A2 are not closed€ surfaces, but A1+ A2 = A so:

€
∫∫∫ ∇ ⋅ (∇ × u)dV = ∫∫ n ⋅ (∇ × u) dA + ∫∫ n ⋅ (∇ × u) dA .
V A1 A2

where n is the same as when the surfaces were joined. However, the bounding curve C for A1 and
A2 is the same, so Stokes theorem produces:
€ ∫∫∫ ∇ ⋅ (∇ × u) dV = ∫ u ⋅ t1 ds + ∫ u ⋅ t 2 ds .
V C C
Here the tangent vectors t1 = nc1 × n and t 2 = nc 2 × n have opposite signs because nc1 and nc2, the
normals to C that are tangent to surfaces A1 and A2, respectively, have opposite sign. Thus, the
two terms on the right side of the last equation are equal and opposite, so
€ €
∫∫∫ ∇ ⋅ (∇ × u)dV = 0 . (i)
V
For an arbitrary closed volume of any size, shape, or location, this can only be true if
∇ ⋅ (∇ × u) = 0 . For example, if ∇ ⋅ (∇ × u) were nonzero at some location, then integration in
small volume centered on this location would not be zero. Such a nonzero integral is not allowed
€
by (i); thus, ∇ ⋅ (∇ × u) must be zero everywhere because V is arbitrary.
€ €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 2.22. Use Stokes’ theorem to prove that ∇ × (∇φ ) = 0 for any single-valued twice-
differentiable scalar φ regardless of the coordinate system.

Solution 2.22. From (2.34) Stokes Theorem€is:

∫∫ (∇ × u) ⋅ ndA = ∫ u ⋅ tds .
A C
Let u = ∇φ , and note that ∇φ ⋅ tds = (∂φ ∂s) ds = dφ because the t vector points along the contour
C that has path increment ds. Therefore:
€∫∫ (∇ × [∇φ ]) ⋅ ndA = ∫ ∇φ ⋅ tds = ∫ dφ = 0 , (ii)
A C C
€ €
where the final equality holds for integration on a closed contour of a single-valued function φ.
For an arbitrary surface A of any size, shape, orientation, or location, this can only be true
if ∇ × ( ∇φ ) = 0 . For example, if ∇ × ( ∇φ ) = 0 were nonzero at some location, then an area
€
integration in a small region centered on this location would not be zero. Such a nonzero integral
is not allowed by (ii); thus, ∇ × ( ∇φ ) = 0 must be zero everywhere because A is arbitrary.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.1. The gradient operator in Cartesian coordinates (x, y, z) is:

∇ = e x (∂ ∂x ) + e y (∂ ∂y ) + e z (∂ ∂z) where e x , e y , and e z are the unit vectors. In cylindrical polar
coordinates (R, ϕ, z) having the same origin, (see Figure 3.3b), coordinates and unit vectors are
related by: R = x 2 + y 2 , ϕ = tan−1 ( y x ) , and z = z; and e R = e x cosϕ + e y sin ϕ ,
€ eϕ = −e x sin ϕ + e y cosϕ , and e z€= e€
z . Determine
€ the following in the cylindrical polar coordinate
system.
a) ∂e R ∂ϕ and ∂€ eϕ ∂ϕ
€ €
€ b) the gradient operator
€ ∇
c) the divergence of the velocity field ∇⋅u
€ d) the Laplacian operator ∇ ⋅ ∇ ≡ ∇ 2
€
e) the advective acceleration term (u⋅∇)u

Solution 3.1. The

€ Cartesian unit vectors do not depend on the coordinates so the unit vectors
from the cylindrical coordinate system can be differentiated when they are written in terms of ex,
ey, and ez.
a) First work with eR, use the given unit vector definition, and proceed with straightforward
differentiation. The variables R and z do not appear in the formula for eR, so
∂e R ∂R = ∂e R ∂z = 0 . However eR does depend on the angle ϕ. Thus,
∂e R ∂
∂ϕ ∂ϕ
= (e x cosϕ + e y sinϕ ) = −e x sinϕ + e y cosϕ = eϕ .
€
Proceed to determine the derivatives of eϕ. Again note that the variables R and z do not appear in
the formula for eϕ, so ∂eϕ ∂R = ∂eϕ ∂z = 0 . However, like eR, eϕ does depend on the angle ϕ.
Thus, €
∂eϕ ∂
∂ϕ ∂ϕ
= (−e x sinϕ + e y cosϕ ) = −e x cosϕ − e y sinϕ = −e R .
€
The third unit vector, ez, is the same as the Cartesian unit vector and does not depend on the
coordinates.
b) Start by constructing the expressions for ex, ey, and ez in terms of eR, eϕ, and ez. This can be
€ $ cosϕ sin ϕ 0($e x ( $e R (
& && & & &
done my inverting the linear system %−sin ϕ cosϕ 0)%e y ) = % eϕ ) to find
& 0 0 1&*&'e z &* &' e z &*
'

e x = e R cosϕ − eϕ sin ϕ , e y = e R sin ϕ + eϕ cos ϕ , and e z = e z (1,2,3)

The next step is to use the coordinate definitions:
€
R = x 2 + y 2 , ϕ = tan−1 ( y x ) , and z = z (4,5,6)
to transform
€ the Cartesian partial derivatives.
€ €
∂ ∂R ∂ ∂ϕ ∂ ∂z ∂ x ∂ 1 % y( ∂ ∂
= + + = − 2 ' − 2 * + (0)
∂x ∂x ∂R ∂x ∂ϕ ∂x ∂z x 2 + y 2 ∂R 1+ ( y x ) & x ) ∂ϕ ∂z
€ €
∂ sin ϕ ∂
= cosϕ −
∂R R ∂ϕ
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

∂ ∂R ∂ ∂ϕ ∂ ∂z ∂ y ∂ 1 $ 1' ∂ ∂
= + + = + 2 & ) + (0)
∂y ∂y ∂R ∂y ∂ϕ ∂y ∂z x 2 + y 2 ∂R 1+ ( y x ) % x ( ∂ϕ ∂z
∂ cosϕ ∂
= sin ϕ +
∂R R ∂ϕ
∂ ∂R ∂ ∂ϕ ∂ ∂z ∂ ∂ ∂ ∂ ∂
€ = + + = (0) + (0) + (1) =
∂z ∂z ∂R ∂z ∂ϕ ∂z ∂z ∂R ∂ϕ ∂z ∂z
∂ ∂ ∂
Now€reassemble the gradient operator ∇ = e x + e y + e z using the cylindrical coordinate
∂x ∂y ∂z
unit vectors and differentiation definitions:
€
& ∂ sinϕ ∂ ) & ∂ cos ϕ ∂ ) ∂
∇ = (e R cos ϕ − eϕ sin ϕ )( cosϕ − + + ( e R sin ϕ + e ϕ cos ϕ )( sin ϕ + + + ez .
' € ∂R R ∂ϕ * ' ∂R R ∂ϕ * ∂z
Collect all of the terms with like unit vectors and differential operators together:
∂ 1 ∂
∇ = e R (cos2 ϕ + sin 2 ϕ ) + e R (−cosϕ sin ϕ + sin ϕ cosϕ )
∂R R ∂ϕ
€
∂ 1 ∂ ∂
eϕ (−sin ϕ cos ϕ + cosϕ sin ϕ ) + eϕ (sin 2 ϕ + cos2 ϕ ) + ez
∂R R ∂ϕ ∂z
The terms in (,)-parentheses are either +1 or 0. When evaluated they produce:
€
∂ 1 ∂ ∂
∇ = eR + eϕ + ez .
∂R R ∂ϕ ∂z
€
c) In cylindrical coordinates, the divergence of the velocity is:
& ∂ 1 ∂ ∂)
∇ ⋅ u = (e R + eϕ + e z + ⋅ (e R uR + eϕ uϕ + e z uz ) .
€ ' ∂R R ∂ϕ ∂z *
Further simplification requires that both the unit vectors and the u's be differentiated.
Completing this task term by term produces:
∂ ∂u ∂e ∂u
e R € ⋅ (e R uR ) = e R ⋅ e R R + uR e R ⋅ R = R ,
∂R ∂r ∂R ∂ R
∂ ∂ u ∂ e
eR ⋅ (eϕ uϕ ) = e R ⋅ eϕ ϕ + uϕ e R ⋅ ϕ = 0 ,
∂R ∂R ∂R
∂ ∂uz ∂e z
€ eR ⋅ (e z uz ) = e R ⋅ e z + uze R ⋅ = 0,
∂R ∂R ∂r
eϕ ∂ e ⋅ e ∂u u ∂e u u
€ ⋅ (e R uR ) = ϕ R R + R eϕ ⋅ R = 0 + R eϕ ⋅ eϕ = R ,
R ∂ϕ R ∂ϕ R ∂ϕ R R
€ e ∂ e ⋅ e ∂ u u ∂ e 1 ∂ u u 1 ∂uϕ
ϕ
⋅ (eϕ uϕ ) = ϕ ϕ ϕ + ϕ eϕ ⋅ ϕ = ϕ
− ϕ eϕ ⋅ e R = ,
R ∂ϕ R ∂ϕ R ∂ϕ R ∂ϕ R R ∂ϕ
€ eϕ ∂ e ⋅ e ∂u u ∂e
⋅ (e z uz ) = ϕ z ϕ + z eϕ ⋅ z = 0
R ∂ϕ R ∂ϕ R ∂ϕ
€ ∂ ∂ u ∂ e
e z ⋅ (e R uR ) = e z ⋅ e R R + uR e z ⋅ R = 0 ,
∂z ∂z ∂z
∂ ∂u ∂e
€ e z ⋅ (eϕ uϕ ) = e z ⋅ eϕ ϕ + uϕ e z ⋅ ϕ = 0 , and
∂z ∂z ∂z
∂ ∂uz ∂e z ∂uz
€ e z ⋅ (e z uz ) = e z ⋅ e z + uze z ⋅ = .
∂z ∂z ∂z ∂ z
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Reassembling the equation produces:

∂uR uR 1 ∂uϕ ∂uz
∇⋅u= + + + .
∂R R R ∂ϕ ∂z
The 1st & 2nd terms on the right side are commonly combined to yield:
1 ∂ 1 ∂uϕ ∂uz
∇⋅u= (RuR ) + + . (10)
€ R ∂ R R ∂ϕ ∂ z
d) The Laplacian operator is ∇ 2 ≡ ∇ ⋅ ∇ , and its form in cylindrical coordinates can be found by
evaluating the dot product. Fortunately, the results of part c) can be used via the following
replacements for the second gradient operator of the dot product:
€ ∂ 1 ∂ ∂
€ uR ↔ , uϕ ↔ , and uz ↔ . (7,8,9)
∂R R ∂ϕ ∂z
Inserting the replacements (7,8,9) into (10) produces:
1 ∂ $ ∂ ' 1 ∂2 ∂2
∇2 = &R ) + 2 2 + 2 .
€ € R ∂R % ∂R ( €R ∂ϕ ∂z
e) Start with the answer to part b) and compute the first dot product to find:
∂ 1 ∂ ∂
u ⋅ ∇ = uR + uϕ + uz
∂R R ∂ϕ ∂z
€
This is the scalar operator applied to u = uR e R + uϕ eϕ + uz e z to find the advective acceleration:
& ∂ 1 ∂ ∂)
(u ⋅ ∇)u = ( uR + uϕ + uz +( uR e R + uϕ eϕ + uze z ) .
€ ' ∂R R ∂ϕ ∂z *
Here the components of u and the unit vectors eR and eϕ depend on the angular coordinate.
& ∂u 1 ∂ uR ∂u ) 1 ∂e R
(u ⋅ ∇)u = e R ( uR R + uϕ + uz R + + uR uϕ +
€ ' ∂ R R ∂ϕ ∂ z * R ∂ϕ
! ∂u 1 ∂ uϕ ∂u $ 1 ∂ eϕ ! ∂u 1 ∂ uz ∂u $
eϕ # uR ϕ + uϕ + uz ϕ & + uϕ2 + e z # uR z + uϕ + uz z &
" ∂R R ∂ϕ ∂z % R ∂ϕ " ∂R R ∂ϕ ∂z %
€ Use the results of part a) to evaluate the unit vector derivatives.
& ∂u 1 ∂uR ∂u ) e
(u ⋅ ∇)u = e R ( uR R + uϕ + uz R + + uR uϕ ϕ +
' ∂R R ∂ϕ ∂z * R
$ ∂uϕ 1 ∂uϕ ∂u ' e $ ∂u 1 ∂uz ∂u '
eϕ & uR + uϕ + uz ϕ ) − uϕ2 R + e z & uR z + uϕ + uz z )
% ∂R R ∂ϕ ∂z ( R % ∂R R ∂ϕ ∂z (
€ Collect components
' ∂u 1 ∂uR ∂u u2 *
(u ⋅ ∇)u = e R ) uR R + uϕ + uz R − ϕ , +
€ ( ∂R R ∂ϕ ∂z R+
$ ∂u 1 ∂uϕ ∂u u u ' $ ∂u 1 ∂uz ∂u '
eϕ & uR ϕ + uϕ + uz ϕ + R ϕ ) + e z & uR z + uϕ + uz z )
% ∂R R ∂ϕ ∂z R ( % ∂R R ∂ϕ ∂z (
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.2. Consider Cartesian coordinates (as given in Exercise no. 1) and spherical polar
coordinates (r, θ, ϕ) having the same origin (see Figure 3.3c). Here coordinates and unit vectors
are related by: r = x 2 + y 2 + z 2 , θ = tan−1 ( )
x 2 + y 2 z , and ϕ = tan−1 ( y x ) ; and
e r = e x cosϕ sin θ + e y sin ϕ sin θ + e z cos θ , eθ = e x cosϕ cos θ + e y sin ϕ cos θ − e z sin θ , and
eϕ = −e x sin ϕ + e y cosϕ . In the spherical polar coordinate system, determine the following items.
a) ∂e€ r ∂θ , ∂e r ∂ϕ , ∂eθ€∂θ , ∂eθ ∂ϕ , and ∂eϕ ∂ϕ €

€ b) the gradient operator ∇ €

€ c) the divergence of the velocity field ∇⋅u
€ d)€the Laplacian
€ ∇ ⋅€∇ ≡ ∇ 2
€
e) the advective acceleration term (u⋅∇)u

Solution€ 3.2. The Cartesian unit vectors do not depend on the coordinates so the unit vectors
from the spherical coordinate system can be differentiated when they are written in terms of ex,
ey, and ez.
a) First work with er, use the given unit vector definition, and proceed with straightforward
differentiation. The variable r doesn’t even appear in the formula for er, so ∂e r ∂r = 0 . However
er does depend on both angles. Thus,
∂e r ∂
= (e x cosϕ sin θ + e y sin ϕ sin θ + e z cos θ ) = e x cos ϕ cosθ + e y sin ϕ cos θ − e z sin θ = eθ
∂θ ∂θ
and, €
∂e r ∂
∂ϕ ∂ϕ
= (e x cosϕ sinθ + e y sinϕ sinθ + e z cosθ ) = −e x sinϕ sinθ + e y cosϕ sinθ = eϕ sinθ .
€
Proceed to determine the derivatives of eθ. Again note that the variable r doesn’t appear in its
formula, so ∂eθ ∂r = 0 . However, like er, eθ does depend on both angles. Thus,
∂eθ ∂
€ = (e x cosϕ cos θ + e y sin ϕ cos θ − e z sin θ ) = −e x cosϕ sin θ − e y sin ϕ sin θ − e z cos θ = −e r
∂θ ∂θ
and,
€ ∂e ∂
θ
∂ϕ ∂ϕ
= (e x cosϕ cosθ + e y sinϕ cosθ − e z sinθ ) = −e x sinϕ cosθ + e y cosϕ cosθ = eϕ cosθ .
€
Now consider eϕ and note that the variables r and θ don’t appear in its formula, so
∂eϕ ∂r = ∂eϕ ∂θ = 0 . However, eϕ does depend on ϕ. Thus,
€ ∂eϕ ∂
∂ϕ ∂ϕ
= (−e x sinϕ + e y cosϕ ) = −e x cosϕ − e y sinϕ . ($)

€ The question now is how to relate the right side of this equation back to er, and eθ [note: because
( )
eϕ ⋅ ∂eϕ ∂ϕ = 0 , ∂eϕ ∂ϕ can only be a linear combination of er and eθ]. Assuming
∂eϕ ∂ϕ = ae r + be€ θ , then ($) and the unit vector definitions require:
acosϕ sin θ + bcosϕ cos θ = −cosϕ ,
€ asin ϕ sin θ + bsin ϕ cosθ = −sin ϕ , and
€
acosθ − bsin θ = 0 .
€ After dividing out common factors, the first two equations are the same: asin θ + bcos θ = −1.
When this simplified € equation is combined with the third equation, we obtain: a = −sin θ and
b = −cosθ . Thus, ∂€eϕ ∂ϕ = −e r sin θ − eθ cos θ .
€
€
€
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

b) Start by constructing the expressions for ex, ey, and ez in terms of e r , eθ , and eϕ . This can be
% cosϕ sin θ sin ϕ sin θ cosθ )%e x ) % e r )
' '' ' ' '
done my inverting the linear system &cos ϕ cosθ sin ϕ cosθ −sin θ *&e y * = &eθ * to find
' −sin ϕ cosϕ € € 0 '+'(€e z '+ '(eϕ '+
(
e x = e r cosϕ sin θ + eθ cosϕ cos θ − eϕ sin ϕ
e y = e r sin ϕ sin θ + eθ sin ϕ cosθ + eϕ cos ϕ (1,2,3)
€ e z = e r cosθ − eθ sin θ .
The next step is to€use the coordinate definitions:
2 2 2
€ r = x + y + z , θ = tan
−1
(
x2 + y2 z ) , and ϕ = tan
−1
( y x) (4,5,6)
€
to transform the Cartesian partial derivatives.
∂ ∂r ∂ ∂θ ∂ ∂ϕ ∂ x ∂ 1 2x ∂ y ∂
= + + = + − 2
∂x ∂x ∂r ∂x ∂θ ∂x ∂ϕ r ∂r 1+ x 2 + y 2 z 2 2z 2
x + y ∂θ x + y ∂ϕ
2 2
€ € ( € )
∂ cos ϕ cosθ ∂ sin ϕ ∂
= cosϕ sin θ + −
∂r r ∂θ r sin θ ∂ϕ
€ ∂ ∂r ∂ ∂θ ∂ ∂ϕ ∂ y ∂ 1 2y ∂ x ∂
= + + = + + 2
∂y ∂y ∂r ∂y ∂θ ∂y ∂ϕ r ∂r 1+ x 2 + y 2 z 2 2z 2
x + y ∂θ x + y ∂ϕ
2 2
( )
€ ∂ sin ϕ cos θ ∂ cos ϕ ∂
= sin ϕ sin θ
+ +
∂r r ∂θ r sin θ ∂ϕ
€ ∂ ∂r ∂ ∂θ ∂ ∂ϕ ∂ z ∂ 1 − x2 + y2 ∂ ∂ sin θ ∂
= + + = + 2 2
= cos θ −
∂z ∂z ∂r ∂z ∂θ ∂z ∂ϕ r ∂r 1+ x 2 + y 2 z z ∂θ ∂r r ∂θ
€
( )
∂ ∂ ∂
Now reassemble the gradient operator ∇ = e x + e y + e z using the spherical coordinate unit
∂x ∂y ∂z
€ vectors and differentiation definitions:
' ∂ cos ϕ cosθ ∂ sin ϕ ∂ *
∇ = (e r cos ϕ sin θ + eθ cos ϕ cosθ − eϕ sin ϕ )) cosϕ sin θ + − ,
€ ( ∂r r ∂θ r sin θ ∂ϕ +
% ∂ sin ϕ cosθ ∂ cos ϕ ∂ (
+(e r sin ϕ sin θ + eθ sin ϕ cosθ + eϕ cos ϕ )'sin ϕ sin θ + + *
& ∂r r ∂θ r sinθ ∂ϕ )
€ % ∂ sin θ ∂ (
+(e r cosθ − eθ sin θ )'cos θ − *.
& ∂r r ∂θ )
€ Collect all of the terms with like unit vectors and differential operators together:
∂
∇ = e r (cos2 ϕ sin 2 θ + sin 2 ϕ sin 2 θ + cos 2 θ )
∂r
€ 1 ∂
+e r (cos 2 ϕ cosθ sin θ + sin 2 ϕ cosθ sin θ − cosθ sin θ )
r ∂θ
1 ∂
€ +e r (−cos ϕ sin ϕ sin θ + sin ϕ cosϕ sin θ )
r sin θ ∂ϕ
∂
€ +eθ (cos 2 ϕ cosθ sin θ + sin 2 ϕ cosθ sin θ − sin θ cosθ )
∂r
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

1 ∂
+eθ (cos 2 ϕ cos2 θ + sin 2 ϕ cos2 θ + sin 2 θ )
r ∂θ
1 ∂
+eθ (−cos ϕ sin ϕ cos θ + sin ϕ cosϕ cos θ )
r sin θ ∂ϕ
∂
€ +eϕ (−sin ϕ cosϕ sin θ + cos ϕ sin ϕ sin θ )
∂r
1 ∂
€ +eϕ (−sin ϕ cosϕ cos θ + cosϕ sin ϕ cos θ )
r ∂θ
1 ∂
€ +eϕ (sin ϕ + cos ϕ )
2 2

r sin θ ∂ϕ
The terms in (,)-parentheses are either +1 or 0. When evaluated they produce:
€
∂ 1 ∂ 1 ∂
∇ = e r + eθ + eϕ .
∂r r ∂θ r sin θ ∂ϕ
€
c) In spherical coordinates, the divergence of the velocity is:
' ∂ 1 ∂ 1 ∂ *
∇ ⋅ u = )e r + eθ + eϕ , ⋅ (e r ur + eθ uθ + eϕ uϕ ) .
€ ( ∂ r r ∂θ r sin θ ∂ϕ +
Further simplification requires that the unit vectors and the u's be differentiated. Completing this
task term by term produces:
∂ ∂u ∂e ∂ u
e r ⋅ (e r€ ur ) = e r ⋅ e r r + ure r ⋅ r = r ,
∂r ∂r ∂r ∂ r
∂ ∂uθ ∂e
e r ⋅ (eθ uθ ) = e r ⋅ eθ + uθ e r ⋅ θ = 0 ,
∂r ∂r ∂r
∂ ∂uϕ ∂eϕ
€ e r ⋅ (eϕ uϕ ) = e r ⋅ eϕ + uϕ e r ⋅ = 0,
∂r ∂r ∂r
eθ ∂ e ⋅ e ∂u u ∂e u u
€ ⋅ (e r ur ) = θ r r + r eθ ⋅ r = 0 + r eθ ⋅ eθ = r ,
r ∂θ r ∂θ r ∂θ r r
eθ ∂ eθ ⋅ eθ ∂uθ uθ ∂eθ 1 ∂uθ uθ 1 ∂uθ
€ ⋅ (eθ uθ ) = + eθ ⋅ = − eθ ⋅ e r = ,
r ∂θ r ∂θ r ∂θ r ∂θ r r ∂θ
eθ ∂ e ⋅ e ∂u u ∂e
€ ⋅ (eϕ uϕ ) = θ ϕ ϕ + ϕ eθ ⋅ ϕ = 0
r ∂θ r ∂θ r ∂θ
eϕ ∂ eϕ ⋅ e r ∂ur ur ∂e ur u
€ ⋅ (e r ur ) = + eϕ ⋅ r = eϕ ⋅ (eϕ sin θ ) = r ,
r sin θ ∂ϕ r sin θ ∂ϕ r sin θ ∂ϕ r sin θ r
€ eϕ ∂ eϕ ⋅ eθ ∂uθ u ∂e u u
⋅ (eθ uθ ) = + θ eϕ ⋅ θ = θ eϕ ⋅ eϕ cosθ = θ ,
r sin θ ∂ϕ r sin θ ∂ϕ r sin θ ∂ϕ r sin θ r tan θ
€ eϕ ∂ 1 & ∂uϕ ∂e )
⋅ (eϕ uϕ ) = ( eϕ ⋅ eϕ + uϕ eϕ ⋅ ϕ +
r sin θ ∂ϕ r sin θ ' ∂ϕ ∂ϕ *
€ 1 ' ∂uϕ * 1 ∂uϕ
= ) − uϕ eϕ ⋅ (e r sin θ + eθ cos θ ), =
r sin θ ( ∂ϕ + r sin θ ∂ϕ
€ Reassembling the equation produces:
∂u 2u 1 ∂uθ u 1 ∂uϕ
∇⋅u= r + r + + θ +
€ ∂r r r ∂θ r tan θ r sin θ ∂ϕ
The 1st & 2nd terms, and the 3trd and 4th terms on the right side are commonly combined to yield:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

1 ∂ 2 1 ∂ 1 ∂uϕ
∇⋅u= 2
r ∂r
( r ur ) +
r sin θ ∂θ
(sin θ uθ ) +
r sin θ ∂ϕ
. (10)
d) The Laplacian operator is ∇ 2 ≡ ∇ ⋅ ∇ , and its form in spherical polar coordinates can be found
by evaluating the dot product. Fortunately, the results of part c) can be used via the following
replacements for the second gradient operator of the dot product:
€ ∂ 1 ∂ 1 ∂
€ ur ↔ , uθ ↔ , and uϕ ↔ . (7,8,9)
∂r r ∂θ r sin θ ∂ϕ
Inserting (7,8,9) into (10), the Laplacian then becomes:
1 ∂$ ∂' 1 ∂ $ ∂ ' 1 ∂2
∇2 = 2 &r2 ) + 2 & sin θ )+ 2 2 .
€ r €∂r % ∂r ( r sin θ€∂θ % ∂θ ( r sin θ ∂ϕ 2
e) Start with the answer to part b) and compute the first dot product to find:
∂ 1 ∂ 1 ∂
u ⋅ ∇ = ur + uθ + uϕ
∂r r ∂θ r sin θ ∂ϕ
€
This is the scalar operator applied to u = ure r + uθ eθ + uϕ eϕ to find the advective acceleration:
' ∂ 1 ∂ 1 ∂ *
(u ⋅ ∇)u = ) ur + uθ + uϕ ,( ure r + uθ eθ + uϕ eϕ ) .
€ ( ∂r r ∂θ r sin θ ∂ϕ +
Here the components of u and € the unit vectors depend on the angular coordinates.
' ∂ur 1 ∂ur 1 ∂ ur * 1 ∂e r 1 ∂e r
(u ⋅ ∇)u = e r ) ur + uθ + uϕ , + ur uθ + ur uϕ +
€ ( ∂r r ∂θ r sin θ ∂ϕ + r ∂θ r sin θ ∂ϕ
% ∂u 1 ∂uθ 1 ∂uθ ( 2 1 ∂eθ 1 ∂eθ
eθ ' ur θ + uθ + uϕ * + uθ + uθ uϕ +
& ∂r r ∂θ r sin θ ∂ϕ ) r ∂θ r sin θ ∂ϕ
€ % ∂u 1 ∂uϕ 1 ∂uϕ ( 1 ∂eϕ 1 ∂eϕ
eϕ ' ur ϕ + uθ + uϕ * + uϕ uθ + uϕ2
& ∂r r ∂θ r sin θ ∂ϕ ) r ∂θ r sin θ ∂ϕ
Use
€ the results of part a) to evaluate the unit vector derivatives.
' ∂u 1 ∂ur 1 ∂ur * eθ e sin θ
(u ⋅ ∇)u = e r ) ur r + uθ + uϕ , + ur uθ + ur uϕ ϕ +
€ ( ∂r r ∂θ r sin θ ∂ϕ + r r sin θ
% ∂u 1 ∂uθ 1 ∂uθ ( 2 e r eϕ cosθ
eθ ' ur θ + uθ + uϕ * − uθ + uθ uϕ
& ∂r r ∂θ r sin θ ∂ϕ ) r r sin θ
€ % ∂uϕ 1 ∂uϕ 1 ∂uϕ ( 2 1
eϕ ' ur + uθ + uϕ * + 0 + uϕ (−e r sin θ − eθ cosθ )
& ∂r r ∂θ r sin θ ∂ϕ ) r sin θ
Collect
€ components
( ∂u 1 ∂ur 1 ∂ur uθ + uϕ +
2 2

(u ⋅ ∇)u = e r * ur r + uθ + uϕ − -+
€ ) ∂r r ∂θ r sin θ ∂ϕ r ,
& ∂u 1 ∂uθ 1 ∂uθ ur uθ uϕ
2 )
θ
eθ ( ur + uθ + uϕ + − cot θ +
' ∂r r ∂θ r sin θ ∂ϕ r r *
€ % ∂uϕ 1 ∂uϕ 1 ∂uϕ ur uϕ uθ uϕ (
eϕ ' ur + uθ + uϕ + + cot θ *
& ∂r r ∂θ r sin θ ∂ϕ r r )
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.3. In a steady two-dimensional flow, Cartesian-component particle trajectories are

given by: x(t) = ro cos (γ (t − to ) + θ o ) and y(t) = ro sin (γ (t − to ) + θ o ) where ro = xo2 + yo2 and
θ o = tan −1 ( yo xo ) .
a) From these trajectories determine the Lagrangian particle velocity components u(t) = dx/dt and
v(t) = dy/dt, and convert these to Eulerian velocity components u(x,y) and v(x,y).
b) Compute Cartesian particle acceleration components, ax = d2x/dt2 and ay = d2y/dt2, and show
that they are equal to D/Dt of the Eulerian velocity components u(x,y) and v(x,y).

Solution 3.3. a) Differentiate as suggested to find:

dx(t) dy(t)
u(t) = = −γ ro sin (γ (t − to ) + θ o ) and v(t) = = γ ro cos (γ (t − to ) + θ o ) .
dt dt
Now use the original trajectory equations to eliminate the trig-functions:
u = −γ y and v = γ x .
b) Again differentiate as suggested to find:
d 2 x(t)
ax (t) = 2
= −γ 2 ro cos (γ (t − to ) + θ o ) = −γ 2 x and
dt
d 2 y(t)
ay (t) = 2
= −γ 2 ro sin (γ (t − to ) + θ o ) = −γ 2 y .
dt
Compute Du/Dt and Dv/Dt from the final two answers of part a):
Du ∂u ∂u ∂u
= + u + v = 0 − γ y(0) + γ x(−γ ) = −γ 2 x and
Dt ∂t ∂x ∂y
Dv ∂v ∂v ∂v
= + u + v = 0 − γ y(γ ) + γ x(0) = −γ 2 y .
Dt ∂t ∂x ∂y
The final equalities match as appropriate: ax = Du/Dt, and ay = Dv/Dt.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.4. In a steady two-dimensional flow, polar coordinate particle trajectories are given
by: r(t) = ro and θ (t) = γ (t − to ) + θ o .
a) From these trajectories determine the Lagrangian particle velocity components ur(t) = dr/dt
and, uθ (t) = rdθ/dt, and convert these to Eulerian velocity components ur(r,θ) and uθ (r, θ ) .
2
b) Compute polar-coordinate particle acceleration components, ar = d 2 r dt 2 − r ( dθ dt ) and
aθ = rd 2θ dt 2 + 2 ( dr dt ) ( dθ dt ) , and show that they are equal to D/Dt of the Eulerian velocity
with components ur(r,θ) and uθ (r, θ ) .

Solution 3.4. a) Differentiate as suggested to find:

dr(t) dθ (t)
ur (t) = = 0 and uθ (t) = r = rγ .
dt dt
These equations are readily interpreted as Eulerian velocity components:
ur = 0 and uθ = γ r .
b) Start with the particle accelerations:
2
ar = d 2 r dt 2 − r ( dθ dt ) = 0 − r(γ )2 = −γ 2 r and aθ = rd 2θ dt 2 + 2 ( dr dt ) ( dθ dt ) = 0 + 2(0)γ = 0 .
Compute Dur/Dt and Duθ/Dt from the forms given in Appendix B and the two answers of part a):
Dur ∂ur ∂u u ∂u u 2 (γ r)2
= + ur r + θ r − θ = 0 + 0 − = −γ 2 r and
Dt ∂t ∂r r ∂θ r r
Duθ ∂uθ ∂uθ uθ ∂uθ ur uθ
= + ur + + = 0 + 0(γ ) + γ (0) + 0 = 0 .
Dt ∂t ∂r r ∂θ r
The final equalities match as appropriate: ar = Dur/Dt, and aθ = Duθ/Dt.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.5. if ds = (dx, dy, dz) is an element of arc length along a streamline (Figure 3.5) and u
= (u, v, w) is the local fluid velocity vector, show that if ds is everywhere tangent to u then
dx u = dy v = dz w .

Solution 3.5. If ds = (dx, dy, dz) and u are parallel, then they must have the same unit tangent
vector t:
€ ds (dx,dy,dz) (u,v,w) u
t= = = = .
ds (dx) 2 + (dy) 2 + (dz) 2 u2 + v 2 + w 2 u
The three components of this equation imply:
dx u dy v dz w
= , = , and = .
ds u ds u ds u
€
But these can be rearranged to find:
ds dx dy dz
= = = .
€ € u u€ v w

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.6. For the two-dimensional steady flow having velocity components u = Sy and v =
Sx, determine the following when S is a positive real constant having units of 1/time.
a) equations for the streamlines with a sketch of the flow pattern
b) the components of the strain rate tensor
c) the components of the rotation tensor
d) the coordinate rotation that diagonalizes the strain rate tensor, and the principal strain rates.
e) How is this flow field related to that in Example 3.5.

Solution 3.6. a) For steady streamlines in two dimensions:

dx dy dy v Sx x y
= or = = = , which implies:
u v dx u Sy y
ydy = xdx → y 2 2 = x 2 2 + const.
Solving for y(x) produces: y = ± x 2 + const . These
€ are hyperbolae
€ that asymptote to the lines y = ±x.
b)
€ Compute the strain rate tensor from its definition:
1 # ∂u ∂u &
Sij = %% i + j ((€ x
2 $ ∂x j ∂x i '
# 1
∂u ∂y + ∂v ∂ x ) &
2(
∂u ∂x
= $1 '
% 2 (∂v ∂x + ∂u ∂y ) ∂v ∂y (
€ " 0 1
S + S )% " 0 S %
2(
= #1 &=# &
$ 2 ( S + S) 0 ' $S 0'
€ c) Compute the rotation tensor from its definition:
∂u ∂ u $ 0 ∂u ∂ y − ∂v ∂x ' $ 0 S − S ' $0 0'
Rij = i − j = % (=% (=% (
€ ∂x j ∂x i &∂v ∂x − ∂u ∂y 0 ) &S − S 0 ) &0 0)
d) From Example 2.4, a θ = 45° coordinate rotation diagonalizes the strain rate tensor. The
$cosθ −sin θ ' 1 $1 −1'
direction cosine matrix is: Cij = % (= % ( , and the rotated strain rate matrix
€ & sin θ cosθ ) 2 &1 1 )
S´ is:
1 % 1 1(% 0 S ( 1 %1 −1( 1 % S S (%1 −1( % S 0 (
S" = CT ⋅ S ⋅ C = & )& ) & )= & )& )=& ).
€ 2 '−1 1*' S 0 * 2 '1 1 * 2 ' S −S *'1 1 * ' 0 −S *
e) When this flow field is rotated 45° in the clockwise direction, it is the same as the flow field in
Example 3.5.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.7. At the instant shown in Figure 3.2b, the (u,v)-velocity field in Cartesian
coordinates is u = A(y 2 − x 2 ) (x 2 + y 2 ) 2 , and v = −2A xy (x 2 + y 2 ) 2 where A is a positive
constant. Determine the equations for the streamlines by rearranging the first equality in (3.7) to
read udy − vdx = 0 = (∂ψ ∂y ) dy + (∂ψ ∂x ) dx and then looking for a solution in the form ψ(x,y) =
const.
€ €
Solution 3.7. Rearrange the two-dimensional streamline condition, dx/u = dy/v, to obtain udy –
€ vdx = 0 as the description of a streamline. Assume this differential equation is solved by the
function ψ(x,y) = const, so that (∂ψ/∂x)dx + (∂ψ/∂y)dy = 0. Comparing the two equations
requires:
u = ∂ψ/∂y , and v = –∂ψ/∂x.
Now use the given velocity field to find:
∂ψ ∂y = A(y 2 − x 2 ) (x 2 + y 2 ) 2 , and ∂ψ ∂x = +2A xy (x 2 + y 2 ) 2 . (a,b)
Integrate (b) treating y as a constant:
∂ψ 2x 2xdx ' −1 *
= Ay 2 2 2
→ ψ − ψ o = Ay ∫ 2 2 2
= Ay) 2 2 ,,
€ ∂ x (x + y ) € (x + y ) ( x + y +
where ψo may depend on y. Differentiate this result with respect to y to determine ψo:
∂ ∂ % −Ay ( −A (−Ay) A(y 2 − x 2 )
(ψ − ψ o ) = ' 2 2 * = 2 2 − 2 2 2 (2y) = 2 2 2 = u .
€ ∂y ∂y & x + y ) x + y (x + y ) (x + y )
This result and equation (a) implies ∂ψo/∂y = 0, so it is at most a constant. Thus, the streamlines
are given by:
Ay
€ ψ (x, y) = const. = − 2 .
x + y2

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.8. Determine the equivalent of the first equality in (3.7) for two dimensional (r,θ)-
polar coordinates, and then find the equation for the streamline that passes through (ro, θo) when
u = (ur, u ) = (A/r, B/r) where A and B are constants.
θ

Solution 3.8. The two-dimensional streamline condition in Cartesian coordinates is dx/u = dy/v,
and is obtained from considering the streamline-tangent vector t:
ds e dx + e dy e u+e v u
t= = x 2 y 2 = x 2 y2 = .
ds (dx) + (dy) u +v u
In two-dimensional polar coordinates this becomes:
ds e dr + e rdθ e u +e u u
t= = r 2 θ = r r 2 θ 2θ = .
ds (dr) + (rdθ ) 2
ur + uθ u
€
Equating components produces two equations:
dr ur rdθ uθ ds dr rdθ
= and = , or = = .
€ ds u ds u u ur uθ
Thus, using the last equality and the given velocity field:
1 dr ur A r A A
= = = → ln(r) = θ + const.
€ r dθ € uθ B r B B
€
The initial condition allows the constant to be evaluated:
A "r% A $A '
ln(ro ) = θ o + const., which leads to ln$ ' = (θ − θ o ) or r = ro exp& (θ − θ o )) .
€B # ro & B %B (

€ € €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.9. Determine the streamline, path line, and streak line that pass through the origin of
coordinates at t = t´ when u = Uo + ωξocos(ωt) and v = ωξosin(ωt) in two-dimensional Cartesian
coordinates where Uo is a constant horizontal velocity. Compare your results those in Example.
3.3 for U o → 0 .

Solution 3.9. (i) For the streamline, time is a constant. Use the first equality of (3.7) to find:
dy v ωξ o sin(ωt)
€ = = = m(t) ,
dx u U o + ωξ o cos(ωt)
where m is the streamline slope. Since m does not depend on the spatial coordinate, this equation
is readily integrated to find straight time-dependent streamlines: y = m(t)x + const. Thus, the
streamline that passes through (0,0) at t = t´ is:
€ ωξ o sin(ωt $)
y= x.
U o + ωξ o cos(ωt $)
(ii) For the path line, use both components of (3.8):
dx dy
= U o + ωξ o cos(ωt) and = ωξ o sin(ωt) ,
dt dt
and integrate in time to find: €
x − x o = U o t + ξ o sin(ωt) and y − y o = –ξ o cos(ωt) .
Determine xo and yo by requiring the path line to pass through the origin at at t = t´:
€ 0 − x o = U o t # + ξ o sin(ω €t #) and 0 − y = –ξ cos(ωt %) .
o o
The final component equations are:
€ €
x = U o (t − t #) + ξ o (sin(ωt) − sin(ωt #)) and y = –ξ o (cos(ωt) − cos(ωt %)) .
These two parametric equations for x(t) and y(t) can be combined to eliminate some of the t-
€ €
dependence:
2 2
( x − U o (t − t #) + ξ o sin(ωt #)) + ( y − ξ o cos(ωt #)) = ξ o2 ,
€ €
which describes a moving circle with center located at (U o (t − t #) − ξ o sin(ωt #),ξ o cos(ωt #)) .
(iii) For the streak line, use the path line results but this time evaluate the constants at t = to
instead of at t = t´ to find:
€
x = U o (t − t o ) + ξ o (sin(ωt) − sin(ωt o )) and y = –ξ o (cos(ωt) − cos(ωt o )) .
Now evaluate these equations at t = t´ to produce € two parametric equations for the streak line
coordinates x(to) and y(to):
x = U o ( t " − t o ) + ξ o (sin(ωt ") − sin(ωt o )) and y = –ξ o (cos(ωt $) − cos(ωt o )) .
€
Some of the to dependence can be eliminated by € combining the equations:
2 2
( x − U o (t # − t o ) − ξ o sin(ωt #)) + ( y + ξ o cos(ωt #)) = ξ o2 ,
which
€ describes a circle with € a to-dependent center located at
(U o (t " − t o ) + ξ o sin(ωt "),−ξ o cos(ωt ")) .
These results differ from those in Example 3.3 by the uniform translation velocity Uo so
€
they can be put into correspondence with a Galilean transformation x´ = x – Uo(t – t´).
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.10. Compute and compare the streamline, path line, and streak line that pass through
(1,1,0) at t = 0 for the following Cartesian velocity field u = (x, –yt, 0).

Solution 3.10. (i) For the streamline, time is a constant. Use the first equality of (3.7) to find:
dy v yt dy dx
= =− → = −t → ln y = −t ln x + const. , or y = const.x–t.
dx u x y x
Evaluating at x = y = 1 and t = 0 requires the constant to be unity, so the streamline is: y = 1.
(ii) For the path line, use both components of (3.8):
dx dy
€ = x and = −yt ,
dt dt
and integrate these in time to find:
x = C1e t and y = C2 exp{−t 2 2} ,
where C1 and C2 are constants.€ Evaluating € at x = y = 1 and t = 0 requires C1 = C2 = 1. Eliminate t
from the y-equation using t = ln(x) to find the path line as:
€ y = exp{−(ln x) 2 2} .
€
(iii) For the streak line, use the path line results but this time evaluate the constants at t = to
instead of at t = 0 to find:
x = exp{t − t o } and y = exp{(t o2 − t 2 ) 2} .
€
Evaluate at t = 0, and eliminate to from the resulting equations to find:
y = exp{+(ln x) 2 2} .
€ €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.11. Consider a time-dependent flow field in two-dimensional Cartesian coordinates

where u = τ t 2 , v = xy τ , and  and τ are constant length and length time scales, respectively.
a) Use dimensional analysis to determine the functional form of the streamline through x´ at time
t´.
b) Find the equation for the streamline through x´ at time t´ and put your answer in
€
€ dimensionless form. €
c) Repeat b) for the path line through x´ at time t´.
d) Repeat b) for the streak line through x´ at time t´.

Solution 3.11. a) The streamline y(x) will depend on x, t, t´, x´= (x´,y´),  , and τ. There are eight
y % x x # y # t t #(
parameters and two dimensions, thus there are six dimensionless groups: = Ψ' , , , , * .
 &   τ τ)
Here there are too many variables and parameters for dimensional analysis to be really useful.
However, this effort provides a reminder to check units throughout € the remainder of the solution.
i) For the streamline, time is a constant. Use the first equality of (3.7) to find:
€2 2
dy v xy τ dy t 2 xdx t x
= = → = → ln y = + const.
dx u τ t 2
y τ  2 2
τ 2 2 2
The initial condition requires, x = x´ and y = y´ at t = t´, and this allows the constant to be
determined, yielding:
# y & (t 2 x 2 − t "2 x "2 )
€ ln% ( = .
$ y"' 2 2τ 2
(ii) For the path line, use both components of (3.8):
dx τ dy xy
= 2 and = ,
dt t dt τ
€ in time and use the initial condition to find:
and integrate the first of these
x = −τ t + const. or x − x # = −τ ( t −1 − t #−1 ) .
Use this result for x(t) in the€ second equation € for y(t):
dy y % % 1 1 ( ( dy % 1 1 1 (
= ' −τ ' − * + x $* or =' − + x "* dt .
€ dt τ & & t € t $) ) y & t " t τ )
The last expression can be integrated to find:
−1
#y& t x" # t & # x" 1 & #t& y #t& +# x "t " &# t &.
ln% ( = −1+ − ln% ( = % + ((t − t ") − ln% ( or = % ( exp-% + 1(% −1(0 .
$ y"' € t" τ $ t " ' $ τ t " '€ $ t "' y" $ t"' ,$ τ '$ t " '/
Now use the final equations for x(t) and y(t) to eliminate t. The equation for x(t) can be
rearranged to find:
t" (x − x ") t "
€ = 1− € ,
t τ
so the equation for y becomes:
y % (x − x ") t " ( +% x "t " (%% (x − x ") t " ( (.
−1

= '1− * exp-' + 1*'''1− * −1**0.

y " & € τ ) -,& τ )&& τ ) )0/
(iii) For the streak line, use the path line results but this time evaluate the integration constants at
t = to instead of at t = t´ to find:
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

−1
y #t& +# x "t &# t &.
x − x # = −τ ( t − t
−1 −1
o ) , and = % ( exp-% o + 1(% −1(0.
y" $ to ' ,$ τ '$ t o '/
Now eliminate to find:
y % (x − x ")t (
−1 +% x " % 1 (x − x ") (−1 (% (x − x ")t (.
€ = '1+ * exp-'' ' + * + 1**' *0.
y" & τ €) -,& τ & t τ ) )& τ )0/
And evaluate at t = t´ to reach:
y % (x − x ") t " (
−1 +% x " % 1 (x − x ") (−1 (% (x − x ") t " (.
= '1+ * exp-'' ' + * + 1**' *0.
€ y" & τ ) -,& τ & t " τ ) )& τ )0/

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.12. The velocity components in an unsteady plane flow are given by u = x (1+ t) and
v = 2y (2 + t) . Determine equations for the streamlines and path lines subject to x = x0 at t = 0.

Solution 3.12. i) For the streamline, time is a constant. Use the first equality of (3.7) to find:
dy v 2y (2 + t) dy 2(1+ t) dx € t)
2(1+
€ = = → = → ln y = ln x + const.
dx u x (1+ t) y (2 + t) x (2 + t)
Use of the initial condition produces:
2(1+ 0)
ln y 0 = ln x 0 + const.,
(2 + 0)
€
so the final answer is:
2(1+t )
" y % 2(1+ t) " x % y " x % (2+t )
ln$ ' = ln$ ' or =$ ' .
€ # y 0 & (2 + t) # x 0 & y0 # x0 &
(ii) For the path line, use both components of (3.8):
dx x dy 2y
= and = ,
dt 1+ t dt 2 + t
and integrate the €these in time to find: €
ln x = ln(1+ t) + const. and ln y = 2ln(2 + t) + const.
Use the initial condition to determine the two constants, and exponentiate both equations:
€ € 2
x = x 0 (1+ t) and y = y 0 (1+ t 2) .
To determine the path line, eliminate t to find:
€
€ 2
y = y 0 (1+ (x − x 0 ) 2x 0 ) .
€ €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.13. Using the geometry and notation of Fig. 3.8, prove (3.9).

Solution 3.13. Before starting this problem, it is worthwhile to note that the acceleration of a
fluid particle is invariant under the specified Galilean transformation so the components of U
cannot be part of the final answer. Thus, transformation errors can be readily detected if terms
are missing in the final results or extra ones have appeared.
Figure 3.8 supports the following vector addition formula: x = Ut + x "o + x " . Thus, the
Cartesian-coordinate transformations in this case are given by: x " = x − (e x ⋅ U) t − x o" ,
y " = y − (e y ⋅ U) t − y "o , z" = z − (e z ⋅ U) t − z"o , and t " = t . The transformation of the spatial
derivatives between the stationary frame of reference, Oxyz, € and the steadily moving frame,
O´x´y´z´ is straightforward mathematics: €
∂ ∂x # ∂ ∂y # ∂ ∂z#€∂ ∂t # ∂ ∂
€ € = + + + = ,
∂x ∂x ∂x # ∂x ∂y # ∂x ∂z# ∂x ∂t # ∂x #
∂ ∂x # ∂ ∂y # ∂ ∂z# ∂ ∂t # ∂ ∂
= + + + = , and
∂y ∂y ∂x # ∂y ∂y # ∂y ∂z# ∂y ∂t # ∂y #
∂ ∂x # ∂ ∂y # ∂ ∂z# ∂ ∂t # ∂ ∂
€ = + + + = ,
∂z ∂z ∂x # ∂z ∂y # ∂z ∂z# ∂z ∂t # ∂z#
where the final equality on each line follows from differentiating the definitions of the moving
€
coordinate variables given above. The time derivative requires more effort
∂ ∂x # ∂ ∂y # ∂ ∂z# ∂ ∂t # ∂ ∂ ∂ ∂ ∂
€ = + + + = −e x ⋅ U − ey ⋅ U − ez ⋅ U + .
∂t ∂t ∂x # ∂t ∂y # ∂t ∂z# ∂t ∂t # ∂x # ∂y # ∂z# ∂t #
∂
The first three equations imply: ∇ = ∇ # and the fourth implies: −U ⋅ ∇ $ + . The velocities will
∂t $
be related by: u = U + u´. Now use these results to assemble the fluid particle acceleration
€
starting in the stationary coordinate system, and converting each velocity and differential
operation to the moving€coordinate system.
∂u ' ∂* €
+ (u ⋅ ∇ )u = ) −U ⋅ ∇ & + ,[U + u&] + ([U + u&] ⋅ ∇ &)[U + u&]
∂t ( ∂t & +
∂U ∂u$
= −U ⋅ ∇ $U + − U ⋅ ∇ $u$ + + (U ⋅ ∇ $)U + (u$ ⋅ ∇ $)U + (U ⋅ ∇ $)u$ + (u$ ⋅ ∇ $)u$
∂t $ ∂t $
∂u$ ∂u$
€ = −U ⋅ ∇ $u$ + + (U ⋅ ∇ $)u$ + (u$ ⋅ ∇ $)u$ = + (u$ ⋅ ∇ $)u$
∂t $ ∂t $
Here most of the simplifications occur because all derivatives of U are zero; it is a constant.
Thus,€ as expected, the form of the fluid particle acceleration is frame invariant for coordinate
systems related by Galilean transformations.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.14. Determine the unsteady, ∂u/∂t, and advective, (u⋅∇)u, fluid acceleration terms for
the following flow fields specified in Cartesian coordinates.
a) u = ( u(y,z,t),0,0)
b) u = Ω × x where Ω = (0,0,Ωz (t))
c) u = A(t)( x,−y,0)
€ d) u = (Uo + uosin(kx – Ωt), 0, 0) where Uo, uo, k and Ω are positive constants
€ €
Solution 3.14. a) Here there is only one component of the fluid velocity; thus
€ ∂u ∂t = (∂ ∂t )( u(y,z,t),0,0) = (∂u ∂t,0,0) , and
[u ⋅ ∇]u = [(u(y,z,t),0,0) ⋅ (∂ ∂x,∂ ∂y,∂ ∂z)]( u(y,z,t),0,0) = [u(∂ ∂x)]( u(y,z,t),0,0) = 0 .
b) Here the fluid velocity has two components: u = Ω × x = (–Ωzy, +Ωzx, 0), so
€ % dΩz dΩz ( dΩz
∂u ∂t = (∂ ∂t )(−Ωz y,Ωz x,0) = ' −y ,x ,0* = (−y, x,0) , and
€ & dt dt ) dt
[u ⋅ ∇]u = [(−Ωz y,Ωz x,0) ⋅ (∂ ∂x,
€ ∂ ∂y,∂ ∂z)](−Ωz y,Ωz x,0)
= [−Ωz y(∂ ∂x) + Ωz x(∂ ∂y)](−Ωz y,Ωz x,0) = (−Ω2z x,−Ω2z y,0) = −Ω2z ( x, y,0) .
€ c) Again the fluid velocity has two components: (Ax, –Ay, 0), so
€ $ dA dA ' dA
∂u ∂t = (∂ ∂t )( Ax,−Ay,0) = & x ,−y ,0) = ( x,−y,0) , and
€ % dt dt ( dt
[u ⋅ ∇]u = [( Ax,−Ay,0) ⋅ (∂ ∂x,∂ ∂y,∂ ∂z)]( Ax,−Ay,0)
= [ Ax(∂ ∂x) − Ay(∂ ∂y)]( Ax,−Ay,0) = ( A 2 x, A 2 y,0) = A 2 ( x, y,0) .
€ d) Here again there is only one component of the fluid velocity; thus
€ ∂u ∂t = (∂ ∂t )(U o + uo sin(kx − Ωt),0,0) = (−Ωuo cos(kx − Ωt),0,0) , and
€[u ⋅ ∇ ]u = [(U o + uo sin(kx − Ωt),0,0) ⋅ (∂ ∂x,∂ ∂y,∂ ∂z)](U o + uo sin(kx − Ωt),0,0) .
= [(U o + uo sin(kx − Ωt))(∂ ∂x)](U o + uo sin(kx − Ωt),0,0)
€ = ((U o + uo sin(kx − Ωt)) kuo cos(kx − Ωt),0,0)
€ (
= kU o uo cos(kx − Ωt) + 12 kuo2 sin[2(kx − Ωt)],0,0 . )
€
€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.15. Consider the following Cartesian velocity field u = A(t)( f (x),g(y),h(z)) where
A, f, g, and h are non-constant functions of only one independent variable.
a) Determine ∂u/∂t, and (u⋅∇)u in terms of A, f, g, and h, and their derivatives.
b) Determine A, f, g, and h when Du/Dt = 0, u = 0 at x = 0, and u is finite for t > 0.
c) For the conditions in b), determine the equation for€the path line that passes through xo at time
to, and show directly that the acceleration a of the fluid particle that follows this path is zero.

Solution 3.15. a) Here there are three components of the fluid velocity; thus
∂u ∂ # dA dA dA & dA
= ( Af , Ag, Ah ) = % f ,g ,h ( = ( f ,g,h ) , and
∂t ∂t $ dt dt dt ' dt
+ % ∂ ∂ ∂ (.
[u ⋅ ∇]u = -A( f ,g,h) ⋅ ' , , *0( Af , Ag, Ah)
, & ∂x ∂y ∂z )/
€ # ∂ ∂ ∂& ) df dg dh ,
= A 2% f + g + h (( f ,g,h ) = A 2 + f ,g ,h . ,
$ ∂x ∂y ∂z ' * dx dy dz -
€ where the partial derivatives have become total derivatives because f, g, and h are only functions
of one variable.
b) For the given velocity field, Du/Dt = 0 implies:
€ dA df dA dg dA dh
f + A2 f =0 , g + A 2g = 0 , and h + A 2h = 0.
dt dx dt dy dt dz
Start with the first equation, assume A2f is not zero, and divide by it to find:
1 dA df
+ = 0.
€ A 2 dt dx €
The first term in this equation € only depends on t while the second one only depends on x, thus,
each must be equal and opposite, and constant (= C). So,
1 dA dA 1 1 df
2
= −C → − 2 = Cdt € → = C(t − t o ) → A = , and = C → f = C(x − x o ) ,
A dt A A C(t − t o ) dx
where to and xo are constants of integration. Similarly, for the other two component directions:
g = C(y − y o ) and h = C(z − zo ) .
Here, we presume to ≤ 0 so that A is finite for t > 0.
€ €
c) The x-component of the path line is defined by:
dx C(x − x o ) dx dt
= u = Af = € → = €→ ln(x − x o ) = ln(t − t o ) + const. → x − x o = U(t − t o ) ,
dt C(t − t o ) x − x o t − to
where U is a constant. Similarly for the other Cartesian directions: y − y o = V (t − t o ) , and
z − zo = W (t − t o ) , where V and W are constants. So defining the constant vector U = (U, V, W),
the path line of interest is:
€
x – xo = U(t – to).
Thus, x(t) is linear function of t, so dx(t)/dt = U, and a = d€2x(t)/dt2 = 0. This exercise illustrates
€ how the unsteady and advective acceleration terms may be equal and opposite.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.16 If a velocity field is given by u = ay and v = 0, compute the circulation around a
circle of radius ro that is centered on at the origin. Check the result by using Stokes’ theorem.

Solution 3.16. In plane polar coordinates, the vector path-length element, ds, on a circle of radius
ro is ds = tds = eθ ro dθ . From Example 2.1, the radial and angular velocity component are:
ur = ucosθ + vsinθ = arsinθcosθ + 0, and u = –usinθ + vcosθ = –arsin2θ ,
θ

where u = ay = arsinθ and v = 0 has been used. Thus, circulation is:

θ = 2π θ = 2π θ = 2π
€ Γ = ∫ u ⋅ ds = ∫θ = 0 u ⋅ eθ ro dθ = ∫θ = 0 uθ ro dθ = − ∫θ = 0 aro2 sin 2 θdθ = −aπro2 .
Now use Stokes' theorem to reach the same result. Start by computing the vorticity:
ex ey ez
€ ∇ × u = ∂ ∂x ∂ ∂y ∂ ∂z = −ae z .
ay 0 0
Insert this into the Stokes' theorem area integral using n = ez, and dA = rdrdθ:
θ = 2π r= r
Γ = ∫∫ (∇ × u) ⋅ ndA = ∫θ = 0 ∫ r= 0o (−ae z ) ⋅ e z rdrdθ = −aπro2 ,
A
€ of the prior circulation calculation.
and this matches the result

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.17. Consider a plane Couette flow of a viscous fluid confined between two flat plates
a distance b apart. At steady state the velocity distribution is u = Uy/b and v = w = 0, where the
upper plate at y = b is moving parallel to itself at speed U, and the lower plate is held stationary.
Find the rates of linear strain, the rate of shear strain, and vorticity in this flow.

Solution 3.17. Here there is only one velocity component: u = Uy/b. The strain rate tensor is:
1 # ∂ui ∂u j & * ∂u ∂y + ∂v ∂x )- * 0
1
2(
∂ u ∂x U 2b-
Sij = %% + (( → Sij = + 1 .=+ ..
2 $ ∂x j ∂x i ' , 2 (∂v ∂x + ∂u ∂y ) ∂ v ∂y / ,U 2b 0 /
Thus, the linear strain rates are both zero, and the shear strain rate is U/2b.
The vorticity vector has one non-zero component:
∂v ∂u U
€ ωz = − = − .
∂x ∂y b

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.18. The steady two-dimensional flow field inside a sloping passage is given in (x,y)-
(
Cartesian coordinates by u = (u, v) = (3q 4h ) 1− ( y h )
2
) ( 1, ( y h) (dh dx)) where q is the volume
flow rate per unit length into the page, and h is the passage's half thickness. Determine the
streamlines, vorticity, and strain rate tensor in this flow away from x = 0 when h = αx where α is
a positive constant. Sample profiles of u(x,y) vs. y are shown at two x-locations in the figure.
What are the equations of the streamlines along which the x- and y-axes are aligned with the
principal axes of the flow? What is the fluid particle rotation rate along these streamlines?
y!

h(x)!
x!

Solution 3.18. For planar flow in Cartesian coordinates, the streamlines are determined from:
dy v ( y h ) ( dh dx )
= = = ( y α x )α = y x .
dx u 1
Separate the differentials and integrate to find: ln(y) = ln(x) + const. Exponentiate to find: y = Cx,
where C is a constant. Thus, the streamlines are straight lines through the origin of coordinates.
The vorticity ωz is determined from:
∂v ∂u ∂ )# 3q &# y 2 & y dh , ∂ )# 3q &# y 2 &,
ω z = − = +% (%1− 2 ( . − +% (%1− (.
∂x ∂y ∂x *$ 4h '$ h ' h dx - ∂y *$ 4h '$ h 2 '-
3q ∂ ) 1 # y 2 & y , 3q ∂ ) 1 # y 2 &,
= + %1− 2 2 ( α . − + %1− 2 2 (.
4 ∂x *α x $ α x ' α x - 4 ∂y *α x $ α x '-
3q ) 2y 4y 3 , 3q ) 2y , 3qy ) 2y 2 1,
= +− 3 + 3 5 . − +− 3 3 . = +−1+ + .
4 * α x α x - 4 * α x - 2α x 3 * α 2 x2 α 2 -
The strain-rate tensor Sij is computed from the following velocity derivatives:
∂u ∂ )" 3q %" y 2 %, 3q ∂ ) 1 " y 2 %, 3q ) 1 3y 2 , 3q ) 3y 2 ,
S11 = = +$ '$1− 2 '. = + $1− 2 2 '. = +− 2 + 3 4 . = + −1+ .
∂x ∂x *# 4h &# h &- 4 ∂x *α x # α x &- 4 * α x α x - 4α x 2 * α 2 x2 -
∂v ∂ )" 3q %" y 2 % y dh , 3q ∂ ) 1 " y2 % y , 3q " 3y 2 %
S22 = = +$ '$1− 2 ' .= + $1− ' α. = $1− ' , and
∂y ∂y *# 4h &# h & h dx - 4 ∂y *α x # α 2 x 2 & α x - 4α x 2 # α 2 x 2 &
1 " ∂u ∂v % 1 ∂ /( 3q +( y 2 +2 1 ∂ /( 3q +( y 2 + y dh 2 3qy / 1 2y 2 2
S12 = S21 = # + & = 1* -* 1− -4 + 1* -* 1− - 4 = 1− −1+ 4,
2 $ ∂y ∂x ' 2 ∂y 0) 4h ,) h 2 ,3 2 ∂x 0) 4h ,) h 2 , h dx 3 4α x 3 0 α 2 α 2 x2 3
where the differentiation details for S12 are available above in the calculation of ωz.
The principle axes occur then the off-diagonal strain rate components are zero. This
occurs when
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

3qy " 1 2y 2 %
S12 = S21 = $− −1+ '= 0 .
4α x 3 # α 2 α 2 x2 &
This equality is satisfied along the x-axis where y = 0, and when the contents of the [,]-brackets
are zero:
1 2y 2
2
+1 = 2 2
or y = ±x (1+ α 2 ) 2
α α x
These streamlines only occur inside flow wedge when α > 1.
The fluid particle rotation rate is ωz/2, so from the results above for ωz:
!ωz $ !ω $ 3q (1+ α 2 ) 2
# & = 0 , and # z & =± .
" 2 %y=0 " 2 %y=± x (1+α 2 ) 2 2α 3 x 2
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.19. For the flow field u = U + Ω × x , where U and Ω are constant linear- and angular-
velocity vectors, use Cartesian coordinates to a) show that Sij is zero, and b) determine Rij.

Solution 3.19. Since no€simplifications are given, all the components of U = (U1, U2, U3) and Ω
= (Ω1, Ω2, Ω3) should be treated as being non-zero. In Cartesian coordinates, the velocity field is
e1 e 2 e 3
u = U + Ω × x = U1e1 + U 2e 2 + U 3e 3 + Ω1 Ω2 Ω3
x1 x 2 x 3
= (U1 + Ω2 x 3 − Ω3 x 2 )e1 + (U 2 + Ω3 x1 − Ω1 x 3 )e 2 + (U 3 + Ω1 x 2 − Ω2 x1 )e 3
a) Use this velocity field result to compute the velocity gradient tensor, and its transpose
(indicated with a superscript "T" below) to sort out which derivatives are zero and which ones
€
are not.
€ #∂u1 ∂x1 ∂u1 ∂x 2 ∂u1 ∂x 3 ' # 0 −Ω3 +Ω2 '
∂ui % % % %
= $∂u2 ∂x1 ∂u2 ∂x 2 ∂u2 ∂x 3 ( = $+Ω3 0 −Ω1 (
∂x j % % %
&∂u3 ∂x1 ∂u3 ∂x 2 ∂u3 ∂x 3 ) &−Ω2 +Ω1 0 %)
T )∂u1 ∂x1 ∂u2 ∂x1 ∂u3 ∂x1 - ) 0 +Ω3 −Ω2 -
∂u j # ∂ui & + + + +
= %% (( = *∂u1 ∂x 2 ∂u2 ∂x 2 ∂u3 ∂x 2 . = *−Ω3 0 +Ω1 .
∂x i $ ∂ x j ' + + +
€ ,∂u1 ∂x 3 ∂u2 ∂x 3 ∂u3 ∂x 3 / ,+Ω2 −Ω1 0 +/
This result can be used to construct the strain rate tensor Sij:
+ 1
−Ω3 + Ω3 ) 12 (+Ω2 − Ω2 )/
2(
# T& 0
#
1 ∂ui ∂u j & 1 ∂u # ∂u & - -
€ Sij = %% + (( = % i + %% i (( ( = , 12 (+Ω3 − Ω3 ) 0 1
2 (−Ω1 + Ω1 ) 0
2 $ ∂x j ∂x i ' 2 % ∂x j $ ∂x j ' ( - 1 -
$ ' . (−Ω2 + Ω2 ) 1 (+Ω1 − Ω1 ) 0 1
2 2
"0 0 0&
$ $
= #0 0 0' .
$0 0 0$
€ % (
b) Similarly for the rotation tensor:
T + 0 −Ω3 − Ω3 +Ω2 + Ω2 /
∂ui ∂u j ∂ui $ ∂ui ' - -
€ Rij = − = − &
& )
) = , +Ω 3 + Ω 3 0 −Ω − Ω 1 0
∂x j ∂ x i ∂x j % ∂ x j ( - -
. −Ω2 − Ω2 +Ω1 + Ω1 0 1
$ 0 −2Ω3 +2Ω2 (
& &
= %+2Ω3 0 −2Ω1 ) .
&−2Ω +2Ω 0 &*
€ ' 2 1

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.20. Starting with a small rectangular volume element δV = δx1δx2δx3, prove (3.14).

Solution 3.20. The volumetric strain rate for a fluid element is:
1 D 1 D 1 D 1 D 1 D
(δV ) = (δx1δx 2δx 3 ) = (δx1 ) + (δx 2 ) + (δx 3 ) .
δV Dt δx1δx 2δx 3 Dt δx1 Dt δx 2 Dt δx 3 Dt
From Section 3.4 in the text, the linear strain rate corresponding to elongation or contraction of a
fluid element in the first direction is:
1 D ∂u
€ (δx1) = S11 = 1 ,
δx1 Dt ∂x1
and this can be immediately extended to the other two directions,
1 D ∂u 1 D ∂u
(δx 2 ) = S22 = 2 and (δx 3 ) = S33 = 3 ,
δx 2 Dt ∂x 2 δx 3 Dt ∂x 3
€
because its geometric derivation (see Figure 3.10) did not rely on any special properties of the
first direction. Substitution of these relationships into the final version of the volumetric strain
rate given above produces:
€ 1 D € ∂ u ∂u ∂u ∂u
(δV ) = S11 + S22 + S33 = 1 + 2 + 3 = i = Sii .
δV Dt ∂x1 ∂x 2 ∂x 3 ∂x i

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.21. Let Oxyz be a stationary frame of reference, and let the z-axis be parallel with the
fluid vorticity vector in the vicinity of O so that ω = ∇ × u = ω ze z in this frame of reference. Now
consider a second rotating frame of reference Ox "y "z" having the same origin that rotates about
the z-axis at angular rate Ωez. Starting from the kinematic relationship, u = (Ωe z ) × x + u$ , show
that in the vicinity of O the vorticity € ω " = ∇ " × u" in the rotating frame of reference can only be
zero when 2Ω = ωz, where ∇ " is the € gradient operator in the primed coordinates. The following
unit vector transformation rules may be of use: e"x = e x cos(Ωt) + e y sin(Ωt) ,
€
e"y = −e x sin(Ωt) + e y cos(Ωt)€
, and e"z = e z .
€
Solution 3.21. The approach here is to compute ω " = ∇ " × u" in the stationary frame of reference
€
and then determine the parameter choice(s) necessary for ω´ to be zero. The vector x must have a
€ €
consistent representation in either frame, so
x = xe x + ye y = x "e"x + y "e"y = x "(e x€cos(Ωt) + e y sin(Ωt)) + y "(−e x sin(Ωt) + e y cos(Ωt)) .
Equating components in the stationary frame of reference produces:
x = x " cos(Ωt) − y " sin(Ωt) , and y = x " sin(Ωt) + y " cos(Ωt) .
The remaining independent variables are the same in either frame: t = t ", and z = z". Thus, spatial
€ derivatives are related by:
∂ ∂x ∂ ∂ y ∂ ∂z ∂ ∂ t ∂ ∂ ∂
€ = + + € + = cos(Ωt) + sin(Ωt) ,
∂x # ∂x # ∂x ∂x # ∂y ∂x # ∂z ∂x # ∂t € ∂x € ∂y
∂ ∂x ∂ ∂ y ∂ ∂z ∂ ∂ t ∂ ∂ ∂
= + + + = −sin(Ωt) + cos(Ωt) , and
∂y # ∂y # ∂x ∂y # ∂y ∂y # ∂z ∂y # ∂t ∂x ∂y
∂ ∂x ∂ ∂ y ∂ ∂z ∂ ∂ t ∂ ∂
€ = + + + = .
∂z# ∂z# ∂x ∂z# ∂y ∂z# ∂z ∂z# ∂t ∂z
Using the above information, the gradient operator in the rotating coordinates can be rewritten in
€
terms of the stationary frame coordinates and unit vectors:
∂ ∂ ∂
∇" = €e"x + e"y + e"z
∂x " ∂y " ∂z"
$ ∂ ∂'
= (e x cos(Ωt) + e y sin(Ωt))&cos(Ωt) + sin(Ωt) )
% ∂x ∂y (
€ % ∂ ∂( ∂
+(−e x sin(Ωt) + e y cos(Ωt))'−sin(Ωt) + cos(Ωt) * + e z
& ∂x ∂y ) ∂z
€ & ∂ ∂ ∂ ∂)
∇ " = (e x cos 2 (Ωt) + e x sin(Ωt)cos(Ωt) + e y sin(Ωt)cos(Ωt) + e y sin 2 (Ωt) +
' ∂x ∂y ∂x ∂y *
€ % ∂ ∂ ∂ ∂( ∂
+' e x sin 2 (Ωt) − e x sin(Ωt)cos(Ωt) − e y sin(Ωt)cos(Ωt) + e y cos 2 (Ωt) * + e z
& ∂x ∂y ∂x ∂y ) ∂z
€ & ∂ ∂ ∂ ∂ )
∇ " = (e x cos 2 (Ωt) + e x sin(Ωt)cos(Ωt) + e y sin(Ωt)cos(Ωt) + e y sin 2 (Ωt) +
' ∂x ∂y ∂x ∂y *
€ % ∂ ∂ ∂ ∂( ∂
+' e x sin 2 (Ωt) − e x sin(Ωt)cos(Ωt) − e y sin(Ωt)cos(Ωt) + e y cos 2 (Ωt) * + e z
& ∂x ∂y ∂x ∂y ) ∂z
€ ∂ ∂ ∂
∇" = e x + e y + e z .
∂x ∂y ∂z
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

This result seems too simple, but should not be a surprising to a routine user of vector calculus.
Now resolve rotating-frame velocity components in the stationary frame of reference:
u" = u − (Ωe z ) × x = ue x + ve y + we z + Ωye x − Ωxe y .
Using the stationary frame components and unit vectors, the vorticity in the rotating frame is:
ex ey ez ex ey ez ex ey ez
ω " = ∇ " × u" = ∂ ∂x ∂ ∂y ∂ ∂z = ∂ ∂ x ∂ ∂y ∂ ∂z + ∂ ∂x ∂ ∂ y ∂ ∂z
€
u + Ωy v − Ωx w u v w Ωy −Ωx 0
ex ey ez
= ∇ × u + ∂ ∂x ∂ ∂y ∂ ∂z = ω ze z + e x (0) + e y (0) + e z (−Ω − Ω) = e z (ω z − 2Ω)
€ Ωy −Ωx 0
Thus, ω´ will be zero when ωz = 2Ω.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.22. Consider a plane-polar area element having dimensions dr and rdθ. For two-
dimensional flow in this plane, evaluate the right-hand side of Stokes’ theorem
∫ ω ⋅ ndA = ∫ u ⋅ ds and thereby show that the expression for vorticity in plane-polar coordinates
1∂ 1 ∂u
is: ω z = (ruθ ) − r .
r ∂r r ∂θ
€
Solution 3.22. Using the element shown with angular width dθ,
€ ur + (!ur/!!)d!
u! + (!u!/!")dr
dr
u! ur

application of Stokes' theorem provides:

( ∂uθ + ( ∂u +
∫ ω ⋅ ndA = ∫ u ⋅ ds → ω z rdθdr = ur dr + *) uθ + dr-( r + dr) dθ − * ur + r dθ -dr − uθ rdθ , or
∂r , ) ∂θ ,
∂u ∂u
ω z rdθdr = ur dr + ruθ dθ + r θ drdθ + uθ drdθ − ur dr − r dθdr − uθ rdθ + ...
∂r ∂θ
∂uθ ∂ur
€ =r drdθ + uθ drdθ − dθdr + ...
∂r ∂θ
where + ... indicates the presence of higher order terms in dr and dθ. Division by rdθdr and
€
passing to the limit where dr and dθ go to zero produces:
€ ∂u 1 1 ∂ur 1 & ∂ ∂u )
ω z = θ + uθ − = ( ( ruθ ) − r + .
∂r r r ∂θ r ' ∂θ ∂θ *

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.23. The velocity field of a certain flow is given by u = 2xy 2 + 2xz 2 , v = x 2 y , and
w = x 2 z . Consider the fluid region inside a spherical volume x2 + y2 + z2 = a2. Verify the validity
of Gauss’ theorem ∫∫∫ ∇ ⋅ udV = ∫∫ u ⋅ ndA by integrating over the sphere.
V A
€ €
€ Solution 3.23. First compute the divergence of the velocity field.
∂u ∂v ∂w ∂ ∂ ∂
∇ ⋅€u = + + = (2xy 2 + 2xz 2 ) + ( x 2 y ) + ( x 2 z) = 2y 2 + 2z 2 + 2x 2 = 2r 2 .
∂x ∂y ∂z ∂ x ∂y ∂z
The volume integral of ∇ ⋅ u is:
r= a
8
∫∫∫ ∇ ⋅ udV = ∫ (2r 2 ) 4πr 2 dr = 5 πa 5 .
V r= 0
€
Now work on the surface integration using spherical coordinates (see Figure 3.3d, and Appendix
B). Here, n = €
e r = e x cosϕ sin θ + e y sin ϕ sin θ + e x cos θ , so
u ⋅ n = ucosϕ sin θ + v sin ϕ sin θ + w cos θ
€
= (2xy 2 + 2xz 2 ) cosϕ sin θ + ( x 2 y ) sin ϕ sin θ + ( x 2 z) cos θ
Unfortunately, this result is in mixed variables so convert everything to spherical polar
€
coordinates
€ using
x = r cosϕ sin θ , y = r sin ϕ sin θ , z = r cosθ .
€
This conversion produces:
u ⋅ n = 2r 3 (cos ϕ sin 2 ϕ sin 3 θ + cos ϕ sin θ cos 2 θ ) cos ϕ sin θ
+r 3 (cos 2 ϕ€
sin ϕ sin 3 θ ) sin ϕ sin
€ θ + r (cos ϕ sin
3 2
€ 2 θ cos θ ) cos θ
= 2r 3 (sin 2 ϕ sin 2 θ + cos2 θ ) cos2 ϕ sin 2 θ + r 3 (sin 2 ϕ sin 2 θ ) cos2 ϕ sin 2 θ + r 3 (cos 2 θ ) cos 2 ϕ sin 2 θ
€
€
= r 3 ( 3sin 2 ϕ sin 2 θ + 3cos 2 θ ) cos 2 ϕ sin 2 θ = r 3 ( sin (2ϕ)(1− cos θ) + 3cos ϕ cos θ(1− cos θ))
3
4
2 2 2 2 2 2

So, the surface integral produces:

€ ϕ = 2π θ = π

€
∫∫ u ⋅ ndA = ∫ ∫
A ϕ= 0 θ = 0
a3 ( sin (2ϕ)(1− cos θ) + 3cos ϕ cos θ(1− cos θ))a sinθdθdϕ
3
4
2 2 2 2 2 2 2

ϕ = 2π θ=π ϕ = 2π θ=π
2
=a 5
∫ 3
4
2
sin (2ϕ )dϕ ∫ (1− cos2 θ ) sinθdθ + a5 ∫ 3cos2 ϕdϕ ∫ cos2 θ (1− cos2 θ ) sin θdθ
ϕ= 0 θ=0 ϕ= 0 θ=0
+1 +1
€ #3 & 2
= a % π ( ∫ (1− β 2 ) dβ + a 5 ( 3π ) ∫ β 2 (1− β 2 )dβ
5

$ 4 ' −1 −1

€ # 3 & 16 4 # 12 12 & 8
= a % π ( + a ( 3π ) = % + (πa 5 = πa 5 ,
5 5

$ 4 ' 15 15 $ 15 15 ' 5
€which is the same as the result of the volume integration.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.24. A flow field on the xy-plane has the velocity components u = 3x + y and v = 2x –
3y. Show that the circulation around the circle (x − 1)2 + (y − 6)2 = 4 is 4π.

Solution 3.24. The circle is centered at (1,6) and its radius is 2. So, if (x,y) is a point on the circle
then x = 1 + 2cosθ, and y = 6 + 2sinθ, where θ is the angle from the horizontal. The velocity
component tangent to the circle will be
uθ = eθ ⋅ u = (−sin θ,cos θ ) ⋅ (u,v) = −usin θ + v cos θ .
Evaluate this velocity component.
uθ = −(3x + y)sin θ + (2x − 3y)cos θ = (−3sin θ + 2cosθ )x − (sin θ + 3cosθ )y
= (−3sin θ + 2cos θ )(1+ 2cosθ ) − (sin θ + 3cos θ )(6 + 2sin θ )
€
= −3sin θ − 6sin θ cosθ + 2cos θ + 4 cos2 θ − 6sin θ − 2sin 2 θ −18cosθ − 6cos θ sin θ
€ = −9sin θ −12sin θ cosθ −16cos θ + 4 cos2 θ − 2sin 2 θ
Now compute the circulation:
2π 2π
Γ= ∫ uθ rdθ = ∫ (−9sin θ −12sinθ cosθ −16cosθ + 4 cos2 θ − 2sin 2 θ )2dθ
θ=0 θ=0
€
= (0 + 0 + 0 + 4 π − 2π )2 = 4 π

€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.25. Consider solid-body rotation about the origin in two dimensions: ur = 0 and uθ =
ω0r. Use a polar-coordinate element of dimension rdθ and dr, and verify that the circulation is
vorticity times area. (In Section 5 this was verified for a circular element surrounding the origin.)

Solution 3.25. Using the element shown with angular width dθ,

ur + (!ur/!!)d!
u! + (!u!/!")dr
dr
u! ur

the circulation is around the element is the sum or four terms:

' ∂u * ' ∂u *
Γ = ∫ u ⋅ ds = ur dr + ) uθ + θ dr,( r + dr) dθ − ) ur + r dθ , dr − uθ rdθ .
( ∂r + ( ∂θ +
Now substitute in the velocity field: ur = 0 and uθ = ω0r.
Γ = (ω 0 r + ω 0 dr)( r + dr) dθ − ω 0 r 2 dθ = 2ω 0 rdrdθ + ω 0 (dr) 2 dθ = (2ω 0 )rdrdθ ,
2
where
€ the final equality holds in the limit as the differential elements become small and (dr) dθ
is negligible compared to rdrdθ. Thus, since the voriticity is 2ω0 and rdrdθ is the area of the
element, the final relationship states: circulation = (vorticity) x (area).
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

$ '
& −Ay + Ax
Exercise 3.26. Consider the steady Cartesian velocity field u = , ,0) .
& ( x 2 + y 2 )β ( x 2 + y 2 )β )
% (
a) Determine the streamline that passes through x = (x o , y o ,0)
b) Compute Rij for this velocity field.
c) For A > 0, explain the sense of rotation (i.e. clockwise or counter clockwise) for fluid elements
for β < 1, β = 1, and β > 1. €
€
β
dy v Ax ( x 2 + y 2 ) x
Solution 3.26. a) Use the definition of a streamline: = = β = − . Use the
dx u −Ay ( x 2 + y 2 ) y
two ends of this extended equality to find: ydy = −xdx , and integrate the resulting differential
relationship to get: y 2 2 = −x 2 2 + const . Evaluate the constant using the required condition:
x 2 + y 2 = x o2 + y o2 . This is a circle with radius€ x o2 + y o2 . Therefore the streamlines are circles.
€ " 0 ∂u ∂ y − ∂v ∂ x 0 &
∂u ∂u $ $
b) The rotation tensor is: Rij = i − j = # ∂ v ∂ x − ∂ u ∂ y 0 0 ' , where the
€ ∂ x€j ∂ xi $ $
$% 0 0 0 $(
second equality comes from putting the specified velocity field into the definition of Rij with
u = (u, v, w) = (u1, u2 , u3 ) = ui as usual. Evaluating the derivatives produces:
" &
$ 0 −x 2 − y 2 + 2 β y 2 − ( x 2 + y 2 − 2 β x 2 ) 0 $
A $ 2 2 $
Rij = β +1 # x + y − 2 β x 2 − (−x 2 − y 2 + 2 β y 2 ) 0 0 '
( x 2 + y2 ) $$ 0 0
$
0 $
% (
" 0 −1 0 &
2A(1− β ) $ $
= β # 1 0 0 '
( x 2 + y2 ) $% 0 0 0 $(
c) The following answers are based on A > 0. For β < 1, fluid particles rotate counter clockwise
(positive ωz). For β = 1, fluid particles do not rotate. For β > 1, fluid particles rotate clockwise
(negative ωz). Interestingly, the streamlines are the same (circular!) for all three possibilities.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.27. Using indicial notation (and no vector identities) show that the acceleration a of a
fluid particle is given by: a = ∂u ∂t + ∇ ( u ) + ω × u where ω is the vorticity.
1
2
2

Solution 3.27. The acceleration of a fluid particle is a = Du Dt ≡ ∂u ∂t + ( u ⋅ ∇) u . Thus, the task

is to prove (u ⋅€∇ )u = ∇ ( u ) + ω × u. Start from the advective acceleration written in index
1
2
2

notation and force the rotation tensor to appear:

∂u $ ∂u ∂u ' ∂u 1 ∂ 2
u j i = u j && i − j )) − u j j = u j Rij +
∂x j
(u j ) .
€ % ∂x j ∂x i ( ∂x i 2 ∂x i
From (3.15), Rij = −εijkω k , so this becomes
∂u 1 ∂ 2 1 ∂ 2 (1 2+
u j i = −εijk u jω k + ( u j ) = εikjω k u j + ( u j ) = ω × u + ∇* u - ,
€ ∂x j 2 ∂x i 2 ∂x i )2 ,
where
€ the final equality follows from the index notation definitions of the cross product (2.21),
2
gradient (2.22), and vector magnitude ( u 2j = u ).
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.28. Starting from (3.29), show that the maximum uθ in a Gaussian vortex occurs
when 1+ 2(r 2 σ 2 ) = exp(r 2 σ 2 ) . Verify that this implies r ≈ 1.12091σ.

Solution 3.28. Differentiate the uθ equation from (3.29) with respect to r and set this derivative
equal to zero.
€
Γ d '1− exp(−r σ ) * Γ ' 1− exp(−r σ ) exp(−r σ ) ' 2r **
2 2 2 2 2 2
d
(uθ (r)) = ) ,= ) )− 2 ,,, = 0 .
dr 2π dr )( r , 2π )− r2
−
r ( σ ++
+ (
Eliminate common factors assuming r ≠ 0.
0 = −1+ exp(−r 2 σ 2 ) + (2r 2 σ 2 ) exp(−r 2 σ 2 ) = −1+ (1+ 2r 2 σ 2 ) exp(−r 2 σ 2 ) .
€ This can be rearranged to:
exp( r 2 σ 2 ) = 1+ 2r 2 σ 2 ,
which
€ is the desired result. When r/σ ≈ 1.12091, then
exp( r 2 σ 2 ) = 3.51289 and 1+ 2r 2 σ 2 = 3.51288 ,
which is suitable numerical€agreement.

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.29. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of
the area of the parallelogram shown is hl(dθ/dt)cosθ when θ depends on time while h and l are
constants.
y!
l!

θ(t)! h!

x!

Solution 3.29. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and
the integrands are simplified, so (3.35) simplifies to:
d
∫ dV = + ∫ A*(t ) b ⋅ n dA .
dt V*(t )
(Here the volume integration is two dimensional, and produces the parallelogram's area, hlsinθ,
which can be time differentiated, d(hlsinθ)/dt, to reach the desired result. However, this is not the
intended solution path for this problem.)
When θ is time-dependent, the parallelogram has three moving sides (a left side of length
h, a right side of length h, and a top side of length l). Thus, the simplified version of (3.35)
reduces to:
d
{ }
∫ dV = ∫ left si de + ∫ right si de + ∫ top si de b ⋅ ndA .
dt V*(t )
Here we note that the left and right sides move identically, so b will be the same. However, n
points in opposite directions on these two sides, so the contributions from these two sides cancel.
The x-coordinates of points on the parallelogram's top side are hcosθ ≤ x(t) ≤ l + hcosθ. The y-
coordinate of the parallelogram's top side is y(t) = hsinθ. Time differentiate the location of any
point on the parallelogram's top side to find b:
b = (dx/dt, dy/dt) = (–hsinθ, hcosθ)(dθ/dt).
And, on the parallelogram's top side, n is ey, so
b ⋅ n = ( h cosθ ) ( dθ dt ) ,
and in two dimensions dA is just dx. Thus, the simplified version of (3.35) is:
l+h cosθ
d dθ dθ
dt
∫ V*(t )
dV = ∫ top si de
b ⋅ ndA = ∫ h cosθ
dt
dx = hl cosθ
dt
,
h cosθ
and this is the desired result.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.30. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of
the area of the triangle shown is 12 b ( dh dt ) when h depends on time and b is constant.
y!
b!

h(t)!

x!

Solution 3.30. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and
the integrands are simplified, so (3.35) simplifies to:
d
∫ dV = + ∫ A*(t ) b ⋅ n dA .
dt V*(t )
(Here the volume integration is two dimensional, and produces the triangle's area, hb/2, which
can be time differentiated, (d/dt)(hb/2) to reach the desired result. However, this is not the
intended solution path for this problem.)
When h is time-dependent, the triangle has two moving sides (a top side of length b, and
a hypotenuse of length [h2 + b2]1/2). Thus, the simplified version of (3.35) reduces to:
d
{ }
∫ dV = ∫ top si de + ∫ hypotenuse b ⋅ ndA .
dt V*(t )
The x-coordinates of the triangle's top side are 0 ≤ x ≤ b. The y-coordinate of the triangle's top
side is y(t) = h(t). Time differentiate the location of any point on the triangles's top side to find b:
b = (dx/dt, dy/dt) = (0, dh/dt).
And, on the triangle's top side, n is ey, so
b ⋅ n = dh dt ,
The x-y coordinates of the triangle's hypotenuse fall on the line y = (h/b)x for 0 ≤ x ≤ b.
The motion of points on the hypotenuse can be described by a purely vertical velocity. So, for
any constant x-location, differentiate the equation of this line to find b:
b = (dx/dt, dy/dt) = (0, (x/b)dh/dt).
Tangent and normal vectors to an x-y curve are (1, dy/dx) and (dy/dx, –1), respectively. Thus, the
outward unit normal on the hypotenuse of the triangle is:

n=
(h b, −1) = (h, −b) , so b ⋅ n = − x ( dh dt ) .
(h / b)2 +1 h2 + b2 h2 + b2
Thus, the simplified version of (3.35) can be written:
b b
d dh x ( dh dt ) $
∫ dV = ∫ b ⋅ ndA + ∫ b ⋅ ndA = ∫ dx − ∫ % 1+ (h / b)2 dx &' ,
dt V*(t )
top si de hypotenuse 0 dt 0
2
h +b 2

where, in two dimensions, dA is just dx on the top side and dA is a path length element along the
hypotenuse, ds = [1 + (dy/dx)2]1/2dx. The factor in [,]-brackets is this path length element in terms
of dx. Perform the integrations to find:
b
d dh 1 dh # x 2 & b dh
∫ dV = b dt − b dt %$ 2 (' = 2 dt ,
dt V*(t ) 0
and this is the desired result.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.31. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of
the area of the ellipse shown is π b ( da dt ) when a depends on time and b is constant.
y!

b!
x!
a(t)!

Solution 3.31. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and
the integrands are simplified, so (3.35) simplifies to:
d
∫ dV = + ∫ A*(t ) b ⋅ n dA .
dt V*(t )
(Here the volume integration is two dimensional, and produces the ellipse's area, πab, which can
be time differentiated, (d/dt)( πab) to reach the desired result. However, this is not the intended
solution path for this problem.)
When a is time-dependent, the contour that defines the ellipse, (y/b)2 + (x/a)2 = 1, is also
time dependent. However, the symmetry of the ellipse allows the analysis to completed in the
first quadrant alone and then multiplied by 4. Thus, the simplified version of (3.35) reduces to:
d
∫ dV = 4 ∫ b ⋅ ndA .
dt V*(t ) first qua drant

The equation for the ellipse in the first quadrant is:

2 2
x = +a 1− ( y b) or y = +b 1− ( x a ) .
The motion of points on this curve can be described by a purely horizontal velocity when a
varies but b is constant. So, for any constant y-location, differentiate the first equation to find b:
! dx $ ! da 2 $ ! x da $
b = # , 0& = # 1− ( y b) , 0 & = # , 0& ,
" dt % " dt % " a dt %
where the final equality comes from changing the independent variable from y to x.
Tangent and normal vectors to an x-y curve are (1, dy/dx) and (–dy/dx, 1), respectively.
Thus, the outward unit normal on the ellipse in the first quadrant is:

n=
(−dy / dx,1) , so b ⋅ n = x(−dy / dx) da .
(dy / dx)2 +1 a (dy / dx)2 +1 dt
Thus, the simplified version of (3.35) can be written:
a
d x(−dy / dx) da $
∫ dV = 4 ∫ b ⋅ ndA = 4 ∫ % 1+ (dy / dx)2 dx &' ,
dt V*(t )
first qua drant 0 a (dy / dx) +1 dt
2

where, in two dimensions, dA is a path length element, ds = [1 + (dy/dx)2]1/2dx, along the first-
quadrant portion of the ellipse. This path length element is the factor in [,]-brackets above.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Simplify the integrand, insert –dy/dx = (bx/a2)/[1 – (x/a)2]1/2, and perform the integration using
the substitution x = asinθ to find:
a
d x(−dy / dx) da 4 da a xb(x / a 2 )
∫
dt V*(t )
dV = 4 ∫
a dt
dx = ∫
a dt 0 1− (x / a)2
dx
0
π 2 π 2
da sin 2 θ da 2 da π da
= 4b
dt
∫ 2
cosθ dθ = 4b
dt
∫ sin θ dθ = 4b
dt 4
= πb ,
dt
0 1− sin θ 0
and this is the desired result.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.32. For the following time-dependent volumes V*(t) and smooth single-valued
integrand functions F, choose an appropriate coordinate system and show that ( d dt ) ∫ V *(t ) FdV
obtained from (3.30) is equal to that obtained from (3.35).
a) V*(t) = L1(t)L2L3 is a rectangular solid defined by 0 ≤ xi ≤ Li, where L1 depends on time while
L2 and L3 are constants, and the integrand function F(x1,t) depends only on the first coordinate
and time. €
2
b) V*(t) = (π/4)d (t)L is a cylinder defined by 0 ≤ R ≤ d(t)/2 and 0 ≤ z ≤ L, where the cylinder’s
diameter d depends on time while its length L is constant, and the integrand function F(R,t)
depends only on the distance from the cylinder’s axis and time.
c) V*(t) = (π/6)D3(t) is a sphere defined by 0 ≤ r ≤ D(t)/2 where the sphere’s diameter D depends
on time, and the integrand function F(r,t) depends only on the radial distance from the center of
the sphere and time.

d x =b(t ) b
∂F db da
Solution 3.32. The two equations are (3.30) ∫ F(x,t)dx =
dt x =a(t )
∫ ∂t
dx +
dt
F (b,t ) −
dt
F ( a,t ) ,
a
d ∂F(x,t)
and (3.35) ∫
dt V *(t )
F(x,t)dV = ∫
∂t
dV + ∫ F(x,t)b ⋅ ndA .
V *(t ) A *(t )
a) Use Cartesian coordinates with the€origin at xi = 0. The cross sectional area of the rectangular
solid, L2L3, is constant, so dV = L2L3dx1. The volume integral proceeds from x1 = 0 (= a) to x1 =
L1(t) (= b) so (3.30) implies:
€ d d x =L1 (t ) L1
∂F dL
∫ F (x,t)dV = ∫ F(x 1,t)L L
2 3 dx 1 = ∫ L2 L3 dx1 + 1 F ( L1,t ) L2 L3 . (a1)
dt V *(t ) dt x1 =0 0 ∂t dt
Now start from (3.35) using a control volume that encloses the rectangular solid. In this case the
only control surface that moves lies at x1 = L1, has outward normal n = e1, and moves with
velocity b = (dL1/dt)ex. First evaluate the left side term from (3.35).
€
d d L1 (t ) L2 L3 d L1 (t )
∫ F(x,t)dV
dt V *(t )
= ∫ ∫ ∫ 1 1 2 3 2 3 dt ∫ F( x1,t)dx1 .
F(
dt x1 = 0 x 2 = 0 x 3 = 0
x ,t) dx dx dx = L L
x1 = 0

Now evaluate the right side terms from (3.35).

L1 (t ) L 2 L 3 L2 L3
d ∂F( x1,t) $ dL '
∫ F(x,t)dV = ∫ ∫ ∫ dx 1 dx 2 dx 3 + ∫ ∫ F(L1,t)& 1 e1 ) ⋅ e1dx 2 dx 3
€ dt V *(t ) x1 = 0 x 2 = 0 x 3 = 0 ∂t x2 = 0 x3 = 0 % dt (
L1 (t )
∂F( x1,t) dL
= L2 L3 ∫ dx1 + L2 L3 1 F(L1,t)
x1 = 0 ∂t dt
Setting the left and right side terms equal produces
d L1 (t ) L1 (t )
∂F( x1,t) dL
L2 L3 ∫ F( x1 ,t)dx 1 = L 2 3 ∫
L dx1 + L2 L3 1 F(L1,t) , (a2)
€ dt x1 = 0 x1 = 0 ∂ t dt
which is identical to (a1).
b) Use cylindrical coordinates with the origin at R = z = 0. The length area of the cylinder, L, is
constant, so dV = L(2πRdR). The volume integral proceeds from R = 0 (= a) to R = d(t)/2 (= b) so
€
(3.30) implies:
d d d (t )/ 2 d /2
∂F d ( d 2) d
∫ F (x,t)dV = 2πL
dt V *(t )
∫
dt R =0
F(R,t)RdR = 2πL ∫ RdR + 2πL
dt
F ( d /2,t ) .
2
(b1)
0 ∂t

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Now start from (3.35) using a control volume that encloses the cylinder. In this case the only
control surface that moves lies at R = d/2, has outward normal n = eR, and moves with velocity b
= (d(d/2)/dt)eR. First evaluate the left side term from (3.35).
d d L d / 2 2π d d /2
∫
dt V *(t )
F(x,t)dV = ∫ ∫ ∫
dt z= 0R = 0 ϕ = 0
F(R,t)RdϕdRdz = 2πL
dt 0
∫ F(R,t)RdR .
Now evaluate the right side terms from (3.35).
L d / 2 2π L 2π
d ∂F(R,t) & d(d /2) ) d
∫ F(x,t)dV
dt V *(t )
= ∫ ∫ ∫ ∂t
Rd ϕ dRdz + ∫ ∫ F(d/2,t)(
' dt
e R + ⋅ e R dϕdz
* 2
z= 0R = 0 ϕ = 0 z= 0ϕ = 0
€
d /2
∂F(R,t) d(d /2) d
= 2πL ∫ RdR + 2πL F(d/2,t)
R= 0 ∂t dt 2
Setting the left and right side terms equal produces
d d /2 d /2
∂F(R,t) d(d /2) d
2πL
dt 0
∫ F(R,t)RdR = 2πL ∫
∂t
RdR + 2πL
dt
F(d/2,t) ,
2
(b2)
R= 0
€
which is identical to (b1).
c) Use spherical coordinates with the origin at r = 0. The sphere expands symmetrically so the
volume element is dV = 4πr2dr. The volume integral proceeds from r = 0 (= a) to r = D(t)/2 (= b)
€ implies:
so (3.30)
d d D(t )/ 2 2
D /2
∂F 2 d ( D 2) % D (2
∫ F (x,t)dV = 4π dt ∫ F (r,t)r dr = 4π ∫ ∂t r dr + 4 π dt F (D /2,t )'& 2 *) . (c1)
dt V *(t ) r =0 0
Now start from (3.35) using a control volume that encloses the sphere. In this case the moving
control surface lies at r = D/2, has outward normal n = er, and moves with velocity b =
(d(D/2)/dt)er. First evaluate the left side term from (3.35).
€
d d D / 2 π 2π d D /2
∫ F(x,t)dV = ∫ ∫ ∫ F(R,t)r 2 dr sin θdθdϕ = 4 π ∫ F(r,t)r 2 dr .
dt V *(t ) dt r= 0 θ = 0 ϕ = 0 dt 0
Now evaluate the right side terms from (3.35).
d D / 2 π 2π
∂F(r,t) 2 π 2π
' d(D /2) * ' D * 2
∫ F(x,t)dV
dt€V *(t )
= ∫ ∫ ∫ ∂t r dr sin θ d θ d ϕ + ∫ ∫ F(d/2,t) )
( dt
e R , ⋅ e R ) , sin θdθdϕ
+ (2+
r= 0 θ = 0 ϕ = 0 θ = 0ϕ = 0

D /2
∂F(r,t) 2 d(D /2) ' D *2
= 4π ∫ r dr + 4 π F(D/2,t)) ,
r= 0 ∂t dt (2+
Setting the left and right side terms equal produces
d D /2 2
D /2
∂F(r,t) 2 d(D /2) % D (2
4π ∫ F(r,t)r dr = 4π ∫ ∂t r dr + 4π dt F(D/2,t)'& 2 *) ,
dt 0
(c2)
€ r= 0
which is identical to (c1).

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.33. Starting from (3.35), set F = 1 and derive (3.14) when b = u and V*(t) = δV → 0 .

Solution 3.33. With F = 1, b = u, and V*(t) = δV with surface δA, (3.35) becomes:
d €
dt δV
∫ dV = 0 + ∫ u ⋅ ndA .
δA
The first integral is merely δV. Use Gauss' divergence theorem on the second term to convert it
to volume integral.
d
€ (δV ) = ∫ ∇ ⋅ udV .
dt δV
As δV → 0 the integral reduces to a product of δV and the integrand evaluated at the center point
of δV. Divide both sides of the last equation by δV and take the limit as δV → 0 :
1 d 1 1
€
lim
δV →0 δV dt
(δV ) =€ lim
δV →0 δV
∫ ∇ ⋅ udV = lim
δV →0 δV
[(∇ ⋅ u)δV + ...] = ∇ ⋅ u = Sii ,
δV
and this is (3.14). €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 3.34. For a smooth single valued function F(x) that only depends on space and an
arbitrarily-shaped control volume that moves with velocity b(t) that only depends on time, show
( )
that ( d dt ) ∫ V *(t ) F(x)dV = b ⋅ ∫ V *(t ) ∇F(x)dV .

Solution 3.34. Start from Reynolds transport theorem (3.35):

d ∂F(x,t)
€ ∫ F(x,t)dV = ∫
dt V *(t ) ∂t
dV + ∫ F(x,t)b ⋅ ndA .
V *(t ) A *(t )
When F does not depend on time, the first term on the right drops out.
d
∫ F(x,t)dV = ∫ F(x,t)b ⋅ ndA .
dt V *(t ) A *(t )
€
When b does not depend on location, it can be taken outside the surface integral.
d $ '
∫ F(x,t)dV = b ⋅ & ∫ F(x,t)ndA) .
dt V *(t )
€ % A *(t ) (
Apply Gauss' theorem to the integral in large parentheses, to reach the desired form:
d % (
dt
∫ F(x,t)dV = b ⋅ ' ∫ ∇F(x,t)dA *.
€ V *(t ) & V *(t ) )

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

3.35. Show that (3.35) reduces to (3.5) when V*(t) = δV → 0 and the control surface velocity b is
equal to the fluid velocity u(x,t).

Solution 3.35. When V*(t) = δV with surface€δA, δV is small, and b = u, δV represents a fluid
particle. Under these conditions (3.35) becomes:
d ∂F(x,t)
∫
dt δV
F(x,t)dV = ∫
∂t
dV + ∫ F(x,t)u ⋅ ndA ,
δV δA
and the time derivative is evaluated following δV. Use Gauss' divergence theorem on the final
term to convert it to a volume integral,
€ ∫ F(x,t)u ⋅ ndA = ∫ ∇ ⋅ ( F(x,t)u)dV ,
δA δV
so that (3.35) becomes:
d '∂F(x,t) * '∂F(x,t) *
∫
dt δV
F(x,t)dV = ∫ ) + ∇ ⋅ ( F(x,t)u),dV = ∫ ) + F(x,t)∇ ⋅ u + (u ⋅ ∇ ) F(x,t),dV ,
€ ∂t
δV ( ∂t
+ δV ( +
where the second equality follows from expanding the divergence of the product Fu.
As δV → 0 the various integrals reduce to a product of δV and the integrand evaluated at
the center point of δV. Divide both sides of the prior equation by δV and take the limit as δV → 0
€
to find:
€ 1 d 1 (∂F(x,t) +
lim
δV →0 δV dt
∫ F(x,t)dV = lim
δV →0 δV
∫ * ∂t
+ F(x,t)∇ ⋅ u + (u ⋅ ∇ ) F(x,t)-dV ,
δV δV ) ,€
1 d 1 -' ∂F(x,t) * 0
lim
δV →0 δV dt
[ F(x,t)δV + ...] = lim /
δV →0 δV .(
) + F(x,t)∇ ⋅ u + (u ⋅ ∇ ) F(x,t),δV + ...2, or
∂t + 1
€ d 1 d ∂F(x,t)
F(x,t) + F(x,t) lim (δV ) = + F(x,t)∇ ⋅ u + (u ⋅ ∇) F(x,t) ,
dt δV →0 δV dt ∂t
where the product rule for derivative has been used on product FδV in [,]-braces on the left.
€
1 d
From (3.14) or Exercise 3.33: lim
δV →0 δV dt
(δV ) = ∇ ⋅ u , so the second terms on both sides of the
€
last equation are equal and may be subtracted out leaving:
d ∂F(x,t)
F(x,t) = + (u ⋅ ∇ ) F(x,t) ,
dt ∂t
and this is (3.5) when€the identification D Dt ≡ d dt is made.

€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.1. Let a one-dimensional velocity field be u = u(x, t), with v = 0 and w = 0. The
density varies as ρ = ρ0(2 − cos ωt). Find an expression for u(x, t) if u(0, t) = U.

Solution 4.1. Here u = u(x,t)ex, and the density field is given, so a solution for u(x,t) might be
found from the continuity equation:
∂ρ ∂ρ ∂ρ ∂u
+ ∇ ⋅ ( ρu) = 0 , or specifically for this problem: + u + ρ = 0.
∂t ∂t ∂x ∂x
The given density field only depends on time so ∂ρ/∂x = 0, and this leads to:
∂u 1 ∂ρ ρ 0 sin(ωt) ' sin(ωt) *
=− =− → u = −) , x + C(y,z,t) .
€ ∂x ρ ∂t ρ 0 (2 − cos(ωt)) € ( 2 − cos(ωt) +
where C is function of integration that does not depend on x. The initial condition requires:
u(0, t) = U = C(y,z,t),
so the final answer for u(x, t) is
€ $ sin(ωt) '
u = U −& )x .
% 2 − cos(ωt) (

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.2. Consider the one-dimensional Cartesian velocity field: u = (α x t, 0, 0 ) where α is a

constant.
a) Find a spatially uniform, time-dependent density field, ρ = ρ(t), that renders this flow field
mass conserving when ρ = ρo at t = to.
b) What are the unsteady (∂u/∂t), advective ( [u ⋅ ∇]u ), and particle (Du/Dt) accelerations in this
flow field? What does α = 1 imply?

Solution 4.2. a) Use the continuity equation and the given velocity field with ρ = ρ(t):
∂ρ dρ α
+ ∇ ⋅ ( ρu) = 0 implies + ρ = 0.
∂t dt t
Separate variables and integrate:
dρ α
∫ ρ = − ∫ t dt ––> ln ρ = −α ln t + C .
€ € −α
Exponentiate and evaluate the integration constant at t = to to find: ρ = ρ o ( t t o ) .
b) For the given flow field:
€ 2
∂u αx €⋅ ∇)u = u ∂u e = $ α x ' α e = α x e , and Du = ∂u + (u ⋅ ∇)u = α (α −1)x e .
= − 2 e x , (u x & ) x x x
∂t t ∂x % t (t t2 € Dt ∂t t2
When α = 1 the unsteady and advective acclerations are non-zero, but they are equal and
opposite so that Du/Dt is zero.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.3. Find a non-zero density field ρ(x,y,z,t) that renders the following Cartesian velocity
fields mass conserving. Comment on the physical significance and uniqueness of your solutions.
a) u = (U sin(ωt − kx),0,0) where U, ω, k are positive constants.
[Hint: exchange the independent variables x,t for a single independent variable ξ = ωt–kx]
b) u = (−Ωy,+Ωx,0) with Ω = constant. [Hint: switch to cylindrical coordinates]
c) u = ( A x,B y,C z) where A, B, C are constants.
€

€ Solution 4.3. a) u = (U sin(ωt − kx),0,0) where U, ω, k are positive constants. Use the expanded
∂ρ ∂u ∂ρ
€ form of the continuity equation for a unidirectional velocity, u = ( u,0,0) : +ρ +u = 0,
∂t ∂x ∂x
and plug in the given velocity field to find:
€ ∂ρ ∂ρ
− ρkU cos(ωt − kx) + U sin(ωt − kx) = 0 .
∂t € ∂x
€
Now follow the hint and change from independent variables (x, t) to ξ, where ξ = ωt–kx:
∂ ∂ξ d d ∂ ∂ξ d d
= =ω , and = = −k ,
€ ∂t ∂t dξ dξ ∂x ∂x dξ dξ
dρ dρ
to find a first-order differential equation: ω − kU sin ξ = ρkU cosξ that can be
dξ dξ
dρ kU cosξ
€
separated and integrated: ∫ ρ = ∫ ω − kU €
sin ξ
dξ –> ln ρ = −ln(ω − kU sin ξ ) + C(y,z) .

Here, the constant C must be used to € make the solution dimensionally sound. Noting that ω/k
has units of velocity, define M = kU/ω. Since the original equation did not contain any y or z
dependence the constant of integration might depend € on these variables. So, the final solution
€
can be obtained by exponentiating the last equation:
ρ (y,z) ρ o (y,z)
ρ(x, y,z,t) = o = ,
1− M sin ξ 1− ( kU ω ) sin(ωt − kx)
where ρo(y,z) is an undetermined function. Thus, this solution is not fully determined; it is not
unique. This is the density field that corresponds to a traveling-wave disturbance in a stationary
medium. In the limit as M → 0 with ρo = constant, this wave becomes an acoustic plane wave
€
with ω/k = the speed of sound and M = the Mach number of the fluid particle motions.
b) u = (−Ωy,+Ωx,0) with Ω = constant. The change to cylindrical coordinates is straight forward
using: x = Rcosϕ & y = Rsin ϕ , and uR = ucosϕ + v sin ϕ & uϕ = −usin ϕ + v cosϕ .
uR = −Ωy cosϕ + Ωx sin ϕ = −ΩRsin ϕ cos ϕ + ΩRcos ϕ sin ϕ = 0 , and
€ uϕ = +Ωy sinϕ + Ωx cosϕ = ΩRsin 2 ϕ + ΩRcos 2 ϕ = ΩR .
€ Thus, u = ΩRe€ € ∂ρ €1 ∂ ∂ρ ∂ρ
ϕ , so the continuity equation is: + ( ρΩR) = 0 or +Ω = 0.
€ ∂t R ∂ϕ ∂t ∂ϕ
The solution of this equation can be obtained from the method of characteristics. The idea is to
€
determine special or characteristic directions or paths in the r-θ-z-t space along which the
€ solution for ρ is easy to find. Start by postulating the existence of such a path
€ €
ρ = ρ( R(t),ϕ (t),z(t),t ) so that the total derivative of ρ with respect to time is:
dρ ∂ρ dR ∂ρ dϕ ∂ρ dz ∂ρ
= + + + .
dt ∂R dt ∂ϕ dt ∂z dt ∂t
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Comparing this equation with the one above shows that the characteristic paths are defined by
dR dt = 0 , dϕ dt = Ω , and dz dt = 0 . Thus, after a single time integration of each equation, the
characteristic paths are determined to be:
R = Ro , ϕ = Ωt + ϕ o , and z = zo .
where the quantities with subscript zero are constants. Along these paths the original equation
€ € €
for ρ becomes dρ/dt = 0, which implies the density is constant along these paths. If the density
at Ro, ϕo, zo and t = 0 is ρ o (Ro ,ϕ o ,zo ) , then the density at all later times can be obtained by
€ € €
substituting for ro, θo, zo from the three equations that specify the characteristic path. Hence, the
solution for the density is: ρ = ρ o (R,ϕ − Ωt,z) .
Here again
€
ρo is an undetermined function so this solution is not fully determined; it is
not unique. In this case, the velocity field corresponds to solid-body rotation about the z-axis at
angular rate Ω, so conservation of mass implies that the density variations must revolve around
€
the z-axis at the angular rate Ω as well.
c) u = ( A x,B y,C z) where A, B, C are constants.
∂ρ
Use the expanded form of the continuity equation: + ρ∇ ⋅ u + u ⋅ ∇ρ = 0 , and plug in the given
∂t
∂ρ " A B C % A ∂ρ B ∂ρ C ∂ρ
€ velocity field to find: − ρ$ 2 + 2 + 2 '+ + + = 0 . This equation is linear in
∂t #x y z & x ∂ x y ∂ y z ∂z
ρ, so try a separation of variables solution: € i.e. ρ = X(x)Y (y)Z(z)T(t) . Putting in the trial
solution, dividing the whole equation by the trial solution, and grouping terms yields:
T " $ A X " A ' $ B Y " B ' $C Z " C '
+ − +& − )+ − =0
T &% x X €x 2 )( % y Y y 2 ( &% z Z z 2 )(
where ( )´ denotes derivative of ( ) with respect to is argument. Because each group of terms
depends on only one of the independent coordinates, this equation can only be satisfied if each
term is equal to a constant, and the 4 constants sum to zero. This means setting:
T" € A X" A B Y" B C Z" C
= d, − 2 = a, − 2 = b , and − = c where a + b + c + d = 0
T x X x y Y y z Z z2
The solution of the first equation is: T(t) = Toe dt where To is a constant, while that of the second
can be found from:
€ € X " ax € 1 ax 2 € " ax 2 %
= + → ln X = + ln x + const., or X(x) = X o x exp# &,
X A x€ 2A $ 2A '
where Xo is a constant. The solutions of the third and fourth equations are similar to that of the
second. Combining the solutions of these equations and condensing the leading product of
constants
€ to C1 = To X oYo Z o produces:
$ ax 2 €by 2 cz 2 '
ρ(x, y,z,t) = C1 xyz exp% + + − (a + b + c)t (
& 2A 2B 2C )
where C1, a, b, c are undetermined constants, and the above restriction on a, b, c, and d has been
€
used to eliminate d. Any particular version of this solution is acceptable as long as C1 ≠ 0. Here,
the density field is zero everywhere that ∇ ⋅ u → ∞ . The constants C1, a, b, and c are not
determined € so again this solution is not fully determined; it is not unique.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Overall, conservation of mass is typically inadequate to fully specify a fluid density or

velocity field; conservation of momentum and initial & boundary conditions are needed for
complete flow field solutions.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.4. A proposed conservation law for ξ, a new fluid property, takes the following form:
d
∫ ρξ dV + ∫ Q ⋅ n dS = 0 , where V(t) is a material volume that moves with the fluid velocity
dt V (t ) A(t )

u, A(t) is the surface of V(t), ρ is the fluid density, and Q = −ργ∇ξ .

a) What partial differential equation is implied by the above conservation statement?
∂ξ 1
b) Use the part-a) result and the continuity equation to show: + u ⋅ ∇ξ = ∇ ⋅ ( ργ∇ξ ) .
∂t ρ

Solution 4.4. a) Start with the given material CV equation, and apply Reynolds Transport Thm.
to the first term on the left side and Gauss’s Divergence Thm. to the other term:
d ∂
∫
dt V (t )
ρξ dV + ∫ Q ⋅ n dS = 0 → ∫ ( ρξ ) dV + ∫ ρξ ( u ⋅ n) dV + ∫ ∇ ⋅ Q dV = 0 .
S(t ) V (t ) ∂ t A(t ) V (t )

Now apply Gauss’s divergence theorem to the remaining surface integral, subsitute in the
specified relationship Q = −ργ∇ξ , and combine all the terms into one volume integral to find:
$∂ '
∫ %&∂ t ( ρξ ) + ∇ ⋅ ( ρξ u) − ∇ ⋅ ( ργ∇ξ )() dV = 0 .
V (t )

Here V(t) is an arbitrary material volume, so the integrand must be zero. Thus, the partial
differential equation implied by the given CV equation is:
∂
( ρξ ) + ∇ ⋅ ( ρξ u) = ∇ ⋅ ( ργ∇ξ ) .
∂t
b) Expand the left side of the part a) result and group the terms that have ξ as a coefficient:
∂ξ ∂ρ ∂ξ # ∂ρ &
ρ + ξ + ξ∇ ⋅ ( ρ u) + ρ u ⋅ ∇ξ = ρ + ρ u ⋅ ∇ξ + ξ % + ∇ ⋅ ( ρ u) ( .
∂t ∂t ∂t $ ∂t '
The contents of the parentheses is zero because of the continuity equation, so dividing by ρ
yields:
∂ξ 1
+ u ⋅ ∇ξ = ∇ ⋅ ( ργ∇ξ ) ,
∂t ρ
which can be recognized as the advection-diffusion equation for a conserved passive scalar.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.5. The components of a mass flow vector ρu are ρu = 4x2y, ρv = xyz, ρw = yz2.
a) Compute the net mass outflow through the closed surface formed by the planes x = 0, x = 1, y
= 0, y = 1, z = 0, z = 1.
b) Compute ∇ ⋅ ( ρu) and integrate over the volume bounded by the surface defined in part a)
c) Explain why the results for parts a) and b) should be equal or unequal.

Solution 4.5. a) The specified volume is a cube, and the mass outflow will have six contributions
€ for each side):
(one
1 1 1 1
∫∫ ρu ⋅ ndA = ∫ ∫ [4 x 2 y ] x=1 dydz − ∫ ∫ [4 x 2 y ] x= 0 dydz
cube surface y= 0z= 0 y= 0z= 0
1 1 1 1
+ ∫ ∫ [ xyz] y=1 dxdz − ∫ ∫ [ xyz] y= 0 dxdz
x= 0z= 0 x= 0z= 0
1 1 1 1
+ ∫ ∫ [ yz 2 ] dxdy −
z=1
∫ ∫ [ yz 2 ] z= 0 dxdy
x= 0y= 0 x= 0y= 0
The integral evaluations are straightforward and unremarkable, and lead to:
1 1 11
∫∫ ρu ⋅ ndA = 2 + 0 + 4 + 0 + 2 + 0 = 4 .
cube surface
€
b) First compute the divergence of the mass flow:
∂ ∂ ∂
∇ ⋅ ( ρu) = ( ρu) + ( ρv ) + ( ρw ) = 8xy + xz + 2yz ,
∂x ∂y ∂z
€
then integrate it in the cubical volume:
1 1 1
& 1 )& 1 ) & 1 )& 1 ) & 1 )& 1 ) 11
∫∫∫ ∇ ⋅ (ρu)dV = ∫ ∫ ∫ (8xy + xz + 2yz)dxdydz = 8(' 2 +*(' 2 +* + (' 2 +*(' 2 +* + 2(' 2 +*(' 2 +* = 4 .
cube € x= 0y= 0z= 0
c) The two answers should be the same because of Gauss' Divergence Theorem.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.6. Consider a simple fluid mechanical model for the atmosphere of an ideal spherical
star that has a surface gas density of ρo and a radius ro. The escape velocity from the surface of
the star is ve. Assume that a tenuous gas leaves the star’s surface radially at speed vo uniformly
over the star’s surface. Use the steady continuity equation for the gas density ρ and fluid velocity
u = (ur , uθ , uϕ ) in spherical coordinates
1 ∂ 2 1 ∂ 1 ∂
2
r ∂r
( r ρur ) +
r sin θ ∂θ
( ρuθ sinθ ) +
r sin θ ∂ϕ
( ρuϕ ) = 0
for the following items

( )
a) Determine ρ when vo ≥ ve so that u = (ur , uθ , uϕ ) = vo 1− ( ve2 vo2 ) (1− ( ro r )), 0, 0 .
b) Simplify the result from part a) when vo >> ve so that: u = (ur , uθ , uϕ ) = (vo , 0, 0) .
c) Simplify the result from part a) when vo = ve .
d) Use words, sketches, or equations to describe what happens when vo < ve . State any
assumptions that you make.

Solution 4.6. a) ur is the only non-zero velocity component so only the continuity equation
simplifies to
1 ∂ 2
r 2 ∂r
(r ρur ) = 0 ,
multiply by r2 and integrate to find: r 2 ρur = C(θ,ϕ ) , where C is the function of integration that
cannot depend on r. Thus,
C(θ,ϕ ) ρ o ro2 −1 2
ρ =€ 2
r ur r
[
= 2 1− (v e2 v o2 )(1− ( ro r)) ] ,
€
where C(θ,ϕ ) = ρ o ro2v o = const. is determined from the boundary conditions: ρ = ρo & v = vo at r
= ro .
2 2
b) When v o >> v€ e , the square root factor in the result of part a) becomes unity, so ρ = ρ o ro r .
−1 2 32
€ c) When v o = v e , the square root factor simplifies to ( ro r) , so ρ = ρ o [ ro r]
d) Here, the gas leaving the star’s surface does not escape the star’s gravitational field, so it must
€fall back to star’s surface. For a tenuous gas, it might be acceptable € to ignore molecular
€collisions so that the velocity field of the outward
€ moving
+
( 2
)
€ gas, ur = v o 1− v e v o (1− ( ro r)) for
2

r < rmax = ro [1− v o2 v e2 ] , is equal and opposite to the velocity field of the inward falling gas so
that u−r = −ur+ . Both outward and inward mass fluxes must balance because the net mass flux
must be zero. Thus, the resulting density field will have € equal parts contributed by inward and
outward flowing stellar gas. Under these conditions, the density field will be double that from
€
part a) with a radial limitation:
€ % ρ o ro2 −1 2 )
ρ=& r
'2
2 [
1− (v 2
e v 2
o )(1− ( ro r]) ) for r < rmax
'
* where rmax = ro [1− v o2 v e2 ] .
'(0 for r ≥ rmax '+

€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.7. Consider the three-dimensional flow field ui = β xi or equivalently u = βrer, where
β is a constant with units of inverse time, xi is the position vector from the origin, r is the
distance from the origin, and êr is the radial unit vector. Find a density field ρ that conserves
mass when:
a) ρ(t) depends only on time t and ρ = ρo at t = 0, and
b) ρ(r) depends only on the distance r and ρ = ρ1 at r = 1 m.
c) Does the sum ρ(t) + ρ(r) also conserve mass in this flow field? Explain your answer.

Solution 4.7. The full continuity equation is: ∂ρ ∂t + (u ⋅ ∇)ρ + ρ∇ ⋅ u = 0 .

a) When ρ depends only on t, (u ⋅ ∇)ρ = 0 so the cont. equation reduces to: d ρ dt + ρ∇ ⋅ u = 0 .
For the given velocity field, ∇ ⋅ u = ∂ui ∂xi = β ∂xi ∂xi = β (∂x1 ∂x1 + ∂x2 ∂x2 + ∂x3 ∂x3 ) = 3β , so
d ρ dt = −3βρ . This equation has the exponential solution: ρ (t) = ρo e−3βt when ρ = ρo at t = 0.
b) When ρ only depends on distance r, ∂ρ/∂t = 0 so the continuity equation reduces to:
(β rêr )⋅ ∇ρ + 3βρ = 0 , or r ( d ρ dr ) = −3ρ . This equation has the power law solution
3
ρ (r) = ρ1 ⋅ (1m r ) when ρ = ρ1 at r = 1 m.
c) The proposed sum solution, ρ(t) + ρ(r), does conserve mass, and this is readily shown by
substitution of ρ(t) + ρ(r) into the full continuity equation. This sum solution conserves mass
because the continuity equation is linear in ρ when u is specified independently of ρ. When ρ
when u are both dependent field variables, the continuity equation is non-linear.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.8. The definition of the stream function for two-dimensional constant-density flow in
the x-y plane is: u = −e z × ∇ψ , where ez is the unit vector perpendicular to the x-y plane that
determines a right-handed coordinate system.
a) Verify that this vector definition is equivalent to u = ∂ψ ∂y , and v = −∂ψ ∂x in Cartesian
coordinates.
€
b) Determine the velocity components in r-θ polar coordinates in terms of r-θ derivatives of ψ.
c) Determine an equation for the z-component of the vorticity in terms of ψ.
€ €
Solution 4.8. a) Start from the definition of the cross product:
ex ey ez
' ∂ψ * ' ∂ψ * ' ∂ψ * ' ∂ψ *
u = −e z × ∇ψ = − 0 0 1 = −e x )− , − e y ) , = e x ) , + e y ) − , .
( ∂y + ( ∂x + ( ∂y + ( ∂x +
∂ψ ∂x ∂ψ ∂y 0
Setting components equal from the extreme ends of this extended equality produces:
u = ∂ψ ∂y , and v = −∂ψ ∂x .
b) In r-θ polar coordinates u = urer + u e , e z × e r = eθ , and e z × eθ = −e r . Plus,
θ θ
€
∂ψ 1 ∂ψ
∇ψ = e r + eθ . Therefore, using the same definition in r-θ polar coordinates produces:
∂r r ∂θ € €
( ∂ψ 1 ∂ψ + ∂ψ 1 ∂ψ
u = ure r + uθ eθ = −e z × ∇ψ = −e z × *€e r + eθ - = −e
€θ + er .
) ∂r r ∂θ , ∂r r ∂θ
€ Matching components implies:
1 ∂ψ ∂ψ
ur = and uθ = − .
r ∂θ ∂r
c) €
The z-component of the vorticity is:
∂v ∂u ∂ & ∂ψ ) ∂ & ∂ψ ) & ∂ 2ψ ∂ 2ψ )
ω z = − = ( − + − ( + = −( 2 + 2 + = −∇ 2ψ .
∂x€ ∂y ∂x ' ∂x€* ∂y ' ∂y * ' ∂x ∂y *

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.9. A curve of ψ (x, y) = C1 (= a constant) specifies a streamline in steady two-

dimensional constant-density flow. If a neighboring streamline is specified by ψ (x, y) = C2 ,
show that the volume flux per unit depth into the page between the streamlines equals C2 – C1
when C2 > C1.
€
Solution 4.9. Start with a picture of the two streamlines and let Q = volume€flow rate (per unit
depth) between them.

∇ψ ψ=C 2

d
ψ=C 1

€
€

2
By definition, Q = ∫ | u | d where d is an increment of length that lies perpendicular to the flow
1
in the stream tube. Again by definition, ∇ψ lies in the direction perpendicular to the lines of
constant ψ. Thus:
€ ∂ψ ∂x dx + ∂ψ ∂y dy
€ d = ∇ψ ⋅ dx = ( ) ( ) = −vdx + udy = dψ , so | u | d = dψ .
| ∇ψ | (∂ψ
2
€ ∂x ) + (∂ψ ∂y )
2
v 2 + u2 |u |
2 2
This means that: Q = ∫ | u | d = ∫ dψ = C2 − C1 , which completes the proof.
1 1 €
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.10. Consider steady two-dimensional incompressible flow in r-θ polar coordinates
where u = (ur , uθ ) , ur = + ( Λ r 2 ) cosθ , and Λ is positive constant. Ignore gravity.
a) Determine the simplest possible uθ .
b) Show that the simplest stream function for this flow is ψ = ( Λ r ) sin θ .
c) Sketch the streamline pattern. Include arrowheads to show stream direction(s).
d) If the flow is frictionless and the pressure far from the origin is p∞, evaluate the pressure p(r,
θ) on θ = 0 for r > 0 when the fluid density is ρ. Does the pressure increase or decrease as r
increases?

1 ∂ 1 ∂u
Solution 4.10. a) Place the given ur = + ( Λ r 2 ) cosθ into ∇ ⋅ u = (rur ) + θ = 0 , to find:
r ∂r r ∂θ
1 ∂ 1 ∂ Λ# & Λ 1 ∂uθ ∂uθ Λ Λ
(rur ) = % + cosθ ( = − 3 cosθ = − , or = + 2 cosθ → uθ = + 2 sin θ ,
r ∂r r ∂r $ r ' r r ∂θ ∂θ r r
where the final constant of integration has been dropped to produce the simplest possible uθ .
b) By definition: ur = (1 r ) (∂ψ ∂θ ) , and uθ = −∂ψ ∂r . Thus,
1 ∂ψ Λ ∂ψ Λ Λ
= + 2 cosθ so = + cosθ → ψ = sin θ + f (r) , and
r ∂θ r ∂θ r r
∂ψ Λ Λ
− = + 2 sin θ so ψ = sin θ + g(θ ) .
∂r r r y!
The simplest stream function is recovered when f = g = 0:
ψ = ( Λ r ) sin θ .
c) Start from the result of b) and switch to Cartesian coordinates:
Λ y
ψ= , which implies x 2 + y 2 = ( Λ ψ ) y , or
2 2 2 2
x +y x +y
2 2
x!
x 2 + ( y − Λ 2ψ ) = ( Λ 2ψ ) .
Thus, the streamlines are circles centered on the y-axis that are
tangent to the x-axis at the origin of coordinates. The stream
direction is determined from the radial velocity ur, which must be
positive when θ is near zero.
d) Use the ordinary Bernoulli-equation to find:
1 1 # Λ2 Λ2 &
p∞ = p(r, θ )θ =0 + ρ ( vr2 + vθ2 ) = p(r, 0) + ρ % 4 cos2 (0) + 4 sin 2 (0)( , or
2 θ =0 2 $r r '
1 Λ2
p(r, 0) = p∞ − ρ 4 .
2 r
The pressure increases as r increases.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.11. The well-known undergraduate fluid mechanics textbook by Fox et al. (2009)
provides the following statement of conservation of momentum for a constant-shape (non-
rotating) control volume moving at a non-constant velocity U = U(t).
d dU
∫ ρurel dV + ∫ ρurel (urel ⋅ n)dA = ∫ ρgdV + ∫ fdA − ∫ ρ
dt V *(t ) dt
dV .
A *(t ) V *(t ) A *(t ) V *(t )
Here urel = u − U(t) is the fluid velocity observed in a frame of reference moving with the
control volume while u and U are observed in a non-moving frame. Meanwhile, (4.17) states
this law as
€ d
€ ∫ ρudV + ∫ ρu(u − U) ⋅ ndA = ∫ ρgdV + ∫ fdA
dt V *(t ) A *(t ) V *(t ) A *(t )
where the replacement b = U has been made for the velocity of the accelerating control surface
A*(t). Given that the two equations above are not identical, determine if these two statements of
conservation of fluid momentum are contradictory or consistent.
€
Solution 4.11. The goal here is to derive one result from the other. Start with the equation from
the undergraduate textbook, substitute using urel = u − U , and put the forces on the left and move
the extra acceleration term to the other side of the equation.
dU d
∫ ρgdV + ∫ fdA = ∫ ρ dt dV + dt ∫ ρ(u − U)dV + ∫ ρ(u − U)(u − U) ⋅ ndA
V *(t ) A *(t ) V *(t ) V *(t ) A *(t )
€
Now expand the terms on the terms on the right side.
∫ ρgdV + ∫ fdA =
V *(t ) A *(t )
€ dU d dU d
∫ ρ
dt
dV + ∫ ρudV −
dt V *(t )
∫ ρdV − U
dt V *(t )
∫ ρdV + ∫ ρu(u − U) ⋅ ndA − U ∫ ρ(u − U) ⋅ ndA
dt V *(t )
V *(t ) A *(t ) A *(t )
Here, U and dU/dt depend on time only – that is, they are uniform over V*(t) and A*(t) – so they
can move inside or outside of the volume and surface integrals over V*(t) and A*(t), but not
through time differentiations. Thus, the first and third terms on the right side are equal and
€ opposite. Rearrange the other terms to find:
d &d )
∫ ρgdV + ∫ fdA = dt ∫ ρudV + ∫ ρu(u − U) ⋅ ndA − U( dt ∫ ρdV + ∫ ρ(u − U) ⋅ ndA+.
V *(t ) A *(t ) V *(t ) A *(t ) ' V *(t ) A *(t ) *
Conservation of mass requires the contents of the [,]-brackets to equal zero. After removing
these terms, the remaining simplified equation,
d
€ ∫ ρudV + ∫ ρu(u − U) ⋅ ndA = ∫ ρgdV + ∫ fdA ,
dt V *(t ) A *(t ) V *(t ) A *(t )
is the same as (4.17). The two statements of conservation of momentum are consistent. The
essential difference between the two CV formulations of conservation of momentum amounts to
a transformation between an inertial and an accelerating frame of reference when mass is also
conserved. €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.12. A jet of water with a diameter of 8 cm and a speed of 25 m/s impinges normally
on a large stationary flat plate. Find the force required to hold the plate stationary. Compare the
average pressure on the plate with the stagnation pressure if the plate is 20 times the area of the
jet.

Solution 4.12. Choose a rectangular CV with one side that covers and the flat stationary plate.

Ajet
Ujet x

Denote the plate area by A, and the jet velocity & area by Ujet and Ajet, respectively. In this
problem only the x-component of the integral momentum equation is required, and x-direction
fluid momentum only enters the CV of its left side. Assume that atmospheric pressure Po acts on
the three sides of the CV that are not in contact with the plate.
−ρU 2jet A jet + 0 = Po A − ∫ PdA = Fx , or
A
π
Fx = −ρU A jet = −(10 kg /m )(25m /s) 2
2
jet
3 3
(0.08m) 2 = −3.142kN
4
2
€ Fx ρU jet (10 kg /m 3 )(25m /s) 2
3
The average pressure is: Pave = = = = 31.25kPa .
20A jet 20 20
€ 1 (10 3 kg /m 3 )(25m /s) 2 P 1
The stagnation pressure is: Ps = ρU 2jet = = 312.5kPa , so ave =
2 2 Ps 10
Here Ps and Pave€are reported as gauge pressures.

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.13. Show that the thrust developed by a stationary rocket motor is F = ρAU2 + A(p −
patm), where patm is the atmospheric pressure, and p, ρ, A, and U are, respectively, the pressure,
density, area, and velocity of the fluid at the nozzle exit.

Solution 4.13. Use the control volume shown, where F is the force that holds the rocket motor in
place.
patm

F p
A ! U

Assuming the flow is steady and that control volume is not moving (b = 0), the horizontal
component of (4.17) becomes:
ρU 2 A = F + patm A − pA , or F = ρU 2 A + A( p − patm ) .

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.14. Consider the propeller of an airplane moving with a velocity U1. Take a reference
frame in which the air is moving and the propeller [disk] is stationary. Then the effect of the
propeller is to accelerate the fluid from the upstream value U1 to the downstream value U2 > U1.
Assuming incompressibility, show that the thrust developed by the propeller is given by
F = ρ A(U 22 −U12 ) 2, where A is the projected area of the propeller and ρ is the density (assumed
constant). Show also that the velocity of the fluid at the plane of the propeller is the average
value U = (U1 + U2)/2. [Hint: The flow can be idealized by a pressure jump of magnitude Δp =
F/A right at the location of the propeller. Also apply Bernoulli’s equation between a section far
upstream and a section immediately upstream of the propeller. Also apply the Bernoulli equation
between a section immediately downstream of the propeller and a section far downstream. This
will show that Δp = ρ (U 22 −U12 ) / 2 .]

Solution 4.14. The CV and nominal propeller-edge streamlines are:

a b
1 2

where the mass flux within the curved propeller-edge streamlines is constant and equal to
m˙ = ρU1 A1 = ρU 2 A2 . Apply the steady constant density Bernoulli equation from "1" to "a", and
from "b" to "2":
p1 + 12 ρU12 = pa + 12 ρU a2 , and pb + 12 ρU b2 = p2 + 12 ρU 22 .
€ Since p1 = p2, and Ua = Ub, these equations become:
pb − pa = + 12 ρ(U 22 − U12 ) .
Now apply€the momentum principle across € the propeller plane:
−ρU a + ρU b = 0 = pa A − pb A + F , or F = A( pb − pa ) = 12 ρA(U 22 − U12 ) .
2 2

where Ua = Ub, and F is the €force applied to the propeller (positive to the right) by the device that
holds it in place.
To find the velocity at the propeller plane, use the stationary rectangular control volume
€ €
shown above, and start with (4.5),
−ρU1 A1 + ρU 2 A2 + ρU1 (A1 − A2 ) + m˙ sides = 0 ,
where the four terms correspond to the inlet, fast flow at the outlet, ordinary speed flow at the
outlet, and stream-wise sides of the CV. Here the inlet flow is uniform at velocity U1. Using the
definition of m˙ , the conservation of mass statement becomes:
€ ρU1 (A1 − A2 ) + m˙ sides = 0 .
where A2 is the fast-flow area on the outlet side of the CV. Now evaluate the horizontal
component
€ of (4.17)
−m˙ U1 + m˙ U 2 + ρU12 (A1 − A2 ) + U1m˙ sides = ( p1 − p2 )A1 + F .
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Substitute in the conservation of mass result and note that p1 = p2 to find:

−m˙ U1 + m˙ U 2 = F , or F = ρUA(U 2 − U1 ) .
where m˙ = ρUA , while U (= Ua = Ub) and A are flow velocity and area inside the curved lines at
the propeller plane. Equate the relationship for F found from the Bernoulli equations with the
one found from conserving mass and momentum to develop an equation for U:
€ €
F = 12 ρA(U 22 − U12 ) = ρUA(U 2 − U1 ) ,
€
1
Use the second equality, cancel common factors, and solve for U to find: U = 2 (U 2 + U1 ) .

€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.15. Generalize the control volume analysis of Example 4.1 by considering the control
volume geometry shown for steady two-dimensional flow past an arbitrary body in the absence
of body forces. Show that the force the fluid exerts on the body is given by the Euler momentum
integral, F j = − ∫ ( ρui u j − τ ij )n i dA , and that 0 = ∫ ρui n i dA .
A1 A1

Solution 4.15. For steady flow through a stationary control volume (b = 0) and no body force,
(4.5) and (4.17) simplify to:
€
∫ ρ(u ⋅€n)dA = 0 → ∫ ρui n i dA = 0 , and
A *(t ) A *(t )

∫ ρu(u ⋅ n) dA = ∫ f(n,x,t)dA → ∫ ρu j ui n i dA = ∫ n iτ ij dA ,
A *(t ) A *(t ) A *(t ) A *(t )
where the second equality in each case is the same as first after conversion to index notation, and
(2.15) or (4.20b)€has been used to write
the surface force f in terms of the stress
tensor τ€ij.
Here the control surface A* is
composed of three pieces,
A* = A1 + A2 + A3,
where A1 is the outer surface, A2 conforms
to the body's surface, and A3 is the
connection between A1 and A2. Here n
(with components ni) is the outward
normal on the combined surface. Thus, n points outward on A1 (shown as n1 in the figure) and
points into the body on A2 (shown as n2 in the figure). Thus, the conservation laws require:
∫ ρui n i dA = 0 and ∫ (ρu j ui − τ ij )n i dA = 0 .
A1 +A 2 +A 3 A1 +A 2 +A 3

Now consider the contribution of each surface individually. In general, there are no
simplifications to be made on A1 and all the integrand terms must be retained. The surface A2
coincides with the surface of the solid body, thus u ⋅ n = uini = 0 on A2. This fact eliminated the
€ A in the conservation
contribution from € of mass equation, and causes the first integrand term to
2
dropout of the conservation of momentum equation. On A3, as the width of the opening goes to
zero, the magnitude of the integrands on € the upper and lower surfaces become equal but the
normal on these two surfaces is opposite; therefore, A3 provides no net contribution to either
conservation law. Thus, the integral laws become:
∫ ρui n i dA = 0 and − ∫ τ ij n i dA + ∫ (ρu j ui − τ ij )n i dA = 0 .
A1 A2 A1

The first of these is sought as part of the solution to this exercise. The second becomes the Euler
momentum integral after considering the lone integral over A2 and orientation of n on A2. If n
pointed outward from the body, this integral would represent the force, F, the fluid applies to the
€ on A , n (= n ) €
body. However 2 2 points inward so –n points outward from the body, thus
F = − ∫ τ ij n i dA .
A2

Therefore the final result (the Euler momentum integral) is: F = − ∫ ( ρu j ui − τ ij ) n i dA .

A1

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.16. The pressure rise Δp = p2 − p1 that occurs for flow through a sudden pipe-cross-
sectional-area expansion can depend on the average upstream flow speed Uave, the upstream pipe
diameter d1, the downstream pipe diameter d2, and the fluid density ρ and viscosity µ. Here p2 is
the pressure downstream of the expansion where the flow is first fully adjusted to the larger pipe
diameter. €
a) Find a dimensionless scaling law for Δp in terms of Uave, d1, d2, ρ and µ.
b) Simplify the result of part a) for high-Reynolds-number turbulent flow where µ does not
matter.
c) Use a control volume analysis to determine Δp in terms of Uave, d1, d2, and ρ for the high
Reynolds number limit. [Hints: i) a streamline drawing might help in determining or estimating
the pressure on the vertical surfaces of the area transition, and ii) assume uniform flow profiles
wherever possible.]
d) Compute the ideal flow value for Δp and compare this to the result from part d) for a diameter
ratio of d1/d2 = ½. What fraction of the maximum possible pressure rise does the sudden
expansion achieve?
d2!
d1 !

Uave! p1! p2!

Solution 4.16. a) This is a dimensional analysis task. Construct the units matrix.
Δp Uave d1 d2 ρ µ
M 1 0 0 0 1 1
L -1 1 1 1 -3 -1
T -2 -1 0 0 0 -1
The matrix has rank three so there are 6 – 3 = 3 dimensionless groups. These can be found by
inspection; thus: Δp ( ρU ave
2
) = f (d1 d2 , ρU ave d1 µ) where f is an undetermined function.
b) At high Reynolds number the viscosity will not be a parameter, so the Reynolds number can
be dropped from the part a) result: Δp ( ρU ave 2
) = f (d1 d2 ) . The remainder of this exercise
involves
€ finding f.
c) Place a cylindrical CV in the duct that
abuts the area change on the upstream
€ side and extends far enough downstream
to fully enclose the flow’s reattachment
zone. For high-Re turbulence, its OK to
Separation assume uniform inflow and outflow
Reattachment profiles. Denote the outflow velocity by
U2 .
π 2 π
Cons. of mass implies: − d1 U ave + d22U 2 = 0 ; (a)
4 4
π π π π
Cons. of horizontal momentum: − d12 ρU ave 2
+ d22 ρU 22 = p1 d22 − p2 d22 . (b)
4 4 4 4
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Here, the cross-stream momentum equation suggests that p1 acts on the flat surfaces of the area
transition because the flow separates at the area junction and is parallel to the x-direction and this
implies that ∂p ∂r = 0 . Use the above equations to determine Δp = p2 − p1. Equation (b) implies:
p2 − p1 = ρd12U ave
2
d22 − ρU 22 , and equation (a) reduces to: U 2 = ( d12 d22 )U ave . Eliminate U2 to
find:
2 2
€ p2 − p1 = ρd12U ave
2
d22 − ρ( d12 d22 ) U ave , or € 2
p2 − p1 = ρU ave (
(d12 d22 ) 1− (d12 d22 ) . )
€ d) For an ideal flow: p2 − p1 = ρU 1
2
2
ave
1 2
− ρU = ρ€U
2 2
1
2
2
ave (1− d
1
4
d 4
2 ) = (15 32)ρU 2
ave
2
ave 1
2 2
2 ( 1
2 2
2 )
From part d): p2 − p1 = ρU ( d d ) 1− ( d d ) = ( 3 16) ρU . Thus, the pressure rise in a
2
ave
€sudden expansion to four times the area with € separated flow is only 40% of that possible in an
ideal flow. €
Abrupt area changes are exceptionally common in automotive piston-engine exhaust
€
systems where they are used for low-frequency noise control. However, they lead to pumping
losses that reduce the engine's horsepower, especially under wide-open-throttle conditions,
because of the principles illustrated in this exercise.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.17. Consider how pressure gradients and skin friction develop in an empty wind
tunnel or water tunnel test section when the flow is incompressible. Here the fluid has viscosity µ
and density ρ, and flows into a horizontal cylindrical pipe of length L with radius R at a uniform
horizontal velocity Uo. The inlet of the pipe lies at x = 0. Boundary layer growth on the pipe’s
walls induces the horizontal velocity on the pipe’s centerline to be UL at x = L; however, the
pipe-wall boundary layer thickness remains much smaller than R. Here, L/R is of order 10, and
ρUoR/µ >> 1. The radial coordinate from the pipe centerline is r.
a) Determine the displacement thickness, δ L* , of the boundary layer at x = L in terms of Uo, UL ,
and R. Assume that the boundary layer displacement thickness is zero at x = 0. [The boundary
layer displacement thickness, δ*, is the thickness of the zero-flow-speed layer that displaces the
outer flow by the same amount as € the actual boundary layer. For a boundary layer velocity
profile u(y) with y = wall-normal coordinate and U = outer flow velocity, δ* is defined by:
∞
δ* = ∫ 0 (1− ( u U ))dy .]
b) Determine the pressure difference, ΔP = PL – Po, between the ends of the pipe in terms of ρ,
Uo, and UL.
c) Assume the horizontal velocity profile at the outlet of the pipe can be approximated by:
€
(
u(r, x = L) = U L 1− ( r R)
n
) and estimate average skin friction, τ w , on the inside of the pipe

between x = 0 and x = L in terms of ρ, Uo, UL, R, L, and n.

d) Calculate the skin friction coefficient, c f = τ w 12 ρU o2 , when Uo = 20.0 m/s, UL = 20.5 m/s, R =
€
€ 1.5 m, L = 12 m, n = 80, and the fluid is water, i.e. ρ = 103 kg/m3.

€ r!
U o! U L!

x = 0! x = L!
2R!

Solution 4.17. a) Apply the principle of conservation of mass. The effective flow area will be
2 2
πR 2 at the inlet (x = 0), but will be π ( R − δL* ) at the outlet (x = L): U oπR 2 = U L π ( R − δL* ) .
(
Solve for δL* = R 1− U o U L . )
b) Start from the steady Bernoulli equation (4.19) for horizontal flow (no body force):
€ 1 1 1 1 1
ρU o2 + Po = ρU L2 + PL ,€and rearrange it to find: PL − Po€= ρU o2 − ρU L2 = ρ(U o2 − U L2 ) .
2 2 2 2 2
€
c) Choose a CV that encloses all the fluid in the tube between x = 0 and x = L, and apply the CV
form of conservation of momentum (4.17). Here the flow is steady and there is no net body
force, thus:
€ R R €
n 2
( )
−2πρ ∫ U o2 rdr +2πρ ∫ U L2 1− ( r R) rdr = ( Po − PL )πR 2 − 2πRLτ w
r= 0 r= 0
Cancel common factors, evaluate the integrals, and substitute in the part b) result for the pressure
difference:
R
€
r= 0
( n
−ρU o2 R 2 + 2 ρ ∫ U L2 1− 2( r R) + ( r R)
2n
)rdr = (P − P )R
o L
2
− 2RLτ w

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

$ R 2 2 R n +2 1 R 2n +2 '
−ρU o2 R 2 + 2 ρU L2 & − n − 2n 2
) = ( Po − PL ) R − 2RLτ w
% 2 R n + 2 R 2n + 2 (
$ 4 1 ' 1
−ρU o2 R 2 + ρU L2 R 2 &1− ) = ρ(U L − U o ) R − 2RLτ w
2 2 2
+
% n + 2 n + 1( 2
2
Divide
€ by ρR and collect terms:
# 4 1 & 1 2 Lτ
−U o2 + U L2 %1− ( = (U L − U o ) − 2 w
2
+
€ $ n + 2 n + 1' 2 ρR
#1 4 1 & 2 1 2 Lτ w
% − + (U − U = −2
$ 2 n + 2 n + 1' L 2 o ρR
Solve for the average
€ shear stress:
ρR + 2 % 8 2 ( 2.
τw = -U o − '1− + *U 0
€ 4L , & n + 2 n + 1) L /
2τ R+ % 8 2 ( U L2 .
d) Use the result of part c) to find: c f = w2 = -1− '1− + * 0 . Evaluate:
ρU o 2L , & n + 2 n + 1) U o2 /
1.5m ) # 2 &# 20.5 & ,
€ 2
8
cf = +1− %1− + (% ( .= 0.00162
2(12m) * $ 82 81'$ 20.0 ' -
The numbers provided here € are approximately applicable to the William B. Morgan Large
Cavitation Channel in Memphis, Tennessee, the world's largest low-turbulence water tunnel.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.18. An acid solution with density ρ flows horizontally into a mixing chamber at
speed V1 at x = 0 where it meets a buffer solution with the same density moving at speed V2. The
inlet flow layer thicknesses are h1 and h2 as shown, the mixer chamber height is constant at h1 +
h2, and the chamber width into the page is b. Assume steady uniform flow across the two inlets
and the outlet. Ignore fluid friction on the interior surfaces of the mixing chamber for parts a) and
b).
a) By conserving mass and momentum in a suitable control volume, determine the pressure
difference, Δp = p(L) – p(0), between the outlet (x = L) and inlet (x = 0) of the mixing chamber in
terms of V1, V2, h1, h2, and ρ. Do not use the Bernoulli equation.
b) Is the pressure at the outlet higher or lower than that at the inlet when V1 ≠ V2?
c) Explain how your answer to a) would be modified by friction on the interior surfaces of the
mixing chamber.

V1! h1!
V2! h2!
x = 0! x = L!

Solution 4.18. a) Choose a stationary control volume (b = 0) that captures the fluid between x =
0 and x = L. Here the inflows and outflows are presumed steady. Thus, conservation of mass
from (4.5) reduces to:
∫ ρu ⋅ n dA = 0 .
A*(t )

At x = 0 (the inlet CV surface), n = –ex but the velocities of both streams are positive. At x = L
(the outlet CV surface), n = +ex and the velocity of the mixed stream is positive. On both
surfaces dA = bdy. Thus, the reduced form of (4.5) can be written:
∫ ρu ⋅ n dA = − ∫ ρub dy + ∫ ρub dy = −ρV1bh1 − ρV2bh2 + ρV3b(h1 + h2 ) = 0 ,
A*(t ) inlet outlet

where V3 is the outlet flow speed. Here, the density ρ and flow width b are constants, so this
result can be simplified to:
V1h1 +V2 h2 = V3 (h1 + h2 ) . (†)
The same control volume should be used to conserve of momentum. Again the flow is
presumed steady, so the horiztonal component of (4.17) reduces to:
∫ ρu(u ⋅ n)dA = ∫ fx (n, x, t)dA ,
A*(t ) A*(t )

where fx is the horizontal surface force on the control volume. The body force term is absent here
because gravity acts vertially. As for conservation of mass with this control volume, only the
inflow and outflow surfaces contribute. Evaluating the reduced form of (4.17) produces:
∫ ρu(u ⋅ n)dA = − ∫ ρu2b dy + ∫ ρu2b dy = −ρV12bh1 − ρV22bh2 + ρV32b(h1 + h2 )
A*(t ) inlet outlet

= ∫ f x (n, x, t)dA = + ∫ p(0)b dy − ∫ p(L)b dy = ( p(0) − p(L)) b(h1 + h2 ),

A*(t ) inlet outlet

where the friction terms on the upper and lower CV boundaries have been neglected. Divide out
the common factor of b, and rearrange this to isolate the pressure difference on the left:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

ρ
p(0) − p(L) = ρV32 −
h1 + h2
(V12 h1 +V22 h2 ) .

Use (†) to eliminate V3,

2
" V h +V h % ρ
p(0) − p(L) = ρ $ 1 1 2 2 ' −
# h1 + h2 & h1 + h2
(V12 h1 +V22 h2 ) ,

and – after some alegbra – this can be simplified to:

h1h2 2
2 ( 1
p(L) − p(0) = ρ V −V2 ) .
(h1 + h2 )
b) Interestingly, the pressure always rises when the two streams have different speeds: p(L) ≥
p(0), and it does not matter which velocity is larger.
c) If friction were included in the calculation the pressure rise would be less.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.19. Consider the situation depicted below. Wind strikes the side of a simple
residential structure and is deflected up over the top of the structure. Assume the following: two-
dimensional steady inviscid constant-density flow, uniform upstream velocity profile, linear
gradient in the downstream velocity profile (velocity U at the upper boundary and zero velocity
at the lower boundary as shown), no flow through the upper boundary of the control volume, and
constant pressure on the upper boundary of the control volume. Using the control volume shown:
a) Determine h2 in terms of U and h1, and
b) Determine the direction and magnitude of the horizontal force on the house per unit depth into
the page in terms of the fluid density ρ, the upstream velocity U, and the height of the house h1.
c) Evaluate the magnitude of the force for a house that is 10 m tall and 20 m long in wind of 22
m/sec (approximately 50 miles per hour).

Solution 4.19. 3. a) Use the control volume formulation of the continuity equation to find that:
h2 h2
Uy U h22 Uh2
Uh1 = ∫ u(y)dy = ∫ dy = = → h2 = 2h1.
0 0 h2 h2 2 2
b) The control volume form of the x-momentum equation is
h1 h2 h1 h2
2 2
− ∫ ρU dy + ∫ ρu dy = ∫ p∞ dy + ∫ − p∞ dy + ∫ + p∞ (n ⋅ e x )dy + Φx .
€ 0 0 0 0 top

where Φx is the force on the fluid per unit depth into the page. It is delivered through the
foundation of the house. Here the flow is assumed inviscid so there's no friction on the lower CV
surface. Evaluating the integrals yields:
€
1 1
−ρU 2 h1 + ρU 2 h2 = p∞ (+h1 − h2 + (h2 − h1 )) + Φx → Φx = − ρU 2 h1 .
3 3
1
Therefore, the force on the house in the +x direction per unit depth into page = + ρU 2 h1 .
3
3 2
c) (Force on the house) = (1.2 kg/m )(22 m/sec) (10 m)(20 m)/3 = 38.7 kN ≈ 8700 pounds. The
€ €
force attempts to push the house in the positive x-direction, that is: downwind.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.20. A large wind turbine with diameter D extracts a fraction η of the kinetic energy
from the airstream (density = ρ = constant) that impinges on it with velocity U.
a) What is the diameter of the wake zone, E, downstream of the windmill?
b) Determine the magnitude and direction of the force on the windmill in terms of ρ, U, D, and
η.
c) Does your answer approach reasonable limits as η → 0 & η → 1?

Solution 4.20. Place a control volume around the stream tube that hits the windmill with vertical
inflow-outflow surfaces well upstream and downstream of the wind turbine.
a) For ρ = const, the volume flux in the stream tube must be constant. Therefore: V = A1U/A2.
The efficiency of the wind turbine, η, implies: (1 – η)ρU2/2 = ρV2/2, or V = U(1 – η)1/2.
Eliminate V with the continuity result to find: A1U/A2 = U(1 – η)1/2 , or E = D/(1 – η)1/4 .
b) For the chosen CV, cons. of horizontal momentum implies:
–ρU2A1 + ρV2A2 = +Φx,
where Φx is force the wind turbine applies to the fluid. Here the flow speed outside the CV is
assumed to be U everywhere. Therefore, the pressure on the CV surface is P∞ everywhere
because of the steady constant-density horizontal flow Bernoulli equation. Thus, the pressure
integration drops out of the momentum equation because P∞ acts on the entire control surface.
Substituting in the results from part a) produces:
π D2 π D2 2
2 πD #
Force on the windmill = −Φ x = ρ U 2 − ρ U 2 (1− η ) $1− (1− η ) %& .
12
12
= ρ U
4 4(1− η ) 4
The force on the windmill points in the downstream direction, and is dimensionally sound
(ρU2D2 is has units of force).
c) This answer approaches reasonable limits. When as η → 0 the force on the wind turbine also
goes to zero. So, when there is no change in airspeed, there should be no force on the wind
turbine. When η → 1 the force on the windmill approaches πρU2D2/4, which precisely balances
the momentum flux in the incoming stream tube.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.21. An incompressible fluid of density ρ flows through a horizontal rectangular duct
of height h and width b. A uniform flat plate of length L and width b attached to the top of the
duct at point A is deflected to an angle θ as shown.
a) Estimate the pressure difference between the upper and lower sides of the plate in terms of x,
ρ, Uo, h, L and θ when the flow separates cleanly from the tip of the plate.
b) If the plate has mass M and can rotate freely about the hinge at A, determine a formula for the
angle θ in terms of the other parameters. You may leave your answer in terms of an integral.

Solution 4.21. a) The question says to estimate the pressure difference. Thus, reasonable
simplifying assumptions should be acceptable. The first of these is to ignore the two-dimensional
character of the flow field and assume uniform horizontal flow at each x location. The second is
to assume steady flow.
So, for the conditions stated (incompressible), the pressure difference between the upper
and lower side the plate can be estimated from the Bernoulli equation applied between a mid-
duct point comfortably upstream of the plate, where the flow speed is Uo and the pressure is po,
and a location x that is connected to the first point by a streamline, where the flow speed and
pressure below the plate are U(x) and p(x).
1 1
po + ρU o2 = p(x) + ρU 2 (x)
2 2
Conservation of mass in the duct requires:
U o h = U(x)h(x) = U(x)( h − x tan θ ) .
Combining these two equations to eliminate U(x) produces:
€
1 1 % U o2 h 2 ( 1 % 1 (
2€
( )
p(x) − po = ρ U o2 − U 2 (x) = ρ'U o2 −
2 &
* =
(h − x tan θ ) 2 ) 2
ρU 2
o '1− *.
& 1− (x /h) 2 tan 2 θ )
When the flow separates from the plate as shown, the pressure above the dotted separating
streamline will be the same as that below it. This pressure is
1 % 1 (
€ pupper = p(x = L cos θ ) − po = ρU o2 '1− *,
2 & 1− (L /h) 2 sin 2 θ )
so that
1 % 1 1 (
p(x) − pupper = ρU o2 '1− −1+ *
€ 2 & 1− (x /h) 2 tan 2 θ 1− (L /h) 2 sin 2 θ )
1 % 1 1 (
= ρU o2 ' 2 2
− 2 2 *
.
2 &1− (L /h) sin θ 1− (x /h) tan θ )
b) The
€ angle θ can be determined requiring the net moment on the plate to be zero. For a small
element of the plate of length dl = dx/cosθ located at horizontal location x, the moment around A
due to gravity is: – x(M/L)g(dx/cosθ). For this same element, the moment due to fluid mechanical
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

pressure is: +(x/cosθ)(p(x) – pupper)b(dx/cosθ). Thus, the angle that solves the problem is specified
by:
x= L cosθ
$ M b '
Net moment on the plate = 0 = ∫ &%−g L cosθ + ( p(x) − pupper ) cos2 θ )(xdx .
x= 0
Using the above relationship for p(x) – pupper, and pressing on with the integration and a little
algebra yields a final transcendental equation:
ML cosθ €1 % 1 2 1 (
0 = −g
2
+ ρU o2bL2 ''
4 ( L h )
2
sin 2{θ
ln 1− ( )
L h }
sin 2
θ +
1− ( L h )
2
sin 2 *
θ
*,
& )
which could be solved for specific values of the various parameters. Although complicated, this
formula does reach the correct limits. When the fluid dynamic force is negligible, the second
term can be ignored, the plate hangs straight down, and the answer θ = π/2 is provided by the
€first term. When the plate's weight is negligible, the first term can be ignored so the plate's angle
of deflection should be very small and this answer is recovered from the formula above when the
terms inside the big parentheses go to zero, which occurs when θ → 0.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.22 A pipe of length L and cross sectional area A is to be used as a fluid distribution
manifold that expels a steady uniform volume flux per unit length of an incompressible liquid
from x = 0 to x = L. The liquid has density ρ, and is to be expelled from the pipe through a slot of
varying width, w(x). The goal of this problem is to determine w(x) in terms of the other
parameters of the problem. The pipe-inlet pressure and liquid velocity at x = 0 are Po and Uo,
respectively, and the pressure outside the pipe is Pe. If P(x) denotes the pressure on the inside of
1
the pipe, then the liquid velocity through the slot Ue is determined from: P(x) − Pe = 2 ρU e2 . For
this problem assume that the expelled liquid exits the pipe perpendicular to the pipe’s axis, and
note that wUe = const. = UoA/L, even though w and Ue both depend on x.
a) Formulate a dimensionless scaling law for w in terms of x, L, A, ρ, Uo, Po, and Pe.
b) Ignore the effects of viscosity, assume all profiles through the€ cross section of the pipe are
uniform, and use a suitable differential-control-volume analysis to show that:
dU d dP
A + wU e = 0 , and ρ U 2 = − .
dx dx dx
c) Use these equations and the relationships stated above to determine w(x) in terms of x, L, A, ρ,
Uo, Po, and Pe. Is the slot wider at x = 0 or at x = L?
€ €

Solution 4.22. a) Create the parameter matrix.

w x L A ρ Uo Po Pe
–––––––––––––––––––––––––––––
M 0 0 0 0 1 0 1 1
L 1 1 1 2 -3 1 -1 -1
T 0 0 0 0 0 –1 -2 -2
There will be 8 – 3 = 5 dimensionless groups
w x A ρU o2 P w # x A ρU 2 P &
By inspection the groups are: , , 2 , , and o , thus: = fn% , 2 , o , o (
L L L Po Pe L $ L L Po Pe '
b) For a differential slice of the pipe located between x and x+dx, cons. of mass implies:
−ρU(x)A + ρU e (x)w(x)dx + ρU(x + dx)A = 0 .
Group the U-terms together
€ € €and € divide by dx:
€
ρA (U(x + dx) − U(x)) dx € + ρU e (x)w(x) = 0 .
Divide by ρ and€take the limit as dx goes to zero to find:
A dU dx + U e w = 0 .
Using the same CV conserv horizontal momentum
€
−ρU 2 (x)A + ρU 2 (x + dx)A = P(x)A − P(x + dx)A
Divide by A & dx, and take the limit as dx goes to zero to find:
€
ρ dU 2 dx = − dP dx .
c) Integrate the two equations found for part b): U = −(U e w A) x + C1, and ρU 2 = −P + C2
€

€
€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

The boundary conditions for U are: U(0) = Uo, and U(L) = 0, and from the problem statement:
wUe = const. = UoA/L, so U(x) = U o (1− x L) .
Noting that P(0) = Po and U(0) = Uo, the integrated momentum equation becomes:
P(x) + ρU 2 (x) = Po + ρU o2 .
Put this together with the COMA result and the exit velocity condition,
€ 2
P(x) − Pe = 12 ρU e2 = 12 ρ(U o A wL) ,
to find: €
2 2
Pe + 12 ρ(U o A wL) + ρU o2 (1− x L) = Po + ρU o2 ,
which can be solved for
€ w(x) to get −1 2

[(
w(x) = ( A L) (Po − Pe ) 1
2 (
ρU o2 ) + 2 1− (1− x L)
2
)] .
€ at x = 0.
The slot is wider

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.23. The take-off mass of a Boeing 747-400 may be as high as 400,000 kg. An Airbus
A380 may be even heavier. Using a control volume (CV) that comfortably encloses the aircraft,
explain why such large aircraft do not crush houses or people when they fly low overhead. Of
course, the aircraft’s wings generate lift but they are entirely contained within the CV and do not
coincide with any of the CV’s surfaces; thus merely stating the lift balances weight is not a
satisfactory explanation. Given that the CV’s vertical body-force term, −g ∫ ρdV , will exceed
CV
4x106 N when the airplane and air in the CV’s interior are included, your answer should instead
specify which of the CV’s surface forces or surface fluxes carries the signature of a large
aircraft’s impressive weight.
€
Solution 4.23. Assume that the airplane is in steady level flight so that the flow is steady in the
frame of reference of the airplane. Therefore, choose a frame of reference attached to the
airplane and select a control volume (CV) that comfortably encloses the aircraft. In this frame of
reference the CV is not moving but is instantaneously aligned above the house. The overall
problem is really three dimensional, but will be simplified here. Therefore, take the CV to be a
cube of volume V = L3 with z = 0 at mean roof level. Retaining the roof-peak in the solution is
not essential since there is no flow through the bottom of the control volume. Denote the speed
and mass of the aircraft by Uac and Mac, let the Cartesian-coordinate fluid velocity with respect

to the control volume be u = (u,v,w) , and choose the front of the control volume to coincide with
the y-z plane.
z
€ y

x
For steady flow of a perfect fluid with density ρ (viscous forces are irrelevant here because of the
enormous Reynolds number), the integral momentum equation in the z-direction is:
− ∫ ρwudS + ∫ ρwudS + ∫ ρw 2 dS − ∫ ρwvdS + ∫ ρwvdS
front back top close side far side

= −g( ρ airV + M ac ) − ∫ PdS + ∫ PdS

top bottom
where g is the acceleration of gravity, P is pressure, and the volume of the aircraft has been
ignored €compared V. There is no flux term on the bottom of the CV because the rooftop is
assumed to be solid and impenetrable.
€ back of the control volume u is approximately equal to U . The
On the front and ac
contribution from sides and top of the control volume will be small since |w|,|v| « Uac at any
reasonable distance from the aircraft. For low-Mach-number subsonic flight, such as during
take-off and landing, the fluid density can be treated as constant. If the length of one side of the
cubical control volume is L, then momentum conservation simplifies to :
L L L L
ρU ac ∫ ∫ (w back − w front ) dydz = −g( ρ airV + M ac ) + ∫ ∫ (Pbottom − Ptop ) dxdy . (a)
0 0 0 0
In the absence of a difference between the front and back vertical velocity on the CV surface, the
weight of the aircraft and air in the CV would be borne by the ground through the pressure
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

difference Pbottom – Ptop. Here we can set Pbottom = Patm + Pextra, where Patm is the atmospheric
pressure that supports the air in the CV and counter acts the pressure on the top of the CV, and
Pextra is the extra pressure on the ground caused by the aircraft. In particular, evaluate (a) when
the aircraft is absent to find a hydrostatic balance,
L L
0 = −gρ airV + ∫ ∫ (Patm − Ptop )dxdy ,
0 0
and subtract this from (a) to find:
L L L L
gM ac = ρU ac ∫ ∫ (w front − w back )dydz + ∫ ∫ Pextra dxdy . (b)
0 0 0 0
€
Thus, the weight of the aircraft is borne by moving air (the first term in b) and an extra pressure
on the ground (the second term in b). In reality, the pressure distribution on the aircraft's wings
causes wfront – wback to be both nonzero and positive (i.e. wfront is an updraft and wback is a
downdraft), and€ this term primarily balances the impressive weight of the aircraft. For a landing
speed of 100 m/s, L ≈ 100 m (~50% larger than the aircraft’s wingspan), gMac = 4 mega-Newton,
we can estimate the vertical velocity difference necessary to support the aircraft:
4 ×10 6 N
w front − w back ≈ ~ 3.3m /s.
(1.2kg /m 3 )(100m /s)(10 4 m 2 )
Such wind speeds are not much of a hazard to most people, buildings, or other structures.
Further consideration of the flow-field produced by a flying aircraft shows that heavy
aircraft do not crush houses and people because the plane's weight ends up being spread over an
€ area. A flying aircraft's pressure distribution may exist for a kilometer or more in
extremely large
all directions, so the supporting area for a plane with a weight of 4 mega-Newtons might easily
exceed 1 km2 or 106 square meters. Therefore the extra pressure on the ground will be at most a
few Pa (less than 0.01% of an atmosphere).
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.24. An inviscid incompressible liquid with density ρ flows in a wide conduit of
height H and width B into the page. The inlet stream travels at a uniform speed U and fills the
conduit. The depth of the outlet stream is less than H. Air with negligible density fills the gap
above the outlet stream. Gravity acts downward with acceleration g. Assume the flow is steady
for the following items.
a) Find a dimensionless scaling law for U in terms of ρ, H, and g.
b) Denote the outlet stream depth and speed by h and u, respectively, and write down a set of
equations that will allow U, u, and h to be determined in terms of ρ, H, and g.
c) Solve for U, u, and h in terms of ρ, H, and g. [Hint: solve for h first.]

Solution 4.24. a) Density is the only

parameter with units of mass. Thus, it %! &! '!
cannot be part of a dimensionless law $"
for U. The remaining parameters (U, g, !" "#$!
#" !
H) can only be assembled into one
2
dimensionless group, Π1 = U gH ,
thus:
U = const ⋅ gH . !
b) Place the
€ CV around the change in liquid level as shown above, ignore friction, set
atmospheric pressure to zero, and assume uniform inflow and outflow.
Cons. of mass: ρUH = ρuh (1)
€ of horizontal momentum
Cons. 2 2 1 2 1
−ρU H + ρu h = P0 H + 2 ρgH − 2 ρgh 2
(2)
Bernoulli from ‘0’ to ‘1’: P0 + 12 ρU 2 + ρgH = 0 + 0 + ρgH (3)
Bernoulli from ‘1’ to ‘2’: € 1 2
0 + 0 + ρgH = 0 + 2 ρu + ρgh (4)
The zeros arise in the Bernoulli
€ equations because point ‘1’ is a stagnation point and the pressure
there is atmospheric pressure. There are four unknowns (U, u, h, P0) so these four equations
€
should be sufficient for a complete solution.
c) Use (4) to find u = 2g(H € − h) . Use this to eliminate u from (1) and (2), and use (3) to
eliminate P0 from (2). The remaining equations are:
ρUH = ρh 2g(H − h) , and −ρU 2 H + ρ2g(H − h)h = − 12 ρU 2 H + 12 ρg(H 2 − h 2 ) .
Divide each
€ equation by ρ. Square the first and divide by H, and simplify the second to get:
1 2 1 2 2
U H = 2g(H − h)h 2 H , and − 2 U H + 2g(H − h)h = + 2 g(H − h ) .
2

€Use the first of these to eliminate

€ U 2 H in the second, and then divide by (H–h) to find:
− gh 2 H + 2gh = + 12 g(H + h) , or h 2 − 32 Hh + 12 H 2 = 0 = (h − H)(h − 12 H) .
€ h = H 2 so (4) implies u = €gH , and these together with (1) produce: U = gH 2 .
Thus,
€

€ €
€ € €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.25. A hydraulic jump is the shallow-water-wave equivalent of a gas-dynamic shock

wave. A steady radial hydraulic jump can be observed safely in one’s kitchen, bathroom, or
backyard where a falling stream of water impacts a shallow pool of water on a flat surface. The
radial location R of the jump will depend on gravity g, the depth of the water behind the jump H,
the volume flow rate of the falling stream Q, and stream’s speed, U, where it impacts the plate.
In your work, assume 2gh << U where r is the radial coordinate from the point where the falling
stream impacts the surface.
a) Formulate a dimensionless law for R in terms of the other parameters.
b) Use the Bernoulli
€ equation and a control volume with narrow angular and negligible radial
Q % 2U 2 (
extents that contains the hydraulic jump to show that: R ≅ 2'
− H*.
2πUH & g )
c) Rewrite the results of part b) in terms of the dimensionless parameters found for part a).

Solution 4.25. a) Units matrix R g H Q U

M 0 0 0 0 0
L 1 1 1 3 1
T 0 -2 0 -1 -1
The matrix has rank two, so the number of dimensionless groups is: (5 parameters) – (2
dimensions) = 3 dimensionless groups. Generate the dimensionless groups by inspection:
R U Q
Π1 = , Π 2 = , and Π 3 =
H gH gH 5
State the dimensionless law: R H = f U ( gH ,Q gH 5 )
b) The pressure is atmospheric on all the water surfaces and observation of kitchen- or bathroom-
sink hydraulic jump€ suggests€that any change of € the thickness of the fast moving layer upstream
of the jump is negligible compared to this distance the water has fallen to reach the bottom of the
€
sink. Thus, the Bernoulli equation for any radial liquid-surface streamline from the point of
impact to the hydraulic jump suggests that the fluid velocity is essentially constant and equal to
U. If h is the thickness of the fast moving water layer upstream of the hydraulic jump, then
conservation of mass implies: h(r) = Q (2πrU) where r is the radial distance from the point of
water impact. The stage is now set for a control volume calculation. Consider a rectangular CV
of small radial extent that encloses the hydraulic jump. Let ho = h(R) = Q (2πRU) , and V be the
velocity of the outlet flow.
Conserve Mass:€ Uho = VH
2 h2 H2
Conserve Horizontal Momentum: −ρU € ho + ρV 2 H = ρg o − ρg
2 2

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Here the conservation of momentum equation specifies a balance between momentum fluxes and
hydrostatic pressure terms. First eliminate V from the second equation using the first.
2 $ Uho ' 2 ho2 H2
−ρU ho + ρ& ) H = ρg − ρg
% H ( 2 2
Divide by ρ, and manipulate this equation to eliminate a common factor of H – ho.
h g h g # U2 &
−U 2 o (H − ho ) = − ( H − ho )( H + ho ) –> U 2 o = ( H + ho ) –> ho %2 −1( = H
H € 2 H 2 $ gH '
Now use the relationship ho = h(R) = Q (2πRU) to eliminate ho and solve for R:
Q $ U2 ' Q $ U2 '
& 2 −1) = H –> R = & 2 − H )
€ 2πRU % gH (€ 2πUH 2 % € g (
c) Putting this result
€ into dimensionless form just takes a little algebra. The final result is:
R 1 Q $ U gH '
= & 2 − )
€ H 2π € gH 5 % gH U (

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.26. A fine uniform mist of an inviscid incompressible fluid with density ρ settles
steadily at a total volume flow rate (per unit depth into the page) of 2q onto a flat horizontal
surface of width 2s to form a liquid layer of thickness h(x) as shown. The geometry is two-
dimensional.
a) Formulate a dimensionless scaling law for h in terms of x, s, q, ρ, and g.
b) Use a suitable control volume analysis, assuming u(x) does not depend on y, to find a single
cubic equation for h(x) in terms of h(0), s, q, x, and g.
c) Determine h(0).

Solution 4.26. a) There appear to be six parameters, but the density is the only parameter that
includes the units of mass so it can be set aside. Thus, there will only be two independent
dimensions, length (L) and time (T).
Therefore, the units matrix is: h x s q g
L 1 1 1 2 1
T 0 0 0 -1 -2
There will be 5 – 2 = 3 dimensionless groups. By inspection: Π1 = h s , Π 2 = x s , and
( 2 3
Π 3 = q 2 gs3 , so h s = fn x s,q gs . )
b) Conserve mass using the differential CV:
−u(x)h(x) + u(x + dx)h(x + dx) − (q€s) dx = 0 ,€or d(uh) dx = q s .
€ Conserve horizontal momentum in same differential CV with hydrostatic pressure forces:
€ −ρu 2 (x)h(x) + ρu 2 (x + dx)h(x + dx) = 12 ρgh 2 (x) − 12 ρgh 2 (x + dx) ,
€ or d(u 2 h) dx = −( g 2)€dh 2 dx .
Integrate these equations and evaluate the constants with u(0) = 0 to get
€
2
(
2 2
uh = qx s , and u h = −( g 2) h − h (0) . )
2 2 2
( 2 2
€ two equations to find: q x hs = −( g 2) h − h (0) , which is a
Eliminate u(x) between these )
cubic equation so h(x) cannot be put into a simple form.
€ of mass implies:
c) Global conservation € q = u(s)h(s) . If Pa = atmospheric pressure, a Bernoulli
streamline from x = 0 to x = s on the flat surface (the liquid surface is not a streamline) produces:
Pa + ρgh(0) = Pa + ρgh(s) + 12 ρu€2 (s) , or u 2 (s) = 2g( h(0) − h(s)) = q 2 h 2 (s) .
2
(
€ of part b) evaluated at x = s, q h(s) = −( g 2) h (s) − h (0) ,
This final equality and the result
2 2
)
represent two equations in two unknowns, h(0) & h(s), that can be solved simultaneously to
determine
€ h(0) in terms of q and g: €
13
h(0) = 3(q 2 4g) = 3h(s).
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.27. A thin-walled pipe of mass mo, length L, and cross sectional area A is free to
swing in the x-y plane from a frictionless pivot at point O. Water with density ρ fills the pipe,
flows into it at O perpendicular to the x-y plane, and is expelled at a right angle from the pipe’s
end as shown. The pipe’s opening also has area A and gravity g acts downward. For a steady
mass flow rate of m˙ , the pipe hangs at angle θ with respect to the vertical as shown. Ignore fluid
viscosity.
a) Develop a dimensionless scaling law for θ in terms of mo,
L, A, ρ, €
m˙ , and g.
b) Use a control volume analysis to determine the force
components, Fx & Fy, applied to the pipe at the pivot point in
€ terms of θ, mo, L, A, ρ, m
˙ , and g.
c) Determine θ in terms of mo, L, A, ρ, m˙ , and g.
d) Above what value of m˙ will the pipe rotate without
stopping? €
€
Solution 4.27. a) The
€ units matrix is:
θ mo L A ρ m˙ g
–––––––––––––––––––––––––––––––––––––
M 0 1 0 0 1 1 0
L 0 0 1 2 € -3 0 1
T 0 0 0 -1 0 -1 -2
This matrix has rank three and there are seven parameters so four dimensionless groups should
be constructed. By inspection the dimensionless groups are: θ, m˙ 2 L gmo2 , L2 A , and ρAL mo ,
( 2 2 2
)
so the dimensionless scaling law is: θ = f m˙ L gmo ,L A, ρAL mo , where f is an undetermined
function.
b) Note that this part of the problem involves steady flow. € Place€the CV around
€ the outside of
the pipe except at the pivot point O where the CV slices through the pipe parallel to the x-y
plane. Assume that the flow € is in the z-direction across this control surface slice at O. Here,
conservation of mass merely requires m˙ = const. The control surface pressure is atmospheric
everywhere outside the pipe, and pressure forces can only act in the z-direction on the control
surface slice at O. Thus, there is no pressure contribution to the reaction forces, Fx and Fy, which
arise from the portion of the € control surface that passes through the pipe structure. The
momentum flux of water at O is perpendicular to the x-y plane so only the outflow momentum
flux is relevant for determining Fx and Fy. For the chosen stationary control volume, this outflow
flux term can be readily evaluated because the water’s discharge velocity and the CV’s outward
normal are parallel. Assume the outflow velocity is uniform too; u = ( m˙ ρA)n , so conservation
of momentum implies:
m˙ % m˙ ( m˙ 2
∫ ρu(u ⋅ n)dA = ρ ρA n' ρA * A = − ρA (e x cosθ + e y sin θ ) = −( mo + ρAL) ge y + Fxe x + Fye y ,
outlet & ) €
where n = −e x cos θ − e y sin θ . Equating x & y components produces the final answers:
Fx = −( m˙ 2 ρA) cosθ , and Fy = (mo + ρAL)g − ( m˙ 2 ρA) sinθ .
€ c) This part of the problem again involves STEADY flow. A small CV placed inside the pipe at
€ its end shows that the force resulting from the flux of fluid discharged from the pipe primarily
acts at end of the pipe. Based on this, equate the gravity- and flow-induced moments on the pipe
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

(2
)
about O so that Fx and Fy are not needed: (mo + ρAL)g( L 2) sin θ = m˙ ρA L . Solve for the
(2
)
angle: sin θ = 2 m˙ ρA (mo + ρAL)g .
( 2
)
d) When 2 m˙ ρA (mo + ρAL)g > 1 there is no answer for θ since sinθ ≤ 1. Thus, when
m˙ > ρA(mo + ρAL)g 2 , no steady€solution is possible, and the pipe will spin without stopping.
€
€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.28. Construct a house of cards, or light a candle. Get the cardboard tube from the
center of a roll of paper towels and back away from the cards or candle several feet so that by
blowing you cannot knock down the cards or blow out the candle unaided. Now use the tube in
two slightly different configurations. First, place the tube snugly against your face encircling
your mouth, and try to blow down the house of cards or blow out the candle. Repeat the
experiment while moving closer until the cards are knocked down or the candle is blown out
(you may need to get closer to your target than might be expected; do not hyperventilate; do not
start the cardboard tube on fire). Note the distance between the end of the tube and the card
house or candle at this point. Rebuild the card house or relight the candle and repeat the
experiment, except this time hold the tube a few centimeters away from your face and mouth,
and blow through it. Again, determine the greatest distance from which you can knock down the
cards or blow out the candle.
a) Which configuration is more effective at knocking the cards down or blowing the candle out?
b) Explain your findings with a suitable control-volume analysis.
c) List some practical applications where this effect might be useful.

Solution 4.28. a) The most effective configuration is when the tube is held a few centimeters in
front of the face and mouth.
b) The difference between the two cases is the greater volume flux through the tube when it is
held a few centimeters in front of the mouth. Since the exit area of the tube does not change, the
volume flux change results entirely from an increased exit velocity. Therefore, the control
volume solution should determine the exit velocity from the tube in terms of other parameters.
In the following, the subscript ‘o’ will refer to conditions at the inlet end of the tube, the
subscript ‘e’ will refer to conditions at the exit or downstream end of the tube, and the subscript
‘a’ will refer to ambient (motionless) conditions some distance from the tube.
Consider a control volume that encloses the air inside the tube, ignore the frictional losses
between the tube and the flowing air, and align the tube with the x-axis (see picture). For the air
speeds and temperature differences under consideration the flow is essentially incompressible
and isothermal which means, ρ ≈ const.

po
Uo pe
Tube Walls

Ao Af Uf Ue
Control Volume
Boundary

Let Af be the cross sectional area of the fast moving air blown out of the mouth, and Uf be the
velocity of this air stream where it enters the tube. When ρ = const, conservation of mass reduces
to conservation of volume implying:
A f U f + (Ao − A f )U o = AeU e (1)

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

The horizontal-component of the momentum equation is then

−ρA f U 2f − ρ(Ao − A f )U o2 + ρAeU e2 = ∫ Po dA − ∫ Pe dA (2)
Ao Ae

In both cases, Ao = Ae = A. Using these relationships and assuming that the inlet and exit pressure
integrals can be approximated by the PoAo and PeAe, produces:
A f U f + (A − A f )U o = AU e (3)
€
(Po − Pe )A + ρA f U 2f + ρ(A − A f )U o2 = ρAU e2 (4)
These are two equations in three unknowns: (Po–Pe), Uo, and Ue. The third equation is
determined by the tube configuration.
€
CASE I.
When the tube is € held against the face, the third relationship is Uo = 0, so (3) implies:
( )
U e = A f A U f = sU f , (5)
where s = Af/A.
CASE II.
When the tube is held away from the mouth. The remaining relationship comes from a pressure
balance outside the tube. The € pressure of the air drawn into the tube from the ambient condition
can be estimated from the steady Bernoulli equation:
Pa + 12 ρU a2 = Pa = Po + 12 ρU o2 (6)
where, of course, Ua ≈ 0. In addition, if the exit velocity is parallel to the tube axis, then the
component of the momentum equation perpendicular to the tube axis suggests that Pa ≈ Pe so that
Pe − Po = 12 ρU o2 . (7)
€
Put (7) into (4), then the equations to be solved for Ue are (3) and
− 12 U o2 A + A f U 2f + (A − A f )U o2 = AU e2 . (8)
where the common factor of ρ has € been divided out. Rearrange (3) and (8) with s = Af/A to find:
U e − sU f 1
Uo = , and ( 2 − s)U o2 = U e2 − sU 2f . (9a,b)
1− s
€ for U in terms of U and s (this involves some algebra).
Eliminate Uo and solve e f

U e s(2s −1) + 2s(1− s) 3

= (10)
€ Uf € (1− s) 2 + s2
Although it is not obvious by looking at it, this result, (10), is always larger than the result for
Case I, (5), especially when s is small. For example, when s = 1/10: Case I –> Ue = 0.100Uf,
while Case II –> Ue = 0.368Uf, almost 4 times larger! Or, when s = 1/4: Case I –> Ue = 0.250Uf,
€
while Case II –> Ue = 0.535U f, more than twice as large!
c) For this problem, the fast-moving air from the mouth and the cardboard tube form a crude
ejector pump. Applications for ejector pumps occur wherever rapid or steady pumping without
moving parts is required for inflation or evacuation of a container or chamber. Examples: air
bags, emergency exit ramps on commercial aircraft, emergency life support (ventilation of the
lungs), and a variety of industrial processes involving corrosive materials. In addition, if the
fast-moving stream carries a fuel and the entrained fluid carries the oxidizer (for example), the
shape (i.e. the area ratio) of the ejector pump can be used to mix reactants to a set the mixture
ratio for combustion.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.29. Attach a drinking straw to a 15-cm-diameter cardboard disk with a hole at the
center using tape or glue. Loosely fold the corners of a standard piece of paper upward so that
the paper mildly cups the cardboard disk (see drawing). Place the cardboard disk in the central
section of the folded paper. Attempt to lift the loosely folded paper off a flat surface by blowing
or sucking air through the straw.
a) Experimentally determine if blowing or suction is more effective in lifting the folded paper.
b) Explain your findings with a control volume analysis.

Solution 4.29. a) It would seem that the only way to lift the paper would be to suck on the straw.
However, blowing is found to be just as successful!
b) This can be explained with a control volume analysis. Choose a control volume that encloses a
portion of the fluid between the paper and the cardboard disk. For simplicity assume that the
flow is axisymmetric, constant density, and inviscid. Also assume that the paper and the
cardboard are perfectly planar surfaces. Denote the distance between the paper and the
cardboard disk by h, the diameter of the cardboard disk by R, and the radius & cross sectional
area of the straw by rs & As.

2rs h

Cardboard
ur control volume
Paper
p∞
r
R
Conservation of mass implies: 2πrhur = UiAs for r >> rs, where r is the radius of the control
volume, ur is the local radial component of the flow at the circular edge of the control volume,
and Ui is the inlet velocity from the straw. Therefore: ur = U i As (2πhr) for r >> rs.
What we need to know is the pressure distribution above the piece of paper in terms of the
parameters of the problem and p∞, the pressure below the piece of paper. The Bernoulli equation
in this case is:
€ + 1 ρu 2 = p + 1 ρU 2
p + 1 ρu 2 = p(r)
∞ 2 R 2 r i 2 i
where the 'i' subscript refers to conditions at the lower tip of the straw. For the paper to remain
suspended we need the following to be true:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

r= R
W p ≤ 2π ∫ [− p(r) + p ]rdr ∞ (†)
r= o
where Wp is the weight of the paper. This pressure integral can be estimated from the vertical
momentum equation for a control volume like that shown above with radius R:
r= R r= R
∫ ρU i (U ie z ⋅ e€z )dA = ρU i2 As = −2π ∫ p(r)rdr − pi As + 2π ∫ p(r)rdr .
As r= rs r= 0

The first and second integrals on the right side come from the cardboard and paper surfaces,
respectively. Rearrange this, use the Bernoulli equation, and then approximate the flow under
the cardboard as purely radial for r > rs.
€ r= R r= R
2π ∫ p(r)rdr = ρU i2 As + 2π ∫ p(r)rdr + pi As
r= 0 r= rs
r= R r= R
2π ∫ p(r)rdr = ρU 2
i As + 2π ∫ (p ∞ )
+ 12 ρuR2 − 12 ρur2 rdr + pi As .
r= 0 r= rs
Now start reconstructing
€ the integral of the pressure difference (†):
r= R r= R
2π ∫ [p 2
∞ − p(r)] rdr = − ρU i As − 2π ∫( 1
2
ρuR2 − 12 ρur2 ) rdr − pi As + p∞ As
€ r= 0 r= rs

Inserting ur = uR ( R r) yields:
r= R
2π ∫ [p ∞ − p(r)] rdr = −ρU i2 As − 12 πρuR2 ( R 2 − rs2 ) + πρuR2 R 2 (ln R − ln rs ) − pi As + p∞ As .
€ r= 0
Collect
€ terms and introduce Wp:
W p ≤ ( p∞ − pi − ρU i2 ) As + 12 πρuR2 ( rs2 + 2R 2 ln( R rs ) − R 2 ) .
2
€ Simplifying further using As = πrs , the Bernoulli relationships, and
uR = U i As (2πhR) = U i rs2 (2hR) produces:
1 % r2 (
W p ≤€− 12 ρU i2 As + 12 πρuR2 R 2 (2ln( R rs ) −1) , or, W p ≤ ρU i2 As'−1+ s 2 (2ln( R rs ) −1)* .
€ 2 & 4h )
€ For a straw diameter rs = 3 mm, h = 1 mm, and R = 80 mm, the factor inside the large
parentheses is about +11.5. With these dimensions and Ui ≈ 20 m/sec, Wp can be as large as a
€ 0.08 N. This strange lifting effect is used in manufacturing processes to pick up computer chips
without touching them. Note also that Ui enters € the final answer as U i2 so it doesn’t matter which
ways the flow goes; suction or blowing through the straw should yield identical results.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.30. A compression wave in a long gas-filled constant-area duct propagates to the left
at speed U. To the left of the wave, the gas is quiescent with uniform density ρ1 and uniform
pressure p1. To the right of the wave, the gas has uniform density ρ2 (> ρ1) and uniform pressure
is p2 (> p1). Ignore the effects of viscosity in this problem.
a) Formulate a dimensionless scaling law for U in terms of the pressures and densities.
b) Determine U in terms of ρ1, ρ2, p1, and p2 using a control volume.
c) Put your answer to part b) in dimensionless form and thereby determine the unknown function
from part a).
d) When the density and pressure changes are small, they are proportional:
p2 − p1 = c 2 ( ρ2 − ρ1 ) for ( ρ2 − ρ1 ) ρ2 << 1 ,
where c 2 = (∂p ∂ρ )s . Under these conditions, U is associated with what common property of the
gas?
U!

p1, ρ1! p2, ρ2!

u1 = 0!

Solution 4.30. a) Use dimensional analysis. There are five parameters but there are only two
independent dimensions ([density] and [velocity] because [density x velocity2] = [pressure]), so
there are three dimensionless groups. Three satisfactory groups are easily found by inspection:
Π1 = ρ1U 2 p1 , Π 2 = ρ1 ρ2 , and Π3 = p1 p2 . Thus, the scaling law is:
ρ1U 2 p1 = fn ( ρ1 ρ2 , p1 p2 ) .
b) Place a moving control volume around the compression wave with vertical surfaces on the left
and on the right of the wave, similar to Example 4.6. The contents of this control volume do not
vary in time so:
Cons. of mass −ρ1UA + ρ2 (U + u2 )A = 0
Cons. of x-momentum −ρ1U 2 A + ρ2 (U + u2 )2 A = p1 A − p2 A
where A is the cross section of the duct, and u2 is the horizontal velocity of the gas on the right of
the wave in a stationary frame of reference. Divide both equations by A, and use the first
2
equation to eliminate U + u2 from the second equation to find: −ρ1U 2 + ρ2 ( ρ1U ρ2 ) = p1 − p2 .
p2 − p1
Solve this equation to find: U= .
ρ1 (1− ρ1 ρ2 )
ρ1U 2 p p −1 Π−1 −1
c) Manipulate the result of part b) to find: = 2 1 , or Π1 = fn(Π 2 , Π3 ) = 3 .
p1 (1− ρ1 ρ2 ) 1− Π 2
d) Start with the result of part b) and insert the small-change approximation:
p2 − p1 c 2 ( ρ2 − ρ1 ) ρ ρ
U= = = c 2 → c as 2 →1
ρ1 (1− ρ1 ρ2 ) ρ1 (1− ρ1 ρ2 ) ρ1 ρ1
Thus, in this limit, U is associated with c, the speed of sound.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.31. A rectangular tank is placed on wheels and is given a constant horizontal
acceleration a. Show that, at steady state, the angle made by the free surface with the horizontal
is given by tan θ = a/g.

Solution 4.31. Make a drawing and add some

dimensions to the tank. Place the control y!
volume around the water inside the rolling
θ! g!
tank. Here the continuity equation merely
ensures that the mass of water in the tank, x!
M = ∫ ρdV , doesn’t change as the tank
!h1
L ! ! h2
V*
accelerates. The horizontal component of the
momentum equation is:
d
€ ∫ ρudV + ∫ ρu( u − at )e x ⋅ ndA = ∫ f x dA
dt V *(t ) S*(t ) S*(t )
where the CV velocity, b, has been set to b = atex, so that db/dt = aex is the specified steady
horizontal acceleration. The body force term does not appear in the above equation since
gravity acts in the vertical direction. At steady state, the fluid velocity will equal the velocity of
the moving tank€so the flux terms will be zero, and the pressure forces will be hydrostatic:
∇p = ρg , where p is the pressure and g is the body force acceleration. In this situation, g has
both horizontal and vertical components. However, the vertical component of the hydrostatic
equation, ∂p ∂y = −ρg , is not changed by the horizontal acceleration. This equation integrates
to: p = A(x) − ρgy where A(x) is the function of integration; it cannot depend on y. Here we
€
know that atmospheric pressure occurs on the surface of the water so the pressure on the left and
right sides of the tank are p1 (y) = po + ρg(h1 − y) and p2 (y) = po + ρg(h2 − y) , respectively.
€
Thus, the horizontal momentum conservation equation becomes:
€ h1 h2 ρgB 2
Ma = ∫ 0 ρg(h1 − y)Bdy − ∫ 0 ρg(h2 − y)Bdy =
2
( h1 − h22 ) ,
€ €
where B is the dimension of the tank into the page, and the CV-surface integration of po leads (as
usual) to no net force. Now, use the two ends of this extended equality and divide by M noting
that M = ρBL (h1 + h2 ) 2 where L is the horizontal length of the tank:
€ g a
a = ( h1 − h2 ) = gtan θ , so tan θ =
L g
€

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.32. Starting from rest at t = 0, an airliner of mass M accelerates at a constant rate
a = ae x into a headwind, u = −uie x . For the following items, assume that: i) the x-component of
the fluid velocity is –ui on the front, sides, and back upper half of the control volume (CV), ii)
the x-component of the fluid velocity is –uo on the back lower half of the CV, iii) changes in M
can be neglected, iv) changes of air momentum inside the CV can be neglected, and v)
€ € In addition, assume constant air density ρ and uniform flow conditions exist
frictionless wheels.
on the various control surfaces. In your work, denote the CV’s front and back area by A. (This
approximate model is appropriate for real commercial airliners that have the engines hung under
the wings).
a) Find a dimensionless scaling law for uo at t = 0 in terms of ui, ρ, a, M, and A.
b) Using a CV that encloses the airliner (as shown) determine a formula for uo(t), the time-
dependent air velocity on the lower half of the CV’s back surface.
c) Evaluate uo at t = 0, when M = 4 x 105 kg, a = 2.0 m/s2, ui = 5 m/s, ρ = 1.2 kg/m3, and A = 1200
m2. Would you be able to walk comfortably behind the airliner?

Solution 4.32. a) The units matrix is:

uo ui ρ a M A
––––––––––––––––––––––––––––––––
M 0 0 1 0 1 0
L 1 1 -3 1 0 2
T -1 -1 0 -2 0 0
This matrix has rank three and there are six parameters so three dimensionless groups should be
constructed. These are found easily by inspection. The final dimensionless scaling law can be
uo # aA1 2 M &
stated: = fn% 2 , 3 2 ( .
ui $ ui ρA '
b) For constant acceleration from rest, the CV’s velocity is b = atex. Conserve mass:
d $ '
( M + M air ) + ρ% ∫ + ∫ + ∫ + ∫ ((u − b) ⋅ ndA = 0
dt & upper back lower back front sides )
€
Drop the time derivative terms and evaluate the first three surface integrations:
A A
ρ(−ui − at)e x ⋅ (−e x ) + ρ(−uo − at)e x ⋅ (−e x ) + ρ(−ui − at)e x ⋅ (e x )A + ρ ∫ (u − b) ⋅ ndS = 0
€ 2 2 sides
A
Simplify to find: ∫ (u − b) ⋅ ndS = (ui − uo ) 2 . (*)
sides
Note that ui must equal uo if the integral over the sides of the CV is not included. Now conserve
€
horizontal momentum noting that there are no external forces on the CV.
d $ '
( Mat + M€air uair ) + ρ% ∫ + ∫ + ∫ + ∫ ((u ⋅ e x )(u − b) ⋅ ndS = 0
dt & upper back lower back front sides )

Drop the air term from the time derivative, set dM/dt = 0, and evaluate the first three surface
integrations
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

A A
Ma − ρui (ui + at) − ρuo (uo + at) + ρui (ui + at)A − ρui ∫ (u − b) ⋅ ndS = 0 .
2 2 sides
Simplify this equation and use (*) to eliminate the integral over the sides of the control surface:
A A A
Ma + ρui (ui + at) − ρuo (uo + at) = ρui ∫ (u − b) ⋅ ndS = ρui (ui − uo ) .
2 2 sides 2
€
Distribute the various multiplications and cancel terms to find a quadratic equation for uo:
$ 2Ma '
uo2 − ( ui − at ) uo − & ui at + ) = 0.
€ % ρA (
Thus: uo = 12 (ui − at) + 12 (ui − at) 2 + 4 ( ui at + 2Ma ρA) , where the plus sign is chosen in front of
the square root to ensure that uo is positive.
# & u # &
c) At t = 0, uo = 12 %% ui +€ ui2 + 8Ma ρA (( , or o = 12 %%1+ 1+ 8( aA1 2 ui2 )( M ρA 3 2 ) ((
€
$ ' ui $ '
# &
= 12 %%5m /s + 25m 2 /s2 + 8(4 ×10 5 kg)(2m /s2 ) (1.2kg /m 3 )(1200m 2 ) ((
$ '
= 35.9 m/s ≈ 130 kph.
€ € behind the airliner.
No, you could not walk comfortably
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.33. A cart that can roll freely in the x-direction deflects a horizontal water jet into its
tank with a vane inclined to the vertical at an angle θ. The jet issues steadily at velocity U with
density ρ, and has cross sectional area A. The cart is initially at rest with a mass of mo. Ignore the
effects of surface tension, and the cart’s rolling friction and wind resistance in your answers.
a) Formulate dimensionless law for the mass, m(t), in the cart at time t in terms of t, θ, U, ρ, A,
and mo.
b) Formulate a differential equation for m(t).
c) Find a solution for m(t) and put it in dimensionless form.

Solution. 4.33. a) Create the parameter matrix

m mo t θ U ρ A
M 1 1 0 0 0 1 0
L 0 0 0 0 1 -3 2
T 0 0 1 0 -1 0 0
There will be: 7 – 3 = 4 dimensionless groups. By inspection these are:
m Ut ρ A3
Π1 = , Π2 = θ , Π3 = , and Π 4 = .
mo A mo
m $ Ut ρ A 3 '
Thus the dimensionless law is: = f &&θ, , ))
€ m o % A m o (
€ €
b) Choose a control volume that contains: i) the cart,€ii) the fluid in the cart, and iii) the tip of the
fluid jet that just hits the deflector plate. The vane is presumed to be entirely inside the CV. Let
b(t) be the cart's velocity in the positive x-direction.
€ dm
Conservation of mass: = ρ(U − b)A (1)
dt
d(mb)
Conservation of x-momentum: − ρU(U − b)A = 0 (2)
dt
Use (1) to find: b = U − (1 ρA)( dm € ( 2
dt ) , and by differentiation: db dt = −(1 ρA) d m dt ) 2

db dm
Expand the left-hand-side of (2), i.e. m + b − ρU(U − b)A = 0 , and eliminate b and db/dt
€ dt dt
using the relationships derived from (1):
€ € 2
$ 1 d 2m ' $ 1 dm ' dm 1 dm d 2 m " dm %
m&− + U−
2 ) & ) − ρU A = 0 , or m 2 + $ ' = 0
% ρA dt ( % € ρA dt ( dt ρA dt dt # dt &
d " dm % dm
c) The result of part a) can be written: $ m ' = 0 . Integrate once to find: m = C1.
dt # dt & dt
Integrate a second time and take the square root: m(t) = 2C
€ € 1t + C2 . Evaluate the constants

€ €
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

using the initial conditions: m(0) = mo, and dm/dt = ρUA at t = 0. The second initial condition
comes from the fact that b(0) = 0 (the cart starts from rest). With these conditions the constants
become: 2C1 = 2mo ρUA , and C2 = mo2 . Thus: m(t) = 2mo ρUAt + mo2 , or in dimensionless form
m(t) ρA 3 2 Ut
= 2 + 1 . The deflection angle of the vane doesn’t enter the solution.
mo mo A1 2
€ € €

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.34. A spherical balloon of mass Mb is filled with air of density ρ and is initially
stationary at x = 0 with diameter Do. At t = 0, an opening of area A is created and the balloon
travels horizontally along the x-axis. The aerodynamic drag force on the balloon is given by
(1/2)ρU2π(D/2)2CD, where: CD is a constant, U(t) is the velocity of the balloon, and D(t) is the
current diameter of the balloon. Assume incompressible air flow and that D(t) is known.
a) Find a differential equation for U that includes: Mb, ρ, CD, D, and A.
b) Solve the part a) equation when CD = 0, and the mass flow rate of air out of the balloon, m , is
π
 at time t.
constant, so that the mass of the balloon and its contents are M b + ρ Do3 − mt
6
c) What is the maximum value of U under the conditions of part b)?

A!
D!
x!

Solution 4.34. a) Enclose the balloon with an accelerating control volume that moves with
velocity b = U(t)ex. Denote the velocity of the air leaving the balloon by u = –Ueex + U(t)ex,
where Ue is the speed of the exiting air stream relative to the balloon and it appears with a minus
sign because it is oriented opposite to the x-direction.
For this control volume, the the total mass inside at any time is Mb + ρ(π/6)D3, and the
only place fluid crosses the control surface is at the balloon's outlet where n = –ex. Thus, from
d
(4.5), ∫ ρdV + ∫ ρ (u − b) ⋅ n dA = 0 , conservation of mass implies:
dt V*(t ) A*(t )

d! π 3$
# M b + ρ D & + ρ (−Ue +U −U ) (−1)A = 0 ,
dt " 6 %
which simplifies to:
π dD
ρ D2 + ρUe A = 0 .
2 dt
d
Similarly, from (4.17), ∫ ρudV + ∫ ρu (u − b) ⋅ ndA = ∫ ρgdV + ∫ fdA , conservation of x-
dt V*(t ) A*(t ) V*(t ) A*(t )

momentum implies:
2
d '! π 3$
* 1 2 !D$
)# M b + ρ D &U , + ρ (−Ue +U ) (−Ue +U −U ) (−1)A = − ρU π # & CD ,
dt (" 6 % + 2 "2%
which simplifies to:
2
d '! π 3$
* 1 2 !D$
#
) bM + ρ D &U , + ρ ( −U e +U ) U e A = − ρU π # & CD ,
dt (" 6 % + 2 "2%
Evaluate the derivative, and use the conservation of mass result to eliminate the Ue, the balloon-
exit speed to find:
2 2
! π 3 $ dU π 2 D 4 ! dD $ 1 2 !D$
# Mb + ρD & =ρ # & − ρU π # & CD .
" 6 % dt 4 A " dt % 2 "2%
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

b) When CD = 0, the final term is absent, and, for the specified mass flow rate, conservation of
mass implies
d! π 3$ π 2 dD
# Mb + ρD & = ρD = −m ,
dt " 6 % 2 dt
Thus, the part a) result for conservation of horizontal momentum simplifies to:
" π % dU m 2
M
$ b + ρ Do
3
− 
mt ' = .
# 6 & dt ρ A
This equation can be separated and integrated:
m 2 dt m "
∫ dU = ∫ or U + const. = −  $% .
ln # M b + (π / 6)ρ Do3 − mt
3
ρ A M b + (π / 6)ρ Do − mt  ρA
The constant can be evaluated using the initial condition U(0) = 0, to find:
m " M b + (π / 6)ρ Do3 %
U= ln $ '.
ρ A # M b + (π / 6)ρ Do3 − mt &
c) Here we see that U increases monotonically with time, so the highest speed occurs at the latest
time that the part b) answer applies. The mass flow from the balloon will stop when the balloon's
contents are exhausted. Denoting this time by tf, it occurs when (π / 6)ρ Do3 − mt  f = 0 . Thus, the
m ! M b + (π / 6)ρ Do3 $ ! (π / 6)ρ Do3 $
final (and highest balloon speed) is: U f = ln # & = Ue ln #1+ & , and
ρA " Mb % " Mb %
this may be substantially higher than Ue when (π / 6)ρ Do3 >> M b .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.35. For time t < 0, a rolling water tank with frictionless wheels, horizontal cross-
sectional area A, and empty mass M sits stationary while filled to a depth ho with water of density
ρ. At t = 0, the outlet of the tank is opened and the tank starts moving to the right. The outlet tube
has cross sectional area a and contains a narrow-passage honeycomb so that the flow speed
through the tube is Ue = gh/R, where R is the specific flow resistivity of the honeycomb material,
g is the acceleration of gravity, and h(t) is the average water depth in the rolling tank for t > 0.
Here, Ue is the leftward speed of the water with respect to the outlet tube; it is independent of the
speed b(t) of the rolling tank. Assume uniform flow at the pipe outlet and use an appropriate
control volume analysis for the following items.
a) By conserving mass, develop a single equation for h(t) in terms of a, A, g, R, and t.
b) Solve the part a) equation for h(t).
c) By conserving horizontal momentum, develop a single equation for b(t) in terms of a, A, M, h,
ρ, g, and R.
d) Determine for b(t) in terms of a, A, M, ho, ρ, g, R, and t. [Hint: use db/dt = (db/dh)(dh/dt)]

y!
g! b(t)!

h(t)!
Ue!
x!

Solution 4.35. a) Enclose the rolling cart within a moving control volume. The mass inside the
control volume is: M + ρAh, and b = b(t)e x so conservation of mass implies:
d d
∫
dt V*
ρ dV + ∫ ρ (u − b)⋅ ndA = 0 →
dt
( M + ρ Ah) + ρ aUe = 0 .
A*
Here flowing water only crosses the control surface at the tube outlet. At this location the water's
velocity is u = −Ue e x + be x and the outward normal is: n = −e x , so (u − b)⋅ n = −Ue e x ⋅ (−e x ) = Ue .
The area of this outlet is a, so the flux integral is simply +ρaUe. To find the equation for h(t),
complete the indicated differentiation and use Ue = gh/R to reach: dh dt = − ( ag AR) h .
b) This equation has the exponential solution: h(t) = ho exp {−agt AR} when h(0) = ho.
c) Use the same control volume and note that only horizontal momentum needs to be conserved.
In addition assume that the mass of water in the outlet pipe is negligible compared M + ρhA so
that the horizontal momentum in the control volume is simply (M + ρhA)b. With frictionless
wheels, there is no horizontal force on the control volume, so horizontal conservation of
momemtum implies:
d d
∫
dt V*
ρu dV + ∫ ρu(u − b)⋅ ndA = 0 →
dt
[(M + ρ Ah)b] + ρ a(–Ue + b)Ue = 0 .
A*

db ! d $
Expand the derivative & flux term to find: (M + ρ Ah) + b # (M + ρ Ah) + ρ aUe & − ρ aUe2 = 0 .
dt " dt %
Cons. of mass requires the terms in [,]-brackets to be zero, so this equation simplifies to:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

2
db ! gh $
(M + ρ Ah) = ρ a # & , where Ue = gh/R has been used to eliminate Ue.
dt "R%
d) Use the hint, and the cons. of mass finding for dh/dt to rewrite the differential equation for b:
2
db dh db " agh % 2 " gh %
(M + ρ Ah) = (M + ρ Ah) $ − ' = ρ aU e = ρ a $ ' .
dh dt dh # AR & #R&
The second and fourth terms of this extended equality simplify to:
db g g ρ Ah g g M
(M + ρ Ah) = − ρ Ah , or ∫ db = − ∫ dh = − ∫ dh + ∫ dh .
dh R R M + ρ Ah R R M + ρ Ah
Perform the integrals and evaluate the integration constant using b(ho) = 0 to reach:
) gM " M + ρ Aho e (
− ag AR)t %
gh (1− e (
− ag AR)t
g(ho − h) gM " M + ρ Ah %
b(h) = + ln $ ' , or b(t) = o + ln $$ '' ,
R ρ AR # M + ρ Aho & R ρ AR # M + ρ Aho &
where the final result comes from substituting in the part a) result for h(t).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.36. Prove that the stress tensor is symmetric by considering first order changes in
surface forces on a vanishingly small cube in rotational equilibrium. Work with rotation about
the no. 3 coordinate axis to show T12 = T21. Cyclic permutation of the indices will suffice for
showing the symmetry of the other two shear stresses.
x2!
dx2 ∂T21
T21 +
2 ∂x1
dx1!
dx1 ∂T12 dx2! T − dx1 ∂T12
T12 − 12
2 ∂x1 centroid! 2 ∂x1
axis!

dx2 ∂T21
T21 −
2 ∂x1
x1!

Solution 4.36. Consider a small cubical volume dV = dx1dx2dx3. All of the surface tractions
lying in the x1-x2 plane act to rotate or counter rotate this cube around the x3 axis. The horizontal
and vertical moment arms from cube's center to each side are dx1/2 or dx2/2, respectively.

Rotational equilibrium about an axis through the cube's center parallel to the x3 axis requires:
! ∂ T dx $ dx ! ∂ T dx $ dx
+ #Y12 + 12 1 & ( dx2 dx3 ) 1 + #T12 − 12 1 & ( dx2 dx3 ) 1
" ∂ x1 2 % 2 " ∂ x1 2 % 2
" ∂ T dx % dx " ∂ T dx % dx
− $T21 + 21 2 ' ( dx1dx3 ) 2 − $T21 − 21 2 ' ( dx1dx3 ) 2 + higher order terms =
# ∂ x2 2 & 2 # ∂ x2 2 & 2
ρ 2 2 dΩ

12
{
dx1dx 2 dx 3 ( dx1 ) + ( dx 2 )} dt
where Ω is the rotation rate. Here, the first 4 terms are written as triple products, (surface
stress)(surface area)(moment arm). The final term is the product of the moment of inertia about
the x3-axis and the rotational acceleration. Canceling terms and dividing by dV produces:
€
ρ 2 2 dΩ
T12 − T21 =
12
dx{
( 1 ) + ( dx2 )
dt
}.
Taking dx1,dx2 → 0, produces τ12 = τ 21. Cyclic permutation of the indices can be used to show
the equality of the other two shear stresses: T23 = T32, and T31 = T13.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.37. Obtain an empty plastic milk jug with a cap that seals tightly, and a frying pan.
Fill both the pan and jug with water to a depth of approximately 1 cm. Place the jug in the pan
with the cap off. Place the pan on a stove and turn up the heat until the water in the frying pan
boils vigorously for a few minutes. Turn the stove off, and quickly put the cap tightly on the jug.
Avoid spilling or splashing hot water on yourself. Remove the capped jug from the frying pan
and let it cool to room temperature. Does anything interesting happen? If something does
happen, explain your observations in terms of surface forces. What is the origin of these surface
forces? Can you make any quantitative predictions about what happens?

Solution 4.37. When the sealed milk jug is allowed to cool, its volume shrinks. This occurs
because the water vapor inside the jug provides an outward-pushing surface pressure of only 3
kPa at room temperature, while the atmosphere in your kitchen provides an inward-pushing
surface force of ~1 atm (101.3 kPa). The plastic of the milk jug is not strong enough to support
the inward-pushing load, so the jug ends up being "crushed" by atmospheric pressure. This
simple experiment demonstrates that pressure produces surface forces that point into the control
volume represented here by the milk jug (opposite the outward normal).
For the experiments conducted in third author's kitchen, the volume of the milk jug was
reduced by about half. This volume change cannot be accounted for by the perfect gas law since
the temperature change from boiling water (373K) to room conditions (298K) is too small to
account for the actual decrease in jug volume. If the jug lacked any strength (like a plastic bag)
and it was perfectly filled only with water vapor when it was removed from the boiling water in
the frying pan, its volume would decrease by a factor of approximately 30 to 40 while it cooled.
The observed volume reduction is not this severe, and is mitigated by the actual strength of the
jug, the air trapped in the jug when it is sealed, and any air that leaks into the jug interior while it
cools. The trapped air does not condense and its presence produces a pressure that adds to the
vapor pressure of water in the jug.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.38. In cylindrical coordinates (R,ϕ,z), two components of a steady incompressible

viscous flow field are known: uϕ = 0 , and uz = −Az where A is a constant, and body force is zero.
a) Determine uR so that the flow field is smooth and conserves mass.
b) If the pressure, p, at the origin of coordinates is Po, determine p(R,ϕ,z) when the density is
constant. €
€
Solution 4.38. a) Incompressible flow implies:
∇ ⋅ u = 0 = (1 R)(∂RuR ∂R) + (1 R)(∂uϕ ∂ϕ ) + ∂uz ∂z (in cylindrical coords).
Substitute in the given uϕ and uz to find: ∂ (RuR ) ∂R − RA = 0 . Integrate in R to get:
uR = AR 2 + B R .
The constant B must be zero for the flow field to be smooth as R → 0 , so
€ uR = AR 2 .
€
b) Convert to Cartesian coordinates. With uϕ = 0 : u = uR cosϕ = Ax 2 and v = uR sin ϕ = Ay 2 .
€
A calculation of the vorticity produces: ∇ × u = 0 , so€the steady Bernoulli equation may be used
to get the pressure:
€
( ) ( )
ρ uR2 + u€z2 → p(R, ϕ, z) = Po −€12 ρ A 2 14 R 2 + z 2
Po = p(R, ϕ, z) + 12 €
€
because u = 0 at the origin.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.39. Consider solid-body rotation of an isothermal perfect gas (with constant R) at
temperature T in (r,θ)-plane-polar coordinates: ur = 0, and uθ = Ωz r , where Ωz is a constant
rotation rate and the body force is zero. What is the pressure distribution p(r) if p(0) = patm? If the
gas is air at 295 K and the container has a radius of ro = 10 cm, what Ωz is needed to produce
p(ro) = 2patm? Ignore gravity.

Solution 4.39. For steady solid body rotation, the unsteady fluid acceleration will be zero and the
viscous stress in the fluid will be zero. In plane polar coordinates, the radial momentum equation
is under these conditions is
u2 1 ∂p
− θ =− .
r ρ ∂r
(see Appendix B.6). Here, both ρ and p can vary, but the flow is presumed isothermal so ρ =
p/RT. Thus, with uθ = Ωz r , the steady radial momentum equation is:
2
2 RT ∂ p 2 r
Ω r=− z , which integrates to: Ωz + const. = RT ln( p) .
p ∂r 2
Exponentiate and evaluate the constant using p(0) = patm to find:
" Ω2 r 2 %
p(r) = patm exp # z & .
$ 2RT '
If the gas is air at 295 K and the container has a radius of ro = 10 cm, then the Ωz needed to
produce p(ro) = 2patm is given by:
" Ω2z r 2 % 2RT ln(2) 2(287)(295)ln(2)
2 = exp # & or Ωz = = = 3426rad / s .
$ 2RT ' r 0.10
This is almost 33,000 rpm.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.40. Solid body rotation with a constant angular velocity, Ω, is described by the
following Cartesian velocity field: u = Ω × x . For this velocity field:
(" ∂ u ∂ u % 2 ∂ u + ∂u
a) Compute the components of: Tij = − pδij + µ *$$ i + j '' − δij k - + µυδij k .
*)# ∂ x j ∂ xi & 3 ∂ xk -, ∂ xk
€
b) Consider the case of Ω1 = Ω2 = 0 , Ω3 ≠ 0 , with p = po at x1 = x2 = 0. Use the differential
momentum equation in Cartesian coordinates to determine p(r), where r 2 = x12 + x 22 , when there
is no body force and ρ = constant. Does your answer make sense? Can you check it with a
simple experiment?
€ €
Solution 4.40. a) The components of the velocity field u = Ω€ × x are u1 = Ω2 x 3 − Ω3 x 2 ,
u2 = Ω3 x1 − Ω1 x 3 , and u3 = Ω1 x 2 − Ω2 x1. The strain rate tensor is for this velocity field is:
) ∂u1 ∂u1 ∂u2 ∂u1 ∂u3 -
+ 2 + € + +
+ ∂ x1 ∂ x 2 ∂ x1 ∂ x 3 ∂x1 +€ ) 0 +Ω3 +Ω2 − Ω2 -
€ 1 # ∂ui € ∂u j & 1 + ∂u2 ∂u1 ∂ u2 ∂u2 ∂u3 + + +
Sij = %% + (( = * + 2 + . = *+Ω3 − Ω3 0 −Ω1 + Ω1 .
2 $ ∂x j ∂x i ' 2 + ∂x1 ∂x 2 ∂x 2 ∂x 3 ∂ x 2 + + +
∂u3 + ,−Ω2 + Ω2 +Ω1 − Ω1 0 /
+ ∂u3 ∂u1 ∂u3 ∂u2
+ + 2
+, ∂x ∂x ∂x ∂x ∂x +/
1 3 2 3 3
"0 0 0&
$ $
= #0 0 0' ,
$0 0 0$
€ % (
and this also shows that ∂uk/∂xk = 0. Thus, all the viscous stress components of Tij are zero, so
Tij = –pδij.
b) When Ω1 = Ω2 = 0 , with Ω3 = const., the fluid velocity is u = Ω3eϕ in cylindrical coordinates
€
with x3 = z. In spite of this simplification, it's even simpler to continue with Cartesian
coordinates. The switch to cylindrical coordinates can be made at the end. Use Cauchy's form
of the momentum equation (4.24)
€ €
∂uj ∂ u j € 1 ∂ Tij
+ ui = gj + .
∂t ∂ xi ρ ∂ xi
The flow is steady, ∂u j ∂t = 0 , and there’s no body force, gj = 0. Combine these facts with the
result of part a) to find:
∂u 1 ∂p
ui j = − .
€ ∂ x i ρ ∂ x j
Evaluate the left side of this equation using the given velocity field:
' ∂ ∂ * ∂p ∂p
ρ(−Ω3 x 2e1 + Ω3 xe 2 ) ⋅ ) e1 + e2 ,(−Ω3 x 2e1 + Ω3 x1e 2 ) = −e1 − e2 .
€ ( ∂x1 ∂x 2 + ∂x1 ∂x 2
This reduces to:
∂p ∂p ∂p
−ρΩ32 x1e1 − ρΩ32 x2 e 2 = e1 − e2 − e3 .
∂ x1 ∂ x2 ∂ x3
€
Considering each component separately along with the boundary condition on x1 = x2 = 0, yields:
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

ρΩ23 2 ρΩ23 2
p(x1, x 2 ) = po +
2
( 1 2)
x + x 2
, or in cylindrical coordinates p(R) = p o +
2
R .
This answer makes sense because the necessary centripetal acceleration of the rotating fluid must
be provided by pressure forces. So, increasing pressure with increasing R is physically
meaningful, even when viewed from a rotating dynamics point of view. This result can be
€ checked with the swirling flow of a liquid in a cylindrical container
€ (see Section 5.1). The height
of the free surface is proportional to the pressure in the fluid, and the free surface of a liquid in
solid body rotation is parabolic. This effect was utilized by astronomers to cast telescope mirror
blanks on rotating tables in the early part of the 1900’s. The radial pressure gradient found
above is also used to separate constituents in fluid mixtures in a centrifuge, a technique
employed in the production of weapons-grade uranium through the separation of isotopes of UF6
(a gas), and in the biological sciences to obtain (the denser) nucleic material from cells.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.41. Using only (4.7), (4.22), (4.36), and (3.12) show that
Du
ρ
Dt
(
+ ∇p = ρg + µ∇ 2u + µυ + 13 µ ∇ (∇ ⋅ u) )
when the dynamic (µ) and bulk (µ ) viscosities are constants.
υ

Solution 4.41. Start by expanding the left side of the momentum equation (4.22), then regroup
the terms. €
∂ ∂ ∂u ∂ρ ∂u ∂
( ρu j ) + ( ρui u j ) = ρ j + u j + ρui j + u j (ρui ) ,
∂t ∂x i ∂t ∂t ∂x i ∂x i
∂u ∂u $ ∂ρ ∂ ' ∂u ∂u
= ρ j + ρui j + u j & + ( ρui )) = ρ j + ρui j
∂t ∂x i % ∂t ∂x i ( ∂t ∂x i
€ The terms inside the large parentheses on the second line are zero because of the continuity
equation (4.7). Now work on the right side of the momentum equation (4.22). Substituting in the
definition of Tij from (4.36) produces:
€ ∂T ∂ ( " 1 % +
ρ g j + ij = ρ g j + *− pδij + 2µ $ Sij − Skkδij ' + µυ Skkδij - .
∂ xi ∂ xi ) # 3 & ,
Now use (3.12), the definition of Sij, to find:
∂T ∂ ( " ∂u ∂u 2 ∂u % ∂u +
ρ g j + ij = ρ g j + *− pδij + µ $$ i + j − k
δij '' + µυ k δij - .
∂ xi ∂ xi *) # ∂ x j ∂ xi 3 ∂ x k & ∂ xk -,
With the summation convention and the definition of the Kronecker δ-function,
∂ ∂( ) ∂( )
∂x i
[( )δij ] = δij =
∂x i ∂x j
where ( ) represents any variable or combination of variables. Using this and distributing ∂ ∂x i
over the terms in [,]-braces yields:
∂τ ij € ∂p ∂ + % ∂ui ∂u j (. ∂ + ∂uk .
∂x i
=− + -µ'' +
∂x j ∂x i -, & ∂x j ∂x i )0/ ∂x j ,
(
**0 + )
- µυ − 3 µ
2
0.
∂x k / €
Combining the modified forms of the right and left sides of (4.22) leads to:
∂u j ∂u j ∂p ∂ + % ∂ui ∂u j (. ∂ + ∂uk .
€ ∂t
ρ + ρui
∂x i
= ρg j − + -µ'' +
∂x j ∂x i -, & ∂x j ∂x i )0/ ∂x j ,
(**0 + ) 2
- µυ − 3 µ 0.
∂x k /
When the dynamic viscosity µ and the bulk viscosity µυ are constants; they can be brought
outside the ∂ ∂x i -differentiations, i.e.
€ ∂u ∂u ∂p ∂ % ∂ui ∂u j ( ∂ ,∂uk /
ρ j + ρui j = ρg j −
∂t ∂x i ∂x j
+µ '' +(
∂ x i & ∂x j ∂ x i )
)
** + µυ − 23 µ . 1
∂x j -∂x k 0
To€convert from index to vector notation, the following definitions and replacements apply:
∂ ∂ #∂ & 2
u j = u , g j = g, = ∇, % (=∇ ,
€ ∂ x j ∂ x i$ ∂ x i'

∂ # ∂ & ∂ # ∂ & ∂ ∂ D
% (= %% (( = ∇∇ (no dot product), and + ui = .
∂x j $ ∂ x i ' ∂x i $ ∂x j ' ∂t ∂x i Dt
€ €
€ €

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Whenever an index is repeated, a summation over that index (i.e. a vector-notation dot product)
is implied. Making term for term replacements in the last version of the momentum equation
produces:
∂u
( )
ρ + ρ(u ⋅ ∇)u = ρg − ∇p + µ∇(∇ ⋅ u) + µ(∇ ⋅ ∇ )u + µυ − 23 µ ∇(∇ ⋅ u)
∂t
Combining terms, moving the pressure gradient to the other side of the equation, and invoking
the definition of D/Dt produces the intended final result:
Du
€ ρ
Dt
( )
+ ∇p = ρg + µ∇ 2u + µυ + 13 µ ∇ (∇ ⋅ u) .

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

4.42. Air, water, and petroleum products are important engineering fluids and can usually be
treated as Newtonian fluids. Consider the following materials and try to classify them as:
Newtonian fluid, non-Newtonian fluid, or solid. State the reasons for your choices and note the
temperature range where you believe your answers are correct. Simple impact, tensile, and shear
experiments in your kitchen or bathroom are recommended. Test and discuss at least five items.
a) toothpaste d) glass g) hot oatmeal j) silly putty
b) peanut butter e) honey h) creamy salad dressing
c) shampoo f) mozzarella cheese i) ice cream

Solution 4.42. One simple way to conduct

rheological tests is with two ordinary plastic
cups that are meant to stack inside each
other. Coat the outside of one of the cups
with liquid or material of interest. Then slide this cup into the second one to make a crude
Couette rheometer. Shear flow of the material of interest can be produced using one's hands to
provide a torque difference and differential rotation of the cups.
The following information is based on such 2-cup shear tests, simple observations, or a
little reading about materials. Unless otherwise noted, all of the following information applies at
room temperature. The results you obtain may be different if, for example, materials like peanut
butter or shampoo are put in a freezer before conducting shear tests.
a) Toothpaste appears to be a viscous Newtonian fluid. (a blue gel was tested)
b) Creamy peanut butter is a Newtonian fluid. However, its viscosity decreases rapidly with
increasing temperature.
c) Shampoos are typically non-Newtonian liquids. In two cup shear tests it can feel like there is
a torsional spring connecting the two cups for very small angular deflections. In addition, rising
air bubbles in shampoo have pointed tails and this indicates there is memory within the fluid.
Moving bubbles in Newtonian fluids do not have pointed tails.
d) Glass is an elastic amorphous solid. Its material properties are continuous functions of
temperature. When it is heated, it softens and becomes a non-Newtonian liquid or a visco-elastic
solid. At high temperatures it becomes a Newtonian liquid.
e) Honey is a viscous Newtonian liquid, at least in the two-cup shear tests.
f) Mozzarella cheese is an elastic solid at room temperature. When heated, it becomes a non-
Newtonian liquid (note the stretchy cheese in pizza advertisements).
g) Oatmeal is not really a continuum material. Hot off the stove, it is approximately a Newtonian
liquid. When it cools, it becomes non-Newtonian. When it is allowed to dry out, it turns into a
solid material that is a lot like cement.
h) Creamy salad dressing may be either a non-Newtonian fluid or a viscous Newtonian fluid,
depending on the brand and the temperature. It’s shear-thinning properties are easily observed
when it is poured from a bottle. The first slug of comes out in big lump but the stream that
follows the slug can be much thinner.
i) Ice cream is effectively a solid when you first get it out of the freezer. At room temperature it
becomes a Newtonian liquid. In between, it may be non-Newtonian depending on its
composition.
j) Silly putty can be considered a visco-elastic solid (when formed into a ball, the ball bounces)
or a non-Newtonian liquid (when a ball of silly putty is left alone on a flat surface, gravity
compresses it and it begins to flow horizontally).
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.43. The equations for conservation of mass and momentum for a viscous Newtonian
fluid are (4.7) and (4.39a) when the viscosities are constant.
a) Simplify these equations and write them out in primitive form for steady constant-density flow
in two dimensions where ui = ( u1 (x1, x 2 ),u2 (x1, x 2 ),0) , p = p(x1, x 2 ) , and gj = 0.
b) Determine p = p(x1, x 2 ) when u1 = Cx1 and u2 = −Cx 2 , where C is a positive constant.

Solution 4.43. a) There are a number of simplifications to make. For steady flow ∂ ∂t → 0 ; for
€
constant density€flow ∇ρ = 0 ; for 2D flow ∂ ∂x 3 → 0 and u3 = 0; and gj = 0 implies g1 = g2 = 0.
€ € €
Thus, the continuity equation is simplified as follows:
steady exp and ρ = const≠0
∂ρ ∂ ∂ ∂u ∂ρ ∂ui 2D ∂u ∂u
+ ( ρui ) = 0 → (ρui ) = 0 → ρ i + ui = 0 → = 0€ → 1 + 2 = 0 .
∂t ∂x i ∂x i ∂x i ∂x i ∂x i ∂x1 ∂x 2
€ €
Following a similar procedure and using this result, the simplified momentum equation is:
∂u j ∂p ∂ 2u j
ρui =− +µ 2 .
€ ∂x i ∂x j ∂x i
The components of this equation are obtained by setting j = 1 and 2:
∂u1 ∂u1 ∂p % ∂ 2 u1 ∂ 2 u1 ( ∂u2 ∂u2 ∂p % ∂ 2 u2 ∂ 2 u2 (
ρu1 + ρu2 =− + µ' 2 + 2 * , and ρu1 + ρu2 =− + µ' 2 + 2 * .
∂x1 ∂x 2 ∂x€1 & ∂x1 ∂x 2 ) ∂x1 ∂x 2 ∂x 2 & ∂x1 ∂x 2 )
b) With u1 = Cx1 and u2 = −Cx 2 , the simplified continuity equation is satisfied. The two
momentum equations become:
∂p ∂p
€ ρC 2 x1 = −€ , and ρC 2 x 2 = − .
∂x1 ∂x 2
€ €
Integrate both to find: p = −ρC 2 x12 2 + f (x 2 ) , and p = −ρC 2 x 22 2 + g(x1 ) , where f and g are
single-variable functions. These two equations for p are consistent when:
2 2
f (x 2 ) = −C x 2 2 + const€ , and g(x1 ) = −C
2 2
( 2
€ x1 2 + const . Thus, p(x1, x 2 ) = const − C x1 + x 2 2 ,
2
) 2

where the constant € is readily determined as the € pressure at the origin of coordinates; thus
( )
p(x1, x 2 ) = p(0,0) − C 2 x12 + x 22 2 .
€ € €

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.44. Simplify the planar Navier-Stokes momentum equations (given in Example 4.9)
for incompressible flow, constant viscosity, and conservative body forces. Cross differentiate
these equations and eliminate the pressure to find a single equation for ωz = ∂v/∂x – ∂u/∂y. What
process(es) might lead to the changes in ωz for fluid elements in this flow?

Solution 4.44. The starting-point equations are:

!∂u ∂u ∂u $ ∂ p ∂ ( ∂u ! 2 $! ∂ u ∂ v $+ ∂ ( ! ∂ v ∂ u $+
ρ # + u + v & = − + ρ gx + *2µ + # µυ − µ &# + &- + *µ # + &- , and
" ∂t ∂x ∂y% ∂x ∂x ) ∂x " 3 %" ∂ x ∂ y %, ∂ y ) " ∂ x ∂ y %,
!∂v ∂v ∂v $ ∂ p ∂ ( ! ∂ u ∂ v $+ ∂ ( ∂ v ! 2 $! ∂ u ∂ v $+
ρ # + u + v & = − + ρ gy + *µ # + &- + *2µ + # µυ − µ &# + &-
" ∂t ∂x ∂y% ∂y ∂ x ) " ∂ y ∂ x %, ∂ y ) ∂ y " 3 %" ∂ x ∂ y %,
In two-dimensional incompressible flow, ∂u/∂x + ∂v/∂y = 0. Thus, these equations simplify to:
!∂u ∂u ∂u $ ∂ p ∂ ( ∂ u + ∂ ( ! ∂ v ∂ u $+
ρ # + u + v & = − + ρ gx + *2µ - + *µ # + &- , and
" ∂t ∂x ∂y% ∂x ∂ x ) ∂ x , ∂ y ) " ∂ x ∂ y %,
!∂v ∂v ∂v $ ∂ p ∂ ( ! ∂ u ∂ v $+ ∂ ( ∂ v +
ρ # + u + v & = − + ρ gy + *µ # + &- + *2µ -
" ∂t ∂x ∂y% ∂y ∂ x ) " ∂ y ∂ x %, ∂ y ) ∂ y ,
When the body force is conservative, g = −∇Φ , where Φ if the body-force potential (a scalar
function). When the viscosity is constant the terms on the right may be rearranged and simplified
with the incompressibility requirement, ∂u/∂x + ∂v/∂y = 0:
!∂u ∂u ∂u $ ∂ p ∂Φ ∂ 2u ∂ ! ∂ u ∂ v $ ∂ 2u
ρ# +u + v & = − − ρ + µ 2 + µ # + &+ µ 2
" ∂t ∂x ∂y% ∂x ∂x ∂x ∂ x "∂ x ∂ y % ∂ y
∂p ∂Φ ! ∂ 2u ∂ 2u $
=− −ρ + µ # 2 + 2 &,
∂x ∂x "∂ x ∂ y %
!∂v ∂v ∂v $ ∂ p ∂Φ ∂ 2v ∂ ! ∂u ∂ v $ ∂ 2v
ρ# +u + v & = − − ρ + µ 2 + µ # + &+ µ 2
" ∂t ∂x ∂y% ∂y ∂y ∂x ∂ y "∂ x ∂ y % ∂ y
∂p ∂Φ ! ∂ 2v ∂ 2v $
=− −ρ + µ # 2 + 2 &.
∂y ∂y "∂ x ∂ y %
Now divide each equation by ρ, apply –∂/∂y to the first equation, apply ∂/∂x to the second
equation and add the equations to together. This causes the body force terms to cancel out.
∂ "∂u ∂v ∂v % ∂ "∂v ∂u ∂u %
$ +u + v '− $ +u + v '
∂x # ∂ t ∂ x ∂ y & ∂y # ∂ t ∂x ∂y&
∂ " 1 ∂p % ∂ " 1 ∂p % ∂ ) 1 " ∂ 2 v ∂ 2 v %, ∂ ) 1 " ∂ 2 u ∂ 2 u %,
= $− ' − $ − ' + µ + $ + '. − µ + $ + '..
∂x # ρ ∂y & ∂y # ρ ∂x & ∂x * ρ # ∂ x 2 ∂ y 2 &- ∂y * ρ # ∂ x 2 ∂ y 2 &-
Expand derivatives on both sides of the equation and use the incompressible flow requirement
and definition ωz = ∂v/∂x – ∂u/∂y to consolidate terms.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

"∂ ∂ ∂%
$ + u + v 'ω z
#∂t ∂x ∂y &
∂ " 1 % ∂p ∂ " 1 % ∂p µ " ∂ 2 ∂ 2 % ∂ " 1 %" ∂ 2 v ∂ 2 v % ∂ " 1 %" ∂ 2u ∂ 2u %
=− $ ' + $ ' + $ + ω
' z + µ $ '$ + ' − µ $ '$ + '.
∂x # ρ & ∂y ∂y # ρ & ∂x ρ # ∂ x 2 ∂ y 2 & ∂x # ρ &# ∂ x 2 ∂ y 2 & ∂y # ρ &# ∂ x 2 ∂ y 2 &
Here we can recognize the term on the left as Dωz/Dt, the time rate of change of vorticity
following a fluid particle. The first two terms on the right are the two-dimensional outcome of
−∇ (1 ρ ) × ∇p , and these represent a torque applied to fluid elements because of mis-alignment
of their density and pressure gradients (aka the baroclinic torque). The third term on the right
represents diffusion of vorticity. The final two terms on the right are the two dimensional
outcome of −µ∇ (1 ρ ) × ∇ 2 u , and these represent a torque applied to fluid elements because of
mis-alignment of the density gradient and net viscous stress.
When the density is constant and the viscosity is zero, then Dωz/Dt = 0 and the vorticity
of fluid elements is constant.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.45. Starting from (4.7) and (4.39b), derive a Poisson equation for the pressure, p, by
taking the divergence of the constant-density momentum equation. [In other words, find an
equation where ∂ 2 p ∂x 2j appears by itself on the left side and other terms not involving p appear
on the right side]. What role does the viscosity µ play in determining the pressure in constant
density flow?
€
Solution 4.45. For constant density flow the continuity and momentum equations are:
2
∂ui ∂u j ∂u 1 ∂p µ ∂ uj
= 0 and + ui j = − + gj + .
∂x i ∂t ∂x i ρ ∂x j ρ ∂x i2
Apply ∂ ∂x j to the momentum equation and swap the order of differentiation in the first and last
terms to find:
€ ∂ #€ ∂ u & ∂ # ∂u j & 1 ∂ 2 p ∂g j µ ∂ 2 ∂u j
%% j (( + u
% i ( = − + + .
€ ∂t $ ∂ x j ' ∂x j $ ∂x i ' ρ ∂x 2j ∂x j ρ ∂x i2 ∂x j
The constant density flow continuity equation, ∂u j ∂x j = 0 , allows first and last terms to be set
to zero. Expand the second term on the left to find:
€ ∂ ∂u j ∂ui ∂u j 1 ∂ 2 p ∂g j
ui + =− + .
∂x i ∂x€j ∂x j ∂x i ρ ∂x 2j ∂x j
Here again the first term is zero for constant density flow, so the final Poisson equation for the
pressure is:
∂2 p % ∂u (% ∂u ( ∂g
€ 2
= − ρ'' i **' j * + ρ j .
∂x j & ∂x j )& ∂x i ) ∂x j
Interestingly, the viscosity is does not appear in this result! Thus, µ can influence p if i) µ
influences ∂u j ∂x i in the region of interest, and/or if ii) µ in enters in a factor that is linear in the
independent spatial variables.
€ Both of these possibilities occur in flows of interest to scientists
and engineers.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.46. Prove the equality of the two ends of (4.40) without leaving index notation or
using vector identities.

Solution 4.46. Equation (4.40) holds for incompressible flow where ∂ui/∂xi = 0. Therefore, start
from the first equality of (4.40), and subtract µ(∂/∂xj) (∂ui/∂xi) = 0 to start building something that
looks like a cross product.
∂ 2u j ∂ 2u j ∂ 2 ui ∂ % ∂ui ∂u j (
( ) j ∂ x 2 ∂ x 2 ∂x ∂x
µ ∇ 2
u = µ = µ − µ = −µ ' − *
∂x i '& ∂x j ∂x i *)
i i j i

The terms in large parentheses on the right are the rotation tensor Rij from (3.13). Thus,
∂ ∂ ∂
(µ∇ 2u) j = −µ ∂x (−εijkω k ) = µεijk ∂x ω k = −µε jik ∂x ω k = −µ(∇ × ω ) j ,
i i i
€
where (3.15), Rij = –εijkωk has been used.
Alternatively, start with the curl of the vorticity and expand it to show both cross
products in terms of the alternating tensor:
€
∂ ∂ ( ∂ + ∂ 2 un
−µ(∇ × ω ) j = −µε jkl ω l = −µε jkl *εlmn un - = −µε jklεlmn .
∂x k ∂x k ) ∂ x m , ∂x k∂x m
Now use epsilon-delta relation (2.19) for the product εijkεlmn, to find:
∂ 2 un ( ∂ 2u ∂ 2u j + ∂ 2u j
−µ(∇ × ω ) j = −µ(δ jmδkn − δ jnδkm ) = −µ** k
− -- = µ 2 = (µ∇ 2u) j .
€ ∂x k∂x m ) ∂x k∂x j ∂x k∂x k , ∂x k

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.47. The viscous compressible fluid conservation equations for mass and momentum
are (4.7) and (4.38). Simplify these equations for constant-density constant-viscosity flow and
where the body force has a potential, g j = − ∂Φ ∂x j . Assume the velocity field can be found
from u j = ∂φ ∂x j , where the scalar function φ depends on space and time. What are the
simplified conservation of mass and momentum equations for φ?
€
Solution 4.47. For constant density, the continuity equation simplifies:
€ ∂ρ ∂ ∂ρ ∂ρ ∂u ∂u ∂u
0= + ( ρui ) = + ui + ρ i = ρ i , or i = 0 .
∂t ∂x i ∂t ∂x i ∂x i ∂x i ∂x i
For the momentum equation with constant density and constant viscosities, the viscous stress
terms are altered:
∂ ) #€∂ui ∂u j &, ∂ )# 2 & ∂ui , )# ∂ 2 ui ∂ 2 u€j &, # 2 & ∂ )∂ui , ∂ 2u j
+µ%% + .
( + +%µυ − µ( . = µ %% + + .
( + % µυ − µ( + .=µ 2 ,
∂x i +* $ ∂x j ∂x i ('.- ∂x j *$ 3 ' ∂x i - +*$ ∂x i∂x j ∂x i2 ('.- $ 3 ' ∂x j *∂x i - ∂x i
where the continuity-equation result has been used to eliminate the first and third terms in the
middle portion of the extended equality. Thus the momentum equation simplifies to:
∂u j ∂u j ∂p ∂ 2u j
€ ρ + ρui =− + ρg j + µ 2
∂t ∂x i ∂x j ∂x i
Direct substitution of ui = ∂φ ∂x i into the simplified continuity equ. produces: ∂ 2φ ∂x i2 = 0 .
∂ ∂φ ∂φ ∂ ∂φ ∂p ∂ 2 ∂φ
Similarly, the momentum equation becomes: ρ +ρ =− + ρg j + µ 2 .
€ ∂t ∂x j ∂x i ∂ x i ∂x j ∂x j ∂x i ∂x j
2
∂φ€ ∂ ∂φ ∂φ ∂ ∂φ 1 ∂ $ ∂φ ' €
Note that = = & ) , rearrange terms, and insert g j = − ∂Φ ∂x j :
∂x i ∂x i ∂x j ∂ x i ∂x j ∂x i 2 ∂x j % ∂x i (
% € 2(
∂ ' ∂φ ρ % ∂φ ( * ∂ ∂ ∂ 2φ
ρ + ' * = − ( p + ρΦ) + µ .
∂x j '& ∂t 2 & ∂x i ) *) ∂x j ∂x € 2
j ∂x i
€
The final term is zero from the continuity equation, put everything on one side, pull out ∂ ∂x j ,
$ 2 '
∂ & ∂φ 1 $ ∂φ ' p
and divide by ρ: + + + Φ)) = 0 . Thus, the simplified equations are
€ ∂x j & ∂t 2 &% ∂x i )( ρ
% (
2 €
∂ 2φ ∂φ 1 $ ∂φ ' p
= 0 , and + & ) + + Φ = f (t) ,
∂x i2 ∂t 2 % ∂ x i ( ρ
€ the Laplace equation and an unsteady Bernoulli equation. Here, f(t) is an undetermined
which is
function of time that may be set by boundary conditions on φ and p. These are the equations for
unsteady potential flow, and can be used to solve fluid flow problems when: i) the fluid's density
€ € is constant, iii) the flow is irrotational, and iv) only a no
is constant, ii) the fluid's viscosity
through-flow boundary condition is required on solid surfaces. The no slip boundary condition
cannot be satisfied (in general) with these equations.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.48. The viscous compressible fluid conservation equations for mass and momentum
are (4.7) and (4.38).
a) In Cartesian coordinates (x,y,z) with g = ( gx , 0, 0 ) , simplify these equations for unsteady one-
dimensional unidirectional flow where: ρ = ρ (x, t) and u = (u(x, t), 0, 0 ) .
b) If the flow is also incompressible, show that the fluid velocity depends only on time, i.e.,
u(x, t) = U(t) , and show that the equations found for part a) reduce to
∂ρ ∂ρ ∂u ∂p
+ u = 0 , and ρ = − + ρ gx .
∂t ∂x ∂t ∂x
c) If ρ = ρo (x) at t = 0, and u = U(0) = Uo at t = 0, determine implicit solutions for ρ = ρ (x, t)
and U(t) in terms of x, t, ρo (x) , Uo, ∂ p ∂ x , and gx.

Solution 4.48. a) For one-dimensional flow: ∂ ∂y = ∂ ∂z = 0 = v = w , so the continuity equation

reduces to:
∂ρ ∂ρ ∂u
+ u + ρ = 0.
∂t ∂x ∂x
Similarly, the Navier-Stokes momentum € equation in the x-direction becomes:
∂u ∂u ∂p ∂ 2u
(
ρ + ρu = − + ρgx + µυ + 43 µ 2 . )
€ ∂t ∂x ∂x ∂x
b) Incompressible flow means ∇ ⋅ u = 0 . Here this implies: ∂u ∂x = 0 , thus the equations for part
a) simplify to:
∂ρ ∂ρ ∂u ∂p
€ + u = 0 , and ρ = − + ρgx .
€ ∂t ∂x ∂t ∂x
€
c) For ∂u ∂x = 0 , u(x,t) = U(t) . Therefore, start with what’s left of the continuity equation and
look for a method-of-characteristics solution, i.e. set ρ = ρ( x(t),t ) so that
€ dρ( x(t),t
€ ) = ∂ρ + $& dx ') ∂ρ = 0 ,
€ € dt ∂t % dt ( ∂x
t
where the characteristic curves are specified € by: dx dt = u(x,t) = U(t) , or x = ∫ o Udt + x o . The
equation above requires ρ to be constant along these curves, so if ρ(x o ,0) = ρ o (x o ) , then
ρ(x,t) = ρ o x − ∫ o Udt €
( )
t
.
€ €
Now, take what’s left of the momentum equation, set u(x,t) = U(t) , divide by ρ and
integrate in time to find: €
t % (
1 ∂p
€ U(t) = ∫ ' − + gx *dt + U o .
0 & ρ ∂x € )

Here the constant has been evaluated with the initial condition. These results for U and ρ are an
implicit solution because each relies on the other. If the density is uniform, then these reduce to
t % (
€ ρ = ρ o , and U(t) = ∫ ' − 1 ∂p + gx *dt + U o .
0 & ρ o ∂x )

€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.49. a) Derive the following equation for the velocity potential for irrotational inviscid
∂ 2φ ∂ 1
compressible flow in the absence of a body force:
∂t 2
+
∂t
(2
) 2
( )
∇φ + ∇φ ⋅ ∇ ∇φ − c 2∇ 2φ = 0
2
where ∇φ = u , as usual. Start from the Euler equation (4.41), use the continuity equation,
assume that the flow is isentropic so that p depends only on ρ, and denote (∂p ∂ρ) s = c 2 .
b) What limit does c → ∞ imply? €
€ c) What limit does ∇φ → 0 imply?
€
Solution 4.49. a) The path to the final equation may not be obvious, so the steps listed here may
€ ∂u 1
not necessarily appear in logical sequence at first glance. Start with + (u ⋅ ∇)u = − ∇p , and
∂t ρ
use the vector identity (B3.9) for the nonlinear advective acceleration to find:
∂u $ u2' 1
+ ∇&& )) + ∇p = −ω × u = 0
∂t % 2 ( ρ €
where the final equality holds in irrotational flow. Take the time derivative of this equation to
find:
∂ 2u ∂ $ u ' ∂ $ 1 '
2
€ + ∇& ) + & ∇p) = 0 (1)
∂t 2 ∂t &% 2 )( ∂t % ρ (
Here we assume that the flow isentropic so that the density is only a function of the pressure,
ρ = ρ( p) ; this means that the final term of (1) can be rewritten:
∂ %1 ( 1 dρ ∂p 1 ∂p % 1 ( ∂p 1 ∂p % 1 ∂p (
' ∇p* €=− 2 ∇p + ∇ = '∇ * + ∇ = ∇' *.
∂t & ρ ) ρ dp ∂t ρ ∂t & ρ ) ∂t ρ ∂t & ρ ∂t )
€ Thus with ∇φ = u , (1) becomes:
& ∂ 2φ 1 ∂ 2 1 ∂p )
∇( 2 + ∇φ + + = 0.
€ ' ∂t 2 ∂t ρ ∂t *
This
€ means the quantity in parentheses is at most a function of time. Thus a suitable redefinition
of φ will produce:
∂ 2φ 1 ∂ 2 1 ∂p
€ 2
+ ∇φ + =0 (2)
∂t 2 ∂t ρ ∂t
Save this result, and start again with the Euler equation, but this time take a dot product with u:
∂ #u & # u2& 1
2

%% (( + u ⋅ ∇%% (( + u ⋅ ∇p = 0
€ ∂t $ 2 ' $ 2 ' ρ
Insert ∇φ = u into the first two terms of this equation and add it to (2)
∂ 2φ ∂ 2 &1 2) 1 ∂p 1
2
+ ∇φ + ∇φ ⋅ ∇( ∇φ + + + u ⋅ ∇p = 0
€ ∂ t ∂ t ' 2 * ρ ∂ t ρ
2
€ Now use the isentropic relationship, dp = c dρ , to eliminate p.
∂ 2φ ∂ 2 &1 2) -
2 1 ∂ρ 1 0
+ ∇ φ + ∇φ ⋅ ∇( ∇ φ + + c . + u ⋅ ∇ ρ 1=0
€ ∂t 2 ∂t '2 * / ρ ∂t ρ 2
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

From the continuity equation, ∂ρ ∂t + ρ∇ ⋅ u + u ⋅ ∇ρ = 0 , the term in {,}-braces is merely −∇ ⋅ u

= −∇ 2φ , thus:
∂ 2φ ∂ 2 &1 2)
2
+ ∇φ + ∇φ ⋅ ∇( ∇φ + − c 2∇ 2φ = 0 €
€ ∂t ∂t '2 *
€ b) As c → ∞, only the final term matters, and the equation simplifies to ∇ 2φ = 0 which
represents incompressible potential flow.
c) When ∇φ → 0, the nonlinear terms in the final result of part a) go to zero faster than the two
€ 2
∂φ
linear ones leaving: 2 − c 2∇ 2φ = 0 which is the linear wave € equation for acoustic disturbances.
∂t
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.50. Derive (4.43) from (4.42).

Solution 4.50. The entire statement of equation (4.42) is:

dx dX dx " d
u= = + = U + ( x1"e1" + x "2e"2 + x "3e"3 )
dt dt dt dt
dx1" dx 2" dx 3" de" de" de"
= U+ e1" + e"2 + e"3 + x1" 1 + x "2 2 + x "3 3 = U + u" + Ω × x " ,
dt dt dt dt dt dt
Thus, we see that dx " dt = u" + Ω × x " . Now time differentiate u, to find:
€ du dU du# d
≡a= + + (Ω × x #)
€ dt dt dt dt
€ dU d dΩ dx #
= + ( u1#e1# + u#2e#2 + u#3e#3 ) + × x# + Ω ×
dt dt dt dt
dU du1# du#2 du#3 de1# de#2 de# dΩ
= + e1# + e#2 + e#3 + u1# + u#2 + u#3 3 + × x # + Ω × (u# + Ω × x #)
dt dt dt dt dt dt dt dt
Here the various terms written in component form may be identified. The second through fourth
terms are the fluid particle acceleration, a´, observed in the non-inertial frame of reference. The
fifth through seventh terms, which involve the time derivatives of the unit vectors, can be written
€ in terms of a cross product:
de" de" de"
u1" 1 + u"2 2 + u"3 3 = Ω × u" ,
dt dt dt
as depicted in Figure 4.7 and described in paragraph below (4.42). With these replacements, the
last equality for the fluid particle acceleration becomes:
dU dΩ
a€= + a " + Ω × u" + × x " + Ω × u" + Ω × (Ω × x ")
dt dt
dU dΩ
= + a " + 2Ω × u" + × x " + Ω × (Ω × x "),
dt dt
which matches (4.43).

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.51. Observations of the velocity u´ of an incompressible viscous fluid are made in a
frame of reference rotating steadily at rate Ω = (0, 0, Ωz). The pressure at the origin is po and g =
–gez.
a) In Cartesian coordinates with u´ = (U, V, W) = a constant, find p(x,y,z).
b) In cylindrical coordinates with u´ = –ΩzReϕ, determine p(R,ϕ,z). Guess the result if you can.

Solution 4.51. The pressure and velocity of an incompressible viscous fluid in a rotating
coordinate system both appear in (4.45).
" D!u! % , dU dΩ /
ρ$ ' = −∇!p + ρ .g − − 2Ω × u! − × x! − Ω × (Ω × x!)1 + µ∇!2 u!
# Dt &O1!2!3! - dt dt 0
a) With U = 0, Ωz = const., u´ = const., g = –gez, , this equation reduces to:
0 = −∇"p + ρ &'−ge z − 2Ω × u" − Ω × (Ω × x")() .
Using Ω = (0, 0, Ωz), u´ = (U, V, W), and x´ = (x,y,z), evaluating the cross product terms leads to:
∇!p = ρ %&−ge z + 2ΩzVe x − 2ΩzUe y + Ω2z (xe x + ye y )'( .
The three components of this equation can be written:
∂p ∂p ∂p
= 2 ρΩzV + ρΩ2z x , = −2 ρΩzU + Ω2z y , and = −ρ g .
∂x ∂y ∂z
Integrating the three equations leads to:
x2 y2
p = 2 ρΩzVx + ρΩ2z + A(y, z) , p = −2 ρΩzUy + Ω2z + B(x, z) , and p = −ρ gz + C(x, y) .
2 2
where A, B, and C are undetermined functions of integration. These three equations are
consistent when:
ρΩ2z 2 2
p(x, y, z) − po = 2 ρΩzVx − 2 ρΩzUy +
2
( x + y ) − ρ gz .
b) When u´ = –ΩzReϕ, the rotation of the flow field precisely reverses the rotation introduced by
the rotating coordinate system. Thus, the fluid is stationary in a non-rotating frame and this
leaves a hydrostatic balance for the pressure field: p = –ρgz + po.
Alternatively, this result can be shown by evaluating the terms in (4.45) with U = 0, and
Ωz = const. For convenience, convert the velocity to Cartesian coordinates: u´ = –ΩzReϕ = Ωzyex
– Ωzxey, then evaluate the remaining terms of (4.45) that involve u´ or Ωz:
" D!u! %
ρ$ ' = −Ω2z xe x − Ω2z ye y , −2Ω × u$ = −2Ω2z xe x − 2Ω2z ye y ,
# Dt &O1!2!3!
−Ω × (Ω × x$) = Ω2z (xe x + ye y ) , and ∇!2 u! = 0 .
Reassemble (4.45), to find:
ρ (−Ω2z xe x − Ω2z ye y ) = −∇#p + ρ %&g − ( 2Ω2z xe x + 2Ω2z ye y ) + Ω2z (xe x + ye y )'( .
This simplifies to: 0 = −∇"p + ρ g , which integrates to p = –ρgz + po when g = –gez.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.52. For many atmospheric flows, rotation of the earth is important. The momentum
equation for inviscid flow in a frame of reference rotating at a constant rate Ω is:
∂u ∂t + (u ⋅ ∇)u = −∇Φ − (1 ρ)∇p − 2Ω × u − Ω × (Ω × x)
For steady two-dimensional horizontal flow, u = (u, v, 0) , with Φ = gz and ρ = ρ(z), show that the
streamlines are parallel to constant pressure lines when the €fluid particle acceleration is
dominated by the Coriolis acceleration (u ⋅ ∇)u << 2Ω × u , and when the local pressure gradient
€
dominates the centripetal acceleration Ω × (Ω × x) << ∇p ρ . [This seemingly strange result
governs just about all large-scale weather phenomena like hurricanes and other storms, and it
allows weather forecasts to be made based on surface pressure measurements alone.]
€
Hints. 1. If Y(x) defines a streamline contour, then dY dx = v u is the streamline slope.
€
2. Write out all three components of the momentum equation and build the ratio v/u.
3. Using hint 1., the pressure increment along a streamline is: dp = (∂p ∂x ) dx + (∂p ∂y ) dY

Solution 4.52. Since the rotation rate is steady, choose the coordinate system so that the z-axis is
parallel to Ω: i.e. Ω = (0,0,Ωz ) . When the flow is steady, Ω × (Ω × x) << ∇p ρ .
€
(u ⋅ ∇)u << 2Ω × u , the viscous terms are neglected, and Φ = gz with ρ = ρ(z), the momentum
equation becomes:
0 = −ge z − (1 ρ)∇p − 2Ωz ue y + 2Ωzve x
€ €
Written out in terms of the coordinate directions, this equation becomes:
€
∂p ∂x = +2 ρΩzv , ∂p ∂y = −2 ρΩz u , and ∂p ∂z = −ρg
The final relationship establishes that the vertical pressure gradient will be hydrostatic and will
€
not be directly influenced by u = (u,v,0) . Let y = Y (x) be the equation of a streamline, so from
(3.7):
€ € €
dY v +(1 2 ρΩz ) ∂p ∂x ∂p ∂x
= = =− .
€ dx u −(1 2 ρΩ z ) ∂ p ∂ y ∂ p ∂ y
€
where the middle equality follows from the horizontal components of the simplified momentum
equation. Using the form on the far left and the far right of the last equation produces:
(∂p ∂x)dx + (∂p ∂y)dY = 0 = (dp) streamline
€
Thus, that p = constant on streamlines, so streamlines and isobaric lines are everywhere parallel.
The situation described in this problem leads to geostrophic flow and is considered further in Ch.
13.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.53. Show that (4.55) can be derived from (4.7), (4.53), and (4.54).

Solution 4.53. Start with (4.53),

∂ ' ! 1 2 $* ∂ ' ! 1 2 $ * ∂ ∂q
) ρ #e + u j &, +
∂ t ( " 2 %+ ∂ xi ( " 2 % +
) ρ #e + u j & ui , = ρ gi ui +
∂ xi
( Tij u j ) − i ,
∂ xi
and work on the left side terms first:
∂ * $ 1 2 '- ∂ * $ 1 2 ' -
,ρ e + u j / + , ρ e + u j ui /
∂t + &% 2 )(. ∂x i + &% 2 )( .
∂* 1 - ∂ * 1 2 - * 1 2 -$∂ρ ∂ '
= ρ ,e + u 2j / + ρui ,e + u j / + ,e + u j /& + ( ρui ))
∂t + 2 . ∂x i + 2 . + 2 .% ∂t ∂x i (
D* 1 -
= ρ ,e + u 2j /.
Dt + 2 .
Here the final equality follows from the definition of D/Dt and because the terms inside [,]-
brackets on the second line are zero because of continuity equation (4.7).
Now consider the term in (4.53) involving the stress tensor, use (4.54),
€
∂ ∂u ∂T " ∂ u ∂u % " ∂p ∂τ %
( Tij u j ) = Tij j + u j ij = $$ − p j + τ ij j '' + $$ −u j + u j ij '' .
∂ xi ∂ xi ∂ xi # ∂ x j ∂ xi & # ∂ x j ∂ xi &
Substituting the left-side-term result and this result into (4.53) produces:
D! 1 $ ! ∂u ∂u $ ! ∂p ∂τ $ ∂ q
ρ # e + u 2j & = ρ gi ui + ## − p j + τ ij j && + ## −u j + u j ij && − i .
Dt " 2 % " ∂xj ∂ xi % " ∂ x j ∂ xi % ∂ xi
which is (4.55).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.54. Multiply (4.22) by uj and sum over j to derive (4.56).

Solution 4.54. The goal here is to develop the partial differential equation that governs the
kinetic (or mechanical) energy, (1 2) ρu 2j , in a fluid flow. Start from (4.22), multiply this by uj
and consider the implied sum over j to find:
∂ ∂ ∂
u j ( ρu j ) + u j ( ρui u j ) = ρu j g j + u j (Tij ) .
€ ∂t ∂ xi ∂ xi
Expand and regroup the terms on the left side,
∂ ∂ ∂u $∂ρ ∂ ' ∂u
u j ( ρu j ) + u j ( ρui u j ) = ρu j j + u j u j & + ( ρui )) + ρu j ui j .
∂t ∂x i ∂t % ∂t ∂x i ( ∂x i
Recognizing that the contents of the [,]-brackets are zero because of (4.7), and simplify the terms
on the left to find:
∂ !u $ ∂ ! uj $
2 2
∂
€ ρ ## j && + ρui ## && = ρu j g j + u j (Tij ) .
∂t " 2 % ∂ xi " 2 % ∂ xi
Now substitute for the stress tensor from (4.27):
∂ !u $ ∂ ! uj $
2 2
∂
ρ ## j && + ρui ## && = ρu j g j + u j (− pδij + τ ij ) ,
∂t " 2 % ∂ xi " 2 % ∂ xi
use the definition of D/Dt and utilize the index-exchange property of δij, to reach
D !u $
2
∂p ∂τ
ρ ## j && = ρu j g j − u j + u j ij ,
Dt " 2 % ∂xj ∂ xi
which is (4.56).
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.55. Starting from ε = (1 ρ ) τ ij Sij , derive the right most expression in (4.58).

Solution. 4.55. First determine τij from (4.27) and (4.37),

" 1 %
τ ij = 2µ $ Sij − Smmδij ' + µυ Smm δij ,
# 3 &
then substitute this into ε = (1 ρ ) τ ij Sij to find:
1 1# # 1 & &
ε≡ τ ij Sij = % 2µ % Sij − Smmδij ( + µυ Smm δij ( Sij
ρ ρ$ $ 3 ' '
2µ # 1 & µ
= % Sij − Smmδij ( Sij + υ Smm δij Sij
ρ $ 3 ' ρ
2µ # 1 &# 1 1 & µ
= % Sij − Smmδij (% Sij − Smmδij + Smmδij ( + υ Smm S jj
ρ $ 3 '$ 3 3 ' ρ
2
2µ # 1 & 2µ # 1 &1 µ
= % Sij − Smmδij ( + % Sij − Smmδij ( Smmδij + υ Smm S jj
ρ $ 3 ' ρ $ 3 '3 ρ
2
2µ # 1 & µ 2µ # 1 &S
= % Sij − Smmδij ( + υ Smm S jj + % Sijδij − Smmδijδij ( mm .
ρ $ 3 ' ρ ρ $ 3 ' 3
The first two terms on the last line look promising since Sjj = Smm = (∂um/∂xm)2. The third term on
the last line is readily determined to be zero since δijδij = δjj = 3. Therefore the contents of the last
set of parentheses is:
1 1
Sijδij − Smmδijδij = S jj − Smm (3) = 0 .
3 3
So,
2 2
2µ & 1 ) 2 µυ & 1 ∂um ) µυ & ∂um )
ε = ( Sij − Smmδij + + Smm S jj = 2ν ( Sij − δij + + ( + ,
ρ€' 3 * ρ ' 3 ∂x m * ρ ' ∂x m *
where ν ≡ µ ρ .

€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.56. For many gases and liquids (and solids too!), the following equations are valid:
q = −k∇T (Fourier's law of heat conduction, k = thermal conductivity, T = temperature),
e = eo + cvT (e = internal energy per unit mass, cv = specific heat at constant volume), and
h = ho + cpT (h = enthalpy per unit mass, cp = specific heat at constant pressure),
where eo and ho are constants, and cv and cp are also constants. Start with the energy equation
∂e ∂e ∂u ∂q
ρ + ρui = − p i + τ ij Sij − i
∂t ∂ xi ∂ xi ∂ xi
for each of the following items.
a) Derive an equation for T involving uj, k, ρ, and cv for incompressible flow when τij = 0.
b) Derive an equation for T involving uj, k, ρ, and cp for flow with p = const. and τij = 0.
c) Provide a physical explanation why the answers to a) and b) are different.

Solution 4.56. a) Put q = −k∇T , e = eo + cvT, and τij = 0 into the energy equation to find:
! ∂T ∂T $ ∂ u ∂ ! ∂T $
ρ cv # + ui & = −p i + #k &.
" ∂t ∂ xi % ∂ xi ∂ xi " ∂ xi %
€
DT ∂ ! ∂ T $
For incompressible flow ∂ui ∂x i = 0 , so the final equation is ρ cv = #k &.
Dt ∂ xi " ∂ xi %
b) This part is a little harder than part a) because the flow is compressible. However, the first
step is nearly the same as part b). Put q = −k∇T , and τij = 0 into the energy equation to find:
€
De ∂ui ∂ % ∂T (
ρ = −p + ' k * . Now manipulate the pressure term using the continuity equation:
Dt ∂ x i ∂x i & ∂ x i )
∂u € p Dρ 1 Dρ D(1 ρ) D( p ρ)
p i =− = − pρ 2 = pρ =ρ .
∂x i ρ Dt ρ Dt Dt Dt
€ De D( p ρ) ∂ % ∂T (
Substitute the final term into the energy equation: ρ = −ρ + ' k * . Collect like
Dt Dt ∂ x i & ∂x i )
€ Dh ∂ $ ∂T '
terms and recall that h = e + p ρ to find: ρ = & k ) . Now use h = ho + cpT to get
Dt ∂x i % ∂x i (
€DT ∂ ! ∂ T $
ρ cp = #k &.
€ Dt ∂ x i " ∂ xi %

c) The results from parts a) and € b) must be different because constant volume (incompressible)
and constant pressure (isobaric) processes involving thermal energy are fundamentally different.
When the material is compressible, it can expand (or shrink) and do (or absorb) work.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.57. Derive the following alternative form of (4.60):

DT Dp ∂ ! ∂T $
ρ cp = αT + ρε + #k & , where ε is given by (4.58) and α is the thermal expansion
Dt Dt ∂ xi " ∂ xi %
coefficient defined in (1.26). [Hint: dh = (∂ h ∂ T ) p dT + (∂ h ∂ p)T dp ]

Solution 4.57. Start from (4.60),

2
De ∂um & 1 ∂um ) & ∂um ) 2 ∂ & ∂T )
ρ = −p + 2µ( Sij − δij + + µυ ( + + (k +,
Dt ∂x m ' 3 ∂x m * ' ∂x m * ∂ x i ' ∂x i *
and use (4.58) so simplify the appearance of the middle terms on the right:
De ∂u ∂ & ∂T )
ρ = − p m + ρε + (k +.
€ Dt ∂x m ∂x i ' ∂x i *
The final two terms on the right of this equation are in the correct form so the remainder of this
effort involves the term on the left and the first term on the right. First put them on the same side
of the equation within the same parentheses:
€ $ De p ∂um ' ∂ $ ∂T '
ρ& + )=ε+ &k ) ,
% Dt ρ ∂x m ( ∂x i % ∂ x i (
then eliminate the divergence of the velocity field using the expanded continuity equation (4.8)
and use the definition of the enthapy per unit mass: h = e + p/ρ.
$ De p Dρ ' $ De D(1 ρ) 1 Dp 1 Dp ' Dh Dp ∂ $ ∂T '
ρ& − 2 €) = ρ& +p + − )=ρ − =ε+ & k ) . (&)
% Dt ρ Dt ( % Dt Dt ρ Dt ρ Dt ( Dt Dt ∂x i % ∂x i (
Now work on the thermodynamic variables. Use the hint to make the choice of T and p as
independent variables, h = h(T,p), so that:
! ∂h $ ! ∂h $ ! ∂h $
€ dh = # & dT + # & dp = c p dT + # & dp . (†)
" ∂T % p " ∂ p %T " ∂ p %T
However, the natural variables for h are s and p, h = h(s,p), so that
# ∂h & # ∂h & 1 # ∂h & # ∂s & 1 # ∂p & # ∂s & 1
dh = % ( ds + % ( dp = Tds + dp → % ( = T% ( + % ( = T% ( + .
$ ∂s ' p $ ∂p ' s ρ $ ∂p 'T $ ∂p 'T ρ $ ∂p 'T $ ∂p 'T ρ
Substitute the final equality for (∂h/∂p)T into (†) to reach:
! ! ∂s $ 1 $ ! T ! ∂ρ $ 1 $
dh = c p dT + #T # & + &dp = c p dT + # − 2 # & + &dp ,
€ " " ∂ p %T ρ % " ρ " ∂ p %T ρ %
2
( )
where the second equality follows from the Maxwell relation: (∂s ∂p)T = − 1 ρ (∂ρ ∂p)T ,which
is derived from the free enthaply (= h – Ts). Now, using (1.26) for α, the enthaply increment is:
" αT 1 % Dh DT Dp
dh = cp dT + $ − + ' dp → ρ = ρ cp + (−αT +1) ,
# ρ ρ& Dt€ Dt Dt
and inserting this into the final equality of (&) produces:
DT Dp Dp ∂ " ∂T % DT Dp ∂ ! ∂T $
ρ cp + (−αT +1) − =ε + $k ' , or ρ cp = αT + ρε + #k &.
Dt Dt Dt ∂ xi # ∂ xi & Dt Dt ∂ xi " ∂ xi %
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.58. Show that (4.68) is true without abandoning index notation or using vector
identities.

Solution 4.58. Equation (4.68) is an identity for the advective acceleration. Start from ui(∂uj/∂xi)
and create the rotation tensor Rij from (3.15) by adding and subtracting ∂ui/∂xj.
∂u $ ∂u ∂u ∂ u ' ∂u
ui j = ui && j − i + i )) = −ui Rij + ui i .
∂x i % ∂x i ∂x j ∂ x j ( ∂x j
Now use (3.15), Rij = –εijkωk, for the first term on the right, and move ui inside the gradient of the
second term on the right to find:
∂u ∂ % 1 2( ∂ %1 2( ∂ %1 2(
ui j€= uiεijkω k + ' ui * = −ε jik uiω k + ' ui * = −(u × ω ) j + ' ui * .
∂x i ∂x j & 2 ) ∂x j & 2 ) ∂x j & 2 )

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.59. Consider an incompressible planar Couette flow, which is the flow between two
parallel plates separated by a distance b. The upper plate is moving parallel to itself at speed U,
and the lower plate is stationary. Let the x-axis lie on the lower plate. The pressure and velocity
fields are independent of x, and fluid has uniform density and viscosity.
a) Show that the pressure distribution is hydrostatic and that the solution of the Navier–Stokes
equation is u(y) = Uy/b.
b) Write the expressions for the stress and strain rate tensors, and show that the viscous kinetic-
energy dissipation per unit volume is µU2/b2.
c) Evaluate the kinetic energy equation (4.56) within a rectangular control volume for which the
two horizontal surfaces coincide with the walls and the two vertical surfaces are perpendicular to
the flow and show that the viscous dissipation and the work done in moving the upper surface
are equal.

y
b

Solution 4.59. In this flow is one dimensional, u = (u(y), 0), and ∂u/∂x = 0. This means that the
continuity equation is satisfied, and that the flow is incompressible. Thus, the x (horizontal) and
y (vertical) components of the constant-viscosity Navier-Stokes' momentum equation are:
∂u ∂u ∂u 1 ∂p & ∂ 2 u ∂ 2 u ) ∂v ∂v ∂v 1 ∂p & ∂ 2v ∂ 2v )
+u +v =− + ν ( 2 + 2 + , and +u +v =− − g − ν( 2 + 2 +.
∂t ∂x ∂y ρ ∂x ' ∂x ∂y * ∂t ∂x ∂y ρ ∂y ' ∂x ∂y *
Here, v is zero, ∂u/∂x = ∂u/∂t = ∂p/∂x = 0, so these equations simplify to:
∂ 2u 1 ∂p
0 = 2 , and 0 = − − g.
∂y ρ ∂y
€ €
a) The simplified vertical mometum equation can be rearranged to: ∂p/∂y = –ρg, which is the
equation of hydrostatics and is readily integrated to find p = po – ρgy, when the density is
constant. The simplified horizontal momentum equation can be integrate twice to find u = Ay +
€ €
B, and the boundary conditions on the lower plate (u = 0 on y = 0) and on the upper plate (u = U
on y = b), allow the constants A (= 0) and B (= U/b) to be determined. The final velocity result is:
u(y) = Uy b.
1 # ∂u ∂u & # (∂u ∂y + ∂v ∂x )&
1
∂u ∂x # 0 U 2b&
b) Strain rate: Sij = %% i + j (( = % 1 2
(=% (
2 $ ∂x j ∂x i ' $ 2 (∂v ∂x + ∂u ∂y ) ∂v ∂y ' $U 2b 0 '
€ " − p µU b %
For incompressible flow, the stress tensor is: Tij = − pδij + 2µ Sij = $ '
$ µU b − p '
# &
€
The viscous dissipation per unit volume is:
ρε = 2µSij Sij = 2µ[ S122 + S21
2
] = 2µ[U 2 4b2 + U 2 4b2 ] = µ U 2 b2
∂ !1 2$ ∂
c) The steady-flow kinetic energy equation, ρui # u j & = ρ gi u + u j
∂ xi " 2 % ∂ xi
(Tij ) , can be rewrittten:
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

∂ ( 12 u 2j ) ∂ (u jTij )
∂uj ∂ (u jTij ) ∂ (u jTij )
ρui = ρ gi u + − Tij
= ρ gi u + − Tij Sij = ρ gi u + + pSii − ρε .
∂ xi ∂ xi ∂ xi ∂ xi ∂ xi
Note that 12 u 2j (∂ ∂x i )( ρui ) = 0 (because of the steady continuity equation) add it the left side and
integrate the two ends of this equation in the stationary control volume described in the statement
of part c) to find:
€ ∫ ( 12 ρu2j )uini dA = ∫ ρ g j u j dV + ∫ u jTij ni dA + ∫ pSii dV − ∫ ρε dV .
A* V* A* V* V*
Here Gauss' divergence theorem has been used to covert the left side term and the second term
on the right to surface integrals. The body force term is zero because the dot product gjuj = 0 for
a horizontal flow with a vertical body force. The term involving the pressure is zero because Sii =
0 in this flow. Evaluating the remaining terms at locations x = 0 and x = L produces:
L b #
ρ b b L
µU 2 &
2 0
−( u(y)
3
∫ [ ] x=0 [ ] x=L
+ u(y)
3
dy = +
0
up)
∫ ([ ] x=0 [ ] x=L ) ∫ [ 21 ] y=b ∫ ∫ %$ b2 ('dxdy .
− up dy +
0
uT dx −
0 0
where the dimension of the control volume into the page, which is common to all terms, has been
divided out, and ρε has been evaluated with the result of part b). There is no contribution to the
surface work term from the lower plate because u = 0 on y = 0. The flow field is independent of
x, so the left-side integrand terms cancel, as do the integrand terms in the first right-side integral.
This leaves a balance of surface work and dissipation rate terms. On the upper plate at y = b, u =
U and T21 = τ21 = µU/b; thus
L b "
L
" µU % µU 2 % µU 2 µU 2
0 = + ∫ U$ 'dx − ∫ ∫ $ 2 'dxdy → 0 = L − 2 Lb = 0,
0 # b & 0 0# b & b b
and the energy balance is verified.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.60. Determine the outlet speed, U2, of a chimney in terms of ρo, ρ 2, g, H, A1, and A2.
For simplicity, assume the fire merely decreases the density of the air from ρo to ρ 2 (ρo > ρ 2) and
does not add any mass to the airflow. (This mass flow assumption isn’t true, but it serves to keep
the algebra under control in this problem.) The relevant parameters are shown in the figure. Use
the steady Bernoulli equation into the inlet and from the
outlet of the fire, but perform a control volume analysis
across the fire. Ignore the vertical extent of A1 compared
to H, and the effects of viscosity.

Solution 4.60. Start from the stagnant air (Uo ≈ 0) on

the suction side of the chimney and use the steady
Bernoulli equation and a CV analysis to follow the flow
all the way to the chimney exit. The effects of viscosity
are ignored throughout this problem.
Suction flow into the fireplace opening: Po = P1 + 12 ρ oU12 (1)
Cons. of mass through the fireplace: ρ oU1 A1 = ρ 2U1a A1 = ρ 2U 2 A2 (2,3)
2 2
Cons. of horiz. mom through the fire: −ρ oU1 A1 + ρ 2U1a A1 = P1 A1 − P1a A1 (4)
1 2 1 2
Outflow from the fire: € P1a + 2 ρ 2U1a = P2 + 2 ρ 2U 2 + ρ 2 gH (5)
Static pressure outside the chimney: € P2 + ρ o gH = Po (6)
All of the velocities and pressures subscripted€ “1” and “1a” need to be eliminated along with Po
and P2 to find a single equation for U2. First, divide (4) by A1 and use (2,3) to substitute for the
unwanted velocities. €
€
$ ρ 2U 2 A2 ' 2 $ ρ 2U 2 A2 ' 2 ρ 2U 22 A22 $ ρ 2 '
−ρ o & ) + ρ 2& ) = P 1 − P1a , or &1− ) = P1 − P1a (7)
% ρ o A1 ( % ρ 2 A1 ( A12 % ρ o (
Use (1), (2,3), and (7) to find:
2
ρ 2U 22 A22 $ ρ 2 ' 1 $ ρ 2U 2 A2 ' ρ 2U 22 A22 $ ρ2 '
&1− ) = P o − ρ o& ) − P 1a , or P o − P1a = &1− ). (8)
€ A12 % ρ o ( 2 % ρ o A1 ( € A12 % 2 ρ o (
Combine (2,3), (5), (6), and (8):
ρ 2U 22 A22 $ ρ 2 ' ρ 2U 22 A22
Po − 2 &1− )+ 2
= Po − ρ o gH + 12 ρ 2U 22 + ρ 2 gH
€ A1 % 2 ρ o ( 2A1 €
Cancel Po’s, divide by ρ2, and collect all the terms with U2 to one side of the equation.
$ A2 $ ρ ' A 2 1 ' ρ − ρo
U 22 & − 22 &1− 2 ) + 22 − ) = 2 gH
€ % A1 % 2 ρ o ( 2A1 2 ( ρ2
Change signs on both sides of the equation, simplify terms involving the area ratio, and solve for
U2 to find:
2( ρ o ρ 2 −1) gH
€ U2 =
1+ ( A22 A12 )(1− ρ 2 ρ o )
Thus, for inviscid flow, U2 increases with increasing g, H, and A1, and with decreasing ρ2 and A2.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.61. A hemispherical vessel of radius R containing an inviscid constant-density liquid

has a small rounded orifice of area A at the bottom. Show that the time required to lower the
level from h1 to h2 is given by
2π $ 2R 3 2 1 52 52 '
& ( h1 − h2 ) − ( h1 − h2 ))
32
t 2 − t1 =
A 2g % 3 5 (

Solution 4.61. Let z denote the height of the

liquid in the vessel. Although there is slight R
€ 2
unsteadiness even when πR >> A, use the
steady Bernoulli equation between the free h1 r
surface and the orifice at the bottom of the dz
tank: h2
2
z
1 # dz & 1 A
patm + ρ% ( + ρgz = patm + ρU e2 , (1)
2 dt$ ' 2
where Ue is the flow speed exiting the tank.
For an incompressible liquid, conservation
of mass reduces to conservation of volume:
€ πr 2 ( dz dt ) = AU e . (2)
Eliminate Ue from (1) and (2) to find:
$ π 2 r 4 '$ dz ' 2 $ π 2 r 4 '1 2 $ dz '
2gz = & 2 −1)& ) , or −& 2 −1) & ) = 2gz ,
€ % A (% dt ( % A ( % dt (
where the minus sign arises because z decreases as the tank empties. When πr2 >> A, this relationship
can be simplified to −πr 2 ( dz dt ) = A 2gz . From the geometry shown, (R − z) 2 + r 2 = R 2 , so
r 2 = 2Rz − z€2 . Thus, the simplified conservation
€ of mass relationship becomes:
−π (2Rz − z )( dz dt ) = A 2gz .
2

Separate the
€ differentials and integrate. €
h2
€ t2
π h2
π & z3 2 z5 2 )
∫ dt = − ∫ (2Rz − z )dz → t2 − t1 = −
12 32
( 2R − + .
t1 A €2g h1 A 2g ' 3 2 5 2 * h1
Evaluate and simplify.
2π $ 2R 3 2 1 52 52 '
& ( h1 − h2 ) − ( h1 − h2 )) .
32
t 2 − t1 =
€ A 2g % 3 5 (

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.62. Water flows through a pipe in a gravitational field as shown in the accompanying
figure. Neglect the effects of viscosity and surface tension. Solve the appropriate conservation
equations for the variation of the cross-sectional area of the fluid column A(z) after the water has left
the pipe at z = 0. The velocity of the fluid at z = 0 is uniform at Vo and the cross-sectional area is A0.

Solution 4.62. Using the z coordinate shown and the steady Bernoulli
equation between the pipe exit and the vertical location z:
1 1
patm + ρVo2 = patm + ρu 2 (z) + ρgz , (1)
2 2
where u(z) is the flow speed at vertical location z. For an incompressible
liquid, conservation of mass reduces to conservation of volume:
Vo Ao = u(z)A(z) , (2)
€ u(z) from (1) and (2) to find:
Eliminate
2
Vo2 = (Vo Ao A(z)) + 2gz .
Solve for A(z) to find:
€
Vo Ao
A(z) = .
2
€ Vo − 2gz

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.63. Redo the solution for the orifice-in-a-tank problem allowing for the fact that in
Fig. 4.16, h = h(t) but ignoring fluid acceleration. Estimate how long it takes for the tank take to
empty.

Solution 4.63. Let h(t) denote the height of the liquid in the vessel. Let the tank and orifice cross
sectional areas be At and Ao, respectively. Although there is slight unsteadiness even when At >> Ao,
use the steady Bernoulli equation between the free surface and the orifice at the bottom of the tank:
2
1 $ dh ' 1
patm + ρ&− ) + ρgh = patm + ρU o2 (1)
2 % dt ( 2
where Uo is the flow speed exiting the tank throught the orifice, and the minus sign appears because
the water level is decreasing. For an incompressible liquid, conservation of mass reduces to
conservation of volume:
€ At (−dh dt ) = AoU o . (2)
Eliminate Uo from (1) and (2) to find:
# At2 &# dh & 2 # At2 &−1 2 1 dh
2gh = % 2 −1(%− ( , or 2g % 2 −1( = − ,
$ A€o '$ dt ' $ Ao ' h dt
which is a differential equation for h(t). The left side is merely a constant. Integrate this equation
from t = 0 when h = h0 to t = tf when h = 0 to find:
# At2 &−1 2 # h1 2 & 0
€ 2g % 2€−1( t f = −% 12
( = 2h0 ,
$ Ao ' $ 1 2 ' h0
which can be rearranged to yield:
12
2h0 # At2 &
tf = % −1( .
€ g $ Ao2 '

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.64. Consider the planar flow of Example 3.5, u = (Ax,–Ay), but allow A = A(t) to
depend on time. Here the fluid density is ρ, the pressure at the origin or coordinates is po, and
there are no body forces.
a) If the fluid is inviscid, determine the pressure on the x-axis, p(x,0,t) as a function of time from
the unsteady Bernoulli equation.
b) If the fluid has viscosities µ and µυ, determine the pressure throughout the flow field, p(x,y,t),
from the x-direction and y-direction differential momentum equations.
c) Are the results for parts a) and b) consistent with each other? Explain your findings.

2
∂u #1 2 p& #1 2 p&
Solution 4.64. a) Use (4.82): ∫ ∂ t ⋅ ds + %$ 2 u + gz + ( = % u + gz + ( , and choose the
ρ '2 $ 2 ρ '1
1

streamline along the x-axis from the origin to the location x:

x
∂u "1 2 p% "1 2 p% x
dA "1 p(x, 0, t) % p
∫ ∂ t $# 2
dx + u + =
' $ u + ' or ∫ x dx + $ A 2 x 2 + '= 0+ o .
0 ρ &x # 2 ρ &0 0 dt #2 ρ & ρ
The integral is elementary, so after some rearrangement the last equation implies:
p(x, 0, t) − po " dA % x2
= −$ + A2 ' .
ρ # dt &2
b) For this velocity field ∇ ⋅ u = 0 , so the flow is incompressible. Thus, the relevant version of
the Navier-Stokes momentum equation is (4.39b) with g = 0:
Du
ρ = −∇p + µ∇ 2 u .
Dt
Using u = (Ax,–Ay) in this equation leads to:
(! dA $ ! dA $ + ∂p ∂p
ρ *# x + A 2 x & e x + # − y + A 2 y & e y - = − e x − e y + 0 .
)" dt % " dt % , ∂x ∂y
Separating this into two component equations and integrating yields:
" dA % x2 " dA % y2
p = −ρ $ + A 2 ' + B(y, t) and p = −ρ $ − + A 2 ' + C(x, t) ,
# dt &2 # dt &2
where B and C are functions of integration. These two equations are consistent and match the
origin condition p(0,0,t) = po when
p(x, y, t) − po " dA % x 2 " dA % y2
= −$ + A2 ' − $− + A2 ' .
ρ # dt & 2 # dt &2
c) The results of parts b) and c) are consistent. Fluid viscosity does not influence the pressure in
this flow field because there is no net viscous force on fluid particles.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.65. A circular plate is forced down at a steady velocity Uo against a flat surface.
Frictionless incompressible fluid of density ρ fills the gap h(t). Assume that h << ro, the plate
radius, and that the radial velocity ur(r,t) is constant across the gap.
a) Obtain a formula for ur(r,t) in terms of r, Uo, and h.
b) Determine ∂ur(r,t)/∂t.
c) Calculate the pressure distribution under the plate assuming that p(r = ro) = 0.

Solution 4.65. a) Choose a cylindrical control volume that fits under the moving disk, and has
radius r and height h. Conservation of mass for this control volume implies:
dM d dh
= 0 = ( ρπr 2 h ) + ρur ⋅ 2πrh , or 0 = r 2 + 2rhur
dt dt dt
In this problem, dh dt = −U o , so the above conservation of mass equation requires: ur = rU o 2h .
∂u rU dh rU o2
b) Differentiate the part a) result: r = − 2o =
€ ∂t 2h dt € 2h 2
c) Use €this result in the unsteady Bernoulli equation. The streamlines are€radial so the unsteady
term can be evaluated on paths directed through the origin from a radial distance r to the radial
distance ro.
€
ur2 (r) p(r) ro ∂ur u 2 (r ) p(ro )
+ =∫ dr + r o + .
2 ρ r ∂t 2 ρ
Now substitute p(r = ro) = 0, and the results of parts a) and b) to find:
r 2U o2 p(r) U o2 $ ro2 r 2 ' ro2U o2
+ = 2 & − )+
€ 8h 2 ρ 2h % 2 2 ( 8h 2
3ρU o2 2 2
Solve for the pressure at distance r: p(r) =
8h 2
(ro − r ) . Such a distribution is reasonable
because it is highest at r = 0 and drops toward the edges of the disk; thus, the pressure gradient
€
pushes fluid out from underneath the disk.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.66. A frictionless, incompressible fluid with density ρ resides in a horizontal nozzle
of length L having a cross sectional area that varies smoothly between Ai and Ao via:
A(x) = Ai + ( Ao − Ai ) f ( x L) , where f is a function that goes from 0 to 1 as x/L goes from 0 to 1.
Here the x-axis lies on the nozzle’s centerline, and x = 0 and x = L are the horizontal locations of
the nozzle’s inlet and outlet, respectively. At t = 0, the pressure at the inlet of the nozzle is raised
to pi > po, where po is the (atmospheric) outlet pressure of the nozzle, and the fluid begins to flow
€ horizontally through the nozzle.
a) Derive the following equation for the time-dependent volume flow rate Q(t) through the
nozzle from the unsteady Bernoulli equation and an appropriate conservation-of-mass
relationship.
Q˙ (t) x= L Ai Q2 (t) $ 1 1 ' $ pi − po '
∫
Ai x= 0 A(x)
dx + & 2 − 2)=&
2 % Ao Ai ( % ρ (
)
b) Solve the equation of part a) when f ( x L) = x L .
c) If ρ = 103 kg/m3, L = 25 cm, Ai = 100 cm2, Ao = 30 cm2, and pi – po = 100 kPa for t ≥ 0, how
long does it take for the flow rate to reach 99% of its steady-state value?
€
Solution 4.66. a) A conservation € of mass equation and the unsteady Bernoulli equation are
needed here. Considering inflow and outflow at the inlet and outlet of the nozzle produces:
volume flux = Q(t) = Ui(t)Ai = Uo(t)Ao, where the ‘o’ and ‘i’ subscripts stand for outlet and inlet,
respectively. Now consider the streamline that follows a horizontal path along the centerline of
the nozzle:
pi U i2 x= L
∂u p U2
+ + gzi = ∫ ⋅ ds + o + o + gzo .
ρ 2 x= 0 ∂t ρ 2
This streamline lies on the x-axis so that ds = e x dx and [u(x,t)] centerline = e xU i (t) Ai A(x) . In
addition, zi = zo, so
€ pi − po x= L ∂ % Ai ( U i2 % Ai2 (
= ∫ 'U i (t) e x * ⋅ e x dx + ' −1* .
ρ € x= 0 ∂t & A(x)
€ ) 2 & Ao2 )
Perform the dot product, set Q(t) = U i (t)Ai , and rearrange:
Q˙ (t) x= L Ai Q2 (t) $ 1 1 ' pi − po
∫
Ai x= 0 A(x)
dx + & 2 − 2)=
2 % Ao Ai ( ρ
.
€
where the over-dot€signifies a time derivative. This is a first-order nonlinear differential
equation for Q(t).
# ( A − Ai ) x &
b) When A(x) =€Ai + ( Ao − Ai ) f ( x L) = Ai %1+ o ( , then
$ Ai L'
x= L$ −1
x= L
Ai ( Ao − Ai ) x ' L
∫ A(x)dx = ∫ &%1+ A L )( dx = β ln(1+ β )
x= 0 x= 0 i

€where β = ( Ao − Ai ) Ai . Thus, the differential equation for the volume flux becomes:
βAi $ 1 1' 2 βAi $ pi − po '
Q˙ (t) + & 2 − 2 )Q (t) = & ).
€ 2L ln(1+ β ) % Ao Ai ( L ln(1+ β ) % ρ (
€ The coefficients are clumsy, but they are constants so the above equation can be rewritten:
Q˙ (t) + BQ 2 (t) = C
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

where B and C are constants. This is a first-order nonlinear differential equation, a Ricatti
equation. Its steady state solution is Qss = C B , so look for a general solution of the form
Q(t) = q(t) + Qss . Substitute this into the differential equation above to find:
q˙ 2BQss
q˙ + B(q 2 + 2qQss + Qss2 ) = C , or q˙ + B(q 2 + 2qQss ) = 0 , or − 2 − =B
€ q q
€ Now let y(t) = 1 q(t) , to find a first-order linear differential equation, y − 2BQss y = B , which has
solution: y(t) = Dexp{+2BQsst} −1 (2Qss ) where D is an undetermined constant, so
€ € 1 1
Q(t) = Qss + q(t) = Qss + = Qss + € .
€ y(t) Dexp{2BQsst} −1 (2Qss )
€Here the fluid starts from rest, so the initial condition Q(0) = 0 allows D to be determined:
D = −1 (2Qss ) , thus
# &
€ C% 2 (
Q(t) = 1−
B % 1+ exp{2 BCt} (
$ '
€
C 2Ai2 Ao2 $ pi − po ' βAi Ai2 − Ao2 % pi − po (
where = Qss = & ) , and BC = ' *.
B Ai2 − Ao2 % ρ ( L ln(1+ β ) 2Ai2 Ao2 & ρ )
€ rate will be reached when: exp{+2 BCt} = 199 . Thus for
c) 99% of the steady flow
β = (Ao − Ai ) Ai = (30cm 2 −100cm 2 ) 100cm 2 = −0.7 , the requisite time is:
€ € ln(1− 0.7) $ 2(0.01m 2 ) 2 (0.003m 2 ) 2 10 3 kg /m 3 '1 2
ln(199) ln(199) 0.25m
t= = & ⋅ ) .
2 BC 2 €2 ) %(0.01m 2 ) 2 − (0.003m 2 ) 2 10 5 Pa (
(−0.7)(0.01m
€ = (2.64665)(42.999)(4.4475x10–4) = 0.051 seconds.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.67. For steady constant-density inviscid flow with body force per unit mass
1 2
g = −∇Φ , it is possible to derive the following Bernoulli equation: p + 2 ρ u + ρΦ = constant
along a streamline.
a) What is the equivalent form of the Bernoulli equation for constant-density inviscid flow that
appears steady when viewed in a frame of reference that rotates a constant rate about the z-axis,
i.e. when Ω = (0,0,Ωz ) with Ωz constant? €
b) If the extra term found in the Bernoulli equation is considered a pressure correction. Where
on the surface of the earth (i.e. at what latitude) will this pressure correction be the largest?
What is the absolute size of the maximum pressure correction when changes in R on a streamline
€
are 1 m, 1 km, and 103 km.

Solution 4.67. a) The steady-flow Bernoulli equation can be derived from a dot product of the
fluid velocity, u, with the momentum equation. For this problem, the flow appears steady in a
frame of reference rotating at a constant rate, so the point of departure for this problem is the
momentum equation (4.45) in a non-interial frame of reference (the primes denote quantities in
the non-inertial frame):
D#u# ( dU dΩ +
ρ = −∇ #p + ρg + µ∇ #2u# − ρ* + × x # + 2Ω × u# + Ω × (Ω × x #)-.
Dt ) dt dt ,
This equation can be simplified for steady (∂/∂t = 0) inviscid (µ = 0) flow in a steadily rotating (
∂Ω ∂t = 0 ) co-ordinate system that does not accelerate ( dU dt = 0 ).
ρ(u# ⋅ ∇)u# = −∇ #p + ρg + µ∇ #2u# − ρ [2Ω × u# + Ω × (Ω × x #)] .
€
The term on the left can be rewritten using a vector identity (B3.9), the body force term can be
replaced with the gradient of the body-force potential g = −∇Φ , and the last term on the right can
€ €
be evaluated when Ω = (0,0,Ωz ) :
€
%1 2(
[ ]
ρ∇ #' u# * + ρ(∇ # × u#) × u# = −∇ #p − ρ∇ #Φ − ρ 2Ω × u# − Ω2z ( x #e#x + y #e#y ) .
&2 ) €
The terms€containing Ω2z can be combined and put under a gradient:
&1 )
−Ω2z ( x $e$x + y $e$y ) = −∇ $( Ω2z ( x $2 + y $2 )+ .
€ '2 *
Collect all the
€ terms involving a gradient together on the left side of the equation:
(ρ 2 1 +
∇ "* u" + p + ρΦ − ρΩ2z ( x "2 + y "2 )- = −2 ρΩ × u" − ρ (∇ " × u") × u"
)2 2 ,
€
Now take the dot product of this equation with u´ noting that this will zero the right side of the
equation because of the properties of the cross product:
)ρ 2 1 ,
€ u" ⋅ ∇ "+ u" + p + ρΦ − ρΩ2z ( x "2 + y "2 ). = 0
*2 2 -
ρ 2 1
This implies that u# + p + ρΦ − ρΩ2z ( x #2 + y #2 ) = constant along a streamline in the rotating
2 2
frame. The sign of the new term can be checked by considering u´ = 0 with Φ = 0 (standard
€
solid body rotation viewed from the inertial frame of reference) and computing the pressure
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

∂p ρ
gradient in the direction perpendicular to the axis of rotation, = + Ω2z r# where
∂r# 2
2 2
r" = x " + y " . This result is correct so the sign of the new term is correct.
b) At any latitude, the local rotation rate at the surface of the earth is Ωzcosθ where θ is zero at
€ ρ
the pole and 90° at the equator. Thus, the pressure correction term is: Ω2z R 2 cos2 θ where R
2
€ represents the distance from the local origin of coordinates. Thus, the largest value of the
correction term will occur at θ = 0, but it will depend on the path length. For example, consider
two points along a streamline where the first point lies at the origin of coordinates and the second
lies a distance R away. The maximum value of the pressure € correction will be:
ρ 2 2 (1.2kg /m )(2π /day) (day /24hr) 2 (hr /3600s) 2 R 2
3 2
Ωz R =
2 2
–9
= 3.2x10 Pa for R = 1 m
= 3.2x10–3 Pa for R = 1 km
= 3.2 kPa for R = 103 km
Given that€ 1 atmosphere of pressure is approximately 100 kPa, this pressure correction is only
likely to be important on streamlines that traverse an appreciable fraction of the earth’s surface.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.68. Starting from (4.45) derive the following unsteady Bernoulli equation for inviscid
incompressible irrotational fluid flow observed in a non-rotating frame of reference undergoing
acceleration dU/dt with its z-axis vertical.
2
∂u $ u2 p dU ') $& u
2
p dU '
∫ ∂t ⋅ ds + & 2 + ρ + gz + x ⋅ dt ) = & 2 + ρ + gz + x ⋅ dt )) .
&
1 % (2 % (1

Solution 4.68. Start from (4.45) with Ω = 0, work in the accelarting frame of reference, and drop
the “primes” from the coordinates.
! Du $ ) dU ,
ρ# & = −∇p + ρ +g − . + µ∇ 2 u (1)
" Dt % * dt -
Here the acceleration of the coordinate system is dU/dt and fluid is frictionless so µ = 0. Thus,
(1) becomes:
& ∂u 1 ) , dU / & dU )
' ∂t 2 ( ) 2
ρ ( + ∇ u − u × ω + = −∇p + ρ .g − 1 = −∇ ( p + ρ gz + ρ x ⋅
* - dt 0 '
+,
dt *
(2)

where the z-axis is vertical, g = –gez, Du/Dt has been expanded using its definition and vector
identities, and the body force and the apparent acceleration have been drawn inside the gradient
on the right using: −ge z = −∇ ( gz ) and − dU dt = −∇ ( x ⋅ dU dt ) . For irrotational flow ω = 0, so
(2) can be rewritten:
∂u $ u2 p dU ')
+ ∇ && + + gz + x ⋅ =0, (3)
∂t % 2 ρ dt )(
when the density is constant. Now integrate this equation along a streamline that starts at location
"1" and ends at location "2" to find
2
∂u $ u2 p dv(t) ') $& u
2
p dv(t) ')
∫ ∂t ⋅ ds + &
& 2 ρ + + gz + x
dt
=
) & 2 ρ + + gz + x
dt ) . (4)
1 % (2 % (1
were ds is the the differential path-length increment tanget to the streamline.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.69. Using the small slope version of the surface curvature 1 R1 ≈ d 2ζ dx 2 , redo
Example 4.7 to find h and ζ(x) in terms of x, σ, ρ, g and θ. Show that the two answers are
consistent when θ approaches π/2.
€
Solution 4.69. Pressure matching between the air just above and the liquid just beneath the free
surface where the deflection is ζ(x) produces:
#1 1& #1 1&
patm = ( p) air = ( p) liquid + σ % + ( = patm − ρgζ + σ % + ( ,
$ R1 R2 ' $ R1 R2 '
where the Laplace pressure must be added to the liquid pressure, and the radii of curvature are
presumed to emanate from points in the air above the liquid. When 1/R1 ≈ d2ζ/dx2 and 1/R2 = 0,
the extreme members of this equality imply:
€ 1 d 2ζ & ρg )
0 = −ρgζ + σ , or − ( +ζ = 0 .
R1 dx 2 ' σ *
Defining δ ≡ σ ρg , the solutions to the differential equation for ζ are:
ζ (x) = A+ exp{+ x δ} + A− exp{−x δ} .
In the present situation,
€ we must have ζ € → 0 as x → ∞ , and z = h at x = 0. These conditions are
satisfied when A+ = 0 and A– = h, so that
€
ζ (x) = h exp{−x δ} = h exp{−x ρg σ } .
€
There is also a slope condition
€ € vertical wall (see Figure 4.20) that allows h to be
at the
determined in terms of the contact angle θ:
# dζ & h
€ % ( = −cot θ = − , or h = − ρg σ cot θ .
$ dx ' x= 0 δ
The exact solution given in Example 4.7 is in implicit form,
−1 2 2 1/2 −1 2 2 1/2
cosh (2δ/ζ ) − (4 − ζ /δ ) − cosh (2δ/h) + (4 − h /δ ) = x/δ ,
and must be simplified for θ approaching π€/2. The complexity here comes from the cosh-inverse
€
function which can be written in terms of a natural logarithm:
€ ( )
cosh−1 (y) = ln y ± y 2 −1 .
When θ → π 2 , then cot θ → 0 . Thus, in this limit h/δ << 1 and ζ/δ << 1. Therefore, the exact
solution can be simplified using:
cosh−1 (2δ ζ ) ≈ ln( 4δ ζ ) + ... and cosh−1 (2δ h) ≈ ln( 4δ h ) + ..., and
€
€ € (4 − ζ 2 /δ 2 )1/2 ≈ 2 + ..., and (4 − h 2 /δ 2 )1/2 ≈ 2 + ...
Substituting these replacements into the exact solution produces:
ln( 4δ ζ ) − 2 − ln( 4δ h ) + 2 + ... = x/δ , or ln( h ζ ) + ... = x/δ .
€ €
Exponentiating both sides of the last expression and solving for z produces
€ €
ζ = h exp{−x/δ} ,
which the same as the small-slope solution. Thus, the exact and small-slope solutions are
€ €
consistent.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.70. An spherical bubble with radius R(t), containing gas with negligible density,
creates purely radial flow, u = (ur(r,t), 0, 0), in an unbounded bath of a quiescent incompressible
liquid with density ρ and viscosity µ. Determine ur(r,t) in terms of R(t), its derivatives. Ignoring
body forces, and assuming a pressure of p∞ far from the bubble, find (and solve) an equation for
the pressure distribution, p(r,t), outside the bubble. Integrate this equation from r = R to r → ∞ ,
and apply an appropriate boundary condition at the bubble's surface to find the Rayleigh-Plesset
equation for the pressure pB(t) inside the bubble:
2
pB (t) − p∞ d 2 R 3 # dR & 4µ dR 2σ
=R 2 + % ( + + ,
ρ dt 2 $ dt ' ρ R dt ρ R
where µ is the fluid's viscosity and σ is the surface tension.

Solution 4.70. The flow of the liquid is purely radial and incompressible so the continuity
equation and radial momentum equation in spherical coordinates are:
1 ∂ 2 ∂ur ∂u 1 ∂p µ # 1 ∂ # 2 ∂ur & 2ur &
2
r ∂r
( r ur ) = 0 and
∂t
+ ur r = −
∂r
+ % %r
ρ ∂r ρ $ r 2 ∂r $ ∂r ' r 2 '
(− (.
The first equation is readily integrated to find: ur = A(t)/r2. The function of integration, A(t), can
be evaluated by requiring ur(R,t) = dR/dt, and this leads to:
R 2 (t) dR
ur (r, t) = 2 .
r dt
Substituting this relationship into the radial momentum equation leads to:
∂ " R 2 (t) dR % " R 2 (t) dR % ∂ " R 2 (t) dR % 1 ∂p µ " 1 ∂ " 2 ∂ " R 2 (t) dR %% 2 " R 2 (t) dR %%
$ 2 '+$ 2 ' $ 2 '=− + $ $r $ '' − $ '' ,
∂t # r dt & # r dt & ∂r # r dt & ρ ∂r ρ $# r 2 ∂r # ∂r # r 2 dt && r 2 # r 2 dt &'&
and this is an equation for the pressure. Carefully evaluate derivatives and simplify the various
terms:
2 2
2R(t) ! dR $ R 2 (t) d 2 R R 4 (t) ! dR $ 1 ∂p µ
2 # & + 2 2
−2 5 # & =− + ( 0!) .
r " dt % r dt r " dt % ρ ∂r ρ
Interestingly, the viscous terms cancel out at this point. Rearrange this equation and integrate in r
from R to ∞:
2 2
∞
dp p∞ − p(R) $ dR ' ∞ dr 2 d 2 R ∞ dr 4 $ dR ' ∞ dr
∫ = = −2R(t) & ) ∫ 2
% dt ( R r
− R (t) ∫
dt 2 R r 2
+ 2R (t) & ) ∫ 5
% dt ( R r
p( R) ρ ρ
2 ∞ ∞ 2 ∞
$ dR ' * 1- 2 d 2 R * 1- $ dR ' * 1 -
= −2R(t) & ) −
,+ /. − R (t) 2 ,
− / + 2R 4 (t) & ) ,+− 4 /.
% dt ( r R dt + r .R % dt ( 4r R
2 2 2
$ dR ' d 2 R 1 $ dR ' d 2 R 3 $ dR '
= −2 & ) − R(t) 2 + & ) = −R(t) 2 − & ) .
% dt ( dt 2 % dt ( dt 2 % dt (
This produces an equation for the pressure p(R) in the liquid at r = R, but the pressure pB(t) inside
the bubble is sought.
Equation (4.93) provides the means to jump from the liquid to the gas side of the bubble's
surface. Here we assume that no mass is transfered and that "1" indicates the inside the bubble
while "2" indicates the outside of the bubble. In this case (4.93) reduces to:
0 = − ( p2 − p1 ) + ((τ rr )2 − (τ rr )1 ) − 2 σ R ,
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

where τrr = 2µ(∂ur/∂r) is the normal viscous stress, and the sign of the surface tension term is set
by the fact that the center of bubble surface curvature lies at the center of the bubble (in region
"1"). Recasting this boundary condition using p2 = p(R) and p1 = pB, leads to
# ∂u & 2σ
pB = p(R) − 2µ % r ( + ,
$ ∂r 'r=R R
where the viscous stress in the gas, (τrr)1, has been neglected. Inserting p(R) and ur from above
leads to:
2
d 2 R 3 ! dR $ ! ∂ ! R 2 (t) dR $$ 2σ
pB = ρ R 2 + ρ # & + p∞ − 2µ # # 2 && +
dt 2 " dt % " ∂r " r dt %%r=R R
2
d 2 R 3 ! dR $ 4µ dR 2σ
= ρ R 2 + ρ # & + p∞ + + ,
dt 2 " dt % R dt R
and this can be rearranged to reach the form provided in the question statement:
2
pB − p∞ d 2 R 3 # dR & 4µ dR 2σ
=R 2 + % ( + + .
ρ dt 2 $ dt ' ρ R dt ρ R
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.71. Redo the dimensionless scaling leading to (4.101) by choosing a generic viscous
stress, µU/l, and then a generic hydrostatic pressure, ρgl, to make p − p∞ dimensionless. Interpret
the revised dimensionless coefficients that appear in the scaled momentum equation, and relate
them to St, Re, and Fr.

Solution 4.71. (i) Viscous scaling of the pressure.

& ∂u )
Start from ρ( + (u ⋅ ∇ )u+ = −∇p + ρg + µ∇ 2u , and render it dimensionless using the
' ∂t *
dimensionless variables: x i = x i l , t ∗ = Ωt , u∗j = u j U , p∗ = ( p − p∞ )l µU , and g∗j = g j g .
∗

The result is:

€ ∂u* U 2 * * * 1 µU * * µU 2
UΩ * +
∂t€ l €
( u ⋅ ∇ )u = − ∇ p + gg* + 2 ∇ * u* ,
€ €ρl l ρl €
2
which can be simplied by dividing all terms by µU/ρl and using ν = µ/ρ:
$ Ωl Ul ' ∂u* $ Ul ' * * * $ gl Ul ' * *2 *
) * + & )(u ⋅ ∇ )u = −∇ p + & 2 )g + ∇ u .
* *
&
€ % U ν ( ∂t % ν ( %U ν (
Here the coefficients are the Strouhal number-Reynolds number product, the Reynolds number,
and the product of the inverse square of the Froude number and the Reynolds number. This
scaling is useful for low Reynolds number flow where it naturally produces a pressure gradient-
€ balance.
viscous force

(ii) Hydrostatic scaling of the pressure.

& ∂u )
Start from ρ( + (u ⋅ ∇ )u+ = −∇p + ρg + µ∇ 2u , and render it dimensionless using the
' ∂t *
dimensionless variables: x i = x i l , t ∗ = Ωt , u∗j = u j U , p∗ = ( p − p∞ ) ρgl , and g∗j = g j g .
∗

The result is:

€ ∂u* U 2 * * * 1 µU 2
UΩ * +
t
(
l €
u ⋅ ∇ )u = − ρgl∇ * p* + gg* + 2 ∇ * u* ,
€ ∂€ €ρl ρl €
which can be simplied by dividing all terms by g and using ν = µ/ρ:
# Ωl U 2 & ∂u* # U 2 & * * * # ν U 2 & *2 *
€
%
$ U gl ' ∂t $ gl '
( )
( * + % ( u ⋅ ∇ u = −∇ p + g + %
* * *
(∇ u .
$ Ul gl '
Here the coefficients are the product of the Strouhal number and the square of the Froude
number, the square of the Froude number, and the product of the inverse of the Reynolds number
and the square of the Froude number. This scaling is useful for low Froude number flow where it
naturally€produces a hydrostatic balance.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.72. A solid sphere of mass m and diameter D is released from rest and falls through
an incompressible viscous fluid with density ρ and viscosity µ under the action of gravity g.
When the z coordinate increases downward, the vertical component of Newton’s second law for
du
the sphere is: m z = +mg − FB − FD , where uz is positive downward, FB is the buoyancy force on
dt
the sphere, and FD is the fluid-dynamic drag force on the sphere. Here, with uz > 0, FD opposes
the sphere’s downward motion. At first the sphere is moving slowly so its Reynolds number is
low, but Re D = ρuz D µ increases with time as the sphere’s velocity increases. To account for
this variation in ReD, the sphere’s coefficient of drag may be approximated as: CD ≅ 12 + 24 Re D .
For the following items, provide answers in terms of m, ρ, µ, g and D; do not use z, uz, FB, or FD.
a) Assume the sphere's vertical equation of motion will be solved by a computer after being put
into dimensionless form. Therefore, use the information provided and the definition t * ≡ ρ gtD µ
d Re D
to show that this equation may be rewritten: *
= A Re 2D + B Re D + C , and determine the
dt
coefficients A, B, and C.
b) Solve the part a) equation for ReD analytically in terms of A, B, and C for a sphere that is
initially at rest.
c) Undo the dimensionless scaling to determine the terminal velocity of the sphere from the part
c) answer as t → ∞ .

g!
z!

D!

m!

µ, ρ!
vz!

Solution 4.72. a) The buoyant and drag forces on the sphere will be:
π 1 π 1 π ! 1 24 $
FB = ρ D 3g and FD = ρuz2 D 2CD = ρuz2 D 2 # + &.
6 2 4 2 4 " 2 Re D %
Thus, the approximate equation for the sphere’s velocity is:
du π 1 π " 1 24 %
m z = mg − ρ D 3g − ρuz2 D 2 $ + '.
dt 6 2 4 # 2 Re D &
Now introduce the dimensionless scalings one at a time. First use uz = µ Re D ρ D to eliminate uz.
2
duz µ d Re D π 3 1 " µ Re D % π 2 " 1 24 %
m =m = mg − ρ D g − ρ $ ' D $ + '.
dt ρ D dt 6 2 # ρ D & 4 # 2 Re D &
Simplify the right side:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

mµ d Re D " π % π " 1 24 %
= $ m − ρ D 3 ' g − µ 2 Re 2D $ + '.
ρ D dt # 6 & 8ρ # 2 Re D &
Switch to the dimensionless time t* using the chain rule for differentiation and t * = ρ gtD µ :
mµ d Re D mµ ! d Re D dt * $ mµ ! d Re D ρ gD $
= # &= # &
ρ D dt ρ D " dt * dt % ρ D " dt * µ %
d Re D ! π $ π ! 1 24 $
= mg *
= # m − ρ D 3 & g − µ 2 Re 2D # + &.
dt " 6 % 8ρ " 2 Re D %
Divide both sides of the final equality by mg, and rearrange the right side:
d Re D π " µ2 % 2 " µ2 % " π ρ D3 %
= − $ ' Re D − 3 π $ ' Re D $1−
+ '.
dt * 16 # ρ mg & # ρ mg & # 6 m &
π " µ2 % " µ2 % π ρ D3
Thus: A=− $ ' , B = −3 π $ ' , and C = 1− .
16 # ρ mg & # ρ mg & 6 m
b) Assuming C is positive, the scaled equation found for part a) can be separated and integrated:
d Re D 2 # 2A Re + B &
∫ A Re2 + B Re = ∫ dt * , or t ∗ + const. = − tanh −1 $ D
'.
D+ C
2 2
D B − 4AC % B − 4AC (
This integrated result can be algebraically inverted:
1 ) # 1 &,
Re D = *−B + B 2 − 4AC tanh %− B 2 − 4AC (t ∗ + const.)(- .
2A + $ 2 '.
The constant can be evaluated with the initial condition ReD = 0 at t = 0 to find:
/ ) B 2 − 4AC # &,31
1 1 B
Re D = 0−B + B 2 − 4AC tanh +− t ∗ + tanh −1 % (.4 ,
2A 12 +* 2 $ B 2 − 4AC '.-15
which is a formal solution to the problem.
c) Terminal velocity will occur when dReD/dt* = 0. Use this fact and the result of part a) to find:
π " µ2 % 2 " µ2 % " π ρ D3 %
0=− $ Re
' T − 3π $ ' T $1−
Re + ',
16 # ρ mg & # ρ mg & # 6 m &
where ReT is the Reynolds number at terminal velocity. Divide by the leading coefficient:
16 " ρ mg %" π ρ D 3 %
0 = ReT2 + 48ReT − $ 2 '$1− '.
π # µ &# 6 m &
Thus, using the quadratic formula and choosing the appropriate sign to ensure ReT > 0, yields:
" " 16 " ρ mg %" π ρ D 3 %% %
1$ 16 " ρ mg %" π ρ D 3 %
2
ReT = −48 + 48 − 4(1) $ − $ 2 '$1− ' ' ' = −24 + 24 2 + $ 2 '$1− '.
2 $# # π # µ &# 6 m && '& π # µ &# 6 m &
" %
µ "$ %
2
µ $ −B " B % C 16 " ρ mg %" π ρ D 3 % '
so: (uz )T = + $ ' − '= −24 + 24 2 + $ 2 '$1− ' .
ρ D $# 2A # 2A & A '& ρ D $# π # µ &# 6 m & '&
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.73. a) From (4.101), what is the dimensional differential momentum equation for
steady incompressible viscous flow as Re → ∞ when g = 0.
b) Repeat part a) for Re → 0 . Does this equation include the pressure gradient?
c) Given that pressure gradients are important for fluid mechanics at low Re, revise the pressure
scaling in (4.100) to obtain a more satisfactory low-Re limit for (4.39b) with g = 0.

Solution 4.73. Equation (4.101) with g = 0 is a scaled equation that involves the Strouhal
number St = Ωl/U, and the Reynolds number Re = ρUl/µ.
∂ u∗ 1
( )
St ∗ + u∗ ⋅ ∇* u∗ = −∇* p∗ + ∇∗2 u∗ .
∂t Re
a) When Re → ∞ , the viscous terms drop out. In dimensional terms, the result is the Euler
equation:
∂u 1
+ ( u ⋅ ∇) u = − ∇p ,
∂t ρ
which does apply to high-speed flows with weak viscous effects.
b) Multiply (4.101) by Re to find:
∂ u∗
( )
ReSt ∗ + Re u∗ ⋅ ∇* u∗ = −Re ∇* p∗ + ∇∗2 u∗ .
∂t
Taking the limit as Re → 0 leads to: ∇ 2 u = 0 , which does not include the pressure gradient and
is not correct.
c) The problem is the assumed scaling for the pressure that went into creating (4.101). As an
alternative, assume that pressure p should be scaled by a viscous stress ~ µU/l; p* = pl/µU. In this
case, the dimensionless version of (4.39b) with g = 0 becomes:
" Ωl % ∂ u∗ " µ % * ∗ " µ % *2 ∗
( ∗ *
$# '& ∗ + u ⋅ ∇ u = −$
U ∂t
) ∗
'∇ p + $
# ρUl &
' ∇ u , or
# ρUl &
∂ u∗ 1 1
St ∗
+ ( u∗ ⋅ ∇* ) u∗ = − ∇* p∗ + ∇*2 u∗ ,
∂t Re Re
where the scalings for all the other dependent variables (and not the pressure) are given by
(4.100). As Re → 0 with this pressure scaling, the dimensional momentum equation becomes:
∇p = µ∇ 2 u ,
which is the correct equation for low-Reynolds number creeping flow.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.74. a) Simplify (4.45) for motion of a constant-density inviscid fluid observed in a
frame of reference that does not translate but does rotate at a constant rate Ω = Ωez.
b) Use length, velocity, acceleration, rotation, and density scales of L, U, g, Ω, and ρ to
determine the dimensionless parameters for this flow when g = –gez and x = (x, y, z). (Hint:
subtract out the static pressure distribution.)
c) The Rossby number, Ro, in this situation is U/ΩL. What are the simplified equations of
motion for horizontal constant-density inviscid flow, u = (u, v, 0), observed in the rotating frame
of reference when Ro << 1.

Solution 4.74. a) For no translation, a constant rotation rate, and inviscid flow, (4.45) simplifies
to:
$ ∂u '
ρ & + (u ⋅ ∇)u ) = −∇p + ρ g − ρ -.2Ω × u + Ω × (Ω × x )/0 ,
% ∂t (
where the primes have been dropped. When Ω = Ωez, g = –gez, and x = (x, y, z), several terms in
the equation can be evaluated:
∂u 1
+ (u ⋅ ∇)u = − ∇p − ge z − 2 (−Ωve x + Ωue y ) + Ω2 ( xe x + ye y ) ,
∂t ρ
where u = (u, v, w) and whole equation has been divided by ρ.
b) The primary point of difficulty here is figuring out how to scale the pressure. One way to
begin is to set u = 0 in the momentum equation, to determine the equation for ps, the static
pressure distribution.
1
0 = − ∇ps − ge z + Ω2 ( xe x + ye y )
ρ
Subtract this equation from the full momentum equation to find:
∂u 1
+ (u ⋅ ∇)u = − ∇( p − ps ) − 2 (−Ωve x + Ωue y ) .
∂t ρ
Now continue the scaling task using p – ps. For this equation, use the following definitions of
dimensionless variables:
xi∗ = xi L , t ∗ = Ωt , u∗j = u j U , and p∗ = ( p − ps ) ρΩUL .
This scaling for the pressure should apply when both fluid motion and rotation of the coordinate
frame play a role. Inserting these into the revised version of (4.45) produces:
∂u∗ U 2 ∗ ∗ ∗ ρΩUL ∗ ∗
ΩU ∗ + (u ⋅ ∇ )u = − ∇ p − 2ΩU (−v∗e x + u∗e y ) .
∂t L ρL
Divide the whole equation by ΩU to find:
∂u∗ U ∗ ∗ ∗
∗
+ (u ⋅ ∇ )u = −∇∗ p∗ − 2 (−v∗e x + u∗e y ) .
∂t ΩL
Thus, the dimensionless parameter that governs this flow is U/ΩL.
c) When U/ΩL = Ro << 1, the non-linear advective accleration terms may be dropped, and this
leaves:
∂u 1 ∂p ∂v 1 ∂p
=− + 2Ωv and =− − 2Ωu .
∂t ρ ∂x ∂t ρ ∂y
These equations are commonly used in the study of large-scale atmospheric flows (see Ch. 13).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.75. From Figure 4.23, it can be seen that CD ∝ 1/Re at small Reynolds numbers and
that CD is approximately constant at large Reynolds numbers. Redo the dimensional analysis
leading to (4.99) to verify these observations when:
a) Re is low and fluid inertia is unimportant so ρ is no longer a parameter.
b) Re is high and the drag force is dominated by fore-aft pressure differences on the sphere and µ
is no longer a parameter.

Solution 4.75. Overall there are 5 parameters D = drag force, U, d, ρ, and µ.

a) When ρ is not a parameter, the dimensional analysis proceeds a follows. The parameter &
units matrix is:
D U d µ
M 1 0 0 1
L 1 1 1 -1
T -2 -1 0 -1

This rank of this matrix is three. There are 4 parameters and 3 independent units, so there will be
1 dimensionless group:
D
Π1 = = const., or D = (const.)µUd .
µUd
Divide both sides of the last equation by 12 ρU 2 d 2π /4 to find:
D (const.)µUd µ const.
1 2 2
= CD = 1 € 2 2 = const. = ,
€ 2 ρU d π /4 2
ρU d π /4 ρUd Re
where the 2, 4, and π have been€absorbed into the undetermined constant.
b) When µ is not a parameter, the dimensional analysis proceeds a follows. The parameter &
units matrix is:
€
D U d ρ
M 1 0 0 1
L 1 1 1 -3
T -2 -1 0 0

This rank of this matrix is three. There are 4 parameters and 3 independent units, so there will be
1 dimensionless group:
D
Π1 = 2 2
= const., or D = (const.) ρU 2 d 2 .
ρU d
Divide both sides of the last equation by 12 ρU 2 d 2π /4 to find:
D (const.) ρU 2 d 2
1 2 2
= C€D = 1 2 2
= const.,
€ 2
ρU d π /4 2
ρU d π /4
where the 2, 4, and π have again
€ been absorbed into the undetermined constant.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.76. Suppose that the power to drive a propeller of an airplane depends on d (diameter
of the propeller), U (free-stream velocity), ω (angular velocity of propeller), c (velocity of
sound), ρ (density of fluid), and µ (viscosity). Find the dimensionless groups. In your opinion,
which of these are the most important and should be duplicated in a model testing?

Solution 4.76. Start with dimensional analysis. The using P = power, the parameter & units
matrix is:
P U d ω ρ µ c
M 1 0 0 0 1 1 0
L 2 1 1 0 -3 -1 1
T -3 -1 0 -1 0 -1 -1

This rank of this matrix is three. There are 7 parameters and 3 independent units, so there will be
P ρUd
4 dimensionless groups. By inspection these are: Π1 = 3 2
= a power coefficient, Π 2 =
ρU d µ
ωd U
= Reynolds number, Π 3 = = Strouhal number, and Π 4 = = Mach number.
U c
At low speeds the Mach number should not matter so, but the Reynolds number will
determine boundary layer characteristics, € and the Strouhal number will set €the effective angle of
attack along the propeller's blades; therefore:
€ P € ωd '
$ ρUd U
3 2
= f& , ) for << 1 .
ρU d % µ U( c
At higher speeds where the impact of compressibility is felt, vicous effects may be irrelevant, so:
P $ ωd U ' U
3 2
= f & , ) for near unity or higher.
ρU d % U c ( €c
€

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.77. A 1/25 scale model of a submarine is being tested in a wind tunnel in which p =
200 kPa and T = 300 K. If the prototype speed is 30 km/hr, what should be the free-stream
velocity in the wind tunnel? What is the drag ratio? Assume that the submarine would not
operate near the free surface of the ocean.

Solution 4.77. Since submarine are not designed for extended operations on the sea surface,
gravity is unimportant. Therefore the coefficient of drag depends only on the Reynolds number
which should be duplicated in model test. Equating Reynolds numbers between the model
(subscript "m") and the full-size submarine (subscript "s") produces:
U m Lm U s Ls L ν
= , or U m = U s s m .
νm νs Lm ν s
Using the numbers from the problem statement leads to:
ρ m = pm Rm Tm =200,000Pa (287m 2 s−2K)(300K) = 2.32kgm−3
−5 −1 −1
€ν m = µm = 1.85 ×10 € kgm −3
s
= 7.97 ×10−6 m 2 s−1, and
ρm 2.32kgm
€ 7.97 ×10−6 m 2 /s
U m = ( 30kph )(25) = 5,980kph !
10−6 m 2 /s
This speed is not reasonable because it is so high, and because it corresponds to a Mach number
€
of nearly 5 in air while the Mach number of the submarine in water will be ~0.006. Thus, all the
critical dimensionless numbers cannot be matched, and this exercise illustrates a leading
€ model testing: Reynolds numbers typically cannot be matched without
difficulty of scale
introducing other (possibly significant) problems.
If the coefficient of drag were the same in both cases (it will not be!), then the force ratio
would be:
Dm
=
[ ]
CD 12 ρU 2 ⋅ Area
m
=
ρ m U m2 Aream 2.32 5,980 2 1
= = 0.147 .
Ds [ ]
CD 12 ρU 2 ⋅ Area
s
ρ s U s2 Areas 10 3 30 2 25 2
which suggests that the tiny model must be very robust! The drag it will experience is 15% of the
drag on the full size submarine but its surface area will be 625 times smaller. This calculation
shows that small size models designed for large Reynolds number testing must be exceptionally
€
sturdy.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.78. The volume flow rate Q from a centrifugal blower depends on its rotation rate Ω,
its diameter d, the pressure rise it works against Δp, and the density ρ and viscosity µ of the
working fluid.
a) Develop a dimensionless scaling law for Q in terms of the other parameters.
b) Simplify the result of part c) for high Reynolds number pumping where µ is no longer a
parameter.
c) For d = 0.10 m and ρ = 1.2 kg/m3, plot the measured centrifugal blower performance data
from the table in dimensionless form to determine if your result for part b) is a useful
simplification. Here RPM is revolutions per minute, Q is in liter/s, and Δp is in kPa.
RPM = 5000 8000 11,000
Q ΔP Q ΔP Q ΔP
0.3 0.54 0.5 1.4 0.9 2.6
1.1 0.51 2.0 1.3 4.1 2.2
1.5 0.48 3.3 1.1 6.2 1.8
2.8 0.37 5.0 0.84 7.7 1.3
3.8 0.24 6.5 0.49 9.5 0.81
d) What maximum pressure rise would you predict for a geometrically similar blower having
twice the diameter if it were spun at 6,500 RPM?

Solution 4.78. a) As in Example 4.17, the six parameters Q, Ω, d, Δp, ρ and µ lead to three
dimensionless groups. These are:
Q Δp ρΩd 2
The flow coefficient = , The head coefficient = , and Re = .
Ωd 3 ρΩ2 d 2 µ
Q $ Δp ρΩd 2 '
Thus, the scaling law is: = Ψ & 2 2, ) , where Ψ is an undetermined function.
Ωd 3 % ρΩ d µ (
b) At high Reynolds number where µ is no longer a parameter, the Reynolds number is no longer
a parameter and this scaling law simplifies to:
Q $ Δp '
3
= Φ& 2 2 ) ,
Ωd % ρΩ d (
where Φ is an undetermined function.
c) The task here is to plot the data provided in dimensionless form. Converting RPM to radians
per second and liters per second to (meters)3 per second, and then moving to dimensionless
parameters leads to the following dimensionless tabulations:

Ω= 523.6 rad/s 837.8 rad/s 1152 rad/s

Coefficients: flow head flow head flow head
0.000573 0.1641 0.000597 0.1662 0.000781 0.1633
0.002101 0.1550 0.002387 0.1544 0.003559 0.1382
0.002865 0.1459 0.003939 0.1306 0.005382 0.1130
0.005348 0.1125 0.005968 0.0997 0.006685 0.0816
0.007257 0.0730 0.007759 0.0582 0.008247 0.0509

These are plotted below, and to follow standard engineering practice the flow coefficent is
plotted on the horizontal axis.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

0.18"
0.16" 5000"RPM"

0.14" 8000"RPM"

11000"RPM"

0.12"
Δp
0.1"
ρΩ2 d 2
0.08"
0.06"
0.04"
0.02"
0"
0" 0.002" 0.004" 0.006" 0.008" 0.01"
Q
Ωd 3

The data provided in the tabulations collapse well eventhough the Reynolds number varies by
more than a factor of two; Re increases in direct proportion to the rotational speed. Thus, the
conclusion from this plot is that the assumption leading to the part b) result is a justified
simplification for this data set.
d) From the first row of entries in the dimensionless tabulations of part c) the average highest
head coefficient value is 0.1645. The data collapse shown then suggests that
Δp
= 0.1645 ,
ρΩ2 d 2
is valid for finding the peak pressure rise of any geometrically similar blower, independent of the
Reynolds number. Thus, the predicted peak pressure rise for a geometrically similar blower
having twice the diameter (0.2 m) and spinning at 6,500 RPM (680.7 rad/s) is:
Δp = 0.1645ρΩ2 d 2 = 0.1645(1.2kg / m 3 )(680.7rad / s)2 (0.2m)2 = 3.66kPa .
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 4.79. A set of small-scale tank-draining experiments are performed to predict the liquid
depth, h, as a function of time t for the draining process of a large cylindrical tank that is
geometrically similar to the small-experiment tanks. The relevant parameters are gravity g,
initial tank depth H, tank diameter D, orifice diameter d, and the density and viscosity of the
liquid, ρ and µ, respectively.
a) Determine a general relationship between h and the other parameters.
b) Using the following small-scale experiment results, determine whether or not the liquid’s
viscosity is an important parameter.
H = 8 cm, D = 24 cm, d = 8 mm H = 16 cm, D = 48 cm, d = 1.6 cm
h (cm) t (s) h (cm) t (s)
8.0 0.00 16.0 0.00
6.8 1.00 13.3 1.50
5.0 2.00 9.5 3.00
3.0 3.00 5.3 4.50
1.2 4.00 1.8 6.00
0.0 5.30 0.0 7.50
c) Using the small-scale-experiment results above, predict how long it takes to completely drain
the liquid from a large tank having H = 10 m, D = 30 m, and d = 1.0 m.

Solution 4.79. a) The required relationship can be obtained from dimensional analysis. The
parameter & units matrix is:
h t g H D d ρ µ
M 0 0 0 0 0 0 1 1
L 1 0 1 1 1 1 -3 -1
T 0 1 -2 0 0 0 0 -1

This rank of this matrix is three. There are 8 parameters and 3 independent units, so there will be
5 dimensionless groups. All are readily
h D
found by inspection: Π1 = , Π 2 = ,
H H
2 12 32
d gt ρg H
Π3 = , Π4 = , and Π 5 = .
H H µ
Thus, the general
€ relationship
€ is: h/H
h # D d gt ρg H &
2 1 2 3 2
= f% , , , (.
€ € H $€H H H µ '
b) If the effects of viscosity are unimportant,
then a plot of h/H vs gt2/H will be the same
for both tests. Such a plot is shown here, and
€ the two data sets are essentially identical.
Thus, there is no influence of the viscosity
for the tested geometry. gt2/H
c) From the experimental data the tank is
empty when gt2/H ≈ 3500. Thus, the time to drain the large tank will be:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

3450H 3500(10m)
t= = = 59.8s ≈ 1.0 minute.
g 9.8ms−2

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.1. A closed cylindrical tank 4 m high and 2 m in diameter contains water to a depth
of 3 m. When the cylinder is rotated at a constant angular velocity of 40 rad/s, show that nearly
0.71 m2 of the bottom surface of the tank is uncovered. [Hint: The free surface is in the form of a
paraboloid of revolution. For a point on the free surface, let h be the height above the
(imaginary) vertex of the paraboloid and r be the local radius of the paraboloid. From Section
5.1, h = ω 02 r 2 2g , where ω0 is the angular velocity of the tank. Apply this equation to the two
points where the paraboloid cuts the top and bottom surfaces of the tank.]

Solution 5.1. Denote the local radius of the paraboloid of revolution as r(h), and let h = 0 at
vertex of this parabaloid. The goal of this problem is to determine πr2(h1), where h1 coincides
with the bottom of the tank.
Start with the geometric constraints. The volume V of water in the tank is: V =
π(1.0m)2(3m) = 3π m3, where the tank radius is R = 1 m. This volume is conserved. Thus, when
the tank is rotating, the following must be true:
h= h 2
V= ∫ π ( R 2 − r 2 (h))dh ,
h= h1

where h2 – h1 = 4 m. From the hint we can write:

h = ω 02 r 2 2g , or r 2 = 2gh ω 02 ,
so that the volume relationship becomes:
€ h2
h= h 2 %
2g ( % 2 2g h 2 ( % g (
V = ∫ π ' R − 2 h *dh = π ' R h − 2 * = π ' R 2 ( h2 − h1 ) − 2 ( h22 − h12 )*
2

h= h1 & € ω 0 ) & € ω 0 2 ) h1 & ω0 )

% g (
= π ( h2 − h1 )' R 2 − 2 ( h2 + h1 )*.
& ω0 )
Substitute in V = 3π, R = 1, ω0 = 40, g = 9.81, and h2 – h1 = 4 m to find:
$ 9.81 ' 1 )1600 # 3 & ,
3π = 4 π &1− (2h1 + 4 )) or h1 = + %1− ( − 4. = 18.39m ,
€ % 1600 ( 2 * 9.81 $ 4 ' -
so the uncovered area at the bottom of the tank is:
πr 2 (h1) = 2πgh ω 02 = 2π (9.81)(18.39) 1600 = 0.0708m 2 .
€ €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.2. A tornado can be idealized as a Rankine vortex with a core of diameter 30 m. The
gauge pressure at a radius of 15 m is −2000 N/m2 (i.e.,, the absolute pressure is 2000 N/m2 below
atmospheric).
(a) Show that the circulation around any circuit surrounding the core is 5485 m2/s. [Hint: Apply
the Bernoulli equation between infinity and the edge of the core.]
(b) Such a tornado is moving at a linear speed of 25 m/s relative to the ground. Find the time
required for the gauge pressure to drop from −500 to −2000 N/m2. Neglect compressibility
effects and assume an air temperature of 25°C. (Note that the tornado causes a sudden decrease
of the local atmospheric pressure. The damage to structures is often caused by the resulting
excess pressure on the inside of the walls, which can cause a house to explode.)

Solution 5.2. At 25°C and one atmosphere, air density is ρair = (101.3 kPa)/(287m2s–2K–1)(298K)
= 1.18 kgm–3.
a) The flow is irrotational outside the Rankine vortex core, so the steady constant density
Bernoulli equation implies at any distance r from the center of vortex core:
1 1 12
ρ airU∞2 + p∞ = ρ airU 2 (r) + p(r) , or U(r) = [2( p∞ − p(r)) ρ air ] ,
2 2
where the "∞" subscript refers to conditions very far from the vortex (U∞ ≈ 0). From equation
(3.28), U(r) = Γ/2πr, so where r = rc = the core radius:
12 12
€ Γ = 2πrc [2( p∞ − pc ) ρ air ] =€2π (15)[2(2000) 1.18] = 5485m 2 s−1 .
b) The radial distance at which the gauge pressure is –500 Pa can be determined from the results
developed for part a) for U(r):
12 −1 2
U(r) = Γ/2πr = = [2( p∞ − p(r)) ρ air ] , or r = (Γ 2π )[2( p∞ − p(r)) ρ air ] .
€
Evaluate to find r, and divide by (r – rc) by 25 m/s to determine the time for the gauge pressure
to drop from −500Pa to −2000Pa.
−1 2 −1 2
2( p∞ − p(r)) ρ air ] = (5485m
r = (Γ 2π )[€ € s 2π )[2(500Pa) 1.18kgm ]
2 −1 −3
= 30.0m , so
time = (30m – 15m)/25ms–1 = 0.60 s.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.3. The velocity field of a flow in cylindrical coordinates (R, ϕ, z) is u = (uR, uϕ, uz) =
(0, aRz, 0) where a is a constant.
(a) Show that the vorticity components are ω = (ωR, ωϕ, ωz) = (–aR, 0, 2az).
(b) Verify that ∇ ⋅ ω = 0 .
(c) Sketch the streamlines and vortex lines in an (R,z)-plane. Show that the vortex lines are given
by zR2 = constant.
€
Solution 5.3. a) The vorticity components in cylindrical coordinates are provided in Appendix B:
1 ∂uz ∂uϕ
ωR = − = 0 – aR = −aR
R ∂ϕ ∂z
∂u ∂u
ωϕ = R − z = 0 – 0 = 0 , and
∂z ∂R
1 ∂ (Ruϕ ) 1 ∂uR
€ ωz = − = 2az – 0 = 2az .
R ∂R R ∂ϕ
€ 1 ∂ (Rω R ) 1 ∂ωϕ ∂ω z 1 ∂ (−aR 2 ) 1 ∂ (0) ∂ (2az)
b) ∇ ⋅ ω = + + = + + = −2a + 2a = 0
R ∂R R ∂ϕ ∂z R ∂R R ∂ϕ ∂z
€ c) The only non-zero velocity component is uϕ, so the streamlines are circles centered on the z
axis. Vortex lines are given by:
dR dz dR dz 1
€ = → = → − ln(R) = ln(z) + const.
ωR ωz −aR 2az 2
Exponentiate the last expression to find:
1 R = const. z , or R2z = const.
Thus the vortex lines asymptote to the plane z = 0 and to the z-axis.
€
z

R
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.4. Starting from the flow field of an ideal vortex (5.2), compute the viscous stresses
τrr, τrθ, and τθθ, and show that the net viscous force on a fluid element, (∂τij/∂xi), is zero.

Solution 5.4. In two dimensions, the ideal vortex flow field is u =(ur, uθ) = (0, Γ/2πr). This flow
field is incompressible since ∇ ⋅ u = (1 r)(∂uθ ∂θ ) = 0 . Therefore, the viscous stress listings for
plane polar coordinates in Appendix B apply:
∂u ! 1 ∂ uθ ur $
τ rr = 2µ Srr = 2µ r = 0 , τ θθ = 2µ Sθθ = 2µ # + & = 0 , and
∂r " r ∂θ r %
€
! ∂ ! u $ 1 ∂ ur $ ! Γ $ Γ
τ rθ = 2µ Srθ = µ # r # θ & + & = µr # 3&
(−2) = −µ 2 .
" ∂ r " r % r ∂θ % " 2π r % πr
For incompressible flow (∂ui/∂xi = 0) with constant viscosity:
∂τ ij ∂ ∂ ! ∂ ui ∂ u j $ ∂ ∂ ui ∂ 2u ∂ 2u
= 2µ Sij = µ ## + && = µ + µ 2j = µ 2j = ∇ 2 u .
∂ xi ∂ xi ∂ xi " ∂ x j ∂ xi % ∂ x j ∂ xi ∂ xi ∂ xi
Using the velocity field given above, and remembering to differentiate unit vectors, this
becomes:
1∂% ∂ ( 1 ∂2 1 ∂ % ∂uθ ( 1 ∂
∇ 2u = ' r ( uθ eθ )* + 2 2 ( uθ eθ ) = ' reθ *+ (−uθ e r )
r ∂r & ∂ r ) r ∂θ r ∂r & ∂r ) r 2 ∂θ
1 ∂uθ ∂ 2 uθ uθ µΓ % 1 2 1 (
= eθ + eθ 2
− 2 eθ = −eθ '− + − * = 0.
r ∂r ∂r r π & r3 r3 r3 )

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.5. Consider the situation depicted in Figure 5.6. Use a Cartesian coordinate system
with a horizontal x-axis that puts the barrier at x = 0, a vertical y-axis that puts the bottom of the
container at y = 0 and the top of the container at y = H, and a z-axis that points out of the page.
Show that, at the instant the barrier is removed, the rate of baroclinic vorticity production at the
interface between the two fluids is:
Dω z 2( ρ 2 − ρ1 )g
= ,
Dt ( ρ 2 + ρ1 )δ
where the thickness of the density transition layer just after barrier removal is δ << H, and the
density in this thin interface layer is assumed to be (ρ1 + ρ2)/2. If necessary, also assume that
fluid pressures match at y = H/2 just after barrier removal, and that the width of the container
€
into the page is b. State any additional assumptions that you make.
y=H
Solution 5.5. At the instant the barrier is removed hydrostatic pressure dy
will persist in each liquid. Let this pressure be p2 for the fluid on the
left and p1 for the fluid on the right. If p1 and p2 match at the center of "2 "1
the square container, then:
p2 = po − ρ 2 g( y − H 2) , and p1 = po − ρ1g( y − H 2) . !
The final answer does not depend on this matching point. y=0
Now consider a direct application of Newton's second law in to a small element, δ wide
and dy tall, located at height y above the bottom of the container and straddling the interface. At
€the instant of interest the € fluid is not yet moving so the forces come from the hydrostatic
pressures. The horizontal equation of motion is:
# ρ 2 + ρ1 & du
% ((bδdy ) = ( p2 − p1 )bdy ,
$ 2 ' dt
where u is the element's horizontal velocity, and b is the dimension of the container into the
page. Because the pressure forces are hydrostatic, there is no vertical pressure force imbalance so
the vertical component of Newton's second law for the same fluid element is:
€ # ρ 2 + ρ1 & dv
% ((bδdy ) = 0.
$ 2 ' dt
Thus, since v is initially zero along the interface, it remains that way until the fluid starts moving.
Returning to the horizontal equation, canceling common factors, and inserting the
relationships for p1 and p2 produces:
# ρ 2 + ρ1 & du €
% ((δ ) = po − ρ 2 g( y − H 2) − po + ρ1g( y − H 2) = ( ρ 2 − ρ1 ) g(−y + H 2) .
$ 2 ' dt
Simplify and solve for du/dt to find:
du 2( ρ 2 − ρ1 ) g
= (−y + H 2) .
€ dt ( ρ2 + ρ1)δ
In this circumstance the z-component of vorticity is ωz = (∂v/∂x) – (∂u/∂y). At the instant of
interest, v = dv/dt = 0, so
d d % ∂u ( ∂ % du ( 2( ρ 2 − ρ1 ) g
€ ω z = '− * = − ' * = .
dt dt & ∂y ) ∂y & dt ) ( ρ 2 + ρ1 )δ
Thus, when the heavier fluid is on the left, positive vorticity directed along the z-axis is produced
when the barrier is removed.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.6. At t = 0 a constant-strength z-directed vortex sheet is created in the x-z plane (y =
0) in an infinite pool of a fluid with kinematic viscosity ν, that is ω(y,0) = ezγδ(y). The symmetry
of the initial condition suggests that ω = ωzez and that ωz will only depend on y and t. Determine
ω(y,t) for t > 0 via the following steps.
a) Determine a dimensionless scaling law for ωz in terms of γ, ν, y, and t.
b) Simplify the general vorticity equation (5.13) to a linear field equation for ωz for this situation.
c) Based on the fact that the field equation is linear, simplify the result of part a) by requiring ωz
to be proportional to γ, plug the simplified dimensionless scaling law into the equation
determined for part b), and solve this equation to find the undetermined function to reach:
γ ' y2 *
ω z (y,t) = exp(− +
2 πνt ) 4νt ,

Solution 5.6. a) This part of the exercise is dimensional analysis. The parameters are: ωz, γ, ν, y,
and t. Use these to create the parameter matrix:
€
ωz γ ν y t
––––––––––––––––––––––––––––
Mass: 0 0 0 0 0
Length: 0 1 2 1 0
Time: -1 -1 -1 0 -1

This rank of this matrix is two so 5 parameters – 2 dimensions = 3 groups. These are readily
constructed by inspection: ∏1 = ωzt, ∏2 = γ[t/ν]1/2, ∏3 = y/[νt]1/2. Thus, a dimensionless scaling
law is: ωzt = Φ(γy/ν, y/[νt]1/2), where Φ is an unknown function.
b) The vorticity equation in an inertial frame of reference is:
∂ω
+ (u ⋅ ∇ )ω = (ω ⋅ ∇)u + ν∇ 2ω .
∂t
Here ω = (0, 0, ωz), so this equation simplifies to:
∂ω z ∂ω z ∂ω z ∂ω z ∂w % ∂ 2ω z ∂ 2ω z ∂ 2ω z (
+u +v +w = ωz + ν' 2 + + 2 *.
∂t € ∂x ∂y ∂z ∂z & ∂x ∂y 2 ∂z )
where u = (u, v, w). However, the exercise statement suggests that ωz only depends on y and t. In
this case the vorticity equation simplifies further.
∂ω z ∂ω ∂w % ∂ 2ω (
€ + v z = ωz + ν ' 2z * .
∂t ∂y ∂z & ∂y )
The boundary conditions on ωz are independent of x and z. Therefore, all the dependent field
variables should all be independent of these directions. This implies ∂u/∂x = ∂w/∂z = 0, so the
incompressible flow continuity equation becomes ∂v/∂y = 0 which implies v = constant. A
constant velocity can be€ removed from the field equations via a Galilean transformation to a
moving coordinate system. Thus, v can be chosen equal to zero without loss of generality, so the
∂ω z % ∂ 2ω z (
middle terms in the vorticity equation drop out completely leaving: = ν' 2 * .
∂t & ∂y )
c) For ωz to be the solution of a linear equation it must be proportional to γ the initial field
strength. Therefore set: ωz = (γ/[νt]1/2)F(y/[νt]1/2), where F is merely another undetermined
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

function. Compute the derivatives that compose the field equation using η = y/[νt]1/2 as the
argument of F.
∂ω z 1 γ γ ) y , ∂ 2ω z νγ 1
=− F( η ) + F ((η)+ − . , and ν 2
= F &&(η) ,
∂t 2 νt 3 νt * 2 νt 3 - ∂y νt νt
where the prime denotes differentiation of F with respect to its argument. Recreating the field
equation produces:
1 γ γ ' y * νγ 1 1 1
€ − F( η ) + F &(η)) − ,= F &&€
(η) , or F ""(η) + ηF "(η) + F(η) = 0 .
2 νt 3
νt ( 2 νt + 3
νt νt 2 2
This last equation can be rewritten, and integrated once to find:
d $ 1 ' 1
& F #(η) + ηF(η)) = 0 → F "(η) + ηF(η) = C ,
€ dη % 2 ( € 2
where C is a constant. The resulting first-order differential equation has an integrating factor of
exp{+η2/4} that leads to:
d +η 2 4 € η

( ) € F(η) = De−η 2 4 + e−η 2 4 C ∫ e +ξ 2 4 dξ .

2
€ e F(η) = Ce +η 4 , so
dη 0
where D is another constant. The initial condition, ωz = 0 for y > 0 at t = 0, requires C = 0. This
leaves:
D % y2 (
€ ω€z (y,t) = exp&− ).
νt ' 4νt *
To match the initial spatial profile, ωz(y,0) = γδ(y), integrate in y and take the limit as t → 0 .
+∞ ) +∞ , ) +∞ D 0 y2 3 , ) +∞ ,
∫ γδ (y)dy = γ = lim + ∫ ω z (y,t)dy . = lim + ∫ exp 1 − 4 dy . = 2Dlim + ∫ exp{−ζ 2 }dζ . = 2D π .
−∞
t →0 *
−∞ € - t →0* −∞ νt 2 4νt 5 - t →0 *
−∞ -
Thus, D = γ 2 π , so the final solution is: €
γ ' y2 *
ω z (y,t) = exp(− +.
€ 2 πνt ) 4νt ,
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.7. a) Starting from the continuity and Euler equations for an inviscid compressible
fluid, ∂ρ ∂t + ∇ ⋅ ( ρu) = 0 and ρ( Du Dt ) = −∇p + ρg , derive the Vazsonyi equation:
D $ω ' $ω ' 1
& ) = & ) ⋅ ∇u + 3 ∇ρ × ∇p ,
Dt % ρ ( % ρ ( ρ
€ when the body force€is conservative: g = −∇Φ . This equation shows that ω/ρ in a compressible
flow plays the nearly same dynamic role as ω in an incompressible flow [see (5.26) with Ω = 0
and ν = 0]. €
b) Show that the final term in the Vazsonyi equation may also be written: (1 ρ )∇T × ∇s .
€
c) Simplify the Vazsonyi equation for barotropic flow.

Solution 5.7. a) Start by taking the curl of the Euler equation:

(∂u 1 + ∂ω € 01 3
∇ × * + (u ⋅ ∇)u = − ∇p + g- → + ∇ × (u ⋅ ∇)u = −∇ × 2 ∇p5 + ∇ × g .
) ∂t ρ , ∂t 1ρ 4
2
The second term can modified by using the vector identity: (u ⋅ ∇)u = ω × u + 12 ∇ u ,
2
∇ × (u ⋅ ∇)u = ∇ × (ω × u) + ∇ × 12 ∇ u = ∇ × (ω × u) ,
€since ∇ × ∇(any scalar) = 0 . This can be further manipulated using the vector identity (B3.10):
∇ × (ω × u) = (u ⋅ ∇ )ω – u(∇€⋅ ω ) + ω (∇ ⋅ u) − (ω ⋅ ∇)u .
€ term on the right is zero because ∇ ⋅ ω = ∇ ⋅ (∇ × u) = 0 , and the third term on
Here the second
€ the right can be rewritten using the continuity equation (4.8):
ω Dρ
€ ∇ × (ω × u) = (u ⋅ ∇ )ω − − (ω ⋅ ∇ )u .
ρ Dt
The term involving the pressure in the curl € of the Euler equation is:
&1 ) & 1) 1 & 1)
−∇ × ( ∇p+ = −(∇ + × ∇p − ∇ × ∇p = −(∇ + × ∇p .
€ ' ρ * ' ρ* ρ ' ρ*
The body force term in the curl of the Euler equation is: ∇ × g = −∇ × (∇Φ) = 0 .
Thus, the curl of the Euler equation becomes:
€ 1 ( ∂ω ω Dρ + 1 ( 1+
* + (u ⋅ ∇ )ω + − (ω ⋅ ∇)u- = − * ∇ - × ∇p
ρ ) ∂t ρ Dt , ρ ) ρ,
€
when each side is divided by ρ. Use of the definition of D/Dt simplifies the left side:
1 Dω D(1 ρ) ' ω * 1' 1*
+ω − ) ⋅ ∇,u = − )∇ , × ∇p .
€ ρ Dt Dt (ρ + ρ ( ρ+
Combine the first to terms, and perform two simple rearrangements to reach the requisite form:
D $ω ' $ω ' 1
& ) = +& ⋅ ∇)u + 3 ∇ρ × ∇p .
€ Dt % ρ ( % ρ ( ρ
b) Start from the thermodynamic property relationship: de = Tds – pd(1/ρ). Therefore:
∇e = T∇s − p∇ (1 ρ ) , and ∇ × ∇e = 0 = ∇ × (T∇s) − ∇ × ( p∇(1 ρ)) = ∇T × ∇s − ∇p × ∇(1 ρ) ,
since ∇ × ∇s = ∇ × ∇(€ 1 ρ) = 0 . Thus, (1 ρ 3 )∇p × ∇ρ = (1 ρ )∇T × ∇s .
c) In barotropic flow, ∇p and ∇ρ will be parallel, so the Vazsonyi equation becomes
€ (D Dt )(ω ρ) = (€ω ρ) ⋅ ∇u
€ €
€ €
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.8. Starting from the unsteady momentum equation for a compressible fluid with
constant viscosities:
Du
ρ
Dt
( )
+ ∇p = ρg + µ∇ 2u + µυ + 13 µ ∇ (∇ ⋅ u) ,
show that
∂u µ 1
∂t
( 2
)
+ ω × u = T∇s − ∇ h + 12 u + Φ − ∇ × ω + µυ + 43 µ ∇ (∇ ⋅ u)
ρ ρ
( )
€
where T = temperature, h = enthalpy per unit mass, s = entropy per unit mass, and the body force
is conservative: g = −∇Φ . This is the viscous Crocco-Vazsonyi equation. Simplify this equation
for steady inviscid non-heat-conducting flow to find the Bernoulli equation (4.78)
€ 2
h + 12 u + Φ = constant along a streamline,
which is
€ valid when the flow is rotational and nonisothermal.

Solution 5.8. Divide the given momentum equation by ρ and use a vector identity (B3.9 with u =
€ 2
F) for the advective acceleration term, (u ⋅ ∇)u = ω × u + 12 ∇ u ,
Du 1 ∂u 2 1 µ 1
+ ∇p =
Dt ρ ∂t ρ ρ ρ
(
+ ω × u + 12 ∇ u + ∇p = g + ∇ 2u + µυ + 13 µ ∇(∇ ⋅ u) . )
Using the second equality, € g = −∇Φ , and the vector identity (B3.13) for the Laplacian of a
vector, ∇ 2u = ∇(∇ ⋅ u) − ∇ × (∇ × u) = ∇(∇ ⋅ u) − ∇ × ω , allows the last equation to be rewritten:
∂u 2 1 µ 1
€
∂t
+ ω × u + 12 ∇ u + ∇p = −∇Φ + (∇(∇ ⋅ u) − ∇ × ω ) + µυ + 13 µ ∇ (∇ ⋅ u) .
ρ ρ ρ
( )
€
Collect and combine terms to see how close this is to the target equation.
€
∂u 1 µ 1
∂t
( 2
) (
+ ω × u = − ∇p − ∇ 12 u + Φ − ∇ × ω + µυ + 43 µ ∇ (∇ ⋅ u) .
ρ ρ ρ
)
€
Comparing this equation with the target equation shows that the term involving the
thermodynamic variable must be modified. Use the property relationship, dh = Tds + (1/ρ)dp, to
deduce:
€ ∇h = T∇s + (1 ρ)∇p → − (1 ρ)∇p = T∇s − ∇h .
Substituting this into the last equation produces:
∂u µ 1
∂t
( 2
) (
+ ω × u = T∇s − ∇ h + 12 u + Φ − ∇ × ω + µυ + 43 µ ∇ (∇ ⋅ u) ,
ρ ρ
)
€
which is the requisite result.
Steady flow implies (∂/∂t)(anything) = 0. Inviscid implies µ = µυ = 0. Inviscid
(frictionless) non-heat conducting flows are isentropic, so ∇s = 0 . Thus, the Crocco-Vazsonyi
€ simplifies to:
equation
( 2
ω × u = −∇ h + 12 u + Φ .)
Take the dot-product of this equation with the streamline€ direction, e s ≡ u u , noting that ω × u
is perpendicular to es.
∂
( 2
) = 0 = −e s ⋅ ∇ h + 12 u + Φ →
e s ⋅ (ω × u€ ) ∂s
(2
h + 12 u + Φ = 0 ,
€
)
1 €
2
where s is the path length along a streamline. Thus, h + 2 u + Φ = constant along a streamline.

€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.9. a) Solve ∇ 2G(x, x #) = δ (x − x #) for G(x,x´) in a uniform unbounded three-

dimensional domain, where δ(x – x´) = δ(x – x´)δ(y – y´)δ(z – z´) is the three dimensional Dirac
delta function.
1 q( x %) 3
b) Use the result €
of part a) to show that: φ (x) = − ∫
4 π all x % x − x %
d x % is the solution of the Poisson

equation ∇ 2φ (x) = q(x) in a uniform unbounded three-dimensional domain.

Solution 5.9. a) First apply a simple shift transformation that places x´ at the origin of
€
coordinates. Define these new coordinates by: X = x − x #, Y = y − y #, Z = z − z# , and set
€
r = x − x # = X 2 + Y 2 + Z 2 . The gradient operator ∇ XYZ in the shifted coordinates X = (X, Y, Z)
is the same as ∇ in the unshifted coordinates (x, y, z), so the field equation for G becomes:
∇€2XYZ G = δ (X)€= δ (x − x %)€.
Integrate this equation inside a sphere of radius r:
€ €
Z=+r Y =+ r 2 −Z 2 X=−+ r 2 −Y 2 −Z 2
€ 2
∫∫∫ ∇ XYZ G dV =€ ∫∫ ∇ XYZ G ⋅ n dA = ∫ ∫2 2 2∫ 2 2 δ (X)dXdYdZ ,
sphere spherical surface Z=−r Y =− r −Z X=− r −Y −Z

where the first equality follows from Gauss' divergence theorem, and the triple integral on the
right side includes the location X = 0 (aka x = x´) so a contribution is collected from the three-
dimensional delta function. Thus, the right side of this equation is unity.
The dot product in the middle portion of the above equation simplifies to ∂G/∂r because
n = er on the spherical surface and e r ⋅ ∇ XYZ = ∂ ∂r . Plus, in an unbounded uniform environment,
there are no preferred directions so G = G(r) alone (no angular dependence). Thus, the integrated
field equation simplifies to
π 2π
# ∂G & ∂G π 2 π ∂G
∫ ∫ %$€∂r ('r 2 sin θdθdϕ = r 2 ∂r ∫ ∫ sin θdθdϕ = 4 πr 2 ∂r = 1,
θ = 0ϕ = 0 θ = 0ϕ = 0
which implies
∂G 1 1 1
= 2
or G = − =− .
∂r 4 πr 4 πr 4π x − x $
€
b) Work in the unshifted coordinate system, and start with the given information about the
Green’s function, ∇ 2G(x, x #) = δ (x − x #) , and multiply both sides by q( x ") to get:
€ ∇ 2 [q( x #)G(x, x #)] = q( x #)δ (x − x #) . (1)
€ 2
The q-function can slide inside the differential operator, ∇ , because q( x ") depends on x´ while
the operator
€ acts on x. Now integrate (1) over all possible€ values of x´:
2 3 2 3
∫ [∇ q( x #)G(x,
€
x #) ]d x # = ∇ ∫ q( x #)G(x, x #)d x # = ∫ q(x #)δ (x − x #)d 3 x # = q(x) . (2)
all x # all x # all x #
€ € the integration can be exchanged
Here, the first equality follows because the ∇ 2 -operation and
since they act on different variables. The final equality follows from the properties of the three-
dimensional Dirac delta-function. The formal solution for φ (x) can now be found by comparing
€
the original equation,
€
∇ 2φ (x) = q(x) , (3)
to the second and fourth terms of (2):
€
% (
∇ & ∫ q( x )G(x, x #)d 3 x #) = q(x) .
2
# (4)
€ ' all x # *

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Here, both statements are true, so we must conclude that

1 q( x #) 3
φ (x) = ∫ q( x #)G(x, x #)d 3 x # = − 4π ∫ d x #, (5)
all x # all x # x − x #

where the final equality in (5) comes from the result of part (a). This conclusion is possible
because solutions of Poisson’s equation with suitably defined boundary conditions are unique.
However, existence and uniqueness proofs are beyond the scope of this text.
€ the results of this primarily-mathematical exercise are useful for
Interestingly,
incompressible ideal fluid flows.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.10. Start with the equations of motion in the rotating coordinates, and prove (5.29)
Kelvin’s circulation theorem for the absolute vorticity. Assume that the flow is inviscid and
barotropic and that the body forces are conservative. Explain the result physically.

Solution 5.10. The general idea is to redo the ordinary derivation of Kelvin's theorem in a
steadily rotating coordinate system where the inviscid momentum equation is:
Du
ρ = −∇p + ρ[g e − 2Ω × u] , ($)
Dt
Here all velocities are observed in the rotating frame so the primes used in Section 4.7 have been
dropped, and g e = g − Ω × (Ω × x ) , as also discussed in Section 4.7.
Start by time differentiating the definition of the circulation (3.18):
€
DΓ D Du D
=
Dt Dt C
∫ u ⋅ dx = ∫ ⋅ dx + ∫ u ⋅ (dx ) , (#)
C Dt C Dt
€
dx is an arc length element of C, a closed contour. The first term of the right-most equality of (#)
simplifies as follows:
Du ) 1 ,
€ ∫ ⋅ dx = ∫ +− ∇p + g e − 2Ω × u. ⋅ dx = − ∫ (2Ω × u) ⋅ dx ,
C Dt C * ρ - C
because the pressure-integral and body force terms are single valued, and will integrate to zero
on a closed contour. The second term in the right-most equality of (#) is zero as shown in Section
5.2. Thus, (#) implies:
€ DΓ
= − ∫ (2Ω × u) ⋅ dx . (@)
Dt C
Consider the product of vectors in the integrand
−2(Ω × u) ⋅ dx = −2εijk Ω j uk dx i = 2Ω jε jik dx i uk = −2Ω ⋅ (u × dx ) .
Now define n to be the unit vector perpendicular to the plane defined by u and dx, and let u⊥ be
€
the component of u perpendicular to dx so that:
−2(Ω × u) ⋅ dx = −2Ω ⋅ nu⊥ dx .
€
In an interval of time dt, u⊥ will locally advect the contour element dx
€ !
through an area δA = u⊥ dxdt . Since n is normal to δA, the directed area
element is nδA = nu⊥ dxdt , and this implies ( D Dt )(nδA) = nu⊥ dx . Thus
€
€ −2(Ω × u) ⋅ dx = −2Ω ⋅ ( D Dt )(nδA) , n dx u
and when
€ integrated around the whole contour this produces: C u
D D
€ − ∫ (2Ω × u) ⋅ dx = − ∫ 2Ω€ ⋅ (ndA) = − ∫ 2Ω ⋅ ndA .
C C Dt Dt C
€
Thus, (@) can be written:
DΓa D
Dt
=
Dt C
∫ (ω + 2Ω) ⋅ ndA = 0 .
€
This means that the circulation developed from vorticity observed in the flow and from
the rotation of the coordinate frame together satisfy Kelvin's theorem in a steadily rotating frame
of reference.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.11. In (R,ϕ,z) cylindrical coordinates, consider the radial velocity uR = −R–1(∂ψ/∂z),
and the axial velocity uz = R–1(∂ψ/∂R) determined from the axisymmetric stream function
Aa 4 # R 2 &# R 2 z 2 &
ψ (R,z) = % (%1− − ( where A is a constant. This flow is known as Hill’s spherical
10 $ a 2 '$ a 2 a 2 '
vortex.
a) For R2 + z2 ≤ a2, sketch the streamlines of this flow in a plane that contains the z-axis. What
does a represent?
€ b) Determine u = uR(R,z)eR + uz(R,z)ez
c) Given ωϕ = (∂uR ∂z) − (∂uz ∂R) , show that ω = AReϕ in this flow and that this vorticity field is
a solution of the vorticity equation (5.13). z
d) Does this flow include stretching of vortex lines?
€Solution 5.11. a) The zero streamline is given by R = 0
and by R2 + z2 = a2.
For R2 + z2 ≤ a2, the others appear as shown to the right.
b) The velocity components are: R
1 ∂ψ 1 Aa 4 % R 2 (% −2z ( ARz
uR = − =− ' *' *= , and
R ∂z R 10 & a 2 )& a 2 ) 5
1 ∂ψ 1 Aa 4 % 2R 4R 3 z 2 2R ( Aa 2 % R2 z2 (
uz = = ' 2 − 4 − 2 2 *= '1− 2 2 − 2 * .
R ∂R R 10 & a a a a ) 5 & a a )
€ c) Compute the components of the vorticity. Here, uϕ = 0 and neither uR
or uz depend on ϕ, so
1 ∂uz ∂uϕ ∂u ∂u AR Aa 2 & R )
€ ωR = − = 0 , ωϕ = R − z = − ( −4 + = AR , and
R ∂ϕ ∂z ∂z ∂R 5 5 ' a2 *
1 ∂ (Ruϕ ) 1 ∂uR
ωz = − = 0.
R ∂R R ∂ϕ
2
Therefore,
€ ω = AReϕ in this € flow. The vorticity equation is: ∂ω ∂t + (u ⋅ ∇)ω = (ω ⋅ ∇ )u + ν∇ ω .
The flow is steady so the ∂/∂t term is zero. For ω = ωϕeϕ = AReϕ, the remaining terms are:
€ (u ⋅ ∇ ) ω e = e u ∂ωϕ = Au e ,
advection term: ( ϕ ϕ ) ϕ R ∂R R ϕ
€
ω ∂ AR ∂e
vortex-stretching term: (ωϕ eϕ ⋅ ∇)(u) = Rϕ ∂ϕ (uR e R + uze z ) = R uR ∂ϕR = AuR eϕ ,
' ∂2 1 ∂ 1 ∂2 *
and viscous term:€ ν∇ 2 (ωϕ eϕ ) = ) 2 + + 2 2 ,(ωϕ eϕ )
( ∂R R ∂R R ∂ϕ +
2 2
€ ∂ ωϕ 1 ∂ωϕ 1 ∂ eϕ
= eϕ + eϕ + ω ϕ
∂R 2 R ∂R R 2 ∂ϕ 2
€ & ∂ 2ωϕ 1 ∂ωϕ ωϕ ) & 1 AR )
= eϕ ( 2 + − 2 + = eϕ (0 + A − 2 + = 0
' ∂R R ∂R R * ' R R *
So, the vorticity equation satisfied by a balance of the advection and vortex stretching terms.
d) Yes, this flow includes stretching of vortex lines.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.12. In (R,ϕ,z) cylindrical coordinates, consider the flow field uR = –αR/2, uϕ = 0, and
uz = αz.
a) Compute the strain rate components SRR, Szz, and SRz. What sign of α causes fluid elements to
elongate in the z-direction? Is this flow incompressible?
b) Show that it is possible for a steady vortex (a Burgers’ vortex) to exist in this flow field by
adding uϕ = (Γ/2πR)[1 – exp(–αR2/4ν)] to uR and uz from part a) and then determining a pressure
field p(R,z) that together with u = (uR, uϕ, uz) solves the Navier-Stokes momentum equation for a
fluid with constant density ρ and kinematic viscosity ν.
c) Determine the vorticity in the Burgers’ vortex flow of part b).
d) Explain how the vorticity distribution can be steady when α ≠ 0 and fluid elements are
stretched or compressed.
e) Interpret what is happening in this flow when α > 0 and when α < 0.

Solution 5.12. a) Using the strain rates from Appendix B,

∂u α ∂u 1 # ∂u ∂u &
SRR = R = − , Szz = z = α , and SRz = % R + z ( = 0 .
∂R 2 ∂z 2 $ ∂z ∂ R '
Therefore, when α > 0, fluid elements elongate in the z-direction. The divergence of the velocity
field is: (1 R)(∂ (RuR ) ∂R) + ∂uz ∂z = (1 R)(−2αR 2) + α = 0 , so the flow is incompressible.
2
b) Here:€ uR = –αR/2, uϕ € = (Γ/2πR)[1 – exp(– €αR /4ν)], and uz = αz. None of these velocities
depends on the angle ϕ. Consider the three components of the Navier-Stokes momentum
equation:
€
(i) Steady R-direction equation with ∂/∂ϕ = 0, from Appendix B:
∂u ∂u u2 1 ∂p ' 1 ∂ ' ∂uR * ∂2 uR uR *
uR R + uz R − ϕ = − + ν) )R , + 2 − 2 ,.
∂R ∂z R ρ ∂R ( R ∂R ( ∂R + ∂z R +
The various terms involving the velocity are:
∂uR αR % α ( α 2 R ∂uR 1 ∂ # ∂uR & α ∂2 uR u α
uR = − '− * = , uz = 0, %R (=− , 2
= 0 , and − R2 = .
∂R
€ 2 & 2) 4 ∂z R ∂R $ ∂R ' 2R ∂z R 2R
So, the R-direction equation becomes:
2
α 2R Γ2 - ' αR 2 *0 1 ∂p
− 2 3/
1− exp)− ,2 = − ,
€ € 4 €4 π R . ( 4ν +€ 1 ρ ∂R €
which can formally be integrated
R 5 ) αR%2 ,2 9
2
7 α 2 R% Γ2 / 7
p(R,z) = −ρ ∫ 6 − 2 31
1− exp+ − .4 :dR% + f (z) .
€ 0 78 4 4 π R% 0 * 4ν -3 7 ;
where f(z) is an unknown function.
(ii) Steady ϕ-direction equation with ∂/∂ϕ = 0, from Appendix B, is
∂uϕ ∂uϕ uR uϕ % 1 ∂ % ∂uϕ ( ∂2 uϕ uϕ ( % ∂ % 1 ∂(Ruϕ ) ( ∂2 uϕ (
u€R + uz + = ν ' 'R * + 2 − 2 * = ν' ' *+ *.
∂R ∂z R & R ∂R & ∂R ) ∂z R ) & ∂R & R ∂R ) ∂z 2 )
The various terms are:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

∂uϕ αR Γ ) 1 / ) αR 2 ,2 α ) αR 2 ,, ∂uϕ
uR =− ++− 2 11− exp+− .4 + exp+− ... , uz = 0,
∂R 2 2π * R 0 * 4ν -3 2ν * 4ν -- ∂z
uR uϕ αΓ . ( αR 2 +1
=− 01− exp* − -3 ,
R 4 πR / ) 4ν ,2
€
∂€% 1 ∂(Ruϕ ) ( νΓ ∂ % 1 ∂ / % αR 2 (2( Γ α 2 R % αR 2 ( ∂2 uϕ
ν ' *= ' 11− exp'− *4* = − exp'− * , and = 0.
∂R & R ∂R ) 2π ∂R '& R ∂R 0 & 4ν )3*) 2π 4ν & 4ν ) ∂z 2
€
So, the ϕ-direction equation becomes:
'
αΓ 1 - ' αR *0 αR
2 ' αR 2 ** αΓ - ' αR 2 *0 Γ α 2 R ' αR 2 *
)) /1− exp) − ,2 − exp)− ,
, − /1− exp) − ,2 = − exp)− ,.
€ 4π ( R . ( 4ν +1 2ν ( 4ν +,+ 4 πR . ( 4ν +1€ 2π 4ν ( 4ν +
Multiply by 4π/Γα, and collect like terms:
1 1 % αR 2 ( αR % αR 2 ( αR % αR 2 (
(1−1) − (1−1) exp'− * − exp'− * = − exp'− * ;
€ R R & 4ν ) 2ν & 4ν ) 2ν & 4ν )
so the ϕ-direction momentum equation is satisfied.
(iii) Steady z-direction equation with ∂/∂ϕ = 0, from Appendix B:
∂u ∂u 1 ∂p & 1 ∂ & ∂uz ) ∂2 uz )
€ uR z + uz z = − + ν( (R + + 2 +.
∂R ∂z ρ ∂z ' R ∂R ' ∂R * ∂z *
The various terms are:
∂uz ∂uz 2 1 ∂ # ∂uz & ∂2 uz
uR = 0 , uz = α z, % R ( = 0 , and = 0.
€ ∂R ∂z R ∂R $ ∂R ' ∂z 2
So, the z-direction equation becomes:
1 ∂p
α 2z = − , or p(z,R) = −ρα 2 z 2 2 + g(R) ,
€ € ρ ∂z €
€
and this determines the unknown function, f (z) = − ρα 2 z 2 2 , in the final equation of (i), so the
pressure is:
€ R 5 % αR-2 (2 9
2
€ ρα 2 % 2 R 2 ( 7 Γ2 / 7
p(R,z) − p(0,0) = − ' z + * + ρ ∫ 6 2 3 11− exp' − *4 :dR- .
€2 & 4) 8 4 π R- 0
0 7 & 4ν )3 7 ;
c) There is only one non-zero component of the vorticity:
1 ∂ 1 ∂uR Γα * αR 2 -
ωz =
R ∂R
( Ruϕ ) −
R ∂ϕ
=+
4 πν
exp,−
+ 4ν .
/.
€
d) When α > 0, the vorticity distribution can be steady because the strain field concentrates the
vorticity at precisely the correct rate to balance the vorticity's outward diffusion.
e) When α > 0, then flow comes inward toward the z-axis along z = 0 and turns vertically upward
€
and downward along the +z and –z axis, respectively. Here the vorticity is concentrated near the
z-axis and diffuses outward against the incoming flow; it decays to zero with increasing R. When
α > 0, the flow comes vertically upward and downward along the –z and +z axis, respectively,
and then turns outward away from the z-axis as it approaches z = 0. Here the vorticity is very
high at large R and it diffuses inward against the outflow.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.13. A vortex ring of radius a and strength Γ lies in the x-y plane as shown in the
figure.
a) Use the Biot-Savart law (5.13) to reach the following formula for the induced velocity along
the x-axis:
Γe 2 π (x cos ϕ − a)adϕ
u(x) = − z ∫ .
4π ϕ =0 # x 2 − 2ax cos ϕ + a 2 %3 2
$ &
b) What is u(0), the induced velocity at the origin of coordinates?
c) What is u(x) to leading order in a/x when x >> a?
z!

Γ!
a! y!

ϕ"
x!

Solution 5.13. a) Start with:

2π
Γ (x − x%)
u(x) =
4π
∫e ϕ ×
x − x%
3
adϕ ,
0

where eϕ = – exsinϕ + eycosϕ, x = xex, x´ = aeR = a(excosϕ + eysinϕ), eϕ × e x = −e z cos ϕ ,

12
eϕ × e R = −e z , and x − x" = #$(x − a cos ϕ )2 + a 2 sin 2 ϕ %& . Insert these relationships, to find:
2π
Γ (−xe z cos ϕ + ae z )
u(x) =
4π
∫# 2 2 2 %3 2
adϕ ,
$(x − a cos
0 ϕ ) + a sin ϕ &
Rearrange the numerator and simplify the denominator of the integrand to find:
Γe 2 π (x cos ϕ − a)adϕ
u(x) = − z ∫ ,
4π 0 # x 2 − 2ax cos ϕ + a 2 %3 2
$ &
b) At the origin, x = 0, so the part a) result simplifies to:
Γe 2 π −a 2 dϕ Γe z
u(0) = − z ∫ = ,
4π 0 #a 2 %3 2 2a
$ &
c) For x >> a, the denominator of the part a) result can be expanded in a two-term power series:
" 2 $−3 2
"# x 2 − 2ax cos ϕ + a 2 $% = x −3 &1− 2a cos ϕ + a ' ≈ x −3 "1+ 3a cos ϕ $ for a x → 0 .
−3 2
&# '%
# x x2 % x
So, the induced velocity becomes:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Γe z 2 π % 3a (
32 ∫
u(x, t) ≈ − (x cos ϕ − a)'1+ cos ϕ * adϕ for a x → 0 .
4π x 0 & x )
Multiplying the integrand factors together and evaluating the integrals leads to:
Γe z Γa 2
u(x, t) ≈ − ( −2 π a + 3π a ) a = − ez
for a x → 0 .
4π x 3 4x 3
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.14. An ideal line vortex parallel to the z-axis of strength Γ intersects the x-y plane at x
= 0 and y = h. Two solid walls are located at y = 0 and y = H > 0. Use the method of images for
the following.
a) Based on symmetry arguments, determine the horizontal velocity u of the vortex when h =
H/2.
b) Show that for 0 < h < H the horizontal velocity of the vortex is:
Γ ' ∞
1 *
u(0,h) = )1− 2∑ 2 ,,
4 πh ( n=1 (nH /h) −1+
and evaluate the sum when h = H/2 to verify your answer to part a).

Solution 5.14. There are two flat solid walls at y = 0 yt

€
and y = H. The combined effect of each requires many
y = 3H
image vortices. t
a) When the vortex is located at x = (0, H/2), the y = 2H t
induced velocities from the two closest image vortices y=H
cancel, and the induced velocities from the next closest t x
y=0 t
image vortices cancel as well. In fact, the total induced
velocity from the whole array of image vortices is zero y = –H
t
because vortex contributions cancel in pairs. Therefore, y = –2H
u(0, H/2) = 0.
b) Construct the induced velocity by adding the contributions from each image vortex. When 0 <
h < H, the first image vortex below the lower solid wall must be located at y = –h and it must
have a strength of –Γ to mimic the effect of the wall at y = 0. Based on the drawing, this induced
velocity will be in the positive x-direction, so
Γ Γ $1 '
u(0,h) = + ... = & + ...)
4 πh 2π % 2h (
Now consider the first image vortex above the upper wall. It must be located at y = 2H – h and it
must have a strength of –Γ to mimic the effect of the wall at y = H. Based on the drawing, this
induced velocity will be in the negative x-direction, so
€
Γ %1 1 (
u(0,h) = ' − + ...*
2π & 2h 2H − 2h )
However, the first upper image vortex spoils the lower wall boundary condition, so another
image vortex must be added at y = –2H + h and it must have a strength of +Γ to mimic the effect
of the wall at y = 0. Based on the drawing, this induced velocity will be in the negative x-
direction, so €
Γ %1 1 1 (
u(0,h) = ' − − + ...*
2π & 2h 2H − 2h 2H )
But, now another vortex with strength +Γ needs to added at y = 2H + h to preserve the upper wall
boundary condition, so
Γ %1 1 1 1 (
€ u(0,h) = ' − − + + ...*.
2π & 2h 2H − 2h 2H 2H )
Continuing this construction leads to:
Γ %1 1 1 1 1 1 1 1 1 (
u(0,h) = ' − − + + − − + + ...*,
2π€& 2h 2H − 2h 2H 2H 2H + 2h 4H − 2h 4H 4H 4H + 2h )

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

which can be simplified by removing terms that are equal and opposite:
Γ %1 1 1 1 1 1 1 (
u(0,h) = ' − + − + − + ...*.
2π & 2h 2H − 2h 2H + 2h 4H − 2h 4H + 2h 6H − 2h 6H + 2h )
The first term is unique while all the others follow in pairs, so
Γ -1 ∞
% 1 1 (0 Γ - ∞
1 0
u(0,h) = / + ∑ ' − + *2 = / 1− 2 ∑ 2 2
2.
€ 2π . 2h n=1& 2nH − 2h 2nH + 2h )1 4 πh /. n=1 n ( H h ) −12 1
where the second equality follows from algebraic manipulations within the big parentheses.
When h = H/2, this formula becomes:
Γ ' ∞
1 *
€ u(0,H /2) =
4 πh (
)1− 2 ∑ 2 ,.
n=1 4n −1+
The sum is 1/2, and this can be found by looking in an appropriate mathematical reference (see
Gradshteyn, I. S., and Ryzhik, I. M., Tables of Integrals, Series, and Products [Academic Press,
New York, 1980], p. 8), or by considering the terms in the sum:
∞
1€ 1 ∞% 1 1 ( 1 +% 1 ( % 1 1 ( % 1 1 ( .
∑ 2 = ∑'& − * = -'1− * + ' − * + ' − * + ...0
n=1 4n −1 2 n=1 2n −1 2n + 1) 2 ,& 3 ) & 3 5 ) & 5 7 ) /
1 + % 1 1( % 1 1( % 1 1( . 1
= -1+ '− + * + '− + * + '− + * + ...0 = .
2 , & 3 3) & 5 5 ) & 7 7 ) / 2
The pair-cancellation of terms is the mathematical signature of the symmetry discussed in part
a), and series of this type are sometimes known as telescoping series. With either evaluation
approach, the net induced velocity is
€ Γ + % 1 (.
u(0,H /2) = -1− 2' *0 = 0 ,
4 πh , & 2 )/
which matches the part a) result.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.15. The axis of an infinite solid circular cylinder with radius a coincides with the z-
axis. The cylinder is stationary and immersed in an incompressible inviscid fluid, and the net
circulation around it is zero. An ideal line vortex parallel to the cylinder with circulation Γ passes
through the x-y plane at x = L > a and y = 0. Here two image vortices are needed to satisfy the
boundary condition on the cylinder’s surface. If one of these is located at x = y = 0 and has
strength Γ. Determine the strength and location of the second image vortex.
y
Solution 5.15. If the net circulation around the
cylinder is zero, then the second image vortex must
(xs,ys)
have strength –Γ. Therefore, the fluid velocity u at any
point will be a sum of velocities induced by the vortex x
at (0, a), the first image vortex at (0, 0), and second ! !
(0,0) (0,L)
image vortex at (x´, y´). This velocity must be tangent !
to the surface of the cylinder. Therefore, determine the (x´,y´)
fluid velocity at the cylinder-surface point (xs, ys).
Using Cartesian unit vectors and the diagram to the right, with the circulation directions of the
vortices shown, leads to:
Γ % ys x ( Γ % ys (L − x s ) (
u(x s, y s ) = '− e x + s e y * + '− e − e *
a ) 2π (L − x s ) 2 + y s2 '& (L − x s ) 2 + y s2 (L − x s ) 2 + y s2 *)
x y
2πa & a
% (
Γ ys − y+ xs − x+
+ ' e − e *
2π (x s − x +) 2 + (y s − y +) 2 '& (x s − x +) 2 + (y s − y +) 2 (x s − x +) 2 + (y s − y +) 2 *)
x y

This can be simplified to:

Γ & ys ys ys − y% )
u(x s, y s ) = (− 2 − 2 2
+ 2
e
2+ x
€ 2π ' a (L − x s ) + y s (x s − x %) + (y s − y %) *
Γ & xs (L − x s ) xs − x% )
+ ( 2− 2 2
− 2
e .
2+ y
2π ' a (L − x s ) + y s (x s − x %) + (y s − y %) *
Here the outward normal to the surface of the cylinder is e r = ( x s a)e x + ( y s a)e y . Therefore to
find (x´, y´), set the dot product of u and er to zero,
€ Γ ' xsys xsys x s (y s − y &) *
u(x s, y s ) ⋅ e r = )− 2 − + ,
2πa ( a (L − x s ) 2 +€y s2 (x s − x &) 2 + (y s − y &) 2 +
Γ ' xsys (L − x s )y s (x s − x &)y s *
+
) 2 − 2 2
− 2 2,
= 0.
2πa ( a (L − x s ) + y s (x s − x &) + (y s − y &) +
After dividing out common factors and cancelling equal and opposite terms this reduces to:
Ly s x #y s − x s y #
− + = 0,
(L − x s ) + y s (x s − x #) 2 + (y s − y #) 2
2 2
€
which must be true for all values of (xs, ys). So, at xs = a and ys = 0, this equation becomes:
−ay #
=0 ,
(a − x #) 2 + y #2
€
which can only be true when y´ = 0, and this leaves a single equation for x´:

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Ly s x #y s
− + = 0.
(L − x s ) + y s (x s − x #) 2 + y s2
2 2

Expand the denominators, divide by ys, and use x s2 + y s2 = a 2 , to find:

L x#
2 2
= 2 .
€ L − 2Lx s + a a − 2x s x # + x #2
Cross multiply and cancel common terms € to reach:
Lx " − (L + a ) x " + La 2 = L( x " − L)( x " − a 2 L) = 0 .
2 2 2

The root that leads to €a vortex location inside the cylinder is: x´ = a2/L. So, the final answers are:
• the second image vortex has strength –Γ, and
• its location is (a2/L, 0).
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.16. Consider the interaction of two vortex rings of equal strength and similar sense of
rotation. Argue that they go through each other, as described near the end of section 5.6.
VC
Solution 5.16. Consider two vortex rings with a similar sense of
rotation as shown in the figure to the right. The radii and speeds of C VB A
the two vortices are equal. The motion at A is the resultant of VB, VD VB
VC, and VD, while the motion at C is the resultant of VA, VB, and VD
VD. Comparing the velocity components on A and C, it is clear that
the net resultant is an enlargement of the vortex on the right, and a VA
contraction of the vortex on the left, as indicated by the vertical
arrows in the lower figure (a). Parts (b) and (c) of the lower figure
show that left-side vortex accelerates to the right and passes
through the right-side vortex. Part c) show that the new right side
vortex grows and slows while the new left-side vortex contracts
and accelerates. This results in part d), which is identical to the
starting condition shown in the first figure except that the ring
vortices have exchanged places. Thus, the entire process starts
D B
over.

(a) (b) (c) (d)

Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.17. A constant density irrotational flow in a rectangular torus has a circulation Γ and
volumetric flow rate Q. The inner radius is r1, the outer radius is r2, and the height is h. Compute
the total kinetic energy of this flow in terms of only ρ, Γ, and Q.

Solution 5.17. For an irrotational vortical flow with

circulation Γ. The velocity in the angular direction will be:
Γ
r1 uθ = .
h r2 2πr
The volumetric flow rate will be:
r2 h r2 h
Γ hΓ & r2 )
Q = ∫ ∫ uθ dzdr = ∫ ∫ dzdr = ln( + .
r1 €
0 r1 0 2πr 2π ' r1 *
The total kinetic energy, KE, of the fluid in the in the torus
r2 h 2 π
1 2 ρΓ 2 r2 h 1 ρΓ 2 h ' r2 * 1
will be: KE = ∫ ∫ ∫ ρuθ dzrdrdθ = ∫ ∫ dzdr = 4π ln) r , = 2 ρΓQ .
r1 0 0 2 € 4 π r1 0 r ( 1+

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.18. Consider a cylindrical tank of radius R filled with a viscous fluid spinning
steadily about its axis with constant angular velocity Ω. Assume that the flow is in a steady state.
(a) Find ∫ A ω ⋅ ndA where A is a horizontal plane surface through the fluid normal to the axis of
rotation and bounded by the wall of the tank, and n is the normal on A.
(b) The tank then stops spinning. Find again the value of ∫ A ω ⋅ ndA .
€
Solution 5.18. a) At steady state the flow inside the tank will be in solid body rotation, and the
vorticity will be twice the rotation rate: ω = 2Ω = constant. Thus, ∫ A ω ⋅ ndA = 2ΩπR 2 .
€
b) Let C be the bounding curve around A, and let ds be an element of C. Then,
2π
∫ A ω ⋅ dA = ∫ C u ⋅ ds = ∫ 0 [uθ ] r= R Rdθ = 0 , since uθ = 0 at r = R because of the no-slip boundary
condition. €

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.19. Using Figure 5.13, prove (5.32) assuming that: (i) the two vortices travel in
circles, (ii) each vortex's speed along it's circular trajectory is constant, and (iii) the period of the
motion is the same for both vortices.

Solution 5.19. When the two vortices travel in circles, with radii h1 and h2, at constant speeds
with the same period T, the circumference of each circular trajectory divided by the induced
speed of the vortex on that trajectory must match:
2π h1 2π h2
=T = ,
Γ 2 2π h Γ1 2π h
where the term on the left applies to the vortex on the left (1), and the term on the right applies to
the vortex on the right (2). This relationship can be simplified to:
h1 Γ 2 = h2 Γ1 . (#)
Now use h = h1 + h2 to substitute for h2 using h2 = h – h1, and solve for h1 to find:
−1
h1 h − h1 "1 1% h h"1 1% Γ2h
= , or h1 $ + ' = , which implies h1 = $ + ' = .
Γ2 Γ1 # Γ 2 Γ1 & Γ1 Γ1 # Γ 2 Γ1 & Γ 2 + Γ1
Γh
The relationship (#) then implies: h2 = 1 for the other vortex.
Γ1 + Γ2
Thus, G is located at the place where the diagonal line connecting the tips of the induced velocity
vectors crosses the line segment that connects the vortices. So, G is halfway between the two
vortices when they are of equal strength, but it lies closer to the stronger vortex when the vortices
are of unequal strength. €Here G is also the "center of vorticity". For example, when G is located
at the origin of coordinates:
2
center of vorticity ≡ ∑ rn Γ n
n=1

% −Γ hΓ + Γ hΓ (
= (−h1, 0)Γ1 + (h2 , 0)Γ 2 = (−h1Γ1 + h2 Γ 2 , 0 ) = ' 2 1 1 2 , 0 * = (0, 0).
& Γ1 + Γ 2 )
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.20. Consider two-dimensional steady flow in the x-y plane outside of a long circular
cylinder of radius a that is centered on and rotating about the z-axis at a constant angular rate of
Ωz. Show that the fluid velocity on the x-axis is u(x,0) = (Ωza2/x)ey for x > a when the cylinder is
replaced by
a) a circular vortex sheet of radius a with strength γ = Ωza, and
b) a circular region of uniform vorticity ω = 2Ωzez with radius a.
c) Describe the flow for x2 + y2 < a2 for parts a) and b). y
Solution 5.20. a) For a circular vortex sheet, an
(acos!,asin!) du
element of the circle of length adθ located at angle θ !
will have a circulation of dΓ = γadθ = Ωza2dθ. The !
distance between the vortex element and the location x
(x, 0) is (x − acos θ ) 2 + a 2 sin 2 θ . Thus, the magnitude
and direction of the induced velocity increment du at
(x, 0) are
(asin θ )e x + (x − acos θ )e y
€ (
dΓ 2π (x − acosθ ) 2 + a 2 sin 2 θ and) (x − acosθ ) 2 + a 2 sin 2 θ
.

The total induced velocity at (x, 0) is determined from integrating over all vortex elements.
+π
Ωz a 2 ' (asin θ )e x + (x − acosθ )e y * Ωz a 2 + π ' (asin θ )e x + (x − acos θ )e y *
u(x,0) = ∫ ) 2 2 2 , d θ = ∫ ) x 2 + a 2 − 2ax cosθ ,dθ .
€ θ =− π 2π ( (x − acos θ ) + a sin € θ + 2π − π ( +
Separate the components and consider them individually.
Ωz a 2e x π asin θdθ Ωz a 2e y π (x − acos θ )dθ
u(x,0) = ∫
2π − π x 2 + a 2 − 2ax cosθ
+ ∫
2π − π x 2 + a 2 − 2ax cosθ
.
€
The x-component of u is zero because it is specified as an integration of an odd function on an
even interval. The remaining y-component integration can be rewritten to reach a tabulated
integral form, and then evaluated:
€ Ω a 2e 1 π & )
x 2 − a2
u(x,0) = z y ∫ (1+ 2 2 +dθ
2π 2x − π ' x + a − 2ax cosθ *
+π
& & 2 ))
=
2
Ωz a e y 1 (
θ+
2 2
2(x − a )
tan−1( ( x 2 + a2 ) − 4a2 x 2 θ ++
tan ++
2π 2x (( 2 ( x 2 + a 2 − 2ax 2 ++
' ( x 2 + a2 ) − 4a2 x 2 (' **− π
+π
Ωz a 2e y 1 & & 2
−1 x − a
2
θ )) Ωz a 2e y 1 & π & π )) Ωz a 2e y
= (θ + 2tan ( 2
tan ++ = ( 2π + 2 − 2(− ++ = .
2π 2x ' ' (x − a) 2 **− π 2π 2x ' 2 ' 2 ** x
+π
(x − acos θ )dθ 2π
[Thus, u(x,0) = ∫ 2 2
= and is independent of a!]
− π x + a − 2ax cos θ x
€ b) For a uniform distribution of vorticity within the circle specified by x2 + y2 < a2, an element of
area dA = rdθdr will have a circulation dΓ = ωzdA = 2Ωzrdθdr. The distance between the vortex
€ element and the location (x, 0) is (x − r cos θ ) 2 + r 2 sin 2 θ . Thus, the magnitude and direction of
the induced velocity at (x, 0) are

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

(r sin θ )e x + (x − r cos θ )e y
(
dΓ 2π (x − r cosθ ) 2 + r 2 sin 2 θ ) and (x − r cosθ ) 2 + r 2 sin 2 θ
.

The total induced velocity at (x, 0) is determined from integrating over all vortex sheet elements.
a +π
2Ωz ' (r sin θ )e x + (x − r cosθ )e y * Ωz a + π ' (r sin θ )e x + (x − r cosθ )e y *
u(x,0) = ∫ ∫ ) ,rdθdr = ∫ ∫ ) x 2 + r 2 − 2rx cosθ ,dθrdr .
€ 0 − π 2π ( (x − r cos θ ) 2 + r 2 sin 2 θ + π 0 −π ( +
€
The angular integration is the same as that completed in part a) with a replaced by r, so
Ωze y a 2π Ωze y 2π a 2 Ωz a 2e y
u(x,0) =
π 0 x
∫ rdr = π x 2 = x .
€
c) In part a) the fluid is stationary for x2 + y2 < a2. In part b), the fluid is in solid body rotation for
x2 + y2 < a2.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 5.21. An ideal line vortex in a half space filled with an inviscid constant-density fluid
has circulation Γ, lies parallel to the z-axis, and passes through the x-y plane at x = 0 and y = h.
The plane defined by y = 0 is a solid surface.
a) Use the method of images to find u(x,y) for y > 0 and show that the fluid velocity on y = 0 is
u(x,0) = Γhe x [π (x 2 + h 2 )] .
b) Show that u(0,y) is unchanged for y > 0 if the image vortex is replaced by a vortex sheet of
strength γ (x) = −u(x,0) on y = 0.
c) (If you have the patience) Repeat part b) for u(x,y) when y > 0.
€
Solution 5.21. a) If the actual vortex is at (0, h) and has circulation Γ, then an image vortex of
€
with circulation –Γ at (0, –h) produces a no-through-flow boundary condition on y = 0. In
Cartesian coordinates, the velocity u at any location (x, y) is the sum of the two induced
velocities:
Γ % −(y − h)e + xe ( Γ % (y + h)e − xe (
x y x y
u(x, y) = ' *+ ' *,
2 2 ' 2 2 * 2 2 ' 2 2 *
2π x + (y − h) & x + (y − h) ) 2π x + (y + h) & x + (y + h) )
and this form can be simplified to:
Γ % −(y − h)e x + xe y (y + h)e x − xe y (
u(x, y) = ' + 2 *. (†)
€ 2π & x 2 + (y − h) 2 x + (y + h) 2 )
When evaluated on y = 0, the velocity is:
Γ % he x + xe y he x − xe y ( Γhe x
u(x,0) = ' 2 2
+ 2 2 *
= .
€ 2π & x + h x + h ) π (x 2 + h 2 )
b) First of all, determine u(0,y) from the results of part a)
Γ % −e x e (
u(0, y) = ' + x *. (%)
€ 2π & y − h y + h )
For a flat vortex sheet lying on y = 0, an element of the length dx´ located at x´ will have a
circulation of dΓ = γ(x´)dx´. The distance between this vortex element and the location (x, y) is
(x − x #) 2 + y 2 . The magnitude
€ and direction of the induced velocity increment du at (x, y) are
−ye x + (x − x #)e y
(
dΓ 2π (x − x %) 2 + y 2 and) (x − x #) 2 + y 2
.

€ When the image vortex is replaced by a vortex sheet, u(x,y) for y > 0 will be the sum of the
induced velocity from the actual vortex and that from the vortex sheet, which is an integral.
€ Γ / −(y − h)e x + xe y +∞& −ye x + (x − x %)e y ) γ ( x %)dx % 2
u(x, y) = 0 € + ∫( 2 +
3. (£)
2π 1 x 2 + (y − h) 2 2
−∞ ' (x − x %) + y * Γ 4
For this part of this exercise, finding u(0,y) from this formula is sufficient, so put x = 0 to find:
Γ / −e x +∞& −ye x − x %e y ) γ ( x %)dx % 2
u(0, y) = 0 + ∫( + 3.
€ 2π 1 y − h −∞' x %2 + y 2 * Γ 4
Comparing this equation with (%), sets the remaining task as showing the equality of
+∞$ −ye − x #e '
γ ( x #)dx # e
∫ & x #2x + y 2 y ) Γ = y +x h .
€ −∞ % (
To do this set, γ(x´) = – u(x´, 0) from part a), and separate the left side into components:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

$ −ye x − x #e y '
+∞
−h hye x +∞ dx # he y +∞ x #dx #
∫ & x #2 + y 2 ) π ( x #2 + h 2 ) dx = π ∫ ( x #2 + y 2 )( x #2 + h 2 ) + π ∫ ( x #2 + y 2 )( x #2 + h 2 ) .
#
−∞ % ( −∞ −∞
The y-component integral is zero because it involves an odd integrand and an even interval. This
leaves the x-component:
hye x +∞ dx # hye x 1 +∞' 1 1 *
€ ∫ 2 2 2
π −∞ ( x # + y )( x # + h ) 2
= 2 2 ∫) 2
π h − y −∞( x # + y 2
− 2 ,dx #
x# + h 2 +
hye x π - 1 1 0 hye - h − y 0 ex
= 2 2/
− 2= 2 x2/ 2= ,
π h − y . y h 1 h − y . hy 1 y + h
where first equality follows from a partial fractions decomposition of the integrand on the left,
+∞ −1
the second equality follows from: I(a) = ∫ −∞ ( x 2 + a 2 ) dx = π a , and the final equality achieves
the
€ desired form. Therefore, the vortex sheet on y = 0 appropriately mimics an image vortex.
c) The effort here is nearly a repeat of part b). This time start from (£) and compare it to (†).
The task here is to show that the following equality is holds:
€+∞$ −ye + (x − x #)e ' γ ( x #)dx # (y + h)e − xe
∫ & (xx− x #) 2 + y 2 y ) Γ = x 2 + (y x+ h) 2 y . (&)
−∞ % (
First set, γ(x´) = – u(x´, 0) from part a), and begin working on L, the left side of (&):
+∞$ −ye + (x − x #)e '
x y −hdx # h +∞ ( ye x − (x − x #)e y ) dx #
L = Lx e x + Ly e y = ∫ & 2 )
= ∫ .
( π ( x # + h ) π −∞ ((x − x #) 2 + y 2 )( x #2 + h 2 )
2 2 2
€ −∞ % (x − x #) + y

Consider the x-component integration first:

hy +∞ dx # hy +∞ ' A( x # − x) + B Cx # + D *
Lx = ∫ = ∫ ( ( x # − x) 2 + y 2 + x #2 + h 2 +dx #,
€ π −∞ ((x − x #) 2
+ y 2
)( x # 2
+ h 2
) π −∞ ) ,
where A, B, C, and D are constants. Some algebra is required to find:
A = −2x E , B = (h 2 + x 2 − y 2 ) E , C = +2x E , and D = (x 2 + y 2 − h 2 ) E ,
where E = (x 2 + y 2 − h 2 ) 2 + 4 x 2 h 2 . The integrations that include A and C are zero because they
€
involve odd integrands and even intervals. The integrals involving B and D can be evaluated
using the I(a) formula above, to determine:
€ € € €
hy # Bπ Dπ & h 3 + hx 2 − hy 2 + yx 2 + y 3 − yh 2
€ Lx = % + ( = Bh + yD = .
π $ y h ' (x 2 + y 2 − h 2 ) 2 + 4 x 2 h 2
The numerator and denominator in can be rearranged:
h(h 2 − y 2 ) + x 2 (h + y) + y(y 2 − h 2 ) (h + y)[ h(h − y) + x 2 + y(y − h)]
Lx = = .
€ (x 2 + y 2 + h 2 ) 2 − 4 y 2 h 2 ( x 2 + y 2 + h 2 + 2yh)[ x 2 + y 2 + h 2 − 2yh]
Further rearrangement shows that the factors in square brackets are equal, so
(h + y)[ x 2 + (y − h) 2 ] h+ y
Lx = 2 = 2 ,
€ ( x + (y + h) )[ x + y + h − 2yh] x + (y + h)2
2 2 2 2

which matches the x-component on the right side of (&).

Now consider the y-component integration:
h +∞ ( x # − x)dx # h +∞ ' F( x # − x) + G Hx # + J *
€ Ly = ∫ = ∫( + + dx #,
π −∞ ((x − x #) 2 + y 2 )( x #2 + h 2 ) π −∞ ) ( x # − x) 2 + y 2 x #2 + h 2 ,
where F, G, H, and J are constants. Some algebra is required to find:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

F = (h 2 + x 2 − y 2 ) E , G = 2xy 2 E , H = (−h 2 − x 2 + y 2 ) E , and J = −x(x 2 + y 2 + h 2 ) E ,

The integrations that include F and H are zero because they involve odd integrands and even
intervals. The integrals involving G and J can be evaluated using the I(a) formula above, to
determine:
€ € € €
h # Gπ Jπ & h 2xyh − x 3 − xy 2 − xh 2 x [ x 2 + y 2 − 2yh + h 2 ]
Ly = % + (=G +J = 2 =− 2
π$ y h' y (x + y 2 − h 2 ) 2 + 4 x 2 h 2 (x + y 2 + h 2 ) 2 − 4 y 2 h 2
x [ x 2 + y 2 − 2yh + h 2 ] x
=− =− ,
(x
+ y + h + 2yh )[ x + y + h − 2yh ]
2 2 2 2 2 2
x + (y + h) 2
2

and the final form matches the y-component on the right side of (&). Thus, the equivalence of a
variable strength vortex sheet and an image vortex is established.
€
 1

Chapter 6 Excercises

Problem 1
Show, by Taylor expansion, that

d3 f fj+2 − 2fj+1 + 2fj−1 − fj−2

3
≈ .
dx 2∆x3
What is order of this approximation?

Solution
Expand around xj :

∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∂ 4 f ∆x4

fj+1 = fj + ∆x + + + + O(∆x5 ) (1)
∂x ∂x2 2 ∂x3 6 ∂x4 24
∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∂ 4 f ∆x4
fj−1 = fj − ∆x + 2
− 3
+ + O(∆x5 ) (2)
∂x ∂x 2 ∂x 6 ∂x4 24
∂f ∂ 2 f 4∆x2 ∂ 3 f 8∆x3 ∂ 4 f (2∆x)4
fj+2 = fj + 2∆x + 2
+ 3
+ + O(∆x5 ) (3)
∂x ∂x 2 ∂x 6 ∂x4 24
∂f ∂ 2 f 4∆x2 ∂ 3 f 8∆x3 ∂ 4 f (2∆x)4
fj−2 = fj − 2∆x + − + + O(∆x5 ) (4)
∂x ∂x2 2 ∂x3 6 ∂x4 24
Subtract equation (2) from equation (1):

∂f ∂ 3 f ∆x3
fj+1 − fj−1 = 2∆x + + O(∆x5 ) (5)
∂x ∂x3 3
and equation (4) from equation (3)

∂f ∂ 3 f 8∆x3
fj+2 − fj−2 = 4∆x + + O(∆x5 ). (6)
∂x ∂x3 3
Multiplying equation (5) by 2 and subtracting it from equation (6) yields

∂ 3 f 6∆x3
fj+2 − 2fl+1 + 2fj−1 − fj−2 = + O(∆x5 ), (7)
∂x3 3
which can be rearranged to give

d3 f fj+2 − 2fj+1 + 2fj−1 − fj−2

3
= + O(∆x2 ).
dx 2∆x3
Thus, this is a second order approximation to the third derivative.
2

Problem 2
Consider the following “backward in time” approximation for the diffusion equa-
tion:
∆tD  n+1 
fjn+1 = fjn + f j+1 + f n+1
j−1 − 2fj
n+1
∆x2
(a) Determine the accuracy of this scheme.
(b) Find its stability properties by von Neumann’s method. How does it compare
with the forward in time, centered in space approximation considered earlier?

Solution
Expand around fjn+1

∂f ∂ 2 f ∆t2
fjn = fjn+1 − ∆t + 2 + O(∆t3 )
∂t ∂t 2

n+1 ∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3

fj+1 = fjn+1 + ∆x + 2
+ + O(∆x4 )
∂x ∂x 2 ∂x3 6
n+1 ∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3
fj−1 = fjn+1 − ∆x + 2
− + O(∆x4 )
∂x ∂x 2 ∂x3 6
Substituting into:
∆tD  n+1 
fjn+1 = fjn + f + f n+1
− 2f n+1
∆x2 j+1 j−1 j

and rearranging the terms gives:

∂f ∂2f ∂ 2 f ∆t ∂ 4 f D∆x2
−D 2 = 2 + 4 + O(∆t2 , ∆x4 ). (1)
∂t ∂x ∂t 2 ∂t 12
The right hand side is the error, so the accuracy is O(∆t, ∆x2 ).
To find stability, write fjn = n eikxj and substitute into the equation given
in the problem statement, using that eik(xj ±∆x) = eikxj e±ik∆x . The result is
∆tD n+1 ik∆x
n+1 = n + ( ) (e − 2 + e−ik∆x ),
∆x2
or
n+1 1 1
= = .
n 1 − (cos k∆x − 1)2 (∆tD/∆x2 ) 1 + 4(∆tD/∆x2 ) sin2 k∆x
2

Thus, we always have

n+1
≤ 1,
n
and the scheme is unconditionally stable.
 3

Problem 3
Approximate the linear advection equation
∂f ∂f
+U =0 U >0
∂t ∂x
by the backward in time method from problem 2. Use the standard second
order centered difference approximation for the spatial derivative.
(a) Write down the finite difference equation.
(b) Write down the modified equation
(c) Find the accuracy of the scheme
(d) Use the von Neuman’s method to determine the stability of the scheme.

Solution
(a) The scheme is:
fjn+1 − fjn n+1
fj−1 n+1
− fj−1
+U =0
∆t 2∆x
(b) First expand around fjn+1 :

∂f ∂ 2 f ∆t2
fjn = fjn+1 − ∆t + 2 + O(∆t3 )
∂t ∂t 2
Rearrange to get
fjn+1 − fjn ∂f ∂ 2 f ∆t
= − 2 + O(∆t2 )
∆t ∂t ∂t 2
Then expand

n+1 ∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3

fj+1 = fjn+1 + ∆x + 2
+ + O(∆x4 )
∂x ∂x 2 ∂x3 6
n+1 ∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3
fj−1 = fjn+1 − ∆x + − + O(∆x4 )
∂x ∂x2 2 ∂x3 6
subtract and divide by 2∆x to get:
n+1 n+1
fj+1 − fj−1 ∂f ∂ 3 f ∆x2
= + 3 + O(∆x4 )
2∆x ∂x ∂t 6
The modified equation is therefore
∂f ∂f ∂ 2 f ∆t ∂ 3 f ∆x2
+U = 2 −U 3
∂t ∂x ∂t 2 ∂x 6
(c) From the modified equation the error is O(∆t, ∆x2 )
(d) Substituting fjn = n eikxj into the discrete equation and using that eik(xj ±∆x) =
eikxj e±ik∆x , we get
n+1 − n U n+1 n+1 ∆tU
= n+1 (eik∆x − e−ik∆x ) ⇒ − 1 + i i sin k∆x = 0
∆t 2∆x n n ∆x
n+1
 1
⇒ n
= ∆tU
 1 + i ∆x i sin k∆x
4

n+1
Thus, the method is unconditionally stable since n ≤ 1.
 5

Problem 4
Consider the following finite difference approximation to the diffusion equation:
∆tD  n 
fjn+1 = fjn + 2 f j+1 − f n+1
j − fj
n−1
+ f n
j−1 .
∆x2
This is the so-called Dufort-Frankel scheme, where the time integration is the
”Leapfrog” method, and the spatial derivative is the usual center difference
approximation, except that we have replaced fjn by (1/2)(fjn+1 +fjn−1 ) . Derive
the modified equation and determine the accuracy of the scheme. Are there any
surprises?

Solution
Writing

∂f ∂ 2 f ∆t2 ∂ 3 f ∆t3 ∂ 4 f ∆t4

fjn+1 = fjn + ∆t + 2 + 3 + 4 + O(∆t5 )
∂t ∂t 2 ∂t 6 ∂t 24
∂f ∂ 2 f ∆t2 ∂ 3 f ∆t3 ∂ 4 f ∆t4
fjn−1 = fjn − ∆t + 2 − 3 + 4 + O(∆t5 )
∂t ∂t 2 ∂t 6 ∂t 24
n ∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∂ 4 f ∆x4
fj+1 = fjn + ∆x + 2
+ 3
+ + O(∆x5 )
∂x ∂x 2 ∂x 6 ∂x4 24
n ∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3 ∂ 4 f ∆x4
fj−1 = fjn − ∆x + − + + O(∆x5 )
∂x ∂x2 2 ∂x3 6 ∂x4 24
and substituting into the finite difference equation gives

∂f ∂2f ∂ 4 f D∆x2 ∂ 3 f ∆t ∂ 2 f D∆t2

−D 2 = 4 − 3 − 2 + O(∆t3 , ∆x3 )
∂t ∂x ∂t 12 ∂t 6 ∂t ∆x2
2 2
∆t ∆t
The accuracy is O(∆t2 , ∆x2 , ∆x 2 ). Notice the ∆x2 term, which must go to zero

as ∆x → 0 and ∆t → 0 for the results to converge. This, the scheme is only

conditionally consistent.
6

Problem 5
The following finite difference approximation is given

1 n ∆tU  n 
fjn+1 = n
(fj+1 + fj−1 )− n
fj+1 − fj−1
2 2∆x

(a) Write down the modified equation

(b) What equation is being approximated?
(c) Determine the accuracy of the scheme
(d) Use the von Neuman’s method to examine under which conditions this
scheme is stable.

Solution
(a) Start by expanding

∂f ∂ 2 f ∆t2
fjn+1 = fjn + ∆t + 2 + O(∆t3 )
∂t ∂t 2
n ∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3
fj+1 = fjn + ∆x + + + O(∆x4 )
∂x ∂x2 2 ∂x3 6
n ∂f ∂ 2 f ∆x2 ∂ 3 f ∆x3
fj−1 = fjn − ∆x + 2
− + O(∆x4 )
∂x ∂x 2 ∂x3 6
Substitute to get

∂f ∂ 2 f ∆t2 1 ∂2f U ∆t  ∂f ∂ 3 f ∆x3 

fjn + ∆t + 2 = (2fjn + ∆x 2
) − 2∆x +
∂t ∂t 2 2 ∂x2 2∆x ∂x ∂x3 3
Or:
∂f ∂f ∂ 2 f ∆t ∂ 2 f ∆x2 ∂ 3 f U ∆x2
+U =− 2 + 2
−
∂t ∂x ∂t 2 ∂x 2∆t ∂x3 6

(b) The equation being approximated is

∂f ∂f
+U =0
∂t ∂x

2
(c) The accuracy is O(∆t, ∆x 2 2
∆t , ∆x ). If ∆x ∼ ∆t, then ∆x /∆t ∼ ∆x and the
method is first order in time and space.
(d) To determine the stability, we substitute fjn = n eikxj into the discrete
equation and use that eik(xj ±∆x) = eikxj e±ik∆x , giving:

1 U ∆t ik∆x
n+1 = n (eik∆x + e−ik∆x ) − n (e − e−ik∆x ).
2 2∆x
Rearrange
n+1 U ∆t
n
= cos k∆x − i sin k∆x.
 h
 7

The absolute value is

n+1 2  U ∆t 2
= cos2 k∆x + sin2 k∆x.
n ∆x
So the scheme is stable if
U ∆t
≤ 1.
∆x
8

Problem 6
Consider the equation
∂f
= g(f ),
∂t
and the second-order predictor-corrector method:

fj∗ = fjn + ∆tg(f n )

∆t
fjn+1 = fjn + (g(f n ) + g(f ∗ )).
2
Show that this method can also be written as:

fj∗ = fjn + ∆tg(f n )

fj∗∗ = fj∗ + ∆tg(f ∗ )
fjn+1 = (1/2)(f n + f ∗∗ ).

That is, you simply take two explicit Euler steps and then average the solution
at the beginning of the time step and the end. This makes it particularly simple
to extend a first order explicit time integration scheme to second order.

Solution
The first equations in each formulation are the same. To show that the second
two equations in the second formulation are the same as the final equation of
the first formulation, substitute the second and first equation into the third one:
1 n
fjn+1 = (f + fj∗∗ )
2 j
1 n
fjn+1 = (f + fj∗ + ∆tg(fj∗ ))
2 j
1 n
fjn+1 = (f + fjn + ∆tg(fjn ) + ∆tg(fj∗ )
2 j
1
fjn+1 = fjn + (∆tg(fjn ) + ∆tg(fj∗ ))
2
 9

Problem 7
Modify the code used to solve the one-dimensional linear advection equation
(Code 1) to solve the Burgers equation:

∂f ∂ f2  ∂2f
+ =D 2
∂t ∂x 2 ∂x

using the same initial conditions. What happens? Refine the grid. How does
the solution change if we add a constant (say 1) to the initial conditions?

Solution
The modified code is:

% one-dimensional NONLINEAR advection-diffusion

% by the FTCS scheme
N=21; nstep=10; L=2.0; dt=0.05;D=0.05; k=1;
dx=L/(N-1); for j=1:N, x(j)=dx*(j-1);end
f=zeros(N,1); fo=zeros(N,1); time=0.0;
for j=1:N, f(j)=0.5*sin(2*pi*k*x(j)); end;

for m=1:nstep, m, time

plot(x,f,’linewidt’,2); axis([0 L -1.5, 1.5]); pause
fo=f;
for j=2:N-1
f(j)=fo(j)-(0.25*dt/dx)*((fo(j+1)+fo(j))^2-...
(fo(j)+fo(j-1))^2)+...
D*(dt/dx^2)*(fo(j+1)-2*fo(j)+fo(j-1));
end;
f(N)=fo(N)-(0.25*dt/dx)*((fo(2)+fo(N))^2-...
(fo(N)+fo(N-1))^2)+...
D*(dt/dx^2)*(fo(2)-2*fo(N)+fo(N-1));
f(1)=f(N);
time=time+dt;
end;
10

1.5

0.5

-0.5

-1

-1.5
0 0.5 1 1.5 2
Running the code for for three different resolutions N = 21, 41, 81 with dt =
0.05, 0.025, 0.0125 up to time 0.25 (for 5, 10, 20) steps, produces the figure above.
Even for the coarsest resolution, the solution is essentially fully converged, so
we have used thicker lines for the low resolution results, since otherwise those
would not be visible.
For the second part of the problem, replace the initial conditions by

for j=1:N, f(j)=1.0+0.5*sin(2*pi*k*x(j)); end;

Taking dt=0.0125, gives a solution that travels to the right.

 11

Problem 8
Modify the code used to solve the two-dimensional linear advection equation
(Code 2) to simulate the advection of an initially square blob with f = 1 diag-
onally across a square domain by setting u = v = 1. The dimension of the blob
is 0.2 × 0.2 and it is initially located near the origin. Refine the grid and show
that the solution converges by comparing the results before the blob flows out
of the domain.

Solution
The modified code is:

% Two-dimensional unsteady diffusion by the FTCS scheme

Nx=32;Ny=32;nstep=20;D=0.025;Lx=2.0;Ly=2.0;
dx=Lx/(Nx-1); dy=Ly/(Ny-1); dt=0.02;
f=zeros(Nx,Ny);fo=zeros(Nx,Ny);time=0.0;
for i=1:Nx,for j=1:Ny,x(i,j)=dx*(i-1);y(i,j)=dy*(j-1);end,end;
u=1.0; v=1.0; x0=0.2;y0=0.2;
for i=2:Nx-1, for j=2:Ny-1
if x(i,j)>x0-0.5*0.2 & x(i,j)<x0+0.5*0.2 &...
y(i,j)>y0-0.5*0.2 & y(i,j)<y0+0.5*0.2
f(i,j)=1.0;
end;end;end
for l=1:nstep,l,time
hold off;mesh(x,y,f); axis([0 Lx 0 Ly -0.5 1.5]);pause
fo=f;
for i=2:Nx-1, for j=2:Ny-1
f(i,j)=fo(i,j)-(0.5*dt*u/dx)*(fo(i+1,j)-fo(i-1,j))-...
(0.5*dt*v/dy)*(fo(i,j+1)-fo(i,j-1))+...
D*dt*( ((fo(i+1,j)-2*fo(i,j)+fo(i-1,j))/dx^2)...
+((fo(i,j+1)-2*fo(i,j)+fo(i,j-1))/dy^2) );
end,end
for i=1:Nx, f(i,1)=f(i,2);end;
for j=1:Ny,f(1,j)=f(2,j);f(Nx,j)=f(Nx-1,j);end;
time=time+dt;
end;

The code is best run interactively by advancing one step at a time and
pressing a key to go to the next step. The figure shows the initial condition on
a 32 × 32 grid and the solution at time 0.5 on the 32 × 32 grid and an 128 × 128
grid. As we run the code further, the solution quickly dampens out. Try lover
D and finer grids.
 12

1.5

0.5

-0.5
2

1.5 2
1.5
1
1
0.5
0.5
0 0

The initial condition on a 32 × 32 grid.

1.5 1.5

1 1

0.5 0.5

0 0

-0.5 -0.5
2 2

1.5 2 1.5 2
1.5 1.5
1 1
1 1
0.5 0.5
0.5 0.5
0 0 0 0

The solution at time 0.5 on the 32 × 32 grid and an 128 × 128 grid.
 13

Problem 9
Derive a second order expression for the boundary vorticity by writing the
stream function at j = 2 and j = 3 as a Taylor series expansion around the value
at the wall (j = 1). How does the expression compare with equation (6.67).

Solution
The derivation is essentially the same as in the text. The stream function, one
mesh block away, can be expressed using a Taylor series expansion around the
boundary point:

∂ψi,1 ∂ 2 ψi,1 ∆y 2 ∂ 3 ψi,1 ∆y 3

ψi,2 = ψi,1 + ∆y + + + O(∆y 4 ).
∂y ∂y 2 2 ∂y 3 6
Similarly, two mesh blocks away, we have

∂ψi,1 ∂ 2 ψi,1 (2∆y)2 ∂ 3 ψi,1 (2∆y)3

ψi,3 = ψi,1 + 2∆y + 2
+ + O(∆y 4 ).
∂y ∂y 2 ∂y 3 6

Multiplying the first equation by 23 = 8, and subtracting it from the second

one, using that ∂ψi,1 /∂y = Uwall and that ωwall = −∂ 2 ψi,1 /∂y 2 gives:

ψi,3 − 8ψi,2 = −7ψi,1 − 6∆yUwall + 2ωwall ∆y 2 + O(∆y 4 ),

which can be solved for the wall vorticity

!
1 3
ωwall = 7ψi,1 + ψi,3 − 8ψi,2 + Uwall + O(∆y 2 ).
2∆y 2 ∆y

This is similar to the first order condition (equation 6.67)

!
2 2
ωwall = ψi,1 − ψi,2 2
+ Uwall + O(∆y)
∆y ∆y

but involves ψi,3 as well as ψi,2 .

14

Problem 10
Modify the vorticity-stream function code used to simulate the two-dimensional
driven cavity problem (Code 3) to simulate the flow in a rectangular 2 × 1
channel with periodic boundaries. Set the value of the vorticity and the stream
function at the top and bottom to zero. As initial conditions place two circular
blobs with radius r = 0.25 and ω = 10 on the centerline of the channel at y = 0.5
and x = 0.6 and x = 1.4. Refine the grid to ensure that the solution converges.
Describe the evolution of the flow as the viscosity is decreased.

Solution
The code, modified as specified in the problem is listed below. Notice that
the viscosity here is 0.01 and that we accommodate the periodic boundary
conditions by extending the grid one grid line in the x-direction. Thus, the
grid spacing is given by h = 2.0/(N x − 2), rather than h = 2.0/(N x − 1). Here,
the 2 is the length of the domain in the x-direction.

% Problem 10 Modified Vorticity-Stream Function Code

Nx=34; Ny=17; MaxStep=200; Visc=0.01; dt=0.005; time=0.0;
MaxIt=100; Beta=1.5; MaxErr=0.001; % parameters for SOR
sf=zeros(Nx,Ny); vt=zeros(Nx,Ny); vto=zeros(Nx,Ny);
x=zeros(Nx,Ny); y=zeros(Nx,Ny); h=2.0/(Nx-2);
for i=1:Nx,for j=1:Ny,x(i,j)=h*(i-1);y(i,j)=h*(j-1);end,end;

for i=1:Nx,for j=1:Ny,

if(sqrt((x(i,j)-0.6)^2+(y(i,j)-0.5)^2)<0.25); vt(i,j)=10;end;
if(sqrt((x(i,j)-1.4)^2+(y(i,j)-0.5)^2)<0.25); vt(i,j)=10;end;
end;end

for istep=1:MaxStep, % Time loop

for iter=1:MaxIt, % solve for the stream function
vto=sf; % by SOR iteration
for i=2:Nx-1; for j=2:Ny-1
sf(i,j)=0.25*Beta*(sf(i+1,j)+sf(i-1,j)...
+sf(i,j+1)+sf(i,j-1)+h*h*vt(i,j))+(1.0-Beta)*sf(i,j);
end; end;
sf(1,2:Ny-1)=sf(Nx-1,2:Ny-1); sf(Nx,2:Ny-1)=sf(2,2:Ny-1);

Err=0.0; for i=1:Nx; for j=1:Ny, % check error

Err=Err+abs(vto(i,j)-sf(i,j)); end; end;
if Err <= MaxErr, break, end % stop if converged
end;

vto=vt;
for i=2:Nx-1; for j=2:Ny-1
vt(i,j)=vt(i,j)+dt*(-0.25*((sf(i,j+1)-sf(i,j-1))*...
 15

(vto(i+1,j)-vto(i-1,j))-(sf(i+1,j)-sf(i-1,j))*...
(vto(i,j+1)-vto(i,j-1)))/(h*h)...
+Visc*(vto(i+1,j)+vto(i-1,j)+vto(i,j+1)+...
vto(i,j-1)-4.0*vto(i,j))/(h^2) );
end; end;
vt(1,2:Ny-1)=vt(Nx-1,2:Ny-1); vt(Nx,2:Ny-1)=vt(2,2:Ny-1);

time=time+dt
subplot(211),contour(x,y,vt,40),axis(’equal’);axis([0 2 0 1]);
subplot(212),contour(x,y,sf),axis(’equal’);axis([0 2 0 1]);
pause(0.01)
end;

0.8

0.6

0.4

0.2

0
0 0.5 1 1.5 2

0.8

0.6

0.4

0.2

0
0 0.5 1 1.5 2

The vorticity and the streamfunction at time 1 are plotted in the figure, showing
that the vorticity patches are in the process of merging. The grid here is very
coarse, for the viscosity used.
16

Problem 11
Derive the discrete pressure equation for a corner point. How does it compare
with the equation for an interior point, (6.91) and a point next to a straight
boundary (6.96)?

Solution
Start with the continuity equation for the cell in the lower left corner. For the
control volume around the pressure node p(2, 2), it is

un+1
i+1/2,j − Ub,j
n+1
vi,j+1/2 − Vi,b
+ = 0. (1)
∆x ∆y
Since the velocities for the left and the bottom boundary are known, we only
substitute the equations for the correction velocities for the two unknown veloc-
ities, through the top and right edge. Thus, the pressure equation for the cell
in the bottom left corner is:
pi+1,j − pi,j pi,j+1 − pi,j
2
+
∆x ∆y 2
∗ ∗
ρ ui+1/2,j − Ub,j
 vi,j+1/2 − Vi,b 
= + . (2)
∆t ∆x ∆y
 17

Problem 12
Modify the velocity-pressure code used to simulate the two-dimensional driven
cavity problem (Code 4) to simulate the mixing of a jet of fast fluid with slower
fluid. Change the size of the domain to 3 and specify an inflow velocity of 1 in
the middle third of the left boundary and an inflow velocity of 0.25 for the rest
of the boundary. For the right boundary specify a uniform outflow velocity of
0.5. Keep other parameters the same. Refine the grid and check the convergence
of the solution.

Solution
The main change is in the boundary conditions. On the left boundary we
specify the inflow by setting u(1, 2 : N y + 1) to different values depending on
whether the grid point is in the jet or not. In general we would specify “gentle”
outflow boundary conditions that allowed the flow to leave the domain as freely
as possible. Here, however, we take a simpler approach and specify the outflow
velocity, u(N x + 1, 2 : N y + 1) = 0.5. Notice that in this approach the total
inflow must be equal to the total outflow. For the tangent velocity, we take all
walls to be stationary and full-slip. We do not check weather the the pressure
has converged (as we really should do). This leads to a slight “sloshing” in the
beginning and we have increased the maximum iterations to 500 for the pressure
to reduce this.

% Starting Jet by the MAC Method

Nx=3*33; Ny=33; Lx=3; Ly=1; MaxStep=500; Visc=0.01; rho=1.0;
MaxIt=500; Beta=1.5; MaxErr=0.001; % parameters for SOR
dx=Lx/Nx;dy=Ly/Ny; un=0;us=0;ve=0;vw=0;time=0.0; dt=0.02;

u=zeros(Nx+1,Ny+2);v=zeros(Nx+2,Ny+1);p=zeros(Nx+2,Ny+2);
ut=zeros(Nx+1,Ny+2);vt=zeros(Nx+2,Ny+1);c=zeros(Nx+2,Ny+2)+1;
c=zeros(Nx+1,Ny+2)+1/(2/dx^2+2/dy^2);
c(2,3:Ny)=1/(1/dx^2+2/dy^2); c(Nx+1,3:Ny)=1/(1/dx^2+2/dy^2);
c(3:Nx,2)=1/(1/dx^2+2/dy^2); c(3:Nx,Ny+1)=1/(1/dx^2+2/dy^2);
c(2,2)=1/(1/dx^2+1/dy^2); c(2,Ny+1)=1/(1/dx^2+1/dy^2);
c(Nx+1,2)=1/(1/dx^2+1/dy^2); c(Nx+1,Ny+1)=1/(1/dx^2+1/dy^2);

Ny1=12; Ny2=23;
u(1,2:Ny+1)=0.25; u(1,Ny1:Ny2)=1.0;
u(Nx+1,2:Ny+1)=0.5;
ut=u;

for is=1:MaxStep % Impose full-slip boundaries

u(1:Nx+1,1)=u(1:Nx+1,2);u(1:Nx+1,Ny+2)=u(1:Nx+1,Ny+1);
v(1,1:Ny+1)=v(2,1:Ny+1);v(Nx+2,1:Ny+1)=v(Nx+1,1:Ny+1);

for i=2:Nx,for j=2:Ny+1 % temporary u-velocity

18

ut(i,j)=u(i,j)+dt*(-0.25*(...
( (u(i+1,j)+u(i,j))^2-(u(i,j)+u(i-1,j))^2 )/dx+...
( (u(i,j+1)+u(i,j))*(v(i+1,j)+v(i,j))-...
(u(i,j)+u(i,j-1))*(v(i+1,j-1)+v(i,j-1)) )/dy )+...
Visc*((u(i+1,j)+u(i-1,j)-2*u(i,j))/dx^2+...
(u(i,j+1)+u(i,j-1)-2*u(i,j))/dy^2));
end,end

for i=2:Nx+1,for j=2:Ny % temporary v-velocity

vt(i,j)=v(i,j)+dt*(-0.25*(...
( (u(i,j+1)+u(i,j))*(v(i+1,j)+v(i,j))-...
(u(i-1,j+1)+u(i-1,j))*(v(i,j)+v(i-1,j)) )/dx+...
( (v(i,j+1)+v(i,j))^2-(v(i,j)+v(i,j-1))^2 )/dy )+...
Visc*((v(i+1,j)+v(i-1,j)-2*v(i,j))/dx^2+...
(v(i,j+1)+v(i,j-1)-2*v(i,j))/dy^2));
end,end

for it=1:MaxIt % solve for pressure

for i=2:Nx+1,for j=2:Ny+1
p(i,j)=Beta*c(i,j)*...
( (p(i+1,j)+p(i-1,j))/dx^2+...
(p(i,j+1)+p(i,j-1))/dy^2 -...
(rho/dt)*( (ut(i,j)-ut(i-1,j))/dx+...
(vt(i,j)-vt(i,j-1))/dy ) )+...
(1-Beta)*p(i,j);
end,end
end
% correct the velocity
u(2:Nx,2:Ny+1)=...
ut(2:Nx,2:Ny+1)-...
(dt/dx)*(p(3:Nx+1,2:Ny+1)-p(2:Nx,2:Ny+1));
v(2:Nx+1,2:Ny)=...
vt(2:Nx+1,2:Ny)-...
(dt/dy)*(p(2:Nx+1,3:Ny+1)-p(2:Nx+1,2:Ny));

time=time+dt % plot the results

uu(1:Nx+1,1:Ny+1)=0.5*(u(1:Nx+1,2:Ny+2)+u(1:Nx+1,1:Ny+1));
vv(1:Nx+1,1:Ny+1)=0.5*(v(2:Nx+2,1:Ny+1)+v(1:Nx+1,1:Ny+1));
w(1:Nx+1,1:Ny+1)=(u(1:Nx+1,2:Ny+2)-u(1:Nx+1,1:Ny+1)-...
v(2:Nx+2,1:Ny+1)+v(1:Nx+1,1:Ny+1))/(2*dx);
hold off,quiver(flipud(rot90(uu)),flipud(rot90(vv)),’r’);
hold on;contour(flipud(rot90(w)),100),axis equal,
plot(20+10*uu(20,1:Ny+1),[1:Ny+1],’k’);
plot([20,20],[1,Ny],’k’);
 19

plot(60+10*uu(60,1:Ny+1),[1:Ny+1],’k’);
plot([60,60],[1,Ny],’k’); pause(0.01)
end

Notice that here we plot the velocity and the vorticity without finding the
physical location of the grid points and instead simply used that the grid is
uniform. This does, however, require us to manipulate the field slightly, since
for 2D plots, Matlab treats the first variable as the vertical one and the second
one as the horizontal one, instead of the other way around as we have assumed.
To convert the data to the convention used here, we rotate the field and then
flip it. If we specify the location of the grid points, this is not needed, for 2D
flow. However, see problem 13 for the same issue with 3D plots.

30

25

20

15

10

10 20 30 40 50 60 70 80 90 100

The velocity and the vorticity field at nearly steady state. In addition to plotting
the velocity vectors and the vorticity, we plot the horizontal velocity along two
vertical lines (at N x = 20, and N x = 60. The velocity is multiplied by 10).
20

Problem 13
Extend the velocity-pressure code used to simulate the two-dimensional driven
cavity problem (Code 4) to three-dimensions. Assume that the third dimension
is unity (as the current dimensions) and take the velocity of the top wall and
the material properties to be the same. Compute the flow on a 93 and 173
grids and compare the results by plotting the velocities along lines through the
center of the domain. How do the velocities in the center compare with the
two-dimensional results?

Solution
The extension of the code to three-dimensional flow is relatively straight forward.
Everything just becomes longer. The biggest difference between 2D and 3D is
that plotting the results becomes more of a challenge. Here we solve the pressure
equation using a fixed number of iterations so it may not be fully converged at
the early stages.

% Driven Cavity by the MAC Method---3D

Nx=16;Ny=16;Nz=16;Lx=1;Ly=1;Lz=1;MaxStep=50;visc=0.1;rho=1.0;
MaxIt=100; Beta=1.5; MaxErr=0.001; % parameters for SOR
dx=Lx/Nx;dy=Ly/Ny; dz=Lz/Nz; time=0.0; dt=0.002;

u=zeros(Nx+1,Ny+2,Nz+2); v=zeros(Nx+2,Ny+1,Nz+2);
w=zeros(Nx+2,Ny+2,Nz+1); p=zeros(Nx+2,Ny+2,Nz+2);
ut=zeros(Nx+1,Ny+2,Nz+2); vt=zeros(Nx+2,Ny+1,Nz+2);
wt=zeros(Nx+2,Ny+2,Nz+1);

c=zeros(Nx+2,Ny+2,Ny+2)+1/(2/dx^2+2/dy^2+2/dz^2);

c(2,3:Ny,3:Nz)=1/(1/dx^2+2/dy^2+2/dz^2);
c(Nx+1,3:Ny,3:Nz)=1/(1/dx^2+2/dy^2+2/dz^2);
c(3:Nx,2,3:Nz)=1/(2/dx^2+1/dy^2+2/dz^2);
c(3:Nx,Ny+1,3:Nz)=1/(2/dx^2+1/dy^2+2/dz^2);
c(3:Nx,3:Ny,2)=1/(2/dx^2+2/dy^2+1/dz^2);
c(3:Nx,3:Ny,Nz+1)=1/(2/dx^2+2/dy^2+1/dz^2);

c(2,2,2) =1/(1/dx^2+1/dy^2+1/dz^2);
c(Nx+1,2,2)=1/(1/dx^2+1/dy^2+1/dz^2);
c(2,Ny+1,2) =1/(1/dx^2+1/dy^2+1/dz^2);
c(Nx+1,Ny+1,2)=1/(1/dx^2+1/dy^2+1/dz^2);
c(2,2,Nz+1) =1/(1/dx^2+1/dy^2+1/dz^2);
c(Nx+1,2,Nz+1)=1/(1/dx^2+1/dy^2+1/dz^2);
c(2,Ny+1,Nz+1)=1/(1/dx^2+1/dy^2+1/dz^2);
c(Nx+1,Ny+1,Nz+1)=1/(1/dx^2+1/dy^2+1/dz^2);

for is=1:MaxStep
 21

u(1:Nx+1,1:Ny+1,1)=-u(1:Nx+1,1:Ny+1,2);
u(1:Nx+1,1:Nz+1,Nz+2)=-u(1:Nx+1,1:Nz+1,Nz+1);
v(1:Nx+1,1:Ny+1,1)=-v(1:Nx+1,1:Ny+1,2);
v(1:Nx+1,1:Nz+1,Nz+2)=-v(1:Nx+1,1:Nz+1,Nz+1);

u(1:Nx+1,1,1:Nz+1)=-u(1:Nx+1,2,1:Nz+1);
u(1:Nx+1,Ny+2,1:Nz+1)=2-u(1:Nx+1,Ny+1,1:Nz+1);
w(1:Nx+1,1,1:Nz+1)=-w(1:Nx+1,2,1:Nz+1);
w(1:Nx+1,Ny+2,1:Nz+1)=-w(1:Nx+1,Ny+1,1:Nz+1);

v(1,1:Ny+1,1:Nz+1)=-v(2,1:Ny+1,1:Nz+1);
v(Nx+2,1:Ny+1,1:Nz+1)=-v(Nx+1,1:Ny+1,1:Nz+1);
w(1,1:Ny+1,1:Nz+1)=-w(2,1:Ny+1,1:Nz+1);
w(Nx+2,1:Ny+1,1:Nz+1)=-w(Nx+1,1:Ny+1,1:Nz+1);

ut=u;vt=v;wt=w;
for i=2:Nx,for j=2:Ny+1,for k=2:Nz+1 % temp. u-velocity
ut(i,j,k)=u(i,j,k)+dt*(-0.25*(...
( (u(i+1,j,k)+u(i,j,k))^2-(u(i,j,k)+u(i-1,j,k))^2)/dx+...
( (u(i,j+1,k)+u(i,j,k))*(v(i+1,j,k)+v(i,j,k))-...
(u(i,j,k)+u(i,j-1,k))*(v(i+1,j-1,k)+v(i,j-1,k)))/dy +...
( (u(i,j,k+1)+u(i,j,k))*(w(i+1,j,k)+w(i,j,k))-...
(u(i,j,k)+u(i,j,k-1))*(w(i+1,j,k-1)+w(i,j,k-1)))/dz)+...
visc*((u(i+1,j,k)+u(i-1,j,k)-2*u(i,j,k))/dx^2+...
(u(i,j+1,k)+u(i,j-1,k)-2*u(i,j,k))/dy^2+...
(u(i,j,k+1)+u(i,j,k-1)-2*u(i,j,k))/dz^2) );
end,end,end

for i=2:Nx+1,for j=2:Ny,for k=2:Nz+1 % temp. v-velocity

vt(i,j,k)=v(i,j,k)+dt*(-0.25*(...
( (u(i,j+1,k)+u(i,j,k))*(v(i+1,j,k)+v(i,j,k))-...
(u(i-1,j+1,k)+u(i-1,j,k))*(v(i,j,k)+v(i-1,j,k)))/dx+...
( (v(i,j+1,k)+v(i,j,k))^2-(v(i,j,k)+v(i,j-1,k))^2)/dy+...
( (v(i,j,k+1)+v(i,j,k))*(w(i,j+1,k)+w(i,j,k))-...
(v(i,j,k)+v(i,j,k-1))*(w(i,j+1,k-1)+w(i,j,k-1)))/dz )+...
visc*((v(i+1,j,k)+v(i-1,j,k)-2*v(i,j,k))/dx^2+...
(v(i,j+1,k)+v(i,j-1,k)-2*v(i,j,k))/dy^2+...
(v(i,j,k+1)+v(i,j,k-1)-2*v(i,j,k))/dz^2) );
end,end,end

for i=2:Nx+1,for j=2:Ny+1,for k=2:Nz % temp. w-velocity

wt(i,j,k)=w(i,j,k)+dt*(-0.25*(...
( (w(i+1,j,k)+w(i,j,k))*(u(i,j,k+1)+u(i,j,k))-...
(w(i,j,k)+w(i-1,j,k))*(u(i-1,j,k+1)+u(i-1,j,k)) )/dx+...
22

( (w(i,j+1,k)+w(i,j,k))*(v(i,j,k+1)+v(i,j,k))-...
(w(i,j,k)+w(i,j-1,k))*(v(i,j-1,k+1)+v(i,j-1,k)) )/dy+...
((w(i,j,k+1)+w(i,j,k))^2-(w(i,j,k)+w(i,j,k-1))^2 )/dz)+...
visc*((w(i+1,j,k)+w(i-1,j,k)-2*w(i,j,k))/dx^2+...
(w(i,j+1,k)+w(i,j-1,k)-2*w(i,j,k))/dy^2+...
(w(i,j,k+1)+w(i,j,k-1)-2*w(i,j,k))/dz^2) );
end,end,end

for it=1:MaxIt % solve for pressure

for i=2:Nx+1,for j=2:Ny+1, for k=2:Ny+1
p(i,j,k)=Beta*c(i,j,k)*...
( (p(i+1,j,k)+p(i-1,j,k))/dx^2+...
(p(i,j+1,k)+p(i,j-1,k))/dy^2+...
(p(i,j,k+1)+p(i,j,k-1))/dz^2 -...
(rho/dt)*( (ut(i,j,k)-ut(i-1,j,k))/dx+...
(vt(i,j,k)-vt(i,j-1,k))/dy+...
(wt(i,j,k)-wt(i,j,k-1))/dz) )+(1-Beta)*p(i,j,k);
end,end,end
end
% correct the velocity
u(2:Nx,2:Ny+1,2:Nz+1)=ut(2:Nx,2:Ny+1,2:Nz+1)-...
(dt/dx)*(p(3:Nx+1,2:Ny+1,2:Nz+1)-p(2:Nx,2:Ny+1,2:Nz+1));
v(2:Nx+1,2:Ny,2:Nz+1)=vt(2:Nx+1,2:Ny,2:Nz+1)-...
(dt/dy)*(p(2:Nx+1,3:Ny+1,2:Nz+1)-p(2:Nx+1,2:Ny,2:Nz+1));
w(2:Nx+1,2:Ny+1,2:Nz)= wt(2:Nx+1,2:Ny+1,2:Nz)-...
(dt/dz)*(p(2:Nx+1,2:Ny+1,3:Nz+1)-p(2:Nx+1,2:Ny+1,2:Nz));

time=time+dt % plot the results

Nzh=Nz/2;
uu(1:Nx+1,1:Ny+1)=0.5*(u(1:Nx+1,2:Ny+2,Nzh)+...
u(1:Nx+1,1:Ny+1,Nzh));
vv(1:Nx+1,1:Ny+1)=0.5*(v(2:Nx+2,1:Ny+1,Nzh)+...
v(1:Nx+1,1:Ny+1,Nzh));
p2d(1:Nx+1,1:Ny+1)= p(2:Nx+2,2:Ny+2,Nzh);
hold off,quiver(flipud(rot90(uu)),flipud(rot90(vv)),’r’);
hold on;contour(flipud(rot90(p2d)),20),axis equal;
axis([1 Nx 1 Ny]); pause(0.01)
end

% ---------- Plot the final results ---------------------

[X,Y,Z]=ndgrid(0.5*dx:dx:1-0.5*dx, 0.5*dy:dy:1-0.5*dy,...
0.5*dz:dz:1-0.5*dz);
quiver3(X,Y,Z,u3(2:Nx+1,2:Ny+1,2:Nz+1),...
v3(2:Nx+1,2:Ny+1,2:Nz+1),w3(2:Nx+1,2:Ny+1,2:Nz+1))
 23

hold on; [X,Y,Z]=meshgrid(0.5*dx:dx:1-0.5*dx,...

0.5*dy:dy:1-0.5*dy, 0.5*dz:dz:1-0.5*dz);
slice(X,Y,Z,flipud(rot90(p(2:Nx+1,2:Ny+1,2:Nz+1))),...
0.5, 0.5,[0.25 0.75])
xlabel(’x’); ylabel(’y’); zlabel(’z’); axis([0 1 0 1 0 1])

We show three figures for the results. The first one shows the velocity and the
vorticity in a plane cutting through the middle of the domain for a 173 grid.
When the code is run, this figure is plotted at every step and shows how the
flow evolves. The second shows a 3D view of the velocity and the pressure
in slices through the domain. Notice that we have used different functions to
provide the location of the grid points, for the different plotting functions and
we needed to flip and rotated the pressure field. In general it would be better
to use more advanced plotting to examine the results, particularly if the flow
was more complicated.
16

14

12

10

2 4 6 8 10 12 14 16

The velocity and the vorticity in a plane cutting through the middle of the
domain.
24

0.8

0.6
z

0.4

0.2

0
1
0.8 1
0.6 0.8
0.4 0.6
0.4
0.2
0.2
y 0 x
0

The velocity plotted using the Matlab function quiver3 and the pressure in
various planes cutting though the domain, using the Matlab function slice.
1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0
-0.2 0 0.2 0.4 0.6 0.8 1 1.2

The velocity plotted along the centerline of the cavity for two different resolu-
tions
 25

Problem 14
Derive equation (6.121),
1
gy fx − gx fy = (gη fξ − gξ fη ).
J

Solution
Use that
1 1
(fξ yη − fη yξ ); and fy = (fη xξ − fξ xη ),
fx =
J J
where J = xξ yη − xη yξ , and substitute into the left hand side:

1    
gy fx − gx fy = (f ξ y η − f η yξ )(gη x ξ − gξ x η ) − (fη x ξ − fξ x η )(gξ y η − gη y ξ )
J2
1  
= fξ gη y η x ξ −f ξ gξ yη xη −f η gη y ξ x ξ +fη gξ yξ x η −f ξ gη yη x ξ +fξ gξ y η x η −fη gη yξ x ξ −f ξ gη y ξ x η
J2
1  
= 2 fξ gη (yη xξ − yξ xη ) + fη gξ (yξ xη − yη xξ )
J
1  
= 2 (yη xξ − yξ xη ) fξ gη − fη gξ
J
1 
= fξ gη − fη gξ )
J
26

Problem 15
Show that the equations for the first derivatives in the mapped coordinates
(equations 6.111) can be written in the so-called conservative form:
1 1
fx = ((f yη )ξ − (f yξ )η ) and fy = ((f xξ )η − (f xη )ξ )
J J

Solution
1 1
1
fx = (fξ yη − fη yξ ) = (f yη )ξ − f yηξ − (f yξ )η + f yξη = ((f yη )ξ − (f yξ )η )
J J J
1 1
1
fy = (fη xξ −fξ xη ) = (f xξ )η −f xξη −(f xη )ξ +f xηξ = ((f xξ )η −(f xη )ξ )
J J J
Since yηξ = yξη and xξη = xηξ .
 27

Problem 16
Derive equation (6.120),

1  
∇2 ξ = q1 (x η y ξξ − yη x ξξ ) − 2q2 (x η yξη − y η x ξη ) + q3 (x η y ηη − y η x ηη ) .
J3
Take f = ξ and use that ξη = 0 and so on.

Solution
The equation listed is actually (6.119) and we will derive that one. The deriva-
tion of (6.120) is similar. We start with

1
fx = ((f yη )ξ − (f yξ )η ) where J = xξ yη − xη yξ .
J
If f = ξ then
1 yη
ξx = ((ξyη )ξ − (ξyξ )η ) = ,
J J
since ξη = 0 and ξξ = 1. Hence
" #

1  yη  y 
η
ξxx = ξx x
= yη − yξ .
J J ξ J η

Similarly " #

1  xη   x 
η
ξyy = ξy y
= − xξ − − xη .
J J η J ξ

Adding them together

" # " #
2 1  yη  y 
η 1  xη  x 
η
∇ ξ = ξxx + ξyy = yη − yξ + − xξ + xη .
J J ξ J η J J η J ξ

Expanding the terms

" #
1 y 
η
y 
η
x 
η
x 
η
∇2 ξ = yη − yξ − xξ + xη =
J J ξ J η J η J ξ

" #
1 yηξ yη2 yηη yη yξ xηη xη xξ xηξ x2η
yη − 2 Jξ − yξ + 2 Jη − xξ + 2 J η + xη − 2 Jξ =
J J J J J J J J J
" #
1
Jyηξ yη − yη2 Jξ − Jyηη yξ + yη yξ Jη − Jxηη xξ + xη xξ Jη + Jxηξ xη − x2η Jξ =
J3
" #
1
(yηξ yη − yηη yξ − xηη xξ + xηξ xη )J − (yη2 + x2η )Jξ + (yη yξ + xη xξ )Jη
J3
28

Now use that J = xξ yη − xη yξ to get

Jξ = xξξ yη + xξ yηξ − xηξ yξ − xη yξξ

Jη = xξη yη + xξ yηη − xηη yξ − xη yξη

Then

∇2 ξ =
"
1
xξ yη (yηξ yη − yηη yξ − xηη xξ + xηξ xη ) − xη yξ (yηξ yη − yηη yξ − xηη xξ + xηξ xη )
J3
−(x2η + yη2 )(xξξ yη + xξ yηξ − xηξ yξ − xη yξξ ) +
#
(xξ xη + yξ yη )(xξη yη + xξ yηη − xηη yξ − xη yξη ) =
"
1
xξ yη2 yηξ − xξ yη yξ yηη − x2ξ yη xηη + xξ yη xη xηξ
J3
−xη yξ yη yηξ + xη yξ2 yηη + xη yξ xξ xηη − x2η yξ xηξ
−x2η yη xξξ − x2η xξ yηξ + x2η yξ xηξ + x2η xη yξξ
−yη2 yη xξξ − yη2 xξ yηξ + yη2 yξ xηξ + yη2 xη yξξ +
xξ xη yη xξη + x2ξ xη yηη − xξ xη yξ xηη − xξ x2η yξη +
#
yξ yη2 xξη + yξ yη xξ yηη − yξ2 yη xηη − yξ yη xη yξη .

The red terms cancel and grouping the surviving terms we have:
"
1
2
∇ ξ = 3 (x2η + yη2 )(xη yξξ − yη xξξ )
J
#
−2(xξ xη + yξ yη )(xη yηξ − yη xηξ ) + (x2ξ + yξ2 )(xη yηη − yη xηη )
" #
1
= 3 q1 (xη yξξ − yη xξξ ) − 2q2 (xη yηξ − yη xηξ ) + q3 (xη yηη − yη xηη ) .
J

where

q1 = x2η + yη2 , q2 = xξ xη + yξ yη , q3 = x2ξ + yξ2

 29

Problem 17
Derive numerical approximations for the velocity-pressure equations for a map-
ping where the grid lines are straight and orthogonal, but unevenly spaced.
That is, x = x(ξ) and y = y(η) only. Assume that ∆ξ = ∆η = 1. How do these
equations compare with (6.85), (6.88), (6.89), (6.90), and (6.91)?

Solution
The various relationships simplify to q1 = yη2 ; q2 = 0; q3 = x2ξ ; and J = xξ yη .
The velocities are therefore
1  1   U
u= U xξ = U xξ =
J xξ yη yη
1  1   V
v= V yη = V yη = .
J xξ yη xξ

The continuity equation is

∂U ∂V ∂u ∂v 1 ∂u 1 ∂v
+ =0 ⇒ yη + xξ =0 ⇒ + = 0.
∂ξ ∂η ∂ξ ∂η xξ ∂ξ yη ∂η

Using that ∆ξ = ∆η = 1 we approximate the derivative between the grid points

as
yj+1 − yj xi+1 − xi
yη ≈ = ∆yj+1/2 and xξ ≈ = ∆xi+1/2 .
1 1
Defining ∆yj = 12 (∆yj+1/2 + ∆yj−1/2 ) and ∆xi = 21 (∆xi+1/2 + ∆xi−1/2 ), the
continuity equation becomes
ui+1/2 + ui−1/2 vj+1/2 + vj−1/2
+ = 0.
∆xi ∆yj

Similarly, the u-momentum equation becomes

∂u 1 ∂u2 1 ∂vu 1 ∂p h 1 ∂ u 
ξ 1 ∂  uη i
+ + =− +ν + .
∂t xξ ∂ξ yη ∂η xξ ∂ξ xξ ∂ξ xξ yη ∂η yη

The discrete version is

un+1 n
i+1/2,j − ui+1/2,j (u2 )ni+1,j − (u2 )ni,j (uv)ni+1/2,j+1/2 − (uv)ni+1/2,j−1/2
+ + =
∆t ∆xi+1/2 ∆yj
 un n
uni+1/2,j − uni−1/2,j 
pi+1,j − pi,j 1 i+3/2,j − ui+1/2,j
h
− +ν −
∆xi+1/2 ∆xi+1/2 ∆xi+1 ∆xi+1
n n n
1  u i+1/2,j+1 − u i+1/2,j u i+1/2,j − uni+1/2,j−1 i
+ − .
∆yj ∆yj+1/2 ∆yj−1/2

Similarly, the v-momentum equation is

∂v 1 ∂uv 1 ∂v 2 1 ∂p h 1 ∂ v 
ξ 1 ∂  vη i
+ + =− +ν + .
∂t xξ ∂ξ yη ∂η yη ∂η xξ ∂ξ xξ yη ∂η yη
30

The discrete version is

n+1 n
vi,j+1/2 − vi,j+1/2 (uv)ni+1,j+1/2 − (uv)ni,j+1/2 (v 2 )ni,j+1 − (v 2 )ni,j
+ + =
∆t ∆xi+1/2 ∆yj
h 1  vn n n n
pi,j+1 − pi,j i+1,j+1/2 − vi,j+1/2 vi,j+1/2 − vi−1,j+1/2 
− +ν −
∆yj+1/2 ∆xi ∆xi+1/2 ∆xi−1/2
n n n n
1  vi,j+3/2 − vi,j+1/2 vi,j+1/2 − vi,j−1/2 i
+ − .
∆yj+1/2 ∆yj+1 ∆yj

The velocities in the advection terms are found by averaging as for the regular
grid. This equation can be solved by the same projection method described in
the book for uniform grids and solved in the same way.
The main observation is that these equations look just like those for the
regular grid, except we divide by the local ∆x and ∆y. Notice that generally
we specify either ∆xi or ∆xi+1/2 and then find the intermediate points by
averaging. Same for ∆yj and ∆yj+1/2 .
 31

Problem 18
When the velocity is high and diffusion is small, the linear advection-diffusion
equation can exhibit boundary layer behavior. Assume that you want to solve
(6.11) in a domain given by 0 ≤ x ≤ 1, that U > 0, and that the boundary
conditions are f (0) = 0 and f (1) = 1. The velocity U is high and the diffusion
D is small so we expect a boundary layer near x = 1.
(a) Sketch the solution for high U and low D.
(b) Propose a mapping function that will cluster the grid points near the x=1
boundary.
(c) Write the equation in the mapped coordinates.

Solution
(a) This is the problem studied in section 6.2.6 and shown in figure 6.9. Here
we assume that we are solving the unsteady problem but at steady state the
solution looks like:
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0 0.5 1 1.5 2
(b): A function like ξ = x2 will cluster the points near the x = 1 part of the
domain.
(c) Start with the advection-diffusion equation

∂f ∂f ∂2f
+U = D 2.
∂t ∂x ∂x
The derivatives transform as
∂f 1 ∂f
=
∂x xξ ∂ξ
and ! !
∂2f ∂ ∂f 1 ∂ 1 ∂f
2
= = .
∂x ∂x ∂x xξ ∂ξ xξ ∂ξ

Substituting into the advection-diffusion equation gives us:

∂f 1 ∂f 1 ∂  1 ∂f 
+U =D
∂t xξ ∂ξ xξ ∂ξ xξ ∂ξ
32

In an actual implementation we usually take ∆ξ = 1 so (xξ )j ≈ ∆xj . Thus, the

discrete equation is:
!
fjn+1 − fjn n
fj+1 n
− fj−1 1 n
fj+1 − fjn fjn − fj−1
n
+U =D −
∆t 2∆xj ∆xj ∆xj+1/2 ∆xj−1/2

where xj+1/2 = 0.5(xj + xj+1 ) and so on, and we have assumed a forward Euler
time discretization.
 33

Problem 19
Derive equation (6.141).

Solution
Start with equation (6.140) and insert (6.133) and (6.134):
n
F̂j+1/2 = F+ (fjn ) + F− (fj+1
n
)=
 n  n  n  n
ρ 0 ρ 0
max(unj , 0)  ρu  +  p+  + min(unj+1 , 0)  ρu  +  p−  .
+ −
ρe j (pu) j
ρe j+1
(pu) j+1

Then use that max(u, 0) = 0.5(u+|u|) and min(u, 0) = 0.5(u−|u|) and equations
(6.135) and (6.134) to get
 n  n
ρ 0
1
F̂nj+1/2 = (unj + |unj |)  ρu  +  21 pnj (1 + Mjn )  +
2 1 n n n
ρe j 2 pj (uj + cj ) j
 n  n
ρ 0
1 n
(u − |unj+1 |)  ρu  +  12 pnj+1 (1 − Mj+1
n
)  .
2 j+1 n n n
ρe j+1 pj (uj+1 − cj+1 ) j+1

Rearranging yields
 n  n !
ρu ρu
n 1 
F̂j+1/2 = ρu2 + p  +  ρu2 + p 
2
ρeu + pu j ρeu + pu j+1
ρnj+1 ρnj
   !    !
0 0
1 1
− |unj+1 |  ρnj+1 unj+1  − |unj |  ρnj unj  −  pnj+1 Mj+1
n  −  pnj Mjn  ,
2 2 n n
n n
ρj+1 ej+1 n n
ρj ej pj+1 cj+1 pnj cnj
or
ρnj+1 ρnj
   !
1  1
F̂nj+1/2 = n
F(fj+1 ) + F(fjn ) − |unj+1 |  ρnj+1 unj+1  − |unj |  ρnj unj 
2 2
ρnj+1 enj+1 ρnj enj
   !
0 0
1  n n  −  pnj Mjn 
− pj+1 Mj+1
2 n n
pj+1 cj+1 pnj cnj
34

Problem 20
Propose a numerical scheme to solve for the unsteady flow over a rectangular
cube in an unbounded domain. The Reynolds number is relatively low, 500-
1000. Identify the key issues that must be addressed and propose a solution.
Limit your discussion to one page and do NOT write down the detailed nu-
merical approximations, but state clearly what kind of spatial and temporal
discretization you would use.

Solution
The key issues and how those control the design of the numerical scheme are
listed below:

• Since the Reynolds number is relatively low, it is safe to assume that the
flow is laminar so no turbulence model is needed.
• Since we are dealing with a cube, we can use rectangular structured grid
where the cube boundaries are aligned with grid lines and impose bound-
ary conditions at the cube surface in the same way we do at a solid exterior
wall. Alternatively, we could use immersed boundaries.
• The simplest way to deal with the unbounded domain is to use stretched
grid, extend it very far from the cube to make the domain very large, and
prescribe inflow at the front, full slip at the side boundaries and outflow
in the back. Since the flow is three dimensional, the effect of the cube is
going to be weak far away, except in the wake behind the cube.
• Since the flow is unsteady, we can use the explicit projection method
described in the text.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.1. a) Show that (7.7) solves (7.5) and leads to u = (U,V).
b) Integrate (7.6) within circular area centered on (x´, y´) of radius r" = (x − x ") 2 + (y − y ") 2 to
show that (7.8) is a solution of (7.6).

Solution 7.1. a) Start with ψ = −Vx + Uy . Insert this into the Laplace equation:
% ∂2 €
2 ∂2 (
∇ ψ = ' 2 + 2 *(−Vx + Uy ) = 0 .
& ∂x ∂ y )
The final equality€ follows because ψ is just a linear function so both of the indicated second
derivatives are zero. The velocity field is found via differentiation.
% ∂ψ ∂ψ ( % ∂ (−Vx + Uy) ∂ (−Vx + Uy) (
u€= ' ,− * = ' ,− * = (U,V ) .
& ∂y ∂ x ) & ∂y ∂x )
b) First apply a simple shift transformation that places x´ = (x´, y´) at the origin of coordinates.
Define these new coordinates by: X = x − x #, Y = y − y # and set r" = x − x " = X 2 + Y 2 . The
gradient operator
€ ∇ XY in the shifted coordinates X = (X, Y) is the same as ∇ in the unshifted
coordinates (x, y), so the (7.6) becomes:
€ ∇ 2XYψ = −Γδ (x − x ')δ (y − y ') = −Γδ ( X)δ (Y ) .
€of radius r´:
Integrate this equation inside a circle € €
€ 2
y= + r ) x= + r ) −y
2

2
∫∫ ∇ ψdA = ∫ ∇ XYψ ⋅ nd = −Γ ∫
XY ∫ δ ( X)δ (Y )dXdY ,
€ circular area circle y=− r ) x=− r ) 2 −y 2

where the first equality follows from Gauss' divergence theorem in two dimensions, and the
double integral on the right side includes the location X = 0 (aka x = x´) so a contribution is
collected from both delta functions. Thus, the double integral on the right side of this equation is
unity. €
The dot product in the middle portion of the above equation simplifies to ∂ψ/∂r´ because
n = er on the circle and e r ⋅ ∇ XY = ∂ ∂r%. Plus, in an unbounded uniform environment, there are
no preferred directions so ψ = ψ(r´) alone (no angular dependence). Thus, the integrated field
equation simplifies to
2π
% ∂ψ ( ∂ψ
€ ∫ '& ∂r$ *)r$dθ = 2πr$ ∂r$ = −Γ ,
θ=0
which implies
∂ψ Γ Γ Γ
=− or ψ = − ln( r&) = − ln (x − x &) 2 + (y − y &) 2 ,
∂r$ €2πr$ 2π 2π
and this is (7.8); thus, (7.8) is a solution of (7.6). Here, the constant from the final integration has
been suppressed (set equal to zero) because it has no impact on the velocity field, which depends
on derivatives of ψ.
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.2. For two-dimensional ideal flow, show separately that: a) ∇ψ ⋅ ∇φ = 0 , b)

2 2 2
−∇ψ × ∇φ = u e z , c) ∇ψ = ∇φ , and d) ∇φ = −e z × ∇ψ .

Solution 7.2. In terms of components, the velocity u = (u, v) = (∂ψ/∂y,€–∂ψ/∂x) = (∂φ/∂x, ∂φ/∂y),
and these equations can be used to complete the various parts of this problem.
€ €' €
∂ψ ∂ψ * ' ∂φ ∂φ *
a) ∇ψ ⋅ ∇φ = ) , , ⋅ ) , , = (−v,u) ⋅ ( u,v ) = −vu + uv = 0 .
( ∂x ∂y + ( ∂x ∂ y +
ex ey ez ex ey ez
2
b) −∇ψ × ∇φ = − ∂ψ ∂x ∂ψ ∂y 0 = − −v u 0 = −e z (−v 2 − u 2 ) = e z u
€ ∂φ ∂x ∂φ ∂y 0 u v 0
2 2
2 % ∂ψ ( 2 % ∂ψ ( 2 2 2 2 % ∂φ ( 2 % ∂φ ( 2
c) ∇ψ = ' * + ' * = (−v) + u = u + v = ' * + ' * = ∇φ
& ∂ x ) & ∂y ) & ∂ x ) & ∂y )
€ ex ey ez
d) −e z × ∇ψ = − 0 0 1
€ ∂ψ ∂x ∂ψ ∂y 0
% ∂ψ ( % ∂ψ (
= −e x '0 − * − ey' − 0* − e z (0 − 0)
& ∂y ) & ∂x )
€ % ∂ψ ( % ∂ψ ( % ∂φ ( % ∂φ (
= e x ' * − e y ' * = e x u + e y v = e x ' * + e y ' * = ∇φ
& ∂y ) & ∂x ) & ∂x ) & ∂y )

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.3. a) Show that (7.14) solves (7.12) and leads to u = (U,V).
b) Integrate (7.13) within circular area centered on (x´, y´) of radius r! = (x − x!)2 + (y − y!)2
to show that (7.15) is a solution of (7.13).
c) For the flow described by (7.15), show that the volume flux (per unit depth into the page)
= ∫ u ⋅ nds computed from a closed contour C that encircles the point (x´, y´) is qs. Here n is
C
the outward normal on C and ds is a differential element of C.

Solution 7.3. a) Start with φ = Ux + Vy . Insert this into the Laplace equation:
% ∂2 ∂2 (
∇ 2φ = ' 2 + 2 *(Ux + Vy ) = 0
& ∂x ∂ y )
The final equality€ follows because φ is a linear function so both of the indicated second
derivatives are zero. The velocity field is found via differentiation.
$ ∂φ ∂φ ' $ ∂ (Ux + Vy) ∂ (Ux + Vy) '
€u = & , ) = & , ) = (U,V ) .
% ∂x ∂ y ( % ∂x ∂y (
b) First apply a simple shift transformation that places x´ = (x´, y´) at the origin of coordinates.
Define these new coordinates by: X = x − x #, Y = y − y # and set r" = x − x " = X 2 + Y 2 . The
€ ∇ XY in the shifted coordinates X = (X, Y) is the same as ∇ in the unshifted
gradient operator
coordinates (x, y), so the (7.13) becomes:
€ ∇ 2XY φ = qsδ (x − x#)δ (y − y#) = qsδ (X)δ (Y ) .
€ € €
Integrate this equation inside a circle of radius r´:
€ 2
y= + r ( x= + r ( −y
2

2
∫∫ ∇ XY φdA = ∫ ∇ XY φ ⋅ nd = m ∫ ∫ δ( X)δ(Y)dXdY ,
circular area circle y=− r ( x=− r ( 2 −y 2

where the first equality follows from Gauss' divergence theorem in two dimensions, and the
double integral on the right side includes the location X = 0 (aka x = x´) so a contribution is
collected from both delta functions. Thus, the right-side double integration is unity.
€ dot product in the middle portion of the above equation simplifies to ∂φ/∂r´ because
The
n = er´ on the circle and e r" ⋅ ∇ XY = ∂ ∂r". Plus, in a unbounded uniform environment, there are
no preferred directions so φ = φ(r´) alone (no angular dependence). Thus, the integrated field
equation simplifies to
2π
" ∂φ % ∂φ
€ ∫ $# ∂ r! '&r!dθ = 2π r! ∂ r! = qs , which implies
θ =0

∂φ q
= s or φ = qs ln(r!) = qs ln (x − x!)2 + (y − y!)2 ,
∂ r! 2 π r! 2π 2π
and this is (7.15). Thus, (7.15) is a solution of (7.13). Here, the constant from the final
integration has been suppressed (set equal to zero) because it has no impact on the velocity field,
which depends on derivatives of φ.
c) For the flux integration, evaluate the dot product of u and n = er´: u ⋅ n = ∇φ ⋅ e r% = ∂φ ∂r% = ur% ,
where (7.21) has been used to reach the last equality. Thus,

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

2π 2π
∂φ $ qs '
∫ u ⋅ nds = ∫ u rdθ = ∫ ∂ r# r# dθ = 2π &% 2π r# )( r# = q ,
C 0
r
0
s

where ∂φ/∂r´ comes from the end of part b).

 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

∂φ 1 2 p
Exercise 7.4. Show that (7.1) reduces to + ∇φ + = const. when the flow is described by
∂t 2 ρ
the velocity potential φ.

∂u
Solution 7.4. Start from (7.1),€ + (u ⋅ ∇ )u + (1 ρ)∇p = 0 , and use the vector identity (B3.9),
∂t
1
( )
2
(u ⋅ ∇)u = ∇ 2 u − u × (∇ × u) to replace the advective acceleration with a gradient and cross
product term:
€ ∂u $ 1 2' 1
+ ∇& u ) − u × (∇ × u) + ∇p = 0.
∂t %2 ( ρ
€ Use the definition of the potential, ∇φ = u , to eliminate u from this equation:
∂ %1 2( 1
∇φ + ∇' ∇φ * − ∇φ × (∇ × ∇φ ) + ∇p = 0 .
€ ∂t &2 ) ρ
The third term is zero because
€ ∇ × ∇φ = 0 for any scalar function φ. The remaining terms can
all be grouped under one gradient operation:
& ∂φ 1 2 p)
€ ∇( + ∇φ + + = 0 .
€ ' ∂t 2 ρ*
Thus, the grouping of terms in the parentheses can at most be a function of time. Let this
function be B(t),
∂φ 1 2 p
€ + ∇φ + = B(t) ,
∂t 2 ρ
Now define a mildly revised potential that includes B(t):
t ∂φ ∂φ $
φ = φ # + ∫ 0 ( B(τ ) − const.) dτ , so that = + B(t) − const.
∂t ∂t
This does not change the € velocity field because ∇φ # = ∇φ = u . Therefore, the remnant of (7.1)
becomes:
€ ∂φ # 1 2 p ∂φ # 1 2 p
+ B(t) − const.+ ∇φ # + €= B(t) , or + ∇φ # + = const.,
∂t 2 ρ ∂t 2 ρ
€
which is the requisite equation when the primes are dropped from φ.

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.5. Consider the following two-dimensional Cartesian flow fields: (i) solid body
rotation (SBR) at angular rate Ω about the origin: (u, v) = (–Ωy, Ωx); and (ii) uniform expansion
(UE) at linear expansion rate Θ: (u, v) = (Θx, Θy). Here Ω, Θ, and the fluid density ρ are positive
real constants and there is no body force.
a) What is the stream function ψSBR(x,y) for solid body rotation?
b) Is there a potential function φSBR(x,y) for solid body rotation? Specify it if it exists.
c) What is the pressure, pSBR(x,y), in the solid body rotation flow when pSBR(0,0) = po?
d) What is the potential function φUE(x,y) for uniform expansion?
e) Is there a stream function ψUE(x,y) for uniform expansion? Specify it if it exists.
f) Determine the pressure, pUE(x,y), in the uniform expansion flow when pUE(0,0) = po.

Solution 7.5. a) Start with the stream-function-based velocity definitions,

u = ∂ψ ∂y = −Ωy and v = −∂ψ ∂x = Ωx ,
and integrate each second equality once to find: ψ = −(Ω 2)y 2 + f1 (x) and ψ = −(Ω 2)x 2 + f2 (y) .
The two equations are consistent when: ψ = −(Ω 2)(x 2 + y 2 ) + const. , and this result makes sense;
the streamlines are circles.
b) There is no potential function for SBR flow because it is everywhere rotational.
c) Start with (u, v) = (–Ωy, Ωx) and use (7.1) simplified for steady flow where ∂/∂t = 0:
∂u 1 ∂p
ui j = − .
∂x i ρ ∂x j
Evaluate the left side of this equation using the given velocity field:
$ ∂ ∂ ' ∂p ∂p
ρ (−Ωye x + Ωxe y ) ⋅ & e x + e y ) (−Ωye x + Ωxe y ) = −e x − ey .
€ % ∂x ∂y( ∂x ∂y
This reduces to:
∂p ∂p
−ρΩ2 xe x − ρΩ2 ye y = −e x − ey .
∂x ∂y
Considering each component separately,
∂p ∂p
ρΩ2 x = and ρΩ2 y = ,
∂x ∂y
and integrate to find: pSBR (x, y) = ρΩ2 ( x 2 2 ) + f (y) and pSBR (x, y) = ρΩ2 ( y 2 2 ) + g(x) , where
f and g are functions of integration. Combining these equations into one by appropriately
choosing f and g, and then applying the boundary condition on x = y = 0 to determine the final
constant, yields:
ρΩ2 2 2 ρΩ23 2
pSBR (x, y) = po +
2
( x + y ) , or in cylindrical coordinates p(R) = po +
2
R .
This answer makes sense because the necessary centripetal acceleration of the rotating fluid must
be provided by pressure forces. So, increasing pressure with increasing R is physically
meaningful, even when viewed from a rotating dynamics point of view. This result can be
€
checked with the swirling flow of a liquid in a cylindrical container (see Section 5.1). The height
of the free surface is proportional to the pressure in the fluid, and the free surface of a liquid in
solid body rotation is parabolic. This effect was utilized by astronomers to cast telescope mirror
blanks on rotating tables in the early part of the 1900’s. The radial pressure gradient found
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

above is also used to separate constituents in fluid mixtures in a centrifuge, a technique widely
employed in the production of weapons-grade uranium through the separation of isotopes of UF6
(a gas), and in the biological sciences to obtain nucleic material from cells.
d) Start with the potential-function-based velocity definitions,
u = ∂φ ∂x = Θx and v = ∂φ ∂y = Θy ,
and integrate each equation once to find: φ = (Θ 2)x 2 + f3 (y) and φ = (Θ 2)y 2 + f4 (x) . The two
equations are consistent when: φ = (Θ 2)(x 2 + y 2 ) + const.
e) There is no stream function for UE flow because it is everywhere not mass conserving in
constant density flow.
f) Use the momentum equation as in part c) or the Bernoulli equation between the origin and the
point (x,y):
pUE (x, y) + 12 ρ (u 2 + v 2 ) = po + 0 , or pUE (x, y) = po − 12 ρΘ2 ( x 2 + y 2 ) .
This answer is, of course, unusual since the flow is everywhere non-mass-conserving.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.6. Determine u and v, and sketch streamlines for a) ψ = A(x2 – y2), and b) φ = A(x2 –
y2).
y
Solution 7.6. a) From the definition of the
stream function: u = ∂ψ ∂y = −2Ay ,
v = − ∂ψ ∂x = −2Ax , and the streamlines
are given by x2 – y2 = const. For large x and
y, these streamlines asymptote to the lines y
= ±x. Assuming € positive A, the signs for u
€ and v indicate that the flow is toward the
origin along y = x in the first quadrant.
Therefore, the streamlines look like the x
upper drawing to the right.

b) From the definition of the

∂φ
potential function: u = = 2Ax , y
∂x
∂φ
and v = = −2Ay . Here the
∂y
streamlines are given by
€
dy v −2Ay y
= = = − which can be
dx u 2Ax x
€ integrated to find: ln(y) = –ln(x) +
const., or xy = const. For large x
and y, these streamlines asymptote x
€ to the lines y = 0 or x = 0 (the
coordinate axes). Assuming positive
A,€the signs of u and v indicate the
flow is away from the origin along
y = 0 (the x-axis) and toward the
origin along x = 0 (the y-axis).
Therefore, the streamlines look like
the lower drawing to the right.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.7. For the following two-dimensional stream and potential functions, find the fluid
velocity u = (u,v), and determine why these are or are not ideal flows.
a) ψ = A(x2 + y2)
b) φ = A(x2 + y2)

Solution 7.7. a) From the definition of the stream function: u = ∂ψ ∂ y = 2Ay ,

v = −∂ψ ∂ x = −2Ax , so ∂u/∂x + ∂v/∂y = 0 and the flow conserves mass. Here, the streamlines are
given by x2 + y2 = const, which are circles. However, ∇ 2ψ = 4A = const. , so (7.4) implies that ωz
= –4A = const. so the given stream function corresponds to solid body rotation at rate –2A about
the z-axis. Thus, this flow is not an ideal flow because fluid elements rotate.
b) From the definition of the potential function: u = ∂φ ∂ x = 2Ax , and v = ∂φ ∂ y = 2Ay , so ∂u/∂x
+ ∂v/∂y = 4A and this flow does not conserve mass. Here the streamlines are given by
dy v 2Ax y
= = = , and this can be integrated to find: ln(y) = ln(x) + const., or y = (const.)x .
dx u 2Ay x
These are straight lines through the origin of coordinates. This potential function corresponds to
uniform expansion in the x-y plane. This is not an ideal flow because it does not conserve mass.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.8. Assume ψ = ax3 + bx2y + cxy2 + dy3 where a, b, c, and d are constants; and
determine two independent solutions to the Laplace equation. Sketch the streamlines for both
flow fields. y

Solution 7.8. Determine the implications of

∇ 2ψ = 0 by evaluating derivatives:
∂ 2ψ ∂ 2ψ
= 6ax + 2by , and = 2cx + 6dy , so
∂x 2 ∂y 2
€ ∂ 2ψ ∂ 2ψ
+ = 0 = 6ax + 2by + 2cx + 6dy .
∂x 2 ∂y 2
€ € x
The first non-trivial solution ψ1 occurs
when 6a + 2c = 0 or c = –3a, and b = d = 0:
€
( )(
ψ1 = a( x 3 − 3xy 2 ) = ax x + 3y x – 3y . )
Thus, the streamlines corresponding to ψ1 = 0 are
three lines specified by x = 0 and y = ± x 3 . The
€ velocity components determined from ψ1 are:
u = ∂ψ1 ∂y = −6axy , and
v = − ∂ψ1 ∂x = −a(3x 2 −€3y 2 ) = −3a(x + y)(x − y) .
Assuming positive a, the streamlines for this flow
field appear in the upper figure to the right.
€
€ y
The second non-trivial solution ψ2
occurs when a = c = 0, and 2b + 6d = 0 or d = –
3d:
( )(
ψ 2 = d (−3x 2 y + y 3 ) = dy y + 3x y – 3x .)
Thus, the streamlines corresponding to ψ2 = 0
are three lines specified by y = 0 and y = ± x 3 .
€ The velocity components determined from ψ2
are: x
u = ∂ψ 2 ∂y = −3dx 2 + 3dy 2 €
= −3d(x + y)(x − y) ,
and v = − ∂ψ 2 ∂x = −d(−6xy) = 6dxy .
Assuming positive d, the streamlines for this
flow field appear in the lower figure to the
€ right.
€
The two flow fields are identical when a = d
and the streamlines are rotated by 30°.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.9. Repeat exercise 7.8. for ψ = ax4 + bx3y + cx2y2 + dxy3 + ey4 where a, b, c, d, and e
are constants.

Solution 7.9. Determine the implications of ∇ 2ψ = 0 by evaluating derivatives:

∂ 2ψ 2 2 ∂ 2ψ
2
= 12ax + 6bxy + 2cy , and 2
= 2cx 2 + 6dxy + 12ey 2 , so
∂x ∂y
∂ 2ψ ∂ 2ψ
+ = 0 =€12ax 2 + 6bxy + 2c(y 2 + x 2 ) + 6dxy + 12ey 2 .
∂x 2 ∂y 2
Therefore,
€ non-trivial solutions occur when € 12a + 2c = 0 or c = –6a, 6b + 6d = 0 or b = –d, and
2c + 12e = 0 or c = –6e. Choosing a and then b as the free parameter, leads to:
ψ1 = a(x 4€− 6x 2 y 2 + y 4 ) = a((x 2 − y 2 ) 2 − 4 x 2 y 2 )
= a( x 2 − y 2 + 2xy )( x 2 − y 2 − 2xy ) = a((x + y) 2 − 2y 2 )((x − y) 2 − 2y 2 )

( )( )(
= a x + (1+ 2)y x + (1− 2)y x − (1− 2)y x − (1+ 2)y , )( )
and y
3 3
ψ 2 = b(x y − xy ) = bxy(x + y)(x − y) .
Consider each solution in turn.
€
The streamlines corresponding to ψ1 =
0 are four lines specified all possible sign
€
combinations of (
y = ± x 1± 2 . The )
velocity components determined from ψ1 are:
u = ∂ψ1 ∂y = −12ax 2 y + 3y 3 , and
x
v = − ∂ψ1 ∂x = −a(4 x 3 −12xy 2 ) .
€
Assuming positive a, the streamlines for this
flow field appear in the upper figure to the
€
right.
€
y

The streamlines corresponding to

ψ2 = 0 are four lines specified by x = 0, y
= 0, y = x, and y = –x. The velocity
components determined from ψ2 are:
x u = ∂ψ 2 ∂y = bx 3 − 3xy 2 , and
v = − ∂ψ 2 ∂x = −3ax 2 y + y 3 .
Assuming positive a, the streamlines for
this flow field appear in the lower figure
€ to the left.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.10. Without using complex variables, determine:

a) The potential φ for an ideal vortex of strength Γ starting from (7.8)
b) The stream function for an ideal point source of strength qs starting from (7.15)
c) Is there any ambiguity in your answers to parts a) and b)? If so, does this ambiguity influence
the fluid velocity?

Solution 7.10. a) The starting point is the stream function for a two-dimensional point vortex of
strength Γ located at x = x´ = (x´, y´):
Γ
ψ = − ln (x − x &) 2 + (y − y &) 2 .
2π
The velocity components are:
∂ψ Γ y − y' ∂φ ∂ψ Γ x − x' ∂φ
u= =− 2 2
= , and v = − = 2 2
= ,
∂y 2π (x − x ') + (y − y ') ∂x ∂x 2π (x − x ') + (y − y ') ∂y
€
where the final equality in each case comes from the definition of the potential function.
Integrate to find φ(x, y) using these final equalities to find
Γ y − y& Γ x − x&
€ φ (x, y) = − ∫ 2
2π (x − x &) + (y − y &) 2
dx + €f (y) , and φ (x, y) = ∫
2π (x − x &) 2 + (y − y &) 2
dy + g(x) ,
where f(y) and g(x) are single-variable functions that appear because of the integrations.
Rearrange the first integral, and create the integration variable tanθ = (y – y´)/(x – x´), noting that
sec2θdθ = –[(y – y´)/(x – x´)2]dx.
€ €
Γ (y − y &) (x − x &) 2 Γ sec 2 θdθ Γ
φ (x, y) = − ∫
2π 1+ [(y − y &) (x − x &)] 2 dx + f (y) = ∫
2π 1+ tan θ 2
+ f (y) =
2π
θ + f (y) .

Use the same integration variable in the second integral to find:

Γ sec 2 θdθ Γ
φ (x, y) = ∫
2π 1+ tan θ 2
+ g(x) =
2π
θ + g(x) .
€
The only way for the two results to be consistent is for f(y) = g(x) = const. Thus,
Γ ' y − y &*
φ (x, y) = tan−1) , + const.
€ 2 π ( x − x &+
b) For this part, the starting point is the potential for a two-dimensional point source of strength
qs located at x = x´ = (x´, y´):
q
€ φ = s ln (x − x")2 + (y − y")2 .
2π
The velocity components are:
∂φ qs x − x" ∂ψ ∂φ qs y − y" ∂ψ
u= = 2 2
= , and v = = 2 2
=− ,
∂ x 2π (x − x") + (y − y") ∂ y ∂ y 2π (x − x") + (y − y") ∂x
where the final equality in each case comes from the definition of the stream function. Integrate
to find ψ(x, y) using these final equalities to find
q x − x" q y − y"
ψ (x, y) = s ∫ 2 2
dy + f (x) , and ψ (x, y) = − s ∫ dx + g(y) ,
2π (x − x") + (y − y") 2π (x − x")2 + (y − y")2
where f(x) and g(y) are single-variable functions that appear because of the integrations. These
integrations are the same as in part a); thus using the same integration variable produces:
q q
ψ (x, y) = s θ + f (x) = s θ + g(y) .
2π 2π
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

The only way for the two results to be consistent is for f(x) = g(y) = const, so
q # y − y" &
ψ (x, y) = s tan −1 % ( + const.
2π $ x − x" '
c) In both cases, there is an undetermined constant in the answer. However, the fluid velocity
field depends on derivatives of φ and ψ, so this constant does not influence the velocity field.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.11. Determine the stream function of a doublet starting from (7.29) and show that the
streamlines are circles having centers on the y-axis that are tangent to x-axis at the origin.

Solution 7.11. The potential for a doublet is:

d cos θ d x
φ= = ,
2π r 2π x + y 2
2

so the velocity field is:

∂φ d ∂ % x ( d % 1 x ( d % y 2 − x 2 ( ∂ψ
u= = ' * = ' − (2x) *= ' 2 2 2*
= , and
∂x 2π ∂x & x 2 € + y 2 ) 2π & x 2 + y 2 (x 2 + y 2 ) 2 ) 2π & (x + y ) ) ∂y
∂φ d ∂ % x ( d % x ( d % −2xy ( ∂ψ
v= = ' 2 2*
= '− 2 2 2
(2y)* = ' 2 2 2*
=− ,
∂y 2π ∂y & x + y ) 2π & (x + y ) ) 2π & (x + y ) ) ∂x
€ where the final equalities for each component follow from the definition of the stream function.
Integrate the v-equation using x2 + y2 = β and 2xdx = dβ to find:
d 2xy d dβ d y d y
€ψ (x, y) = ∫ 2
2π (x + y ) 2 2
dx + f (y) =
2π
y ∫ 2 + f (y) = −
β 2π β
+ f (y) = −
2π x + y 2
2
+ f (y) .

When this result is partial-differentiated with respect y, the result is:

∂ d ∂ & y )
ψ (x, y) = − ( 2 2
+ f (y)+
€ ∂y 2π ∂y ' x + y *
d & 1 y ) d & x2 − y2 )
=− ( 2 − (2y) + f ,(y) + = − ( + f ,(y)+
2π ' x + y 2 (x 2 + y 2 ) 2 * 2π ' (x 2 + y 2 ) 2 *
which matches the u-equation result above when f´(y) = 0. Thus, f = const. and the stream
function is:
d y
€ ψ (x, y) = − + const. y
2π x + y 2
2

Curves of constant ψ are streamlines, so rename const. as ψo,

and rearrange the above equation to determine the streamline
curves in a standard functional form.
€
d y
ψ − ψo = − , or
2π x + y 2
2

2 2
d d d
x2 + y2 + y+ 2
= 2
,
2π (ψ − ψ o ) 16π 2 (ψ − ψ o ) 16π 2 (ψ − ψ o )
€
x
which simplifies to:
2 2
2
% d ( d
x + 'y + * = .
€ & 4 π (ψ − ψ o ) ) 16π 2 (ψ − ψ o ) 2
These are circles of radius a = d 4 π (ψ − ψ o ) that are centered
at y = –a, thus they are tangent to the origin, as shown in the
drawing.
€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.12.
Consider steady horizontal flow at speed U past a stationary source of strength qs located at the
origin of coordinates in two dimensions, (7.30) or (7.31). To hold it in place, an external force
per unit depth into the page, F, is applied to the source.
a) Develop a dimensionless scaling law for F = |F|.
b) Use a cylindrical control volume centered on the source with radius R and having depth B into
the page, the steady ideal-flow momentum conservation equation for a control volume,
<Begin Equation>
∫ ρu (u ⋅ n)dA = − ∫ pn dA + F ,
A* A*
</End Equation>
and an appropriate Bernoulli equation to determine the magnitude and direction of F without
using Blasius Theorem.
c) Is the direction of F unusual in anyway? Explain it physically.

Solution 7.12. a) There are only 4 parameters: F, qs, U, and ρ, and these span all three
dimensions. Thus, there is only one dimensionless parameter, F/ρUqs, so it must be constant.
This implies F = (const.)ρUqs, so rest of this problem merely involves determining the constant
and the direction of F.
b) It is convenient to use x-y and r-θ velocities expressed in polar coordinates. Using the
potential
q q
φ = Ux + s ln x 2 + y 2 = Ur cosθ + s ln r ,
2π 2π
and performing the requisite differentiations produces:
! q q $ " q %
u = (u, v) = #U + s cosθ , s sin θ & , and u = (ur , uθ ) = $U cosθ + s , −U sin θ '
" 2π r 2π r % # 2π r &
Again, choose the control volume to be a cylinder of radius R and span B centered on the point
source. An evaluation of the conservation of mass equation merely validates that a source of
strength m is present in the flow field.
To find the force from the given CV momentum equation, first identify unit vectors and
evaluate dot products. Here, n = e r so that u ⋅ n = ur . Therefore:
θ = 2π θ = 2π
F= ∫ ρ(ue x + ve y )ur BRdθ + ∫ p(e x cosθ + e y sin θ )BRdθ
θ=0 θ=0
The pressure at r can be evaluated via a Bernoulli equation:
€ € "
Uq cosθ q2 % ρUqs cosθ ρ qs2
1 2 1
(
2 2
)
p∞ + 2 ρU = p + 2 ρ u + v = p + 12 ρ $U 2 + s
# πr
+ s2 2 ' , or p = p∞ −
4π r & 2π r
− 2 2.
8π r
€
With this replacement, the x-component of the momentum equation is
θ =2 π !
Fx θ =2 π ! qs $! qs $ ρUqs cosθ ρ qs2 $
= ∫ ρ #U + cosθ &#U cosθ + & R dθ + ∫ # p∞ − − 2 2 & cosθ R dθ
B θ =0 " 2π R %" 2π R % θ =0 " 2π R 8π R %
The only integrand pieces that contribute involve even powers of like trigonometric functions.
Fx qsU θ =2 π qsU θ =2 π qsU θ =2 π
B
=ρ
2π θ =0
∫ (1+ cos θ ) dθ − ρ 2π ∫ cos θ dθ =ρ 2π ∫ dθ = +ρqsU
2 2

θ =0 θ =0
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

The sign of this result is a paradox. It implies that a downstream-pointing force (per unit depth)
is applied to the fluid in the control volume. Thus, if this force were not present, the point source
would move upstream! A point source in a uniform stream is an elementary rocket.
Manipulations similar for the y-component of the momentum equation are similar to
those above with the final finding that: Fy = 0. Interestingly, these results do not depend on the
size of the control volume.
NOTE: The force components must be resolved in a Cartesian coordinate system, since
any net radial force on the CV is accounted for by fluid pressure, and any net angular "force" on
the CV is a torque, not a force.
c) The force points in the positive x-direction, and therefore prevents the source from moving
upstream! Thus, an untethered point source is an elementary a rocket. This rocket-type character
can be visually ascertained by considering a finite-length control volume that follows the
dividing streamline and encloses the source as is shown in the drawing below.
y

x
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.13. Repeat all three parts of Exercise 7.12 for steady ideal flow past a stationary
irrotational vortex located at the origin when the control volume is centered on the vortex. The
stream function for this flow is: ψ = Ur sin θ − ( Γ 2π ) ln(r)

Solution 7.13. a) There are only 4 parameters: F, Γ, U, and ρ, and these span all three
dimensions. Thus, there is only one dimensionless parameter, F/ρUΓ, so it must be constant.
This implies F = (const.)ρUΓ, so rest of this problem merely involves determining the constant
and the direction of F.
b) For this problem, it is convenient to use a mixture of x-y and r-θ velocities expressed in polar
coordinates. The potential for this flow field is φ = Ur cosθ + Γθ 2π . Performing the requisite
differentiations produces:
& Γ Γ ) & Γ )
u = (u,v) = (U − sin θ, cosθ + , and u = (ur ,uθ ) = (U cosθ,−U sin θ + +
' 2πr 2πr * ' 2πr *
€
Choose the control volume to be a cylinder of radius R centered on the point vortex and having
depth B into the page. The equation for conservation of mass is identically satisfied for all
potential flows without sources.
€ To find the force from the given CV € momentum equation, first identify unit vectors and
evaluate dot products. Here, n = e r so that u ⋅ n = ur . Therefore:
θ = 2π θ = 2π
F= ∫ ρ(ue x + ve y )ur BRdθ + ∫ p(e x cosθ + e y sin θ )BRdθ
θ=0 θ=0
The pressure at r can be evaluated via a Bernoulli equation:
€ € (
2UΓsin θ Γ2 + ρUΓ sin θ ρΓ 2
2 2
( 2
)
p∞ + 2 ρU = p + 2 ρ u + v = p + 12 ρ*U 2 −
1 1

) 2πr
+ -
4π 2r 2 ,
, so p = p∞ +
2πr
− 2 2.
8π r
€
Evaluate this at r = R, and place it into the x-component of the momentum equation.
θ = 2π '
Fx θ = 2 π ' Γ * ρUΓ sinθ ρΓ 2 *
= ∫ ρ)U − sin θ ,U cosθRdθ + ∫ ) p∞ + − 2 2 , cos θRdθ = 0
€ B θ=0 ( 2πR + θ=0 ( € 2πR 8π R +
In each case the various trigonometric functions produce a net zero for each term. The y-
component of the equation is:
θ = 2π &
Fy θ = 2 π & Γ ) ρUΓsin θ ρΓ 2 )
€ = ∫ ρ( cos θ +U cosθRdθ + ∫ ( p∞ + − 2 2 + sin θRdθ .
B θ = 0 ' 2πR * θ=0 ' 2πR 8π R *
Evaluate the integrals. The non-trivial ones become:
Fy UΓ θ = 2 π 2 UΓ θ = 2 π 2
B
=ρ ∫
2π θ = 0
cos θdθ + ρ ∫ sin θdθ = ρUΓ .
2π θ = 0
€
Hence, the force on control volume is directed along the positive y-axis (perpendicular to the
uniform stream) for positive Γ. When R → 0, this force remains the same. Therefore, to keep
the vortex stationary, an equal and opposite force must be produced by the hydrodynamic
interaction of €
the point vortex with the incoming flow. This hydrodynamic force is called the lift
(= L = –Fy), because it can support the weight of an object. The final result, which is true for all
two-dimensional lifting bodies, is: L B = −ρUΓ . The minus sign appears here because of the
choice of the positive direction of circulation.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

NOTE: The force components must be resolved in a Cartesian coordinate system, since
any net radial force on the CV is accounted for by fluid pressure, and any net angular "force" on
the CV leads to a torque, not a force.
c) In this case, the flow upstream of the vortex is drawn downward and the flow downstream of
the vortex is deflected upward. This asymmetry in the flow leads to the lift force.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.14. Use the principle of conservation of mass (4.5) and an appropriate control volume
to show that maximum half thickness of the half body described by (7.30) or (7.31) is hmax =
qs/2U.

Solution 7.14. Place a control surface on the streamline that divides the fluid from the source
from the fluid that originates upstream. The velocity u is:
! q q $
u = (u, v) = #U + s cosθ , s sin θ & ,
" 2π r 2π r %
but this simplifies to u = (U, 0) far from the source ( r → ∞ ). Therefore, far downstream of the
source, the volume flux (per unit depth into the page) inside the CV must still be qs and its flow
speed will be U. Therefore, conservation of mass implies:
ρqs = 2ρUhmax € , or hmax = qs/2U.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.15. By integrating the surface pressure, show that the drag on a plane half-body
(Figure 7.7) is zero.

Solution 7.15. For the half-body produced by a free stream of speed U and source of strength m,
the potential and velocity field are:
q q " q %
φ = Ux + s ln x 2 + y 2 = Ur cosθ + s ln r , and u = (ur , uθ ) = $U cosθ + s , −U sin θ ' .
2π 2π # 2π r &
The stream function is given by:
q
ψ = Ur sin θ + s θ ,
2π
so the streamline r(θ) that divides the fluid that originates upstream from the fluid that comes
from the source is given by:
qs q q (π − θ )
= Ur sin θ + s θ , or r(θ ) = s .
2 2π 2πU sin θ
The coefficient of pressure on the dividing streamline will be:
" ur2 + uθ2 % " U 2 cos2 θ + ( q U π r ) cosθ + ( q 2π r )2 +U 2 sin 2 θ %
s s
C p (θ ) = $1− 2 ' = $$1− 2
'
'
# U &r(θ ) # U &r(θ )
" q q2 % q " cosθ q 1%
= $ − s cosθ − 2 s 2 2 ' = − s $ 2 + s 2'
# πUr 4π U r &r(θ ) 2πU # r 2πU r &r(θ )
qs " cosθ q 4π 2U 2 sin 2 θ % " cosθ sin θ sin 2 θ %
=− $2 2πU sin θ + s ' = − $ 2 + '
2 π U # qs ( π − θ ) 2πU qs2 (π − θ )2 & # (π − θ ) (π − θ ) 2 &
The drag force on the body will be:
θ = 2π 2π
Drag = e x ⋅ ∫ −( p − p∞ )ndA = −e x ⋅ ∫θ = 0 12 ρU 2C p (θ )nBdξ = − 12 ρU 2 B ∫ 0 C p (θ )e x ⋅ ndξ ,
surface

where B is the spanwise dimension, n = [∇ψ ∇ψ ] r(θ ) is the outward normal from the dividing
streamline, and dξ is a contour increment along the dividing streamline. Fortunately, some of the
€ intricacies of the geometry cancel out. First consider the dot product:
∇ψ ∂ψ ∂x −v v 1
ex ⋅ n = ex ⋅ €= = =− .
∇ψ 2 2
ur + uθ uθ 1+ ( ur2 uθ2 ) uθ 1+ (1 r) 2 ( dr dθ )
Here v is the vertical velocity component, and the final equality follows from the streamline
condition, dr ur = rdθ uθ , in r-θ polar coordinates (see Exercise 3.8). Now develop dξ in r-θ
polar coordinates:
€ 2
dξ = (dr) 2 + (rdθ ) 2 = rdθ 1+ (1 r) ( dr dθ ) .
Therefore:
€
ν 1 2 ν q
e x ⋅ ndξ = − rdθ 1+ (1 r ) ( dr dθ ) = − rdθ = − s dθ .
€ uθ 1+ (1 r ) ( dr dθ )
2 uθ 2πU
where the final equality follows because v = ( qs 2π r ) sin θ , and uθ = –Usinθ. Thus, the drag
integral becomes:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Drag 1 q 2π q ρU 2 π # sin(2θ ) sin 2 θ &

= 2 ρU 2 s ∫ C p (θ )dθ = − s ∫% + ( dθ .
B 2πU 0 4 π 0 $ (π − θ ) (π − θ ) 2 '
Change the integration variable to β = π – θ, and note that the integrand is not singular because
of the behavior of the sine function for small argument:
Drag qs ρU + π " sin(2 β ) sin 2 β % qs ρU + π " sin(2 β ) 1 cos(2 β ) %
B
=− ∫
4π − π #
$−
β
+ 2 'dβ = −
β &
∫
4π − π #
$−
β
+ 2−
2β 2β 2 &
'dβ .

In the last expression, leave the first term alone, integrate the second term to get –1/2β, and
integrate the third term by parts to reach:
qs ρU /1" 1 + 31
+π +π (
Drag 1 % sin(2 β ) 1 ( −2 +
=− 0$− + cos(2 β )' + ∫ * − − * - sin(2 β )- d β 4 = 0 ,
B 4π 21# 2 β 2 β &− π − π ) β 2) β , , 51
and this is the desired result.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.16. Ideal flow past a cylinder (7.33) is perturbed by adding a small vertical velocity
without changing the orientation of the doublet:
Ua 2 y & a2 )
ψ = −Uγx + Uy − 2 = −Uγr cosθ + U( r − + sin θ .
x + y2 ' r*
a) Show that the stagnation point locations are rs = a and θs = γ/2, π + γ/2 when γ << 1.
b) Does this flow include a closed body?
€ Use the polar-coordinate version of the stream function to find the radial (u ), and
Solution 7.16. r
angular (u ) velocity components.
θ

∂ψ ' a2 * 1 ∂ψ ' a2 *
− = uθ = Uγ cos θ − U)1+ 2 , sin θ , and = ur = Uγ sin θ + U)1− 2 , cos θ .
∂r ( r + r ∂θ ( r +
At a stagnation point both velocity components will be zero, so
a2 a2
0 = Uγ cosθ s − U sin θ s − U 2 sin θ s , and 0 = Uγ sin θ s + U cos θ s − U 2 cosθ s ,
rs rs
€ €
where the subscript s denotes stagnation point variables. These are two non-linear equations in
two unknowns. First eliminate θs by creating tanθs in each equation:
$ a2 ' % 2(
€ 0 = γ − &1+ 2 ) tan €θ s , and 0 = γ tan θ s + '1− a * .
2
% rs ( & rs )
Now eliminate tanθs between these two equations, and solve for rs to find:
rs −1 4 γ
= [1+ γ 2 ] , and tan θ s = .
a 1+ 1+ γ 2
€ €
When γ << 1, these two results can be expanded:
rs 1 γ γ& 1 )
≅ 1− γ 2 + ..., and tan θ s ≅ 1 2
= (1− γ 2 + ...+ .
a€ 4 € 1+ 1+ 2 γ + ... 2 ' 4 *
Thus, to lowest order, the two stagnation point locations are r = a, and θ = γ/2, π + γ/2.
b) No, the flow no longer includes a closed body. The sketch below shows what happens.
€ y
€

x
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.17. For the following flow fields (b, U, Q, and Γ are positive real constants), sketch
streamlines.
a) ψ = b r cos(θ 2) for |θ| < 180°
b) ψ = Uy + (Γ 2π ) ln [ ( x + (y − b) ) − ln( x + (y + b) )]
2 2 2 2

(Q 2π ) ln( x + (y − 2na) ) for |y| < a.

n= +∞ 2 2
c) φ = ∑n=−∞
€
€ Solution 7.17. There are several ways to attack this problem. The most direct is to take a few
derivatives to determine the velocity components and then plot the results. However, detailed
€ plots are not required here (the problem merely says sketch the streamlines); thus, flow directions
can be ascertained by considering limiting values of the independent variables. In particular, for
parts b) and c), very near a point vortex or a point source the streamlines will be circular or
radial, respectively, because the 1/r–factor will cause the point vortex or point source to
dominate the local flow field. Similarly, very far from a point vortex or a point source the
streamlines will be little changed from those that would exist in the absence of the point source
or point vortex. So, in parts b) and c), start with a few appropriate streamlines near the vortex or
point source, and then join them up with a few horizontal parallel segments for |x| large that
represent a parallel stream. In joining-up the streamlines from various parts of the sketches, the
requirement that volume flux (per unit depth) remain constant between streamlines can be useful
in determining where streamlines come closer together (increasing flow speed) and where the
bend away from each other (decreasing flow speed). Here, the Bernoulli equation can also lend
insight; streamlines will converge where pressure decreases and diverge where the pressure
increases. There is an alternative drawing for part b) where the two stagnation points occur on
the y-axis.
a) for b > 0, b)
y
y

x
x

c) y
y=a

y = -a
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.18. Take a standard sheet of paper and cut it in half. Make a simple airfoil with one
half and a cylinder with the other half that are approximately the same size as shown.
a) If the cylinder and the airfoil are dropped from the same height at the same time with the
airfoil pointed toward the ground in its most streamlined configuration, predict which one
reaches the ground first.
b) Stand on a chair and perform this experiment. What happens? Are your results repeatable?
c) Can you explain what you observe?

Tape

Solution 7.18. a) If the airfoil remains aligned in the vertical direction, its drag is less than the
cylinder. Therefore, it should reach the ground first.
b) The airfoil that was tested for this solution manual, reached the ground first about 1 out of 10
ten times. In most cases, the airfoil refused to travel straight down and fluttered off sideways so
that it reached the ground after the cylinder.
c) Consider a frame of reference in which the flow about the airfoil is initially steady. A
symmetric airfoil that is misaligned with respect to a uniform stream feels a lift force and a
pitching moment. These are the result of an uneven pressure distribution on the airfoil surfaces
caused by the fluid being forced to speed up unevenly as it travels around the pitched airfoil.
This uneven velocity leads to an uneven pressure distribution through the Bernoulli equation.
The lift force is proportional to the miss-alignment angle (commonly called the angle of attack).
So, the airfoil is dynamically unstable, that is, any small perturbation in its angle of attack leads
to lift forces that tend to increase the angle of attack. This phenomenon is one of the reasons that
a rear stabilizer – with angle of attack opposite of that of the main wing – is part of the tail
structure of most aircraft.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.19. Consider the following two-dimensional stream function composed of a uniform
horizontal stream of speed U and two vortices of equal and opposite strength in (x,y)-Cartesian
coordinates.
ψ (x, y) = Uy + (Γ 2π ) ln x 2 + (y − b) 2 − (Γ 2π ) ln x 2 + (y + b) 2
a) Simplify this stream function for the combined limit of b → 0 and Γ → ∞ when 2bΓ = C =
(
constant to find: ψ (x, y) = Uy 1− (C 2πU )( x 2 + y 2 ) )
−1

€ to (r,θ)-polar coordinates and find €

b) Switch both components
€ of the velocity using the
simplified stream function.
c) For the simplified stream function, determine where ur = 0.
€
d) Sketch the streamlines for the simplified stream function, and describe this flow.

Solution 7.19. a) The square roots may be simplified for the limit b → 0 :
x 2 + (y ± b) 2 = r 1+ (±2yb + b 2 ) r 2 ≈ r(1± by r 2 )
2 2
( 2
) 2
where r 2 = x 2 + y 2 . Thus: ln x + (y ± b) ≈ ln(r) + ln 1± by r ≈ ln(r) ± by r , so the stream
€
function becomes:
Γ & by ) Γ & by ) Γby & C )
ψ€(x, y) = Uy + ( ln(r) − 2 + − ( ln(r) + 2 + = Uy − 2 = Uy(1− +.
€ 2π ' r * 2π ' r * πr ' Uπr 2 *
€
( 2
)
b) In polar coordinates y = rsinθ so: ψ (r,θ ) = Ur sin θ 1− C Uπr . Thus:
ur = (1 r)(∂ψ ∂θ ) = U cos θ (1− C Uπr ) , and uθ = −(∂ψ ∂r) = −U sinθ (1+ C Uπr 2 )
2

€
c) ur = 0 when θ = ± π 2 , or r = C Uπ .
€ points at θ = 0 and π, and r = C Uπ . In terms of a = C Uπ ,
d) This flow as two stagnation
€ ur = U cosθ (1− a 2 r 2 ) and uθ = −U sin
€ θ (1+ a 2 r 2 ) , which is ideal 2D flow past a round cylinder.
€
Thus the streamlines€ appear as in Figure 7.9.
€ €
€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.20. Graphically generate the streamline pattern for a plane half-body in the following
manner. Take a source of strength qs = 200 m2/s and a uniform stream U = 10 m/s. Draw radial
streamlines from the source at equal intervals of Δθ = π/10, with the corresponding stream
function interval : Δψsource = (qs/2π)Δθ = 10 m2/s. Now draw streamlines of the uniform flow
with the same interval, that is, Δψstream = UΔψ = 10 m2/s. This requires Δy = 1 m, which can be
plotted assuming a linear scale of 1 cm = 1m. Connect points of equal ψ = ψ source + ψstream to
display the flow pattern.

Solution 7.20. Let subscripts "1" and "2" represent the source and the free stream, respectively.
q q q
ψ1 = s θ → Δψ1 = s Δθ = s = 10m 2 / s , and ψ 2 = Uy → Δψ 2 = UΔy = 10m 2 /s .
2π 2π 20
With the above intervals for the stream functions, draw the streamlines for the source and the
uniform stream by connecting values of constant total ψ = ψ source + ψstream. The drawing below
shows this construction where the heavier lines are
€ the streamlines.

!1 = 60 !1 = 50 !1 = 40 !1 = 30
!1 = 70 !1 = 20
!1 = 80
100
90
80
70
60 ! = 10
!1 = 90 1
40
30
!2 = 20
!2 = 10
!1 = 100 !2 = 0 !1 = 0

The radial lines from the point-source location are 18° apart. The point-source stream-function
values are listed around the edge of the drawing, while the free-stream function values are listed
at or near the downstream edge of the drawing.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.21. Consider the two-dimensional steady flow formed by combining a uniform
stream of speed U in the positive x-direction, a source of strength qs > 0 at (x,y) = (–a, 0), and a
sink of strength qs at (x,y) = (+a, 0) where a > 0. The pressure far upstream of the origin is p∞.
a) Write down the velocity potential and the
stream function for this flow field. y!
b) What are the coordinates of the stagnation U! y = h!
points, marked by s in the figure?
c) Determine the pressure in this flow field along +qs! –qs!
the y-axis. s ! s! x!
d) There is a closed streamline in this flow that
defines a Rankine body. Obtain a transcendental
algebraic equation for this streamline, and show
that the half-width, h, of the body in the y- a! a!
direction is given by: h/a = cot(πUh/qs). (The
introduction of angles may be useful here.)

Solution 7.21. a) For both ideal flow functions, there will three terms: the free stream, source at
x = –a, and the sink at x = +a.
q q
Velocity potential: φ (x, y) = Ux + s ln (x + a)2 + y 2 − s ln (x − a)2 + y 2
2π 2π
q q ?
q " y % qs " y %
Stream function: ψ = Uy + s θ1 − s θ 2 =Uy + s tan −1 $ '− tan −1 $ '
2π 2π 2π # x + a & 2π # x−a&
where θ1 and θ2 are angles determined from the x-axis having vertices that lie at x = –a and +a,
respectively. Later on the correct branches of the inverse tangent function will have to be
chosen.
b) First compute the velocities and look for places where both components are zero. Cartesian
velocity components are:
∂φ q ! x+a $ qs ! x−a $
u= =U + s # 2 2&
− # 2 2&
∂x 2π " (x + a) + y % 2π " (x − a) + y %
∂φ qs ! y $ qs ! y $
v= = # 2 2&
− # 2 2&
∂ y 2π " (x + a) + y % 2π " (x − a) + y %
By examination, v is zero on x = 0 (the y-axis) and on y = 0 (the x-axis). However, u is non-zero
on x = 0 (the y-axis). Thus, the stagnation points must occur on y = 0 (the x-axis), when
q ! x+a $ qs ! x−a $ q ! 1 $ qs ! 1 $
u(x, 0) = 0 = U + s # 2 2&
− # 2 2&
=U + s # &− # &
2π " (x + a) + 0 % 2π " (x − a) + 0 % 2π " x + a % 2π " x − a %
Solve for x using the final equality.
q " x−a % q " x+a % q a" 1 % qa
0 = U + s $ 2 2 ' − s $ 2 2 ' = U − s $ 2 2 ' , or x = ± a 2 + s .
2π # x − a & 2π # x − a & π #x −a & πU
! qa $
Thus, the two stagnation points are located at: (x, y) = # ± a 2 + s , 0 & .
" πU %
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

! q ! a $ $
c) Using the information above, (u, v) x=0 = #U + s # 2 2 &, 0 & on the y-axis. Thus, use the
" π "a + y % %
1
( 2
) 2 1
steady ideal flow Bernoulli equation, p(x, y) + 2 ρ u + v = p∞ + 2 ρU , with the above
2

replacements for u and v.

2
! qs ! a $ $
p(x, y) + 2 ρ #U + # 2 2 && = p∞ + 12 ρU 2
1

€ " π " a + y %%
Simplify this expression to find:
2 2
1 2 1
# qs # a && qsU # a & ρ qs2 # a &
p(0, y) − p∞ = + 2 ρU − 2 ρ %U + % 2 2 (( = −ρ % (− % ( .
$ π $ a + y '' π $ a 2 + y 2 ' 2π 2 $ a 2 + y 2 '
d) The streamline that lies on the x-axis divides to become the closed streamline. Upstream of
the source and sink, all the angles are π. Thus the stream-function constant, can be evaluated
q q q q
ψ = Ur sin θ + s θ 2 − s θ1 = Ur sin π + s π − s π = 0
2π 2π 2π 2π
So the equations for the shape of the Rankine body in polar and Cartesian coordinates are:
q q −1 " y % qs " y %
0 = Ur sin θ + s (θ1 − θ 2 ) , or 0 = Uy + s tan $ '− tan −1 $ ',
2π 2π # x + a & 2π # x−a&
where the branches of the inverse tangent function must be chosen correctly. To find an
equation for the half width, h, of the body, a combination of the above formulae is needed. From
the symmetry of the flow, the widest part of the Rankine body will occur at x = 0, so the body
contour will cross the y-axis at (0,±h). Consider the point (0,h). The angles associated with this
point are: θ = π 2 , θ1 = tan−1 ( h a) , and θ 2 = π − tan−1 ( h a) , where the usual range, –π/2 to + π/2,
of the inverse tangent has been chosen. Thus, the polar-coordinate body-shape equation
produces:
€ € qs " −1 " h %
% πUh π "h%
0 = Uh + € $ 2 tan $ ' − π ' , or = − tan −1 $ ' , or
2π # #a& & qs 2 #a&
Now take the cotangent of both sides of the final equation and use the trigonometric identity:
$π ' ! πUh $ !π −1 ! h $
$ ! −1 ! h $$ h
cot& − z) = tan(z) to find: cot # & = cot # − tan # && = tan # tan # && = , and the two
%2 ( " qs % "2 " a %% " " a %% a
ends of this extended equality are the desired result.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.22. A stationary ideal two-dimensional vortex with clockwise circulation Γ is located
at point (0, a), above a flat plate. The plate coincides with the x-axis. A uniform stream U
directed along the x-axis flows past the vortex.
a) Sketch the flow pattern and show that it represents the flow over an oval-shaped body when
Γ/πa > U. [Hint: Introduce the image vortex and locate the two stagnation points on the x-axis.]
b) If the pressure far from the origin just above the x-axis is p∞ show that the pressure p at any
ρΓ 2 a 2 ρUΓa
location on the plate is: p∞ − p = − .
2π (x + a ) π (x 2 + a 2 )
2 2 2 2

c) Using the result of part b), show that the total upward force F on the plate per unit depth into
the page is F = –ρUΓ + ρΓ2/4πa when the pressure everywhere below the plate is p∞.

Solution 7.22. a) The stream function for this flow field is:
ψ = Uy + (Γ 2π ) ln[( ) (
x 2 + (y − a) 2 − ln )]
x 2 + (y + a) 2 ,
where the image vortex is represented by the second natural log function. For y = 0 and Γ/πa > U
the flow field looks like:
y
€

b) The horizontal velocity on y = 0 is:

$∂ψ ' $ Γ y−a Γ y+a ' Γ a
u(x,0) = & ) = &U + 2 2
− 2 2)
=U − .
% ∂y (y= 0 % 2π x + (y − a) 2π x + (y + a) (y= 0 π x 2 + a2
The Bernoulli equation evaluated on y = 0 determines the pressure p on the plate:
2 ' *
1 2 1 ' Γ a * 1 ) Γ a Γ2 a2 ,.
p∞ + ρU = p + ρ)U − , → p − p∞ = ρ 2U −
€ 2 2 ( π x 2 + a2 + 2 ) π x 2 + a2 π 2 ( x 2 + a2 )2 ,
( +
This pressure difference can be mildly simplified to:
ρUΓa ρΓ 2 a 2
p − p∞ = − .
π (x 2 + a 2 ) 2π 2 (x 2 + a 2 ) 2
€
c) The net vertical force F (positive upward) on the plate (per unit depth into the page) is:
+∞( +
+∞
ρUΓa ρΓ 2 a 2
F = ∫ ( p∞ − p)dx = ∫ *− 2 2
+ 2 2 2 2-
dx .
€ −∞ −∞ ) π (x + a ) 2π (x + a ) ,
The integrals can be evaluated with the change of variables x = atanβ, with dx = –asec2β dβ.
ρUΓa 2 + π 2 sec 2 βdβ ρΓ 2 a 3 + π 2 sec 2 βdβ ρUΓ + π 2 ρΓ 2 + π 2 2
F =− 2 ∫ 2
+ 2 4 ∫ 2 2
=− ∫ dβ + 2π 2 a ∫ cos βdβ
€ πa − π 2 1+ tan β 2π a − π 2 (1+ tan β ) π −π 2 −π 2

ρUΓ ρΓ 2 ( π + ρΓ 2
= (π ) + 2 * - = − ρUΓ.
π 2π a ) 2 , 4 πa

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.23. Consider plane flow around a circular cylinder. Use the complex potential and
Blasius theorem (7.60) to show that the drag is zero and the lift is L = ρUΓ. (In Section 7.3, these
results were obtained by integrating the surface pressure distribution.)

Solution 7.23. The complex potential and its derivative for ideal flow past a circular cylinder
with circulation –Γ are:
" a 2 % iΓ dw # a 2 & iΓ
w = U$ z + ' + log z , and = U%1− 2 ( + .
# z & 2π dz $ z ' 2πz
Use these in Blasius' theorem (7.60):
1 $ dw ' 2 1 $ $ a 2 ' iΓ ' 2
D − iL = iρ ∫ & ) dz = iρ ∫ &U&1− 2 ) + ) dz
€ 2 % dz ( 2 € % % z ( 2πz (
$ $ a 2 ' 2 iΓ $ a 2 ' 2 '2
1
) dz = 1 iρ ∫ iΓ Udz
2 Γ
= iρ ∫ &&U &1− 2 ) + U&1− 2 ) − 2 2)
2 % % z ( πz % z ( 4 π z ( 2 πz
1 iΓ
= iρ U(2πi) = −iρUΓ.
2 π
where the last two equalities follow from the residue theorem. Thus, equating real and imaginary
parts produces:
D = 0, and L = ρUΓ.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.24. For the doublet flow described by (7.29) and sketched in Figure 7.6, show u < 0
for y < x and u > 0 for y > x. Also, show that v < 0 in the first quadrant and v > 0 in the second
quadrant.

Solution 7.24. Start from (7.30)

d cos θ d x
φ= = .
2π r 2π x + y 2
2

The velocity field is obtained by direct differentiation.

∂φ d ∂ % x ( d % 1 x ( d % y2 − x2 (
u= = ' 2 * = ' − (2x) *= ' 2 2 2*
, and
∂x 2π ∂x &€ x + y 2 ) 2π & x 2 + y 2 (x 2 + y 2 ) 2 ) 2π & (x + y ) )
∂φ d ∂ % x ( d % x ( d % −2xy (
v= = ' 2 2*
= '− 2 2 2
(2y)* = ' 2 2 2*
,
∂y 2π ∂y & x + y ) 2π & (x + y ) ) 2π & (x + y ) )
€Given the numerator of the final expression for u, clearly u < 0 for y < x and u > 0 for y > x.
Similarly, given the numerator of the final expression for v, it will be negative when x and y have
the same sign (first and third quadrants) and it will be positive when x and y have opposite signs
€ and fourth quadrants).
(second
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.25. Hurricane winds blow over a Quonset hut, that is, a long half-circular cylindrical
cross-section building, 6 m in diameter. If the velocity far upstream is U∞ = 40 m/s and p∞ =
1.003 × 105 N/m, ρ∞ = 1.23 kg/m3, find the force per unit depth on the building, assuming the
pressure inside the hut is a) p∞, and b) stagnation pressure, p∞ + 12 ρ∞U∞2 .

Solution 7.25. If the flow over the hut is ideal, it can be modeled using the potential for flow past
a cylinder without circulation, and compute the velocity components:
# a2 & ∂φ % a2 ( 1 ∂φ & a2 )
φ = U% r + ( cos θ , and ur = = U'1− 2 * cos θ & uθ = = −U(1+ 2 + sin θ .
$ r' ∂r & r ) r ∂θ ' r *
The ideal flow pressure ps on the surface of the hut (r = a) can be determined from the steady
Bernoulli equation:
1 1 1 1
€ 2 2 2
( )
p∞ + ρU 2 = ps +€ ρ ( ur2 + uθ2 ) r= a → ps − p∞ =€ ρ U 2 − [ uθ2 ] r= a = ρU 2 (1− 4 sin 2 θ ) ,
2
where [ uθ ] r= a = −2U sin θ .
a) For ideal flow there is no drag so Fx = 0. When the pressure inside the hut is p∞, the vertical
force Fy (per unit depth) on the hut is
€ π π π
1 1
€ Fy = ∫ ( p∞ − ps )e y ⋅ nadθ = − ρU 2 a ∫ (1− 4 sin 2 θ ) sin θdθ = − ρU 2 a ∫ (−3 + 4 cos2 θ ) sin θdθ
0 2 0 2 0

1 ) 4 ,π 1 ) 8, 5
= − ρU 2 a+3cos θ − cos3 θ. = − ρU 2 a+−6 + . = ρU 2 a.
2 * 3 -0 2 * 3- 3
3
For ρ = 1.23 kg/m , U = 40 m/s, and a = 3m, then Fy = 9.84 kN/m.
b) Again, for ideal flow there is no drag so Fx = 0. When the pressure inside the hut is
p∞ + 12 ρ∞U∞2 , the vertical force Fy (per unit depth) on the hut is
€
π π π
1
Fy = ∫ ( p∞ + 12 ρU 2 − ps )e y ⋅ nadθ = ρU 2 a ∫ ( 4 sin 2 θ ) sin θdθ = 2 ρU 2 a ∫ (1− cos 2 θ ) sin θdθ
0 2 0 0
π
€ ) 1 3 , ) 2, 8
= 2 ρU a+−cosθ + cos θ. = 2 ρU 2 a+2 − . = ρU 2 a.
2

* 3 -0 * 3- 3
3
For ρ = 1.23 kg/m , U = 40 m/s, and a = 3m, then Fy = 15.7 kN/m.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.26. In a two-dimensional ideal flow, a source of strength qs is located a meters above
an infinite plane. Find the fluid velocity on the plane, the pressure on the plane, and the reaction
force on the plane assuming constant pressure p∞ below the plane.

Solution 7.26. The potential function for this flow field is:
φ = ( qs 2π )"$ln x 2 + (y − a)2 + ln x 2 + (y + a)2 %' ,
( ) ( )
# &
The velocity components are:
∂φ qs " x x %
u(x, y) = = $ 2 + ' , and
∂ x 2π # x + (y − a)2 x 2 + (y + a)2 &
∂φ qs " y−a y+a %
v(x, y) = = $ 2 + '.
∂ y 2π # x + (y − a)2 x 2 + (y + a)2 &
On y = 0 these become:
q ! x $
u(x, 0) = s # 2 2 & , and v(x,0) = 0 .
π "x +a %
The pressure on the plane pp can be determined from the steady flow Bernoulli equation:
2
1 "q x % 1 qs2 x2
p∞ = p p + ρ $ s 2 2 ' → € ∞p − p p = ρ .
2 #π x +a & 2 π 2 (x 2 + a 2 )2
The net vertical force F (positive upward) on the plate (per unit depth into the page) will be:
+∞
1 qs2 +∞ x2 1 qs2 +∞ $ 1 a2 '
F = ∫ ( p∞ − p p )dx = ρ 2 ∫ 2 2 2 dx = ρ 2 ∫ & 2 2 − 2 2 2 ) dx ,
−∞ 2 π −∞ (x + a ) 2 π −∞ % x + a (x + a ) (
where the final equality follows from a partial fractions decomposition of the integrand. The
integrals can be evaluated with the change of variables x = atanβ, with dx = –asec2β dβ.
+π 2
1 q2 " sec 2 β sec 2 β % 1 qs2 " π % 1 qs2
F = ρ 2s ∫ $ − ' d β = ρ $ π − '= ρ .
2 π a −π 2 # 1+ tan 2 β (1+ tan 2 β )2 & 2 π 2a # 2 & 4 πa
Interestingly, the force on the plane is positive, independent of the sign of qs.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.27. Consider a two-dimensional ideal

flow over a circular cylinder of radius r = a with
axis coincident with a right angle corner, as shown
in the figure. Assuming that ψ = Axy (with A =
constant) when the cylinder is absent, solve for the
stream function and velocity components.

Solution 7.27. When the cylinder is present, it will

only effect the flow near the origin, so the stream
function far from the origin will be ψo = Axy. And,
the field equation for the stream function, ∇ 2ψ = 0 , is linear, so a solution in the form of a simple
superposition can be sought:
ψ (x, y) = ψ o + ψ1 = Axy + ψ1 = Ar 2 sin θ cos θ + ψ1 (r,θ ) = ( A 2) r 2 sin(2θ ) + R(r)Θ(θ ) ,
where ψ1 is the modification of the stream function necessary near the origin, x = rcosθ, y =
€
rsinθ, and the further assumption that ψ1 can be found via separation of variables in polar
€ coordinates has been made. Placing this two term trial solution 2
into the field equation produces:
1 ∂ $ ∂ψ1 ' 1 ∂ ψ1
0= &r )+ , (&)
r ∂r % ∂r ( r 2 ∂θ 2
and the boundary condition on ψ1 is determined from:
%1 ∂ψ ( %1 ∂ψ o 1 ∂ψ1 (
'& r ∂θ *) = '& r ∂θ + r ∂θ *) = ur (a,θ ) = 0 .
r= a r= a
€
For ψ o = ( A 2) r 2 sin(2θ ) , this implies:
1 %∂ψ1 ( %∂ψ1 ( 2 Aa 2
Aacos(2θ ) + ' * = 0 , or '& ∂θ *) = −Aa cos(2θ ) → ψ1 (a,θ ) = − 2 sin(2θ ) .
€ a & ∂θ )r= a r= a
€ Thus, the form Θ(θ) = sin(2θ) has been determined so that ψ1 = R(r)sin(2θ ) , and the boundary
condition implies R(a) = − Aa 2 2 . Placing the trial solution for ψ1 into (&) produces:
) " % ,
€ € 0 = +1 d $ r dR ' − 4 R. sin(2θ ) ,
2
* r dr # dr & € r -
where the€ partial derivatives have been changed total derivatives because R(r) only depends on
the variable r. This equation is equi-dimensional and therefore has power law solutions of the
form R = Brm. Substituting in this solution form leads to the algebraic equation:
€ m 2 r m−2 − 4r m−2= 0 ,
which implies m = ±2. The positive root reproduces ψo, so the negative root should be chosen
here. The boundary condition allows B to be evaluated:
R(a) = [€ B r 2 ] r= a = B a 2 = − Aa 2 2 → B = − Aa 4 2 .
Therefore:
Aa 2 $ r 2 a 2 '
ψ (x, y) = ψ o + ψ1 = & − ) sin(2θ ) , and
2 % a2 r 2 (
€
1 ∂ψ & r a2 ) ∂ψ & r a2 )
ur = = Aa 2 ( 2 − 3 + cos(2θ ) and uθ = − = Aa 2 ( 2 + 3 + sin(2θ ) .
r ∂θ 'a r * ∂r 'a r *
€
To recover the flow directions shown in the drawing for this exercise, the constant A must be
negative.

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.28. Consider the following two-dimensional velocity potential consisting of two
sources and one sink, all of strength qs:
<Begin Equation>
( )
φ (x, y) = ( qs 2π ) ln (x − b)2 + y 2 + ln (x − a 2 b)2 + y 2 − ln x 2 + y 2 .
</End Equation>
Here a and b are positive constants and b > a.
a) Determine the locations of the two stagnation points in this flow field.
b) Sketch the streamlines in this flow field.
c) Show that the closed streamline in this flow is given by x2 + y2 = a2.

Solution 7.28. a) Here φ only depends on y2 when x = 0. Thus, v = ∂φ ∂y = 0 on y = 0, the x-

axis, so finding the stagnation points means, finding the x-axis locations where u = ∂φ ∂x = 0 ;
!∂φ $ qs ! x−b x − a2 b x $ qs ( 1 1 1+
#" &% = # + − & = * + − -=0
∂ x y=0 2π " (x − b)2 + y 2 (x − a 2 b)2 + y 2 x 2 + y€
2 2
%y=0 2π ) x − b x − a b x ,
€
Working with the final equality produces: x(x − a 2 b) + x(x − b) − (x − b)(x − a 2 b) = 0 , which
simplifies to: x 2 − a 2 = 0 . Thus, the y
two stagnation points are at (±a,0).
b) The two sources and the sink all
lie on the x-axis with the sink at the€
€
origin. The stagnation points lie
between the two sources at x = +a
and to left of the origin at x = –a.
A closed body is obtained, and it is x
shown in part c) that it is circular.
To allow labeling, the streamlines +a2/b
are as shown for y ≥ 0 only. They +a +b
are symmetric about y = 0. –a
c) First convert everything to the
usual polar coordinates, and define:
r1 ≡ (x − b) 2 + y 2 = (r cos θ − b) 2 + r 2 sin 2 θ = r 2 − 2br cos θ + b 2 , and
r2 ≡ (x − a 2 b) 2 + y 2 = (r cos θ − a 2 b) 2 + r 2 sin 2 θ = r 2 − 2(a 2 b)r cosθ + (a 2 b) 2 .
Thus, the potential can be written:
€ φ = ( qs 2π ) ( ln r1 + ln r2 − ln r ) ,
€ so the radial velocity is:
∂φ qs " r − 2br cosθ r − 2(a 2 b)r cosθ 1 %
ur = = $ + − ',
∂ r 2π # r12 r22 r&
∂ ∂ ln(ri ) ∂ri
where the chair rule ln(ri ) = to reach the form shown. To find the location where ur
∂r ∂ri ∂r
= 0, the above equation implies:
rr22 ( r − bcosθ ) + rr12 ( r − (a 2 b)cos θ ) − r12 r22 = 0 ,
or in expanded
€ form:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

(r 2
− 2(a 2 b)r cos θ + (a 2 b) 2 )( r 2 − br cos θ ) + ( r 2 − 2br cos θ + b 2 )( r 2 − (a 2 b)r cosθ )
− ( r 2 − 2br cosθ + b 2 )( r 2 − 2(a 2 b)r cosθ + (a 2 b) 2 ) = 0.
Patience with the multiplications produces at total of 21 terms:
r 4 − br 3 cosθ − 2(a 2 b)r 3 cos θ + 2a 2 r 2 cos 2 θ + (a 4 b 2 )r 2 − (a 4 b)r cosθ
€ r 4 − (a 2 b)r 3 cos θ − 2br 3 cos θ + 2a 2 r 2 cos 2 θ + b 2 r 2 − a 2br cosθ
−r 4 + 2br 3 cosθ − b 2 r 2 + 2(a 2 b)r 3 cos θ − 4a 2 r 2 cos 2 θ + 2a 2br cos θ
−(a 4 b 2 )r 2 + 2(a 4 b)r cosθ − a 4 = 0.
Group like terms according to the power of the cosine starting with cos0θ:
[r 4 + (a4 b2 )r 2 + r 4 + b2r 2 − r 4 − b2r 2 − (a4 b2 )r 2 − a4 ] +
[−br€ − 2(a b)r − (a b)r − (a b)r
3 2 3 4 2 3
− 2br 3 − a 2br + 2br 3 + 2(a 2 b)r 3 + 2a 2br + 2(a 4 b)r] cosθ +

[2a r + 2a r − 4a r ] cos θ = 0
2 2 2 2 2 2 2

Cancel the equal and opposite terms:

[r 4 − a4 ] + [−br 3 − (a2 b)r 3 + a2br + (a4 b)r] cosθ = 0 .
€ Rearrange and factor:
[
r 4 − a 4 + −(b + (a 2 b)) r 3 + a 2 (b + (a 2 b)) r cosθ = 0 . ]
€ r − a + (b + (a b))[−r + a ] r cos θ = 0 .
4 4 2 2 2

(r 2
− a 2 )( r 2 + a 2 ) − (b + (a 2 b))[ r 2 − a 2 ] r cos θ = 0 .
€
(r 2
[ ]
− a 2 ) r 2 + a 2 − (b + (a 2 b)) r cos θ = 0 .
The final equality €
shows that ur = 0 at r = a.
€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.29. Without using complex variables, derive the results of the Kutta–Zhukhovsky lift
theorem (7.62) for steady two-dimensional
irrotational constant-density flow past an
arbitrary-cross-section object by considering the
clam-shell control volume (shown as a dashed
line) in the limit as r → ∞ . Here A1 is a large
circular contour, A2 follows the object’s cross
section contour, and A3 connects A1 and A2. Let
p∞ and Uex€be the pressure and flow velocity far
from the origin of coordinates, and denote the
flow extent perpendicular to the x-y plane by B.

Solution 7.29. In ideal flow, the only surface forces are pressure forces. Therefore, the drag (D)
and lift (L) forces on the body are defined by:
De x + Le y = − ∫ ( p − p∞ )n2 dA = ∫ ( p − p∞ )ndA , (a)
A2 A2
where the free stream velocity is U = Uex, and n is the outward normal on the total control
volume; thus it points into the body on surface A2 so the usual minus sign is missing from the
final equality in (a) because n = –n2. Note that the constant p∞ can be included in the pressure
€
integration because ∫ p∞ndA = 0 .
closed surface
The starting point for the derivation is the integral form of the steady ideal flow
momentum equation applied to the entire clam-shell control volume:
€ ∫ ρu(u ⋅ n)dA = − ∫ ( p − p∞ )ndA
A +A +A 1 A +A +A
2 3 1 2 3

with the implied limit that the separation between the surfaces of A3 goes to zero. When this limit
is taken, the net contribution is zero because the integrands of the two surfaces of A3 will be
identical while the normal vectors will point in opposite directions; thus the integrations on the
€ of A3 cancel out. This leaves:
upper and lower surfaces
∫ ρu(u ⋅ n)dA + ∫ ρu(u ⋅ n)dA = ∫ ρu(u ⋅ n)dA + 0 = − ∫ ( p − p∞ )ndA ,
A1 A2 A1 A1 +A 2
where the first equality follows because u ⋅ n = 0 on the solid surface A2. Now substitute in (a)
for the pressure integration over A2, and rearrange:
∫ ρu(u ⋅ n)dA = − ∫ ( p − p∞ )ndA − ( De x + Le y ) or De x + Le y = − ∫ ( p − p∞ )ndA − ∫ ρu(u ⋅ n)dA .
A1 € A1 A1 A1
€
Thus, the drag and lift on the body can be obtained from integrals over a circular surface (A1) that
is distant from the body. For this circular surface, the area element will be dA = Brdθ, and n = er
= excosθ + eysinθ, so the last equation can be written in component form:
€ 2π € 2π
D L
= − ∫ ( pcos θ + ρuur )rdθ and = − ∫ ( psin θ + ρvur )rdθ .
B 0 B 0
where u = (u, v) = (ur, u ) in Cartesian and polar coordinates, respectively. When r is large the
θ

potential for the flow field can be expanded in inverse powers of r:

q Γ d cosθ
€ φ = Ur cosθ + €s ln(r) − θ + +...
2π 2π 2π r
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

where qs is the net source strength, Γ is the total clockwise circulation, d is the total dipole
strength, etc. for the flow around the body. For a closed body, qs must be zero. Thus, the
velocities on A1 at large r are
∂φ d cos θ ∂φ Γ sin θ ∂φ Γ cos θ
ur = = U cosθ − 2
+ ... , u = =U + + ... , and v = =− + ... .
∂r 2πr ∂x 2π r ∂y 2π r
Using these results, the Bernoulli equation allows the pressure p to be determined at large r:
1 1 1
p + ρ( u 2 + v 2 ) = p∞ + ρU 2 → p − p∞ = ρ(U 2 − u 2 − v 2 ) , or
€ 2 € 2 2
( 2 2 € +
1 ΓU sin θ Γ sin θ Γ 2 cos2 θ
p − p∞ = ρ*U 2 − U 2 − − + ...− + ...-
2 ) π r 4π 2 r 2 4π 2 r 2 ,
€ 1 ( ΓU sin θ Γ 2 +
= ρ* − − 2 2
+ ...-.
2 ) π r 4π r ,
Thus, the drag integral becomes:
2π )
D 1 ) ΓU sin θ Γ2 , ) Γ sin θ ,) d cos θ ,,
= −lim ∫ + ρ+ − − + .... cos θ + ρ +U + + ....+ U cos θ − + .....rdθ .
B €r →∞ 0 * 2 * π r 4π 2r 2 - * 2π r -* 2πr 2 --
where the limit r → ∞ is now explicitly written. Collect terms under the integrand according to
their power of r:
2π ) ,
D 2 ρΓU ρΓU ρΓ 2 cos θ ρUd cosθ
€ r →∞ ∫ +
€ = −lim ρ U r cos θ − sin θ cos θ + sin θ cos θ − 2
− + ....dθ .
B 0* 2π 2π 8π r 2πr -
The first integrand term makes no contribution because the integral of the cosine over one period
is zero. The second and third integrand terms cancel. The fourth and fifth integrand terms are
proportional to 1/r and disappear when the limit is taken, and all remaining integrand terms are
€ even smaller; therefore D = 0.
Now consider the lift integral with the large-r forms for the pressure and velocity
substituted in:
2π )
L 1 ) ΓU sin θ Γ2 , ) Γ cosθ ,) d cos θ ,,
= −lim ∫ + ρ+ − − + .... sin θ + ρ + − .+ U cos θ − + .....rdθ .
B r →∞
0 *2 * π r 4π 2r 2 - * 2π r -* 2πr 2 --
where the limit r → ∞ is now explicitly written. As before, collect terms under the integrand
according to their power of r:
2π ) , 2π ) ,
L ρΓU 2 ρΓU Γ2 ρΓU Γ2
€ =€−lim ∫ + − sin θ − cos 2 θ − 2 sin θ + ....dθ = lim ∫ + + 2 sin θ + ....dθ .
B r →∞ *
0 2π 2π 8π r - r →∞ * 2π
0 8π r -
The first integrand term contributes ρΓU. The second term and the higher order terms are
proportional to 1/r or are even smaller, and disappear when the limit is taken. Thus, the final
results are:
€ D = 0 and L = ρUΓ,
which proves Blasius' theorem.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.30. Pressure fluctuations in wall-bounded turbulent flows are a common source of
flow noise. Such fluctuations are caused by turbulent eddies as they move over the bounding
surface. A simple ideal-flow model that captures some of the important phenomena involves a
two-dimensional vortex that moves above a flat surface in a fluid of density ρ. Thus, for the
following items, use the potential:
Γ & y−h ) Γ & y+h )
φ (x, y,t) = − tan−1( ++ tan−1( +
2π ' x − Ut * 2π ' x − Ut *
where h is the distance of the vortex above the flat surface, Γ is the vortex strength, and U is the
convection speed of the vortex.
a) Compute the horizontal u and vertical v velocity components and verify that v = 0 on y = 0.
€
b) Determine the pressure at x = y = 0 in terms of ρ, t, Γ, h, and U.
c) Based on your results from part b), is it possible for a fast-moving high-strength vortex far
from the surface to have the same pressure signature as a slow-moving low-strength vortex
closer to the surface?

Γ & y−h ) Γ & y+h )

Solution 7.30. a) Start with the given potential, φ (x, y,t) = − tan−1( ++ tan−1( +,
2π ' x − Ut * 2π ' x − Ut *
and differentiate to find:
−1 −1
∂φ Γ ' (y − h) 2 * ' y−h * Γ ' (y + h) 2 * ' y+h *
u= = − )1+ , ) − , + ) 1+ , ) − 2,
∂x 2π ( (x − Ut) 2 + ( (x − Ut) 2
€ + 2π ( (x − Ut) + ( (x − Ut) +
2

Γ % y−h y+h (
= & 2 2
− 2 2
)
2π ' (x − Ut) + (y − h) (x − Ut) + (y + h) *
−1 −1
€ ∂φ Γ ' (y − h) 2 * ' 1 * Γ ' (y + h) 2 * ' 1 *
v= = − )1+ , ) ,+ )1+ , ) ,
∂y 2π ( (x − Ut) 2 + ( x − Ut + 2π ( (x − Ut) 2 + ( x − Ut +
€ Γ % x − Ut x − Ut (
=− & 2 2
− 2 2
)
2π ' (x − Ut) + (y − h) (x − Ut) + (y + h) *
€ At y = 0, the two factors inside the {,}-braces are equal and opposite so v(x, y=0) = 0.
b) The pressure at x = y = 0 can be determined from the Bernoulli equation:
% ∂φ 1 2 2 ( ' ∂φ 1 *
€ ' ρ + ρ (u + v ) + p* = p∞ or p(0,0,t) − p∞ = −) ρ + ρ (u 2 + v 2 ),
& ∂t 2 ) x= y= 0 ( ∂t 2 + x= y= 0
First determine ∂φ ∂t ,
−1 −1
∂φ Γ ' (y − h) 2 * ' y−h * Γ ' (y + h) 2 * ' y+h *
= − )1+ , ) − , ( −U ) + )1+ , ) − ,(−U )
€ ∂t 2π ( (x − Ut) 2 + ( (x − Ut)€ 2
+ 2π ( (x − Ut) 2 + ( (x − Ut) 2 +
€
−UΓ % y−h y+h (
= & − ),
2π ' (x − Ut) 2 + (y − h) 2 (x − Ut) 2 + (y + h) 2 *
€ $ ∂φ ' UhΓ π
and then evaluate it at x = y = 0: & ) = . Now evaluate u and v at x = y = 0:
% ∂t ( x= y= 0 (Ut) 2 + h 2
€ −Γh π
u= 2 2 & v = 0, and put all of this into the Bernoulli relationship:
U t + h2
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

p(0,0,t) − p∞ UhΓ π Γ 2 h 2 2π 2
=− 2 2 − . (1)
ρ U t + h 2 (U 2 t 2 + h 2 ) 2
c) The question here is whether or not U can be replaced with C1U, h can be replaced with C2h
and Γ replaced with C3Γ with the three constants (C1, C2, C3) chosen so that p(0,0,t) remains
unchanged. In other words can (C1, C2, C3) be found so that (2) is identical to (1).
€
p(0,0,t) − p∞ ? C1C2C3UhΓ π C22C32Γ 2 h 2 2π 2
=− 2 2 2 − (2)
ρ C2U t + C12 h 2 (C22U 2 t 2 + C12 h 2 ) 2
For there to be any hope of making (2) a genuine equality, the denominator factors imply that C1
= C2. With this simplification (2) becomes:
p(0,0,t) − p∞ ? C3UhΓ π C32Γ 2 h 2 2π 2
€ =− − (3)
ρ U 2 t 2 + h 2 C12 U 2 t 2 + h 2 2
( )
So, to recover (1), (3) implies: C3 = 1, and C1 = C3; therefore the surface pressure signature of an
ideal moving line vortex is unique.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.31. A pair of equal strength ideal line vortices having axes perpendicular to the x-y
plane are located at x a (t) = ( x a (t), y a (t)) , and x b (t) = ( x b (t), y b (t)) , and move in their mutually-
induced velocity fields. The stream function for this flow is given by:
Γ
ψ (x, y,t) = − (ln x − x a (t) + ln x − x b (t) ) . Explicitly determine x a (t) and x b (t) given that
2π
x a (0) = (−r€o ,0) and x b (0) = (ro,0) . Switching
€ to polar coordinates at some point in your solution
may be useful.
€ Solution 7.31. This problem can quickly get very tedious without clear organization. For
€ notational €
2 2 2 2
simplicity, let (a) = ( x − x a (t)) + ( y − y a (t)) , and (b) = ( x − x b (t)) + ( y − y b (t)) .
Differentiate the stream function to find:
∂ψ Γ ' y − ya y − yb * ∂ψ Γ ' x − xa x − xb *
u(x, y,t) = =− ( + + , and v(x, y,t) = − =+ ( + +.
∂
€ y 2 π ) (a) (b) , € ∂ x 2 π ) (a) (b) ,
This velocity field leads to 4 nonlinear first-order differential equations for point vortex
coordinate positions.
dx a dx Γ % yb − ya ( dy a dy b Γ % xa − xb (
€ =− b = & € ) , and = − = & ).
dt dt 2π ' (x b − x a ) 2 + (y b − y a ) 2 * dt dt 2π ' (x b − x a ) 2 + (y b − y a ) 2 *
The first two equalities yield: x a (t) + x b (t) = constant, and y a (t) + y b (t) = constant.
With the initial conditions, the constants can be evaluated: x a (t) + x b (t) = 0, and y a (t) + y b (t) =
0. Use these two equations to eliminate xb and yb. The remaining two equations for xa and ya:
€ €(
dx a €Γ % −2y a € dy a Γ % 2x a (
= & ) , and = & ).
dt 2π ' (−2x a ) 2 + (−2y a ) 2 * € dt 2π ' (−2x a ) 2€+ (−2y a ) 2 *
Taking a ratio of these implies:
dy a x
= − a , integrate → x a2 + y a2 = const., and initial conditions → x a2 + y a2 = ro2 .
dx a ya
€ €
Hence the vortices move in a circle with radius ro. Based on this, let xa(t) = rocosθ(t) and
evaluate the differential equation for xa(t) in terms of θ(t):
€ dθ € Γ
€ dxa = −ro sin θ dθ = − Γ 2 ro2 − ro2 cos2 θ (t) = − Γsin θ (t) →
Γt
= 2
or θ (t) = + θo .
dt dt 4π ro 4π ro dt 4 πro 4 πro2
The initial conditions set θo = π. With this relationship, reconstruction of the whole solution is
possible:
( ) ( )
x a (t) = −ro cos Γt 4 πro2 , y a (t) =€−ro sin Γt 4 πro2€, and
x b (t) = ro cos(Γt 4πr ) , y b (t) = ro sin(Γt 4πro2 ) .
o
2

Hence, the pair of vortices circles the origin with a constant angular velocity. This type of free
vortex motion€ is seen any time vortices
€ of the same sign interact (for example in the trailing
vortices left by the multiple wing-tips on biplanes, or the multiple tip feathers of birds).
€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.32. Two unequal strength ideal line vortices having axes perpendicular to the x-y
plane are located at x1 (t) = ( x1 (t), y1 (t)) with circulation Γ1, and x 2 (t) = ( x2 (t), y2 (t)) with
circulation Γ2, and move in their mutually induced velocity fields. The stream function for this
flow is given by: ψ (x, y, t) = − ( Γ1 2π ) ln x − x1 (t) − ( Γ 2 2π ) ln x − x 2 (t) . Explicitly determine
x1 (t) and x 2 (t) in terms of Γ1, Γ2, h1, h2, and h, given x1 (0) = (h1, 0) , x 2 (0) = (h2 , 0) , and h2 – h1
= h > 0 [Hint: choose a convenient origin of coordinates, and switch to polar coordinates after
finding the shape of the trajectories.]

Solution 7.32. This problem can quickly get very tedious without clear organization. For
2 2 2 2
notational simplicity, let (1) = ( x − x1 (t)) + ( y − y1 (t)) , and (2) = ( x − x2 (t)) + ( y − y2 (t)) .
Differentiate the stream function to find:
∂ψ 1 # (y − y1 ) (y − y2 ) & ∂ψ 1 # (x − x1 ) (x − x2 ) &
u= =− $Γ1 + Γ2 ' , and v = − =+ $Γ1 + Γ2 '.
∂y 2π % (1) (2) ( ∂x 2π % (1) (2) (
This velocity field leads to 4 nonlinear first-order differential equations for point vortex
coordinate positions.
dx1 Γ # y1 − y2 & dx2 Γ1 # y2 − y1 &
= u(x1, y1, t) = − 2 $ ' , = u(x 2 , y 2 , t) = − $ ',
dt 2π % (x2 − x1 )2 + (y2 − y1 )2 ( dt 2π % (x2 − x1 )2 + (y2 − y1 )2 (
dy1 Γ # x1 − x2 & dy2 Γ1 # x2 − x1 &
= v(x1, y1, t) = 2 $ ' , and = u(x 2 , y 2 , t) = $ '.
dt 2π % (x2 − x1 )2 + (y2 − y1 )2 ( dt 2π % (x2 − x1 )2 + (y2 − y1 )2 (
From these, it is clear that:
2π dx1 2π dx2 y2 − y1 2π dy1 2π dy2 x2 − x1
=− = 2 2
, and − = = ,
Γ 2 dt Γ1 dt (x2 − x1 ) + (y2 − y1 ) Γ 2 dt Γ1 dt (x2 − x1 )2 + (y2 − y1 )2
so:
d " x1 x2 % d"y y %
$ + ' = 0 , and $ 1 + 2 ' = 0 .
dt # Γ 2 Γ1 & dt # Γ 2 Γ1 &
Thus, the contents of the parentheses are constants. The initial conditions then imply:
x1 x2 h1 h2 y y
+ = + , and 1 + 2 = 0 . (&)
Γ 2 Γ1 Γ 2 Γ1 Γ 2 Γ1
Now choose the origin of coordinates so that h1/Γ2 + h2/Γ1 = 0, and use h2 – h1 = h to find:
−hΓ 2 hΓ1
h1 = and h2 = . (%)
Γ1 + Γ 2 Γ1 + Γ 2
With this repositioning, equations (&) become:
x1 x2 y y
+ = 0 , and 1 + 2 = 0 . ($)
Γ 2 Γ1 Γ 2 Γ1
Now return to the differential equations for x1 and y1, and compute the ratio:
1 dy1 Γ
− − 1 x1 − x1
Γ 2 dt dy x − x Γ2 x
=− 1 = 2 1 = = 1 .
1 dx1 dx1 y2 − y1 − Γ1 y − y y1
1 1
Γ 2 dt Γ2
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

where the third equality follows from ($). Integrating the second and final members of this
extended equality and using the initial conditions produces: x12 + y12 = h12 . Similar steps for the
second vortex lead to: x22 + y22 = h22 . Hence the vortices move in circles with radii h1 and h2. In
addition, the distance between them remains constant:
2 2 2
# Γ2 & # Γ2 & # Γ2 &
(x2 − x1 ) + (y2 − y1 ) = x2 %1+ ( + y2 %1+ ( = h2 %1+ ( = h 2 ,
2 2 2 2 2

$ Γ1 ' $ Γ1 ' $ Γ1 '

where the first equality comes from substituting for x1 and y1 from ($), and the final equality
comes from the second equation of (%). Based on this, let x1(t) = h1cosθ(t), y1(t) = h1sinθ(t), x2(t)
= h2cosθ(t), and y2(t) = h2sinθ(t). With these substitutions, the distance between the vortices
remains constant (= h):
(x2 − x1 )2 + (y2 − y1 )2 = (h2 − h1 )2 (cos2 θ + sin 2 θ ) = h 2 ,
so the only remaining task is to find θ(t).
Evaluate the differential equation for x1(t) in terms of θ(t):
dx1 dθ Γ # y1 − y2 & Γ 2 hsin θ
= −h1 sin θ =− 2 $ '= .
dt dt 2π % (x2 − x1 ) + (y2 − y1 ) ( 2π h 2
2 2

Use the second and final terms to reach:

dθ Γ Γ +Γ !Γ +Γ $
= − 2 = 2 2 1 , or θ (t) = # 2 2 1 & t ,
dt 2π hh1 2π h " 2π h %
where the initial conditions have been used to find that the constant of integration is zero. With
this relationship, reconstruction of the whole solution is possible:
hΓ 2 )# Γ + Γ & , hΓ 2 )# Γ + Γ & ,
x1 (t) = − cos+% 2 2 1 ( t . , y1 (t) = − sin +% 2 2 1 ( t . ,
Γ1 + Γ 2 *$ 2π h ' - Γ1 + Γ 2 *$ 2π h ' -
hΓ1 (" Γ + Γ % + hΓ1 (" Γ + Γ % +
x2 (t) = cos*$ 2 2 1 ' t - , and y2 (t) = sin *$ 2 2 1 ' t - .
Γ1 + Γ 2 )# 2π h & , Γ1 + Γ 2 )# 2π h & ,
Hence, the pair of vortices circles the origin with a constant angular velocity. When the two
vortices are of equal strength, this solution reverts to that for Exercise 7.31. In the limit as
Γ1 → −Γ 2 (i.e. the two vortices are of equal and opposite sign), x2 − x1 → const. = h , and the two
Γt
y-coordinates become equal: y1, y2 → − 2 , and this matches the discussion of interacting
2π h
vortices in Section 5.6.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.33. Consider the unsteady potential flow of two ideal sinks located at
x a (t) = ( x a (t),0) and x b (t) = ( x b (t),0) that are free to move along the x-axis in an ideal fluid that
is stationary far from the origin. Assume that each sink will move in the velocity field induced by
the other.
qs " 2 2 2 2%
€ €φ (x, y, t) = − #$ln ( x − xa (t)) + y + ln ( x − xb (t)) + y &' , with qs > 0.
2π
a) Determine x a (t) and x b (t) when x a (0) = (−L,0) and x b (0) = (+L,0)
b) If the pressure far from the origin is p∞ and the fluid density is ρ, determine the pressure p at x
= y = 0 as function of p∞, ρ, qs, and xa(t).
€ € €
Solution 7.33.€a) All the action takes place in the x-direction so ignore the vertical (v) component
of velocity.
! ∂φ $ qs !# x − xa (t) x − xb (t) $
&
u(x, y, t) = # & = − # +
"∂ x % 2π " ( x − xa (t))2 + y 2 ( x − xb (t))2 + y 2 &%

q " xa − xb % " %
' = − qs $ 1 ' = dxa , and
For x = x a (t) and y = 0: u(xa , 0, t) = − s $
2π $# ( xa − xb ) + y 2 '&
2
2π # xa − xb & dt
y=0

qs "$ xb − xa %
' qs " 1 % dxb
€ for x = x b (t) and y = 0: u(xb , 0, t) = − =− $ '= = −u(xa , 0, t) .
2π $# ( xb − xa )2 + y 2 '& 2π # xb − xa & dt
y=0

Thus, dx a dt + dx b dt = 0 , or x a + x b = const. = 0 where final equality comes from the initial

€ conditions. So, dxa dt = −qs ( 4π xa ) or xa = ± const. − qs t 2π . The xa-initial condition
2 2
€ requires the minus sign
x (t) = −xb (t) = − L − qst 2π .
€ and const. = L ; thus, a
1 2
b) Use the unsteady Bernoulli equation: ∂φ ∂t + 2 u + p ρ = p∞ ρ . From the symmetry of the
flow, x = y = 0 is a stagnation point (u = 0), so p = p∞ − ρ∂φ ∂t .
! ∂φ $ qs ( − ( x − xa (t)) ( dxa dt ) ( x − xb (t)) ( dxb dt ) + qs ! dxa dt dxb dt $
# & = − * 2
− 2
- = − # + &
" ∂ t %x=y=0 2π *) ( x − xa (t)) € + y2 ( x − xb (t)) + y 2 -,x=y=0 2π " xa xb %
€
! ∂φ $ q ! dx dt $ q 2 ! 1 $ ρq2
# & = − s # a & = s 2 # 2 & , so p = p∞ − 2 s 2
" ∂ t %x=y=0 π " xa % 4π " xa % 4π xa
Thus, the pressure at the origin becomes very negative as the two sinks approach each other.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.34. Consider the unsteady potential flow of an ideal source and sink located at
x1 (t) = ( x1 (t),0) and x 2 (t) = ( x 2 (t),0) that are free to move along the x-axis in an ideal fluid that
is stationary far from the origin. Assume that the source and sink will move in the velocity field
induced by the other.
qs " 2 2 %
€ € φ (x, y, t) = $ ln ( x − x1 (t)) + y 2 − ln ( x − x2 (t)) + y 2 ' , with qs > 0.
2π # &
a) Determine x1 (t) and x 2 (t) when x1 (0) = (−, 0 ) and x 2 (0) = (+, 0 ) .
b) If the pressure far from the origin is p∞ and the fluid density is ρ, determine the pressure p at x
= y = 0 as function of p∞, ρ, qs, and x1(t).
€ €
Solution 7.34. a) All the action takes place in the x-direction so ignore the vertical (v) component
of velocity.
! ∂φ $ q ! x − x1 (t) x − x2 (t) $
u(x, y, t) = # & = s ## − &
" ∂ x % 2π " ( x − x1 (t))2 + y 2 ( x − x2 (t))2 + y 2 &%

q " x1 − x2 % " %
' = − qs $ 1 ' = dx1 , and
For x = x1 (t) and y = 0: u(x1, 0, t) = s $ −
2π $# ( x1 − x2 ) + y '&
2 2
2π # x1 − x2 & dt
y=0

qs "$ x2 − x1 %
' qs " 1 % dx2
€ for x = x 2 (t) and y = 0: u(x2 , 0, t) = = $ '= = u(x1, 0, t) .
2π $# ( x2 − x1 )2 + y 2 '& 2π # x2 − x1 & dt
y=0

Thus, dx1 dt = dx 2 dt , or x1 − x2 = const. = −2 where final equality comes from the initial
€ conditions. So, dx1 dt = qs ( 4π ) or x1 = ( qs 4π ) t + const. The x1-initial condition requires
const. = –  ; thus, x1 (t) = ( qs 4π ) t −  , and x2 (t) = ( qs 4π ) t +  .
€ 1 2
b) Use the unsteady Bernoulli equation: ∂φ ∂t + 2 u + p ρ = p∞ ρ . From the symmetry of the
1
flow, the vertical velocity component will be zero at the origin, so p = p∞ − ρ∂φ ∂t − 2 ρu 2 .
! ∂φ $ q ( − ( x − x1 (t)) ( dx1 dt ) − ( x − x2 (t)) ( dx2 dt ) + qs ! dx1 dt dx2 dt $
# & = s* − - = # − &
" ∂ t %x=y=0 2π *) ( x − x1 (t))2 + € y2
2
( x − x2 (t)) + y 2 -,x=y=0 2π " x1 x2 %
€
q ! qs 4π  qs 4π  $ qs ! 2qs $ 1
= s ## − && = # &
2π " ( qs 4π ) t −  ( qs 4π ) t +  % 2π " 4π  % ( qs 4π )2 t 2 −  2
qs2 1
=
4π ( qs 4π )2 t 2 −  2
2
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

! ∂φ $ qs !# x − x1 (t) x − x2 (t) $
& qs ! −1 1 $
# & = u(0, 0) = − = # + &
" ∂ x %x=y=0 2π #" ( x − x1 (t)) + y 2 ( x − x2 (t)) + y 2 &%
2 2
2π " x1 (t) x2 (t) %
x=y=0

qs ! −1 1 $ q 1
= ## + && = s (−2) 2
2π " ( qs 4π ) t −  ( qs 4π ) t +  % 2π (qs 4π ) t 2 − 2
qs  1
=−
π ( qs 4π )2 t 2 −  2
qs2 1 qs2  2 1
so p = p∞ − ρ 2 2
− ρ .
2
4π ( qs 4π ) t −  2 2π # q 4π  2 t 2 −  2 %2
(
$ s ) &
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.35. Consider a free ideal line vortex oriented parallel to the z-axis in a 90° corner
defined by the solid walls θ = 0 and θ = 90°. If the vortex passes the through the plane of the
flow at (x, y), show that the vortex path is given by: x–2 + y–2 = constant. [Hint: Three image
vortices are needed at points (−x, −y), (−x, y) and (x, −y). Carefully choose the directions of
rotation of these image vortices, show that dy/dx = v/u = −y3/x3, and integrate to produce the
desired result.]

Solution 7.35. The induced velocity at point (x, y) will include

contributions from the three image vortices. If the first vortex at 2 1
(x, y) has strength +Γ, then the second vortex at (−x, y) and the
third vortex at (x, −y) will have strength –Γ. The fourth vortex at
(−x, −y) will have strength +Γ. Therefore:
# # −y &&
u=
Γ % 1 1
− ey + ex +
1 % e +
x
e (( 4 3
%
2π $ 2x 2y 2 2 % 2 2 x 2 2 y(
(
2 x +y $ x +y x +y ''
which is the contribution of the second, third, and fourth vortices,
respectively. This velocity can be rewritten:
Γ +% 1 y ( % 1 x ( . Γ + x2 y2 .
u= -' − 2 e +
* x ' − + e
2* y 0 = - e x − e y 0 = ue x + ve y
4 π ,& y x + y 2 ) & x x + y ) / 4π ( x 2 + y 2 ) , y
2
x /
The path lines for the location x = (x, y) of the first vortex are:
dx Γ x2 dy Γ y2
=u= , and = v = − .
€ dt 4π ( x 2 + y 2 ) y dt 4π ( x 2 + y 2 ) x
Divide the second equation by the first to find:
dy dy dx Γ y2 Γ x2 y3
= =− = − .
€ dx dt dt 4 π ( x
€
2
+ y 2
) x 4 π ( x 2
+ y 2
) y x 3

Using the two ends of this equality, separate and integrate the equation:
dy dx 1 1
− 3= 3 → 2
+ 2 = const.
y x y x
€
This is the desired result.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.36. In ideal flow, streamlines are defined by dψ = 0, and potential lines are defined
by dφ = 0. Starting from these relationships, show that streamlines and potential lines are
perpendicular.
a) in plane flow where x and y are the independent spatial coordinates, and
b) in axisymmetric flow where R and z are the independent spatial coordinates.
[Hint: For any two independent coordinates x1 and x2, the unit tangent to the curve x2 = f(x1) is
2
t = (e1 + (df dx1 )e 2 ) 1+ (df dx1) ; thus, for a) and b) it is sufficient to show
(t )ψ = const ⋅ (t)φ = const = 0 ]

∂ψ ∂ψ
€ Solution 7.36. a) In two-dimensional ideal flow, dψ = dx + dy = −vdx + udy. Thus a curve
∂x ∂y
€
defined by dψ = 0, has a slope dy dx = v u . Using the hint, the tangent vector to this curve is:
2
(t )ψ = const = (e x + (v
u)e y ) 1+ (v u) = ( ue x + ve y ) u 2 + v 2 .
∂φ ∂φ €
Similarly, dφ = dx
€+ dy = udx + vdy , so a curve defined by dφ = 0, has a slope
∂x ∂y
dy dx = − u v . Using the hint, the tangent vector to this curve is:
€
(t )φ = const = (e x + (−u v)e y ) 1+ (u v) 2 = (ve x − ue y ) v 2 + u 2 .
€
Forming the dot product of the two unit vectors produces:
€
(t )ψ = const ⋅ (t)φ = const =
(ue x + ve y ) ⋅ (ve x − ue y ) = uv − vu = 0 .
€ u2 + v 2 v 2 + u2 v 2 + u2
∂ψ ∂ψ
b) In three-dimensional axisymmetric flow, dψ = dR + dz = Ruz dR − RuR dz. Thus a curve
∂R ∂z
defined by€dψ = 0, has a slope dR dz = uR uz . Using the hint, the tangent vector to this curve is:
2
uz )e R ) 1+ (uR uz ) = ( uze z + uR e R ) uz2 + uR2 .
(t )ψ = const = (e z + (uR
∂φ ∂φ €
Similarly, dφ = dR + dz = uR dR + uz dz , so a curve defined by dφ = 0, has a slope
∂R € ∂z
dR dz = − uz uR . Using the hint, the tangent vector to this curve is:
€
(t )φ = const = (e z + (−uz uR )e y ) 1+ (uz uR ) 2 = ( uR e z − uze R ) uR2 + uz2 .
€
Forming the dot product of the two unit vectors produces:
€ (u e + u e ) (u e − u e ) u u − u u
(t )ψ = const ⋅ (t)φ = const = z z 2 R 2 R ⋅ R z2 z 2 R = z R2 R2 z = 0 .
uz + uR uR + uz uR + uz
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.37. Consider a three-dimensional point source of strength Q (m3/s). Use a spherical
control volume and the principle of conservation of mass to argue that the velocity components
in spherical coordinates are u = 0 and ur = Q/4πr2 and that the velocity potential and stream
θ

function must be of the form φ = φ(r) and ψ = ψ(θ). Integrate the velocity, to show that φ =
−Q/4πr and ψ = −Qcosθ /4π .

Solution 7.37. For a point source of strength Q (m3/s), the tangential velocity is zero because of
symmetry. The radial velocity times the area 4πr2 equals Q so ur = Q/ 4πr2. Therefore,
∂φ 1 ∂ψ Q 1 ∂φ 1 ∂ψ
ur = = 2 = 2
, and uθ = =− = 0.
∂r r sin θ ∂θ 4 πr r ∂θ r sin θ ∂r
The uθ equations, requires φ = φ(r) and ψ = ψ(θ). Using these results and integrating the ur
equations produces:
Q Qcos θ
€ φ (r) = − € and ψ (θ ) = − .
4 πr 4π

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.38. Solve the Poisson equation ∇ 2φ = Qδ (x − x &) in a uniform unbounded three-
dimensional domain to obtain the velocity potential φ = –Q/4π|x – x´| for an ideal point source
located at x´.

Solution 7.38. This exercise is similar € to Exercise 5.9. First apply a simple shift transformation
that places x´ at the origin of coordinates. Define these new coordinates by: X = x − x #,
Y = y − y #, Z = z − z# , and set r = x − x # = X 2 + Y 2 + Z 2 . The gradient operator ∇ XYZ in the
shifted coordinates X = (X, Y, Z) is the same as ∇ in the unshifted coordinates (x, y, z), so the
field equation for φ becomes: €
€ ∇ 2XYZ φ = Qδ (X) = Qδ (x − x &) .
€ € €
Integrate this equation inside a sphere€of radius r:
2 2 2 2 2
z= +r y= + r −z x=−+ r −y −z
2
∫∫∫ ∇ φdV =
XYZ ∫∫ ∇ XYZ φ ⋅ ndA = Q ∫ ∫ ∫ δ(X)dXdYdZ ,
sphere € spherical surface z=−r y=− r 2 −z 2 x=− r 2 −y 2 −z 2

where the first equality follows from Gauss' divergence theorem, and the triple integral on the
right side includes the location X = 0 (aka x = x´) so a contribution is collected from the three-
dimensional delta function. Thus, the right side of this equation is Q (times unity).
€ The dot product in the middle portion of the above equation simplifies to ∂φ/∂r because n
= er on the spherical surface and e r ⋅ ∇ XYZ = ∂ ∂r . Plus, in an unbounded uniform environment,
there are no preferred directions so φ = φ(r) alone (no angular dependence). Thus, the integrated
field equation simplifies to
π 2π π 2π
$ ∂φ ' 2 2 ∂φ ∂φ
∫ ∫€% ∂r (
& ) r sin θ d θ d ϕ = r ∫ ∫ sin θdθdϕ = 4 πr 2 = Q,
θ = 0ϕ = 0 ∂r θ = 0 ϕ = 0 ∂r
which implies
∂φ Q Q Q
= 2
or φ = − =− .
∂r 4 πr 4 πr 4π x − x %
€

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.39. Using (R,ϕ,z) -cylindrical coordinates, consider steady three-dimensional

potential flow for a point source of strength Q at the origin in a free stream flowing along the z-
axis at speed U:
€
Q
φ (R,ϕ,z) = Uz − .
4π R 2 + z 2
a) Sketch the streamlines for this flow in any R-z half-plane.
b) Find the coordinates of the stagnation point that occurs in this flow.
c) Determine the pressure gradient, ∇p , at the stagnation point found in part b).
€ stream surface that encloses the fluid that emerges from the source,
d) If R = a(z) defines the
determine a(z) for z → +∞ .
e) Use Stokes’ stream function to determine an equation for a(z) that is valid for any value of z.
f) Use the control-volume momentum equation, ∫ ρu(u ⋅ n) dS = − ∫ pndS + F where n is the
€ S S
outward normal from the control volume, to determine the force F applied to the point source to
hold it stationary.
g) If the fluid expelled from the source is replaced by a solid body having the same shape, what
is the drag on the front of this body? €

Solution 7.39. a) A simple sketch

R
appears to the right.
b) From symmetry, the stagnation point
will lie on R = 0, thus only the axial
velocity needs to be considered to find
its axial location zs .
z
$ ∂φ ' Qz s
& ) = uz (zs,0) = U + 32
. Solve
% ∂z ( r= 0 4π (z 2 ) s

2 32 3
for zs using z ( )
s= zs to find zs = − Q 4 πU .
c) Take the gradient of the Bernoulli equation or consider the steady ideal-flow momentum
€ 1
equation to find: (u ⋅ ∇ )u = − ∇p . The fluid velocity is zero at a stagnation point, so ∇p = 0 at
ρ
€ €
a stagnation point.
d) Far downstream of the source, all of the fluid will be moving horizontally at speed U. Thus,
the fluid that comes from the source will emerge from a stream tube of radius € a∞ determined
€2
from: πa∞U = Q , which implies: a∞ = Q πU .
e) Set the velocity relationships for the Stokes stream function ψS equal to those from the
potential:
1 ∂ψ S QR ∂φ 1 ∂ψ S Qz ∂φ
€ uR = − =€ 32 = , and uz = =U + 32 = .
R ∂z 4π (R + z )
2 2 ∂R R ∂R 4π (R + z )
2 2 ∂z
Use variable substitutions tan β = R z in the first equation and γ = R 2 + z 2 in the second equation
to integrate these relationships to find:
−1 2 −1 2
€ ψ S = −(Qz 4π )( R 2 + z 2 ) € , and ψ S = UR 2 − (Qz 4 π )( R + z )
+ f (R) 2 2 2
+ g(z)
€ €

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

So, ψ S = UR 2 2 − (Q 4 π ) 1+ z ( )
R 2 + z 2 where the two functions f(r) and g(z) have been chosen
so that ψs = 0 on R = 0 when z < 0. The equation for a(z) is found by setting R = a & ψs = 0:
(
0 = Ua 2 2 − (Q 4 π ) 1+ z )
a 2 + z 2 . This implicit equation for a can be solved to find:
€ 12
a(z) = 1
2 (a − z + z
2
∞
2
z 2 + 2a∞2 ) where a∞ = Q πU .
f) From the symmetry of the flow, only a z-direction force is anticipated and the simplest
€ possible control volume is a sphere of radius a. Thus, using spherical coordinates where
r = R 2 + z 2 denotes the distance from the source and € θ is the polar angle, the vector momentum
€
equation can be simplified:
Fz = ∫ ρuz (u ⋅ e r ) dA + ∫ pe r ⋅ e z dA ,
S S
€ where the replacement n = e r has been made. For the given potential, φ = Ur cosθ − Q 4 πr , the
radial and angular velocities are
∂φ Q 1 ∂φ
€ur = = U cosθ + 2
and uθ = = −U sin θ .
€ ∂r 4 πr r ∂θ
The pressure can be obtained from the Bernoulli equation: €
1 1 % QU cosθ Q2 ( 1
2€
( )
p(r,θ ) + ρ ur2 + uθ2 = p(r,θ ) + ρ'U 2 cos 2 θ +
2 & 2πR 2
+
16π R2 4
+ U 2 sin 2 θ * = p∞ + ρU 2 .
) 2
€
1 ' QU cosθ Q *
2
Thus, p(r,θ ) = p∞ + ρ) − 2
− , , so the pressure integral becomes:
2 ( 2πr 16π 2 r 4 +
π /
€ 1 ) QU cos θ Q2 ,2 1 π
∫ pe r ⋅ e z dS = ∫ 1 p∞ + 2 ρ+− 2πr 2 − 16π 2 r 4 .4(cosθ )2πr 2 sin θdθ = − 2 ρQU ∫ cos2 θ sinθdθ
S θ = 00 * -3 θ=0

€ 1
= − ρQU
3
The terms with odd-powers of the cosine do not contribute to the angular integration on the
€ interval θ = 0 to π. To calculate the flux integral, first find uz and change it into the spherical
coordinate system
€
∂φ Qz Qcosθ
uz = =U + 3 2 =U + .
∂z 4π [R 2 + z 2 ] 4 πr 2
Then proceed with:
π
' Qcos θ *' Q * 2
∫ ρuz (u ⋅ e r )dS = ∫
ρ)U +
4 r 2 ,)
U cos θ +
4 r 2,
2πr sin θdθ
S θ=0 ( π +( π +
€ −1
QU π QU ( β3 + 4
=ρ ∫
2 θ=0
(1+ cos θ ) sin θdθ = − ρ
2

2 )
*β + - = ρQU
3 ,1 3
€ $ 4 1'
Thus, F = ρQU& − )e z = ρQUe z ; again, the odd-powers of the cosine do not contribute. This
% 3 3(
force€points downstream, so if it were not applied the source would move upstream! This result
matches the finding for a source in a free stream in two dimensions.
g) Here, consider a spherical control volume that is very large so that the velocity is nearly Ue z
€
on the surface of the control volume. In this case, the pressure integral will be unchanged as

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

long as the pressure within the body is p∞. However, the flux integral will not include the
contribution from the fluid that emerges from the source. Thus, the force in this case will be
equal to that obtained for part d) minus the momentum flux that is no longer present:
F = (F ) d ) − ( ρUQ)e z = 0
Therefore, there will be no drag on the front of the body when the body is long compared to its
diameter.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.40. In (R, ϕ, z) cylindrical coordinates, the three-dimensional potential for a point
−1 2
source at (0,0,s) is given by: φ = −(Q 4 π )[ R 2 + (z − s) 2 ] .
a) By combining a source of strength +Q at (0,0,–b), a sink of strength –Q at (0,0,+b), and a
uniform stream with velocity Uez, derive the potential (7.85) for flow around a sphere of radius a
by taking the limit as Q → ∞, and b → 0, such that d = −2bQe z = –2πa3Uez = constant. Put your
€
final answer in spherical coordinates in terms of U, r, θ, and a.
b) Repeat part a) for the Stokes stream function starting from
−1 2
ψ = −(Q 4 π )(z − s)[ R 2 + (z − s) 2 ] . €

Solution 7.40. In cylindrical coordinates, the three-dimensional potential for a uniform stream of
U = (0,0,U) , a point source of strength +Q at x = (0,0,−b) , and a point-sink of strength –Q at
€
x = (0,0,+b) is:
Q Q
φ = Uz − + .
2 2 2
€ 4π
€ R + (z + b) 4π R + (z − b)
€ Expand the square roots for b → 0, letting r = R 2 + z 2
1 1 1 1
≈ ≈ ≈ (1  (zb r 2 )) .
r 1± (2zb r ) r(1± (zb r )) r
2 2 2 2
€ R + (z ± b)
Make replacements in the equation € for φ:
Q % zb zb( Qzb d⋅ x
φ = Uz − '1− 2 −1− 2 * = Uz + 3
= U⋅ x − .
€ 4πr & r r ) 2πr 4π | x |3
where the last equality was obtained by using the given definition d = −2bQe z . Using z = rcosθ
(where θ = polar angle) and | d |= 2πUa 3 , the potential becomes:
€ $ |d | ' # a3 &
φ = &Ur + ) cos θ , or φ = Ur%1+ 3(
cosθ .
% 4πr 2 ( €$ 2r '
€
b) In cylindrical coordinates, the Stokes stream function for a uniform stream of U = (0,0,U) , a
point source of strength +Q at x = (0,0,−b) , and a point-sink of strength –Q at x = (0,0,+b) is:
1 Q (z + b) Q (z − b)
€ ψ = UR 2 − € + .
2 4π R 2 + (z + b) 2 4 π R 2 + (z − b) 2
€
As in part a), expand€the square roots for b → 0, letting r = R 2 + z 2€
1 1 1 1
≈ ≈ ≈ (1  (zb r 2 )) .
R 2 + (z ± b) 2 r 1± (2zb r 2 ) r(1± (zb r )) r
€ 2

Make replacements in the equation for ψ: €

1 Q % z + b % zb (( Q % z − b % zb ((
ψ = UR 2 − ' '1− 2 ** + ' '1+ 2 **
€ 2 4 π & r & r )) 4 π & r & r ))
1 Q % z b z 2b zb 2 z b z 2b zb 2 ( 1 Q % 2b 2z 2b (
= UR 2 + ' − − + 3 + 3 + − + 3 + 3 * = UR +
2
'− + 3 *.
2 4π & r r r r r r r r ) 2 4π & r r )
3
Using R = rsinθ and z = rcosθ (where θ = polar angle), and | d |= 2πUa , as given in the problem
statement, the stream function becomes:
€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

1 2 2 2Qb & 1 r 2 cos2 θ ) 1 2 & a 3 ) 2

ψ = Ur sin θ + (− + + = Ur (1− 3 + sin θ .
2 4π ' r r3 * 2 ' r *

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.41. a) Determine the locus of points in uniform ideal flow past a circular cylinder of
radius a without circulation where the velocity perturbation produced by the presence of the
cylinder is 1% of the free stream value.
b) Repeat for uniform ideal flow past a sphere.
c) Explain the physical reason(s) for the differences between the answers for a) and b).

Solution 7.41. a) From (7.34), the velocity components for ideal flow about a cylinder with
radius a are:
# a2 & $ a2 '
ur = U%1− 2 ( cos θ , and uθ = −U&1+ 2 ) sinθ .
$ r ' % r (
Clearly, the velocity magnitude when the cylinder is absent = U. Therefore, the magnitude of the
perturbation is:
2 2
€ u − Ue x = U ( a r ) €cos θ + U ( a r ) sin θ = U ( a r ) .
2 2 2 2 2 2 2 2 2 2

u − Ue x a 2
Therefore, a 1% perturbation implies: 0.01 = = 2 → r = 10a .
U r
b) From€(7.86), the velocity components for ideal flow about a sphere with radius a is:
# a3 & $ a3 '
ur = U%1− 3 ( cos θ , and uθ = −U&1+ 3 ) sin θ .
€$ r ' % 2r (
Clearly, the velocity magnitude when the cylinder is absent = U. Therefore, the magnitude of the
perturbation is:
2 2
€x = U ( a r ) cos θ +€
U 2 ( a 3 2r 3 ) sin 2 θ = U ( a 3 r 3 ) cos 2 θ + 14 sin 2 θ .
2 3 3 2
u − Ue
Therefore, a 1% perturbation implies:
u −Ue x 16
0.01 = = ( a 3 r 3 ) cos2 θ + 14 sin 2 θ → r = 4.64a ( 43 cos2 θ + 14 ) .
€ U
c) For the sphere, the 1%-perturbation distance is smaller because the sphere's projected area in
the direction of the flow (πa2) is smaller than that of the cylinder (2a x span); the sphere's
blockage is smaller. Thus, the flow has an easier time getting around the sphere. This effect also
decreases the 1%-perturbation distance at θ = π/2 in the case of the sphere.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.42. Using the figure for Exercise 7.29

with A3 → 0 and r → ∞ , expand the three-
dimensional potential for a stationary arbitrary-
shape closed body in inverse powers of the
distance r and
€ prove that ideal flow theory predicts
€ zero drag on the body.

Solution 7.42. The figure is reproduced here and a

portion of this solution follows that given for
Exercise 7.29. However, this time the geometry is
three-dimensional and the figure must be
interpreted as a slice through three dimensional control volume so that A1 is a spherical surface.
In ideal flow, the only surface forces are pressure forces. Therefore, the hydrodynamic
force F on the body is defined by:
F = − ∫ ( p − p∞ )n2 dA = ∫ ( p − p∞ )ndA , (a)
A2 A2
where the free stream velocity is U = Uex, and n is the outward normal on the total control
volume; thus it points into the body on surface A2 so the usual minus sign is missing from the
final equality in (a) because n = –n2. Note that the constant p∞ can be included in the pressure
integration because€ ∫ p∞ndA = 0 .
closed surface
The starting point for the derivation is the integral form of the steady ideal flow
momentum equation applied to the entire clam-shell control volume:
€
∫ ρu(u ⋅ n)dA = − ∫ ( p − p∞ )ndA
A +A +A 1 A +A +A
2 3 1 2 3

with the implied limit that the tube denoted by A3 goes to zero. When this limit is taken, the net
contribution of A3 is zero because the surface area of A3 goes to zero. This leaves:
∫ ρu(u€⋅ n)dA + ∫ ρu(u ⋅ n)dA = ∫ ρu(u ⋅ n)dA + 0 = − ∫ ( p − p∞ )ndA ,
A1 A2 A1 A1 +A 2
where the first equality follows because u ⋅ n = 0 on the solid surface A2. Now substitute in (a)
for the pressure integration over A2, and rearrange:

€
∫ ρu(u ⋅ n)dA = − ∫ ( p − p∞ )ndA − F or F = − ∫ ( p − p∞ )ndA − ∫ ρu(u ⋅ n)dA .
A1 A1 A1 A1
€
Thus, the force on the body can be obtained from integrals over a spherical surface (A1) that is
distant from the body.
The pressure can be written in term of the velocity using the Bernoulli equation:
€ 1 € 1 2 1
p∞ + ρU 2 = p + ρ u = p + ρ(u ⋅ u) ,
2 2 2
and this allows the pressure integral to be written:
1 1
( )
− ∫ ( p − p∞ )ndA = − ρ ∫ U 2 − u ⋅ u ndA = ρ ∫ (u ⋅ u)ndA ,
2 A1 2 A1
€ A1
2
where the final equality follows because U is a constant so ∫ U 2ndA = 0 . So, the force on the
closed surface
body can be obtained entirely from considerations of the velocity field:
€ (
F = ρ ∫ 12 (u ⋅ u)n − u(u ⋅ n) dA . )
A1
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Far from the body, the fluid velocity will primarily be U = Uex = constant, so it can be written u
= U + u´, where u´ is the velocity perturbation that comes from the presence of the body.
Inserting this decomposition into the last equation produces:
(
F = ρ ∫ 12 U 2n + (U ⋅ u$)n + 12 (u$ ⋅ u$)n − U(U ⋅ n) − U(u$ ⋅ n) − u$(U ⋅ n) − u$(u$ ⋅ n) dA
A1
)
(
= ρ ∫ (U ⋅ u$)n +
A1
1
2 (u$ ⋅ u$)n − U(u$ ⋅ n) − u$(U ⋅ n) − u$(u$ ⋅ n))dA.
As before the second equality follows by dropping the terms that involve the quadratic factors of
the constant vector U. The five remaining integrand terms all involve the perturbation velocity.
An arbitrary shape closed body may be represented in potential flow by a collection of
€sources and sinks. Thus, for the limit as r → ∞ , the potential may be expanded in inverse powers
of r:
φ φ
φ = U ⋅ x + φ $ = U ⋅ x + 1 + 22 + ...
€ r r
where the coefficients φm may be functions of the angular variables. The first term in this
expansion corresponds to the free stream. The remaining term produce u´. However, the body is
closed so the φ1 term must be zero because it represents the net source strength of the potential.
€
The fluid velocity is the gradient of the potential, therefore the largest term representing the
velocity perturbation will involve another factor of 1/r.
φ
u" ~ 32 + ...
r
Using this scaling of the velocity perturbation with radial distance, the magnitude of the
five integrand terms above can be assessed using dA = r2sinθdθdϕ,
& Uφ φ 2 Uφ Uφ φ2 ) const
F ~ lim ρ ∫ ( 3 2 +€ 26 + 3 2 + 3 2 + + 26 + r 2 sinθdθdφ ∝ lim = 0.
all angles' r 2r r r 2r * r
r →∞ r →∞

Thus, pure potential flow cannot predict drag. This another statement of D'Alembert's paradox.
Interestingly, if a "φ0" term is added to the expanded φ above, non-zero lift and drag are possible.
However, this involves the introduction of vorticity in three-dimensions and elementary
€
treatment of this topic for aircraft wings is postponed to Ch. 14.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.43. Consider steady ideal flow over a hemisphere of constant radius a lying on the y-z
plane. For the spherical coordinate system shown, the potential for this flow is:
φ (r,θ,ϕ ) = Ur(1+ a 3 2r 3 )cosθ where U is the flow velocity far from the hemisphere. Assume
gravity acts downward along the x-axis. Ignore fluid viscosity in this problem.
a) Determine all three components of the fluid velocity on the surface of the hemisphere, r = a, in
' ∂φ 1 ∂φ 1 ∂φ *
€ spherical polar coordinates: (ur ,uθ ,uϕ ) = ∇φ = ) , , ,.
( ∂r r ∂θ r sin θ ∂ϕ +
b) Determine the pressure, p, on r = a.
c) Determine the hydrodynamic force, Rx, on x!
the hemisphere assuming stagnation pressure
is felt everywhere € underneath the hemisphere.
π r!
[Hints: e r ⋅ e x = sin θ cos ϕ , ∫ 0 sin 2 θdθ = π 2 ,
π
!
ϕ
and ∫ 0 sin4 θdθ = 3π 8 ]. θ! z!
d) For the conditions of part c) what density ρh
€ must the hemisphere
€ have to remain on the
surface.
€
y!

( 3
Solution 7.43. a) Differentiate φ (r,θ,ϕ ) = Ur 1+ a 2r cosθ and evaluate on r = a.
3
)
∂φ % a3 ( # a3 &
ur = = U'1− 3 * cos θ –> ur (r = a) = U%1− 3 ( cosθ = 0
∂r & r ) $ a '
1 ∂φ €& a3 ) 3U
uθ = = −U(1+ 3 + sin θ –> uθ (r = a) = − sin θ
r ∂θ ' 2r * 2
€ 1 ∂φ € –>
uϕ = =0 uϕ (r = a) = 0
r sin θ ∂ϕ
€ 1 2 1  2
€ b) The steady Bernoulli equation applies here: p∞ + 2 ρU = p(r = a) + 2 ρ u(r = a) . Use the
p(r = a) − p∞ 1 2 1  2
results of part a) to find: € = 2 U − 2 ρ u(r = a) = 12 U 2 − 98 U 2 sin 2 θ , or
€ ρ
(
p(r = a) − p∞ = 12 ρU 2 1− 49 sin2 θ = 12 ρU€
2 9
4 )
cos 2 θ − 45 ( )
c) The hydrodynamic force will be determined by the pressure-force difference between the top
1 2
€ Here the stagnation pressure is p∞ + 2 ρU , thus:
and bottom of the hemisphere.
Rx = (pressure force pushing up on the flat side) – (pressure force pushing down on the curved side)
€
( )
= p∞ + 12 ρU 2 πa 2 − ∫ p(r = a)(e r ⋅ e x ) dS
hemisphere
π +π 2 €
( )
= p∞ + 12 ρU 2 πa 2 − ∫ ∫
θ = 0ϕ =− π 2
[p 1 2 9
( 2
∞ + 2 ρU 1− 4 sin θ )](cosϕ sinθ )a 2
sin θdϕdθ
π +π 2
€
( 1 2
= p∞ + ρU πa − a
2 ) 2 2
∫ [p ∞
1 2
(
+ ρU 1− sin θ sin θdθ
2
9
4
2
)] 2
∫ cosϕdϕ
θ=0 ϕ =− π 2

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

( 1 2
= p∞ + ρU πa − 2a
2 ) 2 2
∫ [p ∞ ( )]
+ 12 ρU 2 1− 49 sin 2 θ sin 2 θdθ
θ=0
π π

( ) [
= p∞ + 12 ρU 2 πa 2 − 2a 2 p∞ + 12 ρU 2 ] ∫ sin θdθ − 2a (
2 2 1
2
ρU 2 )(− 49 ) ∫ sin 4 θdθ
θ=0 θ=0

€ &π ) 9 & 3π ) 27π

( ) [ '2*
]
= p∞ + 12 ρU 2 πa 2 − 2a 2 p∞ + 12 ρU 2 ( + + a 2 ρU 2 ( + =
4 8
' * 32
ρU 2 a 2

€ where the second-to-last equality has required use of the two integral hints.
d) The hemisphere will remain on the surface when its weight is greater than or equal to the
pressure force and the buoyant force:
€ $ 81 U 2 '
$2 ' 27π $2 '
ρ h & πa 3 )g ≥ ρU 2 a 2 + ρ& πa 3 ) g , and this implies: ρ h ≥ &1+ )ρ .
%3 ( 32 %3 ( % 64 ga (

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.44. The flow-field produced by suction flow into

a round vacuum cleaner nozzle held above a large flat
surface can be easily investigated with a simple experiment,
and analyzed via potential flow in (R, ϕ, z)-cylindrical
coordinates with the method of images.
a) Do the experiment first. Obtain a vacuum cleaner that has
a hose for attachments. Remove any cleaning attachments
(brush, wand, etc) or unplug the hose from the cleaning
head, and attach an extension hose or something with a
round opening (~4 cm diameter is recommended). Find a
smooth dry flat horizontal surface that is a ~0.5 meter or
more in diameter. Sprinkle the central 1/3 of the surface
with a light granular material that is easy to see (granulated sugar, dry coffee grounds, salt, flour,
talcum powder, etc. should work well). The grains should be 1/2 to 1 mm apart on average. Turn
on the vacuum cleaner and lower the vacuum hose opening from ~0.25 meter above the surface
toward the surface with the vacuum opening facing toward the surface. When the hose gets to
within about one opening diameter of the surface or so, the granular material should start to
move. Once the granular material starts moving, hold the hose opening at the same height or lift
the hose slightly so that grains are not sucked into it. If many grains are vacuumed up, distribute
new ones in the bare spot(s) and start over. Once the correct hose-opening-to-surface distance is
achieved, hold the hose steady and let the suction airflow of the vacuum cleaner scour a pattern
into the distributed granular material. Describe the shape of the final pattern, and measure any
relevant dimensions.
Now see if ideal flow theory can explain the pattern observed in part a). As a first
approximation, the flow field near the hose inlet can be modeled as a sink (a source with strength
–Q) above an infinite flat boundary since the vacuum cleaner outlet (a source with strength +Q)
is likely to be far enough away to be ignored. Denote the fluid density by ρ, the pressure far
away by p∞, and the pressure on the flat surface by p(R). The potential for this flow field will be
the sum of two terms:
+Q
φ (R,z) = + K(R,z)
4 π R 2 + (z − h) 2
b) Sketch the streamlines in the y-z plane for z > 0.
c) Determine K(R,z).
d) Use dimensional analysis to determine how p(R) – p∞ must depend on ρ, Q, R, and h.
e) Compute p(R) – p€ ∞ from the steady Bernoulli equation. Is this pressure distribution consistent
with the results of part a)? Where is the lowest pressure? (This is also the location of the highest
speed surface flow). Is a grain at the origin of coordinates the one most likely to be picked up by
the vacuum cleaner?

Solution 7.44. a) The pattern scoured in the granular material is axisymmetric. The scoured
region is a ring that is a little larger than the vacuum cleaner opening. A pile of the granular
material is left directly below the center of the vacuum cleaner opening.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

1. a)
b) z

c) Based on the method of images, K must represent a sink at z = –h, so:

+Q
K(R,z) =
4 π R + (z + h) 2
2

d) 5 parameters - 3 dimensions = 2 groups. Clearly Π1 = R h . A little more effort yields:

Π 2 = ( P(R) − P∞ ) h 4 ρQ 2 . Therefore: P(R) − P∞ = ρQ h ( 2 4
) f ( R h)
e) Use the potential and the
€ Bernoulli equ. There is no velocity very far from the origin and
1
there will be no vertical velocity on z = 0, so: P∞ = P(R) + ρ [ uR2 ] . Here:
€ 2 z= 0

∂φ −QR€ −QR −QR

uR = = 3 2 + 3 2 , so [ uR ] z= 0 = 32 ,
∂R 4 π [ R 2 + (z − h) 2 ] 4 π [ R 2 + (z + h) 2 ] 2π [ R 2 + h 2 ]
2
−€ρQ2 R 2 1 ρQ2 (R h ) .
and P(R) − P∞ = 3 = −
8π 2 [ R 2 + h 2 ] 8π h 1+ ( R h ) 2 3
2 4

€ €
[ ]
This pressure distribution is consistent with results of part a). The radial location of the
minimum pressure (maximum surface flow speed) is found from:
& 2
*
€ d 1 ρQ2 ( 1 ( R h) ( 2R
(P(R) − P∞ ) = 0 = − 2 4 ' 3 −3 4+ 2 .
dR 8π h ( 1+ ( R h ) 2 1+ ( R h ) (, h
) [ ] 2
[ ]
(
Dividing out all the non-zero factors and parameters yields: R 1− 2( R h ) = 0 , which implies: R
2
)
= 0, or R = h 2 . The second answer matches (within experimental accuracy) the radius of the
ring €
on the surface where the scouring was most complete in the experiment. The first answer, R
= 0, is a stagnation point (a pressure maximum) so a grain located there is not likely to be picked
up. In reality, when the vacuum cleaner hose opening€ is held close enough to the surface, even
€
the r = 0 grains are picked up because of the finite size of the grains and because of unsteady
flow processes.
Overall, when the hose opening is comfortably above the surface, the primary difference
between the experiment and the analysis is the size of the sink. In the experiment, air suction
takes place across a finite area whereas the potential flow sink is a point in space. Thus, some
minor differences between theoretical and experimental results are expected, but such differences
might not be easily detected given the simplicity and qualitative nature of the experiments
conducted for part a).
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.45. There is a point source of strength Q (m3/s) at the origin, and a uniform line sink
of strength k = Q/a extending from z = 0 to z = a. The two are combined with a uniform stream U
parallel to the z-axis. Show that the combination represents the flow past a closed surface of
revolution of airship shape, whose total length is the difference of the roots of:
z2 " z % Q
2$
± 1' =
a # a & 4 πUa 2

Solution 7.45. From the symmetry of the flow, the

stagnation points that define the length of the body must U
€ d!
lie on the z-axis. Consider first the stagnation point, P, to ! P
the right of the source and line sink. The fluid velocity, z
Q a
uz, on the z-axis will be that due to the free stream, the
source, and the line sink.
a
Q k
uz (0,z) = U +
4 πz 2
− ∫ 2
dζ .
0 4 π (z − ζ )
Substitute in for the value of k, perform the integration and set the result equal to zero to get:
Q $1 1 ' z2 # z & Q
0 =U + 2& 2
− ), which implies: 2 %$ −1(' = .
€ 4 πz % z z(z − a) ( a a 4 πUa 2
When P is to the left of the source and line sink, the on-axis fluid velocity is:
a
Q k
uz (0,z) = U −
4 πz 2
+ ∫ 2
dζ .
0 4 π (z − ζ )
€ €
Repeating the prior steps produces:
z2 " z % Q
2$
+ 1' = .
€ a # a & 4 πUa 2

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercsie 7.46. Using a computer, determine the surface contour of an axisymmetric half-body
formed by a line source of strength k (m2/s) distributed uniformly along the z-axis from z = 0 to z
= a and a uniform stream. The nose of this body is more pointed than that formed by the
combination of a point source and a uniform stream. From a mass balance, show that far
downstream the radius of the half-body is r = ak πU .

Solution 7.46. The Stokes stream function for this flow is given by:
1 k a (z − ζ )dζ 1 k
ψ = UR 2 −
2
∫ €
4 π 0 (z − ζ ) + R
2 2
= UR 2 +
2 4π( (z − a) 2 + R 2 − z 2 + R 2 ,)
where the final equality follows from evaluating the integral. The body contour will follow the
dividing streamline that starts from the stagnation point located at R = 0 on the negative part of
the z-axis. On the negative z-axis, the difference of square roots above is +a, so the body contour
€ follow the R-z trajectory implicitly specified by:
will
ka 1 k
4π 2
= UR 2 +
4π ( (z − a) 2 + R 2 − z 2 + R 2 . )
For computer evaluation dimensionless variable are best. Therefore, define z* = z/a and R*= R/a,
and find:
2 2 2 2
€ 1 = ΩR* + (z* −1) 2 + R* − z* + R* ,
where Ω = 2πaU/k is the dimensionless flow speed. This implicit relationship is plotted below
for Ω = 2.0. The flow is from left to right, the R* axis is vertical, and the z* axis is horizontal.
€
%"

!#$"

!"
&%" &!#$" !" !#$" %" %#$" '"

The flow becomes uniform with speed U far downstream of the body. If r is the asymptotic
radius of the half body, then a mass balance gives:
Uπr2 = flow out of the line source = ka, which implies: r = ka πU

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.47. Consider the radial flow induced by the collapse of a spherical cavitation bubble
of radius R(t) in a large quiescent bath of incompressible inviscid fluid of density ρ. The pressure
far from the bubble is p∞. Ignore gravity.
a) Determine the velocity potential φ(r,t) for the radial flow outside the bubble.
b) Determine the pressure p(R(t), t) on the surface of the bubble.
c) Suppose that at t = 0 the pressure on the surface of the bubble is p∞, the bubble radius is Ro,
and its initial velocity is −R˙ o (i.e. the bubble is shrinking), how long will it take for the bubble to
completely collapse if its surface pressure remains constant?

1 ∂ % 2 ∂φ (
Solution 7.47.€a) All of the flow will be in the radial direction. Thus: ∇ 2φ = 'r * = 0.
r 2 ∂r & ∂ r )
Away from r = 0, this equation can be integrated directly to find: φ = −A / r , where A may be a
function of time. To evaluate A, match the radial fluid velocity at the surface of the sphere to the
$ ∂φ ' A(t) R 2 R˙
surface velocity of the sphere: & ) = R˙ (t) = 2 . Solve € for A to get: φ = − .
% ∂r ( r= R(t ) R € r
∂φ 1 2 p − p∞ 2RR˙ 2 R 2 R ˙˙ R 4 R˙ 2
b) Use the Bernoulli equation: ρ + ρ u + p = p∞ , to get: =+ + − .
∂t 2 ρ r r 2r 4
€ p(R,t) −€p∞ 3 ˙ 2 ˙˙
Evaluate this on the surface of the bubble r = R(t) to get: = R + RR
ρ 2
c) Set p(R,t) = p∞ to get:€ 0 = 3 R˙ 2 + RR ˙˙ = R−1/ 2 d (R 3€/ 2 R˙ ) , and integrate: R 3 / 2 R˙ = C . Integrate
2 dt
again to find: 2R 5 / 2 5 = Ct + D . Now evaluate the constants from the given information:
€ 25
−Ro3 / 2 R˙ o = C , and 2Ro5 / 2 5 = D . Thus, R(t) = Ro − (5 2) Ro R˙€o t , so the collapse time is:
[ ]
5/2 3/2

t = 2Ro (5 R˙ o ) .€
€
€ € €
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.48. Derive the apparent mass per unit

depth into the page of a cylinder of radius a that
travels at speed U c (t) = dx c dt along the x-axis in
a large reservoir of an ideal quiescent fluid with
density ρ. Use an appropriate Bernoulli equation
and the following time-dependent two-dimensional
€ a 2U c (x − x c )
potential: φ (x, y,t) = − , where xc(t) is location of the center of the cylinder, and the
(x − x c ) 2 + y 2
Cartesian coordinates are x and y. [Hint: steady cylinder motion does not contribute to the
cylinder’s apparent mass; keep only the term (or terms) from the Bernoulli equation necessary to
determine apparent mass].
€
Solution 7.48. In an ideal incompressible fluid, the hydrodynamic loads on the moving cylinder
will occur through pressure forces and the pressure will be set by the unsteady Bernoulli
equation (UBE).
∂φ 1 2 ∂φ 1 2
ρ + ρ ∇φ + ρgy + p = p∞ or p∞ − p = ρ + ρ ∇φ + ρgy
∂t 2 ∂t 2
The force that the fluid applies to the cylinder, Fc, is
θ = 2π
Fc = ∫ surface ( p∞ − p)ndS = ∫θ = 0 ( p∞ − p)e r Badθ
where€B is the length of the cylinder along
€ its axis, e r = e x cosθ + e y sin θ , and θ is an angle
measured from the positive x-axis with its apex at the center of the cylinder. In addition, define
2
2
the distance from € the center of the cylinder as R(t) = ( x − x c (t)) + y so that
( )
€ can be written: φ (R,θ ) = − a 2U c R cosθ where R, θ,
cosθ = ( x − x c (t)) R(t) . Now the potential
and Uc are all functions of time for a fixed (x,y) location. Only the unsteady term in the UBE
will contribute a term that involves dU ˙
€ c dt ≡ U c , thus:
€ p∞ − p = −ρ ( a 2U˙ c R) cos θ + (terms leading to no net force) .
€
Evaluate this one unsteady term on R = a, ( p∞ − p) R= a = −ρaU˙ c cosθ + ..., and this leads to:
F B = − ρa 2U˙ € cos θ (e cosθ + e sin θ )dθ = − ρa 2U˙ (πe + 0e ) = −πa 2 ρU˙ e
θ = 2π
c c ∫θ = 0 x y c x y c x

So, the apparent

€ mass of the cylinder is equal to the mass of the fluid displaced by the cylinder.
The streamlines for this flow
€ when Uc < 0 are shown in Figure 3.2b.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.49. A stationary sphere of radius a and mass m resides in inviscid fluid with constant
density ρ.
a) Determine the buoyancy force on the sphere when gravity g acts downward.
b) At t = 0, the sphere is released from rest. What is its initial acceleration?
c) What is the sphere’s initial acceleration if it is a bubble in a heavy fluid (i.e. when m → 0)?

Solution 7.49. a) The buoyancy force arises from the static pressure acting on the surface of the
sphere. Choose the spherical coordinate system so that ez points opposite gravity and the
hydrostatic pressure is p = po − ρgz . Thus the buoyancy force on sphere will be:
2π π 2π π
FB = − ∫ ∫ ( p − po )e r a 2 sin θdθdϕ = ∫ ∫ ( po − p)(e x cosϕ sin θ + e y sin ϕ sin θ + e z cosθ )a 2 sin θdθdϕ
0 0 0 0
Performing the
€ ϕ integration eliminates two of the coordinate directions, so
π π
FB = 2π ∫ (+ ρgz)e z a 2 cos θ sin θdθ = 2πρga 3e z ∫ cos 2 θ sin θdθ ,
€ 0 0
where p − po = −ρgz , and z = acosθ have been used. The θ-integration is completed via a
straightforward change of variable:
π −1
4
€ FB = +2πρga e z ∫ cos θ sin θdθ = −2πρga e z ∫ β 2 dβ = + πρga 3e z .
3 2 3

€ 0 1 3
du
b) Newton’s second law for the sphere will be: FB − mge z = (m + madded ) s , where madded is 1/2
dt
3
dus FB − mge z (4 /3) ρa g − mg (4 /3) ρa 3 − m
the mass
€ of the displaced fluid. Thus, = = e z = ge
dt m + madded m + (2 /3) ρa 3 (2 /3)ρa 3 + m z
dus €
c) If m → 0, then = 2ge z .
dt
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.50. A sphere of mass m and volume V is attached to the end of a light thin flexible
cable of length L. In vacuum, with gravity g acting, the natural frequencies for small longitudinal
(bouncing) and transverse (pendulum) oscillations of the sphere are ωb and ωp. Ignore the effects
of viscosity, and estimate these natural frequencies when the same sphere and cable are
submerged in water with density ρw. What is ωp when m << ρ wV ?

Solution 7.50. For the longitudinal (bouncing) vibrations, the natural frequency will be
d 2z
determined from the simple linear differential € equation: M dt 2 + σz = 0 where z is the coordinate
of the sphere’s center of mass in longitudinal direction, and σ is the extensional stiffness of the
cable. In a vacuum, there will be no fluid dynamic effects so the longitudinal natural frequency
is ω b = σ M . When the sphere is placed in water,
€ it will have added mass equal to one half the
mass of the displaced fluid, so that small longitudinal oscillations will follow:
d 2z dz
[
M + 12 ρ wV
dt 2 ]
+ γ + σz = 0 ,
dt
€ where γ is the viscous damping coefficient that occurs because of the viscosity of water. The
buoyant force does enter here because it is steady and it does not contribute to the restoring
force. However, the buoyant force does make a difference in the equilibrium length of the cable.
€
When the effects of viscosity are ignored, the in-water natural frequency is seen to be
ω b,w = σ ( M + 12 ρ wV ) = ω b 1+ ρ wV 2 M ,
which is lower than the in-vacuum natural frequency for any ρw, V, and M.
Small transverse (pendulum) oscillations in vacuum will be determined by
2
dθ g
M 2 + M θ€ = 0 , where θ is the angle of departure from the vertical, so that ω p = g L .
dt L
When the sphere is submerged in water, both added mass and buoyancy influence its pendulum
frequency because both contribute to the restoring force when θ ≠ 0. First consider the case
when the sphere sinks in the water ( M > ρ wV ). Here, the differential€equation governing small
€
d 2θ dθ g
changes in θ is M +[1
ρ
2 w
V
dt
]
2
+γ
dt
+ (M − ρwV ) θ = 0 ,
L
where again γ is the viscous € damping coefficient. Therefore, the inviscid natural frequency
# g & M − ρ wV 1− ρ wV M
estimate is ω p,w = % ( 1 = ωp .
€ $ L ' M + 2 ρ wV 1+ ρ wV 2M
If M < ρ wV , pendulum oscillations will still occur but the sphere is now a float and the cable
restrains it from rising to the water surface, thus, the governing differential equation is:
d 2θ dθ g
€
€ [
M + 12 ρ w V
dt
]2
+γ
dt
+ (−M + ρ w V ) θ = 0
L
# g & −M + ρ wV −1+ ρ wV M
so the natural frequency is ω p,w = % ( = ωp .
$ L ' M + 12 ρ wV 1+ ρ wV 2M
€ 1− ρ wV M
Combining the cases produces: ω p,w = ω p , so ω p,w = ω p 2 when M << ρ wV .
1+ ρ wV 2M
€

€ €
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.51. Determine the ideal-flow force on a stationary sphere for the following unsteady
flow conditions.
a) The free stream of velocity Uez is constant but the sphere’s radius a(t) varies.
b) The free stream velocity magnitude changes, U(t)ez, but the sphere’s radius a is constant.
c) The free stream velocity changes direction U(excosΩt+ eysinΩt), but its magnitude U and the
sphere’s radius a are constant.

Solution 7.51. a) The potential for a stationary sphere with a constant radius in a uniform flow
parallel to the z-axis is given by (7.85):
# a3 &
φ = Ur%1+ 3 ( cosθ .
$ 2r '
When the sphere's radius varies, a time dependent source term proportional to 1/r must be added
to this potential to produce the requisite radial velocity at r = a. The strength of this source term
is Q = 4πa2(da/dt), thus the appropriate time dependent potential for this situation is:
€ # a3 & a 2 da
φ = Ur%1+ 3 ( cosθ − , so
$ 2r ' r dt
2
∂φ 3 $ a 2 ' da a $ da ' a 2 d 2 a
= U& 2 ) cos θ − 2 & ) − ,
∂t 2 % r ( dt r % dt ( r dt 2
∂φ $ € a 3 ' 3 a 3 z 2 a 2 z da $ a3 ' a 2 da
∂z
= uz = U&1+ 3 ) − U 5 + 3
% 2r ( 2 r r dt % 2r
(
= U&1+ 3 1− 3cos θ ) + 2 ) 2

( r dt
cos θ ,

∂φ€ % a3 ( a 2 da 1 ∂φ & a3 )
= ur = U'1− 3 * cos θ + 2 and = uθ = −U(1+ 3 + sin θ .
∂r & r ) r dt r ∂θ ' 2r *
€ Now consider a CV that encloses the sphere and expands and contracts with it. In this
case, the control surface velocity will be b = (da/dt)er. The integral momentum equation (4.17)
becomes:
€ d d€ & da )
∫
dt CV
ρudV + ∫ ρu(u − b) ⋅ ndA = ∫
dt CV
ρudV + ∫ ρu( ur − + dA = − ∫ pe r dA + Fs .
' dt *
CS CS CS
Here the control surface is spherical so n = er, the flux integral is zero because ur = da/dt on the
control surface (r = a), and Fs is an externally applied force that holds the sphere stationary. The
ideal flow force on the sphere, FIF, will be equal and opposite to Fs so
€ π 2π
d a π 2π
FIF = −Fs = − ∫ ∫ [ p] r = a e r a 2 sin θdϕdθ − ∫ ∫ ∫ ρur 2 sin θdϕdθdr .
θ = 0ϕ = 0 dt r= 0 θ = 0 ϕ = 0

From the symmetry of the flow, the only non-zero force component is in the z-direction:
π 2π
d a π 2π
(FIF ) z = e z ⋅ FIF = − ∫ ∫ [ p] r = a cosθa 2 sinθdϕdθ − ∫ ∫ ∫ ρuz r 2 sinθdϕdθdr .
€ θ = 0ϕ = 0 dt r= 0 θ = 0 ϕ = 0
The pressure is obtained from the unsteady Bernoulli equation:
∂φ 1 1 1 ∂φ
2
ρ + ρ ∇φ + p = p∞ + ρU 2 , or p = p∞ + ρ U 2 − ∇φ − ρ .
∂t 2 2
(2
) 2

∂t
€On r = a this simplifies to:
1 % 2 % da ( 9 2 2 ( % 3 da d 2a (
2
% da ( 2
[ ] r= a ∞
p = p + ρ 'U − ' * − U sin θ * − ρ' U cos θ − 2' * − a *.
€ 2 '& & dt ) 4€ * ' 2 dt
) & & dt ) dt 2 *)

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

This can be rearranged into terms that do not lead to a net force, those in [,]-brackets below, and
the one term that does, the final term:
, 3 $ da ' 1
2
$ 9 ' d 2 a/ 3 da
[ p] r= a = . p∞ + ρ& ) + ρU 2 &1− sin 2 θ ) + ρa 2 1 − ρ U cosθ .
- 2 % dt ( 2 % 4 ( dt 0 2 dt
Therefore, the pressure contribution to the ideal-flow force is:
π 2π
da π da
− ∫ ∫ [ p] r = a e r a 2 sin θdϕdθ = 3πρUa 2 ∫ cos2 θ sin θdθ = 2πρUa 2 ,
€ θ = 0ϕ = 0 dt θ = 0 dt
which is a drag force when da/dt > 0.
Similarly, the unsteady term's contribution to the ideal-flow force is:
d a π 2π
€ − ∫ ∫ ∫ ρuz r 2 sin θ dϕ dθ dr
dt r=0 θ =0 ϕ =0
d a π # # a3 & a 2 da &
= −2πρ ∫ ∫ %U %1+ 3 (1− 3cos2 θ ) ( + 2 cosθ (sin θ dθ r 2 dr
dt r=0 θ =0 $ $ 2r ' r dt '
d a +1 # # a3 & a 2 da &
= −2πρ ∫ ∫ %U %1+ 3 (1− 3β 2 ) ( + 2 β (d βr 2 dr
dt r=0 β =−1 $ $ 2r ' r dt '
+1
d a) # a3 & a 2 da β 2 , 2 d a 2
= −2πρ ∫ + % 2r 3 (
U
dt r=0 * $
β + β − β 3
)(' r 2 dt 2 .-
+ r dr = −2 πρU
dt
∫ 2r dr
−1 r=0

4π d da
=− ρU ( a 3 ) = −4πρUa 2 .
3 dt dt
This term represents momentum change that occurs within the sphere.
Combining the pressure and internal-momentum-change forces leads to:
da
(FIF ) z = −2πρUa 2 (sphere internal flow included).
dt
which is a thrust force when da/dt > 0; the sphere is pushed upstream when its radius increases.
In this case, momentum change in the sphere's interior provides the decisive contribution to the
ideal flow force. However, if the sphere's interior is considered to be massless or motionless, or
to have zero€or constant momentum, then the unsteady term's contribution is zero, and the ideal-
flow force is just the drag that results from the pressure distribution on the sphere's surface:
da
(FIF ) z = +2πρUa 2 (sphere-internal flow excluded).
dt
b) In this case the sphere's geometry doesn't change, thus the potential is:
# a3 &
φ = U(t)r%1+ 3 ( cosθ ,
€ $ 2r '
and this leads to a ∂φ/∂t term in the pressure integral that is not present in the steady flow case.
The requisite time derivative and the velocity components are:
∂φ dU $ a3 ' ∂φ % a3 ( 1 ∂φ & a3 )
= r&1+ €
) cosθ , = ur = U'1− 3 * cos θ and = uθ = −U(1+ 3 + sin θ .
∂t dt % 2r 3 ( ∂r & r ) r ∂θ ' 2r *
This time consider a fixed-size CV that encloses the sphere. Use the integral momentum
equation (4.17) to find:

€ € €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

2π
π
d a π 2π
(FIF ) z = − ∫ [ p] r = a cosθa 2 sin θdϕdθ −
∫ ∫ ∫ ∫
dt r= 0 θ = 0 ϕ = 0
ρuz r 2 sin θdϕdθdr .
θ = 0ϕ = 0
The flux term considered in part a) is zero here because the sphere doesn't change size; its
surface is motionless (ur = 0 at r = a).
The unsteady term's contribution to the ideal-flow force is
€ d a π 2π 4 π 3 dU
− ∫ ∫ ∫ ρuz r 2 sin θdϕdθdr = − ρa ,
dt r= 0 θ = 0 ϕ = 0 3 dt
and can be obtained by repeating the part a) analysis of this term with a = const. and U = U(t).
To evaluate the pressure term, use the Bernoulli equation
∂φ 1 2
€ ρ + ρ ∇φ + p = const.
∂t 2
from the stagnation point at r = a and θ = π to any other point on the surface of the sphere to find
the pressure p(a,θ) on the surface of the sphere:
$ 3 dU ' 1 $ 9 ' 3 dU 1
ρ& a cosθ ) +€ ρ&U 2 − U 2 sin 2 θ ) + p(a,θ ) = −ρ a + ρU 2 + p(a, π ) , or
% 2 dt ( 2 % 4 ( 2 dt 2
3 dU 1 '9 *
[ p] r= a ≡ p(a,θ ) = p(a, π ) − ρ a (1+ cosθ ) + ρU 2) U 2 sin2 θ , .
2 dt 2 (4 +
€ Only the term involving the cosine leads to a net force:
π 2π π
( 3 dU + dU
− ∫ ∫ [ p] r = a cos θa 2 sinθdϕdθ = −2π ∫ ρ* − a cosθ - cosθa 2 sin θdθ = 2πρa 3 ,
€ θ = 0 ϕ = 0 θ = 0 ) 2 dt , dt
which is a drag force; the flow pushes the sphere downstream when dU/dt > 0.
Thus, as in part a), there are two possible answers:
2 dU
€ (FIF ) z = + πρa 3 (sphere internal flow included), and
3 dt
dU
(FIF ) z = +2πρa 3 (sphere-internal flow excluded).
dt
When sphere's internal flow is neglected, the resulting drag arises from two sources.
€
First, the pressure gradient in the accelerating uniform stream (∂p/∂z = –ρdU/dt) leads to a force
similar to buoyancy with dU/dt replacing the acceleration of gravity, (4/3)πρa3dU/dt, that pushes
€
the sphere toward lower pressure (downstream). Second, the accelerating free stream must go
around the sphere and this leads to an apparent mass effect, (2/3)πρa3dU/dt, that again pushes the
sphere toward lower pressure. The sum of these two effects correctly recovers the result above,
2πρa3dU/dt.
c) Although it is tempting, the switch to a rotating coordinate frame with same origin and z-axis,
such a switch is not necessarily helpful because of the kinematic relationship
[u] inertial frame = Ω × x + [u] rotating frame , where Ω is the angular rotation rate of the rotating frame.
With z-axis vertical, Ω = Ωez, x = (x, y, z), and an inertial-frame velocity of u = U(excosΩt+
eysinΩt), this kinematic relationship becomes:
€ [u] inertial frame = U(e x cosΩt + e y sinΩt) = Ω × x + [u] rotating frame = Ω(− y %e%x + x %e%y ) + u% ,
where the primes denote coordinates, velocities, and unit vectors in the rotating frame; and the
cross product has been computed in the rotating frame. Here, the x´-direction can be chosen to
simplify the inertial-frame velocity: e"x = e x cosΩt + e y sinΩt , and this leaves:
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

u" = [u] inertial frame − Ω(− y "e"x + x "e"y ) = Ue"x − Ω(− y "e"x + x "e"y ) ,
which is only a clear simplification of [u] inertial frame = U(e x cosΩt + e y sinΩt) along the z-axis in
the rotating frame where (x´, y´) = (0, 0).
Therefore,
€ use an inertial frame of reference with the z-axis vertical, so the flow far from
sphere lies in the x-y plane. In this case, the potential can be determined starting from (7.88):
€ % d(t) (
φ = ''U(t) − *
3 * ⋅ x.
& 4 π x )
where the free stream velocity changes direction, U = U(e x cosΩt + e y sinΩt) , and the dipole
strength opposes the free stream and sets the sphere's diameter, d = −2πUa 3 (e x cosΩt + e y sinΩt) .
Using spherical coordinates, € x = e x r cosϕ sin θ + e y r sin ϕ sinθ + e z r cos θ , performing the dot
products, and using the two-angle sum formula € for the cosine function allows the potential to be
written entirely in terms of scalars:
# €
a3 &
€ φ = U% r + 2 ( cosϕ ) sin θ ,
$ 2r '
where ϕ´ = ϕ – Ωt. The radial and angular velocities, and ∂φ/∂t are:
∂φ % a3 ( 1 ∂φ % a3 (
= ur = U'1− 3 * cos ϕ + sin θ , = uθ = U'1+ 3 * cosϕ + cos θ ,
∂r €& r ) r ∂θ & 2r )
1 ∂φ ' a *
3
∂φ % a3 (
= uϕ = −U)1+ 3 , sin ϕ - , and = ΩU' r + 2 * sin ϕ + sin θ .
r sinθ ∂ϕ ( 2r + ∂t & 2r )
€ As in parts a) and b), the ideal € flow force on the sphere, FIF, can be obtained from a
surface integral of the pressure, and a volume integral over the sphere's interior.
π 2π
2 d a π 2π
€ F IF = − ∫ ∫ [ p ] e
r=a r
a sin θ d€ϕ d θ −
dt
∫ ∫ ∫ ρur 2 sin θdϕdθdr .
θ = 0ϕ = 0 r= 0 θ = 0 ϕ = 0

To evaluate the pressure term, use the Bernoulli equation,

∂φ 1 2
ρ + ρ ∇φ + p = const.,
∂t 2
€
from the stagnation point at r = a, θ = π/2, and ϕ´ = π to any other point on the surface of the
sphere to find the pressure p(a,θ,ϕ #) ≡ [ p] r= a on the surface of the sphere:
3 € θ + 1 ρU 2 9 (cos 2 ϕ $ cos2 θ + sin 2 ϕ $) + p(a,θ,ϕ $) = p(a, π /2, π ) .
ρΩUasin ϕ $ sin
2 2 4
Rearrange this and denote p(a, π/2, π) by ps.
€ 9 3
p(a,θ,ϕ #) = ps − ρU 2 (1− cos2 ϕ # sin 2 θ ) − ρΩUasin ϕ # sin θ .
8 2
€
Only the final term on the right leads to a net pressure force:
π 2π
− ∫ ∫ [ p] r = a e r a 2 sin θdϕdθ
θ = 0ϕ = 0
€
π 2π
3
= ρΩUa 2 ∫ ∫ sin ϕ ) sin 2 θ (e x cosϕ sin θ + e y sin ϕ sin θ + e z cosθ ) dϕdθ .
2 θ = 0ϕ = 0

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

The θ-integration renders the z-component zero. For the other two terms, the θ-integration is
π
∫ 0 sin3 θdθ = 4 3 , so the integrated pressure force simplifies to:
π 2π 2π
− ∫ ∫ [ p] r = a e r a 2 sinθdϕdθ = 2 ρΩUa 2 ∫ sin ϕ )(e x cosϕ + e y sinϕ )dϕ .
θ = 0ϕ = 0 ϕ= 0

€ Expand sin ϕ " = sin ϕ cos(Ωt) − cos ϕ sin(Ωt) , and integrate to find:
π 2π
− ∫ ∫ [ p] r = a e r a 2 sinθdϕdθ = 2πρΩUa3 (−e x sin(Ωt) + e y cos(Ωt)) .
€ θ = 0ϕ = 0
€ This force pushes the sphere perpendicular to the instantaneous free stream direction.
To evaluate the sphere's internal-flow contribution to the force, use the volume integral
results from part a) but adjust them for the direction of the free-stream in this problem:
€ a π 2π
4π
∫ ∫ ∫ ρur 2 sin θdϕdθdr = 3 ρa 3U (e x cos(Ωt) + e y sin(Ωt)) , so that:
r= 0 θ = 0 ϕ = 0

d a π 2π 4π 3
− ∫ ∫ ∫ ρur 2 sin θdϕdθdr = − ρa ΩU (−e x sin(Ωt) + e y cos(Ωt)) .
dt r= 0 θ = 0 ϕ = 0 3
This€is the force that results from sphere's internal flow contribution.
So, as in parts a) and b), there are two possible answers:
2
€ FIF = πρΩUa 3 (−e x sin(Ωt) + e y cos(Ωt)) (sphere internal flow included), and
3
FIF = 2πρΩUa 3 (−e x sin(Ωt) + e y cos(Ωt)) (sphere-internal flow excluded).
The direction of the force can be partially explained by the pressure gradient in the accelerating
€uniform stream, ∇p = −ρ dU = − ρΩU (−e sin(Ωt) + e cos(Ωt)) , which leads to a force similar to
x y
dt
€
buoyancy with ΩU (−e x sin(Ωt) + e y cos(Ωt)) replacing the acceleration of gravity. This
buoyancy-mimicking force pushes the sphere toward lower pressure.
€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.52. Consider the flow field produced by a sphere of radius a that moves in the x-
direction at constant speed U along the x-axis in an unbounded environment of a quiescent ideal
fluid with density ρ. The pressure far from the sphere is p∞ and there is no body force. The
velocity potential for this flow field is:
a 3U x −Ut
φ (x, y, z, t) = − .
2 "(x −Ut)2 + y 2 + z 2 $3 2
# %
 
a) If u = (u, v, w) , what is u(x, 0, 0, t) , the velocity along the x-axis as a function of time? [Hint:
consider the symmetry of the situation before differentiating in all directions.]
b) What is p(x,0,0,t), the pressure along the x-axis as a
function of time? z!
c) What is the pressure on the x-axis at x = Ut ± a?
d) If the plane defined by z = h is an impenetrable flat h!
2a !
surface and the sphere executes the same motion, what y!
additional term should be added to the given potential? U!
e) Compared to the sphere's apparent mass in an x!
unbounded environment, is the sphere's apparent mass x = Ut !
larger, the same, or smaller when the impenetrable flat
surface is present?

12 a 3U (x −Ut)
Solution 7.52. a) To save writing let, r(t) = "#(x −Ut)2 + y 2 + z 2 $% , so φ = − .
2 r 3 (t)
 ∂φ a 3U # r 2 (t) − 3(x −Ut)2 &
Compute velocity components from ∇φ = u . Start with: u = =− % (,
∂x 2 $ r 5 (t) '
a 3U
evaluate this on the x-axis (y = z = 0) where r(t) = x −Ut to find: u(x, 0, 0, t) = 3
. Here φ
x −Ut
∂φ ∂φ ∂y 2 ∂φ
depends on y2 so: v = = 2 = 2y 2 ; thus, v(x,0,0,t) = 0. Similarly, φ depends on z2 so
∂y ∂y ∂y ∂y
w(x,0,0,t) = 0. Thus, the fluid velocity on the x-axis has only an x-component:

( 3
u(x, 0, 0, t) = a 3U x −Ut e x . )
∂φ 1  2 ∂φ 1  2
b) The appropriate Bernoulli equation is: ρ + ρ u + p = p∞ , so p − p∞ = −ρ − ρu .
∂t 2 ∂t 2
Time differentiate the potential but utilize the chain-rule and the part a) results:
∂φ ∂φ ∂(x −Ut) ∂φ
= = (−U ) = −Uu .
∂t ∂(x −Ut) ∂t ∂x
3 2
" ∂φ % ρa U
Evaluate on the x-axis: $ρ ' =− 3
, so:
# ∂t &( x,0,0,t ) x −Ut
ρ a 3U 2 1 a 6U 2 1 # 2a 3 a 6 &(
2
p(x, 0, 0, t) − p∞ = + U %
x −Ut
3
− ρ =
2 x −Ut 6 2
ρ % x −Ut 3 x −Ut 6 ( .
−
$ '
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

c) At x = Ut ± a the factor in large (,)-parentheses in the part b) result is unity, so p = p∞ +

(1/2)ρU2 at these locations. This answer can also be reached by noting that the specified
locations are stagnation points on the sphere when it is stationary and the flow moves past it.
d) The method of images applies to this situation, so an image sphere must move along the line
defined by z = 2h and y = 0. Thus, the additional term for the potential is:
a 3U x −Ut
− .
2 "(x −Ut)2 + y 2 + (z − 2h)2 $3 2
# %
e) The impenetrable flat surface requires the fluid to move faster above the real sphere to get out
of its way as it travels so the sphere's apparent mass is larger when the impenetrable flat surface
is present. Here it should be noted that apparent mass is a property of the sphere's geometry, the
geometry of its environment, and the fluid density. It exists independently of the sphere's motion.
When the sphere accelerates both its actual and apparent mass play a role. If the sphere does not
accelerate then neither its actual nor its apparent mass plays a role. However, the sphere does not
loose its actual mass when moving at a constant velocity, so it does not lose its apparent mass
either.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 7.53. In three dimensions, consider a solid object moving with velocity U near the
origin of coordinates in an unbounded quiescent bath of inviscid incompressible fluid with
density ρ. The kinetic energy of the moving fluid in this situation is:
1 2
KE = ρ ∫ ∇φ dV
2 V
where φ is the velocity potential and V is a control volume that contains all of the moving fluid
but excludes the object. (Such a control volume is shown in the figure for Exercise 7.29 when
A3 → 0 and U = 0.)
€
1
a) Show that KE = − ρ ∫ φ (∇φ ⋅ n)dA where A encloses the body and is coincident with its
2 A
surface, and n is the outward normal on A.
€ 1 2
b) The apparent mass, M, of the moving body may be defined by KE = 2 M U . Using this
definition,
€ the result of a), and (7.97) with xs = 0, show that M = 2πa3ρ/3 for a sphere.

Solution 7.53. Start from the given equation, use the

recommended CV, add φ∇ 2φ = 0 to the integrand, and apply€
Green's divergence theorem.
ρ 2 ρ
KE = lim ∫ ∇φ dV = lim ∫ (∇φ ⋅ ∇φ )dV
2 A 3 →0 V 2 A 3 →0 V
€
ρ ρ
= lim ∫ (∇φ ⋅ ∇φ + φ∇ 2φ )dV = lim ∫ ∇ ⋅ (φ∇φ )dV
2 A 3 →0 V 2 A 3 →0 V
3
ρ
= lim ∑ ∫ φ∇φ ⋅ ni dA
2 A 3 →0 i=1 A i
where n1 is the normal on A1, n2 is the normal on A2, and n3 is the normal on A3. When the limit is
taken, the net contribution of A3 is zero because the surface area of A3 goes to zero. This leaves:
ρ ρ
€ KE = ∫ φ∇φ ⋅ n1dA + ∫ φ∇φ ⋅ n2 dA ,
2 A1 2 A2
where n1 points away from the origin on A1 and n2 points inward on A2. By definition, the
volume V enclosed by A1 contains all the moving fluid, thus ∇φ ⋅ n1 = u ⋅ n1 = 0 on A1, so the first
integral in the last equation is equal to zero. Now define the outward normal on A2 as n = – n2,
and drop the subscript € from the body-conforming surface A2 to reach:
ρ
KE = − ∫ φ€ ∇φ ⋅ ndA .
2A
b) The potential for a sphere of radius a moving at velocity U that is instantaneously centered at
the origin of a spherical coordinate system is:
a3 a3
€ φ = − 3 U ⋅ x = − 2 U cosθ ,
2r 2r
where the second equality follows when the direction of the z-axis is chosen to coincide with the
sphere's velocity at the moment of interest. For a sphere, n = er and surface area element is dA =
a2sinθdθdϕ, so the kinetic energy integral becomes:
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

ρ π 2π & ∂φ ) 2 ρ π 2π & a 2 2 )
KE = − ∫ ∫ (φ + a sin θ d θd ϕ = − ∫ ∫ (− U cos θ+ a 2 sin θdθdϕ
2 θ = 0 ϕ = 0 ' ∂r *r= a 2 θ = 0ϕ = 0' 2 *r= a
ρa 3 2 π 2 ρa 3 2 −1 2 ρa 3 21 2 4
=π U ∫ cos θ sin θdθ = −π U ∫ β dβ =π U 3 6
2 θ=0 2 1 2 2 35
1 1 2πa 3 4 2
= 3ρ 6U .
22 3 5
The parentheses in the final equality contain the apparent mass M.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.1. Starting from (8.5) and working in (x,y,z) Cartesian coordinates, determine an
equation that specifies the locus of points that defines a wave crest. Verify that the travel speed
of the crests in the direction of K = (k, l, m) is c = ω/|K|. Can anything be determined about the
wave crest travel speed in other directions?

Solution 8.1. The phase of the waveform in (8.5) is K ⋅ x − ωt . Thus wave crests are specified
by:
K ⋅ x crest − ωt = 2nπ ,
since cos(2nπ) = 1. In (x,y,z) Cartesian coordinates € with K = (k, l, m), this is the equation for a
series of planes:
kx crest + ly crest + mzcrest = 2nπ + ωt , (*)
€
so (8.5) describes plane waves. The unit vector perpendicular to these planes, e ⊥ , is determined
from the gradient of the phase:
∇ (K ⋅ x − ωt ) K
€ e⊥ = = = eK .
∇ (K ⋅ x − ωt ) K €
To determine the travel speed of the crests, time differentiate equation (*) to find:
d d d
k x crest + l y crest + m zcrest = ω ,
dt dt dt
€
since k, l, m, and 2nπ are constants. In terms of a dot product, this last equation is:
d K d d ω
K ⋅ x crest = ω , or ⋅ x crest = e K ⋅ x crest = = c.
€ dt K dt dt K
Thus, the travel speed of wave crests, dxcrest/dt, in the direction of K is c.
And, the phase relationship (8.5) does not provide any information about the travel speed
in other directions.
€ However, it is not possible to determine the wave speed other directions.
Consider the following hypothesis: €
d
x crest = ce ⊥ + de|| ,
dt
where d the crest-parallel wave speed, and e|| is a unit vector that is parallel to the wave crests
with e ⊥ ⋅ e|| = 0 = e K ⋅ e|| . However,
e ⋅ ∇(K ⋅ x − ωt ) = K e|| ⋅ e K = 0 ;
€ ||
thus, there are no changes in the phase of the wave (as represented by ∇(K ⋅ x − ωt ) ) in the crest-
€ parallel direction(s). Hence it is impossible to determine anything about the crest-parallel wave
speed d by observing the wave.
€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.2. For ka << 1, use the potential for linear deep water waves,
φ (z, x,t) = a(ω k )e kz sin(kx − ωt) and the waveform η(x,t) = acos( kx − ωt ) + αka 2 cos[2( kx − ωt )]
to show that:
a) with an appropriate choice of the constant α, the kinematic boundary condition (8.16) can be
satisfied for terms proportional to (ka)0 and (ka)1 once the common factor of aω has been divided
€ €
out, and
b) with an appropriate choice of the constant γ, the dynamic boundary condition (8.19) can be
satisfied for terms proportional to (ka)0, (ka)1, and (ka)2 when ω2 = gk(1 + γk2a2) once the
common factor of ag has been divided out.

Solution 8.2. a) First determine the horizontal (u) and vertical (w) velocities from the given
velocity potential:
∂φ aω kz
u= = e [ k cos(kx − ωt)] = aωe kz cos(kx − ωt) ,
∂x k
∂φ aω kz
w=
∂z
=
k
[ ]
ke sin(kx − ωt) = aωe kz sin(kx − ωt) ,
and the derivatives of the surface waveform:
€
∂η
= ωasin( kx − ωt ) + 2ωαka 2 sin[2( kx − ωt )] ,
∂t
€ ∂η
= −kasin( kx − ωt ) − 2αk 2 a 2 sin[2( kx − ωt )] .
∂x
$ ∂φ ' ∂η ∂η $ ∂φ '
Use these results€and the surface waveform in (8.16), & ) = + & ) :
% ∂z ( z= η ∂t ∂x % ∂x ( z= η
€ − ωt) = ωasin( kx − ωt ) + 2ωαka sin[2( kx − ωt )]
aωe kη sin(kx 2

( )
+ −kasin( kx − ωt ) − 2αk 2 a 2 sin[2( kx − ωt )] aωe kη cos(kx − ωt).
€
To simplify, let β = kx – ωt, and divide out common factors to find:
e kη sin β = sin β + 2αkasin(2β ) − ( kasin β + 2αk 2 a 2 sin(2β ))e kη cos β .
€ Since matching up through terms proportional to ka is necessary, the exponential function should
be expanded through its first-order term, e kη ≅ 1+ kacos β for ka << 1, so

€
(1+ kacos β ) sin β = sin β + 2αkasin(2β ) − (kasin β + 2αk 2 a 2 sin(2β ))(1+ kacos β ) cos β .
0 1
Perform the various multiplications, and group terms that are proportional to (ka) and (ka) .
€ β ) = sin β + ka(2α sin(2β ) − sin β cos β ) + k a (...) .
2 2
sin β + ka(cos β sin
The (ka)0 terms match. The (ka)1 terms will match if:
€ cos β sin β = 2α sin(2β ) − sin β cos β → 2cos β sin β = 2α sin(2β ) ,
and this equation is satisfied when α = 1/2.
€
b) First determine the time derivative of the velocity potential and use the notation of part a):
∂φ aω kz aω 2 kz
€ = e [−ω cos(kx − ωt)] = − e cos β .
∂t k k
After using the Bernoulli equation, the dynamic boundary condition is:
$ ∂φ 1 2 ' aω 2 kη ρ
& + u + gz ) = 0 → − e cos β + ( a 2ω 2e 2kη cos2 β + a 2ω 2e 2kη sin 2 β ) + gη = 0 .
% ∂t 2 ( z= η k 2
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Substitute in the linear approximation for e kη ≅ 1+ kacos β , the part a) result for η = acosβ +
αka2cos(2β), and ω2 = gk(1 + γk2a2) to find:
agk(1+ γk 2 a 2 ) a2
− (1+ kacos β )cos β + gk(1+ γk 2 a 2 )(1+ 2kacos β )
k € 2
+ g( acos β + αka 2 cos(2β )) = 0.
Cancel common factors and simplify:
ka
−(1+ γk 2 a 2 )(1+ kacos β )cos β +(1+ γk 2 a 2 )(1+ 2kacos β ) + (cos β + αkacos(2β )) = 0 .
2
€Perform the various multiplications, and group terms that are proportional to (ka)0, (ka)1 and
(ka)2.
% 1 (
€ −cos β + cos β + ka'−cos 2 β + + α cos(2β )* + k 2 a 2 (−γ cos β + cos β ) + k 3 a 3 (...) = 0 .
& 2 )
The three terms inside the larger parentheses sum to zero when α = 1/2 based on trigonometric
identities, and the k2a2 term is zero when γ = 1.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.3. The field equation for surface waves on a deep fluid layer in two dimensions (x,z)
is: ∂ 2φ ∂ x 2 + ∂ 2φ ∂ z 2 = 0 , where φ is the velocity potential, ∇φ = (u, w) . The linearized free-
surface boundary conditions and the bottom boundary condition are:
(∂φ ∂ z)z=0 ≅ ∂η ∂ t , (∂φ ∂ t )z=0 + gη ≅ 0 , and (∂φ ∂ z )z→−∞ = 0 ,
where z = η(x,t) defines the free surface, gravity g points downward along the z-axis, the
undisturbed free surface lies at z = 0. The goal of this problem is to develop the general solution
for these equations without assuming a sinusoidal form for the free surface as was done in
Section 8.1and 8.2.
a) Assume φ (x, z, t) = Λ(x, t)Z(z) , and use the field equation and bottom boundary condition to
show that φ (x, z, t) = Λ(x, t)exp(+kz) , where k is a positive real constant.
b) Use the results of part a) and the remaining boundary conditions to show:
∂ 2Λ ∂ 2Λ 2
+ gkΛ = 0 and +k Λ = 0.
∂t 2 ∂ x2
c) For a fixed value of k, find Λ(x,t) in terms of four unknown amplitudes A, B, C, and D.
d) For the initial conditions: η = h(x) and ∂η/∂t = h˙ (x) at t = 0, determine the general form of
φ(x, z, t).

Solution 8.3. a) Insert φ (x.z,t) = Λ(x,t)Z(z) into the field equation ∂ 2φ ∂x 2 + ∂ 2φ ∂z 2 = 0 and
1 ∂ 2Λ 1 d 2Z
divide by φ to find: + = 0 . The first term depends only on x and t while the second
Λ ∂x 2 Z dz 2
depends only € on z; thus each must be constant to ensure the equation is satisfied for all possible
(x, z, t). Therefore set the first term equal to –k2 and the €second term equal to +k2. Then, the
equation involving Z(z) is d 2 Z dz 2 − k 2 Z = 0 which has the solution: Z = A+e+kz + A–e–kz, where
€
A± are constants. The bottom boundary condition requires dZ dz → 0 as z → −∞ , so the A– must
be zero. Therefore, φ (x.z,t) = Λ(x,t)e+kz , and the constant A+ has been incorporated into Λ.
b) Substituting the
€ result of part a) into the first surface boundary condition produces:
( )
(∂φ ∂z) z=0 = Λke kz€ = Λk = ∂η€∂t .
z=0
Substituting
€ the result of part a) into the second surface boundary condition produces:
( )
(∂φ ∂t ) z=0 + gη = e kz (∂Λ /∂t) + gη = ∂Λ ∂t + gη = 0
z=0
Take the partial derivative
€ of the last equality with respect to time and use the result of the first
surface boundary condition to eliminate ∂η/∂t to find: ∂ 2 Λ ∂t 2 + gkΛ = 0 . From part a) the
equation involving
€ Λ, is ∂ 2 Λ ∂x 2 + k 2 Λ = 0 .
c) Assume a separation-of-variables form for Λ = X(x)T(t). The two equations derived for part b)
then imply: d 2T dt 2 + gkT = 0 , and 2 2 2
€ d X dx + k X = 0 .
( ) ( )
€ has solutions: sin t gk and cos t gk , while the second has solutions sin(kx)
The first equation
and cos(kx). Allowing for all four possibilities for Λ then implies:
€ €

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

( ) ( )
Λ(x,t) = A# sin(kx)sin t gk + B# sin(kx)cos t gk + C # cos(kx)sin t gk + D# cos(kx)cos t gk . ( ) ( )
Unfortunately, this form hides the wave-propagation character of the solution, but with the use of
trigonometric angle-sum and angle-difference formulae for the sine and cosine functions, the
above form for Λ can be written:
€
( ) ( )
Λ(x,t) = Asin kx − t gk + Bcos kx − t gk + C sin kx + t gk + Dcos kx + t gk , ( ) ( )
where 2A = B´ – C´, 2B = A´ + D´, 2C = B´ + C´, and 2D = –A´ + D´.
d) The above form for Λ is only valid when k ≥ 0, so k may take on any non-negative value.
Thus, the final solution must be of the form:
€ $ A(k)sin kx − t gk + B(k)cos kx − t gk + (
&
+∞
φ (x,z,t) = ∫ %
( ) ( & kz
)e dk .
)
0 &
' C(k)sin kx + t gk + (
D(k)cos kx + t gk &
* ) ( )
Now combine the initial and boundary conditions. The first boundary condition evaluated at t = 0
implies: (∂φ ∂z)(z,t)=0 = (∂η ∂t ) t=0 = h˙ (x) , or
+∞
€
h˙ (x) = ∫ {[ A(k) + C(k)] sin( kx ) + [B(k) + D(k)] cos(kx )}kdk .
0
The second boundary condition evaluated at t = 0 implies: (∂φ ∂t )(z,t)=0 = (−gη) t=0 = −gh(x) , or
€
+∞

€
∫ gk {[−A(k) + C(k)] cos( kx ) + [B(k) − D(k)] sin( kx )}dk
−gh(x) =
0
The final formulae for A, B, C, and D are found from considerations of symmetry about x = 0:
+∞ €
h˙ (x) − h˙ (−x) = ∫ 2k [ A(k) + C(k)] sin(kx ) dk ,
0
€ +∞
h˙ (x) + h˙ (−x) = ∫ 2k [ B(k) + D(k)] cos( kx )dk ,
0
+∞
€ −g( h(x) + h(−x)) = ∫ gk [−A(k) + C(k)] cos( kx ) dk , and
0
+∞
€ −g( h(x) − h(−x)) = ∫ gk [ B(k) − D(k)] sin( kx ) dk
0
and can be€evaluated in formal terms via inverse sine and cosine transforms.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.4. Derive (8.37) from (8.27).

Solution 8.4. Use the velocity components specified in (8.27), and the definition of the stream
function in two-dimensions (x, z):
∂ψ cosh( k(z + H)) ∂ψ sinh( k(z + H))
u= = aω cos(kx − ωt) , and w = − = aω sin(kx − ωt) .
∂z sinh( kH ) ∂x sinh( kH )
Integrate each equation to find:
aω sinh( k(z + H)) aω sinh( k(z + H))
ψ= cos(kx − ωt) + f (x) , and ψ = cos(kx − ωt) + g(z) .
€ k sinh( kH ) € k sinh ( kH )
where f and g are single-variable functions of integration. These are consistent when f(x) = g(y) =
constant, and this constant can be set to zero without loss of generality. The final result is (8.37):
aω sinh( k(z + H))
€ ψ= € cos(kx − ωt) .
k sinh( kH )

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.5. Consider stationary surface gravity waves in a rectangular container of length L
and breadth b, containing water of undisturbed depth H. Show that the velocity potential φ =
Acos(mπx/L)cos(nπy/b)cosh[k(z + H)]e–iωt, satisfies the Laplace equation and the wall boundary
conditions, if (mπ/L)2 + (nπ/b)2 = k2. Here m and n are integers. To satisfy the linearized free-
surface boundary condition, show that the allowable frequencies must be ω2 = gk tanh(kH).
[Hint: combine the two boundary conditions (8.18) and (8.21) into a single equation ∂2φ/∂t2 = −g
∂φ/∂z at z = 0.]

Solution 8.5. Start from φ = Acos( mπx L) cos( nπy b) cosh[ k(z + H)]e−iω t and compute
velocities.
∂φ
u= = −A( mπ L) sin( mπx L) cos( nπy b) cosh[ k(z + H)]e−iω t
∂x
which implies u€= 0 at x = 0 and x = L, so the x-direction BCs are satisfied. Similarly,
∂φ
v= = −A ( nπ b) cos ( mπ x L ) sin ( nπ y b) cosh [ k(z + H )] e−iω t
€ ∂y
which implies v = 0 at y = 0 and y = b, so the y-direction BCs are satisfied. The Laplacian is:
∂ 2φ ∂ 2φ ∂ 2φ & m 2π 2 n 2π 2 )
2
+ 2 + 2 = (− 2 − 2 + k 2 + Acos( mπx L) cos( nπy b) cosh[ k(z + H)]e−iω t ,
∂x ∂y ∂z ' L b *
m 2π 2 n 2π 2
and it equals zero when: k 2 = + 2 .
L2 b
The linearized free surface boundary conditions are:
€
∂η ∂t = ∂φ ∂z and ∂φ ∂t = −gη at z = 0,
which can be combined to get:
€
∂ 2φ ∂t 2 = −g ∂η ∂t = −g ∂φ ∂z at z = 0.
Using the given potential, this requires
€ €
[ A(−iω )2 cos(mπx L) cos(nπy b) cosh[k(z + H)]e−iω t ]z= 0
,
€ = −g[ Ak cos( mπx L) cos( nπy b) sinh[ k(z + H)]e−iω t ] z= 0
or −ω 2 cosh[ kH ] = −gk sinh[ kH ] which can be written: ω 2 = gk tanh[ kH ] .

€
€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.6. A lake has the following dimensions L = 30 km, b = 2 km, and H = 100 m. If the
wind sets up the mode m = 1 and n = 0, show that the period of the oscillation is 32 min.

Solution 8.6. Use the results of Exercise 8.5:

m 2π 2 n 2π 2
k 2 = 2 + 2 and ω 2 = gk tanh[ kH ] ,
L b
and eliminate k to find:
$ m 2π 2 n 2π 2 '1 2 *$ m 2π 2 n 2π 2 '1 2 -
ω 2 = g& 2 + € ) tanh ,& 2 + 2 ) H/.
€ % L b2 ( ,+% L b ( /.
Thus, the frequency ω10 of the m = 1, n = 0 mode is given by:
π $π ' π $ π (100) '
ω10 = g tanh& H ) = (9.81) 3
tanh& 3)
= 3.27 ×10−3 rad /s ,
€ L %L ( 30 ×10 % 30 ×10 (
so its period T10 is:
2π
T10 = = 32 minutes.
€ ω 10

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.7. Fill a square or rectangular cake pan half-way with water. Do the same for a round
frying pan of about the same size. Agitate the water by carrying the two pans (outside) while
walking briskly at a consistent pace on a horizontal surface.
a) Which shape lends itself better to spilling?
b) At what portion of the perimeter of the rectangular pan does spilling occur most readily?
c) Explain your observations in terms of standing wave modes.

Solution 8.7. a) For the pans tested by the third author, the water usually spilled out of the square
or rectangular pan first.
b) The water almost always left the square or rectangular pan from a corner.
c) All of the resonant modes of the rectangular pan have maximum vertical surface displacement
at the pan's corner (this is not true of the circular pan since it has no corners). When side-to-side
and front-to-back waves combine in the rectangular pan they can constructively interfere in the
corners of the pan and cause spilling.

This exercise provides is a simple demonstration of how tides may affect nearly-straight
and jagged coastlines. When an inlet or bay with interior corners (analogous to the square pan) is
excited at its resonant frequency by tidal action (analogous to your random shaking), the tidal
fluctuations can be enormous. The extreme tides (low-to-high change of 14 m or more) of the
Bay of Fundy on southern coast of Newfoundland are an extreme example. Smooth coastlines do
not experience such large tidal swings.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.8. Use (8.27), (8.28), and (8.38), to prove (8.39).

Solution 8.8. Equation (8.38) specifies a double integration,

ρ λ 0 2
Ek = ∫ ∫ (u + w 2 )dzdx ,
2 λ 0 −H
(8.38)

involving the two velocity components given by (8.27):

cosh ( k(z + H )) sinh ( k(z + H ))
u = aω cos(kx − ω t) , and w = aω sin(kx − ω t) . (8.27)
sinh ( kH ) sinh ( kH )
Substituting (8.27) into (8.38) produces:
λ 0
ρ a 2ω 2 2
Ek = 2 ∫ cos (kx − ω t)dx ∫ cosh 2 ( k(z + H ))dz
2 λ sinh (kH ) 0 −H
λ 0
ρ a 2ω 2
+ 2 ∫ sin (kx − ω t)dx ∫ sinh 2 ( k(z + H ))dz
2

2 λ sinh (kH ) 0 −H
The two x-integrations produce λ/2 since the average values of sin2() and cos2() are both 1/2.
This leaves the integrations of the hyperbolic functions.
0
1 0 2k (z+H )
∫ cosh (k(z + H ))dz = 4 ∫ (e
2
+ 2 + e−2k (z+H ) )dz
−H −H
0
e 2kH 2kz 1 0 e−2kH 0
−2kz
=
4
∫e dz + ∫
2 −H
dz +
4
∫e dz
−H −H
0 0
e 2kH ! e 2kz $ H e−2kH ! e−2kz $
= # & + + # &
4 " 2k %−H 2 4 " −2k %−H
e 2kH ! 1 e−2kH $ H e−2kH ! 1 e+2kH $
= # − &+ − # − &
4 " 2k 2k % 2 4 " 2k 2k %
1 1
= ( e 2kH −1+ 4kH − e−2kH +1) = ( e 2kH + 4kH − e−2kH )
8k 8k
0 0
1
∫ sinh 2 (k(z + H ))dz = 4 ∫ (e2k (z+H ) − 2 + e−2k (z+H ) )dz
−H −H
0
e 2kH 1 0 e−2kH 0
= ∫ e 2kz dz − ∫ dz + ∫e −2kz
dz
4 −H 2 −H 4 −H

#e & 2kH 2kz 0

#e & −2kH −2kz 0
e H e
= % ( − + % (
4 $ 2k '−H 2 4 $ −2k '−H
e 2kH # 1 e−2kH & H e−2kH # 1 e+2kH &
= % − (− − % − (
4 $ 2k 2k ' 2 4 $ 2k 2k '
1 1
= ( e 2kH −1− 4kH − e−2kH +1) = ( e 2kH − 4kH − e−2kH )
8k 8k
Thus, wave kinetic energy becomes:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

ρ a 2ω 2 ! λ $ 1 2kH
# & ( e + 4kH − e + e 2kH − 4kH − e−2kH )
−2kH
Ek = 2
2 λ sinh (kH ) " 2 % 8k
ρ a 2ω 2 ! 1 $ 2kH ρ a 2ω 2 ! 1 $! e kH + e−kH $! e kH − e−kH $
= # &
4sinh 2 (kH ) " 8k %
( 2e − 2e −2kH
) = # &#
4sinh 2 (kH ) " k %" 2
&#
%" 2
&
%
ρ a 2ω 2 ! 1 $ ρ a 2ω 2
= # & cosh(kH )sinh(kH ) =
4sinh 2 (kH ) " k % 4k tanh(kH )
Now use the dispersion relationship (8.28)
ω = gk tanh(kH ) (8.28)
and recognize that η 2 = a2/2:
ρ a 2 ( gk tan(kH )) 1
Ek = = ρ gη 2 ,
4k tanh(kH ) 2
and this is (8.39).
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.9. Show that the group velocity of pure capillary waves in deep water, for which the
gravitational effects are negligible, is cg = (3/2)c.

Solution 8.9. For surface waves in the presence of gravity and surface tension, the dispersion
relationship is given by (8.56):
% σk 2 (
ω = k' g + * tanh( kH ) .
& ρ )
When gravity is negligible, this simplifies to:
σk 3
ω= tanh( kH ) .
€ ρ
The group velocity in this case is:
∂ω ∂ 1 & 3σk 2 σk 3 )
cg = =
∂k ∂k
(σk 3 ρ ) tanh( kH ) = ( tanh( kH ) + sech 2 ( kH )+.
€ 2 (σk 3 ρ) tanh( kH ) ' ρ ρ *
For deep water, kH is large enough so that
tanh(kH) ≈ 1, and sech(kH) << 1.
With these simplifications the group velocity becomes:
€ % 3σk 2 ( 3 σk
1
cg ≅ ' *= .
2 (σk 3 ρ ) & ρ ) 2 ρ
For pure capillary waves in deep water, the phase speed c is:
ω 1 σk
c= ≅ σk 3 ρ = ,
€ k k ρ
so the group speed reduces to: cg = (3/2)c.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.10. Assuming deep water, plot the group velocity of surface gravity waves, including
surface tension σ, as a function of λ for water at 20 °C (ρ = 1000 kg/m3 and σ = 0.074 N/m).
1 g 1+ 3σ k 2 ρ g
a) Show that the group velocity is cg = .
2 k 1+ σ k 2 ρ g
σ k2 2
b) Show that this becomes minimum at a wave number given by = −1.
ρg 3
c) Verify that cg min = 17.8 cm/s.

% σk 2 (
Solution 8.10. a) Start from (8.56): ω = k' g + * tanh( kH ) . In deep water tanh(kH) ≈ 1, so
& ρ )
this simplifies to: ω ≅ gk + σk 3 ρ . The group velocity in this case is:
∂ω ∂ 1 & 3σk 2 ) 1 g 1+ 3σk 2 ρg
cg = = gk + σk 3 ρ = (g + += .
∂k ∂k € 2 gk + σk 3 ρ ' ρ * 2 k 1+ σk 2 ρg
This relationship
€ is plotted here; group speed is on the vertical axis in m/s and wavelength is on
the horizontal axis in cm.
€ !#,"

!#+"

!#*"

!#)"

!#("

!#'"

!#&"

!#%"

!#$"

!"
!" %!" '!" )!" +!" $!!" $%!" $'!" $)!" $+!" %!!"

b) The minimum group speed is found by differentiating cg with respect to k, setting this
derivative equal to zero, solving for kmin, and then evaluating cg at kmin.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

% 2(
∂c g ∂ % g 1+ 3σk 2 ρg ( g ' (6σk ρg)( k + σk ρg) − 2 (1+ 3σk ρg) *
3 1 2

0= = ' *= .
∂k ∂k '& 2 k + σk 3 ρg *) 2 ' ( k + σ k 3
ρg)
32
*
& )
Simplify and set x = σk ρg to find:
2

2
0 = (6σk ρg)( k + σk 3 ρg) − 12 (1+ 3σk 2 ρg) = 6x (1+ x ) − 12 (1+ 3x ) or 3x2 + 6x – 1 = 0.
2

€The final quadratic implies: x min = −1+ 2 3 = 0.1547. The corresponding wave number is:
€
kmin = ρgx min σ = 10 3 ⋅ 9.81⋅ 0.1547 0.074 = 143.2 /m .
€ c) Thus:
1 g 1+ 3x min 1 9.81 1+ 3(0.1547)
(c €)
g min = =
2 kmin 1+ x min 2 143.2 1+ 0.1547
= 0.178m /s .
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.11. The energy propagation characteristics of sinusoidal deep-water capillary-gravity

waves follow those of pure gravity waves when corrections are made for the influence of surface
tension. Assume a waveform shape of η(x,t) = acos(kx – ωt) for the following items.
a) Show that the sum of the wave's kinetic and potential energies (per unit surface area) can be
written: E = Ek + E p = 2 ρ a g where g = g + k 2σ ρ .
1 2

b) Determine the wave energy flux EF (per unit length) from

<Begin Equation>
2π ω & 0 )
ω
EF = ∫ ( ∫ p!udz − σ t ⋅ [ u ] z=0 + dt ,
2π 0 ' −∞ *
</End Equation>
where t is the surface tangent vector, and the extra term involving σ represents energy transfer
via surface tension.
c) Use the results of parts a) and b) to show that EF = Ecg, where cg is the group velocity
determined in Exercise 8.10 part a).

Solution 8.11. a) The kinetic and potential energies of sinusoidal deep-water capillary gravity
waves can be written:
# 2 &
ρ λ 0 2 ρg λ η σ λ% # ∂η &
Ek = ∫ ∫ (u + w )dzdx , and E p = λ ∫ ∫ z dz dx + λ ∫ % 1+ %$ ∂x (' −1(( dx ,
2 λ 0 −H
2

0 0 0 $ '
where the second term in the potential energy is the surface tension contribution. Using (8.47)
for u and w in kinetic energy integration leads to:
ρ 2 2 λ 0 2kz
Ek = a ω ∫ ∫ e ( cos2 (kx − ω t) + sin 2 (kx − ω t))dzdx
2λ 0 −∞

ρ 2 2 0 2kz ρ 2 2 ρ 2 $ σ k3 '
= aω ∫e = aω = a & kg + )
2 −∞ 4k 4k % ρ (
ρ ga 2 σ k 2 a 2 ρ $ σ k 2 ' 2
= + = g &1+ )a ,
4 4 4 % ρg (
where the dispersion relationship has been used to eliminate ω2. The potential energy integration
can also be performed for small slopes (ka << 1):
# &
σ λ # 1 # ∂η & &
2 2
ρg λ η σ λ% # ∂η & ( ρg λ 2
Ep = ∫∫
λ 0 0
z dz dx + ∫
λ 0 %$
1+ % (
$ ∂x '
−1
(
dx ≅
λ
∫ η dx +
λ
∫ % 2 %$ ∂x (' (( dx
%
' 0 0 $ '
ρg λ 2 2 σ λ 2 2 2
=
2λ 0
∫ a cos (kx − ω t)dx +
2λ 0
∫ a k sin (kx − ωt)dx
ρ ga 2 σ 2 2 ρ # σ k 2 & 2
= + a k = g %1+ (a
4 4 4 $ ρg '
Thus, the total wave energy per unit area is:
ρ ! σ k2 $ 2 ρ ! σ k2 $ 2 1 2
& a = ρ a g where g = g (1+ k σ ρ g) .
2
E = Ek + E p = g #1+ & a + g #1+
4 " ρg % 4 " ρg % 2
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

b) From the problem statement, the energy flux crossing any fixed x-location is:
2π ω & 0 )
ω
EF = ∫ ( ∫ p!udz − σ t ⋅ [ u ] z=0 + dt .
2π 0 ' −∞ *
The pressure fluctuation p´ is given by (8.30) so the linearized Bernoulli equation implies:
∂φ ∂$ ω ' ω2
p! = −ρ = −ρ & a e+kz sin(kx − ω t)) = ρ a e+kz cos(kx − ω t) .
∂t ∂t % k ( k
where the form for φ is obtained from (8.26) in the limit kH → ∞ . The horizontal velocity in this
case is:
∂φ ∂ # ω +kz &
u= = % a e sin(kx − ω t)( = aω e+kz cos(kx − ω t) ,
∂x ∂x $ k '
and is also given by (8.47). Thus, the vertical integral is:
0 0 % (
ω 2 +kz
∫ p!udz = ∫ '& k e cos(kx − ωt)*) (aω e+kz cos(kx − ωt))dz
ρ a
−∞ −∞
0
ω3 2 2 2 ω
3
= ρa +2kz
cos (kx − ω t) ∫ e dz = ρ a 2
cos2 (kx − ω t)
k −∞ 2k
Before moving to the time average, consider the surface tension term. Here, the unit
vector tangent to the surface is:
1
t=
2
(e x + (∂η ∂x ) e z ) where ∂η/∂x = –aksin(kx – ωt),
1+ (∂η ∂x )
and the fluid velocity at the surface can be obtained from (8.47) evaluated at z = 0:
[u]z=0 = aω e x cos(kx − ω t) + aω e z sin(kx − ω t) .
Thus,
−σ
−σ t ⋅ [ u ] z=0 =
1+ a k2 2 2
sin (kx − ω t)
(aω cos(kx − ωt) − a kω sin (kx − ωt))
2 2

≅ −σ ( aω cos(kx − ω t) − a 2 kω sin 2 (kx − ω t))

and the approximation is valid when ka << 1, which is the case here.
At any location, the time average of cos(kx – ωt) is zero, while the time average of
cos2(kx – ωt) and sin2(kx – ωt) are both 1/2. Thus,
ω 3 σ a 2 kω
EF = ρ a 2 2 + .
4k 2
c) To show that EF = Ecg, where cg is the group velocity determined in Exercise 8.10 part a),
show that ratio EF/E leads to the group velocity.
! 2 $ ! σ k2 $! 1 ! σ k2 $ σ k $
2 ω
3
σ a 2 kω ω # ω + σ k & k #g + &# k #g + &+ &
ρa + 2
ρ % " 2k 2 " ρ % ρ %
EF 4k 2
2 " 2k ρ % "
= = = ,
E ρ ! σ k2 $ 2 ! σ k2 $ ! σ k2 $
g #1+ &a #g + & #g + &
2 " ρg % " ρ % " ρ %
where the third equality comes from using the dispersion relationship (8.56) in the limit of
kH → ∞ . Continue simplifying
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

! g σk σk$ ! 1 3σ k $ ! σ k2 $
k# + + & kg # + & #1+ 3 &
EF " 2k 2 ρ ρ % " 2k 2 ρ g % 1 g" ρg %
= = = .
E σ k2 σ k2 2 k σ k2
g+ 1+ 1+
ρ ρg ρg
The final form is the group velocity specified by Exercise 8.10 part a).
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.12. The effect of viscosity on the energy of linear deep-water surface waves can be
determined from the wave motion’s velocity components and the viscous dissipation (4.58).
a) For incompressible flow, the viscous dissipation of energy per unit mass of fluid is
ε = 2(µ ρ )Sij2 where Sij is the strain rate tensor and µ is the fluid's viscosity. Determine ε using
(8.47).
b) The total wave energy per unit surface area, E, for a linear sinusoidal water wave with
amplitude a is given by (8.42). Assume that a is function time, set dE/dt = –ε, and show that a(t)
€ = a0exp[–2(µ/ρ)k2t] where a0 is the wave amplitude at t = 0.
c) Using a nominal value of µ/ρ = 10–6 m2/s for water, determine the time necessary for an
amplitude reduction of 50% for water-surface waves having λ = 1 mm, 1 cm, 10 cm, 1 m, 10 m,
and 100 m.
d) Convert the times calculated in c) to travel distances by multiplication with an appropriate
group speed. Remember to include surface tension. Can a typhoon located near New Zealand
produce increased surf on the coast of North America? [The circumference of the earth is
approximately 40,000 km].

Solution 8.12. a) First calculate the stress tensor using the velocity components specified by
(8.47): u = aωe kz cos(kx − ωt) and w = aωe kz sin(kx − ωt) .
1 # ∂u ∂u & ) ∂u ∂z + ∂w ∂x ),
1
2(
∂u ∂x
Sij = %% i + j (( = + 1 .
2 $ ∂x j ∂x i ' * 2 (∂w ∂x + ∂u ∂z) ∂w ∂z -
€ € )−sin(kx − ωt) cos(kx − ωt),
= aωke kz + ..
* cos(kx − ωt) sin(kx − ωt) -
Now compute the viscous kinetic-energy dissipation per unit mass:
ε = 2(µ ρ ) Sij2 = 2(µ ρ )( S112 + S122 + S21
2 2
+ S22 ) = 2(µ ρ)a2ω 2k 2e 2kz (2cos2 (kx − ωt) + 2sin2 (kx − ωt))
= 4 (µ ρ€) a 2ω 2 k 2e 2kz .
b) From (8.42) E = (1/2)ρga2. The energy dissipation per unit volume is ρε, so the dissipation per
unit surface area is obtained by integrating ρε through the depth. Therefore:
€ dE 0 0 ( 2kz +0
2 2 2 e E
= − ∫ ρεdz = −4 µa ω k ∫ e dz = −4 µa ω k * - = −2µa 2ω 2 k = −4 µk 2 ,
2 2 2 2kz

dt −∞ −∞ ) 2k ,−∞ ρ
2
where the deep-water dispersion relationship ω = gk has been used to reach the final form.
Integrating the differential equation represented by the extreme ends of this extended equality
produces:
€
E = E 0 exp{−4 (µ ρ ) k 2 t} , or in terms of wave amplitude: a = a0 exp{−2(µ ρ ) k 2 t},
where E0 and a0 are the wave energy and amplitude, respectively, at t = 0.
c) For a factor of two amplitude decrease, the time t1/2 will be:
2(µ ρ ) k 2 t1 2 = ln(2) → t1 2 = ln(2) 2(µ ρ) k 2 = ln(2) λ2 8π 2 (µ ρ ) .
€ €
2
For the given viscosity/density ratio, ln(2) λ2 8π 2 (µ ρ ) = 8.78 ×10 3 ( λ m)
λ = 1 mm, 1 cm, 10 cm, 1 m, 10 m, 100 m.
3 5
t1/2 = €8.78 ms 0.878 s 87.8 s 8.78 ×10 s 8.78 ×10 s 8.78 ×10 7 s
d) The group speeds can be calculated from the results of exercise 8.10. Let d1/2 be the distance
traveled in deep water for a€factor of two amplitude decrease.
€ € €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

λ = 1 mm, 1 cm, 10 cm, 1 m, 10 m, 100 m.

cg = 1.02 m/s 0.311 m/s 0.212 m/s 0.625 m/s 1.98 m/s 6.25 m/s
d1/2 = 8.96 mm 0.273 m 18.6 m 5.49 km 1.74 Mm 549 Mm
(where Mm = mega-meters). The circumference of the earth is only 40 Mm, so a typhoon off the
coast of New Zealand that generates waves with wavelengths of approximately 30 m or longer is
able to produce increased surf on the coast of North America.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.13. Consider a deep-water wave train with a Gaussian envelope that resides near x =
0 at t = 0 and travels in the positive-x direction. The surface shape at any time is a Fourier
superposition of waves with all possible wave numbers:
+∞
$ 1/ 2 '

−∞
% (
η(x,t) = ∫ η˜ (k)exp&i kx − ( g k ) t )dk ,
( ) (†)
where η˜ (k) is the amplitude of the wave component with wave number k, and the dispersion
relation is ω = (gk)1/2. For the following items assume the surface shape at t = 0 is:
1 & x2 )
€ η(x,0) = exp'− 2 + ikd x * .
€ 2πα ( 2α +
Here, α sets the initial horizontal extent of the wave train, with larger α producing a longer wave
train.
a) Plot Re{η(x,0)} for €
|x| ≤ 40 m when α = 10 m and kd = 2π/λd = 2π/10 m–1.
+∞
b) Use the inverse Fourier transform at t = 0, η˜ (k) = (1 2π ) ∫ −∞ η(x,0)exp[−ikx ] dx , to find the
wave amplitude distribution: η˜ (k) = (1 2π ) exp{− 12 (k − kd ) 2 α 2 } , and plot this function for 0 < k <
2kd using the numerical values from part a). Does the dominant contribution to the wave activity
come from wave numbers near kd for the part a) values?
€
c) For large x and t, the integrand of (†) will be highly oscillatory unless the phase
1/ 2 €
Φ ≡ kx − ( g k ) t happens to be constant. Thus, for any x and t, the primary contribution to η will
come from the region where the phase in (†) does not depend on k. Thus, set dΦ/dk = 0, and
solve for ks (= the wave number where the phase is independent of k) in terms of x, t, and g.
d) Based on the result of part b), set ks = kd to find the x-location where the dominant portion of
€ the wave activity occurs at time t. At this location, the ratio x/t is the propagation speed of the
dominant portion of the wave activity. Is this propagation speed the phase speed, the group
speed, or another speed altogether?

Solution 8.13. a) Below is a plot of Re{η(x,0)} vs. x for |x| ≤ 40 m when α = 10 m and kd = 2π/λd
= 2π/10 m–1.
"#")%

"#"$%

"#"&%

"#"'%

"#"(%

"%
!$"% !&"% !'"% !("% "% ("% '"% &"% $"%
!"#"(%

!"#"'%

!"#"&%

!"#"$%
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

b) Start from the given formula for η˜ (k) , and insert the given initial wave form η(x,0):
1 +∞ 1 +∞ 1 ( x2 +
η˜ (k) =
2π −∞
∫ η(x,0)exp[−ikx ]dx = ∫
2π −∞ 2πα
exp)− 2 + ikd x , exp[−ikx ]dx .
* 2α -
€ 1
Complete the square in the exponent, and use the integration variable β =
2α
[ ]
x − α 2i(k d − k)
to find:
€ +∞
1 & 1 )
η˜ (k) = ∫ exp'(− 2α 2 [ x 2 − 2α 2i(kd − k)x + α 4 (kd − k) 2 − α 4 (kd − k) 2 ]*+dx
2π 2πα −∞ €
1 +∞ & 1 2 α 2
(k d − k)
2)
= ∫ exp
2π 2πα −∞ ( 2α 2
' − [ x − α 2
i(k d − k) ] −
2
*dx
+
1 & α 2 (kd − k) 2 ) +∞
1 & α 2 (k d − k) 2 )
= exp'− * 2α ∫ exp{−β 2 }dβ = exp'− *,
2π 2πα ( 2 + −∞ 2π ( 2 +
where the final definite integral over b is π . Below is a plot of η˜ (k) vs. k for when α = 10 m
and kd = 2π/λd = 2π/10 m–1 ≈ 0.628 m–1. Clearly the wave-packet carries most of its energy in
wave numbers near kd, and η˜ (k) is negligible outside 0.2 < k < 1.0.
€
!#(&" € €
!#(%"
€
!#($"

!#("

!#!'"

!#!&"

!#!%"

!#!$"

!"
!" !#$" !#%" !#&" !#'" (" (#$" (#%"
c) From the plot above, only positive wave numbers will matter because η˜ (k) is small away
from kd, so the absolute value can be dropped from k in the formula for the phase. Thus, the
stationary phase wave number ks is determined from:
d d 1 g
dk dk
( 1/ 2

s
)
(Φ) k= ks = kx − (gk ) t k= k = x − € t = 0 .
2 ks
2 2
Solving for ks produces: ks = gt 4 x .
d) Thus, at time t, the dominant portion of the wave packet plotted in part a) will reside at a
1
location x determined
€ from: ks = k d = gt 2 4 x 2 , and this location is: x = 2 g k d t . Therefore, the
propagation speed,
€ x/t, of the dominant portion of the wave packet is the group speed,
1
c g = 2 g kd .
€ €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.14. Show that the vertical component of the Stokes’ drift is zero.

Solution 8.14. Start with the vertical-direction particle velocity equation (8.84b) expanded to
include first order-variations in both fluid velocity components.
dz p (t) $ ∂w ' $ ∂w '
= w(x p ,z p ,t) = w(x 0 ,z0 ,t) + ξ & ) + ζ& ) + ...
dt % ∂x ( x 0 ,z0 % ∂z ( x 0 ,z0
where ξ = xp – x0, and ζ = zp – z0, and (x0, z0) is the fluid element location in the absence of wave
motion. The vertical component of the Stokes' drift will be the time average of this equation over
one wave period (indicated here by angle brackets).
€
dz p (t) $ ∂w ' $ ∂w ' $ ∂w ' $ ∂w '
= w(x 0 ,z0 ,t) + ξ & ) + ζ& ) + ... = 0 + ξ & ) + ζ& ) + ...
dt % ∂x ( x 0 ,z0 % ∂z ( x 0 ,z0 % ∂x ( x 0 ,z0 % ∂z ( x 0 ,z0
kz
From (8.47), w = aωe kz sin(kx − ωt) so aωe sin(kx − ωt) = 0 and this is reflected in the second
equality above. The two time averages involving displacements and velocity gradients can be
€ calculated using the deep-water ka << 1 results for ξ, ζ, u, and w given by (8.46) and (8.47):
€ ξ ≅ −ae kz0 sin(kx 0 − ωt) , ζ ≅ ae kz0 cos(kx 0 − ωt) , and u = aωe kz cos(kx − ωt) ,
€ Substituting in these relationships produces:
plus the result for w given above.
dz p (t) %∂ (
= 0 −ae kz0 sin(kx 0 − ωt)' aωe kz sin(kx − ωt)*
€ dt € & ∂x€ ) x 0 ,z0
%∂ (
+ ae kz0 cos(kx 0 − ωt)' aωe kz sin(kx − ωt)* + ...
& ∂z ) x 0 ,z0
Simplify and consolidate terms.
dz p (t)
= −a 2ωke 2k0 sin(kx 0 − ωt)cos(kx 0 − ωt) + a 2ωke 2kz0 cos(kx 0 − ωt)sin(kx − ωt) + ...
dt
€
1
= a 2ωke 2kz0 [− sin(2kx 0 − 2ωt) + sin(2kx 0 − 2ωt) ] = 0.
2
Thus, the vertical component of Stokes' drift is zero.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.15. Extend the deep water Stokes’ drift result (8.85) to arbitrary depth to derive
(8.86).

Solution 8.15. Start with the horizontal-direction particle velocity equation (8.84a) expanded to
include first order-variations in both fluid velocity components.
dx p (t) $ ∂u ' $ ∂u '
= u(x p ,z p ,t) = u(x 0 ,z0 ,t) + ξ & ) + ζ& ) + ... ,
dt % ∂x ( x 0 ,z0 % ∂z ( x 0 ,z0
where ξ = xp – x0, and ζ = zp – z0, and (x0, z0) is the fluid element location in the absence of wave
motion. The Stokes' drift will be the time average of this equation over one wave period
(indicated here by angle brackets).
€
dx p (t) $ ∂u ' $ ∂u ' $ ∂u ' $ ∂u '
= u(x 0 ,z0 ,t) + ξ & ) + ζ& ) + ... = 0 + ξ & ) + ζ& ) + ...
dt % ∂x ( x 0 ,z0 % ∂z ( x 0 ,z0 % ∂x ( x 0 ,z0 % ∂z ( x 0 ,z0
cosh( k(z + H))
From (8.28), u = aω cos(kx − ωt) so u = 0 because cos(kx − ωt) = 0 , and this
sinh( kH )
€ is reflected in the second equality above. The two time averages involving displacements and
velocity gradients can be calculated using the deep-water ka << 1 results for ξ, ζ, u, and w given
by (8.27) and (8.35): € €
€
cosh( k(z0 + H)) sinh( k(z0 + H))
ξ ≅ −a sin(kx 0 − ωt) , ζ ≅ a cos(kx 0 − ωt) , and
sinh( kH ) sinh( kH )
sinh( k(z + H))
w = aω sin(kx − ωt) .
sinh( kH )
plus
€ the result for u given above. Substituting
€ in these relationships produces:
dx p (t) cosh( k(z0 + H)) %∂ cosh( k(z + H)) (
= 0 −a sin(kx 0 − ωt)' aω cos(kx − ωt)*
dt € sinh( kH ) & ∂x sinh( kH ) ) x 0 ,z0

sinh( k(z0 + H)) %∂ cosh( k(z + H)) (

+ a cos(kx 0 − ωt)' aω cos(kx − ωt)* + ...
sinh( kH ) & ∂z sinh( kH ) )x 0 ,z 0

Simplify and consolidate terms.

dz p (t) 2 cosh 2 ( k(z0 + H)) 2 2 sinh 2 ( k(z0 + H))
= a ωk 2
cos (kx 0 − ωt) + a ωk 2
cos 2 (kx 0 − ωt) + ...
€ dt sinh ( kH ) sinh ( kH )

2
$cosh 2 ( k(z0 + H)) 1 sinh 2 ( k(z0 + H)) 1 '
= a ωk& 2
+ ) + ...
% sinh ( kH ) 2 sinh 2 ( kH ) 2(
a 2ωk
=
2sinh 2 ( kH )
[cosh2 (k(z0 + H)) + sinh2 (k(z0 + H))] + ...
a 2ωk
= cosh(2k(z0 + H)) + ...
2sinh 2 ( kH )
The final result is (8.86).

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.16. Explicitly show through substitution and differentiation that (8.88) is a solution of
(8.87).

Solution 8.16. Equation (8.87) involves time and space differentiations of (8.88). Let
12
γ = ( 3a 4H 3 ) to save a little writing. The necessary derivatives are:
∂η ∂ % 12 ( 2asinh{γ (x − ct)}(−γc) sinh{γ (x − ct)}
∂t ∂t & {
= 'a cosh 2 ( 3a 4H 3 ) (x − ct) * = −
) } 3
cosh {γ (x − ct)}
= 2aγc
cosh 3 {γ (x − ct)}
,

€ ∂η ∂ & a ) 2asinh{γ (x − ct)}(γ ) sinh{γ (x − ct)}

= ( 2 +=− 3
= −2aγ ,
∂x ∂x ' cosh {γ (x − ct)}* cosh {γ (x − ct)} cosh 3 {γ (x − ct)}
€ ∂ 2η ∂ & sinh{γ (x − ct)} ) & sinh 2 {γ (x − ct)} )
2 cosh{γ (x − ct)}
= (−2aγ + = −2a γ ( − 3 +
∂x 2 ∂x ' cosh 3 {γ (x − ct)}* 3
'cosh {γ (x − ct)} cosh 4 {γ (x − ct)}*
€ & cosh 2 {γ (x − ct)} − 3sinh 2 {γ (x − ct)}) & 2 )
2 1− 2sinh {γ (x − ct)}
= −2aγ 2 ( 4 + = −2a γ ( 4 +,
' cosh {γ (x − ct)} * ' cosh {γ (x − ct)} *
and
∂ 3η & 2
2 ∂ 1− 2sinh {γ (x − ct)}
)
= −2a γ ( +
€ ∂x 3 ∂x ' cosh 4 {γ (x − ct)} *
& −4 sinh{γ (x − ct)} cosh{γ (x − ct)} (1− 2sinh 2 {γ (x − ct)}) sinh{γ (x − ct)})
3
= −2aγ ( −4 +
(' cosh 4 {γ (x − ct)} cosh 5 {γ (x − ct)} +*
& 2
(1− 2sinh 2 {γ (x − ct)}) sinh{γ (x − ct)})
3 sinh{γ (x − ct)} cosh {γ (x − ct)}
= 8aγ ( + +
(' cosh 5 {γ (x − ct)} cosh 5 {γ (x − ct)} +*
&sinh{γ (x − ct)}( 3 − cosh 2 {γ (x − ct)}) )
3
= 8aγ ( +
(' cosh 5 {γ (x − ct)} +*
It is now clear that the arguments of all the hyperbolic functions are the same, so, to save writing,
continue with the implicit notation that {} = {γ(x – ct)}. The next step is to start reconstructing
(8.87), the Korteweg-DeVries equation. Consider the first two terms together:
€
∂η ∂η sinh{ } c sinh{ }
+ c0 = 2aγ (c − c 0 ) 3
= a 2γ 0 ,
∂t ∂x cosh { } H cosh 3 { }
where the final equality follows from the propagation speed relationship c = c 0 (1+ a 2H ) . The
third and fourth terms are:
3c 0γa 2 sinh{ } 4c 0 H 2 3 &sinh{ }( 3 − cosh { }) )
2
3 η ∂η € c 0 H 2 ∂ 3η
c0 + =− + aγ ( +
2 H ∂x 6 ∂x 3 H cosh 5 { } 3 (' € {}
cosh 5
+*
& 3c γa 2 4c 0 H 2 , 3a / ) sinh{ } 4c 0 H 2 , 3a / & sinh{ } )
= (− 0 + . 13aγ + − . 1 aγ ( +,
' H 3 - 4H 3 0 * cosh 5 { } 3 - 4H 3 0 'cosh 3 { }*

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

$ 3c γa 2 c 0 2 ' sinh{ } 2 c0
$ sinh{ } '
= &− 0 + 3a γ ) 5
− a γ & )
% H H ( cosh { } H % cosh 3 { }(
c 0 $ sinh{ } '
= 0 − a 2γ& )
H %cosh 3 { }(
Thus, adding together all four terms produces:
∂η ∂η 3 η ∂η c 0 H 2 ∂ 3η 2 c 0 sinh{ } 2 c0
& sinh{ } )
+ c0 + c0 + = a γ − a γ ( + = 0,
€ ∂t ∂x 2 H ∂x 6 ∂x 3 H cosh 3 { } H 'cosh 3 { }*
which shows that (8.88) is a solution of (8.87).

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.17. A sinusoidal long-wavelength shallow-water wave with amplitude A and wave
number k1 = 2π/λ1 travels to the right in water of depth H1 until it encounters a mild depth
transition at x = 0 to a slightly shallower depth H2. A portion of the incident wave continues to
the right with amplitude B and a portion is reflected and propagates to the left.
a) By requiring the water surface deflection and the horizontal volume flux in the water column
to be continuous at x = 0, show that B A = 2 1+ H 2 H1 .( )
b) If a tsunami wave starts with A = 2.0 m in water 5 km deep, estimate its amplitude when it
reaches water 10 m deep if the ocean depth change can be modeled as a large number of discrete
depth changes. [Recommendation: consider multiple steps of H2/H1 = 0.90, since (0.90)59 ≈
0.002 = (10 m)/(5 km)].
c) Redo part b) when the wave's energy flux remains constant at both depths when λ1 = 100 km
(as was done in Example 8.6).
d) Compare the results of parts b) and c). Which amplitude in shallow water is lower and why is
it lower?
A! z! B!

x!
H1! H2!

Solution 8.17. The surface elevation on the left will involve two waves:
{ (
η (x, t) = A exp ik1 x − t gH1 )} + C exp {ik (−x − t
1 gH1 )} for x < 0,
where C is the amplitude of the reflected wave. The surface elevation on the right will only
involve the transmitted wave with amplitude B:
{ (
η (x, t) = B exp ik2 x − t gH 2 )} for x > 0.
To determine the relationships between wave amplitudes, match the surface elevations and the
volume flows at x = 0. Start with the surface elevation:
η (0 −, t) = η (0 +, t) or ( A + C ) exp {−ik1t } {
gH1 = B exp −ik2 t gH 2 . }
Here, the wave frequencies must be the same. This implies:
ω = k1 gH1 = k2 gH 2 and A + C = B . (1,2)
Aω
From (8.51), the horizontal velocity of the incident wave is u =
k1H1
{ (
exp ik1 x − t gH1 )}
so the horizontal volume flux (per unit depth perpendicular to the page) is
Aω
q = uH1 =
k1
{ ( )}
exp ik1 x − t gH1 . Thus, continuity of the volume flux at x = 0 requires:

Aω Cω Bω
q(0 −, t) = q(0 +, t) or − = , (3)
k1 k1 k2
where (1) has been used to eliminate the complex exponentials and the reflected wave carries a
minus sign because it propagates in the negative-x direction. Now, use (2) and (3) to eliminate C
and reach:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

−1 −1
! k1 $ B ! k1 $ ! H2 $
2A = # +1& B or = 2 #1+ & = 2 ##1+ & .
" k2 % A " k2 % " H1 &%
where the final equality follows from (1).
b) For one step with H2/H1 = 0.90, B/A = 1.026334. For 59 such steps, (B/A)59 = (1.026334)59 =
4.635. Thus the wave amplitude will be approximately 9.27 m in 10 m of water. This result must
be approximate since the wave amplitude is now comparable to the water depth and this does not
satisfy the requirements of the linear theory.
c) Follow the approach in Example 8.6. First, determine the wave speed and ka value when H =
5 km:
c = gH = (9.81ms−2 )(5, 000m) = 221.5 ms–1, ka = 2πa/λ = 2π(2.0m)/(100km) = 1.257 ×10 −4 .
Here a/H = 4.0 ×10 −4 and ka are both small enough to use linear wave results. So, from (8.44),
the energy flux, EF, for these shallow water waves is:
ρ ga 2 (10 3 kgm −3 )(9.81ms−2 )(2.0m)2
EF = c= (221.5ms−1 ) = 4.346 MW/m.
2 2
The wave period here is λ/c = (100km)/(221.5 ms–1) = 451.5s, and this will be the same (or
nearly so) when the wave train reaches shallower water where its speed will be
c = gH = (9.81ms−2 )(10m) = 9.90 ms–1.
The wavelength in shallow water will be (9.90 ms–1)(451.5s) = 4.47 km. For constant energy
flux, the wave amplitude at the shallower depth will be:
12 12
! 2EF $ ! 2 ⋅ 4.346 ×10 6 Wm −1 $
a =# & =# 3 −3 −2 −1 &
= 9.46 m,
" ρ gc % " (10 kgm )(9.81ms )(9.90ms ) %
Here again, the amplitude-depth ratio is too large for the linear theory; a nonlinear theory is
needed to better determine the wave shape and its amplitude.
d) The answer to part b) is slightly lower because some wave energy is reflected at each depth
transition, while no such energy reflection is accounted for in part c).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.18. A thermocline is a thin layer in the upper ocean across which water temperature
and, consequently, water density change rapidly. Suppose the thermocline in a very deep ocean
is at a depth of 100 m from the ocean surface, and that the temperature drops across it from 30 to
20 °C. Show that the reduced gravity is gʹ′ = 0.025 m/s2. Neglecting Coriolis effects, show that
the speed of propagation of long gravity waves on such a thermocline is 1.58 m/s.

Solution 8.18. This is the case of an internal wave at the interface between a shallow layer
overlying an infinitely deep fluid (Section 8.7). Here the layer depth is 100 m, the temperature
difference is ΔT = 10°C, and the average temperature is 25°C so the thermal expansion
coefficient α is approximately 2.5 ×10−4 /°C (from Appendix A3). The relative density change
will be Δρ/ρ = αΔT = 2.5 ×10−3 . Thus, the reduced gravity and wave speed can be determined
from (8.115) and (8.116):
€9.81(0.0025) = 0.0245m /s , and c = 0.0245(100) = 1.58m /s .
2
g" = gαΔT =
€

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.19. Consider internal waves in a continuously stratified fluid of buoyancy frequency
N = 0.02 s–1 and average density 800 kg/m3. What is the direction of ray paths if the frequency of
oscillation is ω = 0.01 s–1? Find the energy flux per unit area if the amplitude of the vertical
velocity is wˆ = 1 cm/s and the horizontal wavelength is π meters.

Solution 8.19. Here, N = 0.02 s–1, ω = 0.01 s–1, k = 2π/λ = 2 m–1, wˆ = 1 cm/s, and ρo = 800
3
kg/m
€ . From (8.138), ω = Ncosθ, solve for the angle:
θ = cos−1 (ω N ) = cos−1 (0.01 0.02) = 60°.
To determine the energy flux, calculate the vertical wave number:
€
m = ktanθ = 2tan(60°) = 3.464 m–1,
and use (8.158)
€
ρ 0ωmwˆ 2 % m ( 800(0.01)(3.464)(0.01) 2 % 3.464 (
F= ' e x − e z * = ' e x − e z *
2k 2 & k ) 2(2) 2 & 2 )
= 3.464 ×10−4 (1.732e x − e z )W /m 2 .
The magnitude of the energy flux is:
12
F = 3.464 ×10−4 (1.732 2 + 1) = 6.92 ×10−4 W /m 2 .
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.20. Consider internal waves at a density interface between two infinitely deep fluids,
and show that the average kinetic energy per unit horizontal area is Ek = (ρ2 − ρ1)ga2/4.

Solution 8.20. From Section 8.7, the velocity potentials above and below the interface are:
iωa −kz iωa +kz
φ1 = e exp{i(kx − ωt)} , and φ 2 = − e exp{i(kx − ωt)} .
k k
respectively. The (real) velocity components are:
$ ∂φ ' $ iωa '
u1 = Re& 1 ) = Re& (ik)e−kz exp{i(kx − ωt)}) = −ωae−kz cos(kx − ωt) ,
% ∂x ( % k (
€ €
$ ∂φ ' $ iωa '
u2 = Re& 2 ) = Re&− (ik)e +kz exp{i(kx − ωt)}) = ωae +kz cos(kx − ωt) ,
% ∂x ( % k (
€ $ ∂φ ' $ iωa '
w1 = Re& 1 ) = Re& (−k)e−kz exp{i(kx − ωt)}) = ωae−kz sin(kx − ωt) , and
% ∂z ( % k (
€ $ ∂φ ' $ iωa '
w 2 = Re& 2 ) = Re&− (k)e +kz exp{i(kx − ωt)}) = ωae +kz sin(kx − ωt) .
% ∂z ( % k (
The
€ kinetic energy per unit surface are then involves integrating through the depth:
1 ∞
1 0 2
KE = ρ1 ∫ ( u1 + w1 )dz + ρ 2 ∫ ( u2 + w 22 )dz
2 2

€ 2 0 2 −∞
1 ∞ 1 0
= ρ1 ∫ (ω 2 a 2 cos2 () + ω 2 a 2 sin 2 ())e−2kz dz + ρ 2 ∫ (ω 2 a 2 cos2 () + ω 2 a 2 sin 2 ())e +2kz dz
2 0 2 −∞
ω 2 a 2 ∞ −2kz ω 2 a 2 0 +2kz ω 2a2
= ρ1 ∫ e dz + ρ 2 ∫ e dz = ( ρ1 + ρ2 ).
2 0 2 −∞ 4k
where () = kx – ωt. The dispersion relationship for these waves is (8.95)
%ρ −ρ (
ω 2 = gk' 2 1 * .
€ & ρ 2 + ρ1 )
Use this to eliminate ω from the final form for the kinetic energy per unit surface area to find:
ω 2a2 % ρ 2 − ρ1 ( a 2 ( ρ 2 − ρ1 )ga 2
KE = (ρ1 + ρ2 ) = gk' * ( ρ1 + ρ 2 ) = .
4k € & ρ 2 + ρ1 ) 4k 4

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.21. Consider waves in a finite layer overlying an infinitely deep fluid. Using the
constants given in (8.105) through (8.108), prove the dispersion relation (8.109).

Solution 8.21. Equations (8.105)-(8.108) are:

ia $ ω g ' ia $ ω g ' ia $ ω g ' ia $ ω g ' k
A = − & + ) , B = & − ) , C = − & + ) − & − )e 2kH , and b = i Ce−kH ,
2 % k ω( 2 % k ω( 2 % k ω( 2 % k ω( ω
respectively. Applying BC (8.100)
∂φ ∂φ
ρ1 1 + ρ1gζ = ρ 2 2 + ρ 2 gζ at z = –H.
∂t ∂t €
€ € €
leads to:
( ) ( )
ρ1(−iω ) Ae−kH + Be +kH + ρ1gb = ρ 2 (−iω ) Ce−kH + ρ 2 gb .
Substituting for b € in terms of C produces:
% gk (
−ωρ1 ( Ae−kH + Be +kH ) = Ce−kH '−ωρ 2 + ( ρ 2 − ρ1 )* .
& ω )
€ 2
Now eliminate A, B, and C using (8.105)-(8.107) and let x = ω /gk to ease the algebra. These
steps lead to:
"ρ − ρ $
1 (−(x +1)e + (x −1)e+kH ) = e−kH "#(x +1)e−kH + (x −1)e+kH $%& 2 1 − ρ2 ' .
−kH
ρ€
# x %
This can be rearranged and stated in terms of hyperbolic functions:
(x −1){ x [ ρ1 sinh(kH) + ρ 2 cosh(kH)] − (ρ 2 − ρ1 )sinh(kH)} = 0 ,
which is (8.109).

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.22. A simple model of oceanic internal waves involves two ideal incompressible
fluids (ρ2 > ρ1) trapped between two horizontal surfaces at z = h1 and z = –h2, and having an
average interface location of z = 0. For traveling waves on the interface, assume that the
interface deflection from z = 0 is ξ = ξ o Re{exp(i(ωt − kx))} . The phase speed of the waves is c =
ω/k.
gk( ρ 2 − ρ1 )
a) Show that dispersion relationship is: ω 2 = where g is the
ρ 2 coth(kh2 ) + ρ1 coth(kh1 )
€
acceleration of gravity.
b) Determine the limiting form of c for short (i.e. unconfined) waves, kh1 and kh2 → ∞.
c) Determine the limiting form of c for long (i.e. confined) waves, kh1 and kh2 → 0.
€
d) At fixed wavelength λ (or fixed k = 2π/λ), do confined waves go faster or slower than
unconfined waves?
e) At a fixed frequency, what happens to the wavelength and phase speed as ρ2 – ρ1 → 0?
f) What happens if ρ2 < ρ1?

Solution 8.22. a) For this problem there at two velocity potentials that must be matched at the
moving interface. Based on the form of the interface wave, ξ = ξ o Re{exp(i(ωt − kx))} and the
development given in the chapter, the form of the two potentials can be set:
φ1 (x,z,t) = Z1 (z)e i(ωt−kx ) , and φ 2 (x,z,t) = Z 2 (z)e i(ωt−kx )
Here the boundary conditions and the field equation are:
∇ 2φ = 0 : Z1 (z) = A1+e +kz + A1−€ e−kz , and Z 2 (z) = A2+e +kz + A2−e−kz
∂φ ∂ξ
= on €z = 0: kA1+ − kA1– €= iωξ o = kA2+ − kA2– (1,2)
∂z ∂t
∂φ ∂φ
€ ρ1 1 + ρ1gξ = ρ 2 €2 + ρ 2 gξ on z = 0: €
∂t ∂t
€ iωρ1 ( A1+ + A1− ) + gρ1ξ o = iωρ 2 ( A2+ + A2− ) + gρ 2ξ o (3)
€ $ ∂φ1 ' $ ∂φ '
& ) =& 2) =0 kA1+e +kh − kA1– e−kh = 0 , and kA1+e−kh − kA1– e +kh = 0
1 1 2 2
(4,5)
€ % ∂z ( z= h % ∂z ( z=−h
1 2

The three boundary € conditions yield five equations, enough to determine all four A’s and the
dispersion relationship. First use the (1) & (4), and (2) & (5) to find:
€ ωξ e  kh1 € ωξ e ±kh2
€ A1,± = −i o , and A2,± = +i o
2k sinh( kh1 ) 2k sinh( kh2 )
Plug these into (3), to find:
% ωξ cosh(kh ) ( % ωξ cosh(kh ) (
ωρ1' o 1
* + gρ1ξ o = −ωρ 2 ' o 2
* + gρ 2ξ o
€ & k sinh( 1) ) €
kh & k sinh( 2) )
kh

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Cancel the common factor of ξo, simplify and rearrange:

gk( ρ 2 − ρ1 )
ω2 = .
ρ1 coth(kh1 ) + ρ 2 coth(kh2 )
ω g( ρ 2 − ρ1 ) k
b) Using the results of part a), the phase speed is c = = ± .
k ρ1 coth(kh1 ) + ρ 2 coth(kh2 )
For short waves, kh1 and € kh2 → ∞, and the hyperbolic co-tangent functions both approach unity
so:
ω g( ρ 2 − ρ1 )
c€ = =±
k k( ρ1 + ρ 2 )
c) For long waves, kh1 and kh2 → 0, and the hyperbolic co-tangent functions both approach the
ω g( ρ 2 − ρ1 ) k g( ρ 2 − ρ1 )h1h2
inverse of their arguments: c = = ± =±
€ k ρ1 (1/kh1 ) + ρ 2 (1/kh2 ) ρ1h2 + ρ 2 h1
Note that when ρ1 → 0, the answers to parts b) and c) both recover the ordinary linear water
wave results.
d) For simplicity, take h1 = h2 = h, then for short (unconfined) waves:
€
g( ρ2 − ρ1 ) 1
c=± ,
( ρ1 + ρ2 ) k
and for long (confined) waves:
g( ρ 2 − ρ1 ) g( ρ 2 − ρ1 ) kh
c =± h =± .
ρ1 + ρ 2 ρ1 + ρ 2 k
For k = const., as kh → 0 , the confined waves travel slower than the unconfined waves with the
same wavelength (same wave number k).
e) As ρ2 – ρ1 → 0, the phase speed and wavelength both go to zero.
€
f) If ρ2 €
< ρ1, then the two fluid layers are not stably stratified. The dispersion relationship then
gk( ρ1 − ρ 2 )
requires ω to be imaginary, i.e. ω = ±i . This means that there is an
ρ 2 coth(kh2 ) + ρ1 coth(kh1 )
interface wave solution that is growing exponentially with increasing time. The situation is
unstable, but not much more can be determined from the linearized theory. In reality, the two
fluids will switch places but the linearized theory considered here is not valid throughout this
process. €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.23. Consider the long-wavelength limit of surface and interface waves with
amplitudes a and b, respectively, that occur when two ideal fluids with densities ρ1 and ρ2 (> ρ1)
are layered as shown in Figure 8.28. Here, the velocity potentials – accurate through second
order in kH1 and KH2 – in the two fluids are:
φ1 ≅ A1 (1+ 12 k 2 (z − z1 )2 ) ei(kx−ωt ) and φ2 ≅ A2 (1+ 12 k 2 (z − z2 )2 ) ei(kx−ωt ) for kH1, KH2 << 1,
where b, A1, z1, A2, and z2 are constants to be found in terms of a, g, k, ω, H1 and H2. From
analysis similar to that in Section 8.7, find the dispersion relationship ω = ω(k). When H1 = H2 =
H/2, show that the surface and interface waves are in phase with a > b for the barotropic mode,
and out-of-phase with b > a for the baroclinic mode. For this case, what are the phase speeds of
the two modes? Which mode travels faster? What happens to the baroclinic mode's phase speed
and amplitude as ρ2 − ρ1 → 0 ?
z! a! z! a!
x! x!
ρ1!
H1!

ρ2! H2! b! b!

Barotropic! Baroclinic!
Figure 8.28.

Solution 8.23. The form of the two potentials are given in the problem statement, and the surface
and interface elevations from equilibrium can be taken from (8.101) and (8.102):
η = aei(kx−ωt ) and ζ = bei(kx−ωt ) .
The boundary conditions are:
∂φ2/∂z = 0 on z = –(H1 + H2) (i)
∂φ1/∂z = ∂η/∂t on z = 0 (ii)
∂φ1/∂t + gη = 0 on z = 0 (iii)
∂φ1/∂z = ∂φ2/∂z = ∂ζ/∂t on z = –H1 (iv)
ρ1∂φ1/∂t + ρ1gζ = ρ2∂φ2/∂t + ρ2gζ on z = –H1 (v)
Apply these in turn produces the following algebraic equations:
(i): A2k2(z – z2) = 0 on z = –(H1 + H2), or A2k2(–H1 – H2 – z2) = 0
(ii): A1k2(z – z1) = –iωa on z = 0, or A1k2z1 = iωa
(iii) −iω A1 (1+ 12 k 2 (z − z1 )2 ) + ga = 0 on z = 0, or iω A1 ≅ ga
(iv) A1k2(z – z1) = A2k2(z – z2) = –iωb on z = –H1, or A1k2(H1 + z1) = A2k2(H1 + z2) = iωb
(v) −iωρ1 A1 (1+ 12 k 2 (z − z1 )2 ) + ρ1gb = −iωρ2 A2 (1+ 12 k 2 (z − z2 )2 ) + ρ2 gb on z = –H1, or
−iωρ1 A1 + ρ1gb ≅ −iωρ2 A2 + ρ2 gb
where the second order terms involving [k.(vertical distance)] appearing in (iii) and (v) have
2

been dropped compared to unity in the equations on the right. Noting that (iv) provides two
equations, the first five equations on the right form a non-linear algebraic system that can be
solved to find:
iaω " 1 gkH % " gk 2 H1 %
z2 = –(H1 + H2), z1 = –ω2/k2g, A1 = –iga/ω, A2 = − $ − 2 1 ' , and b = a $1− '.
k # kH 2 ω H 2 & # ω2 &
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Substitute these into the final approximate equation provided by (v):

" −iga % " gk 2 H1 % " −iaω %" 1 gkH1 % " gk 2 H1 %
−iωρ1 $ + ρ
' 1 $ ga 1− ' ≅ −iωρ 2$ ' $ − ' + ρ 2 ga $1− '.
# ω & # ω2 & # k &# kH 2 ω 2 H 2 & # ω2 &
Divide out the common factor of a, and simplify:
g 2 k 2 H1 −ρ2ω 2 # 1 gkH1 & g 2 k 2 H1
−ρ1 ≅ % − ( + ρ 2 g − ρ 2 , or
ω2 k $ kH 2 ω 2 H 2 ' ω2
ρ ω2 # H & g 2 k 2 H1
0 ≅ − 22 + ρ2 g %1+ 1 ( − ( ρ2 − ρ1 )
k H2 $ H2 ' ω2
Multiply –ω2k2H2/ρ2 and rearrange to find:
( ρ2 − ρ1 ) 2 2
0 ≅ ω 4 − gk ( kH 2 + kH1 )ω 2 + k g kH1kH 2 .
ρ2
This is a quadratic equation for ω2. The two solutions are:
" " ρ − ρ % 4H1H 2 %
k 2 g(H1 + H 2 ) $ '.
ω2 = 1± 1− $ 2 1 '
2 $ # ρ & (H + H ) 2 '
# 2 1 2 &
Set H1 = H2 = H/2 to reach:
k 2 gH ! ρ $
ω2 = ##1± 1 && .
2 " ρ2 %
Consider the "+" sign for the barotropic mode. For H1 = H2 = H/2, the relationship between the
surface deflection amplitude a and the interface deflection amplitude b is:
" gk 2 H1 % "$ gk 2 H
% "
' ρ1 ρ2 %
b = a $1− ' = a 1− = a $ '.
# ω 2 & $ k 2 gH 1+ ρ1 ρ2 ' $# 1+ ρ1 ρ2 '&
( )
# &
The factor in parentheses is positive with a value near 1/2 when the densities are nearly equal.
Thus, a and b are in phase. Now consider the "–" sign for the baroclinic mode.
" gk 2 H1 % "$ gk 2 H
% " %
' = a $ − ρ1 ρ2 ' .
b = a $1− ' = a 1−
# ω 2 & $ k 2 gH 1− ρ1 ρ2 ' $# 1− ρ1 ρ2 '&
( )
# &
The factor in parentheses is negative with a magnitude much larger than unity when the densities
are nearly equal. Thus, a and b are out of phase.
The phase speeds of the two modes when H1 = H2 = H/2 are obtained from the dispersion
relationships above:
12
ω ' gH ! ρ1 $*
cp = = ) #1± &, ,
k )( 2 #" ρ2 &%,+
with barotropic mode (corresponding to the "+" sign) traveling faster than the baroclinic mode
(corresponding to the "–" sign). As ρ2 − ρ1 → 0 , the baroclinic mode's phase speed approaches
zero, and its amplitude becomes unbounded. This mode amplitude result is inconsistent with the
long-wavelength & shallow-water approximations, so a more refined theory is needed to truly
explain the baroclinic mode amplitude in this limit.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.1. a) Write out the three components of (9.1) in x-y-z Cartesian coordinates.
b) Set u = (u(y), 0, 0), and show that the x- and y-momentum equations reduce to:
1 ∂p d 2u 1 ∂p
0=− + ν 2 , and 0 = − .
ρ ∂x dy ρ ∂y

Solution 9.1. a) Equation (9.1) is the constant-viscosity Navier-Stokes' momentum equation for
incompressible flow:
Du 1
= − ∇p + ν∇ 2 u ,
Dt ρ
where ν is the kinematic viscosity of the flow. Using u = (u, v, w) and ∇ = (∂ ∂x,∂ ∂y,∂ ∂z) , the
three components of this equation become:
∂u ∂u ∂u ∂u 1 ∂p & ∂ 2u ∂ 2u ∂ 2u )
x: +u +v +w =− + ν( 2 + 2 + 2 + ,
∂t ∂x ∂y ∂z ρ ∂x ' ∂x ∂y€ ∂z *
∂v ∂v ∂v ∂v 1 ∂p & ∂ 2v ∂ 2v ∂ 2v )
y: +u +v +w =− + ν ( 2 + 2 + 2 + , and
∂t ∂x ∂y ∂z ρ ∂y ' ∂x ∂y ∂z *
€ ∂w ∂w ∂w ∂w 1 ∂p & ∂ 2w ∂ 2w ∂ 2w )
z: +u +v +w =− + ν( 2 + 2 + 2 +.
∂t ∂x ∂y ∂z ρ ∂z ' ∂x ∂y ∂z *
b) When € u = (u(y), 0, 0), all the terms involving v and w disappear, so the part a) equations
simplify to:
∂u ∂u 1 ∂p & ∂ 2u ∂ 2u ∂ 2u )
€x: + u +0+0=− + ν( 2 + 2 + 2 +,
∂t ∂x ρ ∂x ' ∂x ∂y ∂z *
1 ∂p
y: 0+0+0+0 =− + ν ( 0 + 0 + 0 ) , and
ρ ∂y
€ 1 ∂p
z: 0+0+0+0=− + ν (0 + 0 + 0) .
ρ ∂z
And, when u depends only on y, then ∂u/∂t = ∂u/∂x = ∂u/∂z = 0 so the part a) equations simplify
further:
1 ∂p & ∂ 2u )
€x: 0=− + ν( 2 + ,
ρ ∂x ' ∂y *
1 ∂p
y: 0=− , and
ρ ∂y
1 ∂p
€z: 0=− .
ρ ∂z
These x- and y-direction equations match those in the problem statement.
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.2. For steady pressure driven flow between parallel plates (see Figure 8.3), there are 7
parameters: u(y), U, y, h, ρ, µ, and dp/dx. Determine a dimensionless scaling law for u(y), and
rewrite the flow-field solution (8.5) in dimensionless form.

Solution 9.2. The parameters are: u(y), U, y, h, ρ, µ, and dp/dx. First, create the parameter
matrix:
u U y h ρ µ dp/dx
––––––––––––––––––––––––––––––––––––––––
Mass: 0 0 0 0 1 1 1
Length: 1 1 1 1 -3 -1 -2
Time: -1 -1 0 0 0 -1 -2

Next, determine the number of dimensionless groups. This rank of this matrix is three so 7
parameters - 3 dimensions = 4 groups, and construct the groups: Π1 = u U , Π 2 = y h ,
2
Π 3 = ρUh µ , and Π 4 = h (dp /dx) Uµ . Now write a dimensionless law:
u # y ρUh h 2 dp &
= f% , , (
U $ h µ U€ µ dx ' €
€ where f is€an unknown function.
When rewritten in dimensionless form, (8.5) is:
u y h 2 dp # y &# y & 1
= −€ % (%1− ( or Π1 = Π 2 − Π 4 Π 2 (1− Π 2 ) .
U h 2Uµ dx $ h '$ h ' 2
In this flow, there is no fluid acceleration so the Reynolds number, Π3, does not appear.

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.3. An incompressible viscous liquid with density ρ fills the gap between two large
smooth parallel walls that are both stationary. The upper and lower walls are located at x2 = ±h,
respectively. An additive in the liquid causes its viscosity to vary in the x2 direction. Here the
flow is driven by a constant non-zero pressure gradient: ∂p ∂x1 = const.
a) Assume steady flow, ignore the body force, set u = ( u1 (x 2 ),0,0) and use
∂ρ ∂ ∂u ∂u ∂p ∂ + % ∂u ∂u (. ∂ +% 2 ( ∂u .
+ ( ρui ) = 0 , ρ j + ρui j = − + ρg j + -µ'' i + j **0 + -'µv − µ* i 0
∂t ∂x i ∂t ∂x i ∂x j € ∂x i -, & ∂x j ∂x i )0/ ∂x j ,& 3 ) ∂x i /
€ 2
(
to determine u1(x2) when µ = µo 1+ γ ( x 2 h ) .)
b) What shear stress is felt on the lower wall?
€ c) What is the€volume flow rate (per unit depth into the page) in the gap when γ = 0?
d) If –1 < γ < 0, will the volume flux be higher or lower than the case when γ = 0?
€

Solution 9.3. a) The continuity equation is satisfied by the form of the velocity field. The j =1-
component of momentum equation simplifies to:
0 = −(∂p ∂x1) + (∂ ∂x 2 )[µ(∂u1 ∂x 2 )] .
Integrate once with ∂p ∂x1 = const. to find: µ(∂u1 ∂x 2 ) = (∂p ∂x1 ) x 2 + C . Divide by µ and
integrate again:
1 # ∂p & (∂p ∂x1 ) x 2 + C dx
€u1 = ∫ % x 2 + C (dx 2 = ∫ 2
€ µ $ ∂x1 € ' (
µo 1+ γ ( x 2 h )
2
)
h 2 ∂p # # x 2 &2 & Ch #x γ&
= ln%%1+ γ % ( (( + tan−1% 2 ( + D.
2γµo ∂x1 $ $ h ' ' µo γ $ h '
The boundary conditions, u1 (±h) = 0 , determine the values of the constants: C = 0, and
D = −( h 2 2γµo )(∂p ∂x1) ln(1+ γ ) , thus:
h 2 ∂p % 1+ γ ( x 2 h ) (
2
€
€ u1 (x 2 ) = − ln' ** .
2γµo ∂x1 '& 1+ γ )
€ b) From the solution of part a) with C = 0: τ w = (µ∂u1 ∂x 2 ) y=−h = −h (∂p ∂x1 )
+h
h 2 ∂p +h % x 22 ( 2h 3 ∂p
c) When γ = 0, the flow profile is parabolic: q = ∫ u1 (x 2 )dx 2 = − ∫' 1− dx
* 2 = −
€ −h 2µo ∂x1 −h & h 2 ) 3µo ∂x1
d) The volume flux will be higher€because the viscosity will be reduced at the wall. Manipulation
of the near-wall viscosity with additives is sometimes used in long piping systems to reduce
pumping power requirements.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.4. An incompressible viscous liquid with density ρ fills the gap between two large
smooth parallel plates. The upper plate at x2 = h moves in the positive x1-direction at speed U.
The lower plate at x2 = 0 is stationary. An additive in the liquid causes its viscosity to vary in the
x2 direction.
a) Assume steady flow, ignore the body force, set u = ( u1 (x 2 ),0,0) and ∂p ∂x1 = 0 , and use the
equations specified in Exercise 8.3 to determine u1(x2) when µ = µo (1+ γ x 2 h ) .
b) What shear stress is felt on the lower plate?
c) Are there any physical limits on γ? If, so specify them.
€ €
€

Solution 9.4. a) For u = ( u1 (x 2 ),0,0) , no body force, and ∂p ∂x1 = 0 in steady incompressible
flow, the continuity equation is automatically satisfied, and the momentum equation for j = 1
simplifies to:
0 = +(∂ ∂x 2 )[µ(∂u1 ∂x 2 )] , or, after integrating once: C = µ(∂u1 ∂x 2 ) ,
€ €
where C is a constant. Now use the specified relationship for the viscosity and integrate to find:
C Cdx 2 Ch
u1 (x 2 ) = ∫ dx 2 = ∫ = ln(1+ γ x 2 h ) + D
µ µo (1+ γ x 2 h ) €µoγ
€
where D is another constant. The boundary conditions u1(0) = 0 and u1(h) = U allow
C = Uγµo ( h ln(1+ γ )) and D = 0 to be determined yielding:
u1 (x 2 ) = U ln(1+ γ x 2 h ) ln(1+ γ ) .
€
b) From part a), the shear stress is constant: τ w = µ(∂u1 ∂x 2 ) = C = Uγµo ( h ln(1+ γ )) .
€ c) Negative viscosities violate the second law of thermodynamics, thus γ > –1 is required.
€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.5. Planar Couette flow is generated by placing a viscous fluid between two infinite
parallel plates and moving one plate (say, the upper one) at a velocity U with respect to the other
one. The plates are a distance h apart. Two immiscible viscous liquids are placed between the
plates as shown in the diagram. The lower fluid layer has thickness d. Solve for the velocity
distributions in the two fluids.

Solution 9.5. For steady viscous flow between infinite parallel plates, the fluid velocity will be
unidirectional: u = (u, 0, 0). For this problem, no pressure gradient is specified so assume it to
be zero. Thus, the horizontal (x1-direction) momentum equation reduces to:
∂τ 21 ∂ x2 = (∂ ∂ x2 ) (µ∂ u ∂ x2 ) = µ∂ 2 u ∂ y 2 = 0 ,
where the last equality follows when the viscosity is constant and x2 = y. Here, the viscosity is
assumed constant within each fluid. This means that the flow profile in each fluid will be piece-
wise linear:
# A + B1 y for 0 ≤ y ≤ d &
u(y) = $ 1 ',
% A2 + B2 y for d ≤ y ≤ h (
where the A’s & B’s are constants and ‘1’ implies the upper fluid layer with viscosity µ1, and ‘2’
implies the lower fluid layer with viscosity µ2. The four constants can be determined from the
four boundary conditions:
€
i) u(0) = 0 (match the speed of the lower boundary)
ii) u(h) = U (match the speed of the upper boundary)
– +
iii) u(d ) = u(d ), and (match flow speeds at the internal fluid-fluid interface)
iv) τ(d–) = τ(d+) (match shear stress at the internal fluid-fluid interface)
where τ is the shear stress in the fluid. These four boundary conditions imply:
A2 = 0, A1 + B1h = U , A2 + B2 d = A1 + B1d , and µ2 B2 = µ1B1
Use the first two equations to eliminate A1 and A2 from the second two equations to find:
B2 d = U − B1 (h − d) , and µ2 B2 = µ1B1.
Eliminate B2 and solve for B1:
€ € €
(µ1 µ2 ) B1d = U − B1(h − d) –> B1 = µ2U [µ2 h + (µ1 − µ2 ) d] .
So, B2 = µ1U [µ2 h +€(µ1 − µ2 ) d ] , and A1 = U (µ€1 − µ2 ) d [µ2 h + (µ1 − µ2 ) d ] with A2 = 0. Thus:
U $ µ1 y for 0 ≤ y ≤ d '
u(y) = % (
2 ) d &( 1
€ µ2 h + (µ1 − µ€ µ − µ2 ) d + µ2 y for d ≤ y ≤ h )
€ €

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.6. Consider plane Poiseuille flow of a non-Newtonian power-law fluid using the
coordinate system and geometry shown in Figure 9.3. Here the fluid's constitutive relationship is
n
given by (4.37): τ xy = m (∂u ∂y) , where m is the power law coefficient, and n is the power law
exponent.
a) Determine the velocity profile u(y) in the lower half of the channel, 0 ≤ y ≤ h/2, using the
boundary condition u(0) = 0.
b) Given that the maximum velocity occurs at y = h/2 and that the flow profile is symmetric
about this location, plot u(y) u(h/2) vs. y (h/2) for n = 2 (a shear thickening fluid), n = 1 (a
Newtonian fluid), and n = 0.4 (a shear thinning fluid).
c) Explain in physical terms why the shear-thinning velocity profile is the bluntest.

Solution 9.6. In plane Poiseuille flow, the upper channel wall does not move. In the lower half of
the channel, the ∂u/∂y > 0, so fractional powers are readily managed.
a) For fully developed constant density flow (so that ∂p/∂x is independent of x) the horizontal
momentum equation reduces to:
∂p ∂τ
0 = − + xy .
∂x ∂y
The vertical momentum equation is simply, ∂p/∂y = 0, so ∂p/∂x does not depend on x or y. Thus,
this horizontal momentum equation can be integrated to find:
∂p
y = τ xy + C ,
∂x
where C is a constant. However, when y = h/2, τxy = 0 so C = (h/2)∂p/∂x, therefore:
n 1n 1n
" h %" ∂p % " ∂u % ∂u # 1 ∂p & # h &
$ − y '$ − ' = τ xy = m $ ' , or = %− ( % − y( ,
# 2 &# ∂x & # ∂y & ∂y $ m ∂x ' $ 2 '
and both equations are written this way because –∂p/∂x is positive. The second equation can be
integrated to find:
1n 1+1 n
# 1 ∂p & 1 #h &
u(y) = − % ( % − y ( +D,
$ m ∂x ' 1+1 n $ 2 '
where D is another constant of integration. The horizontal velocity must go to zero on the lower
channel wall at y = 0, so
1n 1+1 n
" 1 ∂p % 1 "h%
D =$ ' $ ' .
# m ∂x & 1+1 n # 2 &
Thus, the velocity profile is:
" h % , " n % h " h ∂p % ) " % ,
1+1 n
n " 1 ∂p % )" h %
1n 1+1 n 1+1 n 1n
y
u(y) = $ ' +$ ' − $ − y ' . = $ ' $ ' +1− $1− ' ..
n +1 # m ∂x & +*# 2 & # 2 & .- # n +1 & 2 # 2m ∂x & +* # h 2 & .-
b) Evaluate the result of part a) at y = h/2, and form the ratio:
! n $ h ! h ∂p $ ) ! $ ,
1n 1+1 n
y
# & # & +1− #1− & .
u(y) " n +1 % 2 " 2m ∂x % +
* " h 2 % .- )
= *1− (1− η ) ,- ,
1+1 n
= 1n
u(h / 2) ! n $ h ! h ∂p $
# & # &
" n +1 % 2 " 2m ∂x %
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

where η = y (h/2) . The three curves for n = 2, 1, and 0.4 are plotted here:

2.00#
1.80#
1.60#
1.40#
1.20#
η n=2#
1.00#
n=1#
0.80#
n=0.4#
0.60#
0.40#
0.20#
0.00#
0.000# 0.200# 0.400# 0.600# 0.800# 1.000#
u(η)/u(η = 1)

c) From the part a) solution, the shear stress is:

∂p # h &
τ xy = − % − y( .
∂x $ 2 '
The highest magnitude shear stress occurs at the upper and lower wall of the channel. In a shear
thinning fluid, the local viscosity decreases when the shear stress increases. Thus, to maintain
this shear stress profile, the velocity gradient magnitude must increase in regions where the shear
stress magnitude is also large. This leads to a blunter profile for a shear thinning fluid compared
to a Newtonian (or a shear thickening) fluid.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.7. Consider the laminar flow of a fluid layer falling down a plane inclined at an angle
θ with respect to the horizontal. If h is the thickness of the layer in the fully developed stage,
2 2
( )
show that the velocity distribution is u(y) = ( g 2ν ) h − y sin θ , where the x-axis points in the
direction of flow along the free surface, and the y-axis points toward the plane. Show that the
3
( )
volume flow rate per unit width is Q = gh 3ν sin θ , and that and the frictional stress on the
wall is τw = ρghsinθ. €
x
Solution 9.7. In the given coordinates, there is a y
component of gravity, gsin€θ, acting in the x-direction, but stress
the presence of atmospheric pressure on the liquid surface h distribution
ensures that ∂p/∂x = 0. Thus, the steady flow x-direction
momentum equation is:
∂ 2u ! u(y)
0 = +gsin θ + ν 2 .
∂y
Integrating twice produces:
gsin θ 2
u(y) = − y + Ay + B ,
2ν
€ B are constants. The boundary condition on the liquid surface (y = 0) is zero shear
where A and
stress (µ∂u/∂y = 0), and this allows A to be evaluated:
#∂u & # gsin θ &
€ 0 = % ( = %− y + A( = A .
$∂y 'y= 0 $ ν 'y= 0
The boundary condition, on the solid surface (y = h) is zero velocity (u = 0), and this allows B to
be evaluated:
gsin θ 2 gsin θ 2
€ 0 = u(h) = − h + B , or B = h .
2ν 2ν
so the velocity profile is:
gsin θ 2
u(y) = −
2ν
(y − h 2 ).
€ flow rate per unit width is: €
The volume
h gsin θ h 2 gsin θ & h 3 ) gsin θ 3
Q = ∫ 0 u(y)dy = −
2 ν
∫ 0 ( y − h 2
) dy = −
2 ν
(
' 3
− h 3
+=
* 3ν
h .
€
The magnitude of the shear stress is:
∂u & gsin θ )
τ = µ = µ( + y = ρgy sin θ ,
∂y ' ν *
€
and the maximum shear stress, τw = ρghsinθ, occurs at the solid surface.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.8. In two-dimensional (x,y)-coordinates, the Navier-Stokes equations for the fluid
velocity, u = (u, v) , in a constant-viscosity constant-density flow are: ∂u ∂x + ∂v ∂y = 0 ,
∂u ∂u ∂u 1 ∂p # ∂2 u ∂2 u & ∂v ∂v ∂v 1 ∂p # ∂2 v ∂2 v &
+u +v =− + ν % 2 + 2 ( , and +u +v =− +ν % 2 + 2 ( .
∂t ∂x ∂y ρ ∂x $ ∂x ∂y ' ∂t ∂x ∂y ρ ∂y $ ∂x ∂y '
a) Cross differentiate and sum the two momentum equations to reach the following equation for
ω z = ∂v ∂x − ∂u ∂y , the vorticity normal to the x-y plane:
∂ω z ∂ω ∂ω " ∂2ω ∂2ω %
+ u z + v z = ν $ 2z + 2z ' .
∂t ∂x ∂y # ∂x ∂y &
b) The simplest nontrivial solution of this equation is uniform shear or solid body rotation (ωz =
constant). The next simplest solution is a linear function of the independent coordinates: ωz = ax
+ by, where a and b are constants. Starting from this vorticity field, derive the following velocity
field:
b " (ax + by)2 % a ! (ax + by)2 $
u = − $ 2 2 + c' and v = # 2 2 + c& .
2# a +b & 2" a +b %
where c is an undetermined constant.
c) For the part b) flow, sketch the streamlines. State any assumptions you make about a, b, and c.
d) For the part b) flow when a = 0, b > 0, and u = (Uo , 0) at the origin of coordinates with Uo > 0,
sketch the velocity profile along a line x = constant, and determine ∇p .

Solution 9.8. a) Apply –∂/∂y to the x-direction momentum equation, and ∂/∂x to the y-direction
momentum equation to reach:
∂ # ∂u & ∂u ∂u ∂ # ∂u & ∂v ∂u ∂ # ∂u & 1 ∂2 p # ∂2 ∂2 &# ∂u &
− % (− −u % (− −v % (= − ν % 2 + 2 (% ( , and
∂t $ ∂y ' ∂y ∂x ∂x $ ∂y ' ∂y ∂y ∂y $ ∂y ' ρ ∂y∂x $ ∂x ∂y '$ ∂y '
∂ " ∂v % ∂u ∂v ∂ " ∂v % ∂v ∂v ∂ " ∂v % 1 ∂2 p " ∂2 ∂2 %" ∂v %
$ '+ + u $ '+ +v $ '=− + ν $ 2 + 2 '$ ' .
∂t # ∂x & ∂x ∂x ∂x # ∂x & ∂x ∂y ∂x # ∂y & ρ ∂x∂y # ∂x ∂y &# ∂x &
Add these two equations together noting that pressure terms cancel, and that the second and
fourth terms in each equation sum to zero because of the continuity equation, ∂u ∂x + ∂v ∂y = 0 .
∂ # ∂v ∂u & ∂ # ∂v ∂u & ∂ # ∂v ∂u & # ∂2 ∂2 &# ∂v ∂u &
% − ( + u % − ( + v % − ( = ν % 2 + 2 (% − ( .
∂t $ ∂x ∂y ' ∂x $ ∂x ∂y ' ∂y $ ∂x ∂y ' $ ∂x ∂y '$ ∂x ∂y '
Substituting in the definition ω z = ∂v ∂x − ∂u ∂y leads to the final form:
∂ω z ∂ω ∂ω " ∂2ω ∂2ω %
+ u z + v z = ν $ 2z + 2z ' .
∂t ∂x ∂y # ∂x ∂y &
b) To find the velocity field start with: ωz = ax + by = ∂v/∂x – ∂u/∂y. (1)
The part a) equation implies: ua + vb = 0 or v = –(a/b)u. (2)
The continuity equation is also needed: ∂u/∂x + ∂v/∂y = 0. (3)
Use (2) to eliminate v, from (1) and (3) to find:
a ∂u ∂u ∂u a ∂u
ax + by = − − , and − = 0.
b ∂x ∂y ∂x b ∂y
Combine these twice, first to eliminate ∂u/∂y, then to eliminate ∂u/∂x to reach:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

" a 2 % ∂u " a b % ∂u
ax + by = − $ 2 +1' , and ax + by = − $ + ' .
#b & ∂y # b a & ∂x
y2 " a2 % x2 "a b%
Integrate to find: axy + b = − $ 2 +1' u + f (x) , and a + bxy = − $ + ' u + g(y) , where f(x)
2 #b & 2 #b a&
and g(y) are unknown functions. Solve both equations for u and absorb multiplicative constants
into f(x) and g(y).
b " f (x) + 2abxy + b 2 y 2 % b " a 2 x 2 + ab 2 xy + g(y) %
u=− $ ' , and u = − $ '.
2# a2 + b2 & 2# a2 + b2 &
For consistency, the unknown functions must be: f(x) = a x + const. and g(y) = b2y2 + const. so
2 2

b " (ax + by)2 %

u = − $ 2 2 + c ' , and then from (2)
2# a +b &
y!
a a " (ax + by)2 %
v = − u = $ 2 2 + c' .
b 2# a +b & b!
where c is another undetermined constant. a!
c) dysl/dxsl = v/u = –a/b. Thus, if a and b are both
positive constants, the streamlines are negatively- x!
slopped straight lines. In this case, u < 0 and v > 0,
so the arrows point upward and to the left.
d) Evaluate the constants to find: u = Uo − ( b 2 ) y 2 and v = 0, so the
y! U
o! horizontal velocity profile is parabolic. To get the pressure gradient,
evaluate the 2D Navier-Stokes equations using the part b) velocity field:
1 ∂p 1 ∂p 
0=− + ν (−b) , and 0 = − + 0 , so ∇p = (−µ b, 0 ) .
ρ ∂x ρ ∂y
 
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.9. Consider circular Couette flow as described by (9.10) in the limit of a thin gap
between the cylinders. Use the definitions: R = (R1 + R2 ) 2 , h = R2 − R1 , Ω = (Ω1 + Ω2 ) 2 ,
ΔΩ = Ω2 − Ω1 , and R = R + y to complete the following items.
a) Show that uϕ (y) ≅ ΩR + ΔΩR ( y h ) when y << R , h << R , and all terms of order y R and
h R or higher have been dropped. This is the lubrication approximation for circular Couette
flow.
b) Compute the shear stress µ ( duϕ dy) from the part a) flow field and put it terms of µ, R1, R2,
Ω1, and Ω2.
' R ∂ ! uϕ $ 1 ∂ u *
c) Compute the exact shear stress from (9.10) using: τ Rϕ = 2µ ) # &+ R
, , and
( 2 ∂ R R
" % 2R ∂ϕ +
evaluate it at R = R1, and R = R2.
d) What values of h R lead to shear stress errors of 1%, 3% and 10%.

2
Solution 9.9. Here, R1 = R − h / 2 and R2 = R + h / 2 , so R12 = R − Rh + h 2 / 4 and
2
R22 = R + Rh + h 2 / 4 . Plus, R22 − R12 = 2Rh . Use these exact relationships in (9.10) to reach:
0 " 2 h 2 %" 2 h2 %4
2 $ R − Rh + '$ R + Rh + '2
1 2) " 2 h2 % " 2 h 2 %, # 4 &# 4 &2
uϕ (y) = Ω
1+ 2 $
2Rh 2* #
R + Rh +
4&
− Ω
' 1$
#
R − Rh +
4 &-
'. R + (
y − ΔΩ ) R+ y
5.
2
23 26
Simplify terms within the large {,}-brackets.
02# * 4 R 2 h 2 h 4 -42
1 2 h2 & 2
uϕ (y) = 1%ΔΩR + 2ΩRh + ΔΩ ( R + y − ΔΩ ,, R −( ) + //5 .
(
2Rh R + y 32$ ) 4' + 2 16 .62
Perform all the multiplications within the large {,}-brackets and continue with this
simplification:
$ΔΩR 4 + ΔΩ2R 3 y + ΔΩR 2 y 2 + 2ΩR 3h + 4ΩR 2 hy + 2ΩRhy 2 + (
1 & &
uϕ (y) = % 2 2 2 2 2 4).
2Rh R + y &ΔΩ h R + ΔΩ h Ry + ΔΩ h y 2 − ΔΩR + ΔΩ R h − ΔΩ h &
( 2
) 4

' 4 2 4 2 16 *
Cancel equal and opposite terms, and group terms than involve the average, Ω = (Ω1 + Ω2 ) 2 ,
and difference, ΔΩ = Ω2 − Ω1 , rotation rates together.
1 *, 2 # 3 2 # 3h
2 & h2 h 2 y 2 h 4 &.,
uϕ (y) = ( )
+2ΩRh R + y + ΔΩ % 2R y + R % + y2 ( + R y + − (/ .
(
2Rh R + y -, ) $ $ 4 ' 2 4 16 '0,
Remove the large {,}-brackets.
ΔΩR * y 1 # 3h 2 & hy 1 # hy 2 h 3 &-
(
uϕ (y) = Ω R + y + ) , + %
1+ y R + h 2Rh $ 4
+ y2 ( + 2 + 3 % − (/ .
' 4R 2R $ 4 16 '.
−1
(
Expand 1+ y R ) in a power series for y << R :
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

$ y y2 '* y 1 $ 3h 2 ' hy 1 $ hy 2 h 3 '-

( )
uϕ (y) = Ω R + y + ΔΩR &1− + 2 +... ), +
% R R
&
(+ h 2Rh % 4
+ y2 ) + 2 + 3 & − )/ .
( 4R 2R % 4 16 (.
Keeping only linear terms in y R and h R leads to the final series:
# y& #y& # y&
uϕ (y) = ΩR + ΔΩR % ( + ΩR % ( +... ≅ ΩR + ΔΩR % ( ,
$h' $R' $h'
where the final approximate equality follows because h << R .
du ΔΩR (Ω − Ω1 )(R2 + R1 )
b) τ = µ ϕ = =µ 2
dy h 2(R2 − R1 )
u (R) 1 # 2 2 R12 R22 &
c) Here uR = 0, and ϕ = 2 $ Ω R
2 2 − Ω R
1 1 − [ Ω 2 − Ω 1 ] ' . Thus,
R R2 − R12 % R2 (
∂ " uϕ (R) % 1 * R12 R22 - ∂ " uϕ (R) % µ * R12 R22 -
$ ' = + +2 [ 2 1] 3
Ω − Ω . so τ Rϕ = µ R $ ' = + 2 [ 2 1] 2 . .
Ω − Ω
∂R # R & R22 − R12 , R / ∂R # R & R22 − R12 , R /
Evaluating at R = R1, and R = R2 leads to
2
2µ [Ω2 − Ω1 ] 2 µ [Ω2 − Ω1 ] # h & µ [Ω2 − Ω1 ] R # h &
τ Rϕ (R1 ) = R2 = % R + ( ≅ %1+ ( and
R22 − R12 Rh $ 2' h $ R'
2
2µ [Ω2 − Ω1 ] 2 µ [Ω2 − Ω1 ] # h & µ [Ω2 − Ω1 ] R # h &
τ Rϕ (R2 ) = 2 2
R1 = %R − ( ≅ %1− ( .
R2 − R1 Rh $ 2' h $ R'
To first order, the average shear stress is:
1 2µ [Ω2 − Ω1 ] 2 µ [Ω2 − Ω1 ]
τ = (τ Rϕ (R1 ) + τ Rϕ (R2 )) = R2 = R ,
2 R22 − R12 h
which matches the result from the lubrication approximation.
d) To first order, the difference of either exact shear stress from this average shear stress is:
Δτ τ Rϕ (R1or 2 ) − τ h
= =± .
τ τ R
Thus, shear stress errors of shear stress errors of 1%, 3% and 10% occur when h R = 0.01, 0.03,
and 0.10.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.10. Room temperature water drains through a round vertical tube with diameter d.
The length of the tube is L. The pressure at the tube's inlet and outlet is atmospheric, the flow is
steady, and L >> d.
a) Using dimensional analysis write a physical law for the mass flow rate m˙ through the tube.
b) Assume that the velocity profile in the tube is independent of the vertical coordinate,
determine a formula for m˙ , and put it in dimensionless form.
c) What is the change in m˙ if the temperature is raised and the water's viscosity drops by a factor
€
of two?
€
Solution 9.10. a) This part is just dimensional analysis. The solution parameter is m˙ , and
€
boundary condition & material parameters are: ρ = density of the fluid, µ = viscosity of the fluid,
g = acceleration of gravity, d = tube diameter, and L = tube length. The parameter matrix is:
m˙ ρ µ g d L €
–––––––––––––––––––––––––––––––––––
Mass: 1 1 1 0 0 0
Length: 0 -3 -1 1 1 1
€
Time: -1 0 -1 -2 0 0
and it has rank 3. The number of dimensionless groups is: 6 parameters - 3 dimensions = 3.
m˙ gρ 2 d 3 L
These groups are: ∏1 = , ∏2 = 2
, and ∏3 = . Therefore, the scaling law is:
µd µ d
m˙ # gρ d L &
2 3
= F% 2 , (.
µd $ µ d'
b) When the velocity
€ profile
€ in the tube is steady
€ and independent of the vertical coordinate, the
vertical momentum equation is
µ ∂ % ∂uz (
€ 0 = −ρg + 'R *
R ∂R & ∂ R )
where the z-axis points upward (opposite gravity), R is the radial coordinate, and uz is the vertical
velocity. This is the equation for Poiseuille flow with the pressure gradient replaced by ρg.
Noting that downward velocity produces positive m˙ , the results in Section 8.2 imply:
€ 4
π $d' π ρ 2 gd 4 m˙ π ρ 2 gd 3 π
˙
m = ρQ = ρ & ) ρg = or = 2
→ Π1 = Π2.
8µ % 2 ( 128 µ µd 128 µ 128
€ µ
c) The flow rate is inversely proportional to µ and µ = µ(T). Since the ratio room temp = 2 , the
µhotter
flow
€ rate will approximately double (the density € of water falls slightly as T increases toward the
boiling point) but this will likely be a very small effect.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.11. Consider steady laminar flow through the annular space formed by two coaxial
tubes aligned with the z-axis. The flow is along the axis of the tubes and is maintained by a
pressure gradient dp/dz. Show that the axial velocity at any radius R is
1 dp # 2 2 b2 − a2 R&
uz (R) = %R − a − ln ( ,
4 µ dz $ ln(b a) a '
where a is the radius of the inner tube and b is the radius of the outer tube. Find the radius at
which the maximum velocity is reached, the volume flow rate, and the stress distribution.

Solution 9.11. Use € cylindrical coordinates, and assume b

uz = uz(R), where R is the radial coordinate. The axial
(z-direction) momentum equation is: Rmax
a
∂p 1 ∂ # ∂uz &
0= +µ %R (.
∂z R ∂R $ ∂R '
Integrate twice to find:
∂uz R 2 ∂p R 2 ∂p
R = + A , and uz (R) = + Alog R +B .
∂R€ 2µ ∂z 4 µ ∂z
The two boundary conditions are u = 0 at R = a, b and these allow the two constants A and B to
be determined. Alternatively evaluating the above formula for uz(R) at R = a and b produces.
a 2 ∂p b 2 ∂p
€ €uz (a) = 0 = + Alog a +B , and uz (b) = + Alogb +B .
4 µ ∂z 4 µ ∂z
These two equations can be solved to find:
1 $ ∂p ' b 2 − a 2 1 # ∂p &* b 2 − a 2 -
A=− & ) , and B = % (, log(a) − a 2 / ,
€ 4 µ % ∂z ( ln(b /a) € 4 µ $ ∂z '+ln(b /a) .
so that:
1 # ∂p &* 2 2 b2 − a2 -
u(R) = % (,R − a − log( R a)/ .
€ 4 µ€$ ∂z '+ ln(b /a) .
The shear stress is:
∂ u R ∂ p 1 " ∂ p % b2 − a2
τ =µ = − $ ' .
€ ∂ R 2 µ ∂ z 4 µ # ∂ z & R ln(b / a)
The maximum velocity occurs where τ = 0, and this condition implies:
R ∂p 1 $ ∂p ' b 2 − a 2 b2 − a2
0 = max − & ) or R max = .
2µ ∂z 4 µ % ∂z ( Rmax ln(b /a) 2ln(b /a)
The volume flow rate Q is:
, 2 2 2 /
b π % ∂p (.(b − a ) 4 41
Q = 2π ∫ a u(R)RdR = ' * + a −b .
€ 8µ & €
∂z ). ln(b /a) 1
- 0

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.12. A long round wire with radius a is pulled at a steady speed U, along the axis of a
long round tube of radius b that is filled with a viscous fluid. Assuming laminar fully-developed
axial flow with ∂p ∂z = 0 in cylindrical coordinates (R,ϕ,z) with u = (0, 0, w(R)), determine
w(R) assuming constant fluid density ρ and viscosity µ with no body force.
a) What force per unit length of the wire is needed to maintain the motion of the wire?
b) Explain what happens to w(R) when b → ∞. Is this situation physically meaningful? What
€ €
additional term(s) from the equations of motion need to be retained to correct this situation?

Solution 9.12. The steady axisymmetric no-swirl equations are:

1 ∂ ∂w ∂uR ∂uR 1 ∂p µ % 1 ∂ % ∂uR ( ∂ 2 uR uR (
( R)
Ru + = 0 , u R + w = − + ' 'R *+ − *,
R ∂R ∂z ∂R ∂z ρ ∂R ρ & R ∂R & ∂R ) ∂z 2 R 2 )
1 ∂p ∂w ∂w 1 ∂p µ % 1 ∂ % ∂w ( ∂ 2 w (
0=− , and uR +w =− + ' 'R * + *.
ρR ∂ϕ ∂R ∂z ρ ∂z ρ & R ∂R & ∂R ) ∂z 2 )
€Setting u = (0, 0, w(R))
€ automatically satisfies the cons. of mass equation. With ∂p ∂z = 0 , the
other equations further simplify to:
€ 1 ∂p 1 ∂p µ $ 1 ∂ $ ∂w ''
€0 = − =− , and 0 = & & R ))
ρ ∂R ρR ∂ϕ ρ % R ∂R % ∂R€((
Thus, the pressure is constant everywhere because its gradient is zero. For nonzero viscosity and
density, integrate the final equation once to find: C1 = R(∂w ∂R) where C1 is a constant. Divide
by R and integrate again: C1 ln(R) + C2 = w(R) where C2 is a constant. The two constants can be
€ €
determined from the boundary conditions on the wire and tube surfaces:
w(R = a) = U = C1 ln(a) + C2 , and w(R = b) = 0 = C1 ln(b) + C2 .
€ U
Simultaneous solution
€ of these two algebraic equations produces: w(R) = ln(b R)
ln(b a)
a) The viscous
€ shear stress on the wire, τw€ , will act to retard its motion. The viscous force per
unit length of wire will act in the z-direction on the wire and will be
% ∂w ( % U 1 % b (( 2πµU
2πaτ w = 2πaµ' * = 2πaµ' € ' − 2 ** = − .
& ∂R ) R = a & ln(b a) b R & R )) R = a ln(b a)
Thus, the force necessary to maintain the motion will be: = + 2πµUe z ln(b a)
b) When b → ∞, the solution indicates that the force necessary to keep the wire moving will
approach zero. This is not physically meaningful for any finite wire length and time duration
€
because the fluid is viscous. The problem(s) with this limiting situation can be corrected by
including the unsteady terms, ∂ ∂t , or terms involving € uR and ∂ ∂z in the conservation equations.
€

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.13. Consider steady unidirectional incompressible viscous flow in Cartesian

coordinates, u = v = 0 with w = w(x,y) without body forces.
a) Starting from the steady version of (8.1), derive a single equation for w assuming that ∂p ∂z is
non-zero and constant.
2 2
b) Guess w(x,y) for a tube with elliptical cross section ( x a) + ( y b) = 1 .
c) Determine w(x,y) in for a tube of rectangular cross section: –a/2 ≤ x ≤ +a/2,€–b/2 ≤ y ≤ +b/2.
(Hint: find particular (a polynomial) and homogeneous (a Fourier series) solutions for w.)
€ momentum equation in Cartesian coordinates:
Solution 9.13. a) Start with the full z-direction
∂w ∂w ∂w ∂w 1 ∂p & ∂ 2w ∂ 2w ∂ 2w )
+u +v +w =− + ν( 2 + 2 + 2 +.
∂t ∂x ∂y ∂z ρ ∂z ' ∂x ∂y ∂z *
For steady flow with u = v = 0 and w = w(x,y), a lot terms that drop out of the above equation,
∂p # ∂ 2w ∂ 2w &
and it simplifies to: = µ% 2 + 2 ( . Note that the density has dropped out because there is
€ ∂z $ ∂x ∂y '
no fluid acceleration in this type of flow.
b) The boundary condition for this flow is w = 0 on the wall of the tube. For an elliptical cross
2 2
[ 2
section given by ( x a) + ( y b) = 1 , a possible solution is w(x, y) = W o 1− ( x a) − ( y b)
€
2
]
where Wo is a constant that needs to be determined from the equation of motion. Plugging this
∂p $2 2'
guessed solution into the equation found for part a), produces: = −µW o & 2 + 2 ) . If ∂p ∂z is
€ € ∂z %a b (
−1
1 ∂p $ 1 1' * x 2
y -
2
constant then: w(x, y) = − & 2 + 2 ) ,1− 2 − 2 / , and this reduces to the usual Poiseuille
2µ ∂z % a b ( + a b .
€
flow formula when a = b. €
c) The solution for a rectangular cross section is a bit more involved. The equation for w in part
a) is inhomogeneous, so its solution will be made up of a particular solution and a homogeneous
€ w = w + w . Here, the particular solution must satisfy the governing equation, but it
solution: p h
need not satisfy the boundary conditions. The constants in the homogeneous solution can be
used to fix up the solution at the tube walls. Therefore, start by looking for a one-variable
particular solution, i.e. wp = wp(x), and plug this into the field equation:
∂p # ∂ 2w p ∂ 2w p & d 2w p
= µ% 2 + (=µ .
∂z $ ∂x ∂y 2 ' dx 2
1 dp 2
Integrate twice to find: w p (x) = x + C1 x + C2 . The two constants can be chosen to zero-
2µ dz
€ a 2 dp # x2 &
out the velocity at x = ±a/2; w p (x) = − %1− ( . Now, the homogeneous solution is
8µ dz $ (a /2) 2 '
€ ∂ 2wh ∂ 2wh
needed. It solves + = 0 . Assume a variables-separable form: w h (x, y) = X(x)Y (y) ,
∂x 2 ∂y 2
plug it in, and label€the separation constant –k2 to find two ordinary differential equations:
X "" + k 2 X = 0 , and Y "" − k 2Y = 0 . The symmetry of the situation implies that only the cos-part of
X and the cosh-part of Y are needed: w h (x, y) = Acos(kx)cosh(ky) € . To zero-out the velocity at x
€

€ €
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

= ±a/2, the separation constant k must be restricted to a set of special values:

k = kn = (2n −1)π a . Thus, the overall solution takes the following form:
a 2 dp # x2 & ∞ # (2n −1)πx & # (2n −1)πy &
w = w p + wh = − %1− 2(
+ ∑ An cos% ( cosh% (. (1)
8µ dz $ (a /2) ' n=1 $ a ' $ a '
€ Here it is clear than w is zero on x = ±a/2. The An’s are determined by requiring w = 0 on y =
±b/2:
a 2 dp # x2 & ∞ # (2n −1)πx & # (2n −1)πb &
€ %1− 2(
= ∑ An cos% ( cosh% (.
8µ dz $ (a /2) ' n=1 $ a ' $ 2a '
Multiply both sides by cos[(2m–1)πx/a], and integrate from x = –a/2 to x = +a/2:
a 2 dp x= a / 2 # x 2 & # (2m −1)πx & # (2m −1)πb & a
∫ %1−
8µ dz x=−a / 2 $ (a /2) ' $ 2(
cos%
a
( dx = Am cosh%
' $ 2a
(
'2
€
The integral on the left side is a little tedious but can be evaluated by parts:
a 2 dp $ m +1 16a ' $ (2m −1)πb ' a
& (−1) 3 3)
= Am cosh& )
€ 8µ dz % (2m −1) π ( % 2a (2
4a dp $ (−1)
2 m +1 ' $ (2m −1)πb '
Thus: Am = & 3 3)
cosh& ) . This result together with (1) completes
µ dz % (2m −1) π ( % 2a (
the solution.€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.14. A long vertical cylinder of radius b rotates with angular velocity Ω concentrically
outside a smaller stationary cylinder of radius a. The annular space is filled with fluid of
R 2 − a 2 b 2Ω
viscosity µ. Show that the steady velocity distribution is: uϕ = 2 , and that the torque
b − a2 R
exerted on either cylinder, per unit length, equals 4πµΩa2b2/(b2 − a2).

Solution 9.14. Use cylindrical coordinates, and start from equation (8.10),
€
1 % R12 R22 (
uϕ (R) = 2
R2 − R12 '
& [ 2 2 1 1 ] [ 2 1] R )* ,
Ω R 2
− Ω R 2
R − Ω − Ω

and set R1 = a, R2 = b, Ω2 = Ω, and Ω1 = 0 to find:

1 % a 2b 2 ( Ωb 2 R 2 − a 2 R 2 − a 2 b 2Ω
uϕ (R) = 2
b − a2 '
& [Ωb 2
− 0] R − [Ω − 0 ] ) =
R * b2 − a2 R
= 2
b − a2 R
.
€
From Appendix B the shear stress is
∂ % uϕ ( µRb 2Ω ∂ % a 2 ( 2µa 2b 2Ω % 1 (
τ Rϕ = µR ' * = 2 '1− * = ' *.
€ ∂R & R ) b − a 2 ∂R & R 2 ) b 2 − a 2 & R 2 )
If the axial length of the cylinders is h, then
2µa 2b 2Ω $ 1 ' 4 πµa 2b 2 hΩ
torque on the inner cylinder = (τRϕ)R = a(area)(radius) = 2 & ) (2 π ah ) a = ,
€ b − a2 % a2 ( b2 − a2
2µa 2b 2Ω $ 1 ' 4 πµa 2b 2 hΩ
torque on the outer cylinder = (τRϕ)R = b(area)(radius) = 2 & )( 2 πbh )b = .
b − a2 % b2 ( b2 − a2
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.15. Consider a solid cylinder of radius a, steadily rotating at angular speed Ω in an
infinite viscous fluid. The steady solution is irrotational: uθ = Ωa2/R. Show that the work done by
the external agent in maintaining the flow (namely, the value of 2πRuθτrθ at R = a) equals the
viscous dissipation rate of fluid kinetic energy in the flow field.

Solution 9.15. Using the given velocity field, the shear stress is:
∂ %u ( ∂ %1( 1
τ Rϕ = µR ' ϕ * = µΩa 2 R ' 2 * = −2µΩa 2 2 .
∂R & R ) ∂R & R ) R
The work done per unit height = {2πaτ Rϕ uϕ }R = a = 2πa ⋅ 2µΩ ⋅ Ωa = 4 πµa 2Ω2 .
From (4.58) the viscous dissipation rate of kinetic energy per unit volume for an
incompressible € flow is ρε = 2µSij Sij , where ε is the viscous dissipation of kinetic energy per unit
mass. For the given flow€ field there is only one non-zero independent strain component:
R ∂ $ uϕ ' Ωa 2
∂ $1' 1
SRϕ = SϕR = & )= R & 2 ) = −Ωa 2 2 .
€ 2 ∂R % R ( 2 ∂R % R ( R
Therefore:
a4
ρε = 2µSij Sij = 2µ( SR2ϕ + Sϕ2R ) = 4 µΩ2 4 ,
€ R
so the kinetic energy dissipation rate per unit height is:
∞
∞ 1
∫ a ρε2πRdR = 8πµΩ2 a 4 ∫ R 3 dR = 4πµΩ2 a 2 ,
a
€
which equals the work done turning the cylinder.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.16. Redo the lubrication-theory scaling provided in Section 9.3 for the situation when
there is an imposed external time scale τ so that the appropriate dimensionless time is t* = t/τ,
instead of that shown in (9.14). Show that this leads to the additional requirement ρh2/µτ << 1
for the validity of (9.17a). Interpret this new requirement in terms of the viscous diffusion length
ντ . Is the unsteady acceleration term needed to analyze the effect of a 100 Hz oscillation
imposed on a 0.1-mm film of 30-weight oil (ν ≈ 4 x10–4 m2/s)?

Solution 9.16. Use the dimensionless scaling provided by (9.14) except replace that used for the
time variable with t* = t/τ. The remaining members of (9.14),

x* = x/L , y* = y/h = y/εL, u* = u/U , v* = v/εU , and p* = p/Pa, where ε = h/L,

are unchanged. This mildly revised scaling does not influence the continuity equation. However,
it does impact the unsteady term in both momentum equations, but it is only necessary to
consider one. The horizontal momentum equation is the more interesting equation. Substituting
in the dimensionless variables leads to:
ρU ∂ u∗ " ρU 2 % ∗ ∂ u∗ " ερU 2 % ∗ ∂ u∗ " Pa % ∂ p∗ " U % ∂ 2 u∗ " U % ∂ 2 u∗
+ $ ' u + $ ' v = − $ ' ∗ + µ $ 2 ' ∗2 + µ $ 2 2 ' ∗2 .
τ ∂ t ∗ # L & ∂ x ∗ # ε L & ∂ y∗ # L &∂x # L &∂x #ε L &∂ y
Dividing by µU/ε2L2, produces:
ρε 2 L2 ∂ u∗ ∗ ∂u
∗
∗ ∂u
∗
$ 1 ' ∂ p ∗ 2 ∂ 2 u∗ ∂ 2 u∗
µτ ∂ t ∗
+ ( e ) ∂ x∗ ( e ) ∂ y∗ &% Λ )( ∂ x∗ + ε ∂ x∗2 + ∂ y∗2 .
ε 2
R u + ε 2
R v = −

The coefficient of the first term is different than that given in (9.16a), and it must be small for the
validity of (9.17a):
ρε 2 L2 ρ h 2 h 2
= = << 1 or h << [ντ]1/2.
µτ µτ ντ
where ν = µ/ρ. This new requirement states that the diffusion distance, [ντ]1/2, associated with
the time scale of flow unsteadiness τ must much larger than the gap distance h.
For a 100 Hz oscillation (t ~ 10 ms) imposed on a 0.1 mm gap containing 30-weight oil
with ν ≈ 4 x10–4 m2/s:
h2 (10 −4 m)2
= = 0.0025 .
ντ (4 ×10 −4 m 2 s −1 )(0.01s)
This is much less than unity, so the unsteady term is not needed in an elementary lubrication
analysis of this situation. However, if the gap were 10 times larger (1 mm) this conclusion would
be different.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.17. For lubrication flow under the sloped bearing of Example 9.3, the assumed
velocity profile was u(x, y) = −(1 2µ)( dP dx ) y ( h(x) − y)) + Uy h(x) , the derived pressure was
P(x) = Pe + ( 3µUα ho2 L) x(L − x) , and the load (per unit depth) carried by the bearing was
W = µUαL2 2ho2 . Use these equations to determine the frictional force (per unit depth), Ff,
€ lower (flat) stationary surface in terms of W, ho/L, and α. What is the spatially-
applied to the
€ averaged coefficient of friction under the bearing?
€ Solution 9.17. For lubrication flow under the sloped bearing, the velocity field was assumed to
1 ∂p Uy $ ∂u ' h ∂p U
be: u(y) = − y(h − y) + , so the wall shear stress is: τ w = &µ ) = − +µ .
2µ ∂x h % ∂y ( y= 0 2 ∂x h
Integrate this from x = 0 to x = L to find the friction force Ff per unit depth. Here the pressure
( )2
gradient will be: dp dx = 3µUα ho L (L − 2x) , so
€ L L % h (1− αx L) 3µ €Uα U (
F f = ∫ 0 τ w dx = ∫ 0 '− o 2
(L − 2x) + µ *dx
& 2 ho L ho (1− αx L) )
€ µUα 2 L µUL
Both integrals are elementary, so Ff is found to be: F f = − − ln(1− α ) .
4ho hoα
€ µUL $ α '
However, because α is small, only the α0 and α1 terms should be reported so, F f ≅ &1+ ) ,
ho % 2 (
µUαL2 Who
or using W = 2
, Ff ≅ (2 + α ) . €The coefficient of friction is a ratio:
2ho αL
F f ho (2 + α ) 2ho €
≅ ≈ .
W αL αL
Hence, we can € see that it is not possible to obtain a small coefficient of friction since the
€
maximum value of α is ~ho/L. Thus, the purpose of this type of bearing would not necessarily be
to produces a small coefficient of friction. Instead it prevents scuffing and wear since the
€ the bearing pad and the surface from coming into contact.
presence of the oil film prevents
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.18. A bearing pad of total length 2L moves to the right at constant speed U above a
thin film of incompressible oil with viscosity µ and density ρ. There is a step change in the gap
thickness (from h1 to h2) below the bearing as shown. Assume that the oil flow under the bearing
y(h j − y) dP(x) Uy
pad follows: u(y) = − + , where j = 1 or 2. The pad is instantaneously
2µ dx hj
aligned above the coordinate system shown. The pressure in the oil ahead and behind the bearing
is Pe.
a) By conserving mass for the oil flow, find a relationship between µ, U, hj, dP/dx, and an
€
unknown constant C.
b) Use the result of part a) and continuity of the pressure at x = 0, to determine
6µUL(h1 − h2 )
P(0) − Pe =
h23 + h13
c) Can this bearing support a externally applied downward load when h1 < h2?

Solution 9.18. a) Use a CV attached to the bearing pad with vertical flux surfaces that span the
gap. Conservation of mass € implies:
hj hj # &
y(h j − y) dP(x) Uy
∫( u(y) − U ) dy = C → ∫% % −
2µ dx
+
hj
− U ((dy = C
0 0$ '
Performing the integration produces:
1 dP(x) # h j h j &
3 3
hj h 3j dP(x) Uh j
− % − ( + U − Uh j = − − = C.
€ 2µ dx $ € 2 3' 2 12µ dx 2
b) Use part a), solve for dP/dx,
dP(x) 12µ # Uh j & 12µx # Uh j &
= − 3 %C + (, and integrate to find: P(x) − P(0) = − 3 % C + (.
€ dx hj $ 2 ' hj $ 2 '
Evaluate P(x) at x = ±L and note that P(+L) = P(–L) = Pe with P(x) continuous at x = 0:
12µL # Uh & 12µL # Uh &
− 3 %C + 1 ( = + 3 %C + 2 ( .
h1 $ 2 ' h2 $ 2 '
€ €
Uh h # h + h2 &
2 2
Solve for C = − 1 2 $ 13 ' , and then find:
2 % h1 + h23 (
12µL # € Uh & 12µL # Uh h ) h 2 + h22 , Uh1 & 6µUL # ) h12 + h22 , &
P(0) − Pe = 3 %C + 1 ( = 3 %− 1 2 * 13 -+ (= % −h2 * - + 1(
h1 $ 2 ' h1 $ 2 + h1 + h23 . 2 ' h12 $ + h13 + h23 . '
€
6µUL # −h12 h2 − h23 + h13 + h23 & 6µUL # −h12 h2 + h13 & # h1 − h2 &
= % ( = % ( = 6 µ UL % 3 3(
h12 $ h13 + h23 ' h12 $ h13 + h23 ' $ h1 + h2 '
c) When h1 < h2, the pressure at x = 0 is less than Pe. This means that the bearing pad is sucked
toward the surface, so when h1 < h2 the bearing cannot support an external load.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.19. A flat disk of radius a rotates above a solid boundary at a steady rotational speed
of Ω. The gap, h (<< a), between the disk and the boundary is filled with an incompressible
Newtonian fluid with viscosity µ and density ρ. The pressure at the edge of the disk is p(a).
a) Using cylindrical coordinates and assuming that the only non-zero velocity component is
uϕ(R,z), determine the torque necessary to keep the disk turning.
b) If p(a) acts on the exposed (upper) surface of the disk, will the pressure distribution on the
disk’s wetted surface tend to pull the disk toward or push it away from the solid boundary?
c) If the gap is increased, eventually the assumption of part a) breaks down. What happens?
Explain why and where ur and uz might be non-zero when the gap is no longer narrow.

Solution 9.19. a) By analogy with plane Couette flow, set uϕ (R,z) = ΩR(z /h) . The shear stress
on any element of the rotating disk will be: τ = µ(duϕ /dz) = µΩR /h . The torque necessary to
keep the disk turning will be:
r= a r= a
∫ RτdA = 2π (µΩ€/h) ∫ R 3 dR = πµΩa 4 2h .
Torque =
r= 0 r= o
€
b) For the assumed solution used to determine the answer to part a) to be valid, the quadratic
velocity terms in the conservation of momentum equation must approach zero. Thus, the radial
uϕ2 ∂p
momentum equation:
€ −ρ ≈ − suggests that the radial pressure gradient will be very small
R ∂R
but positive because u is non-zero just underneath the rotating disk. Thus, the pressure near the
ϕ

axis of rotation may be slightly lower. This suggests the disk may be mildly drawn downward as
it rotates.
c) If the gap€increases in size, a secondary flow with closed streamlines will develop that has
fluid moving away from the axis of rotation near the spinning disk, downward as the fluid
approaches R = a, toward the axis of rotation near the stationary disk, and upward near the axis
of rotation. This secondary flow will be driven by the radial pressure gradient mentioned above
in part b).
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.20. A circular block with radius a and weight W is released at t = 0 on a thin layer of
an incompressible fluid with viscosity µ that is supported by a smooth horizontal motionless
surface. The fluid layer's initial thickness is ho. Assume that flow in the gap between the block
and the surface is quasi-steady with a parabolic velocity profile:
uR (R,z,t) = − ( dP(R) dR) z( h(t) − z)) 2µ
where R is the distance from the center of the block, P(R) is the pressure at R, z is the vertical
coordinate from the smooth surface, h(t) is the gap thickness, and t is time.
a) By considering conservation of mass, show that: dh dt = (h 3 6µR)(dP dR) .
€
b) If W is known, determine h(t) and note how long it takes for h(t) to reach zero.

Solution 9.20. Consider a circular control volume under the block with radius of R and height
h(t). The integral form of conservation of€mass implies:
2 dh h(t ) h(t ) 1 dP(R) πRρ dP(R) h 3
πR ρ = −2πRρ ∫ 0 uR (R,z,t)dz = 2πRρ ∫ 0 z( h(t) − z)) dz =
dt 2µ dR µ dR 6
3
dh h dP(R)
which simplifies to: = .
dt 6µR dR
€ b) Since h does not depend on R, this equation can be integrated in the radial direction from R to
a:
dh a h3 a dh # a 2 − R 2 & h 3
€
dt R
∫ RdR =
6µ R
∫ dP which implies %
dt $ 2 ' 6µ
(= ( Pe − P(R)) ,
where Pe is the exterior pressure on the circular block (perhaps equal to 1 atm). Assuming quasi-
steady conditions, the weight of the block must be balanced by an integral of the excess-pressure
underneath the block:
€ €
a dh 6πµ a 2 dh 6πµ % 2 a 2 a 4 ( dh 3πµa 4
W = 2π ∫ 0 ( P(R) − Pe )RdR = − ∫
dt h 3 0
( a − R 2
)RdR = − '
dt h 3 & 2
a −
4)
* = −
dt 2h 3
.

Now, treat this result as a first-order nonlinear differential equation for h(t) and integrate in time:
h(t ) 1$ 1 1' 2Wt 1 1 4Wt
∫ ho h −3 dh = − 2 & h 2 − h 2 ) = − 3πµa 4 which implies h 2 (t) = h 2 + 3πµa4 .
€ % o( o

# 4Who2 t &−1 2
Solving for h(t) yields: h(t) = ho %1+ 4(
. For this solution, h(t) → 0 only when t → ∞.
$ 3πµa '
€ €

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.21. Consider the inverse of the previous exercise. A block and smooth surface are
separated by a thin layer of a viscous fluid with thickness ho. At t = 0, a force, F, is applied to
separate them. If ho is arbitrarily small can the block and plate be separated easily? Perform some
tests in your kitchen. Use maple syrup, peanut butter, liquid soap, pudding, etc. for the viscous
liquid. The flat top-side of a metal jar lid or the flat bottom of a drinking glass make a good
circular block. (Lids with raised edges and cups & glasses with ridges or sloped bottoms do not
work well). A flat countertop or the flat portion of a dinner plate can be the motionless smooth
surface. Can the item used for the block be more easily separated from the surface when tilted
relative to the surface? Describe your experiments and try to explain your results.

Solution 9.21. Using a Formica counter-top and the underside of a coffee-cup saucer, a
compelling case can be made for the accuracy of the result of Exercise 9.20. If an upward force
F is applied to the flat-bottomed block, the gap distance as a function of time is:
$ 4Fho2 t '−1 2
h(t) = ho &1− 4)
% 3πµa (
(this is the same as the answer for Exercise 9.20 with W replaced by –F). This formula then
predicts that the time for the plate to rise to z = +∞, t∞, is:
3πµa 4
€ t∞ = .
4Fho2
So, t∞ will increase with increasing a and µ, and decrease with increasing F and ho. Puddles of
peanut butter and pancake syrup both resisted vertical motion of the saucer. The peanut butter
(more viscous) was more effective in preventing the saucer from being lifted. The other
€
parametric dependencies were qualitatively verified as well. In all cases, the adhesion between
the saucer and the countertop failed when a small air-passage opened up under the saucer. This
allowed air, which is much less viscous than the liquids tested, to rush in under the saucer
allowing it to be lifted.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.22. A rectangular slab of width 2L (and depth B into the page) moves vertically on a
thin layer of oil that flows horizontally as shown. Assume
(2
)
u(y,t) = − h 2µ ( dP dx )( y h)(1− y h) where h(x,t) is the instantaneous gap between the slab
and the surface, µ is the oil’s viscosity, and P(x,t) is the pressure in the oil below the slab. The
slab is slightly misaligned with the surface so that h(x,t) = ho (1+ α x L) + h˙ o t where α << 1 and
h˙o is the vertical velocity of the slab. The pressure in the oil outside the slab is Po. Consider the
€
instant t = 0 in your work below.
∂h ∂ h(x,t )
€
a) Conserve mass in an appropriate CV to€show that: +
∂t ∂x 0 (
∫ u(y,t)dy = 0 . )
b) Keeping only linear terms in α, and noting that C and D are constants, show that:
12µ $ h˙ x 2 $ 2αx ' $ 3αx ''
P(x,t = 0) = 3 & o &1− ) + Cx&1− )) + D .
ho % 2€ % L ( % 2L ((
c) State the boundary conditions necessary to evaluate the constants C and D.
d) Evaluate the constants to show that the pressure distribution below the slab is:
6µh˙ L2 2 $ x'
ho (t)
(
P(x,t) − Po =€− 3 o 1− ( x L) &1− 2α ) .
% ) L(
e) Does this pressure distribution act to increase or decrease alignment between the slab and
surface when the slab is moving downward? Answer the same question for upward slab motion.
€

Solution 9.22. a) Use the differential CV shown to the right and

note that the upper boundary is moving at speed
b = (dh dt )e y = h˙ oe y but there is no mass flux through it. The other $""$%%# $""#!"$%%#
boundaries are stationary. Thus, Cons. of Mass, !"#
∂
( )
∫ ρdV + ∫ A ρ(u − b) ⋅ ndA = 0 , implies
∂t CV
€ ∂ h(x,t ) h(x +dx,t )
(ρhBdx ) − ∫ 0 ρu(x, y,t)Bdy + ∫ 0 ρu(x + dx, y,t)Bdy = 0 "#
∂t
∂h 1 ! h(x,t )
€ Divide each term by ρBdx to find: −
∂t dx 0 [ ∫
h(x +dx,t )
u(x + dx, y,t)dy − ]
∫ 0 u(x, y,t)dy = 0
∂h ∂
€ The stuff inside the [,]/dx term is the definition of ∂/∂x, so: +
∂t ∂x ( ∫ 0
h(x,t )
)
u(y,t)dy = 0 .
h h 3 dP h y $ y ' $ y ' h 3 dP
b) First compute the integral:€ ∫0 udy = − ∫ & ) & ) 12µ dx . Put this into the
1−
2µ dx 0 h % h ( % h (
d = −

∂ $ h 3 dP '
result of part a) to find a differential equation for € P(x,t): h˙ o − & ) = 0.
∂x %12µ dx (
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

h 3 ∂P ˙
Integrate once: = ho x + C , and then again with h(x,t) evaluated at t = 0,
12µ ∂x
h˙ x + C 12µ
P = 12µ ∫ 3 o
ho (1+ α x L)
3 dx + D ≈
ho3
∫ h˙o x + C (1− 3α x L)dx + D
( )
€µ % x 2
12 3α x 3 3α x 2 ( 12µ % x 2 % 2α ( % 3α ((
P ≈ 3 ' h˙ o + Cx − h˙ o −C * + D = 3 ' h˙ o '1− x * + Cx'1− x ** + D
ho & 2 L 3 L 2) ho & 2 & L ) & 2L ))
€ Here, C and D could depend on t, but the instant of interest is t = 0 so this dependence is
suppressed.
c) There are two boundary conditions on the pressure at either end of the slab:
€ P(x = −L) = P(x = +L) = Po .
d) Using the results of part b) and c), leads to two equations for the constants C and D.
12µ $ L2 $ 3α ''
P(–L) = Po = 3 & h˙ o (1+ 2α ) − CL&1+ )) + D , and
€ h o % 2 % 2 ((
12µ $ L 2
$ 3α ''
P(+L) = Po = 3 & h˙ o (1− 2α ) + CL&1− )) + D .
ho % 2 % 2 ((
These can be
€ solved to find:
˙ 6µL2 ˙
C = ho Lα and D = Po − 3 ho ,
€ ho
where quadratic terms in α have been dropped. Thus, the pressure distribution is:
12µ $ x 2 $ 2α ' ˙ $ 3α '' 6µL2
P(x)€= 3 & h˙ o &1− x ) + ho Lαx&1− x )) + Po − 3 h˙ o
ho % 2 % € L ( % 2L (( ho
6µL2 ˙ $ x 2 $ 2α ' x $ 3α ' '
= Po + 3 ho & 2 &1− x ) + 2α &1− x ) −1)
ho %L % L ( L % 2L ( (
6µL2 ˙ $ x 2 2α 3 2α '
= Po − 3
ho &1− 2 + 3 x − x + ...)
ho % L L L (
6µL2 ˙ $ x 2 '$ 2α '
= Po − ho &1− 2 )&1− x ) + ...
ho3 % L (% L (
where again quadratic terms in α have been dropped.
e) When the slab is moving downward (so h˙ o is negative), the highest pressure point lies in the
region where the gap is the smallest and this pressure distribution causes a moment on the slab
€
that tends to align the slab with the surface. This is beneficial in lubrication flows because the
oil pressure acts to prevent contact between slab and the surface. When the slab is moving
upward, the lowest pressure lies € in the region where the gap is the smallest and this pressure
distribution causes a moment on the slab that tends to misalign the slab with the surface. These
phenomena are readily apparent when pressing or lifting a flat object from a wet or oily surface.
In particular, it may be nearly impossible to lift the object straight up; however, once one corner
or edge is tilted up, the object is much easier to lift completely.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.23. Show that the lubrication approximation can be extended to viscous flow within
narrow gaps h(x,y,t) that depend on two spatial coordinates. Start from (4.10) and (9.1), and use
Cartesian coordinates oriented so that x-y plane is locally tangent to the center-plane of the gap.
Scale the equations using a direct extension of (9.14):
x* = x/L , y* = y/L , z* = z/h = y/εL , t* = Ut/L , u* = u/U , v* = v/U , w* = w/εU , and p* = p/Pa.
where L is the characteristic distance for the gap thickness to change in either the x or y direction,
and ε = h/L. Simplify these equation when ε2ReL → 0, but µUL/Pah2 remains of order unity to
find:
1 ∂p ∂ 2 u 1 ∂p ∂ 2v 1 ∂p
0≅− + 2 , 0≅− + 2 , and 0 ≅ − .
ρ ∂x ∂ z ρ ∂y ∂ z ρ ∂z

Solution 9.23. In Cartesian coordinates the starting equations (4.10) and the three components of
(8.1) are:
€ € ∂u ∂v ∂w €
+ + = 0,
∂x ∂y ∂z
∂u ∂u ∂u ∂u 1 ∂p & ∂ 2u ∂ 2u ∂ 2u )
x: +u +v +w =− + ν( 2 + 2 + 2 + ,
∂t ∂x ∂y ∂z ρ ∂x ' ∂x ∂y ∂z *
∂v € ∂v ∂v ∂v 1 ∂p & ∂ v ∂ v ∂ 2v )
2 2
y: +u +v +w =− + ν ( 2 + 2 + 2 + , and
∂t ∂x ∂y ∂z ρ ∂y ' ∂x ∂y ∂z *
€ ∂w ∂w ∂w ∂w 1 ∂p & ∂ w ∂ w ∂ 2w )
2 2
z: +u +v +w =− + ν( 2 + 2 + 2 +.
∂t ∂x ∂y ∂z ρ ∂z ' ∂x ∂y ∂z *
First insert the€given scaling into the continuity equation:
∂u ∂v ∂w U ∂u* U ∂v * εU ∂w * U $ ∂u* ∂v * ∂w * '
+ + = *
+ + = & + + ) = 0.
€ ∂x ∂y ∂z L ∂ x L ∂y * h ∂z * L % ∂x * ∂y * ∂z * (
Thus, there are no simplifications of the continuity equation as ε2ReL → 0, so it remains
unchanged.
Now insert the given scaling into the x-momentum equation, to find:
€ ' U ∂ 2 u* U ∂ 2 u* U ∂ 2 u* *
U ∂u* U 2 * ∂u* U 2 * ∂u* εU 2 * ∂u* Pa ∂p*
+ u + v + w = − + ν ) 2 *2 + 2 *2 + 2 *2 , ,
L U ∂t * L ∂x * L ∂y * h ∂z * ρL ∂x * ( L ∂x L ∂y h ∂z +
Simplify using ε = h/L, and collect terms with like coefficients:
U 2 # ∂u* * ∂u
*
* ∂u
*
* ∂u
*&
Pa ∂p* U # ∂ 2 u* ∂ 2 u* & U ∂ 2 u*
% + u + v + w ( = − + ν % + ( + ν .
€ L $ ∂t * ∂x * ∂y * ∂z * ' ρL ∂x * L2 $ ∂x * 2 ∂y * 2 ' h 2 ∂z * 2
Multiply through by ρh2/µU, and use ε = h/L and ReL = ρUL/µ to reach:
$ ∂u* ∂u* ∂u* ∂u* ' P h 2 ∂p* $ 2 * ∂ 2 u* ' ∂ 2 u*
2 ∂ u
ε 2 Re L & * + u* * + v * * + w * * ) = − a + ε & *2 + *2 ) + *2 .
€ % ∂t ∂x ∂y ∂z ( µUL ∂x * % ∂x ∂y ( ∂ z
When ε2ReL → 0, but µUL/Pah2 remains of order unity this equation simplifies to:
P h 2 ∂p* ∂ 2 u* 1 ∂p ∂ 2u
0≅− a + , or in dimensional form: 0 ≅ − + ν .
€ µUL ∂x * ∂z* 2 ρ ∂x ∂z 2

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

The procedure for the y-momentum equation is the same as that for the x-momentum equation
with u replaced by v and ∂p/∂x replaced by ∂p/∂y. The final dimensionless and dimensional
forms are:
P h 2 ∂p* ∂ 2v * 1 ∂p ∂ 2v
0≅− a + , or in dimensional form: 0 ≅ − + ν .
µUL ∂y * ∂z* 2 ρ ∂y ∂z 2
The z-momentum equation is different that the other two because the scaling of w differs
from that of u and v, and the pressure gradient involves z (not x or y). Inserting the given scaling
leads to:
€ €

εU ∂w * εU 2 * ∂w * εU 2 * ∂w * ε 2U 2 * ∂w * Pa ∂p* ' εU ∂ 2 w * εU ∂ 2 w * εU ∂ 2 w * *
+ u + v + w = − + ν ) 2 *2
+ 2 + ,
As ∂t * simplify
L Ubefore, L x * εL= h/L,∂yand
∂using *
h terms
collect ∂z* withρlike *
L ∂zcoefficients:
( L ∂x L ∂y * 2 h 2 ∂z * 2 +
εU 2 $ ∂w * * ∂w
*
* ∂w
*
* ∂w
*'
Pa ∂p* εU $ ∂ 2 w * ∂ 2 w * ' εU ∂ 2 w *
& +u +v +w )=− + ν 2 & *2 + *2 ) + ν 2 *2 .
L % ∂t * ∂x * ∂y * ∂z * ( ρL ∂z* L % ∂x ∂y ( h ∂z
2
€ Multiply through by ρεh /µU, and use ε = h/L and ReL = ρUL/µ to reach:
$ ∂w * ∂w * ∂w * ∂w * ' P h 2 ∂p* $ 2 * ∂ 2w* ' 2 *
4 ∂ w 2∂ w
ε 4 Re L & * + u* * + v * * + w * * ) = − a + ε & *2
+ 2 ) + ε 2
.
€ % ∂t ∂x ∂y ∂z ( µUL ∂z* % ∂x ∂y * ( ∂z *
When ε2ReL → 0, but µUL/Pah2 remains of order unity this equation simplifies to:
Pa h 2 ∂p* 1 ∂p
0≅− *
, or in dimensional form: 0 ≅ − .
€ µUL ∂z ρ ∂z

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.24. A squeegee is pulled across a smooth flat stationary surface at a constant speed U.
The gap between the squeegee and surface is h(x) = hoexp{+αx/L} and this gap is filled with a
fluid of constant density ρ and viscosity µ. If the squeegee is wetted by the fluid for 0 ≤ x ≤ L,
and the pressure in the surrounding air is pe, what is the pressure p(x) distribution under the
squeegee and what force W perpendicular to the surface is needed to hold the squeegee in place?
Ignore gravity in your work.

W! U!
ho! h(x)! pe!
pe!
p(x) = ?!
x!
x = 0! x = L!

Solution 9.24. Follow the solution approach in Example 9.3. Here the pressure gradient will be:
dp(x) 12µ 6µU 12µ 6µU
= − 3 C1 − 2 = − 3 C1e−3α x L − 2 e−2α x L .
dx h (x) h (x) ho ho
Integrate to find the pressure distribution:
12µ L 6µUL 3µUL " 4 %
p(x) = + 3 C1e−3α x L + 2 e−2α x L + C2 = 2 $ C1e−3α x L + e−2α x L ' + C2 .
3hoα 2ho α ho α # 3Uho &
The boundary conditions are p(0) = p(L) = pe, and these lead to two equations for the two
constants C1 and C2.
3µUL ! 4 $ 3µUL " 4 %
pe = 2 # C1 +1& + C2 , and pe = 2 $ C1e−3α + e−2α ' + C2 . (1,2)
ho α " 3Uho % ho α # 3Uho &
Subtract (2) from (1) to find C1:
3µUL " 4 % 3Uho 1− e−2α
0= 2 $ C1 (1− e−3α ) +1− e−2α ' or C1 = − .
ho α # 3Uho & 4 1− e−3α
Substitute this into (1) to find C2:
3µUL " e−2α − e−3α %
C2 = pe − 2 $ '.
ho α # 1− e−3α &
Use these constants to construct p(x):
3µUL " 1− e−2α −3α x L −2α x L e−2α − e−3α %
p(x) − pe = 2 $ − e +e − '.
ho α # 1− e−3α 1− e−3α &
The force (or load) W is determined from the integral:
L
3µ LU L # 1− e−2α −3α x L −2α x L e−2α − e−3α &
W = ∫ ( p(x) − pe )dx = 2 ∫ % − e +e − ( dx .
0 ho α 0 $ 1− e−3α 1− e−3α '
The integration of the exponentials may be done directly:
( L L +
3µ LU * 1− e−2α " e−3α x L % " e−2α x L % e−2α − e−3α L
W= 2 − $ ' $+ ' − [ ]0 -- .
x
ho α *) 1− e−3α # −3α L &0 # −2α L &0 1− e−3α ,
Continue with the evaluation:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

3µ LU " L 1− e−2α −3α L −2α e−2α − e−3α %

W= $
ho2α # 3α 1− e−3α
( e −1 ) −
2α
( e −1) −
1− e−3α
L'
&
3µ LU " L −2
e −e α −3α %
= 2 $ (1− e−2α ) − L '
ho α # 6α 1− e−3α &
2
αµ L U
And, when α << 1, this reduces to W = , which is the same as the result of Example 9.3.
2ho2
This occurs because the two shapes become the same; h(x) = hoexp{+αx/L} ≈ ho(1 +αx/L) when
α << 1.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.25. A close-fitting solid cylinder with net weight W (= actual weight – buoyancy),
length L, and radius a is centered in and may slide along the axis of a long vertical tube with
radius a + h, where h << a. The tube is filled with oil having constant viscosity µ that is pumped
slowly upward at a volume flow rate Q.
a) Use dimensional analysis to find a scaling law for the value of Q that holds the cylinder
stationary when fluid inertia is unimportant.
b) Use the lubrication approximation and assume that the pressure is uniform above and below
the cylinder to determine a formula for the value of Q that holds the cylinder stationary.
h!

W!
L!

2a!

Q!

Solution 9.25. a) There are six parameters (Q, W, L, a, h, µ) so the units matrix (which has rank
3) is:
Q W L a h µ. The 6 – 3 = 3 dimensionless groups are:
M 0 1 0 0 0 1 Π1 = Qµ/LW, Π2 = a/L, and Π3 = h/L, so
L 3 1 1 1 1 -1 the scaling law is: Qµ/LW = f(a/L, h/L),
T -1 -2 0 0 0 -1 where f is an undetermined function.
b) Under the lubrication approximation, the flow in the narrow gap will be
z! parabolic. Choose the z-axis vertically upward, and let y be a coordinate
pa! that is zero on the tube wall and that increases toward the tube centerline
as shown in the drawing to the left. Both the tube and the cylinder are
h! stationary so w(y = 0) = w(y = h) = 0. With these coordinates, the fluid
velocity in the gap will be:
h 2 " dp % y " y % h 2 " p − pb % y " y %
L! w(y) = − $ ' $1− ' = − $ a ' $1− '
y! 2µ # dz & h # h & 2µ # L & h # h &
where pa and pb are the (uniform) pressures in the fluid above and below
the cylinder, respectively. The cylinder's net weight W will be supported
pb!
by shear stress on its sides [area = 2πaL] and the pressure difference
between its ends acting on its round surfaces:
" dw % " h " p − pb %%
W = 2π aLµ $ − ' + ( pb − pa ) π a 2 = 2π aLµ $ − $ a 2
'' + ( pb − pa ) π a ,
# dy &y=h # 2 µ # L &&
or: W = ( pb − pa )"#π ah + π a 2 $% . In addition, all the oil flow must flow past the cylinder. The gap-
center circle has circumference 2π(h + a/2), so
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

h # h 3 # pa − pb &&
Q = 2π ( a + 12 h ) ∫ w(y)dy =2π ( a + 12 h ) % − % (( .
0
$ 12µ $ L ''
Eliminate the pressure difference from last two equations for Q and W to find:
W ( a + 12 h ) h 3 Wh 3
Q= ≅ ,
6µ L !"ah + a 2 #$ 6µ aL
where the second more approximate form is valid as h a → 0 .
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.26. A thin film of viscous fluid is bounded below by a flat stationary plate at z = 0. If
the in-plane velocity at the upper film surface, z = h(x,y,t), is U = U(x,y,t)ex + V(x,y,t)ey, use the
equations derived in Exercise 8.19 to produce the Reynolds equation for constant-density thin-
flim lubrication:
*$ h 3 ' - ∂h
∇ ⋅ ,& )∇p/ = 12 + 6∇ ⋅ ( hU)
+% µ ( . ∂t
where ∇ = e x (∂ ∂x ) + e y (∂ ∂y ) merely involves the two in-plane dimensions.

Solution 9.26. For the specified geometry, solutions for the x- and y-direction equations are
€
given by appropriately evaluated versions of (8.19):
€ h 2 ∂p z % z ( z h 2 ∂p z % z ( z
u≅− ' 1− * + U , and v ≅ − '1− * + V .
2µ ∂x h & h ) h 2µ ∂y h & h ) h
where h, U, and V are functions of x, y, and t.
Now consider a stationary nearly-rectangular control volume of with elemental base area
dxdy and height h(x,y,t). If the flat vertical sides of the CV are perpendicular to the coordinate
€ then an integral statement of conservation
directions, € of mass will include one term involving the
time-variation of h and four flux terms, one for each vertical side:
h(x,y,t ) h(x,y,t ) h(x +dx,y,t ) h(x,y +dy,t )
∂h
ρ dxdy − ρdy ∫ udz − ρdx ∫ vdz + ρdy ∫ udz + ρdx ∫ vdz = 0 .
∂t 0 0 0 0
For constant density, this can be simplified and rearranged:
∂h 1 %h(x +dx,y,t ) h(x,y,t ) ( 1 %h(x,y +dy,t ) h(x,,t ) ( ∂h ∂ $h(x,y,t ) ' ∂ $h(x,,t ) '
+ ' ∫ udz − ∫ udz* + ' ∫ vdz − ∫ vdz* = 0 , or + & ∫ udz) + & ∫ vdz) = 0 .
∂t € dx & 0 0 ) dy & 0 0 ) ∂t ∂x % 0 ( ∂y % 0 (
The two integrations in [,]-brackets in the second equality can be completed using the equations
above for u and v.
h
h 2 ∂p h z & z ) h
z h
h 2 ∂p h z & z ) h
z
€ ∫ udz ≅ − 2µ ∂x ∫ h ('1− h +*dz + U ∫ h dz € ∫ vdz ≅ − ∫ (1− + dz + V ∫ dz
2µ ∂y 0 h ' h *
0 0 0 0 0 h

h 3 ∂p 1 1
h 3 ∂p 1 1
=−
2µ ∂x 0
∫ β (1− β )d β + Uh ∫ βdβ =−
2µ ∂y 0
∫ β (1− β )d β + Vh ∫ βdβ
0 0
3
, and 3
.
h ∂p & 1 1 ) & 1) h ∂p & 1 1 ) & 1)
=− ( − + + Uh( + =− ( − + + Vh( +
2µ ∂x ' 2 3 * ' 2* 2µ ∂y ' 2 3 * ' 2*
h 3 ∂p Uh h 3 ∂p Vh
=− + , =− + .
12µ ∂x 2 12µ ∂y 2
The result of the continuity equation becomes:
∂h ∂ $ h 3 ∂p Uh ' ∂ $ h 3 ∂p Vh '
+ &− + ) + &− + ) = 0.
€ ∂t ∂x % 12µ ∂x € 2 ( ∂y % 12µ ∂y 2 (
Rearrange this equation and use ∇ = e x (∂ ∂x ) + e y (∂ ∂y ) and U = Uex + Vey,
∂h ∂ ∂ ∂ # h 3 ∂p & ∂ # h 3 ∂ p & ∂h % h3 (
12 + 6 (Uh) + 6 (Vh ) = % ( + % ( or 12 + 6∇ ⋅ ( )
hU = ∇ ⋅ ' ∇p* .
∂t ∂x € ∂y ∂x $ µ ∂x ' ∂y $ µ ∂ y ' ∂t &µ )
Here it is worth noting€that when U, V, and h are known, the solution of final equation specifies
the pressure in the fluid film.
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 8.21. Fluid of density ρ and viscosity µ flows inside a long tapered tube of length L and
radius R(x) = (1 – αx/L)Ro where α < 1, and Ro << L.
a) Estimate the volume discharge rate Qv through the tube, for a given pressure difference Δp
sustained between the inlet and the outlet.
b) Discuss the range of validity of your solution in terms of the parameters of the problem.

Solution 8.21. a) If the tube diameter changes slowly enough, the flow will be in equilibrium
with the local diameter of the pipe and the axial velocity profile will be the same as that in an
untapered pipe of the same diameter. For flow in an untapered pipe: Qv = U ave A , and
d 2 dp
U ave = − . For this problem let U ave = U ave (x) since A = A(x) and
32µ dx
R 2 (x) dp €
Qv = U ave (x)A(x) = const , therefore: U ave (x) = − . Eliminate U ave (x) from the last two
8µ dx
€
€ equations and use A(x) = πR 2 (x) to get:
R 2 (x) dp πR 4 (x) dp
€ Qv = − A(x) = − (1)
€ 8µ dx 8µ € dx
Now solve € for −dp dx and integrate in x from 0 to L:
−4
L 8µ Q 8µQv L & x) 8µQv L 1−α −4
−( p(L) − p(0)) = ∫ 0 4
v
dx = 4 ∫0 (
1− α + dx = ∫ ζ dζ
€ πR (x) πRo ' L* πRo4α 1
The integral
€ is elementary and is easily evaluated, yielding:
8µQv L % 1 (
p(0) − p(L) = 4 ' 3
−1* .
€ 3πRo α & (1− α ) )
Inverting this to find Qv produces:
3πRo4α % 1 (−1
Qv = ( p(0) − p(L) )' 3
−1 * .
€ 8 µ L & (1− α ) )
This result should be expanded for α << 1,
$ 1 '−1 (1− α ) 3 1− 3α + 3α 2 − α 3 1 $1− 3α + ...' 1 1− 2α
& 3
−1 ) = 3
= 2 3
= & )≈ (1− 3α )(1+ α ) ≈ .
% (1− α ) ( 1−
€(1− α ) 3α − 3α + α 3α % 1− α + ... ( 3α 3α
Therefore:
πR 4 $ p(0) − p(L) '
Qv = o & )(1− 2α ) ,
€ 8µ % L (
so for a fixed pressure difference Qv is larger when the pipe expands (α < 0) and is smaller when
the pipe contracts (α > 0).
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

b) The flow accelerates or decelerates inside a contracting or expanding pipe, respectively. The
1 dp
above solution will only be valid if fluid acceleration can be ignored compared to or ν∇ 2 u
ρ dx
(recall that these two terms are set equal to obtain the solution for the velocity profile). Hence
∂u 1 dp ∂u
we need only consider one of them, and therefore must have: u << where
€ u is a
∂x ρ€ dx ∂x
typical acceleration term. This requirement implies:
∂U ave R 2 (x) dp ∂ $ Qv ' 1 dp
U ave = & 2 ) <<
∂x 8µ dx ∂x€ % πR (x) ( ρ dx €
2
R (x) dp Q
where the two relationships U ave = − = 2 v has been used outside and inside the
8µ dx πR (x)
€ common factors, evaluate the derivative using R(x) = (1 – αx/L)Ro and
derivative. Cancel
assume R(x) ≈ Ro to find:
€ R 2 (x)Qv % αRo ( Qvα
' *≈ << 1 .
4νπR 3 (x) & L ) 4νπL
€ When this is true, the solution for part a) should be OK.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.28. A circular lubricated bearing of radius a holds a stationary round shaft. The
bearing hub rotates at angular rate Ω as shown. A load per unit depth on the shaft, W, causes the
center of the shaft to be displaced from the center of the rotating hub by a distance εho, where ho
is the average gap thickness and ho << a. The gap is filled with an incompressible oil of viscosity
µ. Neglect the shear stress contribution to W.
a) Determine a dimensionless scaling law for |W|.
b) Determine W by assuming a lubrication flow profile in the
gap and h(θ) = ho(1 + εcosθ) with ε « 1.
c) If W is increased a little bit, is the lubrication action
stabilizing?

Solution 9.28. a) Follow the usual dimensional analysis

approach. The flow is caused by the balance of pressure and
viscous forces, so the fluid inertia (i.e. ρ) doesn't matter.
• Boundary condition & material parameters: µ = viscosity of the fluid
Ω = rotation speed of the shaft
ho = mean gap separation
a = radius of the bearing
ε = accentricity parameter
Solution parameter: W = |W| = magnitude of the applied load per unit depth
• Create the parameter matrix: W Ω ho µ a ε
–––––––––––––––––––––––––––––––––
Mass: 1 0 0 1 0 0
Length: 0 0 1 -1 1 0
Time: -2 -1 0 -1 0 0
This matrix has rank three.
• Determine the number of dimensionless groups: 6 parameters - 3 dimensions = 3 groups
W h
• Construct the dimensionless group: Π1 = , Π 2 = o , and Π 3 = ε .
µΩa a
W $ h '
• Write a dimensionless law: = fn& o ,ε) .
µΩa %a (
b) Because the fluid gap is very €
€ narrow,€consider a new coordinate: z that is zero on the surface
of the shaft and equal to h(θ) at the outer rotating bearing surface, and ξ = aθ = downstream
distance. Place a stationary control volume in the gap that encloses the fluid in the gap between
€
angles θ1 and θ2. Conservation of mass for the gap-flow will imply that:
[∫ h

0
u(z)dz ]
θ =θ 1
= [∫ h

0
u(z)dz ]
θ =θ 2
= C1 = const.
(Note, this situation is subtly different than the sloped bearing worked in lecture; in the
unwrapped geometry, the upper wall is moving but its geometric shape does not move.)
Conservation of momentum in the angular direction in the gap will be satisfied locally by using
the Couette flow€solution for u(z): i.e.
h % z(h − z) dP Ωaz ( h 3 dP Ωah
C1 = ∫ 0 '− + * dz = − +
& 2µa dθ h ) 12µa dθ 2
Treat this result as a differential equation for P(θ):

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

dP −12µaC1 + 6µΩa 2 h −12µaC1 + 6µΩa 2 ho (1+ ε cos θ )

= =
dθ h3 ho3 (1+ ε cosθ ) 3
Even though integration of this equation can be completed exactly, it's tedious and unnecessary
because ε << 1. Therefore, expand for ε << 1 and only keep the linear terms in ε:
dP −12µaC1 + 6µΩa 2 ho 36µaC1 −12µΩa 2 ho
€ = + ε cosθ + ...
dθ ho3 ho3
Integrate from 0 to θ:
−12µaC1 + 6µΩa 2 ho 36µaC1 −12µΩa 2 ho
P(θ ) − P(0) = 3
θ + 3
ε sinθ + ...
€ h o h o
For the pressure to be single valued [i.e. P(2π) = P(0)], we must have C1 = Ωaho/2 (this makes
the linear-in-θ term in the last equation equal to zero). Therefore P(θ ) − P(0) = 6µΩa 2ε sin θ ho2 .
€ Now use this pressure distribution to determine the load on the shaft and ignore any net
contributions from shear stresses. The external load on the shaft, which is equal an opposite to
the fluid load applied to the shaft, may have both x- and y-components, so
2π € 
W = (W x ,W y ) = − ∫ 0 ( P(0) − P(θ )) nadθ ,
where n = (cosθ,sin θ ) is the outward normal from the shaft’s surface.
6µΩa 3ε 2 π ' 6πµΩa 3ε *
W=
ho2
∫ 0 sin θ (cosθ,sin θ ) dθ = )0,+ h 2 ,
€ ( o +
€ This answer matches the dimensional analysis result. However, the external load applied to the
shaft pushes the shaft up! This seemingly odd result, makes sense if you think about what's
going on. For the fluid to pass through the narrowest part of the gap at θ = π where the viscous
€
forces are largest, there must be a favorable (i.e negative) dP/dθ in this region. This means that
the pressure above the shaft (near θ = π/2) must be greater than the pressure below the shaft
(near θ = 3π/2). With high pressure above and low pressure below, the lubricant must be trying
to push the shaft down. In order to maintain shaft position, there must be an upward-directed
externally-applied load on the shaft.
c) If Wy is impulsively increased by a small amount, the shaft will move vertically. The
lubrication action will then push the shaft to the left. Eventually a new steady state will be
reached with the shaft misaligned in the bearing (shaft center to the left) with a larger value of ε
than the original configuration. So, the lubrication action can be considered stabilizing.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.29. As a simple model of small-artery blood flow, consider slowly varying viscous
flow through a round flexible tube with inlet at z = 0 and outlet at z = L. At z = 0, the volume flux
entering the tube is Qo (t) . At z = L, the pressure equals the exterior pressure pe. The radius of the
tube, a(z,t), expands and contracts in proportion to pressure variations within the tube so that (i)
a − ae = γ ( p − pe ) , where ae is the tube radius when the pressure, p(z,t), in the tube is equal to pe,
and γ is a positive
€ constant. Assume the local volume flux, Q(z,t), is related to ∂p ∂z by (ii)
Q = −(πa 4 8µ)(∂p ∂z) .
€ a) By conserving mass, find a partial differential equation that relates Q and a.
b) Combine (i), (ii), and the result of part a) into one partial differential equation
€ for a(z,t).
c) Determine a(z) when Qo is a constant and the flow is perfectly steady.
€

Solution 9.29. a) Conserve mass in a differential disk control volume perpendicular to the flow
direction that has circular inlet and outlet surfaces located at z and z + Δz with radii of a(z,t) and
a(z + Δz,t). There is no flow through the tube wall so:
a(z,t ) a(z +Δz,t )
∂
∫
∂t CV
ρdV − ∫ ρ u(r,t) 2πrdr + ∫ ρu(r,t)2πrdr = 0 .
0 0
r= R(z,t )
Noting that Q(z,t) = ∫ r= 0
u(r,t)2πrdr and that the CV volume is πa 2 (z,t)Δz allows the
equation above to be simplified to:
€ (∂ ∂t )(πa 2 ) + (Q(z + Δz,t) − Q(z,t)) Δz = 0 ,
€ ∂ ∂Q €
or taking the limit as Δz → 0 :
∂t
πa 2 + ( )
∂z
= 0.

€ to find: ∂a ∂z = γ ∂p ∂z . Put this into (ii) to get: Q = −(πa 8γµ)(∂a ∂z) .

4
b) Differentiate (i)
$ 4 '
€ the result of part a): ∂ ( a 2 ) = ∂ & a ∂a ) .
Insert this into
€ ∂t ∂z % 8γµ ∂z (
c) For steady flow, € the part a) result is not needed. From € part b), (i) and (ii) imply:
5
πa
Qo = −(πa 4 8γµ)(∂a ∂z) . Integrate once to find: = −Qo z + C where C is a constant. At z =
€ 40γµ
πae5 % 40γµQo (1 5
L, a = ae so C = + Qo L . Thus the final answer is: a(z) = ' (L − z) + ae5 * .
40γµ & π )
€ €

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.30. Consider a simple model of flow from a tube of toothpaste. A liquid with
viscosity µ and density ρ is squeezed out of a round horizontal tube having radius a(t). In your
work, assume that a is decreasing and use cylindrical coordinates with the z-axis coincident with
the centerline of the tube. The tube is closed at z = 0, but is open to the atmosphere at z = L.
Ignore gravity.
da a
a) If w is the fluid velocity along the z-axis, show that: za + ∫ 0 w(z,R,t)RdR = 0 .
dt
b) Determine the pressure distribution, p(z) – p(L), by assuming the flow in the tube can be
1 dp 2
treated within the lubrication approximation by setting w(z,R,t) = −
4 µ dz
(a (t) − R 2 ) .
€
c) Find the cross-section-average flow velocity w ave (z,t) in terms of z, a, and da dt .
d) If the pressure difference between z = 0 and z = L is ΔP, what is the volume flux exiting the
tube as a function of time. Does this answer partially explain why fully emptying a toothpaste
€
tube by squeezing it is essentially impossible?
€ €
Solution 9.30. a) Choose a CV that encloses the fluid inside the tube from the capped end to an
axial distance z. Conservation of mass implies:
d a da a

dt
( ρπa 2 z) + ∫ 0 ρw(z,R,t)2πRdR = 0 or za + ∫ 0 w(z,R,t)RdR = 0 .
dt
b) Here p = p(z) alone; insert the given velocity profile into the result of part a) and integrate:
da 1 dp a 2 a 4 (t) dp
za =
dt 4 µ dz 0 €
∫ ( a (t) − R 2
) RdR =
16µ dz
.
€
The ends of this equality form a differential equation for the pressure; integrate from z to L:
p(z) − p(L) = −(8µ a 3 )( da dt )( L2 − z 2 ) .
c) Multiply the result
€ of part a) by 2a–2, insert factors of π, and recognize the definition of wave:
2 a 1 a % 2 ( da 2z da
2 ∫0
w(z,R,t)RdR = 2 ∫ 0 w(z,R,t)2πRdR = w ave (z,t) = −' 2 * za = − ,
a € π a & a ) dt a dt
d) Use Q(t) = πa2wave(L,t), and combine the results of parts b) and c) to eliminate da/dt. From
8µ $ da ' 2 " da % a 3ΔP
& )( ) , or $# dt '& 8µL2 ; so
2
part b): p(0) − p(L) = ΔP = − L − 0 = −
€ a 3 % dt (
2L da da % −a 3ΔP ( π a 4 ΔP
Q(t) = πa 2 w ave (L,t) = −πa 2 = −2πLa = −2πLa' 2 *
=
a dt dt & 8µL ) 4 µL
€ Yes, this answer explains a lot about flow€from a toothpaste tube. Here, it is clear that for a
finite ΔP (the pressure one can exert by squeezing), the volume flux from the tube will be
proportional to a 4 . Thus, as the tube empties and its effective cross section decreases, the flow
€
rate from the tube approaches zero. The only hope for better emptying the tube is to decrease L,
and it is an easily observed fact that more toothpaste can be extracted from a nearly-empty tube
by rolling it up from the sealed end.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.31. A large flat plate below an infinite stationary incompressible viscous fluid is set
in motion with a constant acceleration, u˙ , at t = 0. A prediction for the subsequent fluid motion,
u(y,t), is sought.
a) Use dimensional analysis to write a physical law for u(y,t) in this flow.
b) Starting from the x-component of (8.1) determine a linear partial differential equation for
€
u(y,t).
c) The linearity of the equation obtained for part c) suggests that u(y,t) must be directly
proportional to u˙ . Simplify your dimensional analysis to incorporate this requirement.
d) Let η = y/(νt)1/2 be the independent variable, and derive a second-order ordinary linear
differential equation for the unknown function f(η) left from the dimensional analysis.
e) From
€ an analogy between fluid acceleration in this problem and fluid velocity in Stokes’ first
t

[ ( )]
problem, deduce the solution u(y,t) = u˙ ∫ 1− erf y 2 νt $ dt $ and show that it solves the
0
equation of part b).
f) Determine f(η) and – if your patience holds out – show that it solves the equation found in part
d).
g) Sketch the expected€velocity profile shapes for several different times. Note the direction of
increasing time on your sketch.

Solution 9.31. a) The problem parameters are: u˙ , ρ, y, t, and µ. The solution parameter is u.
• Create the parameter matrix:
u u˙ ρ y t µ
––––––––––––––––––––––––––––––––––
€
Mass: 0 0 1 0 0 1
Length: 1 1 -3 1 0 -1
€
Time: -1 -2 0 0 1 -1
This matrix has rank three.
• Determine the number of dimensionless groups: 6 parameters - 3 dimensions = 3 groups
• Construct the dimensionless groups: ∏1 = u/ u˙ t, ∏2 = y/(νt)1/2, ∏3 = ρ u˙ ty/µ
• Write a dimensionless law: u = u˙ tF(y/(νt)1/2, ρ u˙ ty/µ), where F is an unknown function.
b) Assume that u = u(y,t) only!, then ∂u/∂x = ∂2u/∂x2 = 0. In this case the continuity equation
requires ∂v/∂y = 0, so since v = 0 at the
€ surface of the plate it €is zero everywhere. If the plate is
large, there will be no end
€ effects so the
€ pressure gradient must also be zero. This eliminates the
nonlinear terms in the equation, the pressure gradient, and one of the viscous terms. The final
∂u ∂ 2u
result being: =ν 2 .
∂t ∂y
c) If the form of the solution must be linear with respect to u˙ , then ∏3 can not be a parameter
since ∏1 takes care of any linear relationship between u and u˙ . This means: u = u˙ tf(y/(νt)1/2)
€
€
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

d) First convert derivatives, then differentiate being sure to account for the pre-factor of u˙ t.
∂ ∂η ∂ 1 y ∂ η ∂ ∂ ∂η ∂ 1 ∂
= =− =− , and = = .
∂t ∂t ∂η 2 νt ∂η 3 2t ∂η ∂y ∂y ∂η νt ∂η
∂u ∂ η df ∂ 2u ∂ 2 €1 d f
2
Therefore: = [u˙ tf (η)] = u˙ f − u˙ t , and ν 2 = 2 [u˙ tf (η)] = νu˙ t .
∂t ∂t 2t dη ∂y ∂y ν t dη 2
Put these
€ converted derivatives together and cancel € common factors, to find:
2
d f η df
+ − f = 0.
€ dη 2 €2 dη
e) In Stokes' first problem the plate's velocity was impulsively raised from 0 to U and the
velocity field equation and its solution are:
∂u ∂ 2u
∂t
€= ν 2 and u(y,t) = U 1− erf y 2 νt .
∂y [ ( )]
In the current problem, the plate's acceleration is impulsively raised from 0 to u˙ , and the
∂u ∂u ∂u
horizontal acceleration of fluid particles is a(y,t) = + u = + 0 , because ∂u/∂x = 0. Thus,
€ ∂t ∂x ∂t
€
by time differentiating the field equation for u, the field equation for a and its analogous solution
€
is obtained:
∂a ∂ 2a
∂t [
= ν€ 2 and a(y,t) = u˙ 1− erf y 2 νt .
∂y
( )]
t "= t
However, a = ∂u/∂t implies u(y,t) = ∫ t "= 0 a(y, t ")dt ", so the above analogy suggests the velocity
field for the impulsively accelerated plate will be:
€
€ [
t $= t
(
u(y,t) = u˙ ∫ t $= 0 1− erf y 2 νt $ dt $ . )]
To show that
€ this solution satisfies the field equation for u, determine the derivatives
ζ
(
using the definition of the error function, erf(ζ ) = 2 π ∫ 0 exp(−ξ 2 )dξ .)
t €
∂u ∂
∂t
= u˙
∂t
∫ [1− erf ( y )] [ (
2 νt % dt % = u˙ 1− erf y 2 νt )] ,
0

∂u ∂ t
€ ∂ ) 2 t y 2 νt % ,
∂y
= u˙
∂y
∫ [1− erf ( y )]
2 νt % dt % = u˙ * t −
∂y +
∫ ∫ exp(−ξ 2 ) dξdt %-
0 π 0 0 .
€
) 2 t 1 , )/ 1 t exp(− y 2 4νt %) ,/
= u˙ *0 − ∫ exp(− y 4νt %)
2
dt %- = u˙ *− ∫ dt %-,
+ π 0 2 νt % . /+ π 0 νt % /.

u˙ ∂ (* t exp(− y 4νt &) ,* u˙ y t exp(− y 4νt &)

2 2
∂ 2u
ν 2 =ν )− ∫ dt &- = ∫ dt &.
∂y π ∂y *+ 0 νt & *. 2 νπ 0 t &3 2
€ Manipulate the final relationship for the second y-derivative to introduce the integration variable

( )
ξ = y 2 νt $ using dξ = y 2 ν t $−3 2 (−1 2) dt $ = −ydt 4 ν t $3 2 to find:
€

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

y 2 νt
∂ 2u 2˙u t ' −ydt & * 2˙u
ν 2 =−
∂y
∫ exp(− y 4νt &))(
2
32,
=− ∫ exp(−ξ 2 )dξ
π 0 4 ν t& + π ∞

∞ ∞ y 2 νt
2˙u 2˙u 2˙u
=+ ∫ exp(−ξ 2 )dξ = + ∫ exp(−ξ 2 )dξ − ∫ exp(−ξ 2 )dξ
π y 2 νt π 0 π 0

' 2 y 2 νt * ∂u
= u˙ ))1−
( π 0
− ξ d ξ
+
(
∫ ( ) ,, = u˙ 1− erf y 2 νt = ∂t
exp 2
( ))
The final equality follows from the determination of ∂u/∂t above; thus, the velocity solution
obtained by analogy does solve the field equation.
f) Start from the part e) solution and insert the definition of the error function to find:
€ t ( +
2 y 2 νt & 2
u(y,t) = u˙ ∫ *1− ∫ exp(−ξ )d ξ -dt & .
0)* π 0 ,-
Set η" = y ( )
νt " so that dη" = − y 2 νt "3 dt " = −(ν 2y 2 )η"3 dt " and dt " = −2( y 2 νη"3 ) dη" , and use
this in the equation for u(y, t):
€ η '
2 η' 2 2
*0 −2y 2 3
u(y,t) = u˙ ∫ )1− ∫ exp(− ξ )d ξ ,2 3 5
dη/ .
€ € ∞( π 0 € +1 νη/ 4
Multiply by t outside the integral, divide by t inside the integral, and recognize the definition of
η to determine the form of the similarity function:
η ' * dη.
u(y,t)€ 2 η' 2 η
dη.
u˙ t
= f (η) = −2η 2 ∫ )1− ∫ exp(− ξ 2
)d ξ , 3 = −2 η 2
∫ [1− erf (η. 2)] 3 .
∞( π 0 + η. ∞ η.
The first and second derivatives of this function are needed to determine if it solves the part d)
equation. These derivatives are:
df η +
% η$ (. dη$ + % η (. 1 2 % % η ((
€ = −4η ∫ -1− erf ' *0 3 − 2η 2 -1− erf ' *0 3 = ' f −1+ erf ' ** , and
dη ∞, & 2 )/ η$ , & 2 )/ η η& & 2 ))
2
d f 2$ $ η '' 2 $ df 2 −η 2 4 1 '
2
= − 2 & f −1+ erf & )) + & + e ⋅ )
dη η % % 2 (( η % dη π 2(
€ 2$ $ η '' 2 $ 2 $ $ η '' 1 −η 2 4 '
= − 2 & f −1+ erf & )) + & & f −1+ erf & )) + e )
η % % 2 (( η % η % % 2 (( π (
2 2, $ η '/ 2 −η 2 4
= 2 f − 2 .1− erf & )1 + e ,
η η - % 2 (0 η π
where the second equality for the second derivative follows from substituting in the first
derivative result. Use these derivative results to assemble the part d) equation:
d 2 f η df 2 2* $ η '- 2 −η 2 4 $ $ η ''
€ 2
+ − f = 2 f − 2 ,1− erf & )/ + e + & f −1+ erf & )) − f
dη 2 dη η η + % 2 (. η π % % 2 ((
2 $2 '* $ η '- 2 −η 2 4
= 2 f − & 2 + 1),1− erf & )/ + e .
η %η (+ % 2 (. η π
Consider the first term on the right and integrate by parts twice to find:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

2
2 η +
% η$ (. dη$ 2 + % η (. 2 e−η 4 2 + 1 η −η $ 2 4 .
η2
f = −4 ∫ -1− erf ' *0 = -
& 2 )/ η$3 η 2 ,
1− erf ' *0
& 2 )/
− − - ∫e dη$0 .
∞, π η π ,2 ∞ /
The final term on the right can be converted into 1 – erf(η/2) by changing the integration
variable to ξ = η´/2 and breaking the range of integration, ∞ to η, into two intervals, ∞ to 0, and
€0 to2η. Thus, the reassembled part d) differential equation becomes:
2
d f η df 2* $ η '- 2 e −η 4 * $ η '- $ 2 '* $ η '- 2 −η 2 4
2
+ − f = 2 ,1− erf & )/ − + ,1− erf & )/ − & 2 + 1),1− erf & )/ + e
dη 2 dη η + % 2 (. π η + % 2 (. % η (+ % 2 (. η π
= 0.
Here final equality occurs because there is term-by-term cancellation on the right side. Thus, the
function f(η) determined above is a valid y
solution of the part d) equation.
€ g) The velocity disturbance spreads linearly Increasing Time
with time in flow direction, and in proportion
to t in the vertical direction. The contours
shown to the right are for quadratically spaced
time increments. u(y,t)
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.32. a) When z is complex, the small-argument expansion of the zeroth-order Bessel
function J o (z) = 1− 14 z 2 +... remains valid. Use this to show that (9.43) reduces to (9.6) as ω → 0
when dp/dz = Δp/L. The next term in the series is 641 z 4 . At what value of a ν ω is the
magnitude of this term equal to 5% of the second term.
b) When z is complex, the large-argument expansion of the zeroth-order Bessel function
12
J o (z) ≅ ( 2 π z ) cos#$ z − 14 π %& remains valid for |arg(z)| < π. Use this to show that (9.43) reduces to
the velocity profile of a viscous boundary layer on a plane wall beneath an oscillating flow as
ω →∞:
Δp / # ω & ) ω ,2
uz (y, t) = − 1sin(ω t) − exp $−y ' sin +ω t − y .4 ,
ρω L 10 % 2ν ( * 2ν -43
where y is the distance from the tube wall, R = a – y, y << a, and dp/dz = Δp/L.

Solution 9.32. a) Start from (9.43):

/1 # i 3 2 a &, 31
Δp ) # i3 2 R &
uz (R, t) = Re 0i +1− J o %% (( J o %% ((. eiωt 4 , and
21 ωρ L +* $ ν ω ' $ ν ω '.- 51
use the small argument form of Jo for the limit ω → 0 :
0 * $ 32 '
2 - 4
2 , 1− 1 & i R ) +... / 2 0 * iω R 2 - 4
2 Δp , &
4% ν ω ( ) / iωt 2 2
2 Δp , 1+ +... / 22
lim uz (R, t) = limRe 1i ,1− / e 5 = lim Re 1i ,1− 4ν / eiωt 5 .
2 2
ω →0 ω →0
2 ωρ L , 1 $ i3 2 a ' / 2
ω →0
2 ωρ L , 1+ iω a +... / 2
2 , 1− & ) +... / 2 23 ,+ 4ν /. 26
3 + 4 &% ν ω )( . 6
Continue simplifying:
20 Δp * $ iω R 2 iω a 2 '- 24
lim uz (R, t) = limRe 1i ,1− &1+ − +... )/ eiωt 5
ω →0 ω →0 23 ωρ L + % 4ν 4ν (. 26
0 Δp * iω R 2 iω a 2 - iωt 4
= limRe 1i ,− + /e 5
ω →0
3 ωρ L + 4ν 4ν . 6
0 Δp * R 2 a 2 -4 1 $ Δp ' 2
&− )(a − R )
2
= Re 1 ,+ − /5 =
3 ρ L + 4ν 4ν .6 4µ % L (
and this is the same as (9.6) when the pressure gradient is Δp/L.
1 1
To determine when 641 z 4 is 5% of 14 z 2 , set (0.05) z 2 = z 4 and determine z. The
4 64
solution is z = a ν ω = 0.05(64) 4 = 0.894 .
i3 2 R " 1 i % R
b) Here, z = = $− + ' , so
νω # 2 2& ν ω
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

" π % 1/ *( " 1 i % R π *, *( " 1 i % R π *,2

1
cos $ z − ' = exp )i $ − + ' − i - + exp )−i $ − + ' + i -4
# 4 & 2 10 *+ # 2 2& ν ω 4 *. *+ # 2 2& ν ω 4 *.43

1/ *(" i 1 % R π *, *(" i 1 % R π *,2

= 1exp )$ − − ' − i - + exp )$ + ' + i -4
2 10 *+# 2 2& ν ω 4 *. *+# 2 2& ν ω 4 *.43
When ω → ∞ , the first term becomes exponentially small, so
" π% 1 )+" i 1 % R π -+
cos $ z − ' ≅ exp *$ + ' + i . as ω → ∞ .
# 4& 2 ,+# 2 2& ν ω 4 /+
Now use R = a – y in the above expression and collect like factors:
" π% 1 )+ " a − y π % a − y -+
cos $ z − ' ≅ exp *i $$ + '' + . as ω → ∞ .
# 4& 2 ,+ # 2ν ω 4 & 2ν ω /+
or:
" π % eiπ 4 " a(1+ i) % " −(1+ i) %
cos $ z − ' ≅ exp $$ '' exp $$ y '' as ω → ∞ .
# 4& 2 # 2ν ω & # 2ν ω &
So, in this limit:
! i3 2 R $ 2 ν ω eiπ 4 ! a(1+ i) $ ! −(1+ i) $
J o ## && 32
exp ## && exp ## y &&
" ν ω % π i (a − y) 2 " 2ν ω % " 2ν ω %
=
! i3 2 a $ iπ 4 ! $
J o ## && 2 ν ω e a(1+ i)
32
exp ## &&
" ν ω % π i a 2 " 2ν ω %
a ! −(1+ i) $
= exp ## y &&
a−y " 2ν ω %
! −(1+ i) $
≅ exp ## y &&
" 2ν ω %
where the final approximate equality holds when y << a. Now substitute this approximate ratio
of Bessel functions into (9.43) to find:
/1 # −(1+ i)y &, 31 /1 3
iΔp ) iΔp ) iωt # −y & # iy &,1
uz (y, t) = Re 0 +1− exp %% ((. eiωt 4 = Re 0 +e − exp %% (( exp %% iω t − ((.4 .
21 ωρ L +* $ 2ν ω '.- 51 21 ωρ L +* $ 2ν ω ' $ 2ν ω '.-51
Take the real part to reach:
Δp ) # −y & # y &,
uz (y, t) = − +sin(ω t) − exp %% (( sin %%ω t − (. ,
ωρ L +* $ 2ν ω ' $ 2ν ω ('.-
and this is the desired result.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.33. A round tube bent into a U-shape having inner diameter d holds a column of
liquid with overall length L. Initially the column of liquid is pushed upward on the right side of
the U-tube and downward on the left side of the U-tube so that the two liquid surfaces are a
vertical distance 2ho apart. If the liquid has density ρ and viscosity µ, and the column is released
from rest, find and solve an approximate ordinary differential equation that describes the
subsequent damped oscillations of h(t), the liquid height above equilibrium in the right side of
the U-tube, assuming that the flow profile at any time throughout the tube is parabolic. Under
what condition(s) is this approximate solution valid? Will oscillations occur in this parameter
regime?
g!

h(t)!
d!

L!

Solution 9.33. This is the fluid mechanical pendulum with viscous effects included. Use
Newton's second law for the mass of fluid in the water column. The unbalanced gravitational
force tends to decrease h(t) and is –2ρgh(πd2/4) and the wall shear stress opposes the motion.
Thus:
π d 2h π
ρ d 2 L 2 = −π dLτ w − 2 ρ g d 2 h ,
4 dt 4
where τw is presumed to be negative. Here dh/dt is the cross-section-average fluid velocity, so
from the results for steady flow in a round tube:
dh d 2 dp d 2 " 4τ w %
Vave = =− =− $ ',
dt 32µ dz 32µ # d &
where the final equality follows from (9.8). Use this results to eliminate τw from the first
equation to reach:
π 2 d 2h " 8µ dh % π 2
ρ d L 2 = −π dL $ ' − 2 ρ gh d h .
4 dt # d dt & 4
2 2
Divide by the coefficient of d h/dt , and rearrange the terms:
d 2 h 32µ dh 2g
+ + h=0 . ($)
dt 2 ρ d 2 dt L
This equation can be solved by assuming an exponential solution: h(t) = hoe–mt, which leads to an
algebraic equation for m:
" 32ν % 2g 16ν " gd 4 %'
m2 − $ 2 ' m + = 0 or m = 2 $$ −1± 1− ,
# d & L d # 128ν 2 L '&
where ν = µ/ρ. Thus, the general solution is:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

(* , (* ,
16ν t "$ gd 4 %'* 16ν t "$ gd 4 %'*
h(t) = A+ exp ) 2 $ −1+ 1− - + A− exp ) 2 $ −1− 1− -.
+* d # 128ν 2 L '&.* +* d # 128ν 2 L '&.*
The constants can be determined from the initial conditions, h(0) = ho, and dh/dt = 0,
which lead to two algebraic equations:
16ν " gd 4 %' 16ν "$ gd 4 %'
ho = A+ + A− , and 0 = A+ 2 $$ −1+ 1− + A −1− 1− .
128ν 2 L '& d 2 $# 128ν 2 L '&
−
d #
Thus, the constants are:
h " 1 % h " 1 %
A+ = o $$1+ '' , and A− = o $$1− '' ,
2# 1− ψ & 2# 1− ψ &
where ψ = gd4/128ν2L. The final solution is:
h " 1 % (16ν t + h " 1 % (16ν t +
h(t) = o $$1+
2# 1− ψ & * d
( )
'' exp ) 2 −1+ 1− ψ , + o $$1−
- 2# 1− ψ &
( * d
)
'' exp ) 2 −1− 1− ψ , . (&)
-
This approximate solution will be valid when d << L so the nearly the entire water
column is not influenced by either free surface, and when the flow's oscillations are slow enough
so that the full-developed parabolic flow profile is valid. Stated simply, this second requirement
is:
d2
<< 1 ,
νT
where d represents the requisite viscous diffusion distance and T is the time scale of the flow. If
the flow oscillates, the period of this oscillation will scale with [L/g]1/2. Thus, the flow
development time requirement can be stated as:
d2 g
<< 1 .
ν L
Thus, ψ << 1, so the solution provided above suggests that the flow will not oscillate and will
instead be over-damped. In this case, the above solution simplifies to:
# gd 2 &
h(t) ≅ ho exp $− t' ,
% 16ν L (
This solution can be obtained directly from ($) by dropping the acceleration term and solving.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.34. Suppose a line vortex of circulation Γ is suddenly introduced into a fluid at rest at
t = 0. Show that the solution is uθ (r,t) = (Γ 2πr) exp{− r 2 4νt} . Sketch the velocity distribution
at different times. Calculate and plot the vorticity, and observe how it diffuses outward.

Solution 9.34. The solution to this problem is very similar to the decay of a line vortex (see
€
Example 9.8). In two-dimensional (r,θ)-polar coordinates, the governing equation is:
∂uθ + ∂ %1 ∂ (.
= ν- ' (ruθ )*0.
∂t ,∂r & r ∂r )/
The boundary conditions on the velocity u (r, t) are θ

u (r,0+) = 0, u (r,∞) = Γ/2πr, and u (∞, t) = 0.

θ θ θ

In this case the second boundary condition suggests a similarity solution of the form:
€
Γ Γ ' r *
uθ = f (η) = f) ,.
2πr 2πr ( νt +
For this solution form the time and radial derivatives are:
∂uθ Γ df ∂η Γ df ( η + Γ ( η + df ∂ (ruθ ) Γ df ∂η Γ df ( 1 +
= = *− - = − * - , = = * - , and
∂t 2πr dη ∂t 2πr € dη ) 2t , 2πr ) 2t , dη ∂r 2π dη ∂r 2π dη ) νt ,
∂ $1 ∂ ' ∂ $ Γ 1 df ' Γ 1 df Γ 1 d2 f
& (ruθ )) = & ) = − + .
∂r % r ∂ r ( ∂r % 2πr νt dη ( 2πr 2 νt dη 2πr νt dη 2
€ Reassemble the governing equation and divide € out the common factor of Γ/2πr:
$ η ' df ν df ν d 2 f 1 $ 1 df d 2 f '
−& ) =− + 2
= − & − 2 ).
€ % 2t ( d η r ν t d η ν t d η t % η d η dη (
d f $ η 1 ' df
2
Multiply by t and put the second derivative on the left: = &− + ) .
dη 2 % 2 η ( dη
€ df η2 df
Integrate to find: ln = − + ln η + const. Exponentiate = e const.η exp{− η 2 4}
dη 4 dη
and integrate again: $"
€
2
f = A + Bexp{− η 4} . (#'"

The constants
€ A and B can be (#&" €
determined from the boundary (#%"
conditions: f(0) = 1, and f(∞) = 1; (#$"
€ A = 0, and B = 1. Thus, the ("
velocity field is: !#'"

Γ ' r 2 *
!#&"
uθ (r,t) = exp(− +. !#%"
2πr ) 4νt ,
In this flow the z- !#$"
!"
component of the vorticity is the
!" !#)" (" (#)" $" $#)" *" *#)" %" %#)" )"
only non-zero component.
€ ) r2 , ) η2 ,
1∂ 1 ∂ur Γ Γ
ω z (r,t) = ( θ)
ru − = exp * − - = exp *− - .
r ∂r r ∂θ 4 πνt + 4νt . 4 πνt + 4.
The plot above shows ωz (vertical axis) vs. r (horizontal axis) at four different times. With
increasing time, the vorticity ar r = 0 decreases but it spreads outward in the radial direction.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.35. Following Taylor and Green (1937), consider the two-dimensional vortex flow
field with constant density ρ:
u = (u, v) = ( Asin(kx)cos(ky), B cos(kx)sin(ky)) .
a) If the flow is steady and inviscid, and A and B are constants, explicitly determine the pressure,
p(x,y), in terms of x, y, A, ρ, and k from (4.10) and (9.1) in two dimensions.
b) If the flow field is unsteady and viscous (with viscosity µ), A and B are functions of time t,
and A = Ao at t = 0, determine A(t), B(t), and p(x,y,t) so that the given u is an exact solution of
(4.10) and (9.1) in two dimensions.
c) How long does it take for A(t) to fall to Ao/2? Does the parametric dependence of this decay
time follow typical diffusion scaling?

Solution 9.35. a) Here ρ = const. so (4.10) and (9.1) simplify to:

∂u ∂v
+ = 0,
∂x ∂y
∂u ∂u ∂u 1 ∂p µ # ∂2 u ∂2 u &
+u +v =− + % + (,
∂t ∂x ∂y ρ ∂x ρ $ ∂x 2 ∂y 2 '
∂v ∂v ∂v 1 ∂p µ # ∂2 v ∂2 v &
and +u +v =− + % + (.
∂t ∂x ∂y ρ ∂y ρ $ ∂x 2 ∂y 2 '
∂u ∂v
First use the continuity equation to find: + = Ak cos ( kx ) cos ( ky) + Bk cos ( kx ) cos ( ky) = 0 ,
∂x ∂y
which implies B = –A. Without the unsteady & viscous terms, the x- and y-momentum equations
∂u ∂u 1 ∂p ∂v ∂v 1 ∂p
are: u + v = − , and u + v = − . For the given velocity field these imply:
∂x ∂y ρ ∂x ∂x ∂y ρ ∂y
A 2 k sin ( kx ) cos(kx)cos2 ( ky ) − BA cos ( kx ) sin ( kx ) sin 2 ( ky ) = − (1 ρ ) (∂p ∂x ) , and
−ABk sin 2 ( kx ) cos ( ky ) sin(ky) + B 2 cos2 ( kx ) sin(ky)cos(ky) = − (1 ρ ) (∂p ∂y ) .
These equations can be simplified using sin2( ) + cos2( ) = 1, and the relationship B = –A to find:
A2k 1 ∂p A2k 1 ∂p
A 2 k !"sin ( kx ) cos(kx)#$ = sin(2kx) = − and A 2 k !"cos ( ky) sin(ky)#$ = sin(2ky) = − .
2 ρ ∂x 2 ρ ∂y
Integrating both final equalities leads to:
A2 A2
p=ρ cos(2kx) + f (y) and p = ρ cos(2ky) + g(x) ,
4 4
where f and g are single-variable functions of integration. Thus, the pressure field is:
A2
p(x, y) = ρ [ cos(2kx) + cos(2ky)] + po ,
4
where po is an undetermined constant.
b) When the flow field is unsteady and viscous, the continuity equation result from part b), B = –
A, still holds. Plus, the analysis leading to the pressure field of part b) remains the same with the
function A(t) replacing the constant A. So, for the given velocity field, the remnant of the x-
momentum equation is:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

∂u µ " ∂2 u ∂2 u % ! dA $
= $ 2 + 2 ' , or # & sin ( kx ) cos ( ky) = −ν Ak 2 (sin ( kx ) cos ( ky) + sin ( kx ) cos ( ky)) ,
∂t ρ # ∂x ∂y & " dt %
dA
where ν = µ/ρ. Dividing out the common trig-function product leads to = −2ν Ak 2 , which is
dt
readily solved to find:
A(t) = −B(t) = A(0)exp {−2ν k 2 t } .
In this case the pressure field is:
A 2 (0)
p(x, y, t) = ρ exp {−4ν k 2 t } [ cos(2kx) + cos(2ky)] + po ,
4
c) To determine the half-life (t1/2) of the flow, set:
1/2 = exp{–k2ν t1/2}, to reach: t1/2 = ln(2)/k2ν.
Thus, the decay time of this flow does follow typical diffusion scaling. This result also indicates
that Taylor-Green flow with a high wave number decays more quickly than an equivalent flow
with a low wave number.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.36. Obtain several liquids of differing viscosity (water, cooking oil, pancake syrup,
shampoo, etc.). Using an eyedropper, a small spoon, or your finger, place a drop of each on a
smooth vertical surface (a bathroom mirror perhaps) and measure how far the drops have moved
or extended in a known period of time (perhaps a minute or two). Try to make the mass of all the
drops equal. Using dimensional analysis, determine how the drop-sliding distance depends on the
other parameters. Does this match your experimental results?

Solution 9.36. The experiments yield the following three qualitative results. 1) The drops of the
more viscous liquids do not slide as quickly as those of the less viscous liquids. 2) The drops
slow down as time increases. 3) Bigger drops go farther. Water does not behave like oils or
liquid soaps because its surface tension is large enough to influence the motion of the sliding
drop. Here, we wish to use the simplest possible dimensional analysis so surface tension will be
ignored.

• Boundary condition & material parameters: µ = viscosity of the fluid

W = weight of the drop
t = sliding time
Solution parameter: s = sliding distance

• Create the parameter matrix: s t W µ

–––––––––––––––––––––––
Mass: 0 0 1 1
Length: 1 0 1 -1
Time: 0 1 -2 -1
This matrix has rank three.
• Determine the number of dimensionless groups: 4 parameters - 3 dimensions = 1 group
• Construct the dimensionless group: Π1 = Wt µs2
• Write a dimensionless law: s = const Wt µ . This simple law accounts for all three "bathroom
mirror" observations!
€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.37. A drop of an incompressible viscous liquid is allowed to spread on a flat

horizontal surface under the action of gravity. Assume the drop spreads in an axisymmetric
fashion and use cylindrical coordinates (R, ϕ, z). Ignore the effects of surface tension.
∂h 1 ∂
a) Show that conservation of mass implies: +
∂t R ∂R ( h
)
R ∫ o udz = 0 , where u = u(R,z,t) is the
horizontal velocity within the drop, and h = h(R,t) is the thickness of the spreading drop.
b) Assume that the lubrication approximation applies to the horizontal velocity profile, i.e.
2
u(R,z,t) = a(R,t) + b(R,t)z + c(R,t)z€ , apply the appropriate boundary conditions on the upper
and lower drop surfaces, and require a pressure & shear-stress force balance within a differential
g ∂h
control volume h(R,t)RdRdθ to show that: u(R,z,t) = − z(2h − z) .
2ν ∂R
€ ∂h g ∂ $ 3 ∂h '
c) Combine the results of a) and b) to find = & Rh ).
∂t 3νR ∂R % ∂R (
d) Assume a similarity solution: € h(R,t) = A f (η) with η = BR , use the result of part c) and
tn tm
R max (t )
2π ∫ o h(R,t)RdR = V where €Rmax(t) is the radius of the spreading drop and V is the initial
volume of the drop to determine m = 1/8, n = 1/4, and a single nonlinear ordinary differential
equation for f(η) involving € only A, B, g/ν, and η €. You need not solve this equation for f. [Given
that f → 0 as η → ∞, there will be a finite value of η for which f is effectively zero. If this
€
value of η is ηmax then the radius of the spreading drop, R(t), will be: Rmax (t) = ηmax t m B .]

Solution 9.37. a) Consider a stationary “ring” control volume that has inner radius of R, outer
€ €
radius of R + dR, and height h(R,t). There is no flow out through the underside of the CV, but all
three other sides contribute a term to the mass balance. €
h(R,t ) h(R +dR ,t ) ∂h
−2πR ∫ 0 udz + 2π (R + dR) ∫ 0 udz + 2π (R + dR /2) dR = 0
∂t
(inner surface) (outer surface) (top surface)
Divide by 2π, drop the term containing a (dR)2 and divide by dR to find:
∂h 1
€ R + (
∂t dR
h(R +dR,t )
(R + dR) ∫ 0 )
udz − R ∫ 0
h(R ,t )
udz = 0
Take the limit as dR → 0 , recognize the definition of a derivative, and divide by R to obtain:
∂h 1 ∂
+(
∂t R ∂R ) h
R ∫ o udz = 0 .
€
b) Start€with u(R,z,t) = a(R,t) + b(R,t)z + c(R,t)z 2 and apply the no-slip, wall-shear, and free-
surface boundary conditions:
i) no slip: u(R,0,t) = 0 –> a(R,t) = 0,
# ∂u & €
ii) wall
€ shear: µ% ( = τ w –> µb(R,t) = τ w , and
$ ∂z ' z= 0
# ∂u &
iii) free surface: µ% ( = 0 –> b(R,t) + 2c(R,t)h = 0 ,
$ ∂z ' z= h
€ τ€
Therefore: u(R,z,t) = w z(2h − z) . Now consider a creeping-flow force balance in the radial
2hµ
direction €
€ for the differential control volume h(R,t)RdRdθ .

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

h(R,t ) h(R +dR,t )

∫0 p(R)Rdθdz − ∫ 0 p(R + dR)(R + dR)dθdz − τ w RdRdθ = 0
Here take atmospheric pressure as the zero or reference pressure level. The pressure that matters
will be hydrostatic, p(R,t) = ρg( h(R,t) − z) so that the above equation becomes:
€ ρgh 2 (R,t) ρgh 2 (R + dR,t)
Rdθ − (R + dR)dθ − τ w RdRdθ = 0 .
2 2
Divide this€by Rdθ, drop the higher order term, and rearrange to form the definition of a
derivative:
€ 1 $ ρgh 2 (R + dR,t) ρgh 2 (R,t) '
− & − ) = τw
dR % 2 2 (
Take the limit as dR → 0 and perform the radial derivative:
∂ & ρgh 2 ) ∂h
τw = − ( + = − ρgh .
€ ∂R ' 2 * ∂R
Place ν€= µ ρ and this wall-shear-stress-static-pressure-gradient relationship into the velocity
profile formula to simplify it and eliminate τw:
g ∂h
€ u(R,z,t) = − z(2h − z) .
€ 2ν ∂R
c) Use the result of part b) and perform the integration specified in the result of part a)
h
h g ∂h h g ∂h & z 2 z 3 ) gh 3 ∂h
∫ o udz = − 2ν ∂R ∫ o z(2h − z)dz = − 2ν ∂R (2h 2 − 3 + = − 3ν ∂R
€ ' *0
Put this into the result of part a) and bring the constants outside the radial differentiation to find:
∂h g ∂ % 3 ∂h (
− ' Rh * = 0.
€ ∂t 3νR ∂R & ∂R )
d) First consider a CV that entirely encloses the drop and moves with it. Thus, conservation of
d
mass in this case implies: (dt
R (t )
)
2π ∫ 0 max h(R,t)RdR = 0 because there are no fluxes crossing the
€ R (t )
CV boundary. Integrate once in time to find: 2π ∫ 0 max h(R,t)RdR = const . The constant can be
evaluated as V, the volume of the drop, at the initial time (or any other time); thus,
R (t )
€ 2π ∫ 0 max h(R,t)RdR = V .
€ A BR
Now use the proposed similarity solution form, h(R,t) = n f (η) with η = m , and perform the
t t
various differentiations:
€ & −mBR )
∂ A A A A
h(R,t) = −n n +1 f (η) + n f %(η)( m +1 + = −n n +1 f (η) − m n +1 ηf %(η) ,
∂t t t ' t
€
* t €t
4
∂ A B AB ∂ A B
h(R,t) = n f #(η) m = n +m f #(η) , Rh 3 h(R,t) = R 4 n +m f 3 (η) f $(η) , so
∂R t t t ∂r t
€ 1 ∂ # 3 ∂ & A4 B 3 A4 B2
% Rh h(R,t)( = 4 n +m f (η) f *(η) + 4 n +2m ( 3 f 2 (η) f *(η) + f 3 (η) f **(η))
R ∂R $ ∂R ' Rt t
4 2 4 2
A B A B
€ = 4 n +2m € f 3 (η) f #(η) + 4 n +2m ( 3 f 2 (η) f #(η) + f 3 (η) f ##(η)) .
t η t
€ Thus, the field equation determined for part c) implies:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

A A g A4 B2 3 g A4 B2
4 n +2m (
−n n +1 f (η) − m n +1 ηf (η) =
$ 4 n +2m
f (η) f (η) +
$ 3 f 2 (η) f $2 (η) + f 3 (η) f $$(η)) ,
t t 3ν t η 3ν t
which will allow a valid similarity solution when: n + 1 = 4n + 2m . The second relationship for n
and m can be obtained from the conservation of volume requirement:
R max (t ) V R max (t ) A At 2m η max
€ ∫ 0 h(R,t)RdR = 2π = ∫ 0 t n f (η)RdR = t n B 2 ∫ 0 f (η)ηdη , (*)
€
and so this second relationship is 2m = n. Together these two relationships for n and m imply:
n = 1 4 and m = 1 8 , so that the remnant of the field equation is:
1 1 gA 3 B 2 3 gA 3 B 2
€ − f (η) − ηf (η) =
4 8
$
3νη
f (η) f (η) +
$
3ν
( 3 f 2 (η) f $2 (η) + f 3 (η) f $$(η)) (**).

€ The
€ constants A and B can be determined by simultaneous solution of (*) and (**) using
h(Rmax (t),t) = 0 , and (∂h ∂R) R = 0 = 0 as boundary conditions for (**). However, this effort is
beyond the scope of this exercise.
€

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.38. Obtain a clean flat glass plate, a watch, a ruler, and some non-volatile oil that is
more viscous than water. The plate and oil should be at room temperature. Dip the tip of one of
your fingers in the oil and smear it over the center of the plate so that a thin bubble-free oil film
covers a circular area ~10 to 15 cm in diameter. Set the plate on a horizontal surface and place a
single drop of oil at the center of the oil-film area and observe how the drop spreads. Measure
the spreading drop's diameter 1, 10, 102, 103, and 104 seconds after the drop is placed on the
plate. Plot your results and determine if the spreading drop diameter grows as t1/8 (the predicted
drop-diameter time dependence from the prior exercise) to within experimental error.

Solution 9.38. For this kitchen-or-garage experiment, the third author of this textbook used the
glass plate from the front of a document frame and SAE 30-Weight motor oil to obtain the
following measurements.

time (s) Drop Diameter (cm)

1 1.1 ± 0.2
10 1.5 ± 0.1
102 2.0 ± 0.1
103 2.7 ± 0.1
104 ––

Here most of the uncertainty comes from the fact that the spreading droplet was not precisely
circular. The first measurement at t = 1 second was more uncertain than the rest because starting
the experiment and reading the ruler
had be done nearly simultaneously so
catching the diameter at just the right
10
instant in time wasn’t easy. The final
4
measurement at 10 seconds wasn’t
made because the drop had spread so Drop Diameter vs. Time
far that its edge was not distinct.
However, for the experimental results
tabulated above and plotted to the
right, there were no substantial
Drop Diameter (cm)

differences from the theory. Any

differences that may have been found SAE 30 Weight Oil
between the theory and the 1/8th power law slope

experiments are likely caused by

physical processes or disturbances
that were not included in the theory.
Evaporation, surface tension, contact
angle phenomena, foreign matter on
the surface, and surface roughness are
all possible reasons that drop
spreading measurements would be
different from the theoretical
predictions. 1
1 10 100 1000
Time (s)
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.39. A fine stream of a viscous fluid with density ρ and viscosity µ falls slowly at a
constant volume flow rate Q onto the center of a flat horizontal circular disk of radius R. The
fluid flows steadily under the action of gravity g from the center of the disk to its edge in a layer
of thickness h(r), where r is the radial coordinate. For the following items, assume Q is constant,
and apply the approximate boundary condition h(R+) = 0, where R+ is a radial location just
beyond the edge of the disk.
a) Determine a scaling law for h from dimensional analysis.
b) Using the lubrication approximation determine a formula for h(r) that is valid for 0 < r < R.
c) Increasing which parameters increases the thickness of the fluid layer on the disk.

g! Q!
h(r)!

2R!

Solution 9.39. a) There are six parameters (h, r, Q, g, R, ρ, and µ) and all three fundamental
dimensions are present so there will be 7 – 3 = 4 dimensionless parameters. Here h is the solution
parameter so put it in the first group, Π1 = h/R. The second group is also a simple length scale
ratio Π2 = r/R. The other two groups are a Froude number Π3 = Q/[gR5]1/2, and a Reynolds
number Π4 = ρQ/Rµ. Thus: h R = f r R,Q( )
gR 5 , ρQ Rµ is the scaling law.
b) The local velocity profile u(r,z) is assumed parabolic in z when a flow is analyzed with the
lubrication approximation. In this case, the flow is driven by a static pressure gradient and the
shear stress is zero on the free surface on y = h. Thus, from Example 9.5, the steady horizontal
€ g ∂h
velocity profile will be: u(r,z) = − z(2h − z) . The volume flux across the plate surface must
2ν ∂r
be Q, so
z= h
πgr ∂h z= h πgr ∂h ' h 2 h 3 * 2πgr 3 ∂h
Q = 2πr ∫ u(r,z)dz = − ∫ z(2h − z)dz = − )2h − , = − h .
€z= 0 ν ∂r z= 0 ν ∂r ( 2 3+ 3ν ∂r
Treating the two ends of this extended equality as a differential equation leads to:
∫ (1 r)dr = −(2πg 3νQ) ∫ h 3 dh or ln r + C = −(2πg 3νQ)( h 4 4) ,
14
€ where C is a constant. Applying h(R+) = 0 produces C = –ln(R+), and h = !( 6ν Q π g) ln ( R+ r )# .
" $
c) The fluid layer will thicken when µ, Q, or R increase.
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.40. An infinite flat plate located at y = 0 is stationary until t = 0 when it begins
moving horizontally in the positive x-direction at a constant speed U. This motion continues until
t = T when the plate suddenly stops moving.
a) Determine the fluid velocity field, u(y,t) for t > T. At what height above the plate does the
peak velocity occur for t > T? [Hint: the governing equation is linear so superposition of
solutions is possible.]
b) Determine the mechanical impulse I (per unit depth and length) imparted to the fluid while the
T
plate is moving: I = ∫ 0
τ w dt .
c) As t → ∞ , the fluid slows down and eventually stops moving. How and where was the
mechanical impulse dissipated? What is t/T when 99% of the initial impulse has been lost?

# 2 η −ζ 2 &
Solution 9.40. a) For 0 < t < T, the plate moves to the right and u(y, t) = U %1−
$ π 0
∫ e dζ (' ,
y
where η = . Then, at t = T, the plate stops. This is the same as adding an impulsive plate
2 νt
motion to the left to the already moving plate. Since the field equation is linear, the fluid motion
can be obtained from such a superposition of impulsive events as well. Hence, for t > T,
' 2
η * ' 2 T −ζ 2 * 2U T −ζ 2
η η
−ζ 2
€ u(y,t) = U)1−
π 0
∫ e dζ , − U)1− π ∫ e dζ , = π ∫ e dζ
( + ( 0 + η
y
where η is defined above and ηT = . Finding the height of the peak velocity means
2 ν (t − T)
computing
€ ∂u ∂y , setting it equal to zero, and solving for y.
) 2U & e−η T e−η )
2 2
∂u(y,t) 2U & −η T2 ∂ηT −η 2 ∂ηT
= (e −e += ( − +=0
∂y€ π' ∂y ∂y * π ' 2 ν (t − T) 2 νt *
€ find the vertical location of the peak velocity, divide out common factors, move one term to
To
the other side of the equation, take the natural logarithm, and solve for y.
y2 y2 *2ν (t − T)t $ t '-1 2
€ + ln t − T = + ln t –> y = , ln& )/
4ν (t − T) 4νt + T % t − T (.
T T % ∂u ( T U 2µU 1 2
b) I = ∫ τ w dt = µ ∫ ' * dt = −µ ∫ dt = − T . The sign of this answer implies
0 0 ∂y 0
& ) y= 0 πνt πν
that the
€ fluid impeded the motion of the plate.€
c) The input mechanical impulse is dissipated at the surface of the plate. After the plate stops
moving (t > T), the mechanical impulse extracted from the flow, Ie(t), will be:
€ t U t' 1 1* 2µU
Ie (t) = ∫ τ w dt = µ
T
∫ )
πν T ( t − T
− , dt =
t+ πν
( t −T − t + T )
I (t) t −T − t + T t t
Thus, e = 0.99 = 12
= −1 − + 1, or, assuming that t >> T,
I T T T
€ t t t# T& t# T & 1 T T
0.01 = −1 − = %1− 1− ( ≈ %1−1+ + ...( = –> t = 2 = 2500T
.
T T T$ t' T$ 2t ' 2 t (0.02)
€

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Thus, the time for the fluid to slow down is very long time compared to the time that the plate
was actually moving.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.41. Consider the development from rest of plane Couette flow. The flow is bounded
by two rigid boundaries at y = 0 and y = h, and the motion is started from rest by suddenly
accelerating the lower plate to a steady velocity U. The upper plate is held stationary. Here a
similarity solution cannot exist because of the appearance of the parameter h. Show that the
velocity distribution is given by
# y & 2U ∞ 1 # νt & # nπy &
u(y,t) = U%1− ( − ∑ exp%−n 2π 2 2 ( sin% (
$ h ' π n=1 n $ h ' $ h '
Sketch the flow pattern at various times, and observe how the velocity reaches the linear
distribution for large times.
€
∂u ∂ 2u
Solution 9.41. The differential equation is (9.20), = ν 2 , and the boundary and initial
∂t ∂y
conditions on u(y,t) are: u(0,t) = U, u(h,t) = 0, and u(y,0) = 0. The solution is facilitated by
defining u(y,t) = U (1− y h ) + q(y,t) , where q is the deviation of the velocity from the final linear
distribution. The deviation velocity satisfies the same differential equation, but with
€
homogeneous boundary conditions: q(0,t) = q(h,t) = 0, and the initial condition:
q(y,0) = U(1 – y/h).
€A Fourier series solution for q satisfying the two boundary conditions is:
∞
% νt (
q = ∑ An exp&−n 2π 2 2 ) sin( nπy h ) .
n=1 ' h *
The values of the coefficients An are found from the initial condition:
# y& ∞ # nπy &
q(y,0) = U%1− ( = ∑ An sin% (.
$ h ' n=1 $ h '
€
Utilizing the orthogonality of the sine-functions leads to:
2 h # y & # nπy & 2U
An = ∫ U%1− ( sin% (dy = .
€ h 0 $ h' $ h ' π n
Thus, the time-dependent velocity distribution is
# y & 2U ∞ 1 # νt & # nπy &
u(y,t) = U%1− ( − ∑ exp%−n 2π 2 2 ( sin% (.
$ h ' π n=1 n $ h ' $ h '
€
The velocity profiles sketched below for different values of νt/h2 show that the effect of the
upper stationary boundary is initially negligible. Gradually it influences the entire region by
viscous diffusion.
€ y=h
increasing t

y=0 u/U
1
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.42. Two-dimensional flow between flat non-parallel plates can be formulated in
terms of a normalized angular coordinate, η = θ α , where α is the half angle between the plates,
and a normalized radial velocity, ur (r,θ ) = umax (r) f (η) where η = θ α for |θ| ≤ α. Here, u = 0, θ

the Reynolds number is Re = umax rα ν , and Q is the volume flux (per unit width perpendicular
to the page). €
a) Using the appropriate€
versions of (4.10) and (8.1), show€
that f "" + Re αf 2 + 4α 2 f = const.
b) Find f(η) for€symmetric creeping flow, i.e. Re = 0 = f (+1) = f (−1) , and f(0) = 1.
c) Above what value of the channel half-angle will backflow always occur?

€
€

Solution 9.42. a) In two-dimensional (r,θ)-polar coordinates, the governing equations for steady
radial flow, ur (r,θ ) = umax (r) f (η) , are:
1∂
continuity: (rur ) = 0 ,
r ∂r
∂u ∂p , ∂ % 1 ∂ ( 1 ∂u /
€ radial momentum: ρur r = − + µ. ' (rur )* + 2 r2 1, and
∂r ∂r -∂r & r ∂r ) r ∂θ 0
€ 1 ∂p % 2 ∂u (
θ-direction momentum: 0=− + µ' 2 r *.
r ∂θ & r ∂θ )
Integrating the continuity equation
€ once implies: u r = (ν r ) F(η ) , where η = θ α and the factors
of ν and 1/α have been added so that the profile function F(η) will be dimensionless and its
boundary conditions can be € applied at η = ±1. Place this solution form into the momentum
equations to find:
€ €
ν 2F 2 ∂p µ ν 1 ∂p 2ν 1
−ρ 3 = − + 3 2 F '', and 0 = − + µ 3 F '.
r ∂r r α r ∂θ r α
where a prime denotes d/dη. Simplify and isolate the pressure gradients:
1 ∂p ν 2 ' 2 1 * 1 ∂p 2ν 2 1
= 3 ) F + 2 F &&, , and = F '.
€ ρ ∂r r ( α €+ ρ ∂θ r 2 α
Cross differentiate these equations, eliminate the pressure, and remove common factors:
ν2 %2 1 ( 4ν 2 1 1
3 ' F F $ + 3
F $$$* = − 3
F $, or 2 F ### + 2FF # + 4 F # = 0 .
€ r &α α ) r α € α
The final equation can be integrated:
1
F ## + F 2 + 4F = const.
α2 €
€
Now include the maximum velocity in the scaling by redefining the profile function:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

F(η) u
f (η) ≡ = r ,
Fmax umax
where umax = (ν/r)Fmax. This introduces the Reynolds number based on the local half width (αr)
u αr $ ν ' αr
of the channel, Re = max = & Fmax ) = Fmaxα , into the equation:
ν €% r (ν
Fmax 2
2
f ## + Fmax f 2 + 4Fmax f = const., or f "" + α Re f 2 + 4α 2 f = const.
α
b) For creeping flow set Re = 0, and this eliminates the nonlinear term: f """ + 4α 2 f " = 0 . The
€
solutions for f´ will be: f " = C1 cos(2αη) + C2 sin(2αη) . These can be integrated once to find:
f = C1 cos(2αη) + C2 sin(2αη) + C3 where the€unknown constants have been suitably redefined.
€
The constants can be evaluated using the boundary conditions:
f (+1) = 0 = C1 cos(2α ) + C2 sin(2α ) + C3 €
€
f (−1) = 0 = C1 cos(2α ) − C2 sin(2α ) + C3
€ f (0) = 1 = C1 + C3
Here, C1 = (1 2) csc 2 α , C2 = 0, and C3 = 1− (1 2) csc 2 α , so that:
€ 2
€ f (η) = 1+ 1 csc 2 (α )[cos(2αη) −1] , or with a little more manipulation: f (η) = 1− sin (αη)
€ 2 sin 2 (α )
2 2
€ c) For backflow to occur,€f must be negative. This will occur if sin (αη) > sin (α), which can
happen if α > π/2.
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.43. Consider steady viscous flow inside a cone of constant angle θo. The flow has
constant volume flux = Q, and the fluid has constant density = ρ and constant kinematic viscosity
= ν. Use spherical coordinates, and assume that the flow only has a radial component,
u = ( ur (r,θ ),0,0) which is independent of the azimuthal angle ϕ, so that the equations of motion
are:
1 ∂ 2
Cons. of Mass:
r 2 ∂r
(r ur ) = 0
€ & 1 ∂ & ∂u )
∂u 1 ∂p 1 ∂ & ∂u ) 2 )
Cons. of Radial Mom.: ur r = − + ν( 2 (r2 r + + 2 (sin θ r + − 2 ur +
∂r ρ ∂r ' r ∂r ' ∂r * r sin θ ∂θ ' ∂θ * r *
€ 1 ∂p ' 2 ∂u *
Cons. of θ-Mom. 0=− + ν) 2 r ,
ρr ∂θ ( r ∂θ +
For the following items,
€ assume the radial velocity can be
determined using: ur (r,θ ) = QR(r)Θ(θ ) . Define the Reynolds
number of this flow as: Re = Q (πνr) .
€
a) Use the continuity equation to determine R(r).
b) Integrate the θ-momentum equation, assume the constant of
€
integration is zero, and combine the result with the radial mom.
€
equ. to determine a single differential equation for Θ(θ) in
terms of θ and Re. It is not necessary to solve this equation.
c) State the matching and/or boundary conditions that Θ(θ)
must satisfy.

1 ∂ 2 ∂
Solution 9.43. a) Place ur (r,θ ) = QR(r)Θ(θ ) into 2
r ∂r
( r ur ) = 0 , to find: ( r 2 R) = 0 for r ≠ 0.
∂r
Thus: r2R = constant or R ∝1 r 2 , so ur (r,θ ) = Qr−2Θ(θ ) , so Θ is a dimensionless function.
1 ∂p & 2 ∂ur ) p 2ν
b) Rewrite the€θ-mom. equ.: = ν( + , integrate with respect to θ: = ur + f (r) .
ρ ∂θ ' r€∂θ * € ρ r
€ with respect to r to get: 1 ∂p = 2ν ∂ %' ur (* . Use this in the r-
Now set f = 0,€and differentiate
ρ ∂r ∂r & r )
mom. equ. to eliminate
€ the pressure: €
∂ur ∂ % ur ( % 1 ∂ % 2 ∂ur ( 1 ∂ % ∂u ( 2 (
ur = −2ν ' * + ν ' 2 ' r *+ 2 'sin θ r * − 2 ur *
∂r ∂r & r ) & r ∂r & € ∂r ) r sinθ ∂θ & ∂θ ) r )
−2
Now plug in ur (r,θ ) = Qr Θ(θ ) and work to find the differential equation for Θ.
QΘ ∂ $ QΘ ' ∂ $ QΘ ' $ 1 ∂ $ −2QΘ ' Q d $ dΘ ' 2QΘ '
2 & 2 ) = −2ν & 3 ) + ν & 2 & )+ 4 & sin θ )− 4 )
€ r ∂r % r ( ∂r % r ( % r ∂r % r ( r sin θ dθ % dθ ( r (
4
Perform
€ derivatives, cancel opposite terms, divide out the common factor of Q/r , and continue
simplifying.
Q 2Θ 2 QΘ & 2QΘ Q & dΘ d 2Θ ) 2QΘ )
€ −2 5 = +6ν 4 + ν ( 4 + 4 (cos θ + sin θ 2 + − 4 +
r r ' r r sin θ ' dθ dθ * r *
QΘ 2 dΘ d 2Θ
−2 = 6Θ + cot θ +
νr dθ dθ 2
Plug in the Reynolds number definition and rewrite the equation in the ordinary order:
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

d 2Θ dΘ
2
+ cot θ + 6Θ + 2πReΘ 2 = 0
dθ dθ
c) The radial flow speed must be zero at the angle defined by the surface of the cone. Thus, the
first boundary condition is: Θ(θ o ) = 0 . The second matching condition comes from the
requirement that that the volume flux be equal to Q.
θo € θo θo
1
Q = 2π ∫ ur (r,θ )r 2 sin θdθ = 2π ∫ QΘ(θ )sin θdθ , or = ∫ Θ(θ )sin θdθ
€0 0 2π 0
These two conditions will be sufficient for determining the unknown constants that will arise
from integrating the second order differential equation found for part b). In addition there is also
a smoothness constraint, ( dΘ dθ )θ = 0 = 0 , which may be needed to prevent the flow profile from
€ €
having a conical peak at θ = 0, and the requirement that Θ is finite for 0 ≤ θ ≤ θo.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.44. The boundary conditions on obstacles in Hele-Shaw flow were not considered in
Example 9.4. Therefore, consider them here by examining Hele-Shaw flow parallel to a flat
obstacle surface at y = 0. The Hele-Shaw potential in this case is:
z" z%
φ = Ux $1− ' ,
h# h&
where (x, y, z) are Cartesian coordinates and the flow is confined to 0 < z < h and y > 0.
a) Show that this potential leads to a slip velocity of u(x, y → 0) = U ( z h ) (1− z h ) , and determine
the pressure distribution implied by this potential.
b) Since this is a viscous flow, the slip velocity must be corrected to match the genuine no-slip
condition on the obstacle’s surface at y = 0. The analysis of Example (9.4) did not contain the
correct scaling for this situation near y = 0. Therefore, rescale the x-component of (9.1) using:
x* = x/L, y* = y/h = y/εL, z* = z/h = z/εL, t* = Ut/L, u* = u/U, v* = v/εU, w* = w/εU, and p* = p/Pa,
and then take the limit as ε2ReL → 0, with µUL/Pah2 remaining of order unity, to simplify the
resulting dimensionless equation that has
dp " ∂ 2u ∂ 2u %
0≅ +µ$ 2 + 2 '
dx #∂ y ∂ z &
as its dimensional counterpart.
c) Using boundary conditions of u = 0 on y = 0, and u = U ( z h ) (1− z h ) for y >> h. Show that
h
z" z% ∞ " nπ % " nπ % 2U z" z% " nπ %
u(x, y, z) = U $1− ' + ∑ An sin $
h # h & n=1
z ' exp $ −
# h &
y ' , where An = −
# h & h
∫ h $#1− h '&sin $# h
z 'dz ,
&
0
3 3
which implies: An = −8U n π for n = odd, and An = 0 for n = even. [The results here are directly
applicable to the surfaces of curved obstacles in Hele-Shaw flow when the obstacle’s radius of
curvature is much greater than h.]

Solution 9.44. a) From Example 9.4, the u and v velocity components implied by the given
potential are:
∂ z% z( ∂
u = φ = U '1− * and v = φ = 0 .
∂x h & h) ∂y
Neither of these equations shows any dependence on y, so, as y → 0 , the velocity components
are as given above and there is slip at the obstacle surface y = 0. Example 8.2 also shows that
z(h − z) z $ z' 2µU
€ φ =− p = Ux € &1− ) , so p = − 2 x .
2µ h % h (€ h
b) The x-component of (9.1) is:
∂u ∂u ∂u ∂u 1 ∂p & ∂ 2u ∂ 2u ∂ 2u )
+u +v +w =− + ν( 2 + 2 + 2 + .
€ ∂t ∂x ∂y ∂z ρ€∂x ' ∂x ∂y ∂z *
Insert the given scaling into this equation, to find:
U ∂u* U 2 * ∂u* εU 2 * ∂u* εU 2 * ∂u* Pa ∂p* ' U ∂ 2 u* U ∂ 2 u* U ∂ 2 u* *
+ u + v + w = − + ν ) 2 *2 + 2 *2 + 2 *2 , ,
L U ∂t * € L ∂x * h ∂y * h ∂z * ρL ∂x * ( L ∂x h ∂y h ∂z +
Simplify using ε = h/L, and collect terms with like coefficients:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

U 2 # ∂u* * ∂u
*
* ∂u
*
* ∂u
*&
Pa ∂p* U ∂ 2 u* U # ∂ 2 u* ∂ 2 u* &
% * +u * +v + w ( = − + ν + ν % + (.
L $ ∂t ∂x ∂y * ∂z * ' ρL ∂x * L2 ∂x * 2 h 2 $ ∂y * 2 ∂z * 2 '
Multiply through by ρh2/µU, and use ε = h/L and ReL = ρUL/µ to reach:
$ ∂u* ∂u* ∂u* ∂u* ' P h 2 ∂p* 2∂ u
2 *
∂ 2 u* ∂ 2 u*
ε 2 Re L & * + u* * + v * * + w * * ) = − a + ε 2 + 2 + 2
.
€ % ∂t ∂x ∂y ∂z ( µUL ∂x * ∂x * ∂y * ∂z *
When ε2ReL → 0, but µUL/Pah2 remains of order unity this equation simplifies to:
P h 2 ∂p* ∂ 2 u* ∂ 2 u* 1 ∂p ' ∂ 2u ∂ 2u *
0≅− a + + , or in dimensional form: 0 ≅ − + ν ) 2 + 2 ,.
€ µUL ∂x * ∂y * 2 ∂z* 2 ρ ∂x ( ∂y ∂z +
c) The field equation derived in part b) for u is linear. Therefore assume that it can be written as
a sum of two terms, one that provides the correct behavior when y >> h and one that acts to
provide the correct boundary condition at y = 0:
€ z# z& €
u = U %1− ( + υ (y,z) , (i)
h $ h'
where υ = 0 when z = 0 and z = h, υ → 0 when y >> h, and u → 0 when y → 0 . Using –∂p/∂x =
2µU/h2, the part b) field equation becomes:
1 2µU € % 2U ( % ∂ 2υ ∂ 2υ ( ∂ 2υ ∂ 2υ
+ ν ' − * + ν ' + * = 0 , or + = 0.
ρ h 2€ & h 2 ) & ∂y 2 ∂z€2 ) ∂€y 2 ∂z 2
Thus, υ solves a two dimensional Laplace equation. Here, a separation of variable solution,
υ = Y(y)Z(z) is appropriate. Placing this solution form into the two dimensional Laplace
equation and dividing by υ, produces: €
€
1 d 2Y 1 d 2 Z
+ = 0.
€ Y dy 2 Z dz 2
The first term depends only on y and the second term depends only on z, so each must be
constant for the equation to hold throughout the domain. Let the first term equal +k2 and the
second term = –k2, so that the individual Y, and Z equations become:
€ d 2Y
2 d 2Z
2
− k Y = 0 , and 2
+ k 2Z = 0.
dy dz
The solutions to these two equations are: Y(y) = A± exp{±ky} , and Z(z) = Bsin(kz) + C cos(kz) .
To satisfy the boundary conditions υ = 0 when z = 0 and z = h, implies C = 0 and
Bsin(kh) = 0 . The second of these is satisfied € by the (useless) trivial solution B = 0, or by sin(kh)
€
= 0. The later condition leads to k = kn = nπ/h, where € n is a positive integer so that
€
Z(z) = Bsin( nπz h ) . Thus, the separation constant k may take on discrete values.
To satisfy the boundary condition υ → 0 when y >> h, means setting A+ = 0. Therefore,
€
υ(y, z) takes the following form:
∞
€ υ (y,z) = ∑ An sin( k n z) exp{−kn y} . (ii)
€ n=1
to allow for all possible values of k. To satisfy the final boundary condition ( u → 0 when y → 0
), (i) and (ii) imply:
z# z& z# z& ∞
€U %1− ( + υ (0,z) = 0 = U %1− ( + ∑ An sin( k n z) ,
h $ h' h $ h ' n=1 € €

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

and this equation can be used to determine the coefficients An. Following the usual approach for
a Fourier series, multiply it by sin(kmz) and integrate in z from 0 to h. The result is:
h ∞ h ∞
z $ z' h
∫ sin(k m z)U &1−
h % h(
) dz = − ∑ A n ∫ sin(k m z) sin( k n z ) dz = − ∑ An δmn ,
2
0 n=1 0 n=1
where δmn is the Kronecker delta function, so the Am's are given by:
2U h z$ z'
Am = −
h 0
∫ sin(k m z) &1− )dz .
h % h(
(iii)
€
Thus, the final velocity field is:
z# z& ∞ # mπ & * mπ -
u(y,z) = U %1− ( + ∑ An sin% z( exp+− y.
€ h $ h ' m=1 $ h ' , h /
where An is given by (iii). The integral is tedious but can be evaluated by parts to find:
An = −8U n 3π 3 for n = odd, and An = 0 for n = even.
The final
€ answer includes a sum of terms that decay exponentially with increasing y, and
the first of these will have the slowest decay. At a distance of y = h, this slowest decaying term
will be a factor of e– ≈ 0.04 smaller than it is at y = 0.
π
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.45. Using the velocity field (8.49), determine the drag on Stokes’ sphere from the
surface pressure and the viscous surface stresses σrr and σr . θ

Solution 9.45. There are pressure and shear stress contributions to the drag on a moving sphere
at low Reynolds number. The pressure distribution is given by (8.50):
3µaU
p(r,θ ) − p∞ = − cos θ .
2r 2
The pressure drag can be obtained by integrating this result:
θ=π
& 3U )
Fpressure = − ∫ p(r = a,θ )e r ⋅ e z dS = −2πa 2 ∫ µ( + cos 2 θ sin θdθ = 2πµUa
surface θ = 0 ' 2a *
€
The viscous drag can be obtained from surface integrals of the viscous stresses:
Fvicous = − ∫ σ rθ (r = a,θ )eθ ⋅ e z dS + ∫ σ rr (r = a,θ )e r ⋅ e z dS
surface surface
€ θ=π θ=π
= −2πa 2 ∫ σ rθ (r = a,θ )sin 2 θdθ + 2πa 2 ∫ σ rr (r = a,θ )cosθ sin θdθ,
θ=0 θ=0
& 1 ∂ur ∂uθ uθ ) µU sinθ & 3a 3 ) ∂ur & 3a 3a 3 )
where σ rθ = µ( + − +=− ( 3 + , and σ rr = 2µ = 2µU cosθ( 2 − 4 + .
' r ∂θ ∂r r* r ' 2r * ∂r ' 2r 2r *
Thus,
€ at r = a, σr ≠ 0, but σrr = 0, so
θ

θ=π
Fvicous = 3πµUa 3
∫ sin θdθ = 4πµUa .
€ θ =€
0
Thus, one third of the drag comes from pressure forces and two thirds come from the shear
stress. The total drag is the sum of these two contributions:
Fdrag = Fpressure + Fvicous = 2πµUa + 4 πµUa = 6πµUa .
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.46. Calculate the drag on a spherical droplet of radius r = a, density ρʹ′ and viscosity
µʹ′ moving with velocity U in an infinite fluid of density ρ and viscosity µ. Assume Re = ρUa/µ
<< 1. Neglect surface tension.

Solution 9.46. The effort here is similar to that in Section 8.6 for low Reynolds number flow past
a sphere. The main difference being that flow inside and outside the sphere must be considered
simultaneously. In the following solution, dimensionless variables are used without any special
notation. These are: ur/U, u /U, r/a, ψ/(Ua2), (p – p∞)/(µU/a), and τ/(µU/a), so that the surface of
θ

the sphere is r = 1, and the free stream speed is unity. The development leading to (8.48)
suggests that the flow outside the sphere can be determined from:
ψ o = ( Ao r 4 + Bo r 2 + Co r + Do r) sin 2 θ for r > a.
To recover a uniform flow far from the sphere, the first two constants must be A = 0 and B = 1/2.
Similarly, the flow inside the sphere can be represented by
ψ i = ( Ai r 4 + Bi r 2 + Ci r + Di r) sin 2 θ for r < a.
€
Here, Di = 0 to avoid a singularity at r = 0.
Compute the velocities inside and outside the sphere:
, &1 C D ) 0
€ . −2cosθ( + o + 3o + for r > 1.
1 ∂ψ . '2 r r * .
ur = − 2 =- 1 , and
r sin θ ∂θ . & 2 Ci ) .
./ −2cosθ(' Ai r + Bi + r +* for r < 1.2
, & C D ) 0
. sin θ(1+ o − 3o + for r > 1.
1 ∂ψ . ' r r * .
uθ = =- 1.
r sin θ ∂r . & 2 Ci ) .
€ sin θ 4
( i A r + 2B + + for r < 1
./ ' i
r* .2
For a bounded velocity inside the sphere, Ci must be 0. There are four remaining boundary and
matching conditions at r = 1:
1
ur(1+) = 0 and ur(1–) = 0, and these imply + Co + Do = 0 , and Ai + Bi = 0 ;
€ (1,2)
2
u (1+) = u (1–), and this implies 1+ Co − Do = 4 Ai + 2Bi ; and
θ θ (3)
the stress continuity condition:
* ∂ $ u ' 1 ∂ur - * $ ' € -
σr (a+) = σr (a–) or µ,r & θ ) +
θ θ
€/ = µ0,r ∂ & uθ ) + 1 ∂ur / , and this implies
+ ∂r % r ( r ∂θ€.r=1+ + ∂r % r ( r ∂θ .r=1–
$ 1 '
µ sinθ&−1− 2Co + 4Do + 2 + 2Co + 2Do ) = µ* sin θ [ 4 Ai − 2Bi + 2Ai + 22Bi ] .
% 2 (
Fortunately, there
€ are many canceling terms in this last condition and it simplifies to:
6µ Do = 6µ" Ai a . (4)
Simultaneous solution of the four algebraic equations produces:
€ 1 3µ# + 2µ 1 µ" 1 µ 1 µ
Co = − , Do = , Ai = , and Bi = − .
4 µ# + µ 4 µ" + µ 4 µ" + µ 4 µ# + µ
€
The force on the sphere is determined from a surface intergral:
Fi = − ∫ ( pδij − σ ij )n j dA .
sphere' s surface
€ € € €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

In the following, force is made dimensionless by µUa, area is made dimensionless with a2, and
stress is made dimensionless with µU/a. The normal and surface elements are:
n = e r , and dA = [ r 2 sin θdθdϕ ] r=1 .
The dot product of the shear stress and the directed area element is:
, ∂u & ∂u 1 ∂uθ uθ )/
τ ij n j dA = -e r 2 r + eθ ( θ + − +0 sin θdθdϕ .
€ . € ∂r ' ∂ r r ∂r r *1
The θ-component of surface shear stress
3 µ$
eθ 6(Do /a 3 )sin 2 θdθdϕ = eθ sin θdθdϕ .
€ 2 µ + µ $
The r-component includes a pressure contribution which must be found by integrating Stokes'
creeping-flow momentum equation:
% r 2 1 µ# 1 1 3µ# + 2µ ( 2
ψ o =€' + − r* sin θ , and −∇p = ∇ × ∇ × u = −∇ 2u .
& 2 4 µ # + µ r 4 µ # + µ )
In this dimensionless formulation, these give:
∂p 1 , & 1 µ% 1 ) & 1 1 µ% 1 1 3µ% + 2µ 1 ) /
− = 2 .sin θ cosθ(1+ − +
3 + (€ 3
− +4 sin θ cos θ1
€ ∂r r sin θ - ' 2 µ% + µ r * ' 2 4 µ% + µ r 4 µ% + µ r * 0
3µ% + 2µ cosθ
=
µ% + µ r 3
1 ∂p sin 2 θ & 3 µ% 1 ) 2sin θ , ∂ & 1 1 µ% 1 1 3µ% + 2µ 1 )/
− = (− +− . ( + − +1 , or
r ∂θ r sin θ ' 2 µ% + µ r 4 * r -∂r ' 2 4 µ% + µ r 3 4 µ% + µ r *0
€ ∂p 3µ% + 2µ sin θ
− = .
∂θ µ% + µ 2r 2
Integrating for p and setting p∞ = 0 produces:
€
3µ" + 2µ cosθ
p= .
µ" + µ 2r 2
€
Thus, the r-component of the force integrand is:
,$ ∂ur ' / 3 µ2 + 2µ
)e r [ r sin θdθdϕ ] r=11 = − cos θ
2
.& − p + 2 e r sin θdθdϕ .
-% ∂r € ( 0 2 µ2 + µ
So, the force on the liquid sphere is:
π 2π %
3 µ$ + 2µ 3 µ$ (
F = ∫ ∫ '−e r cosθ + eθ * sin θdθdϕ .
€ θ = 0 ϕ = 0 & 2 µ $ + µ 2 µ + µ $)
The unit vectors on the surface of the sphere are functions of the polar and azimuthal angles, θ,
ϕ, respectively. They must be put in terms of the Cartesian unit vectors that do not vary with
position. The necessary relationships are:
€
e r = e x cosϕ sin θ + e y sin ϕ sin θ + e z cos θ , and eθ = e x cosϕ cos θ + e y sin ϕ cos θ − e z sin θ .
Integrating first over j, the x- and y-direction contributions to the force are zero, a result that can
also be deduced from the axial symmetry of the flow. This leaves:
π & 3 µ$ + 2µ 3 µ$ )
€ F = 2πe z ∫ −( € cos2 θ sin θ − sin 3 θ+dθ .
θ = 0 ' 2 µ$ + µ 2 µ + µ$ *
Carrying out the trigonometric integrals leads to:

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

%µ$ + 2µ 2µ$ ( 3µ$ + 2µ

F = −2πe z ' + * = −2πe z .
& µ$ + µ µ + µ$) µ$ + µ
The force on the sphere is a drag force, and in dimensionless form this drag becomes:
3µ# + 2µ 3µ# + 2µ
D = 2πµUa = Ds .
µ# + µ 3µ# + 3µ
€
where Ds = 6πµUa is the Stokes' drag on a solid sphere.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.47. Consider a very low Reynolds number flow over a circular cylinder of radius r =
a. For r/a = O(1) in the Re = Ua/ν → 0 limit, find the equation governing the stream function
ψ(r, θ) and solve for ψ with the least singular behavior for large r. There will be one remaining
constant of integration to be determined by asymptotic matching with the large r solution (which
is not part of this problem). Find the domain of validity of your solution.

Solution 9.47. The Stokes' flow field equation is:

∇p = µ∇ 2u ,
which can be rewritten:
∇p = −µ∇ × ∇ × u.
Applying one more curl, eliminates the pressure gradient:
€ 0 = ∇ × ∇ × ∇ × u.
In two-dimensional (r,θ)-polar coordinates, the relationship between the stream function ψ(r,θ)
and the velocity field is u = −e€z × ∇ψ , so the field equation for the stream function is given by:
€ 0 = ∇ × ∇ × ∇ × (−e z × ∇ψ ) .
This relationship can be converted to coordinate specific derivatives in steps. Start with the
stream function definition.
€
1 ∂ψ ∂ψ
€ −e z × ∇ψ = er − eθ
r ∂θ ∂r
Apply the first curl operation, using the formulae in Appendix B:
.1 ∂ ' ∂ψ * 1 ∂2ψ 1
∇ × (−e z × ∇ψ ) = −e z 0 )r , + 2 2 3.
€ / r ∂r ( ∂r + r ∂θ 2
Apply the second curl operation,
1 ∂ .1 ∂ ( ∂ψ + 1 ∂2ψ 1 ∂ .1 ∂ ( ∂ψ + 1 ∂2ψ 1
∇ × ∇ × (−e z × ∇ψ ) = −e r 0 * r - + 2 23
+ eθ 0 *r - + 2 2 3.
€ r ∂θ / r ∂r ) ∂r , r ∂ θ 2 ∂ r / r ∂r ) ∂r , r ∂θ 2
Apply third and final curl operation,
0 = ∇ × ∇ × ∇ × (−e z × ∇ψ ) =
€ 46 1 ∂ ' ∂ .1 ∂ ' ∂ψ * 1 ∂2ψ 1* 1 ∂ .1 ∂ ' ∂ψ * 1 ∂2ψ 186
ez 5 r) 0 )r ,+ 3, + 0 )r ,+ 39.
67 r ∂r ( ∂r / r ∂r ( ∂r + r 2 ∂θ 2 2+ r 2 ∂θ 2 / r ∂r ( ∂r + r 2 ∂θ 2 26:
For large r, ur ~ cosθ, and u ~ –sinθ, so it is expected that ψ ~ sinθ. Therefore, a trial
θ

solution in the form ψ(r,θ) = f(r)sinθ may be sufficient. The resulting equation for f looks like:
€ 1 ∂ $ ∂ , 1 ∂ $ ∂ψ ' f /' sin θ ∂ $ ∂f ' f
r& . &r ) − 2 sin θ1) − 3 & r ) + sin θ = 0 .
r ∂r % ∂r - r ∂r % ∂r ( r 0( r ∂r % ∂r ( r 4
Divide by sinθ and recognize that this equation is equi-dimensional in r so it has power-law
solutions, f(r) ~ (r/a)m, where the factor of a has been added for later scaling analysis.
Substituting this into the equation for f(r) and dividing by (r/a)m – 4 leaves:
€
m 2 (m − 2) 2 − (m − 2) 2 − m 2 + 1 = 0 .
This can be factored:
(m −1) 2 (m + 1)(m − 3) = 0 .
Thus, m = 1 is a double root which implies that the solutions for f may be proportional to r/a and
€
(r/a)ln(r/a). Therefore:

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

f (r) = c1 (r /a) + c 2 (r /a)ln(r /a) + c 3 (r /a) 3 + c 4 (r /a) –1 .

The growth of r3 is too fast to match the flow at large r, so c3 must be zero. This leaves:
ψ (r,θ ) = (c1 (r /a) + c 2 (r /a)ln(r /a) + c 4 (r /a) –1 ) sinθ .
The boundary € conditions at r = a are:
%1 ∂ψ (
ur = ' = (c1 + c 4 )cosθ = 0 or c4 = –c1, and
& r ∂θ *)r= a
€
& ∂ψ ) 1
uθ = (− + = (c1 + c 2 − c 4 )sin θ = 0 or c2 = –2c1.
' ∂r *r= a a
Therefore: €
%r r r a(
ψ (r,θ ) = c1' − 2 ln − * sinθ .
&a a a r)
€
The final constant is determined by matching this solution to the free-stream at large r/a.
However, for large r/a, ψ ~ (r/a)ln(r/a)sinθ so ur ~ ln(r/a)cosθ and u ~ ln(r/a)sinθ. The
θ

advective acceleration u ⋅ ∇u ~ (a /r)ln(r /a) and the viscous stresses ~ ∇ 2 u ~ (a /r) 2 . The terms
€
neglected ~ terms retained when Re(a/r)ln(r/a) ~ (a/r)2 or (r/a)ln(r/a) ~ 1/Re. So, this solution is
valid for (r/a)ln(r/a) << 1/Re.
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.48. A small neutrally-buoyant sphere is centered at the origin of coordinates in a deep
bath of a quiescent viscous fluid with density ρ and viscosity µ. The sphere has radius a and is
initially at rest. It begins rotating about the z-axis with a constant angular velocity Ω at t = 0. The
relevant equations for the fluid velocity, u = (ur ,uθ ,uϕ ) , in spherical coordinates (r,θ,ϕ ) are:
1 ∂ 2 1 ∂ 1 ∂
2
r ∂r
( r ur ) +
r sin θ ∂θ
(uθ sin θ ) +
r sin θ ∂ϕ
(uϕ ) = 0 , and
∂uϕ ∂u u ∂u u ∂u 1 1 ∂p€
+ ur ϕ + θ ϕ + ϕ € ϕ + ( ur uϕ + uθ uϕ cot θ ) = − +
∂t ∂r r ∂θ r sin θ ∂ϕ r ρr sin θ ∂ϕ
% 1 ∂ %€ ∂uϕ ( 1 ∂ % ∂uϕ ( 1 ∂ 2 uϕ uϕ 2 ∂ur 2cos θ ∂uθ (
2
ν' 2 'r +
* 2 ' sin θ * + − + + *.
& r ∂r & ∂r ) r sin θ ∂θ & ∂θ ) r 2 sin 2 θ ∂ϕ 2 r 2 sin 2 θ r 2 sin2 θ ∂ϕ r 2 sin2 θ ∂ϕ )
€ a) Assume u = ( 0, 0, uϕ ) and reduce these equations to:
∂uϕ 1 ∂p ( 1 ∂ ( ∂u + 1 ∂ ( ∂uϕ + uϕ +
€ =− + ν* 2 *r2 ϕ - + 2 *sin θ -− 2 2 -
∂t ρr sin θ ∂ϕ ) r ∂r ) ∂r , r sin θ ∂θ ) ∂θ , r sin θ ,
b) Set uϕ (r,θ,t) = ΩaF(r,t)sin θ , make an appropriate assumption about the pressure field, and
∂F $ 1 ∂ $ ∂F ' F'
derive the following equation for F: = ν& 2 & r 2 ) − 2 2 ) .
€ ∂t % r ∂r % ∂r ( r (
c) Determine F for t → ∞ for boundary conditions F = 1 at r = a, and F → 0 as r → ∞ .
€
d) Find the surface shear stress and torque on the sphere
€ ∂
Solution €9.48. ) When ur and u are set equal to zero, the equations
θ € become:
€
∂ϕ
(uϕ ) = 0 , and
∂uϕ u ∂uϕ 1 ∂p
+ ϕ =− +
∂t r sin θ ∂ϕ ρr sin θ ∂ϕ
% 1 ∂ % ∂uϕ ( 1 ∂ % ∂uϕ ( 1 ∂€2 uϕ uϕ 2cos θ ∂uϕ (
2
ν' 2 'r *+ 2 ' sin θ *+ 2 2 − + *.
& r ∂r & ∂r ) r sin θ ∂θ & ∂θ ) r sin θ ∂ϕ 2 r 2 sin 2 θ r 2 sin2 θ ∂ϕ )
€ ∂
The remnant of the continuity equation,
∂ϕ
(uϕ ) = 0 , implies that u does not depend on ϕ. Thus,
ϕ

three terms can be dropped from the second equation, so it simplifies to:
€
∂uϕ 1 ∂p ( 1 ∂ ( ∂u + 1 ∂ ( ∂uϕ + uϕ +
=− + ν* 2 *r2 ϕ - + 2 *sin θ - − 2 2 -.
∂t ρr sin θ ∂ϕ€ ) r ∂r ) ∂r , r sin θ ∂θ ) ∂θ , r sin θ ,
b) The symmetry of the flow suggests that p should not depend on the ϕ. Thus, setting
∂p ∂ϕ = 0 and uϕ (r,θ,t) = ΩaF(r,t)sin θ in the result of part a) produces:
€ ∂F & Ωasin θ ∂ & 2 ∂F ) ΩaF ∂ & ∂ sin θ ) ΩaF sin θ )
Ωasin θ = ν( 2 (r ++ 2 (sin θ +− +.
∂t ' r ∂r ' ∂r * r sin θ ∂θ ' ∂θ * r 2 sin 2 θ *
€ Cancel€ the common factor of Ωa, simplify terms, and perform the inner θ-derivative:
∂F % sin θ ∂ % 2 ∂F ( F ∂ F (
sin θ = ν' 2 'r *+ 2 (sin θ cosθ ) − 2 *
€ ∂t & r ∂r & ∂r ) r sin θ ∂θ r sin θ )
Perform the second θ-derivative and note that cos2 θ − sin 2 θ = 1− 2sin 2 θ :

€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

∂F % sin θ ∂ % 2 ∂F ( F F (
sin θ
∂t
= ν' 2 'r *+ 2
& r ∂r & ∂r ) r sin θ
(1− 2sin 2 θ ) − 2 *
r sin θ )
Simplify the second and final terms on the right side, and then divide by sinθ to get:
∂F $ 1 ∂ $ ∂F ' 2F '
= ν& 2 & r 2 ) − 2 ).
€ ∂t % r ∂r % ∂r ( r (
c) For t → ∞ , the flow should be steady ( ∂F ∂t = 0 ), so the result of part b) implies:
∂ # 2 ∂F &
%r ( − 2F = 0 for r ≠ 0. This is an equi-dimensional equation, so it has power-law
∂r $ ∂ r ' €
€ solutions. Thus, substitute F = Ar n to this equation to find n(n + 1) − 2 = n 2 + n − 2 =
€
(n + 2)(n −1) = 0 , which has solutions: n = +1, –2. The second boundary condition, F → 0 as
r → ∞ , requires the choice of n = –2. The first boundary condition, F = 1 at r = a, determines A
€
= a2. Thus for long-time
€ steady rotation of the sphere, F = a 2 r 2 and uϕ = Ωa( a 2 r 2 ) sin θ .
€
€ d) The viscous shear stress is, €
€ r ∂ % uφ ( 3µΩa 3 sin θ
σ φr = 2µ ' *=− 3
, so [σ φr ] r= a = −3µΩsin θ .
2 ∂r & r ) €r €
So,
π 2π
Torque on sphere = ∫ ∫ −3€µΩsin θ ⋅ asinθ ⋅ a 2 sin θdθdϕ ,
θ = 0ϕ = 0
€
where the integrand factors are written (stress)(moment arm)(area element). When evaluated this
produces:
π
on sphere = −6πµΩa 3
Torque € ∫ sin 3 θdθ = −8πµΩa 3 ,
θ=0
where the minus sign indicates that the torque opposes the rotary motion, which is in the positive
ϕ direction.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 9.49. Consider the geometry of a cone and plate rheometer. A flat cone with radius R
and apex angle of θ1, slightly less than π/2, touches a large stationary horizontal flat plate at the
origin and rotates at a constant rate Ω about the vertical z-axis. A viscous fluid with density ρ
and viscosity µ fills the gap between the cone and the plate.
z!
Ω"
R! θ1!
x or y!

a) Assuming the fluid's velocity is steady with a single component: u = eϕuϕ(r,θ), use the
continuity equation and the azimuthal momentum equation for creeping flow in spherical
coordinates, where r = x 2 + y 2 + z 2 , to find:
) 1 # 1+ cosθ & ,
+ sin θ ⋅ ln % ( + cot θ .
2 $ 1 – cosθ '
uϕ (r, θ ) = Ωr sin θ1 + ..
+1 # 1+ cosθ1 & .
+ sin θ1 ⋅ ln % ( + cot θ1 .
*2 $ 1 – cosθ1 ' -
The boundary conditions here are uϕ(r, π/2) = 0, and uϕ(r,θ1) = rΩsinθ1.
- sin θ ∂ & uϕ ) 1 ∂uθ 0
b) Use this to determine the polar-azimuthal shear stress: τθϕ = µ/ ( ++ 2.
. r ∂θ ' sin θ * r sin θ ∂φ 1
c) Simplify the velocity field and shear stress results when θ and θ1 both approach π/2.
d) A torque of 3.0 N-m causes the cone to rotate with an angular velocity of 1.5 rad/s. If the
radius of the cone is R = 6.35 cm and 90° – θ1 = 0.30°, what is the viscosity of the fluid?
€
e) For the conditions in part d) with ρ = 103 kg/m3, compare ρΩ2R2 to τθϕ. Is neglect of fluid
inertia justified here?

Solution 9.49. a) The continuity and azimuthal momentum equation in spherical coordinates are:

1 ∂ 2 1 ∂ 1 ∂
2
r ∂r
( r ur ) +
r sin θ ∂θ
(uθ sin θ ) +
r sin θ ∂ϕ
(uϕ ) = 0 , and
∂uϕ ∂u u ∂u u ∂uϕ 1 1 ∂p
+ ur ϕ + θ ϕ + ϕ + ( ur uϕ + uθ uϕ cot θ ) = − +
∂t ∂r r ∂θ r sin θ ∂ϕ r ρr sin θ ∂ϕ
% 1 ∂ %€ ∂uϕ ( 1 ∂ % ∂uϕ ( 1 ∂ 2 uϕ uϕ 2 ∂ur 2cos θ ∂uθ (
2
ν' 2 'r +
* 2 ' sin θ * + − + + *.
& r ∂r & ∂r ) r sin θ ∂θ & ∂θ ) r 2 sin 2 θ ∂ϕ 2 r 2 sin 2 θ r 2 sin2 θ ∂ϕ r 2 sin2 θ ∂ϕ )
€ a steady single component velocity field, : u = eϕuϕ(r,θ), these simplify to:
For
1 ∂
r sin θ ∂ϕ
(uϕ ) = 0 , and
€
uϕ ∂ uϕ 1 ∂p " 1 ∂ " ∂u % 1 ∂ " ∂ uϕ % uϕ %
=− + ν $ 2 $r2 ϕ ' + 2 $ sin θ '− 2 2 '.
r sin θ ∂ϕ ρ r sin θ ∂ϕ # r ∂ r # ∂ r & r sin θ ∂θ # ∂θ & r sin θ &
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

The reduced continuity equation is satisfied identically by form of the given solution. The
momentum equation can be simplified to creeping flow by dropping the term on the left. In
addition, since velocity field is independent of ϕ, assume that the pressure field is similarly
independent of ϕ so the first term on the right of the reduced azimuthal momentum equation is
also zero. This leaves:
! 1 ∂ ! ∂u $ 1 ∂ ! ∂ uϕ $ uϕ $
0 = 0 + ν # 2 #r2 ϕ & + 2 # sin θ &− 2 2 &
" r ∂ r " ∂ r % r sin θ ∂θ " ∂θ % r sin θ %
subject to boundary condition: uϕ(r,π/2) = 0 and uϕ(r,θ1) = rΩsinθ1.
The equation is linear, so assume a separation of variables solution, uϕ(r,θ) = R(r)Θ(θ) ,
and multiply through by r2. Substituting the trial solution into the twice reduced azimuthal
momentum equation produces:
' 1 ∂ ! 2 ∂ R $* ' 1 ∂ ! ∂Θ $ 1 *
) #r &, + ) # sin θ &− 2 ,= 0 .
( R ∂ r " ∂ r %+ ( Θsin θ ∂θ " ∂θ % sin θ +
Following the separation-of-variable solution approach, set the first set of [,]-braces equal to β,
and the second set equal to –β. The ordinary differential equation that emerges for the r-direction
is:
d ! 2 dR $
#r & = β R which has solutions R = Arγ when β = γ(γ + 1).
dr " dr %
To satisfy the second boundary condition, γ must be 1, so β = 2. Change the variable in the
angular equation to η = cosθ, so that d/dθ = (dη/dθ)(d/dη) = (–sinθ)(d/dη). This leads to:
1 # ∂ &# 2 ∂Θ & 1 ∂ # 2 ∂Θ
& # 1 &
% − (% (η −1) ( = − β or % (1− η ) ( = % − γ (γ +1) (Θ .
Θ $ ∂η '$ ∂η ' 1− η 2 ∂η $ ∂η ' $ 1− η 2 '
This is the Legendre equation degree γ and order 1. It has solutions Pγ1 (η ) and Qγ1 (η ) .
Here γ = 1, so
12 1 2 (1 " 1+ η % η +
and Q11 (η ) = (1− η 2 ) * ln $
P11 (η ) = (1− η 2 ) '+ 2-
) 2 # 1− η & 1− η ,
are the solutions we need. The overall solution is:
uϕ (r, θ ) = ArP11 (cosθ ) + BrQ11 (cosθ ) ,
where A and B are constants. The first boundary condition, uϕ(r,θ = π/2) = 0, leads to A = 0 since
cosθ = 0 at θ = π/2 and P11 (0) = 1 while Q11 (0) = 0 . With A = 0, the second boundary condition,
uϕ(r,θ1) = rΩsinθ1, leads to:
Ωsin θ1
Ωr sin θ1 = BrQ11 (cosθ1 ) or B = 1 .
Q1 (cosθ1 )
Putting all this together leads to:
) 1 # 1+ cosθ & ,
+ sin θ ⋅ ln % ( + cot θ .
+ 2 $ 1 – cosθ ' .,
uϕ (r, θ ) = Ωr sin θ1
+1 # 1+ cosθ1 & .
+ sin θ1 ⋅ ln % ( + cot θ1 .
*2 $ 1 – cosθ1 ' -
which is the desired outcome for the azimuthal velocity.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

b) The approach here is to take the part a) velocity field and plug it into the shear stress formula.
% sin θ % 1+ cosθ ( (
1 1
To simplify the effort set C(θ1 ) = Ωsin θ1 ' ln' * + cot θ1 * , so that
& 2 & 1− cosθ1 ) )
% sin θ %1+ cos θ ( (
uϕ (r,θ ) = rC(θ1 )' ln' * + cot θ * . Continue to the differentiations: ∂uϕ ∂ϕ = 0 , and
& 2 & 1− cos θ ) )
€ ∂ uϕ % ( ∂ %1 1 cos θ (
' * = rC(θ1 ) ' ln(1+ cos θ ) − ln(1− cosθ ) + 2 *
∂θ & sin θ ) ∂θ & 2 2 sin θ )
% −sin θ €
€ ∂ % uϕ ( sin θ sin θ cos2 θ (
' * = rC(θ1 )' − − 2 −2 3 *
∂θ & sin θ ) & 2(1+ cos θ ) 2(1− cos θ ) sin θ sin θ )
€ sin θ ∂ % uϕ ( % −sin θ 2
sin θ 2
cos2 θ (
' * = C(θ1 )' − −1− 2 2 *
r ∂θ & sin θ ) & 2(1+ cosθ ) 2(1− cosθ ) sin θ )
Use some € trigonometric identities for simplification.
sin θ ∂ % uϕ ( % −(1− cosθ ) 1+ cosθ cos 2 θ ( % cos 2 θ ( −2C(θ1 )
' * = C( θ )
1 ' − −1− 2 * = −2C( θ 1 '1+
) 2 *
=
r €∂θ & sin θ ) & 2 2 sin 2 θ ) & sin θ ) sin 2 θ
- sin θ ∂ & uϕ ) 1 ∂uθ 0 2C(θ1 )
Thus, τθϕ = µ/ ( + + 2 = −µ , which does not depend on r!
. r ∂θ ' sin θ * r sin θ ∂φ 1 sin 2 θ
€ c) As θ → π/2, cos(θ) → π/2 – θ, and sin(θ) → 1, thus
' 1 '1+ π /2 − θ * π /2 − θ *
C(θ1 ) ≅ Ω ) ln) 1
,+
1
, ≅ Ω [2(π /2 − θ1 )] ,
€ ( 2 ( 1− π /2 + θ1+ 1 +
( π /2 − θ + µΩ
uϕ (r,θ ) ≅ 2rC(θ1 )(π /2 − θ ) = Ωr* - , and τθϕ = −
) π /2 − θ1 , π /2 − θ1
R
€ 2πµΩR 3
d) The torque necessary to turn the cone will be: M = − ∫ τr 2 dθdr = 2πτR 3 3 = .
r= 0 3(π /2 − θ1 )
Solve this
€ for the viscosity and evaluate: €
3(π / 2 − θ1 )M 3(0.30°π /180°)(3.0Nm)
µ= = = 19.5 kgm-1s-1
2πΩR 3 π (1.50rad / s)(0.0635m)3
2€
e) Here: ρΩ2 R 2 = (103kg/m3)(1.50rad/s)2(0.0635m)2 = 9.07 Pa, while
µΩ (19.5Pa ⋅ s)(1.50rad / s)
τ θϕ = = = 5.59×103 Pa, is more than 500 times higher. Thus,
π / 2 − θ1 0.30°(π /180°)
€neglect of fluid inertia is a good approximation here.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.1. A thin flat plate 2 meters long and 1 meter wide is placed at zero angle of attack
in a low speed wind tunnel in the two positions sketched below.
drag on the plate in position #1
a) For steady airflow, what is the ratio: drag on the plate in position #2 ?
b) For steady airflow at 10 m/sec, what is the total drag on the plate in position #1?
c) If the air flow is impulsively raised from zero to 10 m/sec at t = 0, will the initial drag on the
plate in position #1 be greater or less than the steady-state drag value calculated for part b)?
d) Estimate how long it will take for drag on the plate in position #1 in the impulsively started
flow to reach the steady-state drag value calculated for part b)?

Solution 10.1. a) Here we need only consider the drag coefficient, CD, from the Blasius solution.
CD (L) =
(Drag per unit span on one side of the plate of length L) = 1.328
1
2
ρU 2 L ReL
where ReL= UL/ν. Therefore:
1.328 1
Total Drag on plate "j" = 2(span of plate "j") ⋅ ⋅ ρU 2 L where j = 1 or 2 , so
€ ReL 2
Drag on plate #1 span of plate #1 L1 L1 2
= ⋅ = = 2.
Drag on plate # 2 span of plate # 2 L2 L2 2 2
€ 1.328 1
b) Total Drag on plate #1 = 2(span of plate #1) ⋅ ⋅ ρU 2 L1
ReL1 2
€ 2(2m)1.328 1
= ⋅ (1.21kg /m 3 )(10m /s) 2 (1m)
(10m /s)(1m) (1.5 ×10 m /s) 2
−5 2

= 0.394 N. €
c) When the flow is impulsively started, a temporally-developing boundary layer is produced on
the top and bottom of plate #1. Initially, the shear stress is very high since: τ w ≅ µU(πνt)−1 2 so
€
the initial drag will be greater than the steady-state drag.
d) The flow will have reached steady-state when the temporally developing boundary layer skin
friction has reached the Blasius boundary layer skin friction everywhere on the plate. The
Blasius skin friction is lowest at the trailing edge, so an estimate for € the drag-relaxation time can
be based on matching skin friction coefficients at the plate's trailing edge between the
temporally-developing boundary layer and the Blasius solution. Therefore, with ReL= UL/ν:
4L L
(c f )temporal BL ≈ (c f )spatial BL at x = L –> U2 πνt ≈ 0.664
ReL
, or t ≈ 2
π (0.664) U
≈ 3.0 = 0.30s .
U
Interestingly, this result is independent of the fluid viscosity!

€ € €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.2. Solve the Blasius equations (10.27) through (10.29) with a computer, using the
Runge–Kutta scheme of numerical integration, and plot the results. What value of f !! at η = 0
leads to a successful profile?

Solution 10.2. The Blasius equation is a non-linear third-order differential equation for f(η):
d3 f 1 d2 f
+ f = 0.
dη 3 2 dη 2
The boundary conditions are: df dη →1 as η → ∞, df/dη = f = 0 at η = 0. For a computer
solution using a Runge-Kutta integration scheme, this equation must be reduced to a set of three
first-order differential equations by defining:
€
g(η) = df dη , and h(η) = d 2 f dη 2 .
€ €
The above equation can then be written as three equations:
df dη = g(η) (A)
dg dη = h(η) (B)
€ €dh dη = − fh 2 (C)
subject to:
€
f(0) = g(0) = 0 and g(∞) = 1.
The set of three equations€ are readily integrated via the Runge-Kutta method starting
€
from η = 0 where the initial values of f and g are known. The initial value of h (= f !! at η = 0) is
not known but it can be found by trial and error (0.33205) by looking for the value of h(0) that
produces g = 1 at some suitably large value of η, perhaps 10 or 20.
A simple MatlabTM code that does this based on a trial & error value of h(0) is:

%Compute the Blasius Boundary Layer Profile

clear;
clc;
f0 = 0; %The first known boundary condition
g0 = 0; %The second known boundary condition
h0 = 0.33205; %The value adjusted by trial & error
eta_start = 0; %Starting point for eta
eta_end = 10; %End point for eta
%The next command invokes a Runga-Kutta integration scheme
[eta,f] = ode45(@Blasius,[eta_start eta_end],[f0 g0 h0]);

with the function

function df = blasius(eta,f)
% Solve the Blasius LBL profile equation.
% f(1)=stream function, f(2)=velocity, f(3)=velocity gradient
df = zeros(3,1);
df(1) = f(2); %The equivalent of (A)
df(2) = f(3); %The equivalent of (B)
df(3) = -0.500*f(1)*f(3); %The equivalent of (C)
end

The resulting plot of df/dη vs. η is:

 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

$"
!#,"
!#+"
!#*"
!#)"
df u !#("
=
dη U
!#'"
!#&"
€ !#%"
!#$"
!"
!" %" '" )" +" $!"

η = y U νx

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.3. A flat plate 4 m wide and 1 m long (in the direction of flow) is immersed in
kerosene at 20°C, (v = 2.29 × 10 6 m2/s, ρ = 800 kg/m3) flowing with an undisturbed velocity of
−

0.5 m/s. Verify that the Reynolds number is less than critical everywhere, so that the flow is
laminar. Show that the thickness of the boundary layer and the shear stress at the center of the
plate are δ = 0.74 cm and τ0 = 0.2 N/m2, and those at the trailing edge are δ = 1.05 cm and τ0 =
0.14 N/m2. Show also that the total frictional drag on one side of the plate is 1.14 N. Assume that
the similarity solution holds for the entire plate.

Solution 10.3. ReL = UL/ν = (0.5m/s)(1m)/(2.29x106 m2/s) = 2.18x105 < Recr ~ 106. Thus, the
flow is expected to be laminar everywhere.
At x = 0.5 m, Rex = 1.09x105 so the 99% thickness from (10.30) and the shear stress from (10.31)
are:
δ99 = 4.9x Re1x 2 = 4.9(0.5) 1.09 ×10 5 = 0.742cm , and
τ 0 = 0.332 ρU 2 Re1x 2 = 0.332(800)(0.5) 2 1.09 ×10 5 = 0.201N /m 2 .
At x = 1.0 m, Rex = 2.18x105 so
€ δ99 = 4.9x Re1x 2 = 4.9(1.0) 2.18 ×10 5 = 1.05cm , and
€ τ 0 = 0.332 ρU 2 Re1x 2 = 0.332(800)(0.5) 2 2.18 ×10 5 = 0.142N /m 2 .
The total drag can be obtained from (10.33):
12 −3
CD = 1.33 Re€ x = 2.85 ×10 , so
1
€ D = ρU 2 (Area)CD = 0.5(800)(0.5) 2 (4)(1)(0.00285) = 1.14N .
2
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.4. A fluid with constant density and viscosity flows with a constant horizontal speed
U∞ over an infinite flat porous plate placed at y = 0 through which fluid is drawn with a constant
velocity Vs. For this flow the steady two-dimensional zero-pressure-gradient boundary layer
equations are (7.2) and (10.18) and the boundary conditions are u(y = 0) = 0, v(y = 0) = –Vs, and
u = U∞ for y → ∞ .
a) Assuming u depends only on y, determine u(y) in terms of ν, Vs, U∞, and y.
b) What is the wall shear stress τw? How does it depend on µ?
c)€What parametric change(s) decrease the boundary layer thickness?

Solution 10.4. a) If u = u(y), then ∂u/∂x = 0 = ∂v/∂y and this means that v is a most a function of
x. However, the boundary condition on v at y = 0 does not depend on x, thus v = const. = –Vs.
∂u ∂ 2u
Therefore, the horizontal momentum equation simplifies to: −Vs = ν 2 , which can be
∂y ∂y
integrated to find: −Vs y ν = ln(∂u ∂y ) +C , where C is a constant. This can be rearranged and
integrated again to find: u(y) = D + E exp(−Vs y ν ) . The boundary condition, u(0) = 0 requires D
+ E = 0, while u(∞) = U∞ sets D = U∞. Thus, the velocity€profile is: u(y) = U∞ [1− exp(−Vs y ν )]
€ shear stress is τ w = µ(∂u ∂y ) y= 0 = −µU∞ (−Vs ν ) = ρU∞Vs . It does not depend on µ!
b) The wall
€ layer thickness decreases when Vs increases, and ν decreases. The horizontal
c) The boundary
flow speed does not influence the thickness of this boundary
€ layer.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.5. A square-duct wind tunnel test section of length L = 1 m is being designed to
operate at room temperature and atmospheric conditions. A uniform air flow at U = 1 m/s enters
through an opening of D = 20 cm. Due to the viscosity of air, it is necessary to design a variable
cross-sectional area if a constant velocity is to be maintained in the middle part of the cross-
section throughout the wind tunnel.
a) Determine the duct size, D(x), as a function of x.
b) How will the result be affected if U = 20 m/s? At a given value of x, will D(x) be larger or
smaller than (or the same as) the value obtained in a)? Explain.
c) How will the result be affected if the wind tunnel is to be operated at 10 atm (and U = 1 m/s)?
At a given value of x, will D(x) be larger or smaller than (or the same as) the value obtained in
a)? Explain. [Hint: the dynamic viscosity of air (µ [N·s/m2]) is largely unaffected by pressure.]
d) Does the airflow apply a net force to the wind tunnel test section? If so, indicate the direction
of the force.

Solution 10.5. a) The duct length (1 m) and the flow speed (1 m/s) imply a Reynolds number of:
ReL = UL/ν = (1 m/s)(1 m)/(1.5x10–5m2/s) = 67,000, which much larger than unity but still in the
laminar boundary layer range.
If δ* is the displacement thickness of the boundary layer at position x, then conservation
of mass between the inlet of the duct and location x implies:
2
ρUD2 (0) = ρU ( D(x) − δ * ) .
For constant density and velocity in the duct, the pressure gradient must be zero. Therefore, the
Blasius solution for a flat plate laminar boundary layer can be used, and the above equation
reduces to:
€
D(x) = D(0) + δ * = D(0) + 1.72 νx U = 0.20m + (1.5 ×10−5 m 2 /s)x (1m /s)
= 0.20 + 0.00387 x,
where the final numbers provide D in meters when x is in meters.
b) If U is 20 m/s instead of 1 m/s and the flow remains laminar, the displacement thickness of the
boundary layer will be reduced by a factor of 20 ≈ 4.47. Thus, the required D(x) will be
€
D(x) = D(0) + 1.72 νx U = 0.20m + (1.5 ×10−5 m 2 /s)x (20m /s) = 0.20 + 0.000866 x .
Here, the requisite D(x) is smaller because the boundary layer will be thinner at the higher speed.
c) The factor of ten increase in pressure
€ causes a factor of 10 increase in density. Thus for
constant µ, ν = µ/ρ will decrease by a factor of 10. This implies:
€
D(x) = D(0) + 1.72 νx U = 0.20m + (1.5 ×10−6 m 2 /s)x (1m /s) = 0.20 + 0.00122 x .

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

So, the requisite D(x) will be smaller than the part a) result because the boundary layer will be
thinner at the higher pressure because of the drop in kinematic viscosity (or equivalently, the
increase in Reynolds number).
d) Yes, the airflow applies a force to the wind tunnel via wall shear stress. This force points to
the right.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.6. Use the control volume shown to derive the definition of the momentum
thickness, θ, for flow over a flat plate:
h
u" u% Drag force on the plate from zero to x x
ρU 2θ = ρU 2 ∫ $1− 'dy = = ∫ τ w dx
0 U
# U& unit depth into the page 0
The words in the figure describe the upper and lower control volume boundaries.
y! U!
zero shear stress, !
U! constant pressure, no through flow!
h > δ99!
ho!
no slip, τw ≠ 0!

x!

Solution 10.6. The CV is stationary, so the integral form of the continuity equation is:
h
ρUho = ρ ∫ u(y)dy ,
0
where h is defined in the figure to be greater than the boundary layer thickness. The pressure is
the same everywhere so the integral form of the x-momentum equation is:
h x
€ −ρU 2 ho + ρ ∫ u 2 (y)dy = − ∫ τ w dx .
0 0

Here it is assumed that τw is positive (since µ and ∂u/∂y will be positive for the boundary layer
flow inside the CV), so the right-side integral represents the force the plate applies to the fluid in
the CV. Using the continuity equation, eliminate ho from the x-momentum equation:
h h x
−ρU ∫ u(y)dy + ρ ∫ u 2 (y)dy = − ∫ τ w dx .
0 0 0

Combine the two integrals, multiply by minus one, factor out ρU2, and recognize the definition
of the momentum thickness θ:
h ∞ x
2 u(y) # u(y) & 2
ρ ∫ u(y) (U − u(y)) dy = ρU ∫ %1− ( dy = ρU θ = ∫ τ w dx .
0 U U '
0
$ 0
Since u(y) = U for y > h, it is OK to extend the upper limit of the y-integration to +∞. The
integral of the shear stress on the right side of the last equation corresponds to the force the flow
exerts on the plate (per unit depth into the page). This force is positive and points in the
downstream direction (increasing x) and is therefore a drag force.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.7. Estimate the 99% boundary layer thickness on:

a) a paper airplane wing (length = 0.25 m, U = 1 m/sec),
b) the underside of a super tanker (length = 300 m, U = 5 m/sec), and
c) an airport run way on a blustery day (length = 5 km, U = 10 m/sec).
d) Will these estimates be accurate in each case? Explain.

Solution 10.7. Estimate the various thicknesses from the laminar zero-pressure-gradient
(Blasius) boundary layer solution: δ99 = 4.9 νx U .
a) δ99 = 4.9 (1.5 ×10−5 m 2 s–1 )(0.25m) (1ms−1 ) = 9.5mm
b) δ99 = 4.9 (1.0 ×10−6 m 2 s–1 )(300m) (5ms−1 ) = 3.8cm
c) δ99 = 4.9 (1.5 ×10−5 m€ 2 –1
s )(5000m) (10ms−1 ) = 0.42m
€ d) The estimate for part a) may be fairly accurate, but the estimates for b) and c) will not be
€ accurate. The downstream-distance-based Reynolds numbers in a), b), and c) are 1.7 ×10 4 ,
1.5 ×10 9 , and 3.3 ×10 9 respectively. At the larger two Reynolds numbers, the boundary layer
€
will be turbulent and its 99% thickness will be much greater than the Blasius boundary layer
estimate given above. €
€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.8. Air at 20°C and 100 kPa (ρ = 1.167 kg/m3, ν = 1.5 × 10 5 m2/s) flows over a thin
−

plate with a free-stream velocity of 6 m/s. At a point 15 cm from the leading edge, determine the
value of y at which u/U = 0.456. Also calculate v and ∂u/∂y at this point. [Answer: y = 0.857 mm,
v = 0.384 cm/s, ∂u/∂y = 3012 s–1.]

Solution 10.8. From Table 10.1, η = y U νx = 1.4 when u/U = 0.456. Therefore,
y = 1.4 ν x U = 1.4 (1.5 ×10 −5 )(0.15) 6 = 0.857 mm.
At this wall normal distance, the Blasuis BL formula for v and Table 10.1 produce:
1 νU " € df % 1 1.5 ×10 −5 (6)
v= $− f + η ' = (−0.325 +1.4 ⋅ 0.456) = 0.00384 m/s.
2 x # dη & 2 0.15
Again using Table 10.1, the slope of the velocity profile is:
∂u ∂ (u /U) ∂η
=U = Uf !! U ν x = (6)(0.3074) 6 (1.5×10 −5 )(0.15) = 3012 s–1.
∂y ∂η ∂ y
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.9. An incompressible fluid (density ρ, viscosity µ) flows steadily from a large
reservoir into a long pipe with diameter D. Assume the pipe-wall boundary layer thickness is
zero at x = 0. The Reynolds number based on D, ReD, is greater than 104.
a) Estimate the necessary pipe length for establishing a parabolic velocity profile in the pipe.
b) Will the pressure drop in this entry length be larger or smaller than an equivalent pipe length
in which the flow has a parabolic profile? Why?

Solution 10.9. a) Fortunately, the exercise asks for an estimate. The flow in the entrance length
of a round pipe will accelerate on the pipe's centerline, and the inner wall of the pipe is curved.
Both of these features will cause the boundary layer growth inside the pipe to differ from that of
a Blasius boundary. However, the Blasius solution does account for diffusive boundary layer
growth, so it should be fine for producing an estimate of the entrance length L.
The presence of the pipe walls is felt on the pipe centerline when the wall boundary layer
has attained a thickness of D/2. Therefore, set δ99 ≈ D/2 and estimate δ99 from the Blasius
solution:
D/2 ≈ 5(νL/U)1/2 = 5(LD/ReD)1/2 ,
and solve for L to find: L ≈ 10–2D(ReD).
b) The pressure drop in the entry section will be larger because: 1) wall shear stresses are larger
when BL's are thin, and 2) the central portion of the flow must accelerate to accommodate the
displacement effect of the BL so the velocity at the edge of the boundary layer increases from the
x = 0 to x = L.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.10. A variety of different dimensionless groups have been used to characterize the
importance of a pressure gradient in boundary layer flows. Develop an expression for each of the
following parameters for the Falkner-Skan boundary layer solutions in terms of the exponent n in
Ue(x) = axn, Rex = Uex/ν, integrals involving the profile function f ! , and f !!(0) , the profile slope
at y = 0. Here u(x, y) = Ue (x) f ! ( y δ (x)) = Ue f !(η ) and the wall shear stress
τ w = µ (∂ u ∂ y ) y=0 = (µUe δ (x)) f !!(0) . What value does each parameter take in a Blasius
boundary layer. What value does each parameter achieve at the separation condition?
a) (ν Ue2 ) ( dUe dx ) , an inverse Reynolds number
b) (θ2/ν)(dUe/dx), the Holstein and Bohlen correlation parameter

(
c) µ )
ρτ w3 ( dp dx ) , Patel's parameter
d) (δ * τ w ) ( dp dx ) , Clauser’s parameter

Solution 10.10. a) Here Ue(x) = axn, so dU e dx = nax n−1; thus

ν dU e ν nax n−1 ν n
2
= n
=n = .
U e dx U e ax U e x Re x
The profile function does not€enter here. This parameter is zero for the Blasius boundary layer,
and is –0.0904/Rex at separation.
b) This time the profile function enters through the definition of the momentum thickness.
€ 2
θ 2 dU e 1 % ∞ u % u ( ( dU e δ 2 % ∞ ( 2 dU e
= '∫ '1− *dy * = ' ∫ f .(η)(1− f .(η))dη*
ν dx ν & 0 U e & U e ) ) dx ν &0 ) dx
δ2 % ∞ ( 2 dU e 1 % νx (% ∞ ( 2 nU e
= ' ∫ f .(η)(1− f .(η))dη* = ' *' ∫ f .(1− f .)dη*
ν &0 ) dx ν & U e )& 0 ) x
%∞ (2
= n' ∫ f .(1− f .)dη* .
&0 )
'∞ *2
This parameter is zero for the Blasius boundary layer, and is –0.0904 ) ∫ f "(1− f ")dη, at
(0 +
separation.
€
c) This parameter involves the shear stress and pressure gradient directly. Differentiating the
steady Bernoulli equation without a body force produces dp dx + ρU e dU e dx = 0 , so
€ 32
µ dp µ 1 # dUe & µ # ν x Ue & nUe2
= % − ρU ( = − % ( ρ
ρ %$ µUe f !!(0) ('
32 e
ρτ w3 dx ρ (µUe f !!(0) δ (x)) $ dx ' x
€
µρν 3 4 1 x3 4 # 1 & 2 µ1 4 1 1 1
= −n 32 % 32 34 (U e = −n 32 14
32
ρµ [ f !!(0)] x $ Ue Ue ' ρ [ f !!(0)] x Ue1 4
14

1 1
= −n 32
[ f !!(0)] Re1x 4
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

This parameter is zero for the Blasius boundary layer, and it goes to +∞ at separation since f "" =
0 at separation and for the Falkner-Skan laminar boundary layer solutions –n = +0.0904 > 0.
d) This parameter involves the profile function through the displacement thickness and through
the shear stress. Using the differentiated Bernoulli relation specified in part c) leads to:
δ * dp $ ∞ ' δ (x) $ dUe ' $∞ ' δ 2 (x) € nUe2
= & ∫ (1− u U ) dy ) & −ρUe ) = − & ∫ (1− f *) dη ) ρ
τ w dx % 0 ( µUe f **(0) % dx ( %0 ( µUe f **(0) x
$∞ ' ν x Ue Ue2 n $∞ '
= −n & ∫ (1− f ) dη )
* ρ =− & ∫ (1− f *) dη ).
%0 ( µUe f **(0) x f **(0) % 0 (
This parameter is zero for the Blasius boundary layer, and it goes to +∞ at separation since f "" =
0 at separation and for the Falkner-Skan laminar boundary layer solutions –n = +0.0904 > 0.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Solution 10.11. Consider the boundary layer that develops as a constant density viscous fluid is
drawn to a point sink at x = 0 on and infinite flat plate in two dimensions (x, y). Here Ue(x) = –
UoLo/x, so set η = y νx |U e | and ψ = − νx |U e | f (η) and redo the steps leading to (10.36) to
find f """ − f "2 + 1 = 0 . Solve this equation and utilize appropriate boundary conditions to find
2
&1− αe− 2η ) 3− 2
f " =€
3( − 2η
+ − 2 where€ α= .
'1+ αe * 3+ 2
€
Exercise 10.11. Use the specified Ue to develop the appropriate expressions for ψ and h.
€ €ψ = νx |U | f (η) = νx U o Lo f (η) = νU L f (η) , and
e o o
x
y y U o Lo y
η= = = .
νx |U e | 2
νx U o Lo ν x
The Cartesian
€ velocity components are:
∂ψ df (η) ∂η UL 1 UL
u= = − νU o Lo = − νU o Lo f ' o o = − o o f ' , and
∂y
€ dη ∂ y ν x x
∂ψ df (η) ∂η UL ( y+ νU o L o
v =− = + νU o Lo = νU o L o f ' o o * − 2 - = − ηf ' .
∂x dη ∂x ν ) x , x
The€boundary conditions are:
f " →1 as η → ∞ (far from the wall the boundary layer velocity matches the free-stream),
f " → 0 as η → 0 (no slip at the wall), and
€ ηf # → 0 as η → 0 (no flow through the wall).

€ € ∂u ∂u 1 ∂p ∂ 2u
The horizontal boundary layer momentum equation is (10.9), u + v = − + ν 2 . The
€ € ∂x ∂y ρ ∂x ∂y
€ differentiated
€ Bernoulli equation, dp dx + ρ U e dU e dx = 0 , allows this to be simplified:
∂u ∂u dU e ∂ 2u ∂u ∂u U 2 L2 ∂ 2u
u + v = Ue + ν 2 or u + v = − o 3 o + ν 2 .
∂x ∂y dx ∂y x
∂€ ∂y x ∂y
where the second version € of the equation follows when U e (x) = –U o o/x. Substitute in the
L
velocity components in the final equation to find:
UL ∂ % U L ( νU o L o ∂% UL ( U 2 L2 ∂2 % U L (
€− o o f # '− o o f #* − €ηf # '− o o f #* = − o 3 o + ν 2 ' − o o f #* .
x ∂x & x ) x ∂y & x ) x ∂y & x )
Perform the differentiations and simplify:
UL % UL UL η( νU o L o UL UL 1 U 2 L2 UL UL 1
− o o f #' + o 2 o f # + o o f ## * + ηf # o o f ## o o = − o 3 o − ν o o o o 2 f ### ,
€ x & x x x) x x ν x x x ν x
− f #(+ f # + ηf ##) + ηf f# ## = −1− f ### or − f #(+ f # + ηf ##) + ηf f# ## = − f #2 = −1− f ###.
With a little rearrangement, the final equality then leads to the desired equation: f """ − f "2 + 1 = 0 .
This equation can be solved for f´ as follows. Multiply by f "" and integrate to find:
€
1€ 2 1 3
€ ( f "") − f " + f " + A = 0 .
2 3
The constant A can be evaluated by considering the limit η → ∞ . The first € boundary condition
€
provides f "(∞) = 1, and if f´ is constant then f ""(∞) = 0 . Thus, the last equation becomes
€
€
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

0 – 1/3 + 1 + A = 0 as η → ∞ ,
which requires A = –2/3. Therefore,
1 2 2 1 2 12
( f "") = − f " + f "3 , or f "" = ± [2 − 3 f " + f "3 ] ,
2 3 3 € 3
which can be separated:
df " df " df " 2
1 2
= 1 2
= 1 2
= dη .
€[2 − 3 f " + f "3 ] [(2 + f "€)(1− f ")(1− f ")] (1− f ")[2 + f "] 3
The possibility of the minus sign in front of the square root is dropped because η is always
positive along with 1− f # and 2 + f ". The final equality can be integrated with the substitution
2
F€ = 2 + f ":
2FdF 2dF 2
2
= 2
= dη
€ € (3 − F )F 3 − F 3
€ Now set F = 3 tanh(γ ) , so that dF = 3 (1− tanh 2 (γ )) dγ , thus
2dF 2 3 (1− tanh 2 (γ )) dγ 2 2 % F ( 2
∫ 3 − F 2 €∫ 3 1− tanh2 (γ ) = 3 γ = 3 tanh−1'& 3 *) = 3η + C
=
€
( )
€ % 2 + f $(
2 2
Backtracking all the way to f " produces: η+C = tanh−1' *.
3 3 & 3 )
€ 2 # 2&
The constant C can be evaluated from f "(0) = 0 : C = tanh−1% (
€ 3 $ 3'
€ $ η 3C ' $
2 η 2'
Inverting for f " yields: f " = 3tanh 2 & + ) − 2 = 3tanh & + tanh−1 ) − 2.
€ % 2 2 ( % 2 3 (
To reach the desired form, use the sum€formula for the hyperbolic tangent, and the definition of
the hyperbolic tangent. Then manipulate the exponentials and the square roots.
€ $ η 2 −η 2 '
2
$ '
2
− 2η
€ 2 & e −e 2 ) & 1− e 2 )
$ tanh η 2 + 2 3 ' + +
f " = 3&
( ) ) − 2 = 3& e
η 2
+e − η 2
3 ) − 2 = 3& 1+ e − 2η
3 ) −2
& + . ) & + . )
% (
&1+ tanh η 2 2 3 )
( ) &1+ - e
η 2
−e −η 2
0
2) &1+ - 1− e
− 2η
0
2)
&% , eη 2 + e−η 2 / 3 )( &% ,1+ e− 2η / 3 )(
$ 2 '2 2
&1− e
− 2η
+
3
1+ e−( 2η
)) $ 3 − 3e−
) − 2 = 3&
2η
(
+ 2 1+ e− ) 2η '
) −2
= 3&
& 2) & 3 + 3e− 2η
+ 2 (1− e ) − 2η )
&%1+ e
− 2η
(
+ 1− e− 2η
) 3 )(
% (
2
$ + 3− 2. − '
2η
$ 3+ 2− '
2 &1− - 0e )
= 3&
( 3 − 2 e−) 2η
) − 2 = 3&
, 3+ 2/ ) − 2.
& + 3− 2. − )
& 3+ 2+
% ( 3− 2 )e − 2η
)
( &1+ - 0e
2η
)
&% , 3+ 2/ )(
The final equality is the desired form.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.12. Start from the boundary layer equations, (7.2), (10.9), and (10.10), and
ψ = Ue (x)δ (x) f (η ) , where η = y δ (x) , with δ(x) unspecified, to complete the following items.
a) Show that the boundary-layer profile equation can be written:
f !!! + α ff !! + β (1− f !2 ) = 0 , where α = (δ ν ) d(Ueδ ) dx , and β = (δ 2 ν ) dUe dx .
b) The part a) equation will yield similarity solutions when α and β do not depend on x.
Therefore, assume α and β are constants, set Ue = axn, and show that n = β/(2α – β).
c) Deduce the values of α and β that allow the profile equation to simplify to the Falkner-Skan
profile equation (10.36).

Solution 10.12. a) The given stream function automatically satisfies the continuity equation, so
the next step is to substitute it into (10.9):
∂u ∂u dUe ∂2 u
u + v = Ue +ν 2 .
∂x ∂y dx ∂y
Here, the surface-normal boundary-layer momentum equation (10.10), –∂p/∂y = 0, and the
Bernoulli equation in the free stream have been used to replace the pressure gradient;
− (1 ρ ) (∂p ∂x ) = Ue ( dUe dx ) .
For the given form of the stream function the various velocity components and
derivatives are:
∂ψ 1 ∂ψ y y
u= = Ueδ f " = Ue f " , v = − = −Ue"δ f −Ueδ ! f +Ueδ f ! 2 = −Ue"δ f −Ueδ ! f +Ue f ! ,
∂y δ ∂x δ δ
∂u y ∂u 1 ∂2 u 1
= Ue! f ! −Ue f !! 2 δ ! , = Uee f !! , and 2
= Ue f """ 2 .
∂x δ ∂y δ ∂y δ
where a prime denotes differentiation of a function with respect to its argument. Equation (10.10)
is then reconstructed:
# y & # y &# 1& 1
Ue f !%Ue! f ! −Ue f !! 2 δ !( + % −Ue!δ f −Ueδ ! f +Ue f ! (%Uee f !! ( = UeUe! + νUe f !!! 2 ,
$ δ ' $ δ '$ δ' δ
and this can be simplified by performing the indicated multiplications on the left side, and
canceling equal and opposite terms:
δ! 1
UeUe! f !2 −UeUe! ff !! −Ue2 ff !! = UeUe! + νUe f !!! 2 .
δ δ
Multiply through by δ2/νUe to reach:
δ2 δ2 δδ ! δ2
Ue! f !2 − Ue! ff !! −Ue ff !! = Ue! + f !!! ,
ν ν ν ν
collect all the terms to one side of the equation,
δ2 δ
f + Ue! (1− f !2 ) + (δUe! +Ueδ !) ff !! = 0 ,
!!!
ν ν
and combine the terms in the last set of parentheses:
δ 2 dUe δ d
f !!! +
ν dx
(1− f !2 ) +
ν dx
(δUe ) ff !! = 0 , or f !!! + β (1− f !2 ) + α ff !! = 0 ,
where the given definitions, α = (δ ν ) d(Ueδ ) dx & β = (δ 2 ν ) dUe dx , have been used to reach
the final equation.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

b) When α and β are constants, their defining equations may be used to eliminate δ. Start with
the α-equation:
δ d(Ueδ ) δUe dδ δ 2 dUe Ue dδ 2
α= = + = +β .
ν dx ν dx ν dx 2ν dx
Using the two ends of this extended equality, substitute for δ2 from the definition of β to find a
non-linear equation for Ue:
U dδ 2 U d ! νβ $ U β d 2Ue
α= e +β = e # &+ β = − e +β .
2ν dx 2ν dx " dUe dx % 2 ( dUe dx )2 dx 2
Again using the two ends of this extended equality and simplifying leads to:
2
α − β " dUe % Ue d 2Ue
$ ' = − ,
β # dx & 2 dx 2
and this equation has power-law solutions: Ue = axn. Substituting this trial solution into this
equation produces:
α − β 2 2 2n−2 ax n α −β 1
an x =− an(n −1)x n−2 which simplifies to: n = − (n −1) .
β 2 β 2
Solving this final algebraic equation for n produces the desired relationship: n = β/(2α – β).
β ! n +1 $
c) The part b) result can be rearranged to find: α = # & . Substitute this into the profile
2" n %
equation from part a) to reach:
β # n +1 &
f !!! + β (1− f !2 ) + % ( ff !! = 0 ,
2$ n '
and this equation will match (10.36) when β = n. Thus, the values of α and β that produce the
Falkner-Skan profile equation are:
α = (n + 1)/2 and β = n.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.13. Solve the Falkner-Skan profile equation (10.36) numerically for n = –0.0904, –
0.654, 0, 1/9, 1/3, and 1 using boundary conditions (10.28) and (10.29) and the Runge–Kutta
scheme of numerical integration. Plot the results and compare to Figure 10.8. What values of f !!
at η = 0 lead to successful profiles at these six values of n?

Solution 10.13. The Falkner-Skan profile equation is a non-linear third-order differential

equation for f(η):
2
d 3 f n +1 d 2 f " df %
+ f − n$ ' + n = 0 .
dη 3 2 dη 2 # dη &
The boundary conditions are: df dη →1 as η → ∞, df/dη = f = 0 at η = 0. For a computer
solution using a Runge-Kutta integration scheme, this equation must be reduced to a set of three
first-order differential equations by defining:
f (η ) = f (η ) , f2 (η ) = df dη , and f3 (η ) = d 2 f dη 2 .
€1 €
The above equation can then be written as three equations:
df1 dη = f2 (η ) (A)
df2 dη = f3 (η ) (B)
df3 dη = −(1 2)(n +1) f1 f3 + nf22 − n (C)
subject to:
f1(0) = f2(0) = 0 and f2(∞) = 1.
The set of three equations are readily integrated via the Runge-Kutta method starting
from η = 0 where the initial values of f and g are known. The initial value of f3 (= f !! at η = 0) is
not known but it can be found by trial and error by looking for the value of f3 that produces f2 = 1
at some suitably large value of η, perhaps 10 or 20. These initial values of f !! are : 0.00, 0.1640,
0.33205, 0.51181, 0.75741, 1.232533.

A simple MatlabTM code that does this based on a trial & error value of h(0) is:

%Compute the Falkner-Skan Boundary Layer Profile

clear;
clc;
eta_start = 0;
eta_end = 10;

f0 = 0; %The first known boundary condition

g0 = 0; %The second known boundary condition
n = [-0.0904, -0.0654, 0, 1/9, 1/3, 1]; %Values of n
h0 = [0.00, 0.1640, 0.33205, 0.51181, 0.75741, 1.232533];
%Values adjusted by trial & error

figure; hold all;

for i = 1:length(n);
%The next command invokes a Runga-Kutta integration scheme
[eta, f] = ode45(@(eta, f) FS(eta, f, n(i)), [eta_start
eta_end],[f0 g0 h0(i)]);
xlswrite(['FS_data_n=' num2str(n(i)) '.xlsx'],[eta f])
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

x=(.5*(n(i)+1)).^.5*eta;
plot(x,f(:,2));
end
df u
=
dη U with the function

function df = FS( eta, f, n )

%UNTITLED Falkner-Skan LBL Profile
%f(1)=stream function, f(2)=velocity, f(3)=velocity gradient
%n=1;
df = zeros(3,1);
df(1) = f(2); %same as (A)
df(2) = f(3); %same as (B)
df(3) = -0.500*(n+1)*f(1)*f(3) + n*f(2)^2 - n; %same as (C)
end

η =ηyis:U νx
The resulting plot of df/dη vs.

1"
€
0.9"

0.8"

0.7"

0.6"

0.5"

0.4"

0.3"

0.2"

0.1"

0"
0" 2" 4" 6" 8" 10"

The left most curve corresponds to n = 1, and the right-most curve corresponds to n = –0.0904
with the other monotonically arrayed in between. And – except for the aspect ratio and extent of
the horizontal axis – this plot is identical to Figure 10.8.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.14. By completing the steps below, show that it is possible to derive von Karman's
boundary layer integral equation without integrating to infinity in the surface-normal direction
using the three boundary layer thicknesses commonly defined for laminar and turbulent
boundary layers: i) δ (or δ99) = the full boundary layer thickness that encompasses all (or 99%) of
the region of viscous influence, ii) δ* = the displacement thickness of the boundary layer, and iii)
θ = momentum thickness of the boundary layer. Here, the definitions of the later two involve the
y=δ $
* u(x, y) ' y=δ
u(x, y) $ u(x, y) '
first: δ (x) = ∫ &1− )dy and θ (x) = ∫ &1− )dy , where Ue(x) is the flow
y= 0 % U e (x) ( y= 0 U e (x) % U e (x) (
speed parallel to the wall outside the boundary layer, and δ is presumed to depend on x too.
a) Integrate the two-dimensional continuity equation from y = 0 to δ to show that the vertical
d dU e
€ velocity at the edge of the boundary
€ layer is: v(x, y = δ ) = (U e (x)δ * (x)) − δ .
dx dx
b) Integrate the steady two-dimensional x-direction boundary layer momentum equation from y =
τ0 d 2 δ * (x) dU e2 (x)
0 to δ to show that: =
ρ dx
(
U e (x)θ (x) + ) 2 dx
.
€
d b(x ) # db & # da & b(x ) ∂f (x, y)
[Hint: Use Leibnitz’s rule ∫
dx a(x )
f (x, y)dy = % f (x,b) ( − % f (x,a) ( + ∫
$ dx ' $ dx ' a(x ) ∂x
dy to handle

the fact that

€ δ = δ(x)]
∂u ∂v
Solution 10.14. Start
€ with + = 0 , and integrate from y = 0 to δ to get:
∂x ∂y
δ
∂u
∫ ∂x dy +v(x, y = δ ) = 0
0
where v(x,y=0) =€0. Use Leibnitz’s rule to get the differentiation outside the integral:
δ
∂u d δ dδ
v(x, y = δ ) = − ∫ dy = − ∫ udy + U e (x)
€ 0 ∂x dx 0 dx
Add and subtract Ue(x) within the integral and rearrange the result:
d δ dδ
v(x, y = δ ) = −
dx
∫ (u − U e (x) + U e (x))dy + U e (x) .
dx
€ 0
δ
d d dδ
=− ∫
dx 0
(u − U e (x)) dy − (U eδ ) + U e (x)
dx dx
€ d# δ #
u & & dU e d dU e
= %U e (x) ∫ %1− (dy ( − δ = (U eδ * ) − δ
dx $ 0 $ U e (x) ' ' dx dx dx
€ ∂u ∂u 1 dp ∂ 2u
b) Start with u + v = − + ν 2 , multiply the continuity equation by u and add it to this
∂x ∂y ρ dx ∂y
2
€ ∂u ∂uv 1 dp ∂ 2u
equation to get: + =− + ν 2 . Use the Bernoulli equation substitution for the
∂x ∂y ρ dx ∂y
pressure
€ and integrate this equation from y = 0 to δ to get:
δ
∂u 2 dU τ
∫ ∂x dy + U e (x)v(x, y = δ ) = U eδ dxe − ρw
€ 0

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Use Leibnitz’s rule to get the differentiation outside the integral, substitute from part a) for the
vertical velocity at the edge of the boundary layer, and add and subtract uUe within the integral:
d δ 2 dδ %d dU e ( dU e τ w
dx 0
∫ ( u − uU e + uU e ) dy − U e2
dx
+ U e (x)' (U eδ * ) − δ
& dx dx )
* = U eδ
dx
−
ρ
Invoke the definition of the momentum thickness θ and then the definition of the displacement
thickness δ*:
& δ )
€ − d U 2θ + d (U ∫ ( u − U + U ) dy + − U 2 dδ + U (x)&( d U δ * − δ dU e )+ = U δ dU e − τ w
( ) ( )
e e e e e e
dx dx ' 0 * dx ' dx e dx *
e
dx ρ
d d dδ %d dU e ( dU e τ w
− (U e2θ ) + (−U e2δ * + U e2δ ) − U e2 + U e (x)' (U eδ * ) − δ * = U eδ −
dx dx dx & dx dx ) dx ρ
Now rearrange the result, and group like terms together:
€
τw d 2 d 2 * d dU e2δ dδ dU dU
ρ dx
= ( )
Ueθ +
dx
(
Ueδ − Ue ) dx
( )
U eδ * −
dx
+ U e2
dx
+ U eδ e + U eδ e
dx dx
€
*
The final four terms sum to zero, and the terms containing δ can be expanded to obtain:
2 * *
τw d 2 * dU e 2 dδ 2 dδ dU e d dU e
=
ρ dx
(
Ueθ + δ ) dx
+ Ue
dx
− Ue
dx
− δ *U e =
dx dx
( )
U e2θ + U eδ *
dx
,
€
and the final equality is identical to (10.43), von Karman's boundary layer integral equation.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.15. Derive the von Karman boundary

y = h!
layer integral equation by conserving mass and Ue! Ue!
momentum in a control volume (C.V.) of width dx
and height h that moves at the exterior flow speed
Ue(x) as shown. Here h is a constant distance that is
comfortably greater than the overall boundary layer u(x,y)! !! u(x+dx,y)!
thickness δ.
p(x)! C.V.! p(x+dx)!

Solution 10.15. Use the integral laws as applied to a

moving control volume. Conservation of mass x! x+dx!
implies:
d$ h ' h
& ∫
dx ρBdy ) + ∫ ρ ([ u] x−dx / 2 − U e )e x ⋅ (−e x B)dy
dt % 0 ( 0
h
+ ∫ ρ([u] x +dx / 2 − U e )e x ⋅ (e x B)dy + ρ[(u − U e )e x + ve y ] y= h ⋅ e y Bdx = 0
0
The four terms above correspond to unsteady mass addition to the CV, mass flux on the left side
of the CV, mass flux on the right side of the CV, and mass flux on the top of the CV.
The CV is a constant size and the density is presumed to be constant, so the first term is
€ zero (the time derivative of a constant). Combine terms, perform the dot products, presume dx is
small, and divide by ρBdx to reach:
1 h h
∂u d %h (
∫ ([ u] x +dx / 2
− [ u] x−dx / 2 )
dy + [ v ] y= h
= 0 , or [ v ] y= h
= − ∫ dy = − ' ∫ udy * , (*)
dx 0 0 ∂x dx & 0 )
where the final equality follows from the fact that h is constant.
Now conserve horizontal momentum using the same CV.
d$ h ' h
€ & ∫
dx ρ uBdy ) + ∫ ρ [ u] x−dx / 2 ([ u] x−dx€ / 2 − U e )e x ⋅ (−e x B)dy
dt % 0 ( 0
h
+ ∫ ρ[ u] x +dx / 2 ([ u] x +dx / 2 − U e )e x ⋅ (e x B)dy + ρu[(u − U e )e x + ve y ] y= h ⋅ e y Bdx
0

= −τ 0 Bdx + ([ p] x−dx / 2 − [ p] x +dx / 2 ) hB

where τ0 is the wall shear stress, and p is the pressure. Simplify and combine terms, and divide
by ρBdx to reach:
d #h & 1 h 2 2 1 h
€ % ∫ udy ( +
dt $ 0 ' dx 0
∫ { }
[ u] x +dx / 2 − [ u] x−dx / 2 dy − U e ∫ {[u] x +dx / 2 − [u] x−dx / 2}dy + [uv ] y= h
dx 0
τ0 1
=− +
ρ ρdx
([ p]x−dx / 2 − [ p]x +dx / 2 )h
d dx d d
Use the chain rule on the total time derivative of the unsteady term, = = Ue , noting
dt dt dx dx
that horizontal velocity of the CV (= dx/dt) is Ue; and presume dx is small:
€
d #h & d h 2 d h τ 1 ∂p
U e % ∫ udy ( + ∫ u dy − U e ∫ udy + [ uv ] y= h = − 0 − h.
dx $ 0 ' dx 0 dx 0 € ρ ρ ∂ x

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Two terms on the left cancel, and at y = h, u = Ue and v is given by (*). In addition, the steady
Bernoulli equation implies: dp dx = − ρU e dU e dx . So, with these substitutions, the last equation
can be rewritten:
d h 2 d $h ' τ0 dU e τ0 dU e h
dx€0
∫ u dy − U e &∫
dx % 0
udy ) = −
ρ
+ U e
dx
h = −
ρ
+ U e
dx 0
∫ dy .
(
Rearrange this equation so the skin friction term appears by itself with positive sign. Then
manipulate the integrals to form the definitions of the momentum and displacement thicknesses.
τ0 d h 2 d &h ) dU e h
€
ρ
=−
dx 0
∫ u dy + U e ( ∫ udy + + U e
dx ' 0 dx 0
∫ dy
*
d & h u2 ) d & h u ) dU e h u dU e h
= − (U e2 ∫ 2 dy + + (U e2 ∫ dy + − U e ∫ dy + U e ∫ dy
dx ' 0 U e * dx ' 0 U e * dx 0 U e dx 0
d& h u & u) ) dU e h & u) d * dU e
= (U e2 ∫ (1− +dy = (U e θ ) + U eδ
2
(1− +dy + + U e
dx ' 0 U e ' U e * *
∫
dx 0 ' U e * dx dx
.

The final equality is identical to (10.43), von Karman's boundary layer integral equation.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.16. For the following approximate flat-plate boundary layer profile:
u %sin(πy 2δ ) for 0 ≤ y ≤ δ (
=& ) , where δ is the generic boundary layer thickness, determine:
U '1 for y > δ*
a) the displacement thickness δ*, the momentum thicknesses θ, and the shape factor H = δ * θ .
b) Use the zero-pressure gradient boundary layer integral equation to find: (δ x ) Re1x 2 ,
€ (δ * x )Re1x 2 , (θ x )Re1x 2 , c f Re1x 2 , and CD Re1L 2 for the approximate profile.
c) Compare these results to their equivalent Blasius boundary layer values.€
€
€ €
Solution € boundary layer velocity profile, u = sin$& πy ') = sin$& π ζ ') where
10.16.€a) The sinusoid
U % 2δ ( %2 (
y
ζ= , is a reasonably accurate approximate laminar boundary layer profile. The momentum
δ
thickness for this profile is:
δ 1 € π 2
u$ u' $ π '$ $ π '' 2δ
θ = ∫ &1− )dy = δ ∫ sin& ζ )&1− sin& ζ ))dζ = ∫ sin(ψ )(1− sin(ψ ))dζ
0 U
€ % U( 0
% 2 (% % 2 (( π 0
π 2 π 2
The integration is not too complicated, ∫ 0
sin(ψ )dψ = 1, and ∫ 0
sin 2 (ψ )dψ = π 4 , so
2δ & π ) & 4 − π )
€
θ= (1− + = δ( + = 0.1366δ
π ' 4 * ' 2π *
The displacement thickness for € theπ approximate profile
€ is:
δ 2
$ u' 2δ 2δ $ π ' $ π − 2 '
δ * = ∫ &1− )dy =
% € U( π 0
∫ (1− sin(ψ )) dψ = & −1) = δ&
π %2 ( % π (
) = 0.3634δ ,
0
so that the shape factor H = δ * θ = 2.66 . Substitution into the zero-pressure gradient (ZPG) Von
Karman boundary layer integral relationship produces:
τw dθ 2τ 2(πµU 2δ ) πν dθ ) 4 − π , dδ
€ = U2 , or c f = w2 = 2
= =2 =+ .
€ ρ dx ρU ρ U U δ dx * π - dx
where τ w = µ(∂u ∂y ) y= 0 = πµU 2δ has been obtained directly from the approximate sinusoidal
profile. Now use the fourth and sixth terms in the extended equality to form a differential
equation
€ for δ. €
€ πν & 4 − π ) d δ dδ % π 2 ( ν % 2π 2 ( ν ν
=( + , or δ =' * –> δ 2 = ' * x + const = 23.0 x + const .
Uδ ' π * dx dx & 4 − π ) U &4 − π )U U
Here the implicit assumption is that the boundary layer thickness is zero at x = 0 so the constant
ν 2
drops out. Thus: δ = 23 x = 4.80 ⋅ x ⋅ Re−1 x . Using the results above for the other length
€ U
€ €
scales and parameters produces:
θ 12
b) Rex = 0.654 (0.664)
x
€*
δ
Re1x 2 = 1.743 (1.721)
x
€ δ 12
Rex = 4.80 (5.0)
x
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

πν 1 2 πν
c f Re1x 2 =
Rex = 2
Re1x 2 = 0.654 (0.664)
Uδ U ⋅ 4.80xRe−1x

1 L 0.654 0.654 ν L dx 2 ⋅ 0.654 1 2

CD Re1L 2 = ∫ 12
dx ⋅ Re12
L = ∫ ⋅ Re1L 2 = ReL = 1.309 (1.328)
L 0 Rex L U 0 x Re1L 2
€ c) The Blasius boundary layer results are listed above at the right in parentheses. The sinusoid
velocity profile results are all within 4% of the Blasius values.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.17. An incompressible viscous fluid with kinematic viscosity ν flows steadily in a
long two dimensional horn with cross sectional area A(x) = Aoexp{βx}. At x = 0, the fluid
velocity in the horn is uniform and equal to Uo. The boundary layer momentum thickness is zero
at x = 0.
a) Assuming no separation, determine the boundary layer momentum thickness, θ(x), on the
lower horn boundary using Thwaites method.
b) Determine the condition on β that makes the no-separation assumption valid for 0 < x < L.
c) If θ(x = 0) was nonzero and positive, would the flow in the horn be more or less likely to
separate than the θ(x = 0) = 0 case with the same horn geometry?

Solution 10.17. a) As stated above, use Thwaites method to estimate the boundary layer
momentum thickness. Start with conservation of mass to determine the U(x): U(x)A(x) = U o Ao ,
so U(x) = U o Ao A(x) = U o exp{−βx} . Thus, the Thwaites equation becomes:
0.45ν x 5 0.45ν +6 β x x 5 −5 β x!
θ 2 (x) = ∫ U ( x !)d x ! = e ∫ Uo e € dx!
U 6 (x) 0 Uo6 0

€ $
0.45ν +6 β x e −5 β x
1 ' 0.09ν +6 β x
=
Uo
e & + )=
% −5β 5β ( βUo
e (1− e−5 β x )

θ 2 dU 0.09 +6 βx
b) First compute λ = = e (1− e−5 βx )(−βU oe− βx ) = −0.09(e +5 βx −1) . Separation will
ν dx βU o
have occurred if λ < λseparation = –0.090. Therefore, λ > λseparation at x = L is required to avoid
separation. This means:
−0.09(e +5 βL −1) > λseparation .
€
Convert this requirement to a condition on β:
λ 1 % λseparation ( 0.139
e +5 βL −1 < − separation , or β < ln&1− )≈
€ 0.09 5L ' 0.09 * L
c) If the boundary layer starts with a non-zero momentum thickness it is thicker and therefore
more likely to separate in an adverse pressure gradient.
€ €
 y
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
b
Exercise 10.18. The steady two-dimensional velocity potential for a source of strength q located
a distance b above a large flat surface located at y = 0 is:
x
q xsep 2
φ (x, y) =
2π ( 2 2
ln x + (y − b) + ln x + (y + b) 2
)
a) Determine U(x), the horizontal fluid velocity on y = 0.
b) Use this U(x) and Thwaites method to estimate the momentum thickness, θ(x), of the laminar
boundary layer that develops on the flat surface when the initial momentum thickness θo is zero.
x ξ 5 dξ x 6 (x 2 + 4b 2 )
[Potentially useful information: ∫ 0 2 = ]
(ξ + b 2 ) 5 24b 4 (x 2 + b 2 ) 4
c) Will boundary layer separation occur in this flow? If so, at what value of x/b does Thwaites
method predict zero wall shear stress?
d) Using solid lines, sketch the streamlines for the ideal flow specified by the velocity potential
€
given above. For comparison, on the same sketch, indicate with dashed lines the streamlines you
expect for the flow of a real fluid in the same geometry at the same flow rate.

! ∂φ $ q ! x x $ q! x $
Solution 10.18. a) U(x) = # & = # 2 2
+ 2 2&
= # 2 2&
" ∂ x %y=0 2π " x + (y − b) x + (y + b) %y=0 π " x + b %
b) Use the Thwaites integral without the θo-term.
6 6
2 0.45ν x 5 0.45νπ 6 ( x 2 + b 2 ) x q 5 " ξ %5 0.45νπ ( x 2 + b 2 ) x ξ 5dξ
θ (x) = 6
U (x) 0
∫ U (ξ )dξ = q6 x 6
∫ π 5 $# ξ 2 + b2 '& dξ = qx 6
∫ (ξ 2 + b2 )5
0 0
Evaluate the integral using the given information:
0.45νπ (x 2 + b 2 )6 x 6 (x 2 + 4b 2 ) 0.45νπ (x 2 + b 2 )2 2
θ 2 (x) = 6 4 2 2 4
= 4
(x + 4b 2 )
qx 24b (x + b ) 24b q
c) Compute the correlation parameter λ from the results of part b):
θ 2 dU 0.45π (x 2 + b 2 )2 2 2 q
" b 2 − x 2 % 0.45 " x 2 %" x 2 %
λ= = (x + 4b ) $ ' = $ 4 + '$1− ' .
ν dx 24b 4 q π # (x 2 + b 2 )2 & 24 # b 2 &# b 2 &
Clearly when x > b, λ becomes negative and its magnitude increases with increasing x, therefore
at some x > b boundary layer separation will occur. Within Thwaites method, λ = λsep = –0.090
predicts the point, xzs, where the wall shear stress is zero; therefore, use the prior result and the
quadratic formula to find
0.45 # x zs2 &# x zs2 &
λsep = % 4 + 2 (%1− 2 ( →
24 $ b '$ b '
12
x zs $ 3 9 + 4(4 − 24 λsep 0.45) '
= &− + € ) = 1.35 .
b &% 2 2 )(
€
d) The streamlines for the ideal flow correspond to a
point source above a flat surface. The real flow
streamlines must include boundary layer separation at
€ a value of x that is of order b. Furthermore, the
thickness of the separated flow region will cause
upward shifts in the ideal flow streamlines.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.19. A fluid-mediated particle-deposition process requires a laminar boundary layer

flow with a constant shear stress, τw, on a smooth flat surface. The fluid has viscosity µ and
density ρ (both constant). The flow is steady, incompressible, and two-dimensional, and the flat
surface extends from 0 < x < L. The flow speed above the boundary layer is U(x). Ignore body
forces.
a) Assume the boundary layer thickness is zero at x = 0, and use Thwaites’ formulation for the
shear stress, τ w = (µU θ ) l( λ) with λ = (θ 2 ν )( dU dx ) , to determine θ(x) and U(x) in terms of λ,
ν = µ ρ , x, and τ w µ = constant. [Hint: assume that U θ = A and l( λ) are both constants so
that τ w µ = Al( λ) .]
b) Using
€ the Thwaites integral (10.50) and the results of part a), determine λ.
€
c) Is boundary layer separation a concern in this flow? Explain with words or equations.
€ € € €
€

Solution 10.19. a) For Thwaites method, the shear stress is provided by the correlation
τ w = (µU θ ) l( λ) . Thus, the simplest way to achieve constant τw is for U to be proportional to θ,
i.e. U = Aθ where A is a constant, and l(λ) = constant, where λ = (θ 2 ν )( dU dx ) = constant.
Eliminating U from this system of equations leaves:
€ τw
€
= Al( λ ) , and λ = A(θ 2 ν )( dθ dx ) . (1,2)
µ €
Integrate (2) to find: λνx = Aθ 3 3 where the constant of integration is zero because θ = 0 at x =
0. Now use equation (1) to eliminate A in the relationship for θ. Thus, the solution falls into an
implicit form involving λ and l(λ). €
€ 13 13
€ θ (x) = ( 3λl( λ)νµx τ w ) and U(x) = Aθ (x) = ( 3τ w λνx µ l ( λ))
2 2 2

b) Put the part a) results into the Thwaites integral:

x
23 0.45ν 53
(3λl(λ)νµx τ w ) = 2 6 3 ∫ ( 3τ w2 λνξ µ 2 l 2 ( λ )) dξ .
€ €(3τ w λνx µ 2l 2 (λ)) 0
Perform the integration:
23 0.45ν 53 x 8 3 0.45 3νx
(3λl(λ)νµx τ w ) = 63 (3τ 2
w λν µ 2 l 2 ( λ ))
83
=
8 ( 3τ w2 λνx µ 2 l 2 ( λ))1 3
.
€ (3τ w2 λνx µ 2l 2 (λ))
Use the two ends of the equality and simplify to find:
23 13 0.45 0.45
(3λl(λ)νµx τ w ) (3τ w2 λνx µ 2 l 2 (λ)) 3νx , or λ =
= 3νxλ = = 0.05625
8 8
€
c) Boundary layer separation is not a concern here because λ and τw are both positive and
constant, so the separation condition λ = –0.090 is not approached.
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.20. The steady two-dimensional potential for incompressible flow at nominal
horizontal speed U over a stationary but mildly wavy wall is: φ (x, y) = Ux −Uε exp (−ky) cos ( kx ) ,
where kε << 1. Here, ε is the amplitude of the waviness and k = 2π/Λ, where Λ = wavelength of
the waviness.
a) Use the potential to determine the horizontal velocity u(x, y) on y = 0.
b) Assume that u(x, 0) from part a) is the exterior velocity on the wavy wall and use Thwaites’
method to approximately determine the momentum thickness, θ, of the laminar boundary layer
that develops on the wavy wall when the fluid viscosity is µ, and θ = 0 at x = 0. Keep only the
linear terms in kε and ε/x to simplify your work.
c) Is the average wall shear stress higher for Λ/2 ≤ x ≤ 3Λ/4, or for 3Λ/4 ≤ x ≤ Λ.
d) Does the boundary layer ever separate when kε = 0.01?
e) In 0 ≤ x ≤ Λ, determine where the wall pressure is the highest and the lowest.
f) If the wavy surface were actually an air-water interface, would a steady wind tend to increase
or decrease water wave amplitudes? Explain.
U! y!
ε!
x!
Λ/2! Λ!

Solution 10.20. a) The horizontal velocity u(x,y) is obtained from:

u = ∂φ ∂x = (∂ ∂x )[Ux − Uε exp(−ky ) cos( kx )] = U + Ukε exp(−ky ) sin( kx ) .
when evaluated on y = 0, this becomes: u(x, y = 0) = U (1+ kε sin(kx)) .
b) Evaluate the Thwaites integral with θ = 0 at x = 0 to find:
0.45ν x 5 0.45ν x
5
θ 2€= 6 ∫ U ( x $)d x $ = 6 ∫ (1+ kε sin(kx )) dx
$ $
U (x) 0 U (1+
€ kε sin(kx)) 0
x
0.45ν 0.45ν x
≈ (1− 6kε sin(kx)) ∫ (1+ 5kε sin(kx &)) dx & = (1− 6kε sin(kx))[ x & − 5ε cos(kx &)] 0
U 0 U
€ 0.45ν 0.45νx
= (1− 6kε sin(kx))[ x − 5ε cos(kx) + 5ε] ≈ (1− 6kε sin(kx) + 5(ε x)(1− cos(kx))
U U
'0.45νx *1 2
€ Thus, θ ≈ ) (1− 6kε sin(kx) + 5(ε x)(1− cos(kx)), .
( U +
€ c) First compute the correlation parameter:
θ 2 dU 0.45x
λ= = (1− 6kε sin(kx) + 5(ε x)(1− cos(kx))(Uk 2ε cos(kx)) .
€ ν dx U
This form can be made easier to look at if it is rearranged and the second order terms are
dropped:
λ ≈ 0.45(kx)(kε)cos(kx)(1− 6kε sin(kx) + 5(ε x)(1− cos(kx)) ≈ 0.45(kx)(kε)cos(kx) .
€
Then note that τ w = µ(U θ ) l( λ) and that l(λ) increases monotonically with increasing λ.
Therefore, the shear stress will be highest where λ is highest. Now switch to considerations of
€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

wavelength. Given that cos(kx) < 0 for Λ/2 ≤ x ≤ 3Λ/4 and that cos(kx) > 0 for 3Λ/4 ≤ x ≤ Λ, τw
must be higher for 3λ/4 ≤ x ≤ λ.
d) For separation, set λsep = −0.090 ≈ 0.45(kx)(kε)cos(kx) . This can be solved numerically for kε
= 0.01 to find kx ≈ 21.6, so separation is expected to occur when x is large enough.
e) Use the steady flow Bernoulli equation with the reference condition taken far above the wavy
surface:
1 €2 1
2 2
( 2
)
2
ρU + P∞ = ρ U 2 [1+ kε exp(−ky)sin(kx)] + U 2 [ kε exp(−ky)cos(kx)] + P(x) .
Evaluate on y = 0 and keep only the linear terms in kε to find:
1
P(x) − P∞ = ρU 2 (−2kε sin(kx) + ...) .
€ 2
Thus, the minimum pressure occurs where sin(kx) = 1: kx = π/2 or x = Λ/4 (the wave crest), and
the maximum pressure occurs where sin(kx) = –1: kx = 3π/2 or x = 3Λ/4 (the wave trough).
f) Based on the results
€ provided above, a steady wind will increase the wave amplitude by at
least two mechanisms. The boundary layer shear tends to preferentially pull surface water away
from wave troughs and toward wave crests (a shear stress effect), and the wind-induced pressure
tends to suck wave crests up and push wave troughs down.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.21. Consider the boundary layer that develops in stagnation point flow:
U e (x) = U o x L .
a) With θ = 0 at x = 0, use Thwaites method to determine δ*(x), θ(x), and cf(x).
b) This flow also has an exact similarity solution of the full Navier Stokes equations. Numerical
€ evaluation of the final nonlinear ordinary differential equation produces: c f Rex = 2.4652 ,
where Re x = U e x ν = U o x 2 Lν . Assess the accuracy of the predictions for cf(x) from the
Thwaites method for this flow.

€ & 6
€ Solution 10.21. a) The Thwaites equation is: θ 2 = 0.45ν
x
Uo )
6
U e (x) 0
5
e
2
∫ U (ξ )dξ + θo (' U (x) +* . Here, θ = 0
e
at x = 0 and U e (x) = U o x L , so only the integral term contributes:
2 0.45ν L6 x 5 ξ 5 0.45ν L νL
6 ∫
θ = 6
U o 5 dξ = , or θ = 0.2739 .
U o x 0€ L Uo 6 Uo
€ θ 2 dU e θ 2 U o 0.45
Now compute the correlation parameter λ = = = = 0.075
ν dx ν L 6
The wall shear
€ stress is obtained from: €
µU e µU o x L & U o3 )1 2
τw = l( λ) = (0.3267) = 1.193µx( 3 + ,
θ € 0.2739 νL U o 'νL *
where the numerical value for l(λ) has been obtained by interpolation in the tabulation of laminar
boundary layer functions.
12
€ τw 2(1.193)µx % U o3 ( νL
Thus, the skin friction coefficient is: c f = 1 = 2 2 2 ' 3*
= 2.386 .
2
ρU e ρU o x L &νL ) Uo x 2
2
νL νL
Similarly, δ * = θH( λ) = 0.2739 (2.36) = 0.6464 where again the value of H(λ) has been
Uo Uo
obtained by interpolation. €
b) If c f Rex = 2.4652 is the exact answer, then the Thwaites calculation is acceptably accurate
(3%
€ low) for most engineering purposes.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.22. A laminar boundary layer develops on a large smooth flat surface under the
influence of an exterior flow velocity U(x) that varies with downstream distance, x.
a) Using Thwaites method, find a single integral-differential equation for U(x) if the boundary
layer is to remain perpetually right on the verge of separation so that the wall shear stress, τw, is
zero. Assume that the boundary layer has zero thickness at x = 0.
γ
b) Assume U(x) = U o ( x L) and use the result of part a) to find γ.
c) Compute the boundary layer momentum thickness θ(x) for this situation.
d) Determine the extent to which the results of parts b) and c) satisfy the von Karman boundary
layer integral equation, (10.43), when τw = 0 by computing the residual of this equation.
€
Interpret the meaning of your answer; is Von Karman’s equation well satisfied, or is the residual
of sufficient size to be problematic?
e) Can the U(x) determined for part b) be produced in a duct with cross sectional area
−γ
A(x) = Ao ( x L) ? Explain your reasoning.

Solution 10.22. a) When τw is zero, then λsep = λ ≡ (θ 2 ν ) dU dx . Now plug this into the
Thwaites integral:
0.45ν x 5 νλsep 0.45 dU x 5
θ 2 (x) = 6 ∫ U ( x $)dx $ = , or λ sep = 6 ∫ 0U ( x #)dx # .
U (x) 0 € dU dx U (x) dx
γ −1 5γ +1
γ 0.45 γU o $ x ' U o5 L $ x ' 0.45γ
b) Plugging in U(x) = U o ( x L) produces: λsep = 6 6γ & ) & ) = ,
U o ( x L) L % L ( 5γ + 1 % L ( 5γ + 1
and€for λsep = –0.090, this equation produces: € γ = –0.10. [Note that this value is with in 0.01 of
the exact Falkner-Skan exponent value, –0.0904 for a laminar boundary layer that is perpetually
on the€verge of separation.]
€
c) Evaluate the Thwaites integral with θo = 0 using γ from part b):
2 0.45ν x 5 0.45ν U o5 L ' x *
−0.5+1
0.90νL ' x *
1.1
$ 0.90νL '1 2 $ x ' 0.55
θ = 6
U (x) 0
∫ U ( x $)dx $ = U 6 x L −0.6 −0.5 + 1)( L ,+ = U )( L ,+ , θ (x) = &% U )( &% L )( .
o( ) o o

τw d 2 dU
d) Use = (U θ ) +Uδ * but set τw to zero and recognize that δ * = 3.55θ from the
ρ dx dx
€ Thwaites shape-factor tabulation. Evaluate the two right side€terms of the equation:
# −0.2 12 0.55 & −0.65 12
d d % 2 # x & # 0.90νL & # x & ( U o2 # x & # 0.90€νL &
dx
(U θ ) = dx %U o %$ L (' %$ U (' %$ L (' ( = 0.35 L %$ L (' %$ U ('
2

$ o ' o
1 2 12
* dU # x &−0.1 # 0.90νL & # x & 0.55 # Uo # x & &
−1.1 −0.65
U o2 # x & # 0.90νL &
Uδ = U o % ( 3.55% ( % ( %% −0.1 % ( (( = −0.355 % ( % (
dx $ L' $ Uo ' $ L ' $ L $ L' ' L $ L ' $ Uo '
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

−0.65 12
U 2 " x % " 0.90νL %
Thus, we see that there is a residual = 0.005 o $ ' $ ' which is only 1.4% of either
L # L & # Uo &
of the terms. This residual is acceptable given the approximate nature of Thwaites method.
−γ
e) For an ideal two-dimensional duct with A(x) = Ao ( x L) , the upper and lower walls would
+0.1
only diverge in proportion to € ( x L) . However, the boundary layer displacement thickness
+0.55
grows like ( x L) . Thus, adjustments would have to be made in the duct geometry to account
for boundary layer growth. Such € adjustments would involve greater divergence of the upper and
lower walls, but increased€ wall divergence is likely to produce boundary layer separation. Thus,
−0.10
the part-b) flow speed U(x) = U o ( x L) cannot be produced in a duct with cross sectional area
€ −0.10
A(x) = Ao ( x L) when viscous laminar boundary layers are present on the walls of the duct.

€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.23. Consider the boundary layer that develops on a cylinder of radius a in a cross
flow
a) Using Thwaites method, determine the momentum thickness as a function of ϕ, the angle from
the upstream stagnation point (see drawing).
b) Make a sketch of cf versus ϕ.
c) At what angle does Thwaites method predict vanishing wall shear stress?

Solution 10.23. The potential for 2D ideal flow around a cylinder is: φ (r,θ ) = Ur cosθ (1+ a 2 r 2 ) .
For the specified geometry, θ = π − ϕ and cosθ = −cos ϕ so φ (r,ϕ ) = −Ur cos ϕ (1+ a 2 r 2 ) . Thus,
the exterior velocity, Ue, outside the boundary layer that develops on the cylinder’s surface, is
%1 ∂φ ( €
given by: U e = ' * = [2U sin ϕ ] x= aϕ = 2U sin( x a) , so the Thwaites integral becomes:
& r ∂ϕ
€ )r= a,x= aϕ € €
x ϕ
2 0.45ν 0.45aν
6 ∫ 32U sin (ξ a) dξ =
5 5
θ = 6
2U sin ϕ 0
∫ sin 5 (ϕ ')dϕ '.
(2U sinϕ ) 0
€ initial condition term has been ignored since Ue(0) = 0. The integral of sin5(ϕ) may be
The
looked up or performed via sin5(ϕ) = (1-cos2ϕ)2sin(ϕ) and an appropriate integration variable
switch. The result is:
€
2 0.45aν & 8 2 3 1 5 ) 0.225a 2 & 8 2 3 1 5 )
θ = 6 ( − cos ϕ + cos ϕ − cos ϕ + = 6 ( − cos ϕ + cos ϕ − cos ϕ + ,
2U sin ϕ '15 3 5 * Resin ϕ '15 3 5 *
where Re = Ua/ν (U is the velocity used in all the non-dimensional parameters in this problem).
dU e 2U 0.03cosϕ
b) Here
dx
=
a
cos ϕ , so for λ, we get: λ = 6
sin ϕ
( )
8 −15cosϕ + 10cos 3 ϕ − 3cos5 ϕ . Now,
€
evaluate the skin friction coefficient:
−1 2
µU e 1 νU e l( λ)sin 4 ϕ ( 8 2 3 1 5 +
cf = l( λ) 1 = 2l( λ ) 2 = 8.43 * − cosϕ + cos ϕ − cos ϕ - .
€ θ 2
ρU 2 €U θ Re ) 15 3 5 ,
Before plotting make a table of cf values.
ϕ λ S(λ) c f Re
€ ––––––––––––––––––––––––––––––––––––––––––––
1° 0.075 0.33 0.11
15° 0.074 0.32 1.7
30° 0.072 0.32 € 3.1
60° 0.59 0.31 4.4
90° 0.0 0.22 2.5
100° -0.060 0.11 1.0
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

103° -0.089 ~0.0 ~0.0

5
c f Re
4.5

3.5

2.5

1.5

0.5

0
0 20 40 60 80 100 120
angle (degrees)

The maximum surface shear stress is felt at ϕ ≈ 60°.

c) The angle where Thwaites methog predicts a vanishing shear stress is ϕ ≈ 103°.
Unfortunately, this value is not terribly useful as a prediction of the actual separation point (ϕs ≈
80°-85°) because boundary layer separation causes significant changes to the outer flow around
the cylinder that invalidates a portion of this computation. For this reason, reliable prediction of
flows with separation remains a difficult task.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.24. An ideal flow model predicts the following surface velocity for the suction (i.e.
the upper) side of a thin airfoil with chord c placed in a uniform horizontal air stream of speed
Uo :
15
Ue (x) = 2Uo [ x c ] exp (−x c) .
a) Assuming that x is the coordinate along the foil's suction surface, use Thwaites method to
estimate the momentum thickness θ(x) of the laminar boundary layer that develops on this
surface.
b) Using the results of part a) show that the correlation parameter λ is given by:
0.45 " 5x c 5x %" 5x %
λ= 2$
e −1− '$1− '
125 ( x c ) # c &# c&
c) Does Thwaites method predict boundary layer separation in the range 1/5 < x/c < 1?
d) If a laminar boundary layer is predicted to separate from the surface of this airfoil, suggest at
least two changes that could be made to the foil that would tend to prevent separation.
1.4"
1.2"
1"
Ue (x) 0.8"
Uo 0.6"
Uo!
!
!
0.4"
0.2"
!
!
c!
0"
0" 0.2" 0.4" 0.6" 0.8" 1"
x/c!

Solution 10.24. a) The boundary layer is launched from a stagnation point, so the basic Thwaites
formula is:
x
0.45ν x 5 2 0.45ν
θ2 = 6 ∫ U e dx , or in this case: θ = 6 6 6 5 −6 x c ∫
2 5Uo5 ( x" c) e−5 x" c dx" .
Ue 0 2 Uo ( x c) e 0

Simplify, change to an integration variable ξ = x´/c, and use the hint:

x c
0.45ν ce−6 x c # (1+ 5ξ ) &
x c
2 0.45ν ce−6 x c
65 ∫
θ = ξ exp (−5ξ ) d ξ = 65 %
− exp (−5ξ )( .
2Uo ( x c) 0 2Uo ( x c) $ 25 '0
0.45ν ce−6 x c " (1+ 5 x c) −5x c 1 % 0.45ν ce x c " 5x c
Evaluate: θ2 = 65 $
− e + '= #e − (1+ 5x c)%& .
2Uo ( x c) # 25 25 & 50Uo ( x c)6 5
b) The correlation parameter is λ = (θ2/ν)(dUe/dx), and the missing ingredient from part a) is:
dUe (x) "1 −4 5 1 5 1 % −x c
= 2Uo $ [ x c ] − [ x c ] 'e .
dx # 5c c&
θ 2 dUe (x) 0.45ce x c " 5x c &1 −4 5 1 5 1 ) −x c
So, λ= = 65#
e − (1+ 5x c )$% 2Uo ( [ x c ] − [ x c ] + e , or
ν dx 50Uo ( x c ) ' 5c c*
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

0.45 "#e 5x c − (1+ 5x c)$%(1− 5 ( x c)) which is the desired form.

λ= 2
125 ( x c)
c) From part b), λ = 0 at x/c = 1/5, and λ = (0.45/125)(e5 – 6)(–4) = –2.05 at x/c = 1.
d) Yes, the value of λ at x/c = 1 is less than λsep = –0.090. Thus, the Thwaites method predicts
that a laminar boundary layer will separate downstream of x/c = 1/5 and upstream of x/c = 1.
Four possible changes to the foil that might delay or prevent boundary layer separation
are as follows. (i) The surface could be roughened or a boundary layer trip could be added to
stimulate transition to a turbulent boundary layer. (ii) The thickness of the foil could be reduced
so that the pressure gradient predicted by ideal flow for its upper surface is less adverse aft of x/c
= 1/5. (iii) Compressed air could be jetted downstream along the foil's surface to speed up the
slow moving portion of the boundary layer thereby keeping it attached. (iv) Suction could be
applied to the aft portion of the foil to remove the lower (slow moving) portion of the boundary
layer thereby keeping it attached.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.25. An incompressible viscous fluid flows steadily in a large duct with constant
cross sectional area Ao and interior perimeter b. A laminar boundary layer develops on the duct’s
sidewalls. At x = 0, the fluid velocity in the duct is uniform and equal to Uo, and the boundary
layer thickness is zero. Assume the thickness of the duct-wall boundary layer is small compared
to Ao/b.
a) Calculate the duct-wall boundary layer momentum and displacement thicknesses, θ(x) and
δ*(x) respectively, from Thwaites’ method when U(x) = Uo.
b) Using the δ*(x) found for part a), compute a more accurate version of U(x) that includes
boundary layer displacement effects.
c) Using the U(x) found for part b), recompute θ(x) and compare to the results of part a). To
n
simplify your work, linearize all the power-law expressions, i.e. (1− bδ * Ao ) ≅ 1− nbδ * Ao .
d) If the duct area expanded as the flow moved downstream, would the correction for the
presence of the sidewall boundary layers be more likely to move boundary layer separation
upstream or downstream? Explain.
€
Solution 10.25. a) The Thwaites’ results with a constant velocity and zero momentum thickness
at x = 0 are readily obtained:
2 0.45ν x 5 0.45νx
θ = 6
Uo 0
∫ U o dx =
Uo
, or θ = 0.671 νx U o .

Here, λ is zero because dU/dx = 0 so the shape factor is 2.61 which means that:
δ * = 2.61⋅ θ = 1.75 νx U o .
b) At any downstream € displacement effect of the boundary layer will
distance x in the duct, the
€
decrease the effective flow area of the duct. Therefore a simple mass balance between the duct
entrance (x = 0) and the location x requires:
ρU o€Ao = ρU(x)( Ao − bδ * ) , or U(x) = U o (1− bδ * Ao )
−1

c) Use the revised U(x) found in part b) and Thwaites’ method to find:
x x
0.45ν −5 0.45ν
θ2 = −6 ∫ U 5
o (1− bδ *
Ao ) dx ≈
U
(1− 6b δ *
Ao ) ∫ (1+ 5bδ * Ao )dx
U o (1−€bδ Ao ) 0
6 *
€ o 0
x
0.45ν 0.45ν
=
Uo
( *

0
(
1− 6bδ Ao ) ∫ 1+ 5b(1.75 νx U o Ao dx = ) Uo
(1− 6bδ * Ao )( x + 10bδ * x 3Ao )

€ 0.45νx 0.45νx
≈
Uo
(1− 6bδ * Ao + 10bδ * 3Ao ) =
Uo
(1− 8bδ * 3Ao )

€ where linearization has been used to make the integration and algebra easier. Here, we see that
the revised value of θ(x) will be smaller than that estimated in part a).
d) The displacement effect of the boundary layer acts to increase the flow speed. This means
€ that dU/dx will be larger when the boundary layer’s displacement effect is accounted for, thus the
correlation parameter λ will be larger and further from its “separation” value of –0.090.
Therefore, the predicted location of separation will be further downstream in an expanding duct
when the boundary layer displacement effect is accounted for. In the present situation, λ is zero
without accounting for the boundary layer displacement effect but λ becomes positive because
dU/dx is nonzero and positive. Thus, when the boundary layer displacement effect is included:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

dU(x) d −1 d U ob dδ * U ob *
= U o (1− bδ Ao ) ≈ U o (1+ bδ Ao ) =
* *
= δ ,
dx dx dx Ao dx 2Ao x
and
θ 2 (x) dU(x) 0.45x dU(x) 0.45 bδ *
λ≡
ν dx
≈
Uo
(1− 8b δ *
3Ao)
dx
≈
2
(1− 8b δ *
3Ao)
A0
>0 !
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.26. Water flows over a flat plate 30 m long and 17 m wide with a free-stream
velocity of 1 m/s. Verify that the Reynolds number at the end of the plate is larger than the
critical value for transition to turbulence. Using the drag coefficient in Figure 10.12, estimate the
drag on the plate.

Solution 10.26. ReL = UL/ν = (1.0m/s)(30m)/(1.0x106 m2/s) = 3.0x105 > Recr ~ 106. Thus, the
flow is expected to be turbulent over most of the plate. From Figure 10.12, CD ≈ 0.003 so that:
1
D = ρU 2 (Area)CD = 0.5(10 3 )(1.0) 2 (17)(30)(0.003) = 765N .
2

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.27. A common means of assessing boundary layer separation is to observe the
surface streaks left by oil or paint drops that were smeared across a surface by the flow. Such
investigations can be carried out in an elementary manner for cross-flow past a cylinder using a
blow dryer, a cylinder 0.5 to 1 cm in diameter that is ~10 cm long (a common ball-point pen),
and a suitable viscous liquid. Here, creamy salad dressing, shampoo, dish washing liquid, or
molasses should work. And, for the best observations, the liquid should not be clear and the
cylinder & liquid should be different colors. Dip your finger into the viscous liquid and wipe it
over two thirds of the surface of the cylinder. The liquid layer should be thick enough so that you
can easily tell where it is thick or thin. Use the dry one third of the cylinder to hold the cylinder
horizontal. Now, turn on the blow dryer leaving the heat off and direct its outflow across the
wetted portion of the horizontal cylinder to mimic the flow situation in the drawing for Exercise
10.23.
a) Hold the cylinder stationary, and observe how the viscous fluid moves on the surface of the
cylinder and try to determine the angle ϕs at which boundary layer separation occurs. To get
good consistent results you may have to experiment with different liquids, different initial liquid
thicknesses, different blow-dryer fan settings, and different distances between cylinder and blow
dryer. Estimate the cylinder-diameter-based Reynolds number of the flow you’ve studied.
b) If you have completed Exercise 10.23, do your boundary layer separation observations match
the calculations? Explain any discrepancies between your experiments and the calculations.

Solution 10.27. a) Using a dark shampoo on a white ball-point pen, the 3rd author of this
textbook found that the separation point occurred on the upstream side of the cylinder near ϕ =
90°. The cylinder diameter was 8 mm and flow speed was probably about 10 m/s. Thus, an
estimate of the Reynolds number is:
UD (10m / s)(0.008m)
ReD = ≈ ≈ 5,000
ν 1.5 ×10−5 m 2 / s
b) The blow-drier-and-ball-point-pen separation point results, ϕs near to but less than 90°, match
commonly quoted experimental values for the separation angle (ϕ ≈ 80°-85°). However, they do
not match the Thwaites-method-calculated
€ location of zero shear stress (ϕ ≈ 103°). The primary
difference between the experiments and the Thwaites calculation is the separated flow in the
wake of the cylinder. The experiment includes the separated flow while the potential flow
surface velocity that is input to the Thwaites calculation does not include the separated flow. In
this case the cylinder’s wake makes important changes to the surface flow on the cylinder so the
experimental and calculated boundary layers develop under different flow fields and therefore
separate at different points. In general, Thwaites method is only successful in predicting whether
or not boundary layer separation will occur. Once boundary layer separation has occurred, a
theory that accounts for flow in the separation zone is needed.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.28. Find the diameter of a parachute required to provide a fall velocity no larger
than that caused by jumping from a 2.5 m height, if the total load is 80 kg. Assume that the
properties of air are ρ = 1.167 kg/m3, ν = 1.5 × 10-5 m2/s, and treat the parachute as a
hemispherical shell with CD = 2.3. [Answer: 3.9 m]

Solution 10.28. The fall velocity from a 2.5 m height is [2gh]1/2 = [2(9.81)(2.5)]1/2 = 7.0 m/s.
At steady state, the drag on the parachute equals the load, so that D = mg = (80)(9.81).
Thus, using CD = 2.3, the parachute's cross sectional area should be
D 80(9.81)
Area = 1 2
= = 11.93m 2 , or a diameter = [4(11.93)/π]1/2 = 3.90 m.
CD 2 ρU 2.3(0.5)(1.167)(7) 2

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.29. The boundary layer approximation is sometimes applied to flows that do not
have a bounding surface. Here the approximation is based on two conditions: downstream fluid
motion dominates over the cross-stream flow, and any moving layer thickness defined in the
transverse direction evolves slowly in the downstream direction. Consider a laminar jet of
momentum flux J that emerges from a small orifice into a large pool of stationary viscous fluid
at z = 0. Assume the jet is directed along the positive z-axis in a cylindrical coordinate system.
In this case, the steady, incompressible, axisymmetric boundary layer equations are:
1 ∂ (RuR ) ∂w ∂w ∂w 1 ∂p ν ∂ & ∂w )
+ = 0 , and w + uR =− + ( R +,
R ∂R ∂z ∂z ∂R ρ ∂z R ∂ R ' ∂R *
where w is the (axial) z-direction velocity component, and R is the radial coordinate. Let r(z)
denote the generic radius of the cone of jet flow.
a) Let w(R,z) = (ν z) f (η) where η = R z , and derive the following equation for f:
€ η €
ηf # + f ∫ ηfdη = 0 .
η

€
b) Solve this equation by defining a new function F =
€
∫ ηfdη . Determine constants from the
R = r(z )

€
boundary condition: w → 0 as η → ∞, and the requirement: J = 2πρ ∫ w 2 (R,z)RdR = constant.
R= 0
c) At fixed z, does r(z) increase or decrease € with increasing J?
[Hints: i) the fact that the jet emerges into a pool of quiescent fluid should provide information
about ∂p/∂z, and ii) f (η) ∝ (1+ const ⋅ η 2 )−2 , but try to obtain this result without using it.]
€
Solution 10.29. a) Start with the given boundary layer equations and use w(R,z) = (ν z) f (η)
where η =€R z , ∂ ∂R = (1 z) ∂ ∂η and ∂ ∂z = (−η z) ∂ ∂η in the continuity equation to find:
1 R ∂w 1 R ( νη ν + ν η
uR = − ∫ R dR = − ∫ R* − 2 f ' − 2 f - dR = ∫ η 2 f ' + ηf dη
R 0 ∂z R0 ) z z , €R 0
( )
€ Now start
€ assembling terms of
€ the momentum equation:
∂w ν ' ν νη * ν2 2 ∂w ' ν η 2 *ν
w (
= f ⋅ )− 2 f − 2 f , = − 3 f + ηff , uR ) - ( )
= ) ∫ η f % + ηf dη, 2 f % , and
€ ∂z z ( z z + z ∂R (ηz 0 +z
2
ν ∂ $ ∂w ' ν ∂ $ ν ' ν $ ν ν ' ν
&R ) = & R 2 f *) = & 2 f * + R 3 f **) = 3 ( f * + ηf **) .
R ∂ R % ∂ R ( R ∂ R % z ( ηz % z z ( ηz
€ If the fluid far from the jet is still, then ∂p/∂z€ = 0 in the boundary layer approximation because
∂p/∂R = 0 too. Reassemble the momentum equation:
ν2 'ν η *ν ν2
€ − 3 ( f 2 + ηff %) + ) ∫ (η 2 f % + ηf ) dη, 2 f % = 3 ( f % + ηf %%)
z (ηz 0 +z ηz
Divide out the common factor of ν 2 z 3 :
&1 η 2 ) 1
−( f + ηff $) + ( ∫ (η f $ + ηf ) dη+ f $ = ( f $ + ηf $$)
2

€ 'η 0 * η
This is a single equation€for f without z or R appearing; thus the assumption of a similarity
solution is successful. Multiply both sides by η, expand terms and integrate by parts.
& 2 η ) η
€ 2 2
−ηf − η ff $ + f $(η f − 2 ∫ ηfdη+ + f $ ∫ ηfdη = f $ + ηf $$
' 0 * 0

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

η
2
Combine common terms to find: −ηf − f $ ∫ ηfdη = f $ + ηf $$. This compact form can be further
0

d $ '
η η
d
simplified by noting that: & f ∫ ηfdη) = f * ∫ ηfdη + ηf 2 , and (ηf #) = ηf ## + f # . Thus, one
dη % 0 ( 0 dη
€ η
integration with respect to η yields: ηf # + f ∫ ηfdη = 0 . Here, the constant must be zero based
0
on an evaluation€of the equation at η = 0. €
η
∫
ηfdη , then F " = ηf , and F "" = ηf " + f , so the above equation becomes:
b) If F =
1 1 €
F "" − F " + F "F = 0 , or η 2 F ## + η(F −1) F # = 0 . This last equation is equidimensional so the
η η
2
€ € t € d d 2 d d2 d
substitution η = e will lead to simplification. Here, η = , and η = − , so the
dη dt dη 2 dt 2 dt
€ 2
€ d F dF
equation becomes: 2
+ (F − 2) = 0 . This equation is autonomous; thus we can let
dt dt
€ d 2 F dy dF
dF € the last remnant € of the momentum equation
= y(F) so that = = yy " , and
dt dt 2 dF dt
becomes:€yy " − (F − 2)y = 0 . Therefore when y ≠ 0, y " = −F + 2 , or y = A + 2F − F 2 /2 . Now go
back to the definition of F to determine the constant A:
2
dF dF 1 η
€
€
€ y(F) =
dt
=η
dη
= A + 2F − F 2 /2 = A + 2 ∫ ηfdη +
€ €
η
[
2
∫ ]
ηfdη .
2
When η = 0, then F = 0, so A = 0. Thus, y(F) = 2F − F 2 = η dF dη . Separate and integrate
dη dF dF dF
using a partial fraction decomposition: = 2 =− + to reach:
€ 2η F − 4F 4(F − 4) 4F
1 $ F ' € 12 $ F '1 4 4C 4η 2
ln η1 2 + B = ln& ) . Now exponentiate: C η = & ) and solve for F: F = .
4 %4 −F( %4 − F( (1+ C 4η 2 )
1€ 8C 4
Recall that F " = ηf , so f = F # = 4
4 2 2 . The constant C can be evaluated from:
η (1+ C η )
R = r(z) 2
€ J R = r(z)
2 €ν 2 2
r(z )/ z
2 2€
r(z)/ z
64C 8η
= ∫ w (R,z)RdR = ∫ 2 f (η)RdR = ν ∫ ηf dη = ν ∫ 4 2 4
dη
2πρ R = 0 R= 0 z 0 0 (1+ C η )
€ integral may be performed after a change of variable γ = 1+ C 4η 2 , and by noting that the
The €
upper limit corresponds to η → ∞. Thus:
∞ 4
€ J = 32C 4ν 2 ∫ γ −4 dγ = 32C ν 2 , so C 4 = 3J 2 = 3 Re 2 , where Re = J ρ
2πρ 0 3 64 πν€ ρ 64 π ν
2
Therefore, the final solution is: w(R,z) =
ν ( 3 8π )Re
z (1+ ( 3 64 π ) Re 2 (R /z) 2 ) 2
€ c) From the result of part b),€r(z) will decrease with increasing € J (increasing Re). The laminar jet
narrows as its Reynolds number increases.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.30. A simple realization of a temporal boundary layer involves the spinning fluid in
a cylindrical container. Consider a viscous incompressible fluid (density = ρ, viscosity = µ) in
solid body rotation (rotational speed = Ω) in a cylindrical container of diameter d. The mean
depth of the fluid is h. An external stirring mechanism forces the fluid to maintain solid body
rotation. At t = 0, the external stirring ceases. Denote the time for the fluid to spin-down (e.g. to
stop rotating) by τ.
a) Case I: h >> d. Write a simple laminar-flow scaling law for τ assuming that the velocity
perturbation produced by the no-slip condition on the container’s sidewall must travel inward a
distance d/2 via diffusion.
b) Case II: h << d. Write a simple laminar-flow scaling law for τ assuming that the velocity
perturbation produced by the no-slip condition on the container's bottom must travel upward a
distance h via diffusion.
c) Using partially-filled cylindrical containers of several different sizes (drinking glasses and
pots & pans are suggested) with different amounts of water, test the validity of the above
diffusion estimates. Use a spoon or a whirling motion of the container to bring the water into
something approaching solid body rotation. You'll know when you're close to solid body rotation
because the fluid surface will be a paraboloid of revolution. Once you have this initial flow
condition set-up, cease the stirring or whirling and note how long it takes for the fluid to stop
moving. Perform at least one test when d & h are several inches or more. Cookie or bread
crumbs sprinkled on the water surface will help visualize surface motion. The judicious addition
of a few drops of milk after the fluid starts slowing down may prove interesting.
d) Compute numbers from your scaling laws for parts a) and
b) using the viscosity of water, the dimensions of the
containers, and the experimental water depths. Are the scaling
laws from parts a) and b) useful for predicting the
experimental results? If not, explain why.
(The phenomena investigated here have some important
practical consequences in atmospheric and oceanic flows and
in IC engines where swirl and tumble are exploited to mix the
fuel charge and increase combustion speeds.)

Solution 10.30. For all simple unsteady diffusion problems, the length scale of “diffusion-
penetration” is proportional to the square root of the product of the diffusion constant and time.
For momentum diffusion in fluid flows, ν is the diffusion constant. In the following, let τ be the
time it takes for the swirling fluid to come to rest.
a) If h >> d, it should be possible to ignore the bottom of the container. Therefore, the "stopping"
signal must penetrate a distance of d/2 inward from the walls of the container. So, for an
approximate unsteady diffusion analysis, we have: d/2 ≈ ντ , or τ ≈ d2/4ν .
b) If d >> h, it should be possible to ignore the walls of the container. Therefore, the "stopping"
signal from the no-slip boundary condition on the bottom of the container must penetrate upward
a distance of h from the bottom of the container. So, for an approximate unsteady diffusion
€
analysis, we have: h ≈ ντ , or τ ≈ h2/ν
c) For all standard-kitchen-size containers with water depths of a several cm, the experimental
stopping time was less than a minute.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

d) Using the formulae from part a) or b), a length scale of 5 cm (~two inches), and the kinematic
(0.05m) 2
viscosity of water: τ ≈ = 2500 seconds ≈ 40 minutes!
1.0 ×10−6 m 2 /sec
The expectation here is that the estimated and actual times would match within a factor of two or
so. Clearly something is wrong with the simple diffusion analysis for these simple swirling
flows. The milk drops in the swirling water show that the flow is very three-dimensional and
€ of the flow with the bottom and sides of the pan or cup is very important. In
that the interaction
particular, the viscous boundary layer on the bottom of the container causes inward radial flow
along the bottom and the fluid stops rotating because of a phenomenon called Ekman pumping.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 10.31. Mississippi River boatmen know that when rounding a bend in the river, they
must stay close to the outer bank or else they will run aground. Explain in fluid mechanical terms
the reason for the cross-sectional shape of the river at the bend.

Solution 10.31. The Reynolds number based on mean channel width or radius of curvature is
very large. Thus, the riverbed boundary layer is thin compared with the width or depth of the
river, and most of the flow is inviscid. The primary turning flow around the bend is governed by
∂p ∂r = ρvθ2 r , but v decreases rapidly to zero in the thin boundary layer on the riverbed. Thus,
θ

∂p ∂r in the boundary layer, which is largely the same as that in the inviscid flow, is not
balanced by ρvθ2 r in the boundary layer since [v ]boundary layer < [v ]inviscid flow. Here, ∂p ∂r is too
θ θ

large in the boundary layer flow for it to remain parallel to the flow above it. The excess
€
pressure gradient pushes the boundary layer flow toward the lower pressure which occurs at the
€
riverbank on the inside of the river bend. This induces cross-flow in the boundary layer that
€ boundary layer fluid toward the inner riverbank, and causes a downward
moves € secondary flow
at the outer riverbank. Together these effects set up the secondary flow pattern shown below. In
the Mississippi River, the water transports silt which is scoured from the outer bank and
deposited on the inner bank. This transport accounts for the cross section of the river channel.
This secondary flow phenomenon also explains why small river meanderings tend to grow over
time as the river erodes its outer bank and deposits silt on its inner bank. The "goose necks" of
the San Juan River is southeastern Utah provide an extreme (and impressive) example of river
meandering.

r
!"#!$

boundary
layer
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.1. A perturbed vortex sheet nominally located at y = 0 separates flows of differing
density in the presence of gravity with downward acceleration g. The upper stream is semi-
infinite and has density ρ1 and horizontal velocity U1. The lower stream has thickness h density
ρ2, and horizontal velocity U2. A smooth flat impenetrable surface located at y = –h lies below
the second layer. The interfacial tension between the two fluids is σ. Assume a disturbance
occurs on the vortex sheet with wave number k = 2π/λ, and complex wave speed c, i.e.
[ y ] sheet = f (x,t) = f o Re{e ik(x−ct )}. The four boundary conditions are:
1) u1, v1 → 0 as y → +∞
2) v2 = 0 on y = –h.
3) u1 ⋅ n = u2 ⋅ n = normal velocity of the vortex sheet on both sides of the vortex sheet.
€ ∂2 f
4) p1 − p2 = σ 2 on the vortex sheet (σ = interfacial surface tension)
∂x
a) Following the development in Section 11.3, show that:
€ € 12
ρ1U1 + ρ 2U 2 coth(kh) % (g /k)( ρ 2 − ρ1 ) + σk ρ1ρ 2 (U1 − U 2 ) coth(kh) (
2

c= ±' − 2
* .
€ ρ1 + ρ 2 coth(kh) '& ρ1 + ρ 2 coth(kh) (ρ1 + ρ2 coth(kh)) *)
b) Use the result of part a) to show that the vortex sheet is unstable when:
# ρ 2 &# g ( ρ 2 − ρ1 ) σk & 2
%tanh(kh) + (% + ( < (U1 − U 2 )
€ $ ρ1 '$ k ρ2 ρ2 '
c) Will the sheet be stable or unstable to long wavelength disturbances ( k → 0 ) when ρ2 > ρ1 for
a fixed velocity difference?
d) Will the sheet be stable or unstable to short wavelength disturbances ( k → ∞ ) for a fixed
€
velocity difference?
€
e) Will the sheet ever be unstable when U1 = U2?
f) Under what conditions will the thickness h matter?
€
Solution 11.1. Set up the problem with U1, U2 mean horizontal velocities and u1, u2, v1, and v2,
perturbation velocities due to the small disturbances on the vortex sheet. Above the vortex sheet:
∇ 2φ1 = 0 and below the vortex sheet: ∇ 2φ 2 = 0 . By inspection or separation of variables:
φ1 = ( Ae +ky + Be−ky )e ik(x−ct ) + U1 x (above the vortex sheet)
φ 2 = (Ce +ky + De−ky )e ik(x−ct ) + U 2 x (below the vortex sheet)
€ where the constants must be€evaluated using the boundary conditions (BCs), and c may be
€ complex.
BC 1) requires A = 0.
€ BC 2) requires Cke−kh + Dke +kh = 0 , or that
Ce +ky + De−ky = 2Ce−kh cosh[ k(y + h)]
Thus, φ1 = Be ik(x−ct )−ky + U1 x , and
φ 2 = 2Ce−kh e ik(x−ct ) cosh[ k(y + h)] + U 2 x .
BC 3)€may be linearized since fo is€small (this amounts to ignoring the perturbation velocities
compared to U1 and U2 and requiring the matching to take place on y = 0 instead of y = f(x,t). The
€
linearized version of boundary condition 3) can be evaluated using:
€ ∂f ∂f ∂f ∂f
v1 = U1 + , and v 2 = U 2 + on y = 0.
∂x ∂t ∂x ∂t
Plug in the appropriate derivatives for v1, and v2 and evaluate:

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

−Bk = [U1 (ik) − ikc ] f o , and 2kCe−kh sinh[ kh ] = [U 2 (ik) − ikc ] f o .

The common factor of e ik(x−ct ) has been divided out of all four terms. Solve for A and B:
B = −i[U1 − c ] f o , and C = i[U 2 − c ] f o 2e−kh sinh[ kh ] .
BC 4) €
requires use of the Bernoulli equation
€ above and below the vortex sheet:
∂φ ρ 2 ∂φ ρ 2
ρ€1 1 + 1 u1 + ρ1gy + p1 = Po1 , and ρ 2 2 + 2 u2 + ρ 2 gy + p2 = Po2
€
∂t 2 € ∂t 2
2 2
Linearize these using u j ≅ U j + 2U j u j (j = 1,2 and no summation implied), evaluate the gravity
terms on y = f(x,t) (the other terms are eventually evaluated on y = 0), and set
€ ∂2 f
p1 = p2 +€σ 2 = p2 − k 2σf (x,t)
€ ∂x
∂φ1 ρ1 2 ∂φ ρ
Po1 − ρ1 − [U1 + 2U1u1 ] − ρ1gf (x,t) = Po2 − ρ 2 2 − 2 [U 22 + 2U 2 u2 ] − ρ 2 gf (x,t) − k 2σf (x,t)
∂t 2 ∂t 2
When there is no disturbance on the vortex sheet, the flow is steady and u2 = u1 = 0. Therefore:
€
ρ ρ
Po1 − 1 U12 = Po2 − 2 U 22
2 2
€ Subtract this from the full linearized pressure boundary condition and multiply by –1 to get:
∂φ1 ∂φ
+ ρ1U1u1 + ρ1gf (x,t) = ρ 2 2 + ρ 2U 2 u2 + ρ 2 gf (x,t) + k 2σf (x,t)
ρ1
∂t ∂t
€ for f(x,t) and the potentials, evaluate on y = 0 as appropriate, and
Now plug in the known forms
cancel common terms:
ρ1[B(−ikc + ikU1 ) + gf o ] = ρ 2 [2Ce−kh cosh(kh)(−ikc + ikU 2 ) + gf o ] + k 2σf o
Insert€B and C from above, and divide by fo:
ρ1[k(U1 − c) 2 + g] = ρ 2 [−k coth(kh)(U 2 − c) 2 + g] + k 2σ .
This is a quadratic for c. Divide by k and put the equation in a more standard form:
€
(ρ1 + ρ2 coth(kh))c 2 − 2( ρ1U1 + ρ2U 2 coth(kh))c − σk + ρ1U12 + ρ2U 22 coth(kh) + ( g k )( ρ1 − ρ2 ) = 0 .
Solving for c using the quadratic formula eventually produces:
€ 12
ρ1U1 + ρ 2U 2 coth(kh) %( g k )( ρ 2 − ρ1 ) + σk ρ1ρ 2 (U1 − U 2 ) coth(kh) (
2

c = c r + ic i = ±' − 2
*
€ ρ1 + ρ 2 coth(kh) '& ρ1 + ρ 2 coth(kh) (ρ1 + ρ2 coth(kh)) *)
b) Here we see that ci = 0 (i.e. the right side has no imaginary part) unless the terms inside the
square root together are negative. This means that the vortex sheet is unstable when the terms
inside the [,]-brackets together are negative:
€ 2
(g k )(ρ2 − ρ1 ) + σk − ρ1ρ2 (U1 − U 2 ) coth(kh) < 0, or
2
ρ1 + ρ 2 coth(kh) ( ρ1 + ρ2 coth(kh))
# ρ 2 &# g ( ρ 2 − ρ1) σk & 2
%tanh(kh) + (% + ( < (U1 − U 2 )
$ ρ1 '$ k ρ 2 ρ2 '
c) When ρ2 >€ρ1 and k → 0 , the factor with g in it will be positive and unbounded. Thus, the
inequality will be not be satisfied and the sheet will be neutrally stable.
d) When k → ∞ , the factor with σ in it will be positive and unbounded. Thus, the inequality will
€ and the sheet will be neutrally stable.
be not be satisfied
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

12
%( g k )( ρ 2 − ρ1 ) + σk (
e) When U1 = U2 = U, c = U ± ' * , which specifies neutrally stable flow when
& ρ1 + ρ 2 coth(kh) )
(g k )(ρ2 − ρ1 ) + σk > 0 , and unstable flow when (g k )(ρ2 − ρ1 ) + σk < 0 . Thus, even when the
fluid densities are not stably arranged, non-zero surface tension prevents high wave number
fluctuations from being unstable. In addition, note that when U1 = U2 = ρ1 = 0, we recover the
€ 12
€ dispersion relation for water waves on € [ ]
pool of finite depth: ω = ck = ( gk + σk 3 ρ 2 ) tanh(kh) .
f) Given that tanh(kh) →1 for kh >> 1, the thickness h will only matter when kh << 1 and ρ 2 ρ1
is small enough so that 1 + ρ 2 ρ1 is noticeably different from ρ 2 ρ1 . When considered in terms
of the original answer given for part a), ci ~ [kh]1/2 for kh << 1 (since coth(kh) ~ 1/kh for kh << 1),
€ vortex sheet to have a smaller growth rate,
decreasing h causes a sinusoidal perturbation of the
€ €
thereby making it less unstable.
€
The motivation for this problem is the stability € of a layer of air sandwiched between a
large reservoir of water (below) and a hard surface (above); such as when a layer of air is
injected underneath a flat-bottom ship to reduce skin friction and/or change the ship hull's
mechanical coupling with the water.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.2. Consider a fluid layer of depth h and density ρ2 lying under a lighter infinitely-
deep fluid of density ρ1 < ρ2. By setting U1 = U2 = 0, in the results of the Exercise 11.1, the
following formula for the phase speed is found:
12
% (g /k)( ρ 2 − ρ1 ) + σk (
c = ±' *
& ρ1 + ρ 2 coth(kh) )
Now invert the sign of gravity and consider why drops form when a liquid is splashed on the
underside of a flat surface. Are long or short waves more unstable? Does a professional painter
want interior ceiling paint with high or low surface tension? For a smooth finish should the
€ coats of paint? Assuming the liquid has the properties of water
painter apply thin or thick
(surface tension ≈ 72 mN/m, density ≈ 103 kg/m) and that the lighter fluid is air, what is the
longest neutrally stable wavelength on the underside of a horizontal surface? [This is the
Rayleigh–Taylor instability and it occurs when density and pressure gradients point in opposite
directions. It may be readily observed by accelerating rapidly downward an upward-open cup of
water.]

Solution 11.2. Changing the sign of gravity in the given formula produces:
12
% −(g /k)( ρ 2 − ρ1 ) + σk (
c = ±' * ,
& ρ1 + ρ 2 coth(kh) )
where the symbol "g" is presumed to specify a positive acceleration magnitude. Here, the heavy
liquid with density ρ2 is accelerated into the lighter liquid with density ρ1, so ρ2 – ρ1 > 0. Thus,
the first term in the numerator of the fraction under the square root is negative and while the
second is positive. For€the wave speed to have a non-zero imaginary component and thereby
indicate instability, the first term, which is negative, must dominate. When k is small (long
wavelengths) the first term must dominate, when k is large the second term must dominate. The
wave number kb (presumed positive) that lies at the boundary between instability and neutral
stability is given by: −(g /kb )( ρ 2 − ρ1 ) + σk b = 0 , or kb = g( ρ 2 − ρ1 ) σ , where k < kb implies
instability, and k ≥ kb implies neutral stability. Thus, long wavelengths (small k) are most
unstable.
When the surface tension σ is higher, then kb decreases, and the range of k leading to
€ €
instability is smaller (higher σ increases the realm of stability). Thus, a professional painter
wants interior ceiling paint with a high surface tension so that the paint is less likely to
spontaneously produce ripples if it is applied in a smooth layer. Incidentally, a professional
painter also wants a paintbrush with many fine bristles so that only high-wave-number residual
disturbances are left by brush strokes as the paint is applied.
For small kh, coth(kh) ~ 1/kh, while for large kh, coth(kh) ~ 1. Thus,
12 12
% −(g /k)( ρ 2 − ρ1 ) + σk ( % −(g /k)( ρ 2 − ρ1 ) + σk (
c = ± kh ' * for kh << 1, and c = ±' * for kh >> 1.
& ρ2 ) & ρ1 + ρ 2 )
In a painting scenario, ρ2 corresponds to the paint and ρ1 corresponds to the air, so ρ2 >> ρ1.
Thus, the small-kh formula for c is smaller than the large-kh formula for c, by a factor of [kh]1/2.
This means that instability in a layer of paint on a ceiling develops more slowly when the paint
€ layer is thin. Thus, a smooth finish is more € likely to be obtained via multiple thin coats than one
thick one.
For the given parameters, the longest neutrally stable wavelength will be:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

12 12
2π ' σ * ' 0.072 *
λ = 2π k b = = 2π ) , = 2π ) , ≈ 1.7cm .
g( ρ 2 − ρ1 ) σ ( g( ρ 2 − ρ1 ) + (9.81(10 3 ) +
Of course, considerations of paint viscosity are also important for achieving a high-
quality paint finish. However, it is potentially remarkable how many physical features of paint
application can be deduced from this inviscid analysis.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.3. Inviscid horizontal flow in the half space y > 0 moves at speed U over a porous
surface located at y = 0. Here the fluid density ρ is constant and gravity plays no roll. A weak
vertical velocity fluctuation occurs at the porous surface: [v ] surface = v o Re{e ik(x−ct )} , where vo <<
U.
a) The velocity potential for the flow may be written φ˜ = Ux + φ , where φ leads to [v]surface at y = 0
and φ vanishes as y → +∞. Determine the perturbation
€ potential φ in terms of vo, U, ρ, k, c, and
the independent variables (x,y,t).
b) The porous surface responds to pressure fluctuations in the fluid via: [ p − ps ] y= 0 = −γ [v ] surface ,
€
where p is the pressure in the fluid, ps is the steady static pressure that is felt on the surface when
the vertical velocity fluctuations are absent, and γ is a real material parameter that defines the
porous surface’s flow resistance. Determine a formula for c in terms of U, γ, ρ, and k.
c) What is the propagation velocity, Re{c}, of the surface velocity € fluctuation?
d) What sign should γ have for the flow to be stable? Interpret your answer.

Solution 11.3. a) To match the porous wall boundary condition, the potential must be of the
following form: φ˜ = Ux + φ = Ux + F(y)e ik(x−ct ) . Setting ∇ 2φ = 0 and canceling common factors
produces: −k 2 F + F ## = 0 , so that F = Ae +ky + Be−ky . Here, A must be zero for |φ| = F(y) to vanish
as y → +∞. The second boundary condition implies: (∂φ ∂y ) y= 0 = [v ] surface = v oe ik(x−ct ) , or
−kA = v€o so that φ = −(v o k )e ik(x−ct )−ky . €
€ € ∂φ ρ 2
b) Without gravity, the Bernoulli equation for this flow is: ρ + u + p = Po . Linearize this
€ ∂t 2
∂φ ρ % ∂φ (
€ € with the perturbation potential to get: ρ + 'U 2 + 2U * + p = Po . For flow without
equation
∂t 2 & ∂x )
ρ 2
the perturbation this linearized equation is: U €
+ ps = Po . Subtract these two Bernoulli
2
& ∂φ ∂φ )
equations to find: p − ps = −ρ( + U€ + , thus the linearized dynamic boundary condition is:
' ∂t ∂x *
& ∂φ ∂φ )
−ρ( € + U + = −γ [v ] surface = −γv oe ik(x−ct ) .
' ∂t ∂x * y= 0
Evaluate€ the left side, multiply by –1, and cancel the common exponential factors:
γ
ρ(−ikc + ikU )(−ν o k ) = γv o , solve for: c = c r + ic i = U − i .
ρ
€
c) The phase velocity is U.
d) For the flow to be stable, ci should be negative so γ must be positive. This makes sense too;
€ with γ positive, the exterior flow is allowed to penetrate (or expand) into the porous surface when
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

the local surface pressure is high and this cushioning effect is stabilizing. If γ was negative, then
a local surface pressure maxima would induce an outward vertical flow from the porous surface.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.4. Repeat exercise 11.3 for a compliant surface nominally lying at y = 0 that is
perturbed from equilibrium by a small surface wave: [ y ]surface = ζ (x, t) = ζ o Re {eik ( x−ct ) } .
a) Determine the perturbation potential φ in terms of U, ρ, k, and c by assuming that φ vanishes
as y → +∞, and that there is no flow through the compliant surface. Ignore gravity.
b) The compliant surface responds to pressure fluctuations in the fluid via: [ p − ps ] y=0 = −γζ (x, t) ,
where p is the pressure in the fluid, ps is the steady pressure that is felt on the surface when the
surface wave is absent, and γ is a real material parameter that defines the surface’s compliance.
Determine a formula for c in terms of U, γ, ρ, and k.
c) What is the propagation velocity, Re{c}, of the surface waves?
d) If γ is positive, is the flow stable? Interpret your answer.

Solution 11.4. a) Start with φ˜ = Ux + φ . The x- and t- dependence in the problem will have to
match that of the surface wave, therefore we must have φ˜ = Ux + F(y)e ik(x−ct ), where F(y) must be
determined. Plugging this form for φ into ∇ 2φ = 0 , produces −k 2 F + F ## = 0 after common
factors have been €cancelled. This equation has solutions of growing and decaying exponentials
as y increases. The growing exponential is discarded because it is not meaningful as y → +∞.
€
Thus: φ = Ae−kye ik(x−ct ) where A is a constant that can€be determined from the linearized kinematic
€
boundary condition on the surface:
∂ζ ∂ζ % ∂φ (
U + = ' * → ikUζ o + (−ikc )ζ o = −kA , or A = i(c − U )ζ o
€ ∂ x ∂ t & ∂y ) y= 0
−ky ik(x−ct )
Thus: φ = i(c − U )ζ oe e
b) Without gravity, the Bernoulli equation for this flow is:
€ ∂φ ρ 2 €
€ ρ + u + p = Po .
€ ∂t 2
Linearize this equation with the perturbation potential to get:
∂φ ρ % ∂φ (
ρ + 'U 2 + 2U * + p = Po .
∂t 2 & ∂x )
€
ρ
For flow without the surface perturbation this perturbation Bernoulli equation is: U 2 + ps = Po .
2
Subtract these two Bernoulli equations to find:
€
& ∂φ ∂φ )
p − ps = −ρ( + U + .
' ∂t ∂x *
€
Thus, using the surface's constitutive relationship, the linearized dynamic boundary condition is:
& ∂φ ∂φ )
−ρ( + U + = −γζ (x,t) .
' ∂t ∂x * y= 0
€
Substitute in the result of part a) for φ:

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

−ρi(c − U )(−ikc + ikU )ζ o = −γζ o .

ik(x−ct )
where the common factor of e has been divided out. Reduce this equation to a quadratic
form and solve for c.
2 γ γ
€ (c − U ) = or c = U ± .
€ kρ kρ
γ
c) When γ is positive, the disturbance phase velocity can have two values: c r = U ±
kρ
When γ is negative, the€phase velocity = € U.
d) At this point all that can be said is that the flow is neutrally stable when γ is positive. When γ
is positive, a suction pressure on the surface lifts it up, while an elevated pressure pushes it
€
down. This behavior matches that of gravity. In fact, replacing γ in the surface's constitutive
equation (specified in the statement of part b) by ρg produces:
p − ps = −ρgζ ,
which describes a hydrostatic balance, and
g
cr = U ± ,
€ k
which are the possible phase speeds of surface waves in deep water that is moving at speed U.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.5. As a simplified version of flag waving, consider the stability of a simple
membrane in a uniform flow. Here, the undisturbed membrane lies in the x-z plane at y = 0, the
flow is parallel to the x-axis at speed U, and the fluid has density ρ. The membrane has mass per
unit area = ρm and uniform tension per unit length = T. The membrane satisfies a dynamic
equation based on pressure forces and internal tension combined with its local surface curvature:
∂ 2ζ & ∂ 2ζ ∂ 2ζ )
ρ m 2 = p2 − p1 + T( 2 + 2 + .
∂t ' ∂x ∂z *
Here, the vertical membrane displacement is given by y = ζ (x,z,t) , and p1 and p2 are the
pressures acting on the membrane from above and below, respectively. The velocity potentials
for the undisturbed flow above (1) and below (2) the membrane are φ1 = φ 2 = Ux . For the
€
following items, assume a small amplitude wave is present on the membrane
€
ζ (x,t) = ζ o Re{e ik(x−ct )} with k a real parameter, and assume that all deflections and other
fluctuations are uniform in the z-direction and small enough for€the usual linear simplifications.
In addition, assume the static pressures above and below the membrane, in the absence of
membrane motion, are matched.
€ a) Using the membrane equation, determine the propagation speed of the membrane waves,
Re{c}, in the absence of fluid loading (i.e. when p2 – p1 = ρ = 0).
b) Assuming inviscid flow above and below the membrane, determine a formula for c in terms of
T, ρm, ρ, U, and k.
c) Is the membrane more or less unstable if U, T, ρ, and ρm are individually increased with the
others held constant?
d) What is the propagation speed of the membrane waves when U = 0? Compare this to your
answer for part a) and explain any differences.

ik(x−ct ∂ 2ζ & ∂ 2ζ ∂ 2ζ )
Solution 11.5. a) Plug ζ (x,t) = ζ o Re{e } into ρ m = p2 − p1 + T ( 2 + 2 + with p2 – p1
∂t 2 ' ∂x ∂z *
= ρ = 0) to find: −(kc) 2 ρ mζ = T (−k 2ζ + 0) , or c = ± T ρ m .
b) For this geometry, a separation-of-variables solution will produce the following velocity
€ and below the membrane:
potentials above €
φ1 (x, y,t) = Ae ik(x−ct )−ky + Ux , and φ1 (x, y,t) = Be ik(x−ct )+ky + Ux
€ €
where the boundary conditions at y → ±∞ have already been applied to eliminate the growing
exponential terms in each direction. The linearized kinematic condition on the membrane
∂ζ ∂ζ
€U
requires: + = v(x, y = 0,t) . Apply€this requirement both above and below the
∂x ∂t
membrane,

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

ikUζ o − ikcζ o = −kA , and ikUζ o − ikcζ o = +kB ,

to find: A = −B = i(c − U )ζ o .
The pressure fluctuation on the membrane is found by linearizing the Bernoulli equation and
subtracting out the part that persists when there is no disturbance on the membrane. This leaves:
€ €
' ∂φ1 * ' ∂φ *
€ p1" = p1 − ps = −ρ) + Uu1", , and p"2 = p2 − ps = −ρ) 2 + Uu"2 , ,
( ∂t + ( ∂t +
where ps is the steady pressure on both sides of the membrane when there is no disturbance and
u1", u"2 are the horizontal disturbance velocities above and below the membrane. Thus:
' ∂φ ∂φ *
€ p"2 − p1" = p2 − p€1 = ρ) 1 − 2 + U( u1" − u"2 ),
( ∂t ∂t +
€€ Plug all this into the membrane equation and start working out the algebra to reach a quadratic
equation:
∂ 2ζ ( ∂φ1 ∂φ 2 + ( ∂ 2ζ +
€ ρ m 2 = ρ* − + U( u1' − u2' )- + T* 2 -
∂t ) ∂t ∂t , y= 0 ) ∂x ,
−(kc) ρ mζ o = ρ(2i(−ikc)(c − U )ζ o + 2ikUi(c − U )ζ o ) − k 2Tζ o
2

2
−(kc) 2 ρ m + k 2T = 2 ρk (c − U )
€ 2ρ 2 T
c2 + (c − U ) − = 0
€ kρ m ρm
€ # 2 ρ & 4Uρ T 2 ρU 2
c 2 %1+ ( − c − + =0
$ kρ m ' kρ m ρm kρ m
Now use the quadratic formula
€ and keep simplifying.
4Uρ # 4Uρ & 2 # 2 ρ &# 2 ρU 2 T &
± % ( − 4%1+ (% − (
€ kρ m $ kρ m ' $ kρ m '$ kρ m ρm '
c=
# 2ρ &
2%1+ (
$ kρ m '
2ρ
2Uρ ± kρ m (T ρ m ) +
kρ m
((T ρm ) − U 2 )
c=
€ 2 ρ + kρ m
Now set c o = T ρ m = phase speed of the membrane waves in a vacuum.
2 ρU ± kρ m c o2 + (2 ρ kρ m )(c o2 − U 2 )
€ c=
2 ρ + kρ m
€c) The membrane is more unstable as the flow speed U increases. The membrane is less unstable
as the membrane tension (or c o = T ρ m ) increases. The membrane is more unstable as the fluid
density increases when
€ U > co. The membrane is less unstable as the fluid density increases when
U < co. The influence of the membrane mass is more easily deciphered with an algebraic revision
€ 2Uρ ± Tk 2 ρ m + 2kρ(T − ρ mU 2 )
of the result of part b): c = . Thus, we see that the membrane is
2 ρ + kρ m

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

more unstable with increasing membrane mass when U > [( kρm 2 ρ ) −1](T ρ m ) , but that the
membrane is less unstable with increasing membrane mass when U < [( kρ m 2 ρ ) −1](T ρ m ) .
co
d) When U = 0: c = ± . The waves move slower with a quiescent fluid present than
1+ 2 ρ kρ m €
in a vacuum because of the added mass (inertial loading)
€ provided by the liquid.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.6. Prove that σr > 0 for the thermal instability discussed in Section 11.4 via the
following steps that include integration by parts and use of the boundary conditions (11.38).
a) Multiply (11.36) by Tˆ * and integrate the result from z = –1/2 to z = +1/2, where z is the
2
dimensionless vertical coordinate) to find: σI + I = ∫ Tˆ *Wdz where I ≡ ∫ Tˆ dz ,
1 2 1

# 2 2&
I2 ≡ ∫ % dTˆ /dz€+ K 2 Tˆ (dz , and the limits of the integrations have been suppressed for clarity.
$ '
b) Multiply (11.37) by W* and integrate from z = –1/2 to z = +1/2 to find:
σ € €
Pr
[ 2
J1 + J 2 = RaK 2 ∫ W *Tˆdz where J1 ≡ ∫ dW /dz + K 2 W dz ,] 2

€
[ 2 2
]
2
J 2 ≡ ∫ d 2W /dz 2 + 2K 2 dW /dz + K 4 W dz , and again the limits of the integrations have been
suppressed.
€ c) Combine the results of€ a) and b) to eliminate the mixed integral of W & Tˆ , and use the result
of this combination to show that σi = 0 for Ra > 0. [Note: the integrals I1, I2, J1, and J2 are all
€
positive definite].
€
$ d2 ' ˆ
Solution 11.6. a) Equation (11.36) is &σ + K − 2 )T = W . As directed, multiply by Tˆ * and
2

% dz (
integrate through the thickness of the layer. The term-by-term results on the left side are:
$ * dTˆ '1 2 2
2ˆ
ˆ *ˆ ˆ
2
ˆ * d T ˆ dTˆ * d€
Tˆ dTˆ
(σ + K ) ∫ T Tdz = (σ + K ) ∫ T dz , and − ∫ T dz 2 dz = −&%T dz )( + ∫ dz dz dz = ∫ dz dz ,
2 2

€ −1 2

where the boundary terms in the integration by parts are zero because of the boundary conditions
on Tˆ , and all the integrals are from –1/2 to +1/2. Thus, the remnant of (11.36) is:
€ 2
€ 2 dTˆ
(σ + K ) ∫ T dz + ∫ dz dz = ∫ Tˆ *Wdz ,
2 ˆ

€ which is readily rearranged to

σI1 + I2 = ∫ Tˆ *Wdz ,
2 # 2 2&
where I1 ≡ ∫ Tˆ dz € , and I2 ≡ ∫ % dTˆ /dz + K 2 Tˆ (dz .
$ '
$σ d '$ d
2 2 '
b) Equation (11.37) is & + K €2 − 2 )& 2 − K 2 )W = −RaK 2Tˆ , which expands to:
% Pr dz (% dz (
€ € * $σ 2' 2 $σ 2' d
2
d4 -
,−& + K )K + & + 2K ) 2 − 4 /W = −RaK Tˆ .
2

+ % Pr ( % Pr ( dz dz .
*
As directed, €multiply by W and integrate through the thickness of the layer. The term-by-term
results on the left side are:
$σ ' $σ ' 2
€ −& + K 2 )K 2 ∫ W *W dz = −& + K 2 )K 2 ∫ W dz ,
% Pr ( % Pr (

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

#σ 2 1 12
dW * dW 4 # σ
2
2& * d W #σ 2& * * dW - 2& dW
% + 2K ( ∫ W dz = % + 2K (2,W −∫ dz5 = −% + 2K ( ∫ dz ,
$ Pr ' dz 2 $ Pr '3+ dz /.−1 2 dz dz 6 $ Pr ' dz
and
4
* d W
$ * d 3W '1 2 dW * d 3W
−∫ W dz = −&W ) + ∫ dz
€ dz 4 % dz 3 (−1 2 dz dz 3
$ dW * d 2W '1 2 d 2W * d 2W d 2W
2

= +& 2 )
−∫ dz = − ∫ dz.
% dz dz (−1 2 dz 2 dz 2 dz 2
where all the boundary terms in the various integrations by parts are zero because of the
boundary conditions on W, and all the integrals are from –1/2 to +1/2. Thus, the remnant of
(11.37) multiplied by –1 is:
€ 2 2
#σ 2& 2 #σ 2& dW d 2W
dz = RaK 2 ∫ W *Tˆdz ,
2
% + K (K ∫ W dz + % + 2K ( ∫ dz + ∫
$ Pr ' $ Pr ' dz dz 2
which is readily rearranged to
σ
J1 + J 2 = RaK 2 ∫ W *Tˆdz
Pr
€
[ 2 2
]
2
[ 2
where J1 ≡ ∫ dW /dz + K W dz , and J 2 ≡ ∫ d 2W /dz 2 + 2K 2 dW /dz + K 4 W dz .
2
] 2

c) The complex conjugate of the part a) result is:

€ σ * I1 + I2 = ∫ TˆW * dz ,
€ Use this to eliminate the mixed € integral in the result of part b) to find:
σ
J1 + J 2 = RaK 2 (σ * I1 + I2 ) .
Pr
€
The imaginary part of this equation is:
# J1 & #J &
σ i % ( = −RaK 2 I1σ i , or σ i % 1 + RaK 2 I1 ( = 0 .
$ Pr ' $ Pr '
€
When Ra > 0 (the flow is top heavy), the contents of the final parentheses is positive definite so
the final equation requires σi = 0.
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.7. Consider the thermal instability of a fluid confined between two rigid plates, as
discussed in Section 11.4. It was stated there without proof that the minimum critical Rayleigh
number of Racr = 1708 is obtained for the gravest even mode. To verify this, consider the gravest
odd mode for which
W = Asin(q0 z) + Bsinh(qz) + C sinh(q* z) . (1)
(Compare this with the gravest even mode structure: W = Acosq0z + Bcoshqz + Ccoshq*z.)
Following Chandrasekhar (1961, p. 39), show that the minimum Rayleigh number is now
17,610, reached at the wave number Kcr = 5.365.
€
Solution 11.7. From Section 12.4 the six roots for the vertical wave number q are:
12 12
[ (
±iqo = ±iK(s −1)1 2 , ±q = ±K 1+ s 1+ i 3 2 ) ] [ ( ) ]
, and ±q* = ±K 1+ s 1− i 3 2 , (2)
13
with s ≡ ( Ra K 4 ) . The boundary conditions on W are:
dW # d 2 2
&2
€ W = =% −K ( € W = 0 at z = ± 1/2. (3)
€ dz $ dz 2 '
€ From the mode shape specification, equation (1), the first, second, and fourth derivatives of the
mode shape are:
dW
€ = Aq0 cos(q0 z) + Bqcosh(qz) + Cq* cosh(q* z) , (4)
dz
d 2W
2
= −Aq02 sin(q0 z) + Bq 2 sinh(qz) + Cq*2 sinh(q* z) , and (5)
dz
€ d 4W
4
= Aq04 sin(q0 z) + Bq 4 sinh(qz) + Cq*4 sinh(q* z) . (6)
dz
From (5) and€ (6), 2
#d 2
2
& # d4 2 d
2
4
&
% 2 − K ( W = % 4 − 2K + K (W
$ dz €' $ dz dz 2 '
2 2 2
= A(q02 + K 2 ) sin(q0 z) + B(q 2 − K 2 ) sinh(qz) + C (q*2 − K 2 ) sinh(q* z).
Application of all three boundary conditions (3) at either z = +1/2 or z = –1/2 leads to:
# sin(q0 /2) sinh(q /z) sinh(q* /2) &#A&
% (% (
€ % q0 cos(q0 /2) qcosh(q /2) q* cosh(q* /2) (%B( = 0 .
2 2 2
% q 2 + K 2 sin(q /2) q 2 − K 2 sinh(q /2) q*2 − K 2 sinh(q* /2)(%$C('
$( 0 ) 0 ( ) ( ) '
For nontrivial solutions, the determinant of the 3x3 matrix has to be zero. Dividing by the first
row of the matrix produces:
# 1 1 1 &#A&
€ % (% (
* *
%q0 cot(q0 /2) qcoth(q /2) q coth(q /2)(%B( = 0 . (7)
2 2 2
% q +K
$( 0
2 2
) (q − K ) (q − K ) ('%$C('
2 2 *2 2

From (2) and definition of s given above:

q02 = K 2 (s −1) ,
2 2
(q02 + K 2 ) = (K 2 (s −1) − K 2 ) = K 4 s2
€

€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

2 2 2
(q 2
( [
− K 2 ) = K 2 1+ 12 s(1+ i 3) − K 2 ] ) [ ]
= 14 K 4 s(1+ i 3) , and
2 2
(q*2
[
− K 2 ) = 14 K 4 s(1− i 3) ]
So, (6) becomes
€ # &# &
% 1 1 1 (% A(
* *
€ %q0 cot(q0 /2) qcoth(q /2) q coth(q /2)(%B( = 0 . (8)
% 2 2 (
%$ 1 1
4
1+ i 3 [ 1
4 ]
1− i 3 ('%$C(' [ ]
after cancelling the common factor of K4s2 from the bottom row. Expanding the determinant, and
using (2) and the definition of s given above produces a relationship between Ra and K which is
sketched below.
€
105

Ra

K
2 4 6
This relationship has a minimum of Racr = 17,610 at K = 5.36. The critical Ra for the gravest
even mode is 1708, signifying that it is more unstable.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.8. Consider the centrifugal instability problem of Section 11.6. Making the narrow-
gap approximation, work out the algebra of going from (11.50) to (11.51).

Solution 11.8. The perturbation equations (11.50) are:

∂uR 2Uϕ uϕ 1 ∂p ( u + ∂u $ dU U ' $ u '
− =− + ν * ∇ 2 uR − R2 - , ϕ + & ϕ + ϕ )uR = ν &∇ 2 uϕ − ϕ2 ) ,
∂t R ρ ∂R ) R , ∂t % dR R( % R (
∂uz 1 ∂p ∂ ∂ u
=− + ν∇ 2 uz , and (RuR ) + z = 0 . (1)
∂t ρ ∂z ∂R ∂z
Substituting in
€ €
uR = u(R)eσt cos kz , uϕ = v(R)eσt cos kz , uz = w(R)eσt sin kz , and p ρ = pˆ (R)eσt cos kz ,
the equation set€(1) becomes €
# d # d 1& 2 σ& U dpˆ
ν % % + ( − k − (u + 2 ϕ v = ,
€ € $ dR $
€ dR R ' ν ' €R dR
# d # d 1& σ & # dU U &
ν % % + ( − k 2 − (v − % ϕ + ϕ (u = 0 ,
$ dR $ dR R ' ν ' $ dR R'
€ ## d 1 & d σ&
ν %% + ( − k 2 − ( w = −kpˆ , and
$$ dR R ' dR ν'
€ " d 1%
$ + 'u = −kw . (2)
# dR R &
Eliminating w between € the third and fourth equations of set (2) produces:
ν ## d 1 & d σ &# d 1 &
2 %%
+ ( − k 2 − (% + ( u = pˆ .
k $€ $ dR R ' dR ν '$ dR R '
Inserting this equation for pˆ into the first equation of set (2), and working on the algebra
eventually leads to:
ν # d # d 1 & 2 σ &# d # d 1 & 2 & U
€ 2% % + ( − k − (% % + ( − k (u = 2 ϕ v . (3)
€ k $ dR $ dR R ' ν '$ dR $ dR R ' ' R
The second equation of set (2) and equation (3) are a pair of equations relating u and v.
Using the radius of the outer cylinder R2, define new dimensionless variables and
parameters:
€ r = R/R2, k 2 = K 2 R22 , and ω = σR22 ν ,
so that the relevant equation pair becomes:
" d " d 1% 2
%" d " d 1 % 2
% BK 2 " 1 AR22 %
$ $ + ' − K − ω '$ $ + ' − K ' u = 2 $ + 'v , and
# dr # dr r & € &# dr # dr r€& & ν # r2 B &
" d " d 1% 2
% A 2
$ $ + ' − K − ω 'v = 2 R2 u ,
# dr # dr r & & ν
where
€ U = Ar + B/r. It is convenient to make the transformation
ϕ

AR22
2 u → u,
€ ν
so that the equations take the more convenient forms:
" d " d 1% 2
%" d " d 1 % 2
% 2" 1 %
$ $ + ' − K − ω '$ $ + ' − K ' u = −TaK $ 2 − κ 'v , and
# dr # dr r & € &# dr # dr r & & #r &

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

" d " d 1% 2
%
$ $ + ' − K − ω 'v = u ,
# dr # dr r & &
Ω2 R 4 (1− µ)(1− µ η )
2
AB A 2 1− µ η 2 Ω R
where Ta = −4 2 R22 = 4 1 2 1 2 , κ = − R2 = , µ= 2, η= 1,
ν ν (1− η )
2 B 1− µ Ω1 R2
€
1− µ η 2 1− µ
A = −Ω1η 2 2
, and B = Ω1R12 .
1− η 1− η 2
€ The no-slip and boundary conditions€at the walls require: € €
u = v = 0 and (d/dr)u = 0 at r = η and 1,
where the last of the three conditions is equivalent to w = 0 (see the final equation of set (1)).
€ €
Now consider the narrow gap approximation that is valid when
R2 − R1 << 12 (R2 + R1 ) .
When this is true, d/dr >> 1/r so
d 1 d B % r − R1 (
+ ≅ , and A + 2 ≅ Ω1'1− (1− µ) *.
dr €r dr r & R2 − R1 )
Now convert the independent radial coordinate (R) to one (x) that starts on the inner cylinder
using the gap dimension, d = R2 − R1, as the length scale, and let k = K/d, and w = σd2/ν to find:
€ $ d2 '$ d 2 ' 2Ω1d 4 2
& 2 − K − ω€)& 2 − K ) u =
2 2
K (1− (1− µ)x )v , and
% dx (% dx ( ν
€ $ d2 2
' 2Ad 4
& 2 − K − ω )v = u,
% dx ( ν
€ 2Ω d 2K 4
as the relevant equation set. By the further transformation u → 1 u , these equations
ν
become:
€ 2
$d 2
'$ d 2 2
'
& 2 − K − ω )& 2 − K ) u = (1+ αx )v , and
% dx (% dx (€
$ d2 2
' 2
& 2 − K − ω )v = −TaK u ,
% dx (
where €
4 AΩ
Ta = − 2 1 d 4 , and α = –(1 – µ).
ν
€ as (11. 51).
These are the same equations

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.9. Consider the centrifugal instability problem of Section 11.6. From (11.51) and
(11.53), the eigenvalue problem for determining the marginal state (σ = 0) is
2 2
(d 2 dR 2 − k 2 ) uˆ R = (1+ αx)uˆϕ , (d 2 dR 2 − k 2 ) uˆϕ = −Tak 2uˆ R , (11.92,93)
with uˆ R = duˆ R dR = uˆϕ = 0 at x = 0 and 1. Conditions on uˆϕ are satisfied by assuming solutions
of the form
∞
€ uˆϕ€= ∑ Cm sin(mπx) . (11.94)
m=1
€ €
Inserting this into (11.92), obtain an equation for uˆ R , and arrange so that the solution satisfies the
four remaining conditions on uˆ R . With uˆ R determined in this manner and uˆϕ given by (11.94),
(11.93) leads to an eigenvalue
€ problem for Ta(k). Following Chandrasekhar (1961, p. 300), show
that the minimum Taylor number is given € by (11.54) and is reached at kcr = 3.12.
€ €
€ into (11.92) to find:
Solution 11.9. Continuing the effort from Exercise 11.8, insert (11.94)
∞
2
(d 2 dx 2 − K 2 ) u = (1+ αx)v = (1+ αx) ∑ Cm sin(mπx) , (5)
m=1
where a switch has been made to the dimensionless variables of Exercise 11.8. Now arrange that
the solution satisfies the four remaining boundary conditions on u. With u determined in this
fashion and v given by (11.94), (11.93) will lead to an equation for Ta.
€
The solution of (5) is straightforward. The general solution can be written in the form:
$ A1(m ) coshKx + B1(m ) sinhKx + A2(m ) x coshKx + B2(m ) x sinhKx (
∞
Cm & &
u= ∑ 2 2 2 2
% 4αmπ ), (6)
m=1 (m π + K ) & + (1+ αx)sin(mπx) + 2 2 cos(m πx) &
' m π + K2 *
where the constants are determined by the boundary conditions u = (d/dx)u = 0 at x = 0 and 1.
These conditions lead to the four equations:
4αmπ
€ A1(m ) = − 2 2 , KB1(m ) + A2(m ) = −mπ ,
m π + K2
4αmπ
A1(m ) coshK + B1(m ) sinhK + A2(m ) coshK + B2(m ) sinhK = (−1) m +1 2 2 , (7)
m π + K2
A1(m )K sinhK + B1(m€ )
K coshK + A2(m ) (coshK € + K sinhK ) + B(m ) (sinhK + K coshK ) = (−1) m +1 (1+ α )mπ
2
The solution of these equations is:
4αmπ
€ A1(m ) = − 2 2
m π + K2
€ mπ
B1(m ) = {K + β m (sinhK + K coshK ) − γ m sinhK}
Δ
mπ
A2(m ) = − € {sinh 2 K + β m K (sinhK + K coshK ) − γ m K sinhK }
Δ
m π
B2(m ) =
€ {(sinhK coshK − K ) + β mK 2 sinhK − γ m (K coshK − sinhK )} (8)
Δ
where:
€ 4α
Δ = sinh 2 K − K 2 , β m = 2 2 2{
(−1) m +1 + coshK }, and
m π +K
€ 4α
γ m = (−1) m +1 (1− α ) + 2 2 K sinhK .
m π + K2
€
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Substituting u from (6) and v from (11.94) into (11.93) produces:

∞
∑ Cn ( n 2π 2 + K 2 ) sin nπx
n=1

& A1(m ) coshKx + B1(m ) sinhKx + A2(m ) x coshKx + B2(m ) x sinhKx * (9)
∞
C ( (
= TaK 2 ∑ 2 2 m 2 2 ' 4αmπ +.
m=1 (m π + K ) ( + (1+ αx)sin(mπx) + 2 2 2
cos(mπx) (,
) m π +K
Multiply this equation by sin(nπx) and integrate from x = 0 to x = 1, to obtain a system of linear
homogeneous equations for the constants Cm. The requirement that these constants are not all
zero leads to the equation:
€

* n +1$ 2K ' .
,,[1+ (−1) coshK ] A1 + [(−1) sinhK ] B1 + (−1) &coshK − 2 2
n +1 (m ) n +1 (m )
2
sinhK ) A2(m ) ,
nπ % n π +K ( ,
2 2 2
+ /
n π +K , $ n +1 2K n +1 ' (m ) ,
2{
+ &(−1) sinhK − 2 2 1+ (−1) coshK })B2
where ,
- % n π +K ( ,0
1 1 % 3 δ
(
+αx nm + δnm − ( n 2π 2 +0K 2 ) 2nm = 0 if m + n is even and m ≠ n + (10)
2 2 + K Ta +
x nm = & 1/4 if m = n ).
+ 4nm % 2 1 ( +
+ 2 2
& 2 2 2
− ) if m + n is odd +
'n − m 'm π + K π 2 (n 2 − m 2 ) * *
€
On using the first two equations of (7), equation (10) simplifies to:
nπ % 4mπα 2K (
2 [
2 2 2
& 2 2 (−1) m +n −1] − 2 2 2 [
(−1) n +1 A2(m ) sinhK + (−1) n +1 B2(m ) coshK + B2(m ) ])
n π +K € m π +K
' n π +K *
1 1 3 δ
+αx nm + δnm − ( n 2π 2 + K 2 ) 2nm = 0 (11)
2 2 K Ta

After substituting for the constants A2(m ) and B2(m ) given by (8), (11) becomes:
4mnπ 2α
2 [
€ (−1) m +n −1]
(n π + K )(m π + K )
2 2 2 2 2

€ % € )
'(sinhK coshK − K)[1+ (1+ α )(−1) m +n ] '
2Kmnπ 2 ' '
− 2 2 &+(sinhK − K coshK)[(−1) n +1 + (1+ α )(−1) m +1 ]*
(n π + K )(sinh K − K ) ' 4kα sinhK
2 2 2
'
'− sinhK + K(−1) ][(−1) −1] '
( m 2π 2 + K 2 [
m +1 m +n
+
1 1 3 δ
+αx nm + δnm − ( n 2π 2 + K 2 ) 2nm = 0 (12)
2 2 K Ta
A first approximation to the solution of (12) is obtained by setting the (1,1)-element of
the determinant zero. This implies:
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

1 2 2 3 δ nm 1 1 2Kπ 2 (2 + α )
(π + K ) K 2Ta 4 2 π 2 + K 2 sinh2 K − K 2 [(sinhK coshK − K) + (sinhK − K coshK)]
= α + −
2 ( )( )
and this simplifies to:
3

Ta =
2 (π 2 + K 2 )
€
{ [
2 + α K 2 1−16Kπ 2 cosh 2 (K /2) π 2 + K 2 2 (sinhK + K)
( ) ]}
A plot of Ta(2 + α) as a function of K from this solution shows that the minimum value of the
Taylor number is:
Tacr = 3430 (2 + α ) , and is reached at Kcr = 3.12.
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.10. For a Kelvin–Helmholtz instability in a continuously stratified ocean, obtain a

globally integrated energy equation in the form
1 d ∂U
2 dt
∫ ( u 2 + w 2 + g 2 ρ 2 ρ 02 N 2 ) dV = − ∫ uw dV .
∂z
(As in Figure 11.25, the integration in x takes place over an integer number of wavelengths.)
Discuss the physical meaning of each term and the mechanism of instability.

Solution 11.10.€ From (11.57), multiplying the perturbation equations for the Kelvin-Helmholtz
instability by u, w, and g 2 ρ ρ 02 N 2 and add them together
$ ∂u ∂u ∂U 1 ∂ p ' $ ∂w ∂w ρ 1 ∂p ' g 2 ρ $ ∂ρ ∂ρ N2 '
u& + U + w + ) + w & + U + g + ) + & + U − ρ 0 w ) = 0.
% ∂t ∂x ∂z ρ 0 ∂ x ( % ∂t ∂x ρ 0 ρ 0 ∂z ( ρ 02 N 2 % ∂t ∂x g (
This produces: €
#∂ ∂ &# u 2 w 2 g2ρ 2 & ∂U 1 # ∂p ∂p &
% + U (% + + 2 2 ( + uw + % + (=0
€ $ ∂t ∂x '$ 2 2 2ρ0 N ' ∂z ρ 0 $ ∂x ∂z '
The final term can be rewritten:
#∂ ∂ &# u 2 w 2 g2ρ 2 & ∂U 1 # ∂ ( pu) ∂ ( pw) & # ∂u ∂w &
% + U (% + + 2 2 ( + uw + % + ( − p% + ( = 0 .
$€∂t ∂x '$ 2 2 2ρ0 N ' ∂z ρ 0 $ ∂x ∂z ' $ ∂ x ∂z '
After global integration over an integer number of x-direction wavelengths, the terms involving
U(∂/∂x) will be zero, and the third term, which is equal to ρ −1 0 ∫ ∇ ⋅ ( pu)dV , transforms into a
−1
€surface integral, ρ 0 ∫ pu ⋅ ndA , which vanishes because no flow crosses the channel walls and an
there is an equal in-flux and out-flux across the open boundaries on the CV (see Figure 11.25).
The last term vanishes because of the continuity equation. € Thus, the resulting CV equation is:
1 d ∂U
€
2 dt
∫ ( u 2 + w 2 + g 2 ρ 2 ρ 02 N 2 ) dV = − ∫ uw dV .
∂z
This equation shows that the rate of change of the sum of the kinetic and potential
energies is equal to the energy extracted from the correlation of the stream-wise (u) and vertical
(w) velocity fluctuations with the average shear (∂U/∂z). However, in order for the perturbations
€ from the mean shear, they must be anti-correlated so that uw is negative on
to extract energy
average when ∂U/∂z is positive. This is typical of shear instabilities. And, when u and w
represent turbulent fluctuations the average correlation of u and w is called a Reynolds shear
stress (see Ch. 12).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.11. In two-dimensional (x,y)-Cartesian coordinates, consider the inviscid stability of

$ S + y for y > 0'
horizontal parallel shear flow defined by two linear velocity gradients: U(y) = % − (,
& S y for y ≤ 0)
where S+ and S– are real constants. Assume an infinitesimal velocity perturbation with vertical
component v = f (y)exp {ik(x − ct)} , where k is positive real but ω may be complex.
fU !!
a) Use the Rayleigh equation f !! − k 2 f − = 0 with f (y) →€0 as y → ∞ to find f(y).
U −c
b) Require the pressure perturbation associated with v to be continuous across y = 0, and
determine a single equation for the disturbance phase speed c in terms of the other parameters.
c) For what values of S+, S–, and k, is this flow€stable, unstable,
€ or neutrally stable?
+ –
d) What is special about the case S = S ?
y!
U(y)!

x!

Solution 11.11. a) Here d2U/dy2 = 0, so the Rayleigh equation simplifies to d2f/dy2 – k2f = 0
which has solutions A±exp(±ky). To satisfy the given boundary conditions for k positive & real,
the decaying exponential must be chosen for y > 0 and y < 0, so f(y) = Aexp(–k|y|).
∂ ∂v #
b) The continuity equation is (U(y) + u#) + = 0 . For v = A exp {ik(x − ct) − k y } , this
∂x ∂x
implies: u = − ∫ (∂ v ∂ y) dx = (−1 ik ) ( df dy) exp {ik(x − ct)} = (−A ik ) ( ∓k ) exp {ik(x − ct) − k y }
where the upper & lower signs go with y > 0 & y < 0, respectively. The linearized horizontal
€ ∂u ∂ u ∂U 1 ∂ p"
momentum equation for the disturbance is: +U(y) + v =− . Evaluate the
∂t ∂x ∂y ρ ∂x
derivatives on the left side and integrate in x to find:
( −ikc " ik % + p/
* ( ∓k ) + S ± y $# '& ( ∓k ) + S ± - ∫ A exp {ik(x − ct) − k y } dx = − , or
) −ik −ik , ρ
"#c ( ∓k ) − S ± y ( ∓k ) + S ± $% ∫ A exp {ik(x − ct) − k y } dx = − p' .
ρ
Evaluate this equation above and below y = 0 and set the two results equal to find:
−ck + S + = +ck + S − or c = (S+ – S–)/2k.
c) Here c is always real; this flow is neutrally stable in all cases.
d) When S+ = S–, then c = 0, and this implies zero phase speed for the disturbance.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.12. Consider the inviscid stability of a constant vorticity layer of thickness h
between uniform streams with flow speeds U1 and U3. Region 1 lies above the layer, y > h/2
with U(y) = U1. Region 2 lies within the layer, |y| ≤ h/2, U(y) = 12 (U1 + U 3 ) + (U1 − U 3 )( y h ) .
Region 3 lies below the layer, y < –h/2 with U(y) = U3.
fU ""
a) Solve the Rayleigh equation, f "" − k 2 f − = 0 , in each region, then use appropriate
U −c
boundary and matching conditions to obtain: €
f1 (y) = ( Acosh( kh 2) + Bsinh( kh 2))e−k ( y−h 2) for y > +h/2,
€ Acosh( ky ) + Bsinh( ky )
f 2 (y) = for |y| ≤ h/2,
f 3 (y) = ( Acosh( kh 2) − Bsinh( kh 2))e +k ( y +h 2) for y < –h/2.
where f defines
€ the spatial extent of the disturbance: v " = f (y)e ik(x−ct ) and u" = −( f " ik )e ik(x−ct ) ,
and A and€B are undetermined constants.
∂u# ∂u# ∂U 1 ∂p#
b) The linearized
€ horizontal momentum equation is: +U + v# =− .
∂t ∂x ∂y ρ ∂x
Integrate this equation with respect to x, require € the pressure to be€continuous at y = ± h/2, and
simplify your results to find two additional constraint equations:
U −U
(c − U1 ) f1#(+h /2) = (c − U€1 ) f 2#(+h /2) + 1 3 f 2 (+h /2) , and
h
U1 − U 3
(c − U 3 ) f 3#(−h /2) = (c − U 3 ) f 2#(−h /2) + f 2 (−h /2)
h
c) Define c o = c − 12 (U1 + U 3 ) (this is the phase speed of the disturbance waves in a frame of
€
reference moving at the average speed), and use the results of parts a) and b) to determine a
single equation
€ for co:
# U1 − U 3 & 2 2
€ 2
co = %
$ 2kh '
{ ( ( kh −1) − e } −2kh

[This part of this problem requires patience and algebraic skill.]

d) From the result of part c), co will be
real for kh >> 1 (short wave
disturbances), so the € flow is stable or
neutrally stable. However, for kh << 1
(long wave disturbances), use the result
of part c) to show that:
$ U − U3 ' 4
c o ≅ ±i& 1 ) 1− kh + ...
% 2 ( 3
e) Determine the largest value of kh at
which the flow is unstable.

€ Solution 11.12. Here U "" = 0 in all three regions, so the Rayleigh equation,
fU ""
f "" − k 2 f − = 0 , reduces to: f "" − k 2 f = 0 in all three regions. The solutions of this
U −ω k
equation are€growing and decaying exponentials, or hyperbolic functions. Thus,
f1 (y) = Ce−ky + De +ky for y > +h/2,
€
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

f 2 (y) = Acosh( ky ) + Bsinh( ky ) for |y| ≤ h/2,

f 3 (y) = Ee−ky + Fe +ky for y < –h/2.
Where A, B, C, D, E, and F are all constants. The disturbance must disappear as y → ±∞. This
means that D = E = 0. To ensure continuity of f at y = ± h/2, the following must be true:
€
Ce−kh 2 = Acosh( kh 2) + Bsinh( kh 2) and Fe−kh 2 = Acosh( kh 2) − Bsinh( kh 2) .
€
Using these two equations to eliminate C and F produces:
f1 (y) = ( Acosh( kh 2) + Bsinh( kh 2))e−k ( y−h 2) for y > +h/2,
€ f 2 (y) = Acosh( ky ) + Bsinh € ( ky ) for |y| ≤ h/2,
+k ( y +h 2)
f 3 (y) = ( Acosh( kh 2) − Bsinh( kh 2))e for y < –h/2.
b) The only € x-variation in the whole problem enters through the assumed form of the
ik(x−ct )
disturbance:€ v " = f (y)e and u" = −( f " ik )e ik(x−ct ) , so and x-integration is the same as dividing
1 ∂p% p% ' ∂u% ∂u% dU * 1 ' ∂u% ∂u% dU *
€
by +ik. Therefore: − ∫ dx = − = ∫ ) +U + v% ,dx = ) +U + v% , , and
ρ ∂x ρ ( ∂t ∂x dy + ik ( ∂t ∂x dy +
there
€ is no integration constant€ since the pressure fluctuation must disappear when u" and v "
disappear. Now plug in the relationships for u" and v " and require continuity of pressure at y = ±
h/2:
€ (cf1"− U1 f1"+ f1U1") y= +h 2 = (cf 2" − U 2 f 2" + f 2U 2" ) y= +h 2 , and € €
(cf 2" − U 2 f 2" +€ f 2U 2" )€y=−h 2 = (cf 3" − U 3 f 3" + f 3U 3" ) y=−h 2
Using these two equations and U1(y) = U1 for y > h/2, U 2 (y) = 12 (U1 + U 3 ) + (U1 − U 3 )( y h ) for |y|
≤ h/2, and € U3(y) = U3 for y < –h/2 produces:
€ (c − U1 ) f1#(+h /2) = (c − U1 ) f 2#(+h /2) + U1 − U 3 f 2 (+h /2) , and
€ h
U1 − U 3
(c − U 3 ) f 3#(−h /2) = (c − U 3 ) f 2#(−h /2) + f 2 (−h /2)
h
(U + U 3 )
c) First evaluate
€ rewrite the equations from part b) in terms of c o = c − 1 :
2
$ ΔU ' $ ΔU ' ΔU
€ &c o − ) f1*(+h /2) = & c o − ) f 2*(+h /2) + f 2 (+h /2) , and
% 2 ( % 2 ( h
# ΔU & # ΔU & ΔU
%c o + ( f 3)(−h /2) = % c o − (€
f 2)(−h /2) + f 2 (−h /2)
$ 2 ' $ 2 ' h
where ΔU € = U1 – U3. These equations can be made a little more tidy:
$ ΔU ' ΔU
&c o − )( f1*(+h /2) − f 2*(+h /2)) = f 2 (+h /2) , and
€ % 2 ( h
# ΔU & ΔU
%c o + (( f 3)(−h /2) − f 2)(−h /2)) = f 2 (−h /2) .
$ 2 ' h
Now evaluate € the functions and derivatives in the various regions:
f 2 (+h 2) = Acosh( kh 2) + Bsinh( kh 2) , f 2 (−h 2) = Acosh( kh 2) − Bsinh( kh 2) ,
f " (+h 2)€= −kAcosh( kh 2) − kB sinh( kh 2) , f " (+h 2) = kA sinh( kh 2) + kBcosh( kh 2) ,
1 2

f 2" (−h 2) = −kAsinh( kh 2) + kB cosh( kh 2) , and f 3" (−h 2) = kA cosh( kh 2) − kBsinh( kh 2)

€Stuff these evaluations into the modified€ part b) results to get:
€ €
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

$ ΔU '
&c o − )(−kAcosh( kh 2) − kBsinh( kh 2) − kAsinh( kh 2) − kBcosh( kh 2))
% 2 (
ΔU
=
h
( Acosh(kh 2) + Bsinh(kh 2)),
and
# ΔU &
%c o + (( kAcosh( kh 2) − kBsinh( kh 2) + kAsinh( kh 2) − kBcosh( kh 2))
€
$ 2 '
ΔU
=
h
( Acosh(kh 2) − Bsinh(kh 2)).
Start simplifying these by collecting terms while noting that e ±z = cosh(z) ± sinh(z)
$ ΔU '
−kh& c o − )( A + B)e
+kh 2
= ΔU ( Acosh( kh 2) + Bsinh( kh 2))
€ % 2 (
# ΔU &
+kh% c o + (( A − B)e
+kh 2
= ΔU (€Acosh( kh 2) − Bsinh( kh 2))
$ 2 '
Isolate A
€ and B in each equation.
* $ ΔU ' +kh 2 $ kh '- * $ ΔU ' +kh 2 $ kh '-
,kh& c o − )e + ΔU cosh& )/A + ,kh&c o − )e + ΔU sinh& )/B = 0
€+ % 2 ( % 2 (. + % 2 ( % 2 (.
* # ΔU & +kh 2 # kh &- * # ΔU & +kh 2 # kh &-
,kh% c o + (e − ΔU cosh% (/A + ,−kh% c o + (e + ΔU sinh% (/B = 0
+ $ 2 ' $ 2 '. + $ 2 ' $ 2 '.
€These are two homogeneous linear algebraic equations. They will have a non-trivial solution
when their determinant is zero.
* $ ΔU ' +kh 2 $ kh '- * $ ΔU ' +kh 2 $ kh '-
€ ,kh& c o − )e + ΔU cosh& )/ × ,−kh&c o + )e + ΔU sinh& )/ −
+ % 2 ( % 2 (. + % 2 ( % 2 (.
* $ ΔU ' +kh 2 $ kh '- * $ ΔU ' +kh 2 $ kh '-
,kh& c o − )e + ΔU sinh& )/ × ,kh& c o + )e − ΔU cosh& )/ = 0
+ % 2 ( % 2 (. + % 2 ( % 2 (.
The remaining job is nothing less than a straight out algebraic effort. First carefully multiply the
factors in square brackets together.
$ (ΔU ) ' kh
2
$ ΔU ' +kh 2 $ kh '
€ 2 2
0 = −2k h && c o −2
))e + kh&c o − )e ΔU sinh& )
% 4 ( % 2 ( %2(
$ ΔU ' +kh 2 $ kh ' 2 $ kh ' $ kh '
−kh& c o + )e ΔU cosh& ) + 2(ΔU ) sinh& ) cosh& )
% 2 ( %2( %2( %2(
$ ΔU ' +kh 2 $ kh ' $ ΔU ' +kh 2 $ kh '
+kh& c o − )e ΔU cosh& ) − kh& c o + )e ΔU sinh& )
% 2 ( %2( % 2 ( %2(
Collect and simplify all the terms with hyperbolic functions.

$ 2'
2$ e − e−kh '
€2 2 & 2 (ΔU ) ) kh
+kh
$ ΔU ' +kh $ ΔU ' +kh
0 = −2k h & c o − )e + kh& c − )e ΔU − kh& c + )e ΔU + 2 (ΔU ) & )
Merge the two 4 terms.
% middle ( % o 2 ( % o 2 ( % 4 (
$ (ΔU ) ' kh
2
2 2$ e
+kh
− e−kh '
2 2 2
0 = −2k h && c o − ))e − kh (ΔU ) e +kh + 2(ΔU ) & )
% 4 ( % 4 (
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Solve for c o2 and continue simplification.

2 2 2
(ΔU ) (ΔU ) (ΔU )$ 1− e−2kh ' $ ΔU ' 2 2 2
) ( k h − 2kh + 1− e )
2 −2kh
c o = − + 2 2 & )=&
4 2kh k h % 4 ( % 2kh (
2
€Make the last few adjustments to reach the final form: 2 # ΔU &
co = %
2
( ( kh −1) − e
−2kh
.
$ 2kh ' ( )
–2kh
d) Here
€ we need to expand e out to the third order term:
e−2kh = 1− 2kh + 12 (2kh) 2 − 16 (2kh) 3 + ....
# ΔU & 2 # 2 2 1 2 €1 & # ΔU & 2 # 2 2 4 3 3 &
c o2 = % 3
( % k h − 2kh + 1−1+ 2kh − (2kh) + (2kh) + ...( = % ( %−k h + k h + ...(
$ 2kh ' $ 2 6 ' $ 2kh ' $ 3 '
12
€ # ΔU &# 4 &
Thus: c o = ±i% (%1− kh + ...(
$ 2 '$ 3 '
€ e) The limiting value of kh for instability will occur when the imaginary part of c o2 just reaches
2
zero, i.e. when ( kh −1) − e−2kh = 0 or kh −1 = e−kh which can be solved numerically via iteration
on a hand calculator:€kh ≈ 1.2785. For kh larger than this value the flow will be neutrally stable.
€
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.13. Consider the inviscid instability of parallel flows given by the Rayleigh equation
# d 2vˆ & d 2U
(U − c)% 2 − k 2vˆ ( − 2 vˆ = 0 , (11.95)
$ dy ' dy
where the y-component of the perturbation velocity is v = vˆ (y)exp{ik(x − ct)} .
(i) Note that this equation is identical to the Rayleigh equation (11.81) for the stream
function amplitude ˆ
€ φ, as it must because v (y) = −ikφ . For a flow bounded by walls at y1
and y2, note that the boundary conditions are identical in terms of φ and vˆ .
€
(ii) Show that if c is an eigenvalue of (11.95), then so is its conjugate c* = cr – ici. What
aspect of (11.95) allows this result to be valid?
€
(iii) Let U(y) be antisymmetric, so that U(y) = –U(–y). Demonstrate
€ that if c(k) is an
eigenvalue, then –c(k) is also an eigenvalue. Explain the result physically in terms of the
possible directions of propagation of perturbations in such an antisymmetric flow.
(iv) Let U (y) be symmetric so that U(y) = U(–y). Show that in this case vˆ is either symmetric
or antisymmetric about y = 0.
[Hint: Letting y → – y, show that the solution vˆ (−y) satisfies (11.95) with the same
eigenvalue c. Form a symmetric solution S(y) = vˆ (y) + vˆ (−y) = S(−y) € , and an antisymmetric
solution A(y) = vˆ (y) − vˆ (−y) = −A(−y) . Then write A[S-eqn] – S[A-eqn] = 0 where S-eqn
indicates the differential equation (11.95) in terms € of S. Canceling terms this reduces to (SAʹ′ –
ASʹ′)ʹ′ = 0, where the prime (ʹ′) indicates
€ a y-derivative. Integration gives SAʹ′ – ASʹ′ = 0, where the
constant of integration is zero because of the boundary conditions. Another integration gives S =
€
bA, where b is a constant of integration. Because the symmetric and antisymmetric functions
cannot be proportional, it follows that one of them must be zero.]
Comments: If v is symmetric, then the cross-stream velocity has the same sign across the
entire flow, although the sign alternates every half wavelength along the flow. This mode is
consequently called sinuous. On the other hand, if v is antisymmetric, then the shape of the jet
expands and contracts along the length. This mode is now generally called the sausage instability
because it resembles a line of linked sausages.

Solution 11.13. (i) Since the stream function is ψ = φexp[ik(x – ct)], the cross stream velocity is:
∂ψ
v =− = −ikφ exp[ik(x − ct)] = vˆ exp[ik(x − ct)] so that vˆ = −ikφ .
∂x
Thus, vˆ and φ will satisfy identical equations and boundary conditions.
(ii) The complex conjugate of (11.95) is:
*
# d 2vˆ * 2 *
& d 2U * €
€ (U − c )% 2 − k v ( − 2 vˆ = 0 ,
ˆ
€ $ dy ' dy
and this equation is identical to (11.95) except that c* replaces c and vˆ * replaces vˆ . The
boundary conditions on vˆ * and vˆ are also identical, namely vˆ * = vˆ = 0 at y1 or y2. Thus, if vˆ is
an eigenfunction with€eigenvalue c for some k, then vˆ * is and eigenfunction with eigenvalue c* =
cr – ici for the same k. This property is only possible since€(11.95) does€not directly involve the
imaginary root i.
€ € € € €
(iii) When the basic flow is antisymmetric, U(y) = –U(–y), so that all derivatives of U change
€
sign when y is oppositely directed. That is, U´(y) = –U´(–y) and U´´(y) = –U´´(–y), where a prime
denotes a derivative. Changing y into –y, (11.95) becomes:
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

# d 2vˆ 2
& d 2U # d 2vˆ 2
& d 2U
ˆ
(−U − c)% 2 − k v ( + 2
ˆ
v =0 , or (U + c)% 2 − k v ( − 2 vˆ = 0 .
ˆ
$ dy ' dy $ dy ' dy
Considering the second of these two equations, all terms in are the same as in (11.95), except that
c has been replaced by –c. Thus, if c is an eigenvalue, then –c is also an eigenvalue for the same
k. This means that for each propagating mode that moves in the positive x-direction there is a
€
corresponding €
mode that moves in the negative x-direction. This is obvious from the following
sketch of the antisymmetric basic flow, and from the observation that the two directions of
propagation are "similar" with respect to the basic velocity.
y

(iv) When the basic flow is symmetric, U(y) = U(–y), then U´´(y) = U´´(–y), so changing y into –
y in (11.95) produces:
# d 2vˆ & d 2U
(U − c)% 2 − k 2vˆ ( − 2 vˆ = 0 ,
$ dy ' dy
so that vˆ (−y) satisfies (11.95) with the same eigenvalue c. By adding and subtracting the two
solutions, we get symmetric and anti-symmetric solutions:
S(y) = vˆ (y) + vˆ (−y) = S(−y) and A(y) = vˆ (y) − vˆ (−y) = −A(−y) .
€
Obviously S and A satisfy the differential equation, therefore:
€ ) # d 2S & d 2U , ) # d2A & d 2U ,
A+(U − c)% 2 − k 2 S ( − 2 S. − S+(U − c)% 2 − k 2 A( − 2 A. = 0 ,
€ * $ dy ' dy€ - * $ dy ' dy -
since the contents of both sets of {,]-brackets are zero. Canceling terms and dividing by U – c
leaves:
d 2S d2A d # dS dA & dS dA
€A 2 − S 2 = 0 or % A − S ( = 0 which integrates to A − S = const.
dy dy dy $ dy dy ' dy dy
Evaluating the constant term at the wall, it turns out to be zero. So, divide the final equation by
the product SA to find:
1 dS 1 dA
€ =€ which integrates to: ln(S) = ln(A) + const.
€ , or S = (const.)A .
S dy A dy
However, symmetric and anti-symmetric functions cannot be proportional, so S or A must be
zero. If both are zero this is the trivial case. And, the final relationship does not require both S
and A to be zero if one is zero, because € the division step with the
€ first-derivative equation is not
€ when either A or S is zero.
valid
The unstable waves on a symmetric jet are the either sinuous or sausage-like, but not a
combination of both.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.14. Derive (11.88) starting from the incompressible Navier-Stokes momentum
equation for the disturbed flow:
∂ ∂ 1 ∂ ∂2
( i i) ( j j ) ( i i)
U + u + U + u U + u = − ( P + p) + ν (U i + ui ), (11.96)
∂t ∂x j ρ ∂x i ∂ x j ∂x j
where Ui and ui represent the basic flow and the disturbance, respectively. Subtract the equation
of motion for the basic state from (11.96), multiply by ui and integrate the result within a
stationary volume having stream-wise control surfaces chosen to coincide with the walls where
€
no-slip conditions are satisfied or where ui → 0, and having a length (in the stream-wise
direction) that is an integer number of disturbance wavelengths.

€ ∂U i ∂U i 1 ∂P ∂ 2U i
Solution 11.14. The basic flow momentum equation is: +Uj =− +ν .
∂t ∂x j ρ ∂x i ∂ x j ∂x j
∂ui ∂U i ∂ui ∂ui 1 ∂p ∂ 2 ui
Subtracting this from (11.96) produces: + uj +Uj + uj =− +ν .
∂t ∂x j ∂x j ∂x j ρ ∂x i ∂x j ∂x j
Multiply by ui and integrate each term over the€CV.
& ∂u ∂ui ∂ui ∂U i ui ∂p ∂ 2 ui )
i
∫ i ∂t i j ∂x
' u + u U + u u
i j
∂x j
+ u u
i j
∂x j
= −
ρ ∂x i
+ u iν *dV .
∂x j ∂x j +
(%)
CV ( € j
Consider each term individually.
∂u ∂ $1 ' d 1
∫ ui ∂ti dV = ∫ ∂t &% 2 ui2 )(dV = dt ∫ 2 ui2 dV .
CV CV CV
€
Here the final equality follows because the CV is stationary and the volume integration leaves no
spatial dependence.
$ 1 ∂u 2 ' $ 1 ∂ (u 2U ) '
∂u€i 2 ∂U j i j 1
∫ i j ∂x
u U dV = ∫ & 2 j ∂x
& U i
+ u i )
) dV = ∫ && ))dV = ∫ ui2U j n j dA = 0
CV j CV % j ∂x j ( CV % 2 ∂x j ( CS 2

∂u $ 1 ∂u 2 ∂u ' $ 1 ∂ (u u ) '
2
1
∫ ui u j ∂x i dV = ∫ && 2 u j ∂x i + ui2 ∂x j ))dV = ∫ && 2 ∂xi j ))dV = ∫ 2 ui2 u j n j dA = 0
CV j CV % j j( CV % j ( CS

€ Here CS is the control surface, and nj are the components of the outward normal vector from the
control surface. And, ∂uj/∂xj = 0 = ∂Uj/∂xj has been used along with Gauss' theorem, and the fact
that velocity fluctuations are zero on the stream wise control surfaces (no slip), or equal on the
€stream-normal control surfaces where n changes sign between the inlet and the outlet (periodic)
j
leading to cancellation.
∂p $ ∂ ( pu ) ∂u '
∫ ui ∂x dV = ∫ & ∂x i + p ∂x i )dV = ∫ pui n i dA = 0 .
CV i CV % i i( CS

∂u ∂ $ ∂u ' ∂u ∂u ∂u ∂u ∂u ∂u ∂ u
∫ ui ∂x 2i dV = ∫ ∂x && ui ∂x i ))dV − ∫ ∂x i ∂x i dV = ∫ ui ∂x i n j dA − ∫ ∂x i ∂x i dV = − ∫ ∂x i ∂x i dV .
CV j CV j % j( CV j j CS j CV j j CV j j

Here again incompressibility,

€ Gauss' theorem, and the no slip and periodic boundary conditions
lead to the simplifications. Thus, (%) becomes:
d 1 2 ∂U i ∂u ∂u
€ ∫ ui dV = − ρ ∫ ui u j dV − ν ∫ i i dV ,
dt CV 2 CV ∂x j CV ∂x j ∂x j

which is (11.88).

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 11.15. The process of transition from laminar to turbulent flow may be driven both by
exterior flow fluctuations and nonlinearity. Both of these effects can be simulated with the
simple nonlinear logistic map x n +1 = Ax n (1− x n ) and a computer spread-sheet program. Here, xn
can be considered to be the flow speed at the point of interest with A playing the role of the
nonlinearity parameter (Reynolds number), x0 (the initial condition) playing the role of an
external disturbance, and iteration of the equation playing the role of increasing time. The
€
essential feature illustrated by this problem is that increasing the nonlinearity parameter or
changing the initial condition in the presence of nonlinearity may fully alter the character of the
resulting sequence of xn values. Plotting xn vs. n should aid understanding in parts b) through e).
a) Determine the background solution of the logistic map that occurs when x n +1 = x n in terms of
A.
Now, get into a spread sheet program, and set up a column that computes x n+1 for n = 1 to 100 for
user selectable values of x0 and A for 0 < x0 < 1, and 0 < A < 4.
b) For A = 1.0, 1.5, 2.0, 2.9, choose a few different x0’s and€numerically determine if the
background solution is reached by n = 100. Is the flow stable for € these values of A, i.e. does it
converge toward the background solution?
c) For the slightly larger value, A = 3.2, choose x0 = 0.6875, 0.6874, and 0.6876. Is the flow
stable or oscillatory in these three cases? If it is oscillatory, how many iterations are needed for
it to repeat?
d) For A = 3.5, is the flow stable or oscillatory? If it is oscillatory, how many iterations are
needed for it to repeat? Does any value of x0 lead to a stable solution?
e) For A = 3.9, is the flow stable, oscillatory, or chaotic? Does any value of x0 lead to a stable
solution?

Solution 11.15. a) Setting xn = xn+1 = x in x n +1 = Ax n (1− x n ) , produces: x = Ax(1− x) . Thus for x

≠ 0, the equilibrium or steady-flow solution for x is x = 1− (1 A) .
b) For A = 1.0, the solution is not converged but it is heading monotonically toward x = 0. This
behavior is very much like a viscosity-dominated flow. For values of A above 1.0 up to about
€ € This “flow” is also stable and
2.0, the steady flow value of x is reached in a monotonic fashion.
€
laminar. The following plot is for A = 1.1, xo = 0.5.
!#)"

!#("

!#'"

!#&"

!#%"

!#$"

!"
!" %!" '!" )!" *!" $!!"

For values of A between 2.0 and 2.9, the iteration process still converges but there are
oscillations that become more persistent as A increases. The signature of an instability is
beginning to show. The plot below is for A = 2.9, xo = 0.9
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

$"
!#,"
!#+"
!#*"
!#)"
!#("
!#'"
!#&"
!#%"
!#$"
!"
!" %!" '!" )!" +!" $!!"

c) For A = 3.2, choose x0 = 0.6875, the equilibrium solution is maintained as shown in the
following plot.
!#+"

!#*"

!#)"

!#("

!#'"

!#&"

!#%"

!#$"

!"
!" %!" '!" )!" +!" $!!"

However, for A = 3.2 with x0 = 0.6874, or 0.6876 the iteration process becomes oscillatory with a
period of 2 iterations as shown on the two following plots. This value of A models a transitional
flow situation where both laminar and oscillatory flow states are possible.
!#," !#,"

!#+" !#+"

!#*" !#*"

!#)" !#)"

!#(" !#("

!#'" !#'"

!#&" !#&"

!#%" !#%"

!#$" !#$"

!" !"
!" %!" '!" )!" +!" $!!" !" %!" '!" )!" +!" $!!"

d) For A = 3.5 the flow is always oscillatory no matter what value of x0 is chosen, and the period
is 4 iterations. This “flow” is on its way to becoming turbulent because there are no stable
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

equilibrium solutions and the period of oscillation is longer than for A = 3.2. The plot below is
for A = 3.2 with x0 = 0.25.
$"
!#,"
!#+"
!#*"
!#)"
!#("
!#'"
!#&"
!#%"
!#$"
!"
!" %!" '!" )!" +!" $!!"

e) For A = 3.9, fluctuations occur no matter what value of x0 is chosen, but these fluctuations are
chaotic and do not repeat; the oscillation period is too large to be defined. This “flow” is
effectively turbulent. The plot below is for A = 3.9 with x0 = 0.25
$"
!#,"
!#+"
!#*"
!#)"
!#("
!#'"
!#&"
!#%"
!#$"
!"
!" %!" '!" )!" +!" $!!"
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.1. Determine general relationships for the second, third, and four central moments
(variance = σ2, skewness = S, and kurtosis = K) of the random variable u in terms of its first four
ordinary moments: u , u 2 , u 3 , and u 4 .

Solution 12.1. a) The variance = σ2 is the second central moment, so:

€€ 1 N N
€ − u ) 2 = 1 ∑ ( u 2 (n) − 2u(n)u + u 2 ) = u 2 − 2u u + u 2 = u 2 − u 2 .
σ 2 = € ∑ ( u(n)
N n=1 N n=1
The skewness = S is the third central moment, so
1 N 3 1 N 3
S = ∑ ( u(n) − u ) = ∑ ( u (n) − 3u 2 (n)u + 3u(n)u 2 − u 3 )
€ N n=1 N n=1
= u 3 − 3u 2 u + 3u u 2 − u 3 = u 3 − 3u 2 u + 2u 3 .
The kurtosis = K is the fourth central moment, so
1 N 4 1 N
K = ∑ ( u(n) − u ) = ∑ ( u 4 (n) − 4u 3 (n)u + 6u 2 (n)u 2 − 4u(n)u 3 + u 4 )
€ N n=1 N n=1
= u 4 − 4u 3 u + 6u 2 u 2 − 4u u 3 + u 4 = u 4 − 4u 3 u + 6u 2 u 2 − 3u 4 .

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.2. Calculate the mean, mean square, variance, and rms value of the periodic time
series u(t) = U + U 0 cos(ωt ) , where U , U0 and ω are positive real constants.

Solution 12.2. The given time series is periodic so time averaging over one period will yield the
desired results. €
€
1 2π ω 1 % U0 (2 π ω
Average: u= ∫ (U + U 0 cos(ωt))dt = 2π ω '&U t + ω sin(ωt)*) = U .
(2π ω ) 0 ( ) 0

1 2π ω 2
Mean Square: u 2 = ∫ (U + U 0 cos(ωt)) dt
(2π ω ) 0
€ 1 2π ω 2
= ∫
(2π ω ) 0
(U + 2U 0U cos(ωt) + U 02 cos2 (ωt))dt

€ 1 + 2 2U 0U U 02 % sin(2ωt) (.
2π ω
U 02
2
= -U t + sin(ω t) + ' t + *0 = U + .
(2π ω ) , ω 2 & 2ω )/0 2
2 1 2π ω 2
2π ω
U2
Variance: (u − u ) = ∫ (U + U 0 cos(ωt) − U ) dt = ∫ U 02 cos 2 (ωt)dt = 0 .
(2π ω ) 0 0 2
€ U 02
rms value: u2 = U 2 + .
2
And for
€ completeness,
2 U
Standard deviation: (u − u ) = 0
€ 2
The rms value and the standard deviation are not equal unless u = U = 0 .

€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.3. Show that the autocorrelation function u(t)u(t + τ ) of a periodic series u =
Ucos(ωt) is itself periodic.

Solution 12.3. The given time series is periodic so time averaging over one period will yield the
desired results. €
U 2 2π ω
u(t)u(t + τ ) = ∫ cos(ωt)cos(ω (t + τ ))dt
(2π ω ) 0
U 2 2π ω
= ∫ cos(ωt)[cos(ωt)cos(ωτ ) − sin(ωt)sin(ωτ )]dt
(2π ω ) 0
2π ω 2π ω
U2 2 U2
= cos(ωτ ) ∫ cos (ωt)dt − sin(ωτ ) ∫ cos(ωt) sin(ωt)dt
(2π ω ) 0 (2π ω ) 0

U2 ' 1 2π * U2
= cos(ωτ )) ,−0 = cos(ωτ )
(2π ω ) (2 ω + 2
And, since cos(ωt) is periodic, then u(t)u(t + τ ) is periodic.

€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.4. Calculate the zero-lag cross-correlation u(t)v(t) between two periodic series u(t)
= cos ωt and v(t) = cos(ωt + φ) by performing at time average over one period = 2π/ω. For values
of φ = 0, π/4, and π/2, plot the scatter diagrams of u vs v at different times, as in Figure 12.8.
Note that the plot is a straight line if φ = 0, an ellipse if φ = π/4, and a circle if φ = π/2; the
€
straight line, as well as the axes of the ellipse, are inclined at 45° to the uv-axes. Argue that the
straight line signifies a perfect correlation, the ellipse a partial correlation, and the circle a zero
correlation.

Solution 12.4. The given time series is periodic so time averaging over one period will yield the
desired results.
U 2 2π ω
u(t)v(t) = ∫ cos(ωt)cos(ωt + φ )dt
(2π ω ) 0
U 2 2π ω
= ∫ cos(ωt)[cos(ωt)cos φ − sin(ωt)sin φ ]dt
(2π ω ) 0
2π ω 2π ω
U2 2 U2
= cos φ ∫ cos (ωt)dt − sin φ ∫ cos(ωt) sin(ωt)dt
(2π ω ) 0 (2π ω ) 0

U2 ' 1 2π * U2
= cos φ) , − 0 = cos φ
(2π ω ) (2 ω + 2
The scatter diagrams are obtained by placing sample points from different times in a two-
dimensional (u,v)-coordinate plane. The locus of sample points is obtained by eliminating t from
using the equations for u and v:
€
v = cos(ωt + φ ) = cos ωt cos φ − sin ωt sin φ = ucos φ − 1− u 2 sin φ .
Use the two ends of this extended equality to find:
v − ucos φ = − 1− u 2 sin φ , or v 2 − 2uv cos φ + u 2 cos2 φ = (1− u 2 )sin 2 φ .
Further€simplify:
v 2 − 2uv cos φ + u 2 = sin 2 φ .
This quadratic relationship can be cast in a standard form by switching to sum, v + u, and
€ €
difference, v – u, coordinates:
$1− cos φ ' 2 $1+ cos φ ' 2 2 2
& € )(v + u) + & )(v − u) = sin φ = 1− cos φ ,
% 2 ( % 2 (
which implies:
(v + u) 2 (v − u) 2
+ =1 when φ ≠ 0 or π,
€ 2(1+ cos φ ) 2(1− cos φ )
v=u when φ = 0, and
v = –u when φ = π.
The first possibility is the equation for an ellipse having major and minor axes rotated 45° from
€
the u and v axes. The other two possibilities are just straight lines.
When φ = 0 then v = u so the resulting distribution of sample points is a straight line with
unity slope.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

In this case, each value of u is linked to a single and equal value of v. This is perfect correlation
between u and v.
(v + u) 2 (v − u) 2
When φ = π/4, then the locus of possible points becomes + = 1, which is:
2+ 2 2− 2
v

In this case, each value of u is linked to a two values of v. Here positive v is more likely with
positive u, and negative v is more likely with negative u. Thus, this situation corresponds to
partial correlation between u and v.
When φ = π/2, then the locus of possible points becomes, which is a circle:
v

In this case, each value of u is linked to a two values of v. Here positive v is equally likely with
positive or negative u, and negative v is equally likely with positive or negative u. Thus, this
situation corresponds to no correlation between u and v.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.5. If u(t) is a stationary random signal, show that u(t) and du(t) dt are uncorrelated.

Solution 12.5. a) For a stationary signal u(t), use the time-average definition of the correlation
and evaluate the integral:
€
1 +Δt 2 du 1 +Δt 2 du 2 1
lim
Δt →∞ Δt
∫ u(t) dt dt = Tlim
→∞ 2Δt
∫ dt dt = Δtlim
→∞ 2Δt
[u2 (Δt /2) − u2 (−Δt /2)] = 0 .
−Δt 2 −Δt 2
2
The final equality occurs because u is stationary and remains finite at ±∞ while the divisor of
the [,]-brackets goes to infinity.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.6. Let R(τ) and S(ω) be a Fourier transform pair. Show that S(ω) is real and
symmetric if R(τ) is real and symmetric.

Solution 12.6. Start with:

1 +∞ −iωτ
S(ω ) = ∫ e R(τ )dτ ,
2π −∞
and decompose into real and imaginary parts:
1 +∞
S(ω ) =
2 π
∫ (cosωτ − isinωτ )R(τ )dτ .
−∞
€
Since sinωτ is odd and R(τ) is even in τ, the integral over the imaginary part is zero. Thus,
1 +∞ 1 +∞
S(ω ) = ∫
2π −∞
cos(ωτ )R(τ )dτ = ∫ cos(−ωτ )R(τ )dτ = S(−ω ) ,
2π −∞
€
which clearly shows that:
(1) S(ω) is real because it can be computed from the integral of the two real functions R(τ) and
cos(ωτ), and
€
(2) S(ω) is symmetric about ω = 0 because S(ω) = S(–ω).
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.7. Compute the power spectrum, integral time scale, and Taylor time scale when
R11 (τ ) = u12 exp(−ατ 2 ) cos(ω oτ ) , assuming that α and ωo are real positive constants.

u12 +∞ −ατ 2 u12 +∞ −ατ 2 i(ω o −ω )τ

Solution 12.7. Se (ω ) = ∫ e cos(ω o τ )exp {−iωτ } d τ = ∫ e (e + e i(−ω o −ω )τ ) dτ
€ 2π −∞ 4 π −∞
The exponents of the two terms in the integrand are:
& i(ω ∓ ω o ) (ω ∓ ω o ) 2 (ω ∓ ω o ) 2 )
−ατ 2 − i(ω ∓ ω o )τ = −α (τ 2 + τ+ − +
€ ' α 4α 2 4α 2 *
2
& i(ω ∓ ω o ) ) (ω ∓ ω o ) 2
= −α (τ + + − ,
' 2α * 4α
& i(ω ∓ ω o ) )
where the top sign belongs with the first term. Let β = α ( τ + + be the new integration
' 2α *
variable:€
u12 & (ω − ω o ) 2 ) +∞ − β 2 u12 & (ω + ω o ) 2 ) +∞ − β 2
Se (ω ) = exp'− * ∫ e dβ + exp'− * ∫ e dβ .
4π α ( 4α € +−∞ 4π α ( 4α +−∞
The integral in both terms is π , so the energy spectrum is:
u12 , & (ω − ω o ) 2 ) & (ω + ω o ) 2 )/
Se (ω ) = .exp'− * + exp'− *1 .
€ 4 πα - ( 4α + ( 4α +0
The integral time scale € can be determined from (12.18) and (12.20) evaluated at ω = 0:
1 ∞ 2π π u12 + ω2 . π + ω2 .
Λ t = 2 ∫ R11 (τ )dτ = 2 Se (0) = 2 2exp,− o / = exp,− o / .
€ u# 0 2u1 u1 4 πα - 4α 0 2 α - 4α 0
The Taylor time scale defined in (12.19) involves the second derivative of R11 at τ = 0, which is
tedious to determine directly. However the given R11 is relatively easy to expand for τ << 1:
R 2 2 2 2
( 2
) 2
( [
€11 (τ ) = u1 exp(−ατ ) cos(ω oτ ) = u1 (1− ατ + ...) 1− 2 (ω oτ ) + ... = u1 1− α + 2 ω o τ + ... ,
1 1 2
] 2
)
and this is much easier to differentiate:
# d 2 R11 (τ ) &
% 2
$ dτ 'τ = 0
2
[
( = −u1 2α + ω o .
2
]
€
So,
12
+ % d 2 R11 (τ ) ( . −2u12 −1 2
λt = -−2R11 (0) '
& dτ
2 * 0
)τ = 0 /
=
−u12 [2α + ω o2 ]
= [α + 1
2
ω 2
o ] .
, €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.8. There are two formulae for the energy spectrum Se(ω) of the stationary zero-mean
signal u(t) :
2
1 +∞ 1 +T 2
Se (ω ) = ∫ R11(τ )exp{−iωτ }dτ and Se (ω ) = Tlim
2π −∞ →∞ 2πT
∫ u(t)exp{−iωt}dt .
−T 2
€ Prove these two are identical without requiring the existence of the Fourier transform of u(t) .

Solution 12.8. Start with the second formula and use t´ and t as the integration variables.
€ € 2
1 +T 2 1 +T 2 +T 2
Se (ω ) = lim
T →∞ 2πT
∫ u(t)exp{ }
−iω t dt = lim
T →∞ 2πT
∫ ∫ u(t ()u(t)exp{−iω (€t ( − t)}dt (dt .
−T 2 −T 2 −T 2

Alter the t´ integration by introducing a new integration variable τ = t´ – t. The region of

integration is a square centered at t´ = t = 0.
t´
€
T
t+
=
t´

!
t+
=
t´
!

t
T
!
t–
=
t´

t– t+

–T

The t integration goes from t– to t+. For τ < 0 (the case corresponding to the dashed line),
t– = –T/2 – τ, and t+ = T/2, so t+ – t– = T + τ = T – |τ|.
For τ > 0,
t– = –T/2 , and t+ = T/2 – τ, so t+ – t– = T – τ = T – |τ|.
The τ integration goes from –T to +T. Thus, the arguments in the integrand and the limits of
integration are altered:
1 τ = +T t= t+ −iωτ 1 τ = +T ) t= t + ,
Se (ω ) = lim ∫ ∫ u(t + τ )u(t)e dτ dt = lim ∫ + ∫ u(t + τ )u(t)dt .e−iωτ dτ .
T →∞ 2πT T →∞ 2πT
τ =−T t= t − τ =−T * t= t − -
From the figure, the difference t+ – t– = T – |τ|. Therefore,
1 τ = +T (T − τ ) ) 1 t= t+ ,
Se (ω ) = lim ∫ + ∫ u(t + τ )u(t)dt .e−iωτ dτ ,
T →∞ 2π T * T − τ t= t−
€ τ =−T -
so contents of the large parentheses is R11(τ) for T >> τ when u(t) is uncorrelated with itself at
large time lags, τ >> Λt. This leaves:
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

1 τ = +T ( τ + 1 +∞
Se (ω ) = lim ∫ *1− - R11 (τ )e−iωτ dτ = ∫ R11 (τ )e−iωτ dτ ,
T →∞ 2π T 2 π
τ =−T ) , −∞

where the final equality holds when T >> τ and R11 (τ ) → 0 for τ >> Λt.

€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.9. Derive the formula for the temporal Taylor microscale λt by expanding the
definition of the temporal correlation function (12.17) into a two term Taylor series, and
determining the time shift, τ = λt, where this two term expansion equals zero.

Solution 12.9. Expand the temporal autocorrelation:

# dR (τ ) & τ 2 # d 2 R11 (τ ) &
R11 (τ ) = R11 (0) + τ % 11 ( + % ( + ...
$ dτ 'τ = 0 2 $ dτ 2 'τ = 0
The autocorrelation function is even, so the terms involving odd derivatives are zero. Thus, the
appropriate two-term expansion is:
τ 2 # d 2 R11 (τ ) &
€ R11 (τ ) = R11 (0) + % ( + ...
2 $ dτ 2 'τ = 0
Truncate the expansion, evaluate it at t = λt, and set it equal to zero to find:
λ2 $ d 2 R11(τ ) '
R11 ( λt ) = 0 = R11 (0) + t & ) + ...
€ 2 % dτ 2 (τ = 0
Now solve for λt to reach:
% d 2 R11 (τ ) ( % 1 d 2 R11 (τ ) ( % d 2 r11 (τ ) (
λ2t = −2R11(0) ' 2 * = −2 ' 2 * = −2 ' 2 * ,
€ & dτ )τ = 0 & R11 (0) dτ )τ = 0 & dτ )τ = 0
which duplicates (12.19).

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.10. When x, r, and k1 all lie in the stream-wise direction, the wave number spectrum
S11 (k1 ) of the stream-wise velocity fluctuation u1 (x) defined by (12.45) can be interpreted as a
distribution function for energy across stream-wise wave number k1. Show that the energy-
weighted mean-square value of the stream-wise wave number is:
1 +∞ 1 & d2 ) 2
€ k12 ≡ 2 ∫ k12 S11 (k1 )dk1€= − 2 ( 2 R11 (r)+ , and that λ f = 2 k1 .
u −∞ u ' dr *r= 0

Solution 12.10. The relationship between the stream-wise autocorrelation function and the
stream-wise spatial power spectrum is the Fourier inverse of € (12.45):
€ +∞
R11 (r) = ∫ S11 (k1 )e +ik1 r dk1 .
−∞
Differentiate this twice with respect to r, and evaluate at r = 0:
" d 2 R11 (r) % " +∞ " d 2 +ik r % % " +∞ 2 +ik1 r
% +∞

$ 2 ' = $ ∫ S11 (k1 )$ 2

e 1
' dk1 ' = $ ∫ −k1 S11 (k1 )e dk1 ' = ∫ −k12 S11 (k1 )dk1 .
# dr & r= 0 # −∞ € # dr & & r= 0 # −∞ & r= 0 −∞
Divide by the mean-square fluctuation u"2 and multiply by –1 to get the desired form:
1 # d 2 R11 (r) & 1 +∞ 2
− 2% 2 ( = k S (k )dk1 = k12
2 ∫ 1 11 1
€ u $ dr ' r= 0 u −∞
€
For, the specified geometry, R11 is the longitudinal autocorrelation function, so from (12.39)
1 # d 2 R11 (r) & # d 2 f (r) & 2
− 2% 2 ( = % 2 (
= 2,
€ u $ dr ' r= 0 $ dr ' r= 0 λ f
and this can be combined with the mean-square wave number result to find
λ f = 2 k12 .
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.11. In many situations, measurements are only possible of one velocity component
at one point in a turbulent flow, but the flow has a non-zero mean velocity and moves past the
measurement point. Thus, the experimenter obtains a time history of u1 (t) at fixed point. In
order to estimate spatial velocity gradients, Taylor’s frozen-turbulence hypothesis can be
∂u 1 ∂u1
invoked to estimate a spatial gradient from a time derivative: 1 ≈ − where the “1”-axis
∂x1 U1 ∂t
€
must be aligned with the direction of the average flow, i.e. Ui = (U1, 0, 0). Show that this
approximate relationship is true when ui ui U1 << 1, p ~ ρu12 , and Re is high enough to neglect
the influence of viscosity. €

Solution 12.11. The place to start is with the “1”-component of the Navier-Stokes equation for
€ € & ∂ 2u ∂ 2u ∂ 2u )
∂u1 ∂u1 ∂u1 ∂u 1 ∂p
an incompressible fluid: + u1 + u2 + u3 1 = − + ν ( 21 + 21 + 21 + .
∂t ∂x1 ∂x 2 ∂x 3 ρ ∂x1 ' ∂x1 ∂x 2 ∂x 3 *
Rearrange this equation to isolate the terms with ∂u1 ∂x1 and ∂u1 ∂t .
∂u ∂u ∂u ∂ u 1 ∂p & ∂ 2u ∂ 2u ∂ 2u )
u1 1 + = − 1 − u2 1 − u3 1 − + ν ( 21 + 21 + 21 +
€ ∂x1 ∂t ∂x 2 ∂x 3 ρ ∂x1 ' ∂x1 ∂x 2 ∂x 3 *
Set u1 = U1 + u1" in the two terms, and move € parts of€ them to the other side of the equation:
∂u1# ∂u1# ∂u1 ∂u1 ∂u1 & ∂U1 ∂U1 1 ∂p ) & ∂ 2 u1 ∂ 2 u1 ∂ 2 u1 )
+ U1 + = − u1# − u2 − u3 −( + U1 + + + ν( + + +
∂t€ ∂x1 ∂x1 ∂x 2 ∂x 3 ' ∂t ∂x1 ρ ∂x1 * ' ∂x12 ∂x 22 ∂x 32 *
€ Divide by U1, and note that when Ui = (U1, 0, 0), then u2 = u"2 and u3 = u"3 :
∂u1# 1 ∂u1# u1# ∂u1 u#2 ∂u1 u#3 ∂u1 1 & ∂U1 ∂U1 1 ∂p ) ν & ∂ 2 u1 ∂ 2 u1 ∂ 2 u1 )
+ =− − − − ( + U1 + ++ ( + + +.
€ ∂x1 U1 ∂t U1 ∂x1 U1 ∂x 2 U1 ∂x 3 U1 ' ∂t ∂x1 ρ ∂x1 * U1 ' ∂x12 ∂x 22 ∂x 32 *
When Re is high the terms multiplied by ν will € be small € and can be dropped. Now decompose
the pressure into average and fluctuating parts:
∂u1# 1 ∂u1# u# ∂u u# ∂u u# ∂u 1 & ∂U ∂U 1 ∂p 1 ∂p# )
€ + = − 1 1 − 2 1 − 3 1 − ( 1 + U1 1 + + +
∂x1 U1 ∂t U1 ∂x1 U1 ∂x 2 U1 ∂x 3 U1 ' ∂t ∂x1 ρ ∂x1 ρ ∂x1 *
The first three terms inside the parentheses can be simplified using the “1”-direction RANS
∂U ∂U 1 ∂p ∂ u1%u%j
equation for this situation: 1 + U1 1 = − − ,
€ ∂t ∂x1 ρ ∂x1 ∂x j
∂u1# 1 ∂u1# u# ∂u u# ∂u u# ∂u 1 & ∂ u#u# 1 ∂p# )
+ = − 1 1 − 2 1 − 3 1 − ((− 1 j + +.
∂x1 U1 ∂t U1 ∂x1 U1 ∂x 2 U1 ∂x 3 U1 ' ∂x j ρ ∂x1 +*
When p" ~ ρu1"2 ,€then every term on the right side contains a factor like u"i u"i U1 , either inside
or outside of a differentiation, thus
€ ∂u1 1 ∂u1
+ ≅ 0 when u"i u"i U1 → 0
∂x1 U1 ∂t
€ €

€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.12. a) Starting from (12.33), derive (12.34) via an appropriate process of Reynolds
decomposition and ensemble averaging.
b) Determine an equation for the scalar fluctuation energy = 12 Y "2 , one half the scalar variance.
c) When the scalar variance goes to zero, the fluid is well mixed. Identify the term in the
equation from part b) that dissipates scalar fluctuation energy.
€ density flow by dividing by ρ :
Solution 12.12. a) First simplify (12.33) for constant m

∂ ∂ ∂ % ∂ ˜ ( ∂Y˜ ∂ ˜ ∂ % ∂Y˜ (
ρ ˜
Y + ρ ˜
( m ) ∂x ( m j ) ∂ x ' m m ∂x *
Yu˜ = ' ρ κ Y * → + ( j ) ∂x ' m ∂x ** .
Yu˜ = 'κ
∂t j j & j ) ∂t ∂x j j & j)

Insert the Reynolds decomposition for the fluid velocity and the mass fraction, u˜ j = U j + u j and
Y˜ = Y + Y " , into the constant-density passive-scalar conservation equation:
€ ∂ ∂ ∂ % ∂ (
∂t
(Y + Y #) +
∂x j
((Y + Y #)(U j + u j )) = ''κ m
∂x j & ∂x j€
(Y + Y #)** . (&)
)
€ Average the equation, and drop terms that involve first-order averages of the fluctuating
components, Y " = 0 = u j ,
€ ∂ ∂ ∂ % ∂ (
∂t
(Y + Y #) +
∂x j
((Y + Y #)(U j + u j )
) = ' κ
∂x j '& ∂x j
m (Y + Y #) ** ,
)
€ ∂ ∂ ∂ % ∂ (
∂t
(Y + Y #) +
∂x j
(Y U j + Y u j + Y #U j + Y #u j ) = ''κ m
∂ x j & ∂x j
(Y + Y #)** ,
)
€ ∂Y ∂ ∂ % ∂Y (
+
∂t ∂x j
(Y U j + Y #u j ) = ''κ m
∂x j & ∂x j *)
* , or (†)

€ ∂Y ∂Y ∂ & ∂Y )
+Uj = ((κ m − Y %u j ++ .
∂t ∂x j ∂x j ' ∂x j *
where ∂Uj/∂xj = 0 has€been used to reach the final form which is identical to (12.34).
b) Generate an equation for the scalar fluctuation, by subtracting (†) from (&) to produce:
& #)
€∂Y + ∂ (Y #U j + Y u j + Y #u j − Y #u j ) = ∂ ((κ m ∂Y ++ , or
#
∂t ∂x j ∂x j ' ∂x j *
∂Y # ∂Y # ∂Y ∂Y # ∂ ∂ & ∂Y # )
∂t
+Uj
∂x j
+ uj
∂x j
+ uj −
∂ x j ∂x j
( j ) ∂x ((κ m ∂x ++,
Y u
# =
j ' j*
€j = 0 and ∂uj/∂xj = 0 have been used to reach the second version. Multiply the lower
where ∂Uj/∂x
equation by Y´ to find:
& )
€ Y " ∂Y " + U jY " ∂Y " + u jY " ∂Y + u jY " ∂Y " − Y " ∂ (Y "u j ) = Y " ∂ (κ m ∂Y " + or
∂t ∂x j ∂x j ∂x j ∂x j ∂x j (' ∂x j +*
∂ 1 2 ∂ 1 2 ∂Y ∂ 1 2 ∂ ∂ & ∂Y # )
( ) j ∂x ( 2 ) j ∂x j ∂x ( 2 ) ∂x ( j ) ∂x ((κ m ∂x ++ ,
∂t 2
Y # + U Y # + u Y # + u Y # − Y # Y u
# = Y #
j j j j j ' j*

and€average the result to find:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

& ∂Y # )
∂
∂t
( Y# ) + U
1
2
2
j
∂
∂x j
( Y # ) + u Y # ∂∂xY + u ∂∂x ( Y # ) − Y # ∂∂x (Y #u ) = Y # ∂∂x
1
2
2
j
j
j
j
1
2
2

j
j ((κ m
j '
+.
∂x j +*
The final two terms on the left are zero because Y " = u j = 0 , and the term on the right can be
rearranged to produce:
€ ∂ 1 2 ∂ 1 2 ∂Y ∂ % ∂Y # ( ∂ % ∂ 1 2( ∂Y # ∂Y #
∂t 2 ( )
Y # + U j
∂x j 2
Y # ( )
+ u j Y # = Y # '
'κ m *
* = '
'
∂x j € ∂ x j & ∂x j ) ∂x j & ∂x j
κ m 2 ( )
Y # ** − κ m
) ∂x j ∂x j
.

The terms in this equation for the scalar-fluctuation energy are interpreted as follows:
∂ 1 2 ∂ 1 2 ∂Y ∂ & ∂ 1 2) ∂Y # ∂Y #
€ ∂t 2
Y # +( )
U j
∂x j 2
Y # ( )
= −u j Y # + (
(κ m
∂x j ∂x j ' ∂x j 2 ( )
Y # ++ − κ m
* ∂x j ∂x j
.

Time rate of change Production diffusive transport dissipation

following the mean flow
c) The final term in the last equation dissipates scalar fluctuations. It is always negative and it is
largest
€ where the scalar gradients are largest.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.13. Measurements in an atmosphere at 20 °C show an rms vertical velocity of wrms =

1 m/s and an rms temperature fluctuation of Trms = 0.1°C. If the correlation coefficient is 0.5,
calculate the heat flux ρ cp wT ! .

Solution 12.13. The heat flux will be:

$ wT # ' −3 2 −2 −1 −1
ρCp wT # = ρCp w rmsTrms& ) = (1.2kgm )(1004m s K )(1ms )(0.1°C)(0.5)
% w rmsTrms (
= 60.24Wm−2
where the factor in large parentheses is the correlation coefficient between vertical velocity
fluctuations and temperature fluctuations.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.14. a) Compute the divergence of the constant density Navier-Stokes momentum
∂u ∂u 1 ∂p ∂ 2 ui
equation i + u j i = − +ν to determine a Poisson equation for the pressure.
∂t ∂x j ρ ∂x i ∂ x j ∂x j
∂ 2G −1
b) If the equation = δ (x j − x˜ j ) has solution: G(x j , x˜ j ) = , then use the
∂x j ∂x j 4 π (x j − x˜ j ) 2
€ ρ 1 ∂2
result from part a) to show that: P(x j ) =
4π
∫
x!
(
(x j − x! j ) ! j∂ x!i
2 ∂x
)
UiU j + ui u j d 3 x! in a turbulent

€ €
flow.

∂ui ∂u 1 ∂p ∂ 2 ui
Solution 12.14. a) Start with + uj i = − +ν and take the divergence (i.e.
∂t ∂x j ρ ∂x i ∂ x j ∂x j
∂
apply to both sides of the equation). This produces:
∂x i
2 2
€ ∂ ∂ui + ∂u j ∂ui + u ∂ ∂ui = − 1 ∂ p + ν ∂ ∂ui
j
∂t ∂x i ∂x i ∂ x j ∂x j ∂x i ρ ∂x i∂x i ∂x j ∂x j ∂x i
€ The first, third, and final terms are zero in an incompressible flow, leaving
∂2 p ∂u ∂u
= −ρ j i .
€ ∂x i∂x i ∂x i ∂x j
b) This is merely an application of a Green’s function solution in an unbounded domain. Start
∂ 2G ∂u ∂ u
with = δ (x j − x˜ j ) and multiply by −ρ j i and integrate over all x˜ -space:
∂x j ∂x j € ∂x˜ i ∂x˜ j
2
∂u ∂u ∂ G 3 ∂u ∂u ∂u ∂u
−ρ ∫ ! j i d x! = −ρ ∫ δ (x j − x! j ) j i d 3 x! = −ρ j i ,
all x !
∂ xi ∂ x! j ∂ x j∂ x j all x! ∂ x!i ∂ x! j € ∂ xi ∂ x j
€ where the final equality is achieved € through the sifting property of the Dirac δ-function. Inside
the first integral, the differentiations on the x-coordinates can be brought outside the integral
because the integral involves the x˜ -coordinates. Thus:
∂2 ∂u ∂u ∂u ∂u
− ∫ ρ j i Gd 3 x! = −ρ j i
∂ x j∂ x j all x! ∂ x!i ∂ x! j ∂ xi ∂ x j
€
A comparison of this equation with the result part a) then requires:
∂u ∂u ρ ∂ u j ∂ ui 1
p = −ρ ∫ ! j i Gd 3 x! = ∫ d 3 x!
all x
∂ x!i ∂ x! j 4π all x! ∂ x!i ∂ x! j (x j − x! j )2
Here, we note that:
∂ ∂ ∂ ∂
ui u j = u j ui + ui uj = uj ui
∂x j ∂x j ∂x j ∂x j
for incompressible flow. Thus
∂2 ∂u ∂ u ∂ ∂ui ∂u j ∂ui
ui u j = j i + u j = ,
∂x i∂x j ∂x i ∂ x j ∂x j ∂x i ∂x i ∂ x j
€
so that the pressure equation becomes:

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

ρ 1 ∂2
p= ∫ all x˜ ( ui u j ) d 3 x˜
4π (x j − x˜ j ) ˜ i∂x˜ j
2 ∂x

Now insert the Reynolds decomposition, u j = U j + u"j , for the velocity factors inside the integral:
ρ 1 ∂2
p= ∫ all x˜ (U iU j + U i u j + uiU j + u'i u'j ) d 3 x˜ .
€ 4π (x j − x˜ j ) ∂x˜ i∂x˜ j
2

€ equation:
Take the ensemble average of this
ρ 1 ∂2
p=P=
4π
∫ all x˜
(x j − x˜ j ) ∂x˜ i∂x˜ j
2
(U iU j + ui'u'j ) d 3 x˜ .
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.15. Starting with the RANS momentum equation (12.30), derive the equation for the
kinetic energy of the average flow field (12.46).
1
Solution 12.15. Start with the definition: E ≡ 2 U iU i , and the RANS equation for Ui:
∂U i ∂U 1 ∂p ∂U i ∂
∂t
+Uj i =−
∂x j ρ ∂x i
[
− g 1− α (T − T0 ) δi3 + ν ] −
∂x j ∂x j ∂ x j
u(i u(j .

Take the dot product of this equation € with Ui:

∂U i ∂U i 1 ∂p ∂U i ∂
Ui
∂t
+ U iU j
∂x j
= − Ui
ρ ∂x i
[
− gU i 1− α (T − T0 ) δi3 + νU i] ∂x j ∂x j
− Ui
∂x j
u(i u(j
€
Recognize E where it occurs:
∂E ∂E 1 ∂p ∂U i ∂
+Uj = − Ui + νU i − Ui u&i u&j .
∂t ∂x j ρ ∂x i ∂x j ∂x j ∂x j
€
Now
€ get to work on the remaining terms. The pressure term is fairly easy to get into shape:
1 ∂p 1 ∂ 1 ∂U ∂ % pU ( ∂ % pU (
− Ui =− ( pU i ) + p i = − ' i * = − ' j * ,
€ ρ ∂x i ρ ∂x i ρ ∂x i ∂x i & ρ ) ∂x j & ρ )
where the final equality comes from replacing the summed-over index "i" with "j". The body
force term is also easily managed:
ρ
€ [ ] [
−gU i 1− α (T − T0 ) δi3 = −gU 3 1− α (T − T0 ) = −gU 3 . ] ρ0
The viscous term requires more effort:
∂U i ∂ $ ∂U i ' ∂ $ ∂ U i ∂U j ∂U j ' ∂ ∂ ∂U j
νU i = νU i && )) = νU i && + − )) = 2νU i Sij − νU i
∂x j ∂x j € ∂x j % ∂x j ( ∂x j % ∂ x j ∂x i ∂x i ( ∂x j ∂x j ∂x i
∂ ∂ % ∂U j ( ∂ ∂ ∂U i
= 2νU i
∂x j
Sij − νU i ''
∂x i & ∂x j )
** = 2νU i
∂x j
Sij = 2ν
∂x j
(U i Sij ) − 2νSij
∂x j
€ ∂ ∂
= 2ν
∂x j
( U i Sij ) − 2νSij ( Sij − Ωij ) = 2ν
∂x j
(U iSij ) − 2νSij Sij
∂U j 1 # ∂ U j ∂U i & 1 # ∂U j ∂U i &
where = % + (+ % − ( ≡ Sij + Rij with Sij and Rij equal to the symmetric
∂x i 2 %$ ∂x i ∂x j (' 2 %$ ∂x i ∂x j ('
€ anti-symmetric parts of the mean velocity gradient tensor, respectively. Because of their
and
symmetry, their tensor product is zero: Sij Rij = 0 . The final form for the viscous term occurs
€ €
€ because the indices 'i' and 'j' can be exchanged since both are summed over and Sij = S ji . The
∂ ∂ ∂U
Reynolds stress term produces: −U i
∂x j
u$i u$j = −
∂x j
( )
U i u$i u$j + u$i u$j i
∂x j
€
Now reassemble the E equation: €
∂E ∂E ∂ % pU j ( ∂ ∂ ∂U i ρ
∂t
+Uj
∂x j
=− '
∂x j & ρ )
* + 2ν
∂x j
(U i Sij ) − 2νSij Sij −
∂x j
( )
U i u,i u,j + u,i u,j
∂x j
− gU 3
ρ0
€
Rearrange the € terms to achieve the final form:
∂E ∂E ∂ ' pU j * ∂U i ρ
+Uj = )− + 2νU i Sij − U i u&i u&j , = + u&i u&j − 2νSij Sij − gU 3 .
€ ∂t ∂x j ∂ x j ( ρ + ∂x j ρ0

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

The terms inside the (,)-parenthesis transfer mean kinetic energy from one location in the flow to
another. The first term on the right side is the turbulence production term; it is typically negative
in this equation. The second term on the right side is the mean-flow kinetic energy dissipation
rate. The final term represents kinetic to potential energy conversion.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.16. Derive the RANS transport equation for the Reynolds stress correlation (12.35)
via the following steps.
a) By subtracting (12.30) from (4.86), show that the instantaneous momentum equation for the
fluctuating turbulent velocity ui is:
∂ui ∂U ∂u ∂u 1 ∂p ∂ 2u ∂
+ uk i + U k i + uk i = − + ν 2i + gαT 'δi3 + ui uk .
∂t ∂x k ∂x k ∂x k ρ 0 ∂x i ∂x k ∂x k
Du j Dui ∂ ∂ ∂
b) Show that: ui + uj = ( ui u j ) + U k ( ui u j ) + (uiu j uk ) .
Dt Dt ∂t ∂x k ∂x k
c) Combine
€ and simplify the results of parts a) and b) to reach (12.35)

Solution 12.16. a) Start from (4.86):

€
∂u˜ i ∂ 1 ∂p˜ ∂ 2 u˜
+ ( u˜ j u˜ i ) = − − g 1− α (T˜ − T0 ) δi3 + ν 2i ,
[ ]
∂t ∂x j ρ 0 ∂x i ∂x j
and insert the Reynolds decompositions (12.24) to find:
∂ ∂ 1 ∂ ∂2
∂t
(U i + ui ) + ((U j + u j )(U i + ui )) = − [ ]
(P + p) − g 1− α (T + T & − T0 ) δi3 + ν 2 (U i + ui ) ,
€ ∂x j ρ 0 ∂x i ∂x j
and subtract (12.30),
∂U i ∂U i 1 ∂P 1 ∂ ' ∂U i *
∂t
+Uj
∂x j
=−
ρ 0 ∂x i
[
− g 1− α (T − T0 ) δi3 + ] ))
ρ 0 ∂x j ( ∂x j
− ρ 0 ui u j ,, ,
€ +
to reach:
∂ui ∂ 1 ∂p ∂ 2 ui ∂
∂ t
+
∂ x
( u U
j i + U u
j i + u j i)
u = −
ρ ∂ x
+ g α T &δ i3 + ν
∂ x 2
+
∂ x
ui u j .
€ j 0 i j j
Now use the fact that ∂Uj/∂xj = 0 and ∂uj/∂xj = 0, and switch the summed-over index from
"j" to "k".
∂ui ∂U ∂u ∂u 1 ∂p ∂ 2u ∂
€ + uk i + U k i + uk i = − + gαT &δi3 + ν 2i + ui uk .
∂t ∂x k ∂x k ∂x k ρ 0 ∂x i ∂ x k ∂x k
b) Use "k" as a subscript to denote the dot product within the fluid particle acceleration,
D ∂ ∂
= + u˜ k ,
Dt ∂t ∂x k
€
Du j Dui ∂u ∂u ∂u ∂u ∂ ∂
so that: ui + uj = ui j + ui u˜ k j + u j i + u j u˜ k i = ( ui u j ) + u˜ k (ui u j )
Dt Dt ∂t ∂x k ∂t ∂x k ∂t ∂x k
∂ ∂ ∂ ∂ ∂
= (€ui u j ) + (U k + uk ) ( ui u j ) = ( ui u j ) + U k ( ui u j ) + uk (ui u j )
∂t ∂x k ∂t ∂x k ∂x k
€ ∂ ∂ ∂
= ( ui u j ) + U k ( ui u j ) + (uiu j uk )
∂t ∂x k ∂x k
where the final equality follows because ∂uk/∂xk = 0.
c) The result of part a) can be written:
Dui ∂U 1 ∂p ∂ 2u ∂
€ + uk i = − + gαT &δi3 + ν 2i + ui uk .
Dt ∂x k ρ 0 ∂x i ∂x k ∂ x k
Use this twice to develop the following two equations:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Dui ∂U i u j ∂p ∂ 2 ui ∂
uj + u j uk =− + u j gαT δi3 + νu j 2 + u j
& ui uk , and
Dt ∂x k ρ 0 ∂x i ∂x k ∂x k
Du j ∂U j u ∂p ∂ 2u ∂
ui + ui uk =− i + ui gαT &δ j 3 + νui 2j + ui u j uk .
Dt ∂x k ρ 0 ∂x j ∂x k ∂x k
Add€these and average using the part b) result to find:
∂ ∂ ∂ ∂U i ∂U j
∂t
(
ui u j + U k) ∂x k
(
ui u j + )∂x k
( )
ui u j uk + u j uk
∂x k
+ ui uk
∂x k
€
u ∂p ui ∂p ∂ 2u ∂ 2u
=− j − + ν u j 2i + ν ui 2j + gα T 'u jδi3 + gα T 'uiδ j 3 .
ρ 0 ∂x i ρ 0 ∂ x j ∂x k ∂x k
Together the viscous terms allow some simplification:
∂ 2u ∂ 2u ∂ # ∂ui & ∂ui ∂u j ∂ # ∂u j & ∂u j ∂ui ∂ ui u j
2
∂u ∂u
u j 2i + ui 2j = u
% j ( − + u
% i ( − = 2
−2 i j .
€ ∂x k ∂x k ∂x k $ ∂ x k ' ∂x k ∂ x k ∂x k $ ∂ x k ' ∂x k ∂ x k ∂x k ∂x k ∂x k
Thus, the final Reynolds-stress transport equation is:
∂ ∂ ∂ ∂U i ∂U j
∂t
( )
ui u j + U k
∂x k
(
ui u j +) ∂x k
( )
ui u j uk = −u j uk
∂x k
− ui uk
∂x k
€
1 % ∂p ∂p ( ∂ 2 ui u j ∂u ∂u
= − '' u j
ρ 0 & ∂x i
+ ui ** + ν
∂x j ) ∂x k2
∂x k ∂x k
(
+ 2ν i j + gα T -u jδi3 + T -uiδ j 3 , )
and this matches (12.35).

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.17. In two dimensions, the RANS equations for constant-viscosity constant-density
turbulent boundary-layer flow are:
∂U ∂V ∂U ∂U 1 ∂ ∂ $ ∂U ' ∂
+
∂x ∂y
=0, U
∂x
+V
∂y
≅−
ρ ∂x( )
P + ρ u 2 + &ν
∂y % ∂y
− uv ) , and 0 ≅ −
( ∂y
P + ρ v2 ,( )
where x & y are the streamwise and wall-normal coordinates, U & V are the average streamwise
and wall-normal velocity components, u & v are the streamwise and wall normal velocity
fluctuations, P is the average pressure, and an overbar denotes a time average.
a) Assume that the fluid velocity Ue(x) above the turbulent boundary layer is steady and not
turbulent so that the average pressure, Pe, at the upper edge of the boundary layer can be
determined from the simple Bernoulli equation: Pe + 12 ρUe2 = const. Use this assumption, the
given Bernoulli equation, and the wall-normal momentum equation to show that:
1 ∂P dUe ∂v 2
− = Ue + .
ρ ∂x dx ∂x
b) Use the part a) result, the continuity equation, and the streamwise momentum equation to
derive the turbulent-flow von Karman boundary-layer momentum-integral equation:
τw d 2 * dU e d ∞ 2 2
= (Ue θ ) +Ue δ
ρ dx
+
dx dx 0
(
∫ v − u dy , )
∞"
U% ∞
U" U%
where: δ * = ∫ $1− ' dy , and θ = ∫ $1− ' dy . In practice, the final term is typically small
0 # Ue & 0 Ue # Ue &
enough to ignore, but the efforts here should include it.

Solution 12.17. a) Integrate the wall-normal momentum equation in the y-direction to find:
P + ρ v!2 = f (x) = Pe (x) , where f(x) is a function of integration. The equality f(x) = Pe(x) follows
when P is evaluated at the edge of the boundary layer where v´ = 0. Differentiate this equation in
the x-direction and use the Bernoulli equation for Pe(x):
∂ ∂P dUe 1 ∂P dUe ∂v#2
∂x
( )
P + ρ v"2 = e = −ρUe
∂x dx
, or −
ρ ∂x
= Ue
dx
+
∂x
.

b) Multiply the continuity equation by U, add this to the horizontal momentum equation, and
insert the part a) result for ∂P/∂x to reach:
∂U 2 ∂UV dUe ∂ 2 ∂ $ ∂U '
∂x
+
∂y
−Ue −
dx ∂x
( )
v# − u#2 = &ν
∂y % ∂y
− u#v#) .
(
Integrate this equation in the vertical direction from y = 0 to y = ∞. Here, ∫(∂UV/∂y)dy = UV with
UV = UeVe as y → ∞ and UV = 0 on y = 0. Plus, ∫(∂/∂y)(ν∂U/∂y – u!v! )dy = ν∂U/∂y – u!v! with
∂U/∂y = u!v! = 0 as y → ∞ , and ν∂U/∂y = τw/ρ & u!v! = 0 on y = 0. Therefore, the horizontal
∞$ '
∂U 2 dU ∂ τ
momentum equation becomes:
0
( )
∫ &% ∂x −Ue dxe − ∂x v#2 − u#2 )(dy +UeVe = − ρw . (†)
∞
The continuity equation for the average flow implies: Ve = − ∫ (∂U ∂x )dy , so
0
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

∞
∂U d % ∞ ( dU ∞
UeVe = −Ue ∫ dy = − 'Ue ∫ Udy * + e ∫ Udy .
0 ∂x dx & 0 ) dx 0
Combine this with (†), rearrange the integrals on the left side, and pull ∂/∂x outside the integrals
(where it changes to d/dx):
d ∞ 2 dUe ∞ τ
dx 0
∫ U (−UU e − v" 2
+ u)
" 2
+ ∫
dx 0
(U −Ue )dy = − w .
ρ
∞"
U% ∞
U" U%
Change the sign of all the terms and form δ * = ∫ $1− ' dy & θ = ∫ $1− ' dy from the
0 # Ue & 0 Ue # Ue &
∞
τ d d dUe
remaining integrals to reach: w = (Ue2θ ) +
ρ dx dx
∫ (v!
0
2
)
− u!2 dy +Ue δ *
dx
.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.18. Staring from (12.38) and (12.40), set r = re1 and use R11 = u 2 f (r) , and
R22 = u 2 g(r) , to show that F(r) = u 2 ( f (r) − g(r))r−2 and G(r) = u 2 g(r) .

Solution 12.18. For homogeneous isotropic turbulence, the spatial correlation function is given
€
by (12.40):
€ € €
Rij = F(r)ri rj + G(r)δij .
When r = (r1, r2, r3) = (r, 0, 0), the longitudinal correlation given by (12.38) is:
u 2 f (r) = R11 = F(r)r1r1 + G(r) = F(r)r 2 + G(r) .
where u 2 is the velocity variance
€ (it is independent of direction), and r1 = r in this case. When r
= (r1, r2, r3) = (r, 0, 0), the lateral correlation given by (12.38) is:
€ u 2 g(r) = R22 = F(r)r2 r2 + G(r) = 0 + G(r) .
€ because r2 = r in this case. This equation implies:
G(r) = u 2 g(r) ,
so the longitudinal correlation result becomes:
€
u 2 f (r) = F(r)r 2 + u 2 g(r) .
Solve this for F(r) to find:
€
u2
F(r) = 2 ( f (r) − g(r)) .
€ r

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.19. a) Starting from Rij from (12.39), compute ∂Rij ∂rj for incompressible flow.
b) For homogeneous-isotropic turbulence use the result of part a) to show that the longitudinal,
f (r) , and transverse, g(r) , correlation functions are related by g(r) = f (r) + ( r 2)( df (r) dr) .
c) Use part b), and the integral length scale and Taylor
€ microscale definitions to find 2Λ g = Λ f
and 2 λg = λ f .
€ € €
∂ ∂ €
Solution 12.19. a) Rij = ui (x) u j (x + r) = 0 because the fluctuating velocity field is
€ ∂r j ∂r j
incompressible.
∂ ∂ % rr (
b) Rij = 0 = u 2 &( f (r) − g(r)) i 2j + g(r)δij ) . At this point with r = r12 + r22 + r32 , so the
∂r j € ∂r j ' r *
differentiations & summations can get completely out of hand unless they are completed in a
systematic manner. The following relationships are needed:
∂ r ∂g(r) r ∂ # 1 &€ 2 rj rj ∂r
€ ( f (r) − g(r)) = ( f $ − g$) j , = g# j , % 2 ( = − 3 = −2 4 , and i = δij ,
∂r j r ∂r j r ∂r j $ r ' r r r ∂r j
where a prime denotes differentiation of the function with respect to its argument. First rewrite
the main equation in terms of a dot product so that the operations and the various terms are more
obvious:
€ # € € €
r12 r1r2 rr &
*( f (r) − g(r)) 2 + g(r) ( f (r) − g(r)) 2 ( f (r) − g(r)) 1 23 *
* r r r *
#∂ ∂ ∂ &* r2 r1 r22 r2 r3 *
0=$ '$ ( f (r) − g(r)) 2 ( f (r) − g(r)) 2 + g(r) ( f (r) − g(r)) 2 ' .
% ∂r1 ∂r2 ∂r3 (* r r
2
r *
* ( f (r) − g(r)) 3 1 r r r r r
*% ( f (r) − g(r)) 2 3 2
( f (r) − g(r)) 2 + g(r)**
3
r2 r r (
The outcome will be a column vector, but only the first entry of this vector is needed to
determine an equation relating f and g because all three directions are equivalent in homogenous
isotropic turbulence. Using the equations above, the first entry of the column vector will be:
€ r r2 r rr r rr r r
0 = ( f " − g") 1 12 + ( f " − g") 2 2 21 + ( f " − g") 3 3 21 − 2( f − g)r12 14 − 2( f − g)r1r2 24
rr r r r r r r
r3 2r1 r1 r1 r1
− 2( f − g)r1r3 4 + ( f − g) 2 + ( f − g) 2 + ( f − g) 2 + g"
r r r r r
Combine like terms:
r r r r r r
0 = ( f " − g") 1 − 2( f − g) 12 + 4( f − g) 12 + g" 1 = f " 1 + 2( f − g) 12
r r r r r r
€ Continue combining, and divide out the common factor of r1/r to reach a single equation
involving f and g:
2 r
€ 0 = f " + ( f − g) , or g = f + f " .
r 2
r df
Therefore, f − g = −
2 dr
€ €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

+ r df (r) ri rj $ r df (r) ' . + r df (r) $ ri rj '.

Rij (r) = u"2 ,− 2
+ & f (r) + )δ ij / = u " 2
, f (r)δ ij + &δ ij − )/ .
- 2 dr r % 2 dr ( 0 - 2 dr % r 2 (0
∞ ∞ ∞
& r ) & 1 d f) 1 ∞ 1∞
c) Λ g = ∫ g(r)dr = ∫ ( f + f %+dr = ∫ ( f + (rf ) − + dr = [ rf ] 0 + ∫ f (r)dr = 2Λ f
0 0
' 2 * 0
' 2 dr 2* 2 2 0
€ For the Taylor length scale, compute two derivatives of g: g" = f " + 1 f " + r f "", so
2 2
3 1 r r
€ g"" = f "" + f "" + f """ = 2 f "" + f """. Now evaluate at r = 0 to find g""(0) = 2 f ""(0) because f
2 2 2 2
and g are even functions f """(0) = 0 . Finally, use the definition of the Taylor scales and the
current results, €

2
$d 2 f ' 2 2
$ d 2g' 2 € 2 1 1
€ λ f = −2 & 2 ) = and λg = −2 & 2 ) = = = λ2f , to find: λg = λf .
% dr €(r= 0 f **(0) % dr (r= 0 g**(0) 2 f **(0) 2 2

€ € €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.20. In homogeneous turbulence: Rij (rb − ra ) = ui (x + ra )u j (x + rb ) = Rij (r) , where

r = rb − ra .
a) Show that (∂ui (x) ∂x k )(∂u j (x) ∂x l ) = −(∂ 2 Rij ∂rk∂rl ) r= 0 .
b) If the flow is incompressible and€
isotropic, show that
€ 2 2
−(∂u1 (x) ∂x1 ) = − 12 (∂u1 (x) ∂x 2 ) = +2(∂u1 (x) ∂x 2 )(∂u2 (x) ∂x1 ) = u 2 ( d 2 f dr 2 ) r= 0
[Hint:
€ expand f(r) about r = 0 before taking any derivatives.]

€Solution 12.20. a) Start with Rij (rb − ra ) = ui (x + ra )u j (x + rb ) and take the divergence with
respect to the r-variables.
∂ ∂ ∂ ∂
Rij (rb − ra ) = Rij (r) = ui (x + ra ) u j (x + rb ) = ui (x + ra ) u j (x + rb )
∂rb,k ∂rk ∂rb,k ∂x k
€
∂2 ∂2 ∂ ∂ ∂ ∂
Rij (rb − ra ) = − Rij (r) = ui (x + ra ) u j (x + rb ) = ui (x + ra ) u j (x + rb )
∂ra,l∂rb,k ∂rl∂rk ∂ra,l ∂rb,k ∂x l ∂x k
€Now take the limit as the r-variables go to zero:
∂2 ∂ ∂ ∂u ∂u
− Rij (0) = ui (x) u j (x) = i j .
€ ∂rl∂rk ∂x l ∂x k ∂x l ∂x k
b) First choose i = j = k = l = 1, and evaluate the formula from part a) using the results of
Exercise 12.18 part b):
2
2 * 2 -
€ # ∂u1 & = − ∂ R (0) = −u 2 ∂ + f (r) + r df (r) #1− r1 &. .
2

% ( 11 % (
$ ∂x1 ' ∂ 2 r1 ∂ 2 r1 , 2 dr $ r 2 '/
r 2 d 2 f (0) r2
Use the hint and expand f(r) about r = 0: f (r) = 1+ + ... = 1+ f ""(0) + ...,
2 dr 2 2
€ # ∂u1 & 2 2 ∂
2 +
r2 r2 # r12 & .
% ( = −u 2
, 1+ f **(0) + f **(0)%1− 2 ( + .../ .
$ ∂x1 ' ∂ r1 - 2 2 $ r ' 0
€
Simplify before starting the differentiation.
2
# ∂u1 & 2 ∂
2 +
2 r12 . ∂ + r .
% ( = −u 2
, 1+ r f **(0) − f **(0) + .../ = − u*2 ,2rf **(0) 1 − r1 f **(0) + .../
$ ∂x€1 ' ∂ r1 - 2 0 ∂r1 - r 0
# ∂u1 & 2 2 ∂
% ( = −u {2r1 f **(0) − r1 f **(0) + ...} = −u 2 f **(0) .
∂
$ 1'x ∂ r1
€Second choose i = j = 1 and k = l = 2, and use the same expansion
# ∂u1 & 2 2 ∂
2 +
2 r12 . 2 ∂ + r .
% ( = −u 2 ,1+ r f **(0) − f **(0) + .../ = −u ,2rf **(0) 2 + .../ = −2u 2 f **(0)
€ $ ∂x 2 ' ∂ r2 - 2 0 ∂r2 - r 0
Third choose, i = k = 1 and j = l = 2, but this time the expansion is different because δij = 0.
# ∂u1 &# ∂u2 & ∂2 ∂ 2 * r df (r) # r1r2 &- 2 ∂
2
* r1r2 - u2
% (% (=− R12 (0) = −u 2 + %− 2 (. = u + f 00(0) + .... = f 00(0)
€ $ ∂x 2 '$ ∂x1 ' ∂r1∂r2 ∂r1∂r2 , 2 dr $ r '/ ∂r1∂r2 , 2 / 2
Putting these three cases together produces the desired result:

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

2 2
−(∂u1(x) ∂x1 ) = − 12 (∂u1(x) ∂x 2 ) = +2(∂u1 (x) ∂x 2 )(∂u2 (x) ∂x1 ) = u 2 ( d 2 f dr 2 ) r= 0 .

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.21. The turbulent kinetic energy equation contains a pressure-velocity correlation,
K j = p(x)u j (x + r) . In homogeneous isotropic turbulent flow, the most general form of this
correlation is: K j = K(r)rj . If the flow is also incompressible, show that K(r) must be zero.

Solution 12.21. Compute the divergence of Kj:

€
€ ∂ ∂
K j = p(x) u j (x + r) = 0 .
∂r j ∂r j
because the flow is incompressible. Now insert the homogeneous-isotropic form of Kj.
∂ dK rj ∂r dK
0= (K(r)rj ) = rj + K j = r + 3K .
€ ∂ r j dr r ∂ r j dr
Use the first and last parts of this equality to find:
1 dK 3
=− .
€ K dr r
Integrate this equation:
const
ln(K) = −3ln(r) + const , and then exponentiate to reach: K = 3 .
r
However, const must be zero because € K must remain finite as r → 0. Thus, K(r) = 0 in
incompressible homogeneous-isotropic turbulence.
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.22. The velocity potential for two-dimensional water waves of small amplitude ξo
on a deep pool can be written:
ωξ o +kx 2
φ (x1, x 2 ,t) = e cos(ωt − kx1 )
k
where x1 and x2 are the horizontal and vertical coordinates with x2 = 0 defining the average free
surface. Here, ω is the temporal radian frequency of the waves and k is the wave number.
a) Compute the two-dimensional velocity field: u = (u1,u2 ) = (∂φ ∂x1 ,∂φ ∂x 2 ) .
€
b) Show that this velocity field is a solution of the two-dimensional continuity and Navier-Stokes
equations for incompressible fluid flow.
c) Compute the strain rate tensor Sij = 12 (∂ui ∂x j + ∂u j ∂x i ) .
€
d) Although this flow is not turbulent, it must still satisfy the turbulent kinetic energy equation
that contains an energy dissipation term. Denote the kinematic viscosity by ν, and compute the
€ rate in this flow: ε = 2ν Sij Sij where the over bar implies a time average
kinetic energy dissipation
over one wave period = 2π/ω. Only time averages of even powers of the trig-functions are non-
zero, for example: cos2 (ωt − kx ) = sin 2 (ωt − kx ) = 1 2 while cos(ωt − kx ) = sin(ωt − kx ) = 0 .
e) The original potential represents a lossless flow and does not include any viscous effects.
Explain how this situation can occur when the kinetic-energy dissipation rate is not zero.
€ €
ωξ
Solution 12.22. a) Using the prescribed potential φ (x1, x 2 ,t) = o e +kx 2 cos(ωt − kx1 ) , the velocity
k
$ ∂φ ∂φ '
) = ωξ oe 2 (sin(ωt − kx1 ),cos(ωt − kx1 )) .
+kx
field is obtained by differentiation: ui = & ,
% ∂x1 ∂x 2 (
∂u ∂u2
b) For the continuity equation: 1 + € = ωξ oe +kx 2 cos(ωt − kx1 ) ⋅ [−k + k ] = 0
∂x1 ∂x 2
€ ∂ui ∂ui 1 ∂p ∂ 2 ui
The NS-momentum equation for incompressible flow, + uj =− − gδi2 + ν ,
∂t ∂x j ρ ∂x i ∂x j ∂x j
€ ∂ & ∂φ p 1 2 )
can be rewritten in mixed notation: ( + + ∇φ + gx 2 + = −ω × u + ν∇ × ω using vector
∂x i ' ∂ t ρ 2 *
identities where ω = ∇ × u is the vorticity. For € the above potential flow, ω = ∇ × u = 0 , so the
modified left side of the NS-momentum equation is all that’s left. Thus,
∂φ p 1 2
€ + + ∇φ + gx 2 = fn(t)
€ ∂t ρ 2 €
but the temporal “function” of integration can be combined with φ or set to zero without loss of
generality. When this is done, all that’s left is a Bernoulli equation that determines p from φ.
Therefore, the potential provided in this problem does produce a solution of the full NS
€
equations provided that fluctuating stresses occur at the water surface. For example, the
fluctuating surface pressure can be obtained from the above Bernoulli equation evaluated on the
water surface:
&∂φ p 1 2) ω 2ξ o p ω 2ξ o2
0 = ( + + ∇φ + + gξ = − sin(ωt − kx1 ) + s + + gξ o sin(ωt − kx1 )
' ∂t ρ 2 *x 2 = 0 k ρ 2

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

where ps is the surface pressure. The usual dispersion relationship for deep water waves,
p ω 2ξ o2
ω 2 = gk , allows the first and final terms to cancel, leaving s = − . Thus, the normal stress
ρ 2
∂u
on the surface will be: σ 22 = − ps + 2µ 2 = − ps + 2µωkξ o cos(ωt − kx1 ) , and the shear stress on
∂x 2
€ $ ∂u ∂u '
the surface will be: σ12 = µ& 1 + 2 ) = 2µωkξ o € sin(ωt − kx) . The work rate per unit span of the
% ∂x 2 ∂x1 (
flow will be: €σ12 u1 + σ 22 u2 = 2µω 2 kξ o2 − psωξ o cos(ωt − kx) . Thus the average rate of surface-
stress work input will be: σ12 u1 + σ 22 u2 = 2µω 2 kξ o2 .
c) Use the results of part a) and compute all four derivatives:
€
∂u1 ∂u1
€ = −kωξ oe +kx 2 cos(ωt − kx1 ) = kωξ oe +kx 2 sin(ωt − kx1 )
∂x ∂x 2
€1
∂u2 ∂u2
= kωξ oe +kx 2 sin(ωt − kx1 ) = kωξ oe +kx 2 cos(ωt − kx1 )
∂x1 ∂x 2
1
Now assemble
€ the 2 × 2 matrix for Sij = 2 (∂€ui ∂x j + ∂u j ∂x i ) :
%−k cos(ωt − kx1 ) k sin(ωt − kx1 ) ( 2
% 1 −tan(ωt − kx1 )(
S€ij = ωξ oe +kx 2 & )
€t − kx )*
= −k φ & )
' k sin(ωt − kx ) k cos(ω
1 1 '−tan(ωt − kx )
1 −1 *
€
where φ is the velocity potential defined in the problem statement.
d) 2νSij Sij = 2νk φ (1+ tan (ωt − kx1 ) + tan 2 (ωt − kx1 ) + 1)
4 2 2

€ = 4νk 4 φ 2 (1+ tan 2 (ωt − kx1 )) = 4νk 2ω 2ξ o2e +2kx 2 = 2ν Sij Sij = ε ≠ 0.
e) The turbulent kinetic energy equation contains an energy transfer term on the other side of the
€ equation that includes the viscosity:
. ( 1 −tan(ωt − kx1 )+1
€∂ 2ν u S = 2ν ∂ 0−k 2φωξ e +kx 2 {sin(ωt − kx ) cos(ωt − kx )})
∂x j
( i ij ) ∂x j 0/
o 1 1
*−tan(ωt − kx1 ) −1
,3
-32

∂ - 3 2' 0 *0 '∂ ∂ *' 0 *

= −2ν /k φ ( 2 +2 = 2ν ( +( 2 2 +2kx 2 +
€ ∂x j . )−tan (ωt − kx1 ) −1,1 ) ∂x1 ∂x 2 ,) kω ξ o e ,
= 2ν (2k 2ω 2ξ o2e +2kx2 ) = 4νk 2ω 2ξ o2e +2kx 2 = ε !
Thus, the kinetic energy that is dissipated is transferred from somewhere else. The most likely
source
€ is the work input from the surface stresses (see part b). The total rate of energy
dissipation per unit span can be obtained by integrating ε through the depth:
€ 0 0
4 µk 2ω 2ξ o2
∫ ρε dx 2 = 4 µ k 2 2 2
ω ξ o ∫ e +2kx 2
=
2k
= 2µkω 2ξ o2 .
−∞ −∞
This is precisely the rate at which work (per unit depth) is done by the surface forces. Thus, the
energy flow for ideal water waves can be understood using the turbulent kinetic energy equation.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.23. A mass of 10 kg of water is stirred by a mixer. After one hour of stirring, the
temperature of the water rises by 1.0 °C. What is the power output of the mixer in watts? What is
the size η of the dissipating eddies?

Solution 12.23. Start from a thermodynamic description of the power delivered to the water.
mCpΔT (10kg)(4200m 2 s−2K −1 )(1K)
Power = = = 11.67W
Δt 3600s
The kinetic energy dissipation rate per unit mass is:
Power 11.67W
=ε = = 1.167m 2 s−3 .
mass 10kg
€
Thus, the Kolmogorov scale is:
14 14
η = (ν 3 ε) = ((1.0 ×10−6 m 2 s−1 ) 3 1.167m 2 s−3 ) = 3.04 ×10−5 m .
However, the actual €dissipation scale is larger by about a factor of 30, so the dissipation length
scale is approximately one millimeter.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.24. In locally isotropic turbulence, A.N. Kolmogorov determined that the wave
14
number spectrum can be represented by S11 (k) (ν 5ε ) = Φ( kν 3 4 ε 1 4 ) in the inertial-subrange
and dissipation-range of turbulent scales, where Φ is an undetermined function.
a) Determine the equivalent form for the temporal spectrum Se (ω ) in term of the average kinetic
energy dissipation rate ε , the fluid’s
€ kinematic viscosity ν, and the temporal frequency ω.
b) Simplify the results of part a) for the inertial range of scales where ν is dropped from the
dimensional analysis.
€
c) To obtain the€ results for parts a) and b) an implicit assumption has been made that leads to the
neglect of an important parameter. Add the missing parameter and redo the dimensional analysis
of part a).
d) Use the missing parameter and ω to develop an equivalent wave number. Insist that your
result for Se only depend on this equivalent wave number and ε to recover the minus-five-thirds
law.

Solution 12.24. a) First layout the units of the various€quantities using square brackets to denote
“units of”.
length 2 1 length 2 length 2 length 2 1
[ e ]
S (ω ) = 2
⋅ = , [ ]
ε = 3
, [ ]
ν = , and [ω ] = .
time frequency time time time time
Four parameters and two independent units means there should be two dimensionless groups.
By inspection these are:
S (ω ) € ν € Se (ω ) % ν(
€ Π1 = e and Π 2 = ω , so = fn'ω € *
ν ε ν & ε)
b) If ν must drop out of the dimensional analysis then fn( ) = const ⋅ ( )−2 , so:
Se (ω ) % ν (−2 $ε '
€ € = const'ω * , or Se (ω ) = const& 2 )
ν & ε )€ %ω (
c) The implicit assumption is that the flow has € zero average velocity, or that the observer moves
at the average velocity. Thus, for a stationary observer, the average flow speed U must also be a
parameter. When this is added to the parameters used for part a), there will be another
€ €
dimensionless group, Π 3 = U νω , leading to:
Se (ω ) % ν U (
= fn'ω , * (&)
ν & ε ων )
d) Wave number€ has units of (length)–1, so ω/U has the right units to be a surrogate wave
number. If Se can only depend on ω through ω/U, then its normalization changes:
+∞ +∞
€ ∫ Se (ω )dω = u&2 = ∫ Se (ω U )d (ω U ) .
−∞ −∞
Se (ω U )
This normalization relationship, requires Se (ω ) = , so (&) becomes:
U
Se (ω U) % ν U (
€ = fn'ω , *
νU & ε ων )
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

To ensure that ω and U always occur as a ratio, the two dimensionless arguments in (&) can be
% ν (1 2 % U (−1 ω ν 3 4
used to create a dimensionless surrogate wave number: 'ω * ' * = . In addition,
& ε ) & ων ) U ε1 4
the divisor of Se (ω U) must not contain U or ω alone. Again using the dimensionless arguments
in (&) it can be converted to a dimensionally-equivalent form:
1 1 % U (% ν € (1 2 1 ω1 2ν 1 4 1
⇒ ' *'ω * = 32 12 14
= 54 14
€ νU νU & ων )& ε ) ν ω ε ν ε

Thus, the dimensionless law becomes:

Se (ω U ) %ω ν 3 4 (
€ = fn' 14 *
.
ν 5 4ε 1 4 &U ε )
When ν is eliminated from this expression, Kolmogorov’s –5/3 power law is recovered:
Se (ω U ) % ω ν 3 4 (−5 3 2 3 %ω (
−5 3

= const' 14 *
→ Se (ω U ) = const ⋅ ε ⋅ ' * .
ν 5 4ε 1 4 € &U ε ) &U )

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.25. Estimates for the importance of anisotropy in a turbulent flow can be developed
by assuming that fluid velocities and spatial derivatives of the average-flow (or RANS) equation
are scaled by the average velocity difference ΔU that drives the largest eddies in the flow having
a size L, and that the fluctuating velocities and spatial derivatives in the turbulent kinetic energy
(TKE) equation are scaled by the kinematic viscosity ν and the Kolmogorov scales η and uK [see
(12.50)]. Thus, the scaling for a mean velocity gradient is: ∂U i ∂x j ~ ΔU L , while the mean-
2 2
square turbulent velocity gradient scales as: (∂ui ∂x j ) ~ ( uK η) = ν 2 η 4 , where the “~” sign
means “scales as”. Use these scaling ideas in parts a) and d) below.
a) The total energy dissipation rate in a turbulent € flow is 2νSij Sij + 2ν Sij# Sij# , where
1 # ∂U ∂U j & 1 $ ∂u€ ∂u ' Sij" Sij"
Sij = %% i + (( and Sij" = && i + j )) . Determine how the ratio depends on the
2 $ ∂x j ∂x i ' 2 % ∂x j ∂x i ( Sij Sij
outer-scale Reynolds number: ReL = ΔU ⋅ L ν . €
b) Is average-flow or fluctuating-flow energy dissipation more important?
c) Show that the turbulent kinetic energy dissipation rate, ε = 2ν Sij$ Sij$ can be written:
€ € €
€ % ∂u ∂u ∂ 2 (
ε = ν' i i + ui u j * .
&∂x j ∂x j ∂x i∂x j )
d) For homogeneous isotropic turbulence, the second € term in the result of part c) is zero but it is
∂2 ∂ui ∂ui
non-zero in a turbulent shear flow. Therefore, estimate how ui u j depends on
€ ∂x i∂x j ∂x j ∂ x j
ReL in turbulent shear flow as means of assessing how much impact anisotropy has on the
turbulent kinetic energy dissipation rate.
e) Is an isotropic model for the turbulent dissipation appropriate at high ReL in a turbulent shear
flow? €

Solution 12.25. a) Use the scaling ideas in the problem statement and the relationship
η ∝ L ⋅ Re−3 4 to find:
2
Sij" Sij" (ν η2) ν 2 L2 L4 1 ν2 L4 34 4
~ 2
= 2 4 4
= 2 2 4 L ( ReL ) = ReL
= Re−2
Sij Sij (ΔU L) (ΔU) L η (ΔU) L η
€ b) Thus since ReL >> 1 in turbulent flow, the fluctuating-flow energy dissipation is more
important.
c) Start with the definition of the dissipation rate and the strain rate tensor:
€ 2 & 2 2)
ν & ∂ui ∂u j ) ν (& ∂ui ) ∂ui ∂u j & ∂u j ) +
ε = 2ν Sij$ Sij$ = (( + + = ( + +2 +( +
2 ' ∂x j ∂x i +* 2 ((' ∂x j +* ∂x j ∂x i ' ∂x i * +
' *
The first and last terms under the final overbar are the same, so
% ∂u ( 2 ∂u ∂u
ε = ν '' i ** + ν i j .
€ & ∂x j ) ∂x j ∂x i
The second term can be manipulated using the requirement of fluctuating flow incompressibility:

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

∂ui ∂u j ∂ $ ∂u j ' ∂ 2u j ∂ $ ∂u j '

ν =ν u
& i ) i − u + =ν & ui )
∂x j ∂ x i ∂x j % ∂x i ( ∂x j ∂x i ∂x j % ∂ x i (
∂2 ∂ % ∂ui ( ∂2 ∂2
=ν
∂x j ∂ x i
( ui u j ) − 'u j *=
∂x j & ∂ x i ) ∂x j ∂ x i
( ui u j ) =
∂x j ∂x i
ui u j .

+%€ ( 2 .
- ∂ui ∂2
Thus, ε = ν '' * + u u 0.
-& ∂x j *) ∂x i∂x j i j 0
, € /
d) Use the simple scaling ideas for the ratio to find:
∂2
ui u j 2 2
€ ∂ x ∂ x ( ΔU L) & ΔUL ) η 4 −3 / 4 4
+ 4 = ReL ( ReL ) = ReL .
i j 2 −1
2 ~ 2 2
=(
∂u ∂x ( i j (ν η ) ' ν * L
)
e) This implies that anisotropy will not be important at high Reynolds number. Therefore, an
isotropic dissipation model should be suitable for engineering purposes.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.26. Determine the self-preserving form of the average stream-wise velocity Uz(z,R)
of a round turbulent jet using cylindrical coordinates where z increases along the jet axis and R is
the radial coordinate. Ignore gravity in your work. Denote the density of the nominally-quiescent
reservoir fluid by ρ.
a) Place a stationary cylindrical control volume around the jet's cone of turbulence so that
circular control surfaces slice all the way through the jet flow at its origin and at a distance z
downstream where the fluid density is ρ. Assuming that the fluid outside the jet is nearly
stationary so that pressure does not vary in the axial direction and that the fluid entrained into the
volume has negligible x-direction momentum, show
d/2 D/2
J0 ≡ ∫ 0
ρ0U 02 2π RdR = ∫ 0
ρU z2 (z, R)2π RdR ,
where J0 is the jet's momentum flux, ρ0 is the density of the jet fluid, and U0 is the jet exit
velocity.
b) Simplify the exact mean-flow equations
∂U z 1 ∂ " %
+ ( RU R ) = 0 , and U z ∂U z +U R ∂U z = − 1 ∂ P + ν ∂ $ R ∂U z ' − 1 ∂ RuzuR − ∂ Ruz2 , ( ) ( )
∂z R ∂R ∂z ∂R ρ ∂z R ∂R # ∂R & R ∂R ∂z
when ∂P/∂z ≈ 0, the jet is slender enough for the boundary layer approximation ∂/∂R >> ∂/∂z to
be valid, and the flow is at high Reynolds number so that the viscous terms are negligible.
c) Eliminate the average radial velocity from the simplified equations to find:
∂U # 1 R ∂U & ∂U 1 ∂
U z z − $ ∫ R z dR ' z = −
∂z % R 0 ∂z ( ∂R R ∂R
Ruz uR ( )
where R is just an integration variable.
d) Assume a similarity form: U z (z, R) = UCL (z) f (ξ ) , −uz uR = Ψ(z)g(ξ ) , where ξ = R δ (z) and f
and g are undetermined functions, use the results of parts a) and c), and choose constant values
12
appropriately to find U z (z, R) = const. ( J 0 ρ ) z −1 f ( R z ) .
e) Determine a formula for the volume flux in the jet. Will the jet fluid from the nozzle be diluted
with increasing z?

R!

d! Uz(z,R)!
z!

UCL(z)!

Solution 12.26. a) Use the stationary CV shown above, consider only the steady mean flow, and
ignore turbulent fluctuations. In this case the CV momentum equation is:
∫ ρUz (U ⋅ n)dA = − ∫ Pn ⋅ e z dA ,
Surface Surface

since there are no shear stresses on any of the CV boundaries. When the fluid entrained into the
volume has negligible z-direction momentum, only the inlet (z = 0) and outlet (z) surfaces
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

provide a contribution, and if the pressure does not vary in the z-direction, then the above
equation reduces to:
d 2 D2
− ∫ "#ρU z2 $% 2π R dR + ∫ "#ρU z2 $% 2π R dR ≅ 0 .
0 z
0 0
where D is a radial location that is comfortably outside the jet's cone of turbulence. [Here it must
be noted that this equation is approximate. The jet entrains reservoir fluid and induces the
reservoir fluid to move. Thus, the assumption of negligible pressure gradient is not precisely
correct.] Noting that the first integral on the left is the jet's momentum flux and that
!"ρU z2 #$ = ρ0U 02 , the above equation simplifies to:
0
d 2 D2

∫ ρ0U 02 2π R dR ≡ J 0 ≅ ∫ ρU z2 2π R dR ,
0 0
where J0 is the jet's momentum flux.
b) The mean-flow continuity equation must be retained as is. However, the mean-flow z-
direction momentum equation can be simplified.
∂U ∂U 1 ∂ P ν ∂ " ∂U z % 1 ∂ ∂
U z z +U R z = −
∂z ∂R
+ $R
ρ ∂z R ∂R # ∂R & R ∂R
'− (
Ruz uR − )
∂z
Ruz2 . ( )
When ∂P/∂z ≈ 0, then
∂U ∂U ν ∂ " ∂U z % 1 ∂ ∂
U z z +U R z ≅ +
∂z ∂R
$R '−
R ∂r # ∂ R & R ∂ R
(
Ruz uR −
∂z
) ( )
Ruz2 .

If the jet is slender enough for the boundary layer approximation ∂/∂R >> ∂/∂z to be valid, then
∂U ∂U ν ∂ " ∂U z % 1 ∂
U z z +U R z ≅ +
∂z ∂R
$R '−
R ∂R # ∂R & R ∂R
(
Ruz uR . )
For high Reynolds number flow, the viscous terms are negligible, so
∂U ∂U 1 ∂
U z z +U R z ≅ −
∂z ∂R R ∂R
(
Ruz uR . )
c) Use the continuity equation to eliminate UR from the problem:
∂U z 1 ∂ (RU R ) 1 R ∂U
+ = 0 → U R = − ∫ R z dR
∂z R ∂R R 0 ∂z
so that the boundary layer RANS equation becomes:
∂U # 1 R ∂U & ∂U 1 ∂
U z z − $ ∫ R z dR ' z = −
∂z % R 0 ∂z ( ∂R R ∂R
(
Ruz uR ) (3*)

where R is an integration variable.

d) Use the similarity solution: U z (z, R) = UCL (z) f (ξ ) , −uz uR = Ψ(z)g(ξ ) , and ξ = R δ (z) , and
evaluate each term in (3*), converting R to ξ wherever possible
∂U ∂ $ −Rδ ! ' δ!
U z z = UCL f (UCL f ) = UCL f &UCL ! f +UCL f ! ⋅ 2 ) = UCLUCL ! f 2 −UCL2
ff !ξ ;
∂z ∂z % δ ( δ
" 1 R ∂U z % ∂U " δ 2 R R ∂U z ( R +% 1 " 1 ξ ∂U z %
# ∫ R dR & R = # ∫ d * -&UCL f . = # ∫ ξ d ξ &UCL f .
$R 0 ∂z ' ∂ R $ r 0 δ ∂ z ) δ ,' δ $ξ 0 ∂ z '
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

'U " ξ U δ" ξ * U U" f " ξ U 2 f " δ" ξ

= ( CL ∫ fξdξ − CL ∫ f ξ" 2 dξ +UCL f " = CL CL ∫ fξdξ − CL ∫ f ξ" 2 dξ
) ξ 0 ξ δ 0 , η 0 η δ 0

U U" f " ξ 2
U f " δ" ( 2 ξ
ξ +
= CL CL ∫ fξdξ − CL *( fξ ) 0 − 2 ∫ fξdξ -
ξ 0 η δ) 0 ,
€ %U U " f " U f " δ "(
2 ξ
2 δ"
= ' CL CL + 2 CL * ∫ fξdξ − UCLξff " ; and
& ξ ξ δ)0 δ
€ 1 ∂ 1 ∂ 1 ∂ RS ∂
R ∂r
(
Ruz uR = ) δ (R δ) ∂ (R δ)
((R δ )uz uR =) δξ ∂ξ
( )
ξ uz uR = − CL (ξ g) ,
δξ ∂ξ
€ Now reassemble (3*) and cancel terms.
δ " &U U " f " U 2 f " δ ") ξ δ " RSCL ∂
UCLUCL " f 2 − UCL2
ff ξ" − ( CL CL + 2 CL 2
+ ∫ fξdξ + UCLξff " = (ξg)
δ ' ξ ξ δ*0 δ δξ ∂ξ
&U U " f " U 2 f " δ ") ξ RSCL ∂
UCLUCL " f 2 − ( CL CL + 2 CL + ∫ fξdξ = (ξg)
' ξ ξ δ*0 δξ ∂ξ
2
€ Multiply by δ and divide by UCL to find:
$δUCL # ' 2 $δUCL # ' f#ξ $ RS ' 1 ∂
& ) f −& + 2δ #) ∫ fξdξ = & 2CL ) (ξg) .
€ % UCL ( % UCL (ξ 0 % UCL ( ξ ∂ξ
The three functions
€ outside the [,]-brackets only depend on the similarity variable ξ.
Thus, the flow will show (the simplest) self-similarity when:
δUCL# δUCL
# RSCL
€ = C1 , + 2δ # = C2 , and 2
= C3 , (a,b,c)
UCL UCL UCL
where C1, C2, and C3 are constants. Equations (a) and (b) require:
" C − C1 %
2δ " = C2 − C1 , or δ = $ 2 ' (z − zo ) .
€ € # € 2 &
It is easiest to choose C2 – C1 = 2, and to assume zo = 0, so that δ = z. Thus (a) implies:
UCL! C
=€ 1 , or lnUCL = C1 ln z + C4 which is the same as UCL = C5 z C1 .
UCL z
The constraint found in part a) requires:
∞ ∞
D/2 2 R $R'
J 0 ≡ 2πρ ∫ U z (z, R)RdR = 2πρδ ∫ UCL f (ξ ) d & ) = 2πρδ UCL ∫ f 2 (ξ )ξ dξ .
2 2 2 2 2
0
0 δ %δ ( 0

Substituting in the required similarity forms for δ and UCL leads to:
∞
J 0 ≡ 2πρx 2C52 x 2C1 ∫ f 2 (ξ )ξdξ .
0
2
Here the integral is just a number and 2πρC is just a product of constants, therefore 2+2C1, the
5
exponent of x, must be zero, and this implies C1 = –1 and C5 = const. J 0 ρ , where "const." is a
dimensionless constant.€When all this is drawn together, the final form for the self-similar round
jet velocity field is:
€ 12
J 0 ρ ) z −1 f ( R z ) .
U z (R, z) = const. (€
e) The volume flux Q in the jet will be:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

D2 12 12
2
∞ # J 0 & −1 2 ∞ #J &
Q = 2π ∫ U z R dR = 2πUCLδ ∫ f ξ d ξ = 2π const. % ( z z ∫ f ξ d ξ ∝ % 0 ( z .
0 0 $ρ' 0 $ρ'
Thus, the volume flux in the jet increases linearly with downstream distance, so the jet fluid will
be diluted with increasing x.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.27. Consider the turbulent wake far from a two-dimensional body placed
perpendicular to the direction of a uniform flow. Using the notation defined in the Figure, the
result of Example 12.5 may be written:
<Begin Equation>
+∞ (
FD l U(x, y) " U(x, y) % v 2 − u 2 +
ρUo2
= θ = ∫ U $#1− U '& + U 2 -- dy ,
*
−∞ *
) o o o ,
</End Equation>
where θ is the momentum thickness of the wake flow (a constant), and U(x,y) is the average
horizontal velocity profile a distance x downstream of the body.
a) When ΔU << Uo, find the conditions necessary for a self-similar form for the wake’s velocity
deficit, U(x, y) = Uo − ΔU(x) f (ξ ) , to be valid based on the equation above and the steady two-
dimensional continuity and boundary-layer RANS equations. Here, ξ = y/δ(x) and δ is the
transverse length scale of the wake.
b) Determine how ΔU and δ must depend on x in the self-similar region. State your results in
appropriate dimensionless form using θ and Uo as appropriate.

∂U ∂V
Solution 12.27. a) The steady boundary layer RANS equations are: + = 0 , and
∂x ∂y
∂U ∂U ∂ uv
U +V =− . First combine the momentum & continuity equations to eliminate V:
∂x ∂y ∂y
∂U % y ∂U ( ∂U €∂ uv
U −&∫ dy ) =−
∂x ' 0 ∂ x * ∂y ∂y
€ where y is an integration variable. Now place the assumed self-similar form for the horizontal
velocity, U(x, y) = U o − ΔU(x) f (ξ ) , and an equivalent form for the Reynolds shear stress,
u"v " = RSCL (x)g(ξ ) , into this equation:
€
∂ (U − ΔUf ) & y ∂ (U o − ΔUf ) ) ∂ (U o − ΔUf ) ∂ (RSCL g)
(U o − ΔUf ) o −'∫ dy * =−
€ ∂x (0 ∂x + ∂y ∂y
€ Evaluate the derivative and work to put as much of the equation as possible in terms of the
similarity variable ξ.
& y ) -y& y ) 0& f $) g$
€ (U o − ΔUf )(−ΔU $f + ΔUf $ 2 δ $+ − . ∫ ( −ΔU $f + ΔUf $ 2 δ $+ dy 1(−ΔU + = −RSCL
' δ * /0' δ * 2' δ* δ
δ$ f$ y 2 f $ δ$
y
g$
−(U o − ΔUf )ΔU $f + (U o − ΔUf )ΔUf ξ$ − ΔUΔU $ ∫ fdy + (ΔU ) ∫ f ξ$ dy = −RSCL .
δ δ 0 δ δ 0 δ
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Convert the integrations to the similarity variable:

ξ
δ$ 2 δ$ ξ g$
−(U o − ΔUf )ΔU $f + (U o − ΔUf )ΔUf ξ$ − ΔUΔU $f $ ∫ fdξ + (ΔU ) f $ ∫ f ξ$ dξ = −RSCL .
δ 0 δ 0 δ
Evaluate the second integral by parts, and cancel the opposite terms to find:
δ$ ξ
2 δ$ ( ξ + g$
−(U o − ΔUf )ΔU $f + U oΔUf ξ$ − ΔUΔU $f $ ∫ fdξ − (ΔU ) f $ * ∫ fdξ - = −RSCL .
€ δ 0 δ )0 , δ
Rearrange each term and place coefficients that depend only on x in square brackets.
ξ
& δ $) 2 & 2 δ $) & −RSCL )
−[U oΔU $] f + (U oΔU +ξf $ + [ΔUΔU $] f − (ΔUΔU $ + (ΔU ) + f $ ∫ fdξ = ( g$
€ ' δ* ' δ* 0 ' δ +*
When ΔU « Uo, only the first two terms on the left matter:
% δ #( %−RS (
[U oΔU #] f + 'U oΔU *ξf # ≈ ' CL *g# .
& δ) & δ )
€
€
2
% U ΔU #δ ( %U δ #( %−RS (
Now divide by (ΔU ) and multiply by δ: ' o 2 * f + ' o *ξf # ≈ ' CL
2 g
* #. This scaling of
'& (ΔU ) *) & ΔU ) '& (ΔU ) *)
€ suggests that similarity will be achieved when the factors in square brackets
the turbulent stresses
are constants.
€ U oΔU #δ U oδ " −RSCL
€2 = C1 , = C2 , and 2 = C3
.
(ΔU ) ΔU (ΔU )
b) When ΔU « Uo, the origin constraint determined from the formula provided in the problem
FD b ΔU(x)δ (x) +∞
statement becomes: 2
≈ ∫ f (ξ )dξ . Thus, similarity requires the product ΔUδ to
€ ρU o €U o −∞ €
€be independent of x. If ΔUδ = C4, then the first similarity requirement from part a) implies
ΔU # C1 1
3 =
→ 2 ∝x
, or ΔU ∝ x −1 2 and δ ∝ x +1 2 .€
(ΔU ) U C
o 4 (ΔU )
€
Here, C4 has units
€ but Uo and the fundamental length scale θ are all that’s needed to put these
laws into proper dimensionless form:
€ −1 2 €
€ U o = const1 ⋅ ( x θ ) , and δ = const 2 ⋅ (θx) +1 2 .
ΔU €

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.28. Consider the two-dimensional shear layer that forms between two steady
streams with flow speed U2 above and U1 below y = 0, that meet at x = 0, as shown. Assume a
self-similar form for the average horizontal velocity:
U(x, y) = U1 + (U 2 − U1 ) f (ξ ) with ξ = y δ (x) .
a) What are the boundary conditions on f(ξ) as y → ±∞?
∂U ∂V ∂U ∂U ∂ 2U
b) If the flow is laminar, use + = 0 and U +V = ν 2 with δ (x) = νx U1 to
€ ∂x ∂y ∂€x ∂y ∂y
obtain single equation for f(ξ). There is no need to solve this equation.
∂U ∂V ∂U ∂U ∂
c) If the flow is turbulent, use:
∂ x
+
∂ y
= 0 and U
€ ∂ x
+V
∂ y
=−( )
∂ y
uv with
2
€ €
−uv = (U 2 − U1 ) g(ξ ) to obtain a single equation involving f and g. Determine how δ must
depend on x for the flow to be self-similar.
d) Does the laminar or the turbulent
€ mixing layer grow € more quickly as x increases?
€

Solution 12.28. a) As y → –∞, f(ξ)→ 0; and as y → +∞, f(ξ)→ 1.

b) First combine the equations to eliminate V:
∂U & y ∂U ) ∂ U ∂ 2U
U −'∫ dy * =ν 2 .
∂x (−∞ ∂x + ∂y ∂y
where y is an integration variable. Now plug in the similarity form for U(x,y) with ΔU = U2 – U1:
yδ $ ( y yδ $ + 1 1
−(U1 + ΔUf )ΔUf 2 + ) ∫ ΔUf $ 2 dy ,ΔUf $ = νΔUf $$ 2
$
€ δ *−∞ δ - δ δ
Now substitute in δ (x) = νx U1 and convert to ξ in place of y:
1 2 1 (ξ + U
−(U1 + ΔUf )ΔUξf % + (ΔU ) f % ) ∫ ξf d% ξ , = νΔUf %% 1
€ 2x 2x *−∞ - νx
Cancel the
€ common factor of ΔU x and integrate by parts inside the {,}-braces:
ξf % f %( ξ +
−(U1 + ΔUf ) + ΔU )ξf − ∫ fdξ , = U1 f %%
€ 2 2* −∞ -
ξ
€ ξf $ ΔU f $
Simplify and divide by U1 to find: −
2
−
U1 2 −∞
∫ fdξ = f $$ .
∂€U & y ∂U ) ∂U ∂ 2U
c) Start with U −'∫ dy * = ν 2 and put in the similarity forms for U(x,y) and the
∂x (−∞ ∂x + ∂y ∂y
Reynolds shear stress: €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

yδ $ ( y yδ $ + 1 2 1
−(U1 + ΔUf )ΔUf $
δ 2
+ ) ∫ ΔU f $
δ 2
dy ,ΔUf $ = (ΔU ) g$ .
δ δ
*−∞ -
Convert to ξ in place of y, and rearrange the results:
δ$ 2 δ$ )ξ , 2 1
−(U1 + ΔUf )ΔU ξf $ + (ΔU ) f $* ∫ f ξ$ dξ - = (ΔU ) g$ .
€ δ δ +−∞ . δ
Integrate by parts and simplify:
ξ
δ$ 2 δ$ 2 1
−U1ΔU ξf $ − (ΔU ) f $ ∫ fdξ = (ΔU ) g$ .
€ δ δ −∞ δ
2
Divide by (ΔU ) δ and regroup the terms on the left:
U ) ΔU ξ ,
− 1 δ $+ξf $ + f $ ∫ fdξ . = g$ .
€ ΔU * U1 −∞ -
€ neither U1 nor ΔU depend on x, so the requirement for self-similarity means that δ ∝ x .
Here,
d) The turbulent shear layer ( δ ∝ x ) grows more quickly with increasing x than the laminar shear
layer ( δ ∝ x 1/2).
€
€
€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.29. Consider an orifice of diameter d that emits an incompressible fluid of density ρo
at speed Uo into an infinite half space of fluid with density ρ . With gravity acting and ρ > ρo, the
∞ ∞

orifice fluid rises, mixes with the ambient fluid, and forms a buoyant plume with a diameter D(z)
that grows with increasing height above the orifice. Assuming that the plume is turbulent and
self-similar in the far-field (z >> d), determine how the plume diameter D, the mean centerline
velocity Ucl, and the mean centerline mass fraction of orifice fluid Ycl depend on the vertical
coordinate z via the steps suggested below. Ignore the initial momentum of the orifice fluid. Use
both dimensional and control-volume analysis as necessary. Ignore streamwise turbulent fluxes
to simplify your work. Assume uniform flow from the orifice.
a) Place a stationary cylindrical control volume around the plume with circular control surfaces
that slice through the plume at its origin and at height z. Use similarity forms for the average
vertical velocity U z (z, R) = Ucl (z) f ( R z ) and nozzle fluid mass fraction
Y (z, R) = ( ρ∞ − ρ ) ( ρ∞ − ρo ) = Ycl (z)h ( R z ) to conserve the flux of nozzle fluid in the plume, and
D2
find: m o = ∫ ρ U dA = ∫ ρoY (z, R)U z (z, R)2π R dR .
source o o 0

b) Conserve vertical momentum using the same control volume assuming that all entrained fluid
enters with negligible vertical momentum, to determine:
D2
−∫ ρ U 2 dA +
source o o
∫ 0
ρ (z, R)U z2 (z, R)2π R dR = ∫ volume
g [ ρ∞ − ρ (z, R)] dV ,
where ρ = Y ρo + (1− Y )ρ∞ .
c) Ignore the source momentum flux, assume z is large enough so that YCL << 1, and use the
results of parts a) and b) to find: Ucl (z) = C1. 3 B ρ∞ z and (( ρ∞ − ρo ) ρ∞ ) Ycl (z) = C2 3 B 2 g3ρ∞2 z 5 ,
where C1 and C2 are dimensionless constants, and B = ∫ ( ρ∞ − ρo )gUo dA .
source

UCL(z)f(R/z)"
D(z)"

z!
g!
ρ∞"

R!
d!

Solution 12.29. Use the similarity forms provided,

U z (z, R) = Ucl (z) f ( R z ) and Y (z, R) = ( ρ∞ − ρ ) ( ρ∞ − ρo ) = Ycl (z)h ( R z )
to complete this exercise. Since the z-dependences of two functions are sought, it will be
necessary to consider both conservation of mass and vertical momentum.
a) Choose a conical control volume with a flat end at a height of z above the orifice that encloses
the plume. Ignoring turbulent fluxes, conservation of orifice fluid then implies:
d2 D2
m! o = ρoUoπ = ∫ ρoY (z, R)w(z, R)2π R dR .
4 0

This equation provides a constraint on the possible z-dependence of the Ycl and Ucl. Substitute in
the trial self-similar forms for Y and Uz, and change the integration variable to ξ = R/z.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

d2 D2 D 2z
ρoUoπ = 2πρoYcl (z)Ucl (z) ∫ h ( R z ) f ( R z ) R dR = 2πρoYcl (z)Ucl (z)z 2 ∫ h(ξ ) f (ξ )ξ d ξ
4 0 0

Since D is presumed proportional to z, the integral is just a pure number, so we can reduce the
above formula to:
U o d 2 = const ⋅ z 2Ycl (z)U cl (z) (1)
b) Now conserve momentum with the same control volume, and assume that all of the fluid
entrained though the conical sides of the control volume either enters horizontally and
contributes no vertical momentum or that it enters slowly enough so that its vertical momentum
contribution is negligible.€ In addition, assume the conical control volume boundary to be stress
free. Under these conditions, the integral vertical-momentum equation becomes:
D2
− ∫ ρoUo2 dA + ∫ 0
ρ (z, R)wU z2 (z, R)2π R dR = − ∫ volume gρ (z, R)dV − ∫ surface
P ( n ⋅ e z )dA .
source
c) The first integral on the right results from the gravitational body force, and the second term is
the pressure integral. The pressure on the conical control surface will be the hydrostatic pressure:
P = Po – ρ∞gz. If we assume that the flow is basically upward within the plume, the radial
momentum equation can be used to find ∂P/∂r ≈ 0 so the foregoing pressure equation can be
used at any radial location. So, use the static pressure relationship and Gauss' Theorem to
transform the pressure integral
− ∫ surface P (n ⋅ e z )dA = − ∫ volume ∇P ⋅ e z dV = + ∫ volume ρ∞ gdV .
and then combine it with the body force integral to find:
D2
− ∫ ρoUo2 dA + ∫ 0
ρ (z, R)U z2 (z, R)2π R dR = ∫ volume
g [ ρ∞ − ρ (z, R)] dV .
source
€
When the source momentum is negligible, the first term on the left can be dropped. With this, the
assumed self-similarity relations produce:
D2
∫ {ρ ∞ −Ycl (z)h(R z) ( ρ∞ − ρo )}Ucl2 (z) f 2 (R z)2π R dR ≅ ∫ gYcl (z)h(R z) ( ρ∞ − ρo ) dV
0 volume
Far above the plume origin a lot of mixing will have taken place and Ycl << 1, so ignore it inside
the first integral.
D2 z D2
ρ∞Ucl2 (z) ∫ 0 f 2 (R z)2π R dR ≅ g ( ρ∞ − ρo ) ∫ Ycl (ζ ) ∫
0 0
h(R ζ )2π R dR dζ
As before, change the variable of integration to ξ = R/z (or ξ = R/ζ in the radial integrations
where ζ is the vertical direction integration variable) and divide out the common factor of 2π:
ρ∞ z 2U cl2 (z) ∫ 0
D 2z z
f 2 (ξ )ξdξ = g( ρ∞ − ρ o ) ∫ 0 ζ 2Ycl (ζ ) [∫ D 2ζ
0 ]
h(ξ )ξdξ dζ .
On both sides, the ξ-integrations yield pure numbers, therefore:
z
ρ∞ z 2U cl2 (z) = const ⋅ g( ρ∞ − ρ o ) ∫ 0 ζ 2Ycl (ζ )dζ .
Now€substitute for ζ 2Ycl (ζ ) from (1):
1 z
ρ∞ z 2U cl2 (z) = const ⋅ g( ρ∞ − ρ o )U o d 2 ∫ 0
dζ .
U cl (ζ )
€
Assume U€cl(z) has a power law form: U cl = Az −b , integrate and solve for A and b.
z b +1
ρ∞ A 2 z 2−2b = const ⋅ g( ρ∞ − ρ o )U o d 2
€ A(b + 1)
This relationship implies: €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

& g( ρ∞ − ρ o )U o d 2 )1 3
A = const ⋅ ( + , and 2 – 2b = b +1 or b = 1/3.
' ρ∞ *
π
Use the definition of the buoyancy flux, B = ∫ (ρ
∞ − ρ o )gU o dA =
4
g( ρ∞ − ρ o )gU o d 2 , and (1) to
evaluate Ycl(z). The final results being:
€
ρ − ρo
U cl (z) = C1 3 B ρ∞ z , and Ycl (z) = C2"U o d 2 3 ρ∞ Bz 5 or ∞ Ycl (z) = C2 3 B 2 g 3 ρ∞2 z 5 ,
ρ∞
€
where the C’s are dimensionless constants. It is also possible to obtain the result for Ucl entirely
from dimensional analysis. For high Reynolds number flow far above the orifice, the parameters
€ are: Ucl, B, ρ∞, z (“High€ Re” implies that µ doesn't matter; “far above” implies severe dilution &
complete "memory loss" by the flow, therefore d and € m˙ don't matter).
o
The parameter matrix is:
Ucl B ρ
∞ z
––––––––––––––––––––––
Mass: 0 1 1 €0
Length: 1 1 -3 1
Time: -1 -3 0 0

Determine the number of dimensionless groups: 4 parameters - 3 dimensions = 1 group

Construct the dimensionless groups: Π1 = B ρ∞U cl3 z
13
Write a dimensionless scaling law: Π1 = const , or U cl = const ⋅ ( B ρ∞ z) .
The mean centerline mass fraction of orifice fluid can be obtained by conserving orifice fluid,
this amounts to finding Ycl(z)
€ from (1) and the relationship for Ucl.
€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.30. Laminar and turbulent boundary layer skin friction are very different. Consider
skin friction correlations from zero-pressure-gradient (ZPG) boundary layer flow over a flat plate
placed parallel to the flow.
τ 0.664
Laminar boundary layer: C f = 1 0 2 = (Blasuis boundary layer)
2
ρU Rex
Turbulent boundary layer: (see correlations in §12.9)
Create a table of computed results at Rex = Ux/ν = 104, 105, 106, 107, 108, and 109 for the laminar
and turbulent skin friction coefficients, and the friction force acting on 1.0 m2 plate surface in
sea-level air at 100 m/s and in water at 20 m/s assuming laminar and turbulent flow.

Solution 12.30. Merely calculate and tabulate results for air: ρ = 1.2 kg/m3 & ν = 1.5×10–5 m2/s,
and for water: ρ = 103 kg/m3 & ν = 10–6 m2/s. The forces below are in Newtons, and the
0.455
turbulent skin friction numbers were obtained from White's formula: C f ≅ 2
.
[ln(0.06Re x )]
Cf Force, air, Force, air, Force, Force,
log Re Cf laminar turbulent laminar turbulent water, lam. water, turb.
4 6.64E-03 1.11E-02 3.98E-01 6.67E-01 € 1.33E+01 2.22E+01
5 2.10E-03 6.01E-03 1.26E-01 3.61E-01 4.20E+00 1.20E+01
6 6.64E-04 3.76E-03 3.98E-02 2.26E-01 1.33E+00 7.52E+00
7 2.10E-04 2.57E-03 1.26E-02 1.54E-01 4.20E-01 5.14E+00
8 6.64E-05 1.87E-03 3.98E-03 1.12E-01 1.33E-01 3.74E+00
9 2.10E-05 1.42E-03 1.26E-03 8.51E-02 4.20E-02 2.84E+00

The laminar results fall faster with increasing Re, and the turbulent skin friction numbers are
always higher than the laminar ones. The forces in water are also much larger than those in air at
the same Reynolds number in the same flow regime.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.31. Derive the following logarithmic velocity profile for a smooth wall:
U + = (1 κ ) ln y + + 5.0 by starting from U = ( u* κ ) ln y + + const. and matching the profile to the
edge of the viscous sublayer assuming the viscous sublayer ends at y = 10.7 v/u*.

Solution 12.31. In the viscous sublayer the velocity profile is linear:

€ € U(y) = u*2 y ν .
At the presumed edge of the viscous sublayer (y = 10.7l = 10.7ν/u*), the velocity is:
ν

10.7 ν u*
U(y = 10.7lν ) = u*2 = 10.7u* .
€ ν
Fitting this condition at the edge of the logarithmic region, U = ( u* κ ) ln y + + const., leads to:
10.7u* = ( u* κ ) ln(10.7) + const., or const. = 10.7u* − ( u* κ ) ln(10.7) .
Thus, the logarithmic € law becomes:
U 1 + const. 1 1 1 1
= ln y + = ln y + + 10.7 − ln(10.7)
€ = ln y + + 10.7 − 5.8 = ln y + + 5 ,
u κ u* κ κ κ κ
€ * €
where the final numerical values have been rounded to one significant digit and a value of κ =
0.41 has been assumed.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.32. Derive the log-law for the mean flow profile in a zero-pressure gradient (ZPG)
flat plate turbulent boundary layer (TBL) through the following mathematical and dimensional
arguments.
a) Start with the law of the wall, U u* = f ( yu* ν ) or U + = f (y + ) , for the near-wall region of the
U −U $ y'
boundary layer, and the defect law for the outer region, e = F & ) . These formulae must
u* %δ (
+
overlap when y → +∞€and y/δ → 0. In this€matching or overlap region, set U and ∂U ∂y from
both formula equal to get two equations involving f and F.
b) In the limit as y+ → +∞, the kinematic viscosity must drop out of the equation that includes
€
df/dy+. Use this fact, to show that U u* = AI ln( yu* ν ) + BI as y+ → +∞ where AI and BI are
€
constants for the near-wall or inner boundary layer scaling.
c) Use the result of part b) to determine F(ξ ) = −AI ln(ξ ) − BO where ξ = y/δ, and AI and BO are
constants for the wake flow or outer boundary layer scaling.
€
d) It is traditional to set AI = 1/κ, and to keep BI but to drop its subscript. Using these new
requirements determine the two functions, fI and FO, in the matching region. Which function
€ number of the flow?
explicitly depends on the Reynolds

Ue % y(
Solution 12.32. a) Set U from both formulae equal: f I ( yu* ν ) = *
− FO ' * .
u &δ )
u* $ yu* ' 1 $ y'
Set ∂U ∂y from both formula equal: f I#& ) = − FO# & ) .
ν % ν ( δ %δ (
+
b) In the limit as y → +∞, the kinematic viscosity € must drop out of the second equation in part
−1
$ yu* ' $ yu* ' yu* df 1
€ a); thus: f I"& ) → const ⋅ & ) as → ∞ . This means: I+ = AI + , or
% ν ( %€ν ( ν dy y
+ +
f I (y ) = AI ln y + BI
c) Plug this result into the second equation from part a):
€ u* €$ yu* ' u* df I u*€ ν AI 1 $ y'
f I#& )= = AI = = − FO# & )
€ ν % ν ( ν dy + ν yu* y δ %δ (
# y& # y&
Now use the last equality and one integration to find: FO % ( = −AI ln% ( − BO
$δ ' $δ '
d) Now use the € first equation from part a), plus the results of parts b) and c) with AI = 1/κ and BI
= B, to find:
# yu* & 1 # yu* & Ue , 1 # y & / Ue # y&
fI % ( = ln% ( + B =€ * − .− ln% ( − BO 1 = * − FO % (
$ ν ' κ $ ν ' u - κ $δ ' 0 u $δ '
Work with the middle equality and introduce some δ’s on the left side:
1 % y δu* ( Ue 1 % y ( 1 $ y ' 1 $ δu* ' U 1 $ y'
ln' * + B = *
+ ln ' * + B O → ln& ) + ln& ) + B = *e + ln& ) + BO
κ & δ €ν ) u κ &δ ) κ %δ ( κ % ν ( u κ %δ (
1 % δu* ( Ue
Thus, BO = ln' * + B − * , so
κ & ν ) u
€ €

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

# yu* & 1 # yu* & # y& 1 # yu* & 1 # δu* & Ue

fI % ( = ln% ( + B , and FO % ( = − ln% ( + ln% ( + B − *
$ ν ' κ $ ν ' $δ ' κ $ ν ' κ $ ν ' u
in the overlap region. The outer function FO explicitly depends on the Reynolds number through
δu*
the term containing .
ν
€ €

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.33. For zero pressure gradient, the Von Karman boundary layer integral equation
simplifies to Cf = 2dθ/dx. Use this fact, and (12.90) to determine Cf and numerically compare this
Cf result to Cf obtained from (12.93). Do the results match well for 106 < Rex < 109? What
difference does the choice of log-law constants make? Consider (κ,B) pairs representative of the
nominal modern values for pipes (0.41, 5.2) and boundary layers (0.38, 4.2).

Solution 12.33. Compute the skin friction by differentiating the momentum thickness correlation
(12.90):
d( !U x $ +
−0.15 −0.15
dθ !U x $
Cf = 2 = 2 *0.016x # e & - = 0.032 ( 0.85) # e & = 0.0272 Re−0.15
x .
dx dx *) "Reν % -, " ν %
x
The evaluation of (12.93) is a bit more involved and requires use of (12.90), and (12.91)
rewritten in terms of Reynolds numbers:
! 7.11κ $ ! 7.11κ $
Reθ = 0.016 Re 0.85x , and Reδ* = Reθ exp " % = 0.016 Re 0.85
x exp " 0.85
%.
# ln(Reθ ) & # ln(0.016 Re x ) &
Thus, (12.93) becomes:
2.0 2.0
Cf = 2
= 2
.
"#κ −1 ln ( Reδ* ) + 3.30$% " , & 7.11κ ) / $
2κ −1 ln . 0.016 Re 0.85
x exp ' 0.85
* 1 + 3.303
2# - ( ln(0.016 Re x ) + 0 3%
The constant B doesn't matter, but κ does enter the formulation for Cf. For clarity, the
comparison should be made graphically.

4.00E'03%
3.50E'03%
3.00E'03%
2.50E'03%
2.00E'03%
1.50E'03% (12.90).%
1.00E'03% (12.93),%0.38%
5.00E'04% (12.93),%0.41%

0.00E+00%
1.00E+06% 1.00E+07% 1.00E+08% 1.00E+09%

Given that (12.90) is for a zero-pressure-gradient (ZPG) turbulent boundary layer (TBL), it
matches the result from (12.93) to within ±5% or so using the ZPG TBL value of κ. The match
between (12.90) and (12.93) using the pipe-flow value of κ is not as good; the difference is 10%
or more throughout the range of Rex considered here.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.34. Prove (12.90) and (12.91) by considering a stationary control volume that
resides inside the channel or pipe and has stream-normal control surfaces separated by a distance
dx and stream-parallel surfaces that coincide with the wall or walls that confine the flow.

Solution 12.34. The two calculations are nearly the same.

For (12.90) consider a slab-like control volume that has thickness dx in the stream-wise
direction, width B perpendicular to the mean flow, and height h that spans the inside of the
channel. For steady fully developed flow, conservation of axial momentum implies:
h h
− ∫ "#ρu 2 (y)$% B dy + ∫ "#ρu 2 (y)$% B dy = p(x)Bh − p(x + dx)Bh − 2τ w Bdx ,
x x+dx
0 0

where τw is the wall shear stress, and the factor of 2 in the final term arises because of the shear
stress on the upper and lower walls of the channel. Here the flux terms on the left side are equal
& opposite, and the pressure at x+dx may be expanded in a Taylor series to find:
0 = p(x)Bh − ( p(x) + ( dp dx ) dx ) Bh − 2τ w Bdx , or ( dp dx ) dxBh = −2τ w Bdx .
Divide by Bhdx to reach (12.90):
dp dx = −2 τ w h .
For (12.91) consider a disk-like cylindrical control volume that has thickness dx and
spans the inside of the pipe of diameter d. For steady fully developed flow, conservation of axial
momentum implies:
2 2
" 2 $ " 2 $ d d
− ∫ #ρu (R)% dA + ∫ #ρu (R)% dA = p(x)π − p(x + dx)π − π dτ w dx ,
pipe area
x
pipe area
x+dx 4 4
where τo is the wall shear stress. Here again, the flux terms on the left side are equal & opposite,
and the pressure at x+dx may be expanded in a Taylor series to find:
2 2 2
d d d
0 = p(x)π − ( p(x) + ( dp dx ) dx ) π − π dτ w dx , or ( dp dx ) dxπ = −π dτ w dx .
4 4 4
Divide by (πd2/4)dx to reach (12.91):
dp dx = − 4τ w d .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.35. The log-law occurs in turbulent channel, pipe, or boundary layer flows and
should be absent in laminar flows in the same geometries. The extent of the log-law is governed
by Reτ = δ+ = δ/lν = δu*/ν, where δ is the channel half-height (h/2), pipe radius (d/2), or full
boundary-layer thickness, as appropriate for each flow geometry.
a) For laminar channel flow, show that Reτ = (3 / 2)Re h , and compute Reτ at an approximate
transition Reynolds number of Reh ~ 3,000.
b) For laminar pipe flow, show that Reτ = 2 Re d , and compute Reτ at an approximate transition
Reynolds number of Red ~ 4,000.
c) For the Blasius boundary layer, show that Reτ ≅ 2.9 Re1x 4 , and compute Reτ at a transition
Reynolds number of Rex ~ 106.
d) If mean profile measurements are made in a wall-bounded turbulent flow at Reτ ~ 102, do you
expect the profiles to display the log-law? Why or why not?
e) Repeat part d) when Reτ > 103.
f) The log-law constants (κ and B) are determined from fitting (12.88) to experimental data. At
which Reτ are κ and B most likely to be accurately determined: 102, 103, or 104?

Solution 12.35. a) For laminar channel flow, the pressure gradient, channel height, and viscosity
determine the wall shear stress and the average velocity:
dp 2τ w h 2 dp 12µU ave dp 2τ w 2 ρu*2
=− and U ave = − , so that =− = = .
dx h 12µ dx h2 dx h h
where τ w = ρu*2 . Use the two ends of the final extended equality to find:
3 2 2 2 3
ρu*2 ρ ( h 2 ) ρ u* ( h 2 ) 12µU ave ρ ( h 2 ) 3
⋅ 2
= 2
= Reτ2 = ⋅ = Re h .
h2 µ µ h2 µ2 2
Thus, Reτ = (3 / 2)Re h which produces Reτ = 67 at Reh = 3,000.
b) For laminar pipe flow, the pressure gradient, diameter, and viscosity determine the wall shear
stress and the average velocity:
dp 4τ w d 2 dp 32µU ave dp 4τ w 4 ρu*2
=− and U ave = − , so that =− = = .
dx d 32µ dx d2 dx d d
Use the two ends of the final extended equality to find:
3 2 2 2 3
2 ρu*2 ρ ( d 2 ) ρ u* ( d 2 ) 2 32µU ave ρ ( d 2 )
⋅ = = Re τ = ⋅ = 2 Re d .
d 2 2µ 2 µ2 d2 2µ 2
Thus, Reτ = 2 Re d which produces Reτ = 89 at Red = 4,000.
c) For laminar boundary layer flow, the flow speed U, downstream distance x, and viscosity
determine the wall shear stress and the boundary layer thickness:
0.332 ρU 2 νx x
τw = 12
and δ ~ 5.0 = 5.0 1 2 .
Re x U Re x
Use the first relationship to find:
0.332U 2 2 δ
2
0.332U 2 δ 2
u*2 = which implies: u* = Re 2
τ = .
Re1x 2 ν2 Re1x 2 ν 2
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Now use the relationship for the overall-boundary layer thickness δ to eliminate δ on the right
side.
2
2 0.332U 2 1 ! x $ 2 Re x
2
Reτ = 12 2 # 5.0 12 & = 0.332(5.0) 32
= 8.3Re1x 2
Re x ν " Re x % Re x
Thus, taking a square root leads to: Reτ = 2.9 Re1x 4 which produces Reτ = 92 at Rex =106.
d) At Reτ ~ 102, the flow is likely transitionally turbulent so there will not be an extensive log-
law region, if one appears at all. At such a low Reynolds number, the outer flow is unlikely to be
fully independent of the viscosity.
e) At Reτ > 103, the outer flow should be fully turbulent so a distinct log-law region is expected.
f) Fitting is always better when there is more data to fit, and the extent of the log-law increases
with increasing Reynolds number. Thus, data from the highest Reynolds number, Reτ = 104,
should produce the most accurate log-law constants.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.36. A horizontal smooth pipe 20 cm in diameter carries water at a temperature of 20

°C. The drop of pressure is dp/dx = –8 N/m2 per meter. Assuming turbulent flow, verify that the
thickness of the viscous sublayer is ≈ 0.25 mm. [Hint: Use dp/dx as given by (12.97) to find τw =
0.4 N/m2, and therefore u* =0.02 m/s.]

Solution 12.36. From (12.96), ∂p/∂x = –4τw/d, therefore:

d ∂p (0.2m) τ 0.4Pa
τw = − =− (−8Pa / m) = 0.4Pa , so u* = w = 3 −3
= 0.02ms −1 .
4 ∂x 4 ρ 10 kgm
From Section 12.9, the viscous sublayer thickness is:
5lν = 5(ν u* ) = 5(1.0 ×10−6 m 2 s−1 0.02ms−1 ) = 2.5 ×10−4 m = 0.25mm .

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.37. The cross-section averaged flow speed Uav in a round pipe of radius a may be
written:
volume flux 1 a 2 a
U av ≡
area
=
π a2 0
∫ U(y)2π r dr = a 2 ∫ U(y)(a − y)dy ,
0
where r is the radial distance from the pipe's centerline, and y = a – r is the distance inward from
the pipe's wall. Turbulent pipe flow has very little wake, and the viscous sublayer is very thin at
high Reynolds number; therefore assume the log-law profile, U(y) = (u* κ ) ln ( yu* ν ) + B , holds
throughout the pipe to find
U av ≅ u* #$(1 κ ) ln ( au* ν ) + B − 3 2κ %& .
Now use the definitions: C f = τ w 12 ρU av2 , Re d = 2U av a ν , f = 4C f = the Darcy friction factor, κ
= 0.41, B = 5.0, and switch to base-10 logarithms to reach (12.105).

Solution 12.37. Start with given relationships and integrate:

2 a 2 a - u ' yu * 0
U av = 2 ∫ U(y)(a − y)dy ≅ 2 ∫ / * ln) * , + u* B2(a − y)dy
a 0 a 0 .κ ( ν + 1
2ν a - u* a ' yu* * u* y ' yu* *0 2u B a
= 2 ∫ / ln) ,−
a κ 0 . ν ( ν + ν ( ν +1
ln) ,2dy + *2
a
∫ [a − y ]dy
0
a
2ν 2 au* ν - au* 0 2u* B - y20
= 2 ∫
a κu* 0 /. ν
ln(β ) − β ln(β )2dβ + 2 /ay − 2
1 a . 2 10
au* ν
2ν 2 - au* β2 β20 2u B - a2 0
= 2 / (β ln β − β ) − ln(β ) + 2 + *2 /a 2 − 2
a κu* . ν 2 4 10 a . 21
2ν 2 - au* ' au* au* au* * 1 ' au* * ' au* * 1 ' au* * 0
2 2

= / ) ln − , − ) , ln) ,+ ) , 2 + u* B
a 2κu* . ν ( ν ν ν + 2( ν + ( ν + 4( ν + 1

2ν 2 - 1 ' au* * ' au* * 3 ' au* * 0

2 2
' 1 ' au * 3 *
= 2 / ) , ln) ,− ) , 2 + u* B = u* ) ln) * , − + B,.
a κu* . 2 ( ν + ( ν + 4 ( ν + 1 ( κ ( ν + 2κ +
The second part of this exercise is primarily algebraic. Start by rewriting the ratio Uav/u* in terms
of f , and rewriting the ratio u*a/ν in terms of Red and f :
1
€ U av U av U av 1 8 au 2aU av 8
f 1 12
= = = = 12
, and * = = f Re d .
u* τw ρ 1
C f U av2 1f f ν 2ν 32
2 8
€ €
Put these into the result of part a) and recall that ln(...) = ln(10)log10(...).
U av 1 $ au* ' 3 8 ln(10) $ 1 12 ' 3
= ln& )− +B → 12
= log10 & f Re d ) − + B.
u* κ % ν ( 2κ f€ κ % 32 ( 2κ
Rearrange to reach the desired form, and evaluate using the numbers provided above.
1 ln(10) 1 $ 3 ln 32 '
10 ( d) ) = 1.986log10 ( f Re d ) −1.020 .
12 12
12
= log f Re + & − + B −
€ f 8κ 8 % 2κ κ (
This is the desired result.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.38. The cross-section averaged flow speed Uav in a wide channel of full height b
may be written:
b2
2
U av ≡ ∫ U(y)dy ,
b 0
where y is the vertical distance from the channel's lower wall. Turbulent channel flow has very
little wake, and the viscous sublayer is very thin at high Reynolds number; therefore assume the
log-law profile, U(y) = (u* κ ) ln ( yu* ν ) + B , holds throughout the channel to find
U av ≅ u* #$(1 κ ) ln ( bu* 2ν ) + B −1 κ %& .
Now use the definitions: C f = τ w 1
2
ρU av2 , Re b = U av b ν , f = 4C f = the Darcy friction factor, κ
= 0.41, B = 5.0, and switch to base-10 logarithms to reach: f −1 2 = 2.0 log10 ( Re b f 1 2 ) − 0.59 .

Solution 12.38. Start with given relationships and integrate:

b2 b2
2 2 ) u* # yu* & ,
U av = ∫ U(y)dy ≅ ∫ + ln % ( + u* B. dy
b 0 b 0 *κ $ ν ' -
b2 b2 bu 2 ν
2ν u* # yu* & 2u B 2ν * 2u B b
=
bκ
∫ ln %
ν $ ν '
( dy + * ∫ dy =
b 0 bκ 0
∫ ln ( β ) d β + *
b 2
0

2ν bu 2 ν 2ν # bu # bu & bu &
= [ β ln β − β ]0 * + u* B = % * ln % * ( − * ( + u* B
bκ bκ $ 2ν $ 2ν ' 2ν '
# 1 # bu & 1 &
= u* % ln % * ( − + B (.
$ κ $ 2ν ' κ '
The second part of this exercise is primarily algebraic. Start by rewriting the ratio Uav/u* in terms
of f , and rewriting the ratio u*b/2ν in terms of Reb and f :
1
U av U av U av 1 8 bu* bU av 8 f 1 12
= = = = , and = = f Re b .
u* τw ρ 1
C f U av2 f12 1
f 2ν 2ν 32
2 8
€ €
Put these into the result of part a) and recall that ln(...) = ln(10)log10(...).
U av 1 ! bu* $ 1 8 ln(10) ! 1 12 $ 1
= ln # &− + B → 12
= log10 # f Re b & − + B .
u* κ " 2ν % κ f κ " 32 % κ
Rearrange to reach the desired form, and evaluate using the numbers provided above.
1 ln(10) 1 " 1 ln 32 %
f 12
=
8κ
log10 f (
12
Re b + $ −
8# κ
)+ B −
κ &
(
' = 1.986 log10 f 1 2 Re d − 0.589 . )
This is the desired result.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.39. For laminar flow, the hydraulic diameter concept is successful when the ratio
( f ⋅Uav dh ν ) ( f ⋅Uav d ν ) is near unity. Show that this ratio is 1.5 when the duct is a
duct round pipe

wide channel.

Solution 12.39. For a channel with height b and width w, the hydraulic diameter from (12.103)
is:
bw b
( dh )channel = 4 =2 , and this simplifies to: (dh)channel ≈ 2b when w >> b.
2b + 2w 1+ b w
1 L
The friction factor is appears in (12.102): Pu − Pd = ρU av2 ⋅ ⋅ f , so:
2 dh
U d 2(Pu − Pd )dh2 2dh2 # dp &
f ⋅ av h = = %− ( .
ν µU av L µU av $ dx '
Evaluate this for laminar channel and round-pipe flows using the results from section 9.2:
" U av dh % 2(2b)2 " dp %
$f ⋅ ' = $ − ' = 12 ⋅ 8 = 96 , and
# ν &channel " b 2 dp % # dx &
µ $− '
# 12µ dx &
" U av dh % 2d 2 " dp %
$f ⋅ ' = $ − ' = 32 ⋅ 2 = 64 .
# ν & pipe " d dp % # dx &
2
µ $− '
# 32µ dx &
Thus, the ratio is: ( f ⋅U av dh ν ) ( f ⋅U av d ν) = 96/64 = 1.5.
channel round pipe
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.40. a) Rewrite the final friction factor equation in Exercise 12.38 in terms of the
channel's hydraulic diameter instead of its height b.
b) Using the friction factor-Reynolds number ratio given in Exercise 12.39, evaluate (12.107) for
a wide channel.
c) Are the results of parts a) and b) in good agreement?

Solution 12.40. a) For a channel with height b and width w, the hydraulic diameter from
(12.103) is:
bw b
( dh )channel = 4 =2 , and this simplifies to: (dh)channel ≈ 2b when w >> b.
2b + 2w 1+ b w
The final result of Exercise 12.38 is f −1 2 = 2.0 log10 ( Re b f 1 2 ) − 0.59 , which can be rewritten:
"1 %
f −1 2 = 2.0 log10 $ Re 2b f 1 2 ' − 0.59 = 2.0 log10 ( Re dh f 1 2 ) −1.19 .
#2 &
b) For this problem, (12.107) implies
$$ f ⋅ Re ' '
−1 2 & pipe d −1 2
fchannel,turb ≅ 2.0 log10 && )
) ⋅ Re dh ⋅ fchannel,turb ) − 0.80
& f ⋅ Re )
%% channel dh (laminar (
$ 1 −1 2 '
= 2.0 log10 & ⋅ Re dh ⋅ fchannel,turb ) − 0.80
% 1.5 (
= 2.0 log10 ( Re dh ⋅ fchannel,turb
−1 2
) −1.15.
c) The only difference between the answers for parts a) and b) is 0.04 in the final subtracted
constant. Thus, the agreement is good and this suggests that (12.107) provides a useful empirical
correction for estimating friction factors in non-circular ducts.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.41. a) Simplify (12.114) when the roughness Reynolds number is large Reks >> 1 to
CfR show that CfR is independent of ν in the fully rough regime.
b) Reconcile the finding of part a) with the results in Figure 12.25 which appear to show that CfR
depends on Rex for all values of Reks.
c) For this fully rough regime, compare CfR computed from (12.114) with the empirical formula
−2.5
provided in Schlichting (1979): C fR = ( 2.87 +1.58⋅ log10 (x / ks )) .

Solution 12.41. a) Eq. (12.114) is:

2 1 " ( H 3.5) Re x C fR 2 %' 2Π

≅ ln $$ R +B+ ,
C fR κ # 1+ 0.26 Re ks C fR 2 & ' κ
log10(x/ks)
where Rex = Uex/ν and Reks = Ueks/ν. When Reks >> 1, the '1' in the denominator may be ignored,
and this allows the argument of the natural log function to be simplified:
2 1 " ( H 3.5) Re x C fR 2 %' 2Π 1 "$ ( H R 3.5) C fR 2 x %' 2Π
≅ ln $$ R + B + = ln $ +B+ ,
C fR κ # 0.26 Re ks C fR 2 & ' κ κ # 0.26 '
ks & κ
where Rex/Reks = x/ks.
b) Figure (12.25) provides curves of CfR at constant Reks vs. Rex. The ratio x/ks is not held
constant on any of the plotted curves. Thus, the reader must examine points on different curves
having the same value of x/ks. For example, consider x/ks = 102, where CfR = 1.16 for Rex/Reks =
105/103, 106/104, and 107/105.
c) A plot of (12.114) (solid line) and Schlichting's formula (dashed line) vs. log10(x/ks) looks like:
0.1$

0.01$

0.001$
2$ 3$ 4$ 5$ 6$

Here, as in Example 12.10, HR = 1.3. The agreement between the two empirical formulae is
generally good. At x/ks = 102, (12.114) is 3% higher than Schlichting's formula. At x/ks = 106,
(12.114) is 6% lower Schlichting's formula.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.42. Perhaps the simplest way to model turbulent flow is to develop an eddy
viscosity from dimensional analysis and physical reasoning. Consider turbulent Couette flow
with wall spacing h. Assume that eddies of size l produce velocity fluctuations of size l(∂U ∂y )
2
so that the turbulent shear stress correlation can modeled as: −uv ∝ l 2 (∂U ∂y ) . Unfortunately, l
cannot be a constant because it must disappear near the walls. Thus, more educated guessing is
needed, so for this problem assume ∂U/∂y will have some symmetry about the€channel centerline
(as shown) and try: l = Cy for 0 ≤ y ≤ h/2 where C is a positive dimensionless constant and y is
the vertical distance measured from the lower wall. € With this turbulence model, the RANS
equation for 0 ≤ y ≤ h/2 becomes:
€ ∂U ∂U 1 dp 1 ∂τ xy ∂U % (2
2 2 ∂U
U +V =− + where τ xy = µ + ρC y ' *
∂x ∂y ρ dx ρ ∂y ∂y & ∂y )
Determine an analytic form for U(y) after making appropriate simplifications of the RANS
equation for fully developed flow assuming the pressure gradient is zero. Check to see that your
final answer recovers the appropriate forms as y → 0 and C → 0. Use the fact that
€ €
U(y = h /2) = U o 2 in your work if necessary.

Solution 12.42. In Couette flow V will be zero because the fluid is confined and ∂/∂x = 0 because
the flow is homogeneous in this direction. Thus, with zero pressure gradient, the RANS BL
1 ∂τ xy
equation becomes very simple: 0 = . Integrate this equation from lower wall to find:
ρ ∂y
τ xy (y) − τ w = 0 where τw is the wall shear stress. Now substitute in the model equation for the
shear stress to find a quadratic equation for the velocity gradient:
2
∂U €2 2 $ ∂U ' ∂U −µ ± µ 2 + 4 ρτ wC 2 y 2
µ + ρC y & ) − τ w = 0 , which implies = .
€ ∂y % ∂y ( ∂y 2 ρC 2 y 2
Here, a positive velocity gradient is expected for the lower half of the channel so the “+” sign
should be chosen. A bit of work or use of a table of integrals produces:
2 -+ 5
4 ρτ wC 2 y 2 -/7
2 2 2
€ 1 τ w 41− 1+ 4 ρτ wC y µ€ % y(
U(y) = + ln,2C ρτ w ' * + 1+ 0
C ρ 43 2C ρτ w ( y µ) -. &µ) µ2 -176
Now check to see that this model outcome has the appropriate behavior. As y → 0, the
velocity should be linear in y. To find this behavior in the equation above, expand the square
roots using the Taylor series 1+ ε = 1+ 12 ε + ... for ε << 1.
€
1 τ w 11− (1+ 2 ρτ wC 2 y 2 µ 2 + ...) + % y( 2 ρτ wC 2 y 2 .4
U(y) = 3 + ln,2C ρτ w ' * + 1+ + ... /6 . ($)
C ρ2 2C ρτ w ( y µ) - & µ) µ2 05
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Simplify the first term, expand the natural logarithm using ln(1+ ε) = ε + ... for ε << 1, and ignore
terms involving y2 or higher powers:
1 τw , & y) & y )/ τ w
U(y) ≅ .−C ρτ w( + + 2C ρτ w ( +1 = y,
C ρ- ' µ€* ' µ *0 µ
and this is the expected form. As C → 0, the same result is recovered. In addition, the fact:
U(y = h /2) = U o 2 , can be used to link τw and Uo by evaluating ($) at y = h/2. And finally, when
4 ρτ wC 2 y 2 €
2
>> 1 a log-layer is produced by this turbulence model.
µ
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.43. Incompressible, constant-density-and-viscosity, fully-developed, pressure-

gradient-driven, turbulent channel flow is often used to test turbulence models for wall-bounded
flows. Thus, for this flow, investigate the following simplified mixing-length model for the
Reynolds shear stress: −u#v # = βy τ w ρ (∂U ∂y ) for 0 ≤ y ≤ h/2 where y is measured from the
lower wall of the channel, β is a positive dimensionless constant, τw = wall shear stress (a
constant), and ρ = fluid density. 
a) Use this turbulence
€ model, the fully-developed flow assumption U = U(y)eˆx , the assumption
of a constant downstream pressure gradient, and the x-direction RANS mom. equ.,
∂U ∂U 1 ∂P & ∂ 2U ∂ 2U ) ∂ ∂
U
∂x
+V
∂y
=−
ρ ∂x
+ ν( 2 + 2 + −
' ∂x ∂y €* ∂x
( )
u,2 −
∂y
( )
u,v ,

u (! 2ν $ ! β u* y $ 2y +
to find: U(y) = * *#1+ & ln #1+ & − - for 0 ≤ y ≤ h/2 where u* = τ w ρ .
β )" β u*h % " ν % h,
€ velocity profile have the proper gradient at y = 0 and y = h/2?
b) Does this
c) Show that this velocity profile returns to a parabolic flow profile as β → 0 .
d) How should the constant β be determined?

€
Solution 12.43. a) When the flow is fully developed (U = U(y)ex) the only non-zero field
gradient in the flow direction is ∂P/∂x, so the Reynolds-averaged x-direction momentum
equation simplifies to:
1 ∂P ∂ 2U ∂ 1 dP ∂ $ ∂U ∂U '
0=−
ρ ∂x
+ν 2 −
∂y ∂y
( )
u"v" ≅ − + &ν
ρ dx ∂ y % ∂ y
+ β uτ y )
∂y (
where the second approximate equality follows from the given turbulence model with
uτ = τ w ρ . Here all the vertical gradients are switched to total derivatives because y is the only
independent variable. Integrate the last form of the equation once in the y-direction:
1 dP ∂U ∂U
y + const = ν + β uτ y .
€ ρ dx ∂y ∂y
Evaluating this equation at y = 0 determines the constant: const = ν ( dU dy ) y= 0 = τ w ρ .
dU y ( dP dx ) + τ w
Rearrange the equation to find: = . The parametric format of this equation
dy ρ (ν + βuτ y )
can be simplified by using a CV to get a simple € relationship between τw and dP/dx. Conserve
horizontal (x) momentum in a stationary rectangular CV that encloses all the fluid in the channel
between x and x + Δx. Here the horizontal velocity profile is steady and unchanged between x
€
and x + Δx, so the unsteady and flux terms are zero. Thus COMOx simplifies to:
0 = P(x)h − P(x + Δx)h − 2τ wΔx , which is a balance of pressure forces on the vertical CV sides
and skin-friction forces on the horizontal CV sides. When this CV equation is rearranged and
the limit as Δx → 0 is taken, it becomes dP dx = −2τ w h . Thus, the differential equation for
U(y) can be rewritten:
€
dU τ w (1− 2y h ) τ w (1− 2y h ) τ w (−2y h +1)
= = =
€ dy ρ (ν + β uτ y€) ρ (ν + 12 β uτ h ( 2y h )) 12 ρβ uτ h ( 2y h + 2ν β uτ h )
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Setting η = 2y h , and integrating produces:

τ −η + 1 τ ) 1+ 2ν βuτ h ,
U(y) = w ∫ dη = w ∫ +−1+ .dη
ρβuτ η + 2ν βuτ h ρβuτ * η + 2ν βuτ h -
€ u & 2y & 2ν ) & 2y 2ν ))
= τ (− + (1+ + ln( + ++ + const
β ' h ' βuτ h * ' h βuτ h **
€
The constant can be evaluated by requiring U(0) = 0.
uτ & & 2ν ) & 2ν )) uτ & 2ν ) & 2ν )
0= (−0 + (1+ + ln( ++ + const or const = − (1+ + ln( +
β '€ ' βuτ h * ' βuτ h ** β ' βuτ h * ' βuτ h *
u & 2y & 2ν ), & 2y 2ν ) & 2ν )/)
Thus, U(y) = τ (( − + (1+ ln
+. ( + + − ln ( +1++ , or
β ' h ' βuτ h *- ' h βuτ h * ' βuτ h *0*
€ €
uτ & 2y & 2ν ) & βuτ y ))
= (− + (1+ + ln(1+ ++
β ' h ' βuτ h * ' ν **
€ Here it should be noted that as the Reynolds number of the flow increases the importance of the
turbulence model increases. This fact can be ascertained by noting how U(y) depends on uτ h ν .
dU τ w (1− 2y h ) τ h
b)€Yes; from above, = equals w at y = 0, and equals 0 at y = .
dy ρ (ν + βuτ y ) µ 2
c) Use a two term expansion of the natural log function for small β in the answer€for part a) and
complete the algebraic multiplications of terms
u ' 2y ' 2ν * € ' βu y ** €
lim U(y) =€lim τ ) − + )1+ , ln)1+ τ ,,
β →0 β →0 β
( h ( βuτ h + ( ν ++
uτ ' 2y ' 2ν *' βuτ y 1 ' βuτ y * **
2
)
= lim )− + )1+ ,)) − ) , + ...,,,,
β →0 β
( h ( β uτ h + ( ν 2 ( ν + ++
€ ' 2 2 *
u 2y βu y 1 ' βu y * 2ν βuτ y 2ν 1 ' βuτ y *
= lim τ ))− + τ − ) τ , + − ) , + ...,,
β →0 β
( h ν 2 ( ν + βuτ h ν βuτ h 2 ( ν + +
Cancel like terms and simplify:
€
u2h y ' y * u2h y ' y * h 2 dP y ' y *
= lim τ )1− − β (...), = τ )1− , = − )1− ,
β →0 ν h ( h + ν h ( h+ 2µ dx h ( h +
€
The final equality is the parabolic flow velocity profile for pressure gradient driven flow between
parallel plates; it follows from the prior expression via the substitution:
dP dx = −2τ w h = −2 ρ uτ2 h .
€
d) The constant β would need to be determined by fitting a sample profile to an experimental
result.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.44. The model equations for the two-equation "k-ε" turbulence model, (12.124) and
(12.126), include 5 empirical constants. One of these, Cε2, can be estimated independently of the
others by fitting a solution of the model equations to experimental results for the decay of the
turbulent kinetic energy, e , downstream of a random grid placed at the inlet of a wind-tunnel
test section. The development of this estimate is further simplified by use of a coordinate system
that translates with the average flow velocity in the wind-tunnel. In these translating coordinates
Ui = 0, and e and ε are both functions of time t alone.
a) Simplify (12.124) and (12.126) for random grid turbulence when Ui = 0.
b) Assume e (t) follows a power-law solution, e = eot −n , where eo and n are positive constants,
and determine a formula for the model constant Cε2 in terms of n.
c) The experimental value of n is approximately 1.3, so the part b) formula then predicts Cε2 =
1.77, which is below the standard value (Cε2 = 1.92 from Launder & Sharma, 1974). Provide at
least two reasons that justify this discrepancy.

Ut!

U!

random grid! grid turbulence!

Solution 12.44. a) When Ui = 0 and e & ε are both functions of time t alone, all the terms with
spatial derivatives (∂/∂xj) drop out. Thus, the simplified model equations are:
2
de
= −ε and
dε
= −Cε 2
(ε)
,
dt dt e
where d/dt has replaced ∂/∂t because t is the only independent variable.
2
d 2e 1 " de %
b) Use the first equation to eliminate ε from the second to find: − 2 = −Cε 2 $ − ' ,
dt e # dt &
−n
Insert the suggested solution, e = eot , simplify & divide out common factors, and solve for Cε2.
1 1
−eo (−n)(−n −1)t −n−2 = −Cε 2 −n (+n)2 eo2 t −2n−2 → (n)(n +1) = Cε 2 (n)2 , or Cε 2 = 1+ .
eot n
c) Two reasons that justify the difference are as follows.
(i) The actual flow includes spatial variations in e and ε so the time-only simplified model
equations do not precisely apply to grid turbulence.
(ii) The five model constants have been tuned to produce acceptable outcomes in many different
flows and this requires trade-offs to be made between accuracy and breadth of applicability. For
example, while Cε2 = 1.77 might be appropriate for decaying turbulent flows (like grid
turbulence), fitting the "k-ε" turbulence model to other flows involving mean-flow shear,
bounding surfaces, and turbulence production may require a larger value for this model constant.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.45. Derive (12.127) from (12.35) with constant density using the definition
equalities in (12.128), (12.129), and (12.131).

Solution 12.45. Start from (12.35) with constant density (α = 0):

∂ ui u j ∂u u ∂u u u ∂U j ∂U 1 " ∂ p ∂p % ∂u ∂uj ∂2
+U k i j + i j k = −ui uk − u j uk i − $$ ui + uj '' − 2ν i + ν 2 ui u j
∂t ∂ xk ∂ xk ∂ xk ∂ xk ρ # ∂ x j ∂ xi & ∂ xk ∂ xk ∂ xk
Rearrange the terms so that the first four are the same as in (12.127):
∂ ui u j ∂u u ∂U j ∂U ∂u ∂uj 1 " ∂ p ∂p % ∂2 ∂u u u
+U k i j = −ui uk − u j uk i − 2ν i − $$ ui + uj '' + ν 2 ui u j − i j k
∂t ∂ xk ∂ xk ∂ xk ∂ xk ∂ xk ρ # ∂ x j ∂ xi & ∂ x k ∂ xk
The fifth term is εij as defined by (12.128). The remaining terms on the right may be rewritten to
extract a ∂/∂xk differentiation from everything but the pressure-rate-of-strain tensor.
1 " ∂p ∂p % ∂2 ∂u u u
− $$ ui + uj '' + ν 2 ui u j − i j k
ρ # ∂xj ∂ xi & ∂ x k ∂ xk
1" ∂ ∂u ∂ ∂u % ∂ " ∂ %
= − $$ ui p − p i + u j p − p j '' + $ν ui u j − ui u j uk '
ρ #∂ x j ∂ x j ∂ xi ∂ xi & ∂ x k # ∂ x k &
1" ∂ ∂u ∂ ∂u % ∂ " ∂ %
= − $$ ui pδ jk − p i + u j pδik − p j '' + $ν ui u j − ui u j uk '
ρ # ∂ xk ∂ x j ∂ xk ∂ xi & ∂ x k # ∂ x k &
∂ " ∂ up u p % 1 " ∂u ∂u %
= $$ν ui u j − ui u j uk − i δ jk − j δik '' + $$ p i + p j ''
∂ xk # ∂ xk ρ ρ & ρ # ∂xj ∂ xi &
= M ij + N ij
The first equality above follows from the product rule of differentiation and the fact that spatial
differentiation and averaging are independent operations that can be performed in either order.
Thus, the rearranged version of (12.35) for constant density is:
∂ ui u j ∂u u ∂U j ∂U
+U k i j = −ui uk − u j uk i − εij + M ij + N ij ,
∂t ∂ xk ∂ xk ∂ xk
where (12.129) has been used for Mij and (12.131) has been used for Nij. This final equation is is
(12.127)
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.46. Derive (12.132) by taking the divergence of the constant-density Navier-Stokes
momentum equation, computing its average, using the continuity equation, and then subtracting
the averaged equation from the instantaneous equation.

Solution 12.46. Start with the constant density NS-momentum equation with the nonlinear term
written in flux form [see (4.23)]:
∂ u!i ∂ 1 ∂ p! ∂ 2 u!i
+ ! !
( i j) ρ ∂x ∂x ∂x ,
u u = − + ν
∂t ∂ x j i j j

where as in (12.24) a tilde indicates the full value of a dependent field variable. Take the
divergence of this equation (i.e. apply ∂/∂xi to both sides of the equation) to reach:
∂ ∂ u!i ∂ ∂ 1 ∂ 2 p! ∂ 2 ∂ u!i
+ ! !
( i j) ρ ∂x∂x ∂x ∂x ∂x
u u = − + ν
∂ t ∂ xi ∂ xi ∂ x j i i j j i

The first and final terms are zero in an incompressible flow, leaving
∂ 2 p! ∂ ∂
∂ xi∂ xi
= −ρ
∂ xi ∂ x j
(u!iu! j ) .
Insert the Reynolds decomposition, u!i = Ui + ui and p! = P + p where a capital letter indicates and
average field variable, and a lower case letter indicates a turbulent fluctuation, to find:
∂ 2 (P + p) ∂ ∂
∂ xi∂ xi
= −ρ
∂ xi ∂ x j
((Ui + ui )(U j + u j )) , or
∂ 2P ∂2p ∂ ∂
+
∂ xi∂ xi ∂ xi∂ xi
= −ρ
∂ xi ∂ x j
(UiU j +Uiu j + uiU j + uiu j ) . ($)

Average this equation to reach:

∂ 2P ∂ ∂
∂ xi∂ xi
= −ρ (
∂ xi ∂ x j
)
UiU j − ui u j .

Subtract this from ($),

∂2p ∂ ∂
∂ xi∂ xi
= −ρ (
∂ xi ∂ x j
)
Ui u j + uiU j + ui u j − ui u j .

Performing the xj-differentiation on the first two term on the right side:
∂2p ∂ " ∂Ui ∂u ∂u ∂U j % ∂ ∂
∂ xi∂ xi
= −ρ $$
∂ xi # ∂ x j
u j +Ui j + i U j + ui
∂xj ∂xj
'' − ρ (
∂ x j & ∂ xi ∂ x j
)
ui u j − ui u j .

The second and fourth terms inside the large parentheses are zero for incompressible flow.
Performing the xi-differentiation on remaining terms in the large parentheses leads to:
∂2p ∂U ∂ u ∂ u ∂U j ∂ ∂
∂ xi∂ xi
= −ρ i j − ρ i
∂ x j ∂ xi ∂ x j ∂ xi
−ρ (
∂ xi ∂ x j
)
ui u j − ui u j .

Since both indices are summed over, i and j can be exchanged in the second term on the right
thereby making it identical to the first term on the right. Making this change leads to the final
form:
∂2p ∂U ∂ u ∂ ∂
∂ xi∂ xi
= −2 ρ i j − ρ
∂ x j ∂ xi
(
∂ xi ∂ x j
)
ui u j − ui u j ,

and this is (12.132).

Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.47. Using the Green's function given in Exercise 12.14 and the properties of
homogeneous turbulence, formally solve (12.132) and then use (12.131) to reach (12.133).

Solution 12.47. From Exercise 12.14, the Green's function solution of the Poisson equation
∂2 p ∂xi2 = f (x j ) , is:
1 1 1 1
p(x j ) = − ∫ f (y j )d 3 y = − ∫ f (y)d 3 y ,
4π all y (x j − y j ) 2 4π all y x−y
where the second form merely involves changes in notation. For the present purposes, the
inhomogeneous term in (12.132) is:
∂U ∂ u ∂ ∂
f = −2 ρ i j − ρ
∂ x j ∂ xi ∂ xi ∂ x j
ui u j − ui u j , ( )
thus the formal solution of (12.132) is:
1 1 # ∂Ui ∂ u j ∂ ∂ &
p(x) = − ∫ %
4π all y x − y %$
−2 ρ
∂ y j ∂ yi
− ρ
∂ yi ∂ y j
u u
i j − u u
i j (
'
3
(d y . ( )
Using this formal solution, the pressure-rate-of-strain tensor Nij may be written:
p # ∂u ∂u & 1 ∂ui 1 # ∂U k ∂ ul ∂ ∂ & 3
N ij ≡ %% i + j (( = ∫ % 2 +
ρ $ ∂x j ∂xi ' 4π ∂x j all y x − y $ ∂ yl ∂ yk ∂ yk ∂ yl
u u
k l − uk l (d y
u
'
( )
1 ∂u j 1 # ∂U k ∂ ul ∂ ∂ &
∫ %2 +
4π ∂xi all y x − y $ ∂ yl ∂ yk ∂ yk ∂ yl
uk ul − uk ul ( d 3 y.
'
( )
Combine terms with like integrands:
p # ∂u ∂u & 1 ∂U k # ∂ui ∂u j &# ∂ ul & d 3 y
N ij ≡ %% i + j (( = ∫ % + (% (
ρ $ ∂x j ∂xi ' 2π all y ∂ yl %$ ∂x j ∂xi ('$ ∂ yk ' x − y

1 # ∂u ∂u &# ∂ 2 u u & d 3 y
+
4π
∫ %% ∂x i + ∂x j ((%$ ∂ y ∂k yl (' x − y
all y $ j i ' k l

1 # ∂u ∂u j &# ∂ 2 uk ul & d 3 y
−
4π
∫ %% ∂x i
+ (% (
∂xi ('$ ∂ yk∂ yl ' x − y
.
$ j all y

The final term is zero because the average of the fluctuating strain rate (the first factor inside the
last integral) is zero. This leaves:
1 ∂U k # ∂ui ∂u j &# ∂ ul & d 3 y 1 # ∂u ∂u &# ∂ 2 u u & d 3 y
N ij = ∫ % + ( % ( + ∫ % i + j (% k l (
2π all y ∂ yl %$ ∂x j ∂xi ('$ ∂ yk ' x − y 4π all y %$ ∂x j ∂xi ('$ ∂ yk∂ yl ' x − y
. (&)

For homogeneous turbulence, ∂Uk/∂yl will be independent of position. Thus, it can be taken
outside the integrand in the first term and this simplification leads to (12.133):
1 ∂U k " ∂u ∂u %" ∂ u % d 3 y 1 " ∂u ∂u %" ∂ 2 u u % d 3 y
j
N ij = ∫ $ i
+ ' $ l
' + ∫ $ i + j '$ k l '
2π ∂ xl all y $# ∂x j ∂xi '&# ∂ yk & x − y 4π all y $# ∂x j ∂xi '&# ∂ yk∂ yl & x − y
. (12.133)

When the turbulence is inhomogeneous, then (&) is correct and (12.133) represents a local
approximation.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.48. Turbulence largely governs the mixing and transport of water vapor (and other
gases) in the atmosphere. Such processes can sometimes be assessed by considering the
conservation law (12.34) for a passive scalar.
a) Appropriately simplify (12.34) for turbulence at high Reynolds number that is characterized
by: an outer length scale of L, a large-eddy turnover time of T, and a mass-fraction magnitude of
Yo. In addition, assume that the molecular diffusivity κm is at most as large as ν = µ ρ = the
fluid’s kinematic viscosity.
b) Now consider a simple model of how a dry turbulent wind collects moisture as it blows over a
nominally flat water surface (x1 > 0) from a dry surface (x1 < 0). Assume the mean velocity is
steady and has a single component with a linear gradient, U j = (Sx 2€,0,0) , and use a simple
gradient diffusion model: −u jY # = ΔUL(0,∂Y ∂x 2 ,0) , where ΔU and L are (constant) velocity and
length scales that characterize the turbulent diffusion in this case. This turbulence model allows
the turbulent mean flow to be treated like a laminar € flow with a large diffusivity = ΔUL (a
turbulent diffusivity). For the simple boundary conditions: Y (x j ) = 0 for x1 < 0, Y (x j ) = 1 at x2
€
= 0 for x1 > 0, and Y (x j ) → 0 as x 2 → ∞ , show that
∞ ∞ $ S '1 3
( 1
9
3
Y (x1, x 2 , x 3 ) = ∫ exp − ζ dζ ) ( 1 3
)
∫ exp − ζ d€ζ where ξ = x 2& ΔULx€ ) for x1,x2 > 0.
9
%
ξ 0 1(
€ €

€ €

∂Y ∂Y ∂ & ∂Y )
Solution 12.48. a) Start with (12.34): +Uj = ((κ m − u jY %++ , but not all the terms
∂t ∂x j ∂x j ' ∂x j *
are needed at high Reynolds number. Using the length, time, and mass-fraction scales, set
* * *
Y * = Y Yo , U j = U j T L , t * = t T , x j = x j L , and u jY " = u jY " ( LYo /T ) , and insert these into
(12.34) to find: €
*
Yo ∂Y L Yo * ∂Y * Yo ∂ 2Y * 1L ∂ *
*
+ U j * = κm 2 * * − Yo * u jY % .
€ € T ∂t T L ∂x€ j L ∂x j ∂ x j L T ∂x j
€ €
Divide by Yo and multiply by T to find that the terms on the left side are of order unity along with
the second term on the right.
∂Y * * ∂Y
*
T ∂ 2Y * ∂ *
€ *
+ U j *
= κ m 2 * *
− * u jY %
∂t ∂x j L ∂x j ∂ x j ∂x j
Given that κm is at most as large as ν, the coefficient of the first term on the right side is at most
as large as 1/Re, where Re is the Reynolds number = L2 νT ; thus, this term can be dropped
compared to the three that are of order unity.
€
b) For steady mean flow, the approximate Reynolds-averaged scalar transport equation becomes:
∂Y ∂
Uj €=−
∂x j ∂x j
(
u jY $ .)

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Insert the given mean velocity U1 = Sx2, and turbulent diffusion model −u2Y # = ΔUL(∂Y ∂x 2 ) :
∂Y ∂ 2Y
Sx 2 = ΔUL 2 . (@)
∂x1 ∂x 2
$ S '1 3 €
Introduce the similarity variable ξ = x 2 & ) , and note that:
% ΔULx1 (
13 23
∂Y dY ∂ξ ξ dY€ ∂Y dY ∂ξ % S ( dY ∂ 2Y $ S ' d 2Y
= =− , = =' * , and =& ) .
∂x1 dξ ∂x1 3x1 dξ ∂x 2 dξ ∂x 2 & ΔULx1 ) dξ ∂x 22 % ΔULx1 ( dξ 2
Thus, the approximate € scalar transport equation becomes:
Sx 2 dY % S ( 2 3 d 2Y 1 2 dY d 2Y
− ξ = ΔUL' * , which simplifies to − ξ =
€ 3x1 dξ€ & ΔULx1 ) dξ
2
€ 3 dξ dξ 2
after some manipulation of the factors. Integrate and exponentiate this last equation to find:
dY $ 1 ' ξ
= Aexp& − ξ 3 ) , and integrate again to reach Y (ξ ) = A ∫ 0 exp(− 19 ζ 3 ) dζ + B ,
€ dξ % 9 ( €
where ζ is just an integration variable. The boundary conditions are:
(i) Y (x j ) = 0 for x1 < 0,
€ (ii) Y (x j ) = 1 at x2€= 0 for x1 > 0, and
(iii) Y (x j ) → 0 as x 2 → ∞ .
Condition (i) can be set€immediately since Y = 0 is a solution of the field equation (@).
0
Condition (ii) implies: Y€(0) = A ∫ 0 exp(− 19 ζ 3 ) dζ + B = 1 , or 0+B = 1 for x1 > 0.
∞
Condition (iii) implies: Y€(∞) = A ∫ 0 exp€(− 19 ζ 3 ) dζ + B = 0 , or 0+B = 1 for x1 > 0.
€ −1

( ∞
Therefore, B = 1 and A = − ∫ 0 exp(− 19 ζ 3 ) dζ , so
€
)
ξ ∞ ξ ∞
(− 19 ζ 3 ) dζ ∫ 0 exp(− 19 ζ 3 ) dζ − ∫ 0 exp(− 19 ζ 3 ) dζ ∫ξ exp(− 9 ζ 3 ) dζ
1
∫ 0 exp
Y (ξ ) = 1− ∞ € = ∞ = ∞ .
∫ 0 exp(− 19 ζ 3 )dζ ∫ 0 exp(− 19 ζ 3 )dζ ∫ 0 exp(− 19 ζ 3 ) dζ
Thus, the final € form is:
'0 for x1 < 0, x 2 > 0 +
)∞ ∞ ) $ S '1 3
Y (x1, x 2 , x 3 ) = ( where ξ = x 2 & ) .
exp(− 19 ζ 3 )dζ ∫ exp(− 19 ζ 3 )dζ x1, x 2 > 0,
)* ξ∫
€ for % ΔULx1 (
0
)
-

€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.49. Estimate the Monin–Obukhov length in the atmospheric boundary layer if the
surface stress is 0.1 N/m2 and the upward heat flux is 200 W/m2.

Solution 12.49. First compute the friction velocity:

τ 0.1Pa
u* = w = 3 −3
= 0.289ms −1 .
ρ 1.2 kgm
The heat flux determines wT "
Q 200Wm−2
wT " = = −3 2 −2 −1
= 0.166ms−1K .
ρCp 1.2kgm (1004m s K )
Thus, the the Monin–Obukhov
€ length is:
3
u* (0.289ms−1 ) 3
LM = = = 10.6m .
€ καgwT $ (0.41)(3.4 ×10−3 K −1 )(9.81ms−2 )(0.166ms−1K)

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 12.50. Consider one-dimensional turbulent diffusion of particles issuing from a point
source. Assume a Gaussian Lagrangian correlation function of particle velocity
r(τ ) = exp{− τ 2 t c2 } ,
where tc is a constant. By integrating the correlation function from τ = 0 to ∞, find the integral
time scale Λt in terms of tc. Using the Taylor theory, estimate the eddy diffusivity at large times
t/Λt >> 1, given that the rms €
fluctuating velocity is 1 m/s and tc = 1 s.

Solution 12.50. Start from the definition of the integral time scale:
∞ ∞ ' τ2* ∞ ' τ2* ∞
π
( tc + ( tc +
( )
Λ t = ∫ r(τ )dτ = ∫ exp)− 2 ,dτ = ∫ exp) − 2 , dτ = t c ∫ exp −β 2 dβ = t c
2
= 0.886t c .
0 0 0 0
From (12.129),
DT ≅ u 2 Λ t = (1ms−1 ) 2 ⋅ 0.886(1s) = 0.886m 2 s−1 .
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 13.1. The Gulf Stream flows northward along the east coast of the United States with a
surface current of average magnitude 2 m/s. If the flow is assumed to be in geostrophic balance,
find the average slope of the sea surface across the current at a latitude of 45°N. [Answer: 2.1 cm
per km]

Solution 13.1. Given v = 2 m/s, and f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1. For
geostrophic balance:
1 ∂p ∂η
fv = =g ,
ρ ∂x ∂x
so the sea-surface slope is:
∂η fv (1.03×10 −4 s−1 )(2ms−1 )
= = = 2.1×10 −5 ,
∂x g 9.81ms−2
which is equivalent to an eastward surface rise of 2.1 cm per km.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 13.2. A plate with water (ν = 10−6 m2/s) above it rotates at a rate of 10 revolutions per
minute. Find the depth of the Ekman layer, assuming that the flow is laminar.

Solution 13.2.
Given Ω = 10 rpm = 20π rad./min. = 1.047 s–1, the Coriolis frequency is f = 2Ω = 2.09 s–1. From
(13.28), the Ekman layer thickness is:
δ = 2ν f = 2(10 −6 m 2s−1 ) (2.09s−1 ) = 0.978 ×10 −3 m ≅ 1 mm .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 13.3. Assume that the atmospheric Ekman layer over the earth’s surface at a latitude of
45ºN can be approximated by an eddy viscosity of ν V = 10 m2/s. If the geostrophic velocity
above the Ekman layer is 10 m/s, what is the Ekman transport across isobars? [Answer: 2203
m2/s]

Solution 13.3.
Given ν V = 10 m2/s, f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1, and U = 10 m/s,
the Ekman layer height from (13.28) is
δ = 2ν f = 2(10m 2s−1 ) (1.03×10 −4 s−1 ) = 440.7m .
Using the equation at the top of page 720, the net transport perpendicular to the geostrophic
stream is:
1
Uδ = 0.5(10ms –1 )(440.7m) = 2, 203m 2s –1 .
2
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 13.4. a) From the set (13.45) – (13.47), develop the following equation for the water
surface elevation η(x,y,t):
∂ ) ∂2 # ∂2 ∂2 &,
* 2 + f 2 − gH % 2 + 2 (-η (x, y, t) = 0
∂t + ∂t $ ∂x ∂y '.
b) Using η (x, y, t) = η̂ exp {i(kx + ly − ω t)} show that that the dispersion relationship reduces to ω
= 0 or (13.82).
c) What type of flows have ω = 0?

Solution 13.4. The starting-point equation set is:

∂η ! ∂u ∂v $ ∂u ∂η ∂ v ∂η
+H# + &=0, − fv = −g , + fu = −g . (13.45, 13.46, 13.47)
∂t "∂ x ∂ y % ∂t ∂ x ∂t ∂y
Apply ∂/∂t to (13.46), multiply (13.47) by f, and add the results to find:
" ∂2 % " ∂2η ∂η %
'. (a)
2
$ 2 + f ' u = −g $ +f
# ∂t & # ∂x∂t ∂y &
Multiply (13.46) by –f, apply ∂/∂t to (13.47), and add the results to find:
" ∂2 % " ∂2η ∂η %
'. (b)
2
$ 2 + f ' v = −g $ −f
# ∂t & # ∂y∂t ∂x &
Apply ∂/∂x to (a) and ∂/∂y to (b), and add the results to reach:
" ∂2 %" ∂u ∂v % " ∂2 % 1 ∂η %
2 "
" ∂2 ∂2 % ∂η
$ 2 + f 2
'$ + =
' $ 2 + f '$ − ' = −g $ 2 + 2'
, (c)
# ∂t &# ∂x ∂y & # ∂t &# H ∂t & # ∂x ∂y & ∂t
where the first equality follows from (13.45). Using the final equality, multiply by –H, and
collect all the terms to left side of the equation to find:
" ∂2 %
2 " ∂η %
" ∂2 ∂2 % ∂η
$ 2 + f '$ ' − gH $ 2 + 2'
=0. (d)
# ∂t &# ∂t & # ∂x ∂y & ∂t
Factor out the common differentiation operation ∂/∂t to reach:
∂ ) ∂2 # ∂2 ∂2 &,
* 2 + f 2 − gH % 2 + 2 (-η (x, y, t) = 0 .
∂t + ∂t $ ∂x ∂y '.
b) Substitute the trial solution η (x, y, t) = η̂ exp {i(kx + ly − ω t)} into the part a) differential
equation to find:
{ }
−iω −ω 2 + f 2 − gH (−k 2 − l 2 ) η̂ = 0 .
For nonzero η̂ , this equation has solutions
ω = 0 and ω 2 = f 2 + gH ( k 2 + l 2 ) = f 2 + gHK 2 ,
where K2 = k2 + l2, and the final equation is (13.82).
c) The specification ω = 0 implies steady flow, and the steady flow solutions of (13.45) – (13.47)
are geostrophic flows.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 13.5. Find the axis ratio of a hodograph plot for a semidiurnal tide in the middle of the
ocean at a latitude of 45°N. Assume that the mid-ocean tides are rotational surface gravity waves
of long wavelength and are unaffected by the proximity of coastal boundaries. If the depth of the
ocean is 4 km, find the wavelength, the phase velocity, and the group velocity. Note, however,
that the wavelength is comparable to the width of the ocean, so that the neglect of coastal
boundaries is not very realistic.

Solution 13.5. Given f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1, ω = 2 rev./day =
1.45 x10–4 s–1, and H = 4 km, the axis ratio (as given at the top of page 733) is
axis ratio = ω/f = 1.45/1.03 = 1.414.
At the given depth the phase speed is:
c = [gH]1/2 = [(9.81 m2/s)(4,000m)]1/2 = 198 m/s,
and the group speed is the same because it is a shallow water wave. The wavelength is:
λ = c(period) = (198 m/s)(12 hrs)(3600 s/hr) = 8554 km.
This wavelength is long enough so that the neglect of coast boundaries is an unrealistic
approximation.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 13.6. An internal Kelvin wave on the thermocline of the ocean propagates along the
west coast of Australia. The thermocline has a depth of 50 m and has a nearly discontinuous
density change of 2 kg/m3 across it. The layer below the thermocline is deep. At a latitude of
30°S, find the direction and magnitude of the propagation speed and the decay scale
perpendicular to the coast.

Solution 13.6. The given information is: f = –2Ωsin30° = –0.73x10–4 s–1, H = 50 m, and Δρ = 2
kg/m3. This is the case of a shallow layer of lighter water, overlying a deep sea. From equation
(8.115), the internal gravity wave speed is:
Δρ 2kgm −3
c = g!H = g −1
H = (9.81ms ) 3 −3
(50m) = 0.99ms−1 .
ρo 10 kgm
These waves propagate southward along the west coast of Australia. The decay scale
perpendicular to the coast is the Rossby radius:
c 0.99ms−1
Λ= = −4 −1
= 1.36 ×10 4 m = 13.6km .
f 0.73×10 s
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 13.7. Derive (13.96) for the vertical velocity w from (4.10), (13.48), (13.49), (13.51),
(13.95) by eliminating all other dependent variables.

Solution 13.7. The five starting-point equations are:

∂u ∂v ∂w
+ + =0, (4.10)
∂x ∂y ∂z
∂u 1 ∂p# ∂v 1 ∂p#
− fv = − , + fu = − , (13.48, 13.49)
∂t ρ0 ∂x ∂t ρ0 ∂y
∂ρ " N2
− ρ0 w = 0 , and (13.51)
∂t g
∂w 1 ∂p# gρ #
=− − . (13.95)
∂t ρ0 ∂z ρ0
Apply ∂/∂t to (13.48), multiply (13.49) by f, and add the results to find:
" ∂2 % 1 " ∂2 p) ∂p) %
'. (a)
2
$ 2 + f ' u = − $ +f
# ∂t & ρ0 # ∂x∂t ∂y &
Multiply (13.48) by –f, apply ∂/∂t to (13.49), and add the results to find:
" ∂2 % 1 " ∂2 p) ∂p) %
'. (b)
2
$ 2 + f 'v = − $ −f
# ∂t & ρ0 # ∂y∂t ∂x &
Apply ∂/∂x to (a) and ∂/∂y to (b), and add the results to reach:
" ∂2 %" ∂u ∂v % " ∂2 %
2 " ∂w % 1 " ∂2 ∂2 % ∂p)
2
$ 2 + f '$ + ' = $ 2 + f '$ − ' = − $ 2 + 2 ' , (c)
# ∂t &# ∂x ∂y & # ∂t &# ∂z & ρ0 # ∂x ∂y & ∂t
where the first equality follows from (4.10). Now apply ∂/∂t to (13.95), and use (13.51) to
substitute for ∂ρ´/∂t:
∂2 w 1 ∂2 p# g ∂ρ # 1 ∂2 p# g $ N 2 ' 1 ∂2 p#
= − − = − − & ρ 0 w ) = − − N 2w . (d)
∂t 2 ρ0 ∂z∂t ρ0 ∂t ρ0 ∂z∂t ρ0 % g ( ρ0 ∂z∂t
Use the definitions ∇ 2H ≡ ∂2 ∂x 2 + ∂2 ∂y 2 to rewrite the final equality of (c), and rewrite the ends
of the extended equality (d):
" ∂2 %
2 ∂w 1 2 ∂p) " ∂2 % 1 ∂2 p(
$ 2 + f ' = ∇ H , and $ 2 + N 2
' w = (e,f)
# ∂t & ∂z ρ0 ∂t # ∂t & ρ0 ∂z∂t
Apply ∂/∂z to (e) and ∇ 2H to (f), and add the results to find:
" ∂2 % 2
2 ∂ w
" ∂2 % 1 2 ∂2 p) 1 2 ∂2 p)
2 2
$ 2 + f ' 2 + ∇H $ 2 + N ' w = ∇H − ∇H =0.
# ∂t & ∂z # ∂t & ρ0 ∂z∂t ρ0 ∂z∂t
Rearrange the terms noting that ∇ 2H + ∂2 ∂z 2 ≡ ∇ 2 :
∂2 w ∂2 w
∇ 2 2 + f 2 2 + N 2 ∇ 2H w = 0 ,
∂t ∂z
which is (13.96).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 13.8. Using the dispersion relation m2 = k2(N2 − ω2)/(ω2 − f2) for internal waves, show
(N 2 − f 2 )km
that the group velocity vector is given by !"cgx , cgz #$ = 2 2 3 2 2 2 2 2 1 2 [ m, −k ] .
(m + k ) (m f + k N )
[Hint: Differentiate the dispersion relation partially with respect to k and m.] Show that cg and c
are perpendicular and have oppositely directed vertical components. Verify that cg is parallel to
u.]

Solution 13.8. Start with m2 = k2(N2 − ω2)/(ω2 − f2), and algebraically rearrange it to find:

ω2 = f2 + k2(N2 − ω2)/m2. (1)

ω
Here we note that c = (cx, cz) = (k, m) , so the components of the group velocity must be
k + m2
2

found next.
To find cgx, differentiate (1) with respect to the horizontal wave number k:
∂ω 2k 2 k 2 # ∂ω &
2ω = 2 ( N − ω 2 ) − 2 % 2ω (.
∂k m m $ ∂k '
Set cgx = ∂ω/∂k, to find:
2k k2
2ω cgx = 2 ( N 2 − ω 2 ) − 2 ( 2ω cgx ) ,
m m
or ω cgx ( m + k ) = k ( N 2 − ω 2 ) ,
2 2

k (N 2 −ω2 )
or cgx = . (2)
ω (m2 + k 2 )
From here ω can be eliminated from (2) using (1) to find:
km 2 ( N 2 − f 2 )
cgx = .
2 32 2 12
(m + k ) (m f + k N )
2 2 2 2

To find cgy, differentiate (1) with respect to the vertical wave number m:
∂ω k 2 # ∂ω & 1
( − 2k ( N − ω ) 3 .
2 2 2
2ω = 2 % −2ω
∂m m $ ∂m ' m
Set cgz = ∂ω/∂m, and rearrange the last equation to find:
k2 (N 2 −ω2 )
cgz = − . (3)
mω ( m 2 + k 2 )
Here again, ω can be eliminated from (3) using (1) to find:
k 2m ( N 2 − f 2 )
cgz = − .
2 32 2 12
( m 2
+ k ) ( m 2 2
f + k 2
N )
(N 2 − f 2 )km
Thus we can write: cg = ( cgx , cgz ) = ( m, −k ) .
(m 2 + k 2 )3 2 (m 2 f 2 + k 2 N 2 )1 2
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

To show that cg and c are perpendicular, it is sufficient to show that their dot product is
zero:
(N 2 − f 2 )km ω
cg ⋅ c =2 2 32 2 2 2 2 12 (
m, −k ) ⋅ 2 (k, m) = 0 .
(m + k ) (m f + k N ) k + m2
Here the sign of cz is opposite that of cgz , so it follows that the vertical components of c and cg
are oppositely directed.
Finally, cg and u are parallel because:
cgx m u
=− = ,
cgz k w
where (13.111) has been used to reach the final equality.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 13.9. Suppose the atmosphere at a latitude of 45°N is idealized by a uniformly stratified
layer of height 10 km, across which the potential temperature increases by 50°C.
a) What is the value of the buoyancy frequency N?
b) Find the speed of a long gravity wave corresponding to the n = 1 baroclinic mode.
c) For the n = 1 mode, find the westward speed of nondispersive (i.e., very large wavelength)
Rossby waves. [Answer: N = 0.01279 s−1; c1 = 40.71 m/s; cx = −3.12 m/s]

Solution 13.9. The given info. is f = 2Ωsin45° = 2(0.73x10–4 s–1)(0.7071) = 1.03 x10–4 s–1, H = 10
km, and ΔT = 50 °C. At absolute temperature T, the coefficient of thermal expansion for a
perfect gas is 1/T.
a) Therefore the buoyancy frequency is given by:
g dρ dT g dT 9.81ms−2 50°
N2 = − = gα = = = 1.635 ×10 −4 s−2 , so N = 0.01279 s–1.
ρo dz dz T dz 300K 10 4 m
b) From equation (13.71), the internal gravity wave speed corresponding to the nth normal mode
is cn = NH/nπ, so
c1 = NH/π = (0.01279 s–1)(104 m)/π = 40.71 m/s.
c) From Section 13.13, the westward speed of a non-dispersive Rossby wave (corresponding to
the n = 1 mode) is:
cx ≅ − β c12 f 2 = −(2 ×10 −11 s−1m −1 )(40.71ms−1 )2 (1.03×10 −4 s−1 )2 = −3.12ms−1 .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 13.10. Consider a steady flow rotating between plane parallel boundaries a distance L
apart. The angular velocity is Ω and a small rectilinear velocity U is superposed. There is a
protuberance of height h << L in the flow. The Ekman and Rossby numbers are both small: Ro
<< 1, E << 1. Obtain an integral of the relevant equations of motion that relates the modified
pressure and the stream function for the motion, and show that the modified pressure is constant
on streamlines.

Solution 13.10. For Rossby and Ekman numbers both much less than unity (Ro << 1 and E <<
1), the Taylor-Proudman theorem gives
∇p
2Ω × u = −
ρ
for the momentum equation, where p includes the hydrostatic pressure. And, for incompressible
flow, we must have: ∇ ⋅ u = 0 . The boundary conditions w = 0 on z = 0 and L require w = 0
everywhere and ∂u/∂z = 0. Thus, we can set u = (u, v), and since all streamlines are in z =
constant planes we can set u = −e z × ∇ψ where ψ is the 2D stream function. Here Ω = ezΩ, so
the momentum equation becomes
∇p ∇p
2Ωe z × (−e z × ∇ψ ) = − , or −2Ω%&e z ( e z ⋅ ∇ψ ) − ∇ψ (e z ⋅ e z )'( = − .
ρ ρ
In the second equation, the first term in (,)-parentheses is zero because ez is perpendicular to the
plane of the flow. Thus, the above equation reduces to:
∇p
2Ω∇ψ = − .
ρ
Since all streamlines fall in z = constant planes, the above equation can be written:
∇( p + 2 ρΩψ ) = 0 ,
which implies p + 2ρΩψ = constant in z = constant planes. And, since ψ is constant on
streamlines, p is constant on streamlines, too.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 13.11. Consider an atmosphere of height H that initially contains quiescent air and N
different cyclonic disks of height H and radius Ri inside which the air rotates at rate Ωi. After
some time, the various cyclonic disks merge into one because of the reverse energy cascade of
geostrophic turbulence. Show that the radius Rf and rotation rate Ωf of the single final disk is

<Begin Equation>
2
N
R 2f = ∑ Ωi2 Ri4
i=1
∑
N

i=1
Ωi2 Ri2 and, Ω2f = (∑ N

i=1
Ωi2 Ri2 ) ∑
N

i=1
Ωi2 Ri4 .
</End Equation>

by conserving energy and enstrophy. How are these answers different if all of the energy but
only a fraction ε (0 < ε < 1) of the enstrophy is retained after the merging process? Assume the
relevant horizontal area is the same at the start and end of the disk-merging process.

Solution 13.11. This exercise may solved similarly to Example 13.14. For an atmosphere of
height H, the kinetic energy KE of rotating disks of air with radii Ri undergoing solid body
rotation with rates Ωi is:
π π
KE = ∑ ρo HΩi2 Ri4 = ρo HΩ2f R 4f .
i 4 4
where ρo is the altitude-averaged density and the second equality applies to the final fully-
merged rotating disk of air. The vorticity inside each of the initial disks is 2Ωi, so conservation of
enstrophy requires
1 π R 2f
Enstropy = ∑ π Ri2 (2Ωi )2 = (2Ω f )2 .
A i A
where A is the relevant horizontal area for averaging the square of the vorticity. Simplify the two
equations by dividing out common factors:
∑Ωi2 Ri4 = Ω2f R 4f and ∑ Ri2Ωi2 = R 2f Ω2f .
i i
Simultaneous solution of these two equations leads to:
2
R =∑ Ω R
2
f
N

i=1
2
i i
4
∑
N

i=1
Ω R is , Ω =
2
i i
2 2
f (∑ N

i=1
ΩR 2
i i
2
) ∑ i=1
N
Ωi2 Ri4 .
When only a fraction ε of the enstropy is retained, then two equations for simultaneous
solution are:
∑Ωi2 Ri4 = Ω2f R 4f and ε ∑ Ri2Ωi2 = R 2f Ω2f ,
i i
and these lead to:
2
N
R 2f = ∑ Ωi2 Ri4
i=1 ( N
)
ε ∑ Ωi2 Ri2 is , Ω2f = ε ∑i=1 Ωi2 Ri2
i=1 ( N
) ∑
N

i=1
Ωi2 Ri4 .
Thus, when enstropy is lost the final rotating disk is larger and rotates slower than when enstropy
is conserved, as was found in Example 13.14 for two cyclonic disks.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.1. As an extension of Example 14.1, consider a sphere with radius a that moves
along the x-axis on a trajectory given by xp(t) = xp(t)ex in a fluid moving with uniform velocity: u
= Uex + Vey. Determine a formula for the mechanical power, W, necessary to overcome the
aerodynamic drag force on the sphere in terms of a, U, V, xp, ρ = the density of the air, and CD =
the drag coefficient of the sphere. If dxp/dt is constant, under what conditions is W reduced by the
presence of non-zero u?

Solution 14.1. The drag force magnitude will depend on the square of the relative velocity
between the sphere and the fluid. The drag force direction will be parallel to the relative velocity
of the sphere and the fluid. The relative velocity is the fluid velocity observed when riding on the
sphere. If the fluid velocity is u = (U,V), then Vrel = (U − dx p dt ) e x +Ve x , so the drag force vector
is:
π a2 2
D=ρ CD (U − dx p dt ) +V 2 "#(U − dx p dt ) e x +Ve x $% .
2
When the sphere is stationary, then D points in the same direction as U. When the fluid is
stationary D points in the opposite direction of dxp/dt; it opposes the motion of the sphere.
From Newton's third law, the force applied to the fluid by the motive mechanism of the
sphere to overcome the drag force is –D. The power output of this force is:
# dx & π a2 2
W = −D ⋅ % p e x ( = −ρ CD (U − dx p dt ) +V 2 (U − dx p dt ) ( dx p dt ) .
$ dt ' 2
When the fluid velocity is zero, then the power is:
π a2 3
Wo = ρ CD ( dx p dt ) .
2
To find out when non-zero u reduces W, first determine the boundary where W = Wo
π a2 3 π a2 2
ρ CD ( dx p dt ) = −ρ CD (U − dx p dt ) +V 2 (U − dx p dt ) ( dx p dt ) .
2 2
U V
Divide out common factors, and define normalized fluid velocities: U! = and V! =
dx p dt dx p dt
2
to reach: 1 = − U! −1 + V! 2 U! −1 .
( ) ( )
Rearrange and square to eliminate the square root, then solve for V!
1 1 1
U! −1
(
2
)
(
2
= U! −1 + V! 2 which implies: V! 2 =
)
U! −1
2
− U! −1 =
(
2

U! −1
)
( )
( )
2 (
1− U! −1
( ) ).
4

Take the square root and factor inside of the last set of parentheses:
12
! 1 (" ! 2 %"
! 2 %+ 1 12
"2U! − U! 2 %"U! 2 + 2U! + 2% .
)( )
V =±
U! −1 )#
( &#) (
*$1− U −1 '$1+ U −1 '- = ±
&, ) U! −1
# &# &
( ) (
When plotted, this relationship leads to:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

10$
8$
6$
4$
2$
V!
0$
!2$
!4$
!6$
!8$
!10$
0$ 0.25$ 0.5$ 0.75$ 1$ 1.25$ 1.5$ 1.75$ 2$
U!
where the area inside the four-lobed contour is where W < Wo.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.2. Consider the elementary aerodynamics of a projectile of mass m with CL = 0 and
CD = constant. In Cartesian coordinates with gravity g acting downward along the y-axis, a set of
equations for such a projectile's motion are:
dV dVy 1
m x = −Dcos θ , m = −mg − Dsin θ , tan θ = Vy Vx , and D = ρ(Vx2 + Vy2 ) ACD ,
dt dt 2
where Vx and Vy are the horizontal and vertical components of the projectile's velocity, θ is the
angle of the projectile's trajectory with respect to the horizontal, D is the drag force on the
€ projectile,2 ρ is€the air density, and A is€ projectile's frontal €area. Assuming a shallow trajectory,
where Vx >> Vy2 and mg >> Dsin θ , show that the distance traveled by the projectile over level
2m % ρACDVo2 cos θ o sin θ o (
ground is: x ≅ ln'1+ * if it is launched from ground level with speed
ρACD & mg )
€
€ of Vo at an angle of θo with respect to the horizontal. Does this answer make sense as CD → 0 ?

€
€

Solution 14.2. Insert the approximations in the horizontal and vertical equations of motion to
find:
dV 1 dVy
m x ≅ − ρVx2 ACD , and m ≅ −mg .
dt 2 dt
Here it has been assumed that the trajectory is shallow enough so that cosθ is approximately
unity on the projectile's trajectory. The initial conditions are Vx = Vocosθo and Vy = Vosinθo at t =
0. Both equations€can be integrated once to find:
€
1 ρACD 1 ρACD
− ∫ 2 dVx ≅
Vx 2m
∫ dt or V = 2m t + C1 , and m ∫ dVy = −mg ∫ dt or Vy ≅ C2 − gt .
x
where C1 and C2 are constants that can be evaluated with the initial conditions to reach:
1 ρACD 1
≅ t+ , and Vy ≅ Vo sin θ o − gt .
Vx 2m Vo cosθ€o €
€ €
To find the approximate position of the projectile, these equations for the velocity components
must be integrated again using Vx = dx/dt and Vy = dy/dt with initial conditions x = 0 and y = 0 at
t = 0. The horizontal coordinate equation can€ be rearranged and integrated,
€ dt
1 dt ρACD 1
= ≅ t+ , Vo cosθ o ∫ ≅ ∫ dx ,
Vx dx 2m Vo cosθ o ( ρACDVo cosθ o 2m)t + 1
to find:
2m % ρACDVo cos θ o (
x≅ ln'1+ t * + C3 .
€ ρ ACD & 2m )
€
The initial condition leads to C3 = 0. The vertical coordinate equation can be integrated directly:

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

dy 1
Vy = ≅ Vo sin θ o − gt or y ≅ Vo t sin θ o − gt 2 + C4 .
dt 2
and the initial condition again leads to C4 = 0.
The time, tf, the projectile is in flight over level ground can be determined from the
vertical coordinate equation by looking for the second time y = 0:
€ 1 € 2V sin θ o
y = 0 ≅ Vo t f sin θ o − gt 2f , which implies: tf = 0 , or t f ≅ o .
2 g
Substituting this time into the horizontal coordinate equation determines the projectile's
horizontal location when it hits the ground:
€ 2m % ρACDVo2 cos (
€θ o sin θ o * .
x≅ ln'1+
ρACD & mg )
When CD → 0 the frictionless mechanics solution should be recovered. Using the
expansion of the natural logarithm, ln(1 + ε) = ε + ... for ε << 1, the above equation becomes:
2m ρACDVo2 cosθ o sin θ o 2Vo2 cosθ o sin θ o
€ x≅ = ,
€ ρACD mg g
which is correct when the projectile flies in vacuum.
Although the main result of this exercise is approximate, it does incorporate the
physically important feature that aerodynamic drag decreases projectile travel distances in still
air. €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.3. As a model of a two-dimensional airfoil's trailing edge flow consider the potential
n
φ (r,θ ) = (Ud n )( r d ) cos( nθ ) in the usual r-θ coordinates (Figure 3.3a). Here U, d, and n are
positive constants, the fluid has density ρ, and the foil's trailing edge lies at the origin of
coordinates.
a) Sketch the flow for n = 3/2, 5/4, and 9/8 in the angle range |θ| < π/n, and determine the full
€
included angle of the foil's trailing edge in terms of n.
b) Determine the fluid velocity at r = d and θ = 0.
c) If p0 is the pressure at the origin of coordinates and pd is the pressure at r = d and θ = 0,
( )
determine the pressure coefficient: C p = ( p0 − pd ) 12 ρU 2 as a function of n. In particular, what
is Cp when n = 1 and when n > 1?

Solution 14.3. a) The velocity components in this flow are:

€
∂ Ud r n−1 & r ) n−1
ur = φ = n n cos(nθ ) = U( + cos(nθ ) , and
∂r n d ' d*
1 ∂ Ud r n−1
& r ) n−1
uθ = φ= ( )
−n sin(n θ ) = −U ( + sin(nθ ) .
r ∂θ n dn ' d*
The angular €velocity is zero at: θ = 0, ± π/n. The radial velocity is positive on θ = 0, but negative
on θ = ± π/n. Thus, the full included angle of the foil's trailing edge is 2(π – π/n) = 2π(n – 1)/n,
and this angle decreases toward zero as n →1.
€
For n = 3/2, the full included angle of the foil's trailing edge is = 2π(3/2 – 1)/(3/2) = 2π/3 (120°).
y
€

120°
x

For n = 5/4, the full included angle of the foil's trailing edge is = 2π(5/4 – 1)/(5/4) = 2π/5 (72°).
y

72°
x

For n = 9/8, the full included angle of the foil's trailing edge is = 2π(9/8 – 1)/(9/8) = 2π/9 (40°).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

40°
x

b) From the velocity results provided above in part a), the fluid velocity at r = d and θ = 0 is:
" d % n−1 $ d ' n−1
ur (d,0) = U$ ' cos(0) = +U , and uθ (d,0) = −U& ) sin(0) = 0 .
# d& %d(
c) To determine the coefficient of pressure, use the steady-flow Bernoulli equation without the
body-force term:
$ 1 2 ' $ 1 2 '
&% p + 2 ρ ( ur + uθ )€)( = &% p + 2 ρ( ur + uθ ))(
2 2
€ ,
r= 0 r= d ,θ = 0
and evaluate:
1 $ 2$ r ' '
2n−2 2n−2
2$ r ' 1
p0 + lim ρ&&U & ) 2
cos (nθ ) + U & ) sin 2 (nθ ))) = pd + ρ (U 2 + 0) .
€r →0 2 % % d ( % d( ( 2
Simplify and rearrange:
1 % % r ( 2n−2 ( $ r ' 2(n−1)
2
p0 − pd = ρU ''1− lim' * ** , or C p = 1− lim& ) .
2 r →0% d (
€ & r →0& d ) )
Therefore:
$1 for n > 1(
& &
€ C p = %0 € for n = 1).
&−∞ for n < 1&
' *
This result shows that there is a stagnation point at a foil's trailing edge when the trailing-edge
included angle is non-zero.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.4. Consider an airfoil section in the xy-plane, the x-axis being aligned with the chord
line. Examine the pressure forces on an element ds = (dx, dy) on the surface, and show that the
net force (per unit span) in the y-direction is
c c
Fy = − ∫ 0 pu dx + ∫ 0 pl dx
where pu and p1 are the pressure on the upper and the lower surfaces and c is the chord length.
Show that this relation can be rearranged in the form
Fy # x&
€ Cy = 2
= ∫ C p d% ( ,
(1/2) ρU c $c'
1
( 2
)
where C p = ( p0 − p∞ ) 2 ρU , and the integral represents the area enclosed in a Cp vs x/c
diagram, such as Figure 14.8. Neglect shear stresses. [Note that Cy is not exactly the lift
coefficient, since the airstream
€ is inclined at a small angle α with respect to the x-axis.]
€ Solution 14.4. Here's a nominal drawing of a foil with the x-axis along the chord line of length c.
y
c
ds !
x

dx
For a small element ds = (dx, dy) = |ds|(cosθ, sinθ) of the foil's upper surface, the y-direction
force per unit span is
dFy = −( pu | ds |)cosθ = − pu dx .
Thus, if pu and p1 are the pressure on the upper and the lower surfaces and c is the chord length,
then
c c c c
Fy = ∫€dFy = − ∫ pu dx + ∫ pl dx = − ∫ ( pu − p∞ )dx + ∫ ( pl − p∞ )dx .
0 0 0 0
This can be recast in terms of a vertical force coefficient by dividing by 12 ρU 2c :
F c
(p − p ) c
(p − p ) c
& x) c & x) & x)
Cy = 1 y 2 = − ∫ 1 u 2 ∞ dx + ∫ 1 l 2∞ dx = − ∫ C p,u d( + + ∫ C p,l d( + = ∫ C p d( + .
€ 2 ρU c 0 2
ρU c 0 2 ρU c 0 'c* 0 'c* 'c*
€

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.5. The measured pressure distribution over a section of a two-dimensional airfoil at
4° incidence has the following form:
Upper Surface: Cp is constant at −0.8 from the leading edge to a distance equal to
60% of chord and then increases linearly to 0.1 at the trailing edge.
Lower Surface: Cp is constant at −0.4 from the leading edge to a distance equal to
60% of chord and then increases linearly to 0.1 at the trailing edge.
Using the results of Exercise 14.4, show that the lift coefficient is nearly 0.32.

Solution 14.5.The distribution of the pressure coefficient is sketched below.

Cp
E !"%
x

A B
–!"#

–!"$ D C
0.6c 0.4c
Using the result of Exercise 14.3,
# x& 1
Cy = ∫ C p d% ( = ABCD + BCE = (0.4)(0.6) + (0.4)(0.4) = 0.32 .
$c' 2
The coefficient of lift is:
CL = Cy cosα = Cy cos 4° = Cy (0.997564) = 0.3192 ,
which€is nearly 0.32.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.6. Zhukhovsky transformation z = ζ + b2/ζ transforms a circle of radius b, centered

at the origin of the ζ-plane, into a flat plate of length 4b in the z-plane. The circulation around the
cylinder is such that the Kutta condition is satisfied at the trailing edge of the flat plate. If the
plate is inclined at an angle α to a uniform stream U, show that
" −iα b 2 +iα % iΓ
(i) The complex potential in the ζ-plane is w = U $ζ e + e ' +
# ζ & 2π
(
ln ζ e−iα , where Γ =)
4πUbsinα. Note that this represents flow over a circular cylinder with circulation, in
which the oncoming velocity is oriented at an angle α.
(ii) The velocity components at point P (−2b, 0) in the ζ-plane are !" 43 U cos α, 94 U sin α #$
(iii) The coordinates of the transformed point Pʹ′ in the xy-plane are [−5b/2, 0].
(iv) The velocity components at [−5b/2, 0] in the xy-plane are [Ucosα, 3Usinα].

Solution 14.6. The three complex planes for this problem appear as:

!´-plane !-plane z-plane

! = !´ei" z = !#+ b2/!

U P P´
(-2b,0) (-5b/2,0)
" "
U

(i) From equation (14.11), the circulation will be Γ = 4 πUbsin α . In the ζ´-plane choose U to
flow parallel to the real axis. Then, the transformation ζ = ζ #e iα rotates the ζ´-plane counter-
clockwise to produce the ζ-plane. The complex potential is:
w = U (ζ " + b 2 ζ€") + (iΓ 2π ) ln ζ "
= U (ζe−iα + b 2e +iα ζ€) + (iΓ 2π ) ln ζe−iα
= U (ζe−iα + b 2e +iα ζ ) + (iΓ 2π )[ln ζ − iα ]
(ii) The complex velocity is:
dw
= U (e−iα − b 2e +iα ζ 2 ) + iΓ 2π ζ .
dζ
€
At point P,
# dw &
% ( = U (e − b e (−2b) ) + iΓ 2π (−2b)
−iα 2 +iα 2

$ dζ 'P€
- 1 0
= U/ cosα − isin α − (cosα − isin α )2 − i(4 πUbsin α ) 4 πb
. 4 1
3 5 3 9
= U cosα − i sin α − iU sin α = U cos α − i sin α = u − iv
4 4 4 4
3 9
so: u = U cosα , and v = U sin α .
4 4
2
(iii) Based on the transformation, z = ζ + b ζ , the transformed point is
€

€ €
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

zP" = −2b + b 2 (−2b) = −5b /2 .

(iv) The complex velocity at (–5b/2, 0) in the z-plane is:
−1 −1
dw dw dζ dw # dz & dw # b 2 & dw # ζ 2 &
= = % ( = %1− 2 ( = % (
dz dζ€dz dζ $ dζ ' dζ $ ζ ' dζ $ ζ 2 − b 2 '
#3 9 &# (−2b) 2 &
= % U cos α − i sin α (% ( = U cos α − i3sin α = u − iv.
$4 4 '$ (−2b) 2 − b 2 '
where ζ = (–2b, 0) has been used as the corresponding point in the ζ-plane.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.7. In Figure 14.13, the angle at Aʹ′ has been marked 2β. Prove this. [Hint: Locate the
center of the circular arc in the z-plane.]

Solution 14.7. From the discussion following Figure 14.12, point C is located on the imaginary
axis in the z-plane at –2bcot2β.

!-plane z-plane

z = !"+ b2/!
A a A´ $ B´
# B –2b & D

C –2bcot2#

From triangle A´DC

2b
tan γ = = tan2β , so γ = 2β.
2bcot 2β
Plus, the full angle at A´ is π/2 so
π
δ +θ = , (i)
2
and A´DC is a right €
triangle so:
π
δ+γ = = δ + 2β . (ii)
2
Comparing (i) and (ii), requires θ €
= 2β .

€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.8. Ideal flow past a flat plate inclined at angle α with respect to a horizontal free
stream produces lift but no drag when the Kutta condition is applied at the plate's trailing edge.
However, pressure forces can only act in the plate-normal direction and this direction is not
perpendicular to the flow. Therefore, to achieve zero drag, another force must act on the plate.
This extra force is known as leading-edge suction and its existence can be assessed from the
potential for flow around the tip of a flat plate that is coincident with the x-axis for x > 0. In two-
dimensional polar coordinates, this velocity potential is φ = 2U o ar cos(θ 2) where Uo and a are
velocity and length scales, respectively, that characterize the flow.
a) Determine ur and u , the radial and angular-directed velocity components, respectively.
θ

b) If the pressure far from the origin is p∞, determine the pressure p at any location (r, θ).
€
c) Use the given potential, a circular control volume of radius ε centered at the origin of
coordinates, and the control volume version of the ideal flow momentum equation,
∫ ρu(u ⋅ n)dξ = − ∫ pndξ + F , to determine the force F (per unit depth into the page) that holds
C C
the plate stationary when ε → 0 . Here, n is the outward unit normal vector to the control volume
surface, and dξ is the length increment of the circular control surface.
d) If the plate is released from rest, in what direction will it initially accelerate?
€
€

Solution 14.8. a) Directly differentiate the potential φ = 2U o ar cos(θ 2) .

12
ur = ∂φ ∂r = 2U o a cos(θ 2) ⋅ r−1 2 2 = U o ( a r) cos(θ 2) , and
12
uθ = (1 r)(∂φ ∂θ ) = (1 r)2U o ar [−sin(θ 2) ⋅ (1 2)] = −U o ( a r) sin(θ 2)
b) Use the Bernoulli equation to determine€p = p(r, θ) :
€ p∞ = p + 2 ρ( ur + uθ ) = p + 2 ρU o ( a r)(cos (θ 2) + sin (θ 2)) = p + 2 ρU o ( a r) .
1 2 2 1 2 2 2 1 2

Thus,
€ the pressure does not depend on the angle θ:
p − p∞ = − 12 ρU o2 ( a r) ,
but it does become very negative as r approaches zero.
€
c) To evaluate the Cartesian components of F, the Cartesian components of the velocity field are
needed. Convert the part a) results using: ux = ur cosθ − uθ sin θ , and uy = ur sin θ + uθ cos θ , to
find: €
12
ux = U o ( a r) [cos(θ 2) cos θ + sin(θ 2) sin θ ] , and
12
uy = U o (€a r) [cos(θ 2) sinθ − sin(θ€2) cosθ ] .
€
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

These can be simplified to: u x = U o ( a r)1 2 cos(θ 2) , and uy = U o ( a r)1 2 sin(θ 2) , using
trigonometric identities. Here n = e r = e x cosθ + e x sin θ , so u ⋅ n = u ⋅ e r = ur . Thus, the CV
equation can be written in terms of x- and y-components:
2π 2π
∫ 0 ρ(uxe€x + uye y )urεdθ = − ∫ 0 p(e x cosθ€+ e x sinθ )εdθ + Fxe x + Fye y .
where dξ = εdθ. The pressure
€ does not depend on θ, and
€ the integrals of cosθ and sinθ from 0 to
2π are both zero so the pressure integration drops out leaving:
2π
€ ∫0 ( 12
) 12
ρ U o (a ε) cos(θ 2) U o ( a ε) cos(θ 2)εdθ = Fx , and
2π
∫0 ( 12
) 12
ρ U o (a ε) sin(θ 2) U o ( a ε) cos(θ 2)εdθ = Fy .
The ε contour-radius factors cancel, so the integrands can be simplified to:
€ ρU 2 a 2 π cos 2 θ 2 dθ = F , and ρU 2 a 2 π sin θ 2 cos θ 2 dθ = F .
o ∫0 ( ) x o ∫0 ( ) ( ) y
Use the double-angle
€ trigonometric identities to evaluate the integrals to find:
2π 1 2π 1
ρU o2 a ∫ 0
2
(cosθ + 1)dθ = πρU o2 a = Fx , and ρU o2 a ∫ 0
2(sinθ )dθ = 0 = F .
y
The€fact that ε does not appear in either€ final answer suggests that the horizontal force, known as
leading edge suction, found in part d) will exist for any ε, even ε → 0 . This force is applied to
the tip of the plate and arises from the singularity in the potential at r = 0.
€ Interestingly, the real-world effects of €this singularity are exploited for both natural and
anthropogenic flight. The leading edges of real airfoils are rounded so that the flow remains
€
attached to the foil's surface and does not have a singularity. Attached flow passing around a
finite-radius-of-curvature leading edge produces a leading-edge suction force of the size
predicted here that increases the lift and reduces the drag of real airfoils. The only superfluous
part of the above discussion and analysis is the limit ε → 0 .
d) Fx holds the plate stationary by pushing it to the right. Therefore, if Fx is not applied, the plate
will accelerate to the left.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.9. Consider a cambered Zhukhovsky airfoil determined by the following

parameters: a = 1.1, b = 1.0, and β = 0.1. Using a computer, plot its contour by evaluating the
Zhukhovsky transformation. Also plot a few streamlines, assuming an angle of attack of 5°.

Solution 14.9. In the case the point Q at the center of the circle in the ζ-plane lies at:
(Qx ,Qy ) = (b − cos β ,asin β ) ≅ (−0.0945,0.1098) .
Thus, horizontal and vertical coordinates of the circle are given by:
(ζ x ,ζ y ) = (Qx + acos θ,Qy + asin θ ) ,
or as a complex
€ variable:
ζ = ζ x + iζ y = Qx + acosθ + i(Qy + asin θ )
where 0 ≤ θ ≤ 360°. Using
€ the Zhukhovsky transformation:
b2
z = x + iy = ζ +
€ ζ
then leads to:
b 2ζ b 2ζ
x = ζ x + 2 x 2 and y = ζ y − 2 y 2 .
€ ζx + ζy ζx + ζy
where (x, y) specifies the coordinates of the foils surface.
Use of a spread sheet program for evaluating and plotting with a = 1.1, b = 1.0, and β =
0.1 (radians), for 1° steps of θ produces the following plot for the foil shape:
€ €
&%

"#$%

"%
!'#$% !'% !&#$% !&% !"#$% "% "#$% &% &#$% '% '#$%

!"#$%

in the z-plane.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.10. A thin Zhukhovsky airfoil has a lift coefficient of 0.3 at zero incidence. What is
the lift coefficient at 5° incidence?

Solution 14.10. From (14.12), the lift coefficient of a Zhukhovsky airfoil is:
CL = 2π (α + β ) .
If CL = 0.3 when α = 0, then β = 0.3/2π. Therefore when α = 5°,
# 5° 0.3 &
CL = 2π % π+ ( = 0.848 .
€ $180° 2π '

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.11. The simplest representation of a three-dimensional aircraft wing in flight is the
rectangular horseshoe vortex.
a) Calculate the induced downwash at the center of the wing.
b) Assuming the result of part a) applies along the entire wingspan, estimate CDi , the lift-induced
coefficient of drag, in terms of the wing’s aspect ratio: AR = s2/A, and the wing’s coefficient of
( )
lift CL = L 12 ρU 2 A , where A is the planform area of the wing.
c) Explain why the result of part b) appears to surpass the performance
€ of the optimal elliptic lift
distribution.
€

Solution 14.11. a) Use the result of exercise 14.10. For the rectangular horseshoe vortex, each of
Γ Γ
the trailing vortices induces a vertical velocity of w = − (cos90° − cos180°) = − .
4 π (b /2) 2πb
Γ
Thus, the total downwash velocity will be twice this: w total = − .
πb
b) If this downwash is constant along the whole span, the wing’s lift will be decreased because
€
−w total Γ
the incoming stream will be rotated downward through the small angle ε = = , so the
U πbU
€
Γ Γ Γ2
lift-induced drag will be: Di ≅ Lε = L = ρUΓb = ρ . The coefficient of drag will be:
πbU πbU π
2
Di 2
ρΓ π 2 2 2
ρUΓ % ρUΓb € ( S CL2
CD,i = 1 = = = ' * =
2
ρU 2 S 12 ρU 2 S 12 πρ 2U 4 S & 12 ρU 2 S ) 2πb 2 2πAR
c) The induced € drag coefficient of part b) is half as large as that obtained from the optimal
elliptic lift distribution on a finite wing because the variation of the induced downwash along the
wingspan was not taken into account. The two trailing vortices induce larger downwash
€ near the wing tips than the estimate produced from the arithmetic in part a). In fact,
velocities
the answer to part a) is the minimum downwash along the wing.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.12. The circulation across the span of a wing follows the parabolic law
( 2
)
Γ = Γ0 1− (2y s) . Calculate the induced velocity w at midspan, and compare the value with
that obtained when the distribution is elliptic.

dΓ
€ ( 2
)
Solution 14.12. From Γ = Γ0 1− (2y s) , determine
dy
= −8Γ0 y s2 . Then, from (14.13)

1 +s 2
dΓ dy 8Γ0 +s 2
ydy 2Γ0 +s 2 2Γ0 , & y1 − s 2 )/
w(y1 ) = ∫ = ∫ =
4 π −s 2 dy y1 − y 4 πs2 −s 2 y − y1 πs2
[ 1 ( 1)]−s 2 πs2
y + y ln y − y = .s + y ln
1 ( +1.
- ' y1 + s 2 *0
This downwash € distribution varies along the€span and at midspan, y = 0, it is:
1
2Γ0
w(0) = .
πs
€ In contrast, the constant downwash for an elliptic distribution, Γ 2s , is lower.

€
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.13. An untwisted elliptic wing of 20-m span supports a weight of 80,000 N in a
level flight at 300 km/hr. Assuming sea level conditions, find (i) the induced drag and (ii) the
circulation around sections halfway along each wing.

Solution 14.13. (i) An untwisted wing with an elliptic area is expected to have an elliptic
circulation distribution. The induce drag in this case is given by (14.25):
2L2 2(8 ×10 4 N) 2
Di = = = 1.22kN ,
πρU 2 s2 π (1.2kgm −3 )(83.3ms−1 ) 2 (20m) 2
where 300 km/hr has been converted to 83.3 m/s.
(ii) Also from (14.25), the maximum circulation Γ1 is
π
Di = € ρΓ12 and this implies Γ1 = 8Di ρπ = 8(1.22 ×10 3 N) π (1.2kgm −3 ) = 50.9m 2 s−1 .
8
Halfway along each wing:
$ $ 2(±s /4) ' 2 '1 2 m2 $ $ 1' '
2 12
m2
Γ(±s /4)
€ = Γ &
1&1− & )) ) = 50.9 &1− )
& &% 2 )( ) = 44.1 .
€ % % s ( ( s % ( s

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.14 . A wing with a rectangular planform (span = s, chord = c) and uniform airfoil
section without camber is twisted so that its geometrical angle, αw, decreases from αr at the root
2
(y = 0) to zero at the wing tips (y = ± s/2) according to the distribution: α w (y) = α r 1− (2y s) .
a) At what global angle of attack, αt, should this wing be flown so that it has an elliptical lift
distribution? The local angle of attack at any location along the span will be αt + αw. Assume the
two-dimensional lift curve slope of the foil section is K.
b) Evaluate the lift and the lift-induced drag forces on the wing€ at the angle of attack determined
in part a) when: αr = 2°, K = 5.8 rad.–1, c = 1.5 m, s = 9 m, the air density is 1 kg/m3, and the
airspeed is 150 m/s

Solution 14.14. a) For an elliptical lift distribution without camber (β = 0), only the first two
terms of the finite wing lifting line equation (14.20) are needed:
∞ $
K cKn '
Ucα = ∑&1+ )Γn sin(nγ ) .
2 n=1% 4ssin γ (
The first two terms of this non-traditional Fourier series “solution” are obtained by matching the
left side of the equation to the right side evaluated at n = 1.
K $ cK ' cK
€ Ucα = &1+ )Γ1 sin(γ ) = Γ1 sin(γ ) + Γ1
2 % 4ssin γ ( 4s
2
For the wing twist specified in the problem statement, α = α t + α w = α t + α r 1− (2y s) . Now,
switch to the angle coordinate, y = −( s 2) cos γ , of the lifting line equation so that
α = α t + α r sin€
γ , and plug this into the last equation:
K cK
Uc (α t + α r sin(
€ γ )) = Γ1 sin(γ ) + Γ1.
2 4s
€
Require equality of the constant and sinγ terms on both sides of the equation:
€
K cK K
Ucα t = + Γ1 , and Ucα r = Γ1.
€ 2 4s 2
Eliminate Γ1 and solve for αt; α t = Kcα r (4s) . Note that this angle is positive so the wing must
be pitched slightly upward. For the parameters given in part b) the value of αt is 0.48°.
πs πsc
b) The lift will be: € L = ρUΓ1 = ρU 2Kα € r , so
€ 4 8
π (9m)(1.5m)
L= (1kg /m 3 )(150m /s) 2 (5.8 /rad.)(2°⋅ π /180°) = 24,150 N
8
π π
The lift-induced
€ drag will be: Di = ρΓ12 = ρU 2c 2K 2α r2 , so
8 32
π 3 2
€ Di = (1kg /m )[(150m /s)(1.5m)(5.8 /rad.)(2° ⋅ π /180°)] = 204 N
32
At a flight speed of 150 m/s, overcoming this drag force requires ~30 kW (40 hp).
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.15. Consider the wing shown in Figure 14.25. If the foil section is uniform along the
span and the wing is not twisted, show that the three-dimensional lift coefficient, CL,3D is related
to the two-dimensional lift coefficient of the foil section, CL,2D, by: CL,3D = CL,2D (1+ 2 Λ) ,
where Λ = s2/A is the aspect ratio of the wing.

Solution 14.15. The wing shown in Figure 14.24 has an elliptical planform. Thus, with a uniform
€ and constant downwash w. The
foil section and no twist, it will have an elliptical lift distribution
lift force L3D of the three-dimensional wing can be calculated in terms of an integral of the
circulation Γ(y) at each span location y. Starting from (14.15) and (14.17) with β = 0, this is:
+s 2 +s 2
K ' w* K 2' w *+s 2 K ' w*
L3D = ρU ∫ Γ(y)dy = ρU ∫ c(y))α − , dy = ρU )α − , ∫ c(y)dy = ρU 2 )α − , A .
2

−s 2 2 −s 2 ( U+ 2 ( U +−s 2 2 ( U+
The final two equalities follow because the downwash velocity w is constant for an elliptical lift
distribution, and the integral of the cord c(y) over the span is the planform area A. The two ends
of this extended equality imply:
€ L3D % w(
CL,3D = 2
= K 'α − * . (&)
(1 2)ρU A & U )
Here, the downwash velocity is given by (14.24), w = Γ1/2s, and Γ1 is related to the total lift force
L3D of the three-dimensional wing by (14.21):
πs
€ L3D = ρUΓ1 .
4
Therefore (&) becomes:
& 1 4L3D ) & A L3D ) & 1 )
CL,3D = K (α − + = K (α − 2 2 +
= K (α − CL,3D + ,
' 2sU€πsρU * ' πs (1 2) ρU A * ' πΛ *
where Λ = s2/A is the wing's aspect ratio. Solving for CL,3D produces:
Kα CL,2D CL,2D
CL,3D = ≅ = ,
€ 1+ (K πΛ ) 1+ ( 2π πΛ ) 1+ ( 2 Λ )
where the angle of attack α is measured from the zero-lift orientation of the wing, and the second
equality is approximate because the foil section's two-dimensional lift-curve slope K in viscous
flow may be a little smaller than its inviscid value K = 2π.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.16. The wing-tip vortices from large heavy aircraft can cause a disruptive rolling
torque on smaller lighter ones. Lifting line theory allows the roll torque to be estimated when the
small airplane’s wing is modeled as a single linear vortex with strength Γ(y) that resides at x = 0
between y = –s/2 and y = +s/2. Here, the small airplane’s wing will be presumed rectangular
(span s, chord c) with constant foil-shape, and the trailing vortex from the heavy airplane’s wing
will be assumed to lie along the x-axis and produce a vertical velocity
€ distribution at x = 0 given
Γ"
by: w(y) =
2πy
[1− exp(− y )] . To simplify your work for the following items, ignore the
trailing vortices (shown as dashed lines) from the small airplane’s wing and assume U >> w.
+s 2
a) Determine a formula for the rolling moment, M = ∫−s 2 ρUyΓ(y)dy , on the small aircraft’s
€ wing in terms of Γ´, s, c,  , the air density ρ, the flight speed of the small aircraft U, and the lift-
curve slope of the small aircraft’s wing section K = dCL,2D dα , where α is the small-aircraft-
wing angle of attack.
€
b) Calculate M when€ ρ = 1.2 kg/m3, U = 150 m/s, K = 6.0/rad, b = 9 m, c = 1.5 m, Γ´ = 50 m2/s,
and s/(2  ) = 1. Comment on the magnitude of this torque.
€
z!
w(y)! y!
€
!´! U

x!

Solution 14.16. a) Use the second lifting line equation, Γ(y) = 12 UcK (α + w(y) U ) , and the
Γ"
specified vertical velocity, w(y) =
2πy
[1− exp(− y )] to determine the roll torque on the small
aircraft’s wing:
€
+s 2 1 +s 2 ) Γ' ,
2
M = ∫ −s 2 ρUyΓ(y)dy = ρU ca2D ∫ −s 2 +α +
2 * 2πUy
[ ]
1− exp(− y ) . ydy .
-
€
The α-term inside the integral has the wrong symmetry and doesn’t contribute to the moment so:
ρUcKΓ# +s 2 ρUcKΓ#
M=
€ 4π '
∫ −s 2[ ]
1− exp(− y ) dy =
4π ( [
0
) +s 2
y − e +y  ]−s 2 + [ y + e−y  ] 0
ρUcKΓ# s s * ρUcKΓ#
=
4π (
)− + + e
2
−s 2
+ + e−s 2 − , =
2 + 4π
(s − 2 + 2e−s 2 )
ρUcsΓ# ' 2 −s 2 *
€ = )1− [1− e ],+
4π ( s
€ (1.2kg /m 3 )(150m /s)(1.5m)(6.0 /rad)(9m)(50m 2 /s)
b) B =
4π
( )
1− [1− e−1 ] = 21.3 kN-m

€ This roll torque is large enough to cause loss of pilot control for a small aircraft, even if it is felt
for only a short period of time.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.17. Consider the ideal rectilinear horseshoe vortex of a simple wing having span s.
Use the (x, y, z) coordinates shown for the following items.
a) Determine a formula for the induced vertical velocity w at (x, y, 0) for x > 0 and y > 0.
b) Using the results of part a), evaluate the induced vertical velocity at the following three
locations (s, 0, 0), (0, s, 0), and (s, s, 0)
c) Imagine that you are an efficiency-minded migrating bird and that the rectilinear horseshoe
vortex shown is produced by another member of your flock. Describe where you would choose
to center your own wings. List the coordinates of the part b) location that is closest to your
chosen location.

Solution 14.17. The induced vertical velocity w at (x, y, 0) for x > 0 and y > 0 will be the result
of the sum of the induced velocities from (1) the port-side vortex located at y = –s/2, (2) the
bound vortex of the wing located at x = 0, and (3) the starboard-side vortex located at y = +s/2.
All three vortices have strength Γ. Using the angles in the planform drawing provided. The sum
of induced velocity from vortices (1) and (2) and (3) is:
Γ[cosψ11 − cosψ11 ] Γ[cosψ 21 − cosψ 22 ] Γ[cosψ 31 − cosψ 32 ]
w(x, y,0) = − − + ,
4 π ( y + s 2) 4 πx 4 π ( y − s 2)
where the signs are determined by the
y!
(x, y)! sense of rotation of each vortex. The
!22!
various angles can be specified in terms of
€ !32! the coordinates:
!31! (3)!
+s/2!
(2)
cosψ11 = (X − x) (X − x) 2 + (y + s /2) 2 ,
! !21!
x! cosψ12 = −x x 2 + (y + s /2) 2 ,
!12! !11! cosψ 21 = (y + s /2) x 2 + (y + s /2) 2 ,
(1)!
–s/2! €
cosψ 22 = (y − s /2) x 2 + (y − s /2) 2 ,
X! €
cosψ 31 = x x 2 + (y − s /2) 2 , and
€
cosψ 32 = −(X − x) (X − x) 2 + (y − s /2) 2 .
In the limit that the trailing vortices are long,€ X → ∞ , the first and last cosines above simplify to
+1 and –1. Thus, €
Γ % −1− x x 2 + (y + s /2) 2 y + s€/2 y − s /2 1+ x x 2 + (y − s /2) 2 (
w= ' − + + *
4 π '& y + s /2 x €x 2 + (y + s /2) 2 x x 2 + (y − s /2) 2 y − s /2 *)
b) Evaluate w at (s, 0, 0)
Γ % −1− s s2 + (s /2) 2 s /2 −s /2 1+ s s2 + (s /2) 2 (
w= ' − + + *
€ 4 π '& s /2 s s2 + (s /2) 2 s s2 + (s /2) 2 −s /2 *)
Γ %−1−1 1+ (1/2) 2 1/2 1/2 1+ 1 12 + (1/2) 2 (
w= ' − − + *
4 πs '& 1/2 1+ (1/2) 2 1+ (1/2) 2 −1/2 *)
€ Γ % 4 1 1 4 ( Γ Γ
w=
4 πs &'−2 −
5
−
5
−
5
−2−
5)
* =−
πs
[
1+ 5 = − [ 3.236]
πs
]
€ Evaluate w at (0, s, 0); here ψ22 = ψ21 = 0 so there is no contribution from vortex (2).
Γ % −1 1 ( Γ % −2 ( Γ %1(
w= ' + * = ' + 2* = + ' *
€ 4 π & s + s /2 s − s /2 ) 4 πs & 3 ) πs & 3)

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Evaluate w at (s, s, 0)
Γ % −1− s s2 + (3s /2) 2 s + s /2 s /2 1+ s s2 + (s /2) 2 (
w= ' − + + *
4 π '& 3s /2 s s2 + (3s /2) 2 s s2 + (s /2) 2 s /2 *)
Γ %−1−1 1+ (3/2) 2 3/2 1/2 1+ 1 1+ (1/2) 2 (
w= ' − + + *
4 πs '& 3/2 1+ (3/2) 2 1+ (1/2) 2 1/2 *)
€ Γ % 2 4 3 1 4 ( Γ %1 5 13 ( Γ
w= '− − − + + 2 + * = ' + − * = + [1.151]
4 πs & 3 3 13 13 5 5 ) πs & 3 4 12 ) πs
€ c) An efficiency minded bird would like to place its wings in a region of upward induced
velocity so that its lift-induced drag is minimized by a beneficial interaction with the vortex
system of a companion migrating bird. From the results of part b), the rectilinear horseshoe
€
vortex does produce upward induced velocities for (0, s, 0), and (s, s, 0), but the magnitude is
greater for (s, s, 0). Thus, the coordinate location (s, s, 0) is the preferred one, and this choice is
verified by observations of real migrating birds that fly in Λ-shaped formations.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.18. As an airplane lands, the presence of the ground changes the plane’s
aerodynamic performance. To address the essential features of this situation, consider uniform
flow past a horseshoe vortex (heavy solid lines below) with wingspan b located a distance h
above a large flat boundary defined by z = 0. From the method of images, the presence of the
boundary can be accounted for by an image horseshoe vortex (heavy dashed lines below) of
opposite strength located a distance h below the boundary.
a) Determine the direction and the magnitude of the induced velocity at x = (0, 0, h), the center
of the wing.
b) Assuming the result of part a) applies along the entire wingspan, estimate L and Di , the lift
and lift-induced drag, respectively, in terms of b, h, Γ, and ρ = fluid density.
c) Compare the result of part b) to that obtained for the horseshoe vortex without a large flat
surface: L = ρUΓb and Di = ρΓ 2 π . Which configuration has more lift? Which € one has less
drag? Why?

€ €

Solution 14.18. a) Five vortices will contribute to the induced velocity at x = (0,0,h) . Number
these as follows:
1 = starboard wingtip vortex
2 = starboard wingtip image vortex
3 = port wingtip vortex €
4 = port wingtip image vortex
5 = main wing image vortex
The geometrical layout ensures that the induced velocities will not point in the same direction, so
work through each one individually using the Biot-Savart induced velocity law from exercise
4.10:
e × eRΓ
ui = ω (cosθ1 − cosθ 2 )
4π
where i is the vortex number (not a coordinate index), eω points along the vortex direction, e R
points along a line perpendicular to the vortex that intersects the point of interest, and θ1 & θ2 are
the polar angles between the eω and the line connecting the ends of the vortex to the point of
€
interest. Here the first four vortices extend from x = 0 to +∞, so the cosine terms above are:
€ €
cos(π/2) – cos(π) = +1
u1 = −e zΓ ( 4 π (b /2)) = −e zΓ (2πb)
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Γ $ b /2 2h '
u2 = &e z + e y )
4 π (b /2) 2 + 4h 2 &% (b /2) 2 + 4h 2 (b /2) 2 + 4h 2 )(
u3 = −e zΓ ( 4 π (b /2)) = −e zΓ (2πb)
Γ % b /2 2h (
u4 = 'e z − ey *
€ 4 π (b /2) 2 + 4h 2 '& (b /2) 2 + 4h 2 (b /2) 2 + 4h 2 *)
€ Here the extra factor in [,] brackets for the second and fourth vortex is merely a unit vector
rotated to the induced velocity direction. The fifth, vortex induces a velocity that opposes the
on-coming free stream and its angular extent is different from the four other vortices.
€
−Γe x % b /2 −b /2 ( −Γe
x b /2
u5 = ' − *=
4 π (2h) '& (b /2) 2 + 4h 2 (b /2) 2 + 4h 2 *) 2π (2h) (b /2) 2 + 4h 2
The sum of these five terms is:
Γe Γ % b /2 ( Γe x b /2
utotal = − z + 'e z *−
€ πb 2π (b /2) 2 + 4h 2 '& (b /2) 2 + 4h 2 *) 2π (2h) (b /2) 2 + 4h 2
Γ % 4h 2 ( Γ b /2
utotal = − & 2 2
)e z − e x = −uze z − ux e x
πb ' (b /2) + 4h * 4 πh (b /2) 2 + 4h 2
€ b) The lift force will be: L = ρ(U − w x )Γb , and the induced drag force will be:
wz Γ2 ' 4h 2 *
Di = Lε = ρ(U − w x )Γb = ρΓbw z = ρ ( +
€ U − wx π ) (b /2) 2 + 4h 2 ,
Note, that as h/b
€ → 0, the induced drag disappears.
c) For constant speed flight close to the ground surface, there is less lift and less drag for a fixed
value of Γ. The lowering of the lift occurs because the induced velocity from the main wing’s
€
image vortex slows the on-coming stream at the location of the real wing. The induced drag is
lower because the image tip vortices produce upwash that partially counter acts the downwash
from the actual wingtip vortices.
For actual aircraft, there are two effects that more than offset the apparent reduction in lift
found in part b) and mentioned here.
(i) Wing circulation Γ increases as the aircraft approaches the ground because the downwash
decreases, and less downwash means an increase in the wing’s angle of attack.
(ii) With a constant engine-throttle setting, the loss of induced drag as h/b → 0 causes the aircraft
to mildly accelerate, thereby increasing U. The aircraft’s passengers may even feel this mild
acceleration, and it may lead to a steady-state speed that prevents the aircraft from descending
further and touching down. When this happens the aircraft is said to be flying with or in ground
effect. Once the aircraft is low enough for ground effect to be apparent, the pilot must typically
reduce the wing's CL by using its spoilers or by lowering the engine throttle setting in a
controlled fashion to achieve a safe smooth landing. Observant airline passengers will notice that
commercial airliners sometimes fly at very low altitudes over the landing runway for an
unexpectedly long period of time before touching down. This apparent delay in touching down is
merely the time necessary for pilot to adjust the aircraft's trim to continue its decent while flying
in ground effect. [Spoilers are flaps on the top of the wing that spoil the airflow on the suction
side of the wing; they are used to increase form drag and reduce lift in a controlled manner.]
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.19. Before modifications, an ordinary commercial airliner with wingspan s = 30 m

generates two tip vortices of equal and opposite circulation having Rankine velocity profiles (see
(3.28)) and a core size σo = 0.5 m for test-flight conditions. The addition of wing-tip treatments
(sometimes known as winglets) to both of the aircraft's wing tips doubles the tip vortex core size
at the test condition. If the aircraft's weight is negligibly affected by the change, has the lift-
induced drag of the aircraft been increased or decreased? Justify your answer. Estimate the
percentage change in the induced drag.

Solution 14.19. For the conditions stated, the lift-induced drag of the aircraft has decreased by
the addition of winglets. This contention is supported by an approximate analysis of the kinetic
energy of the vortex flow behind the aircraft.
In an unbounded nominally-quiescent fluid medium, an increase in length dl of a single
semi-infinite line vortex increases the kinetic energy of the fluid by, dKE, where:
1
[ ] ∞
dKE = ρ ∫ 0 uθ2 2πrdr dl .
2
A Rankine vortex is defined by:
' Γ 2πσ 2 ) r for r ≤ σ *
uθ (r) = (( +
€ ) Γ 2πr for r > σ ,
where σ is the core size, and r is the distance perpendicular to the vortex axis. Thus, the kinetic
energy increment for Rankine vortex is:
1 /σ & Γ ) 2 2
2
€
R
& Γ )2
dKE = ρ1 ∫ ( + r 2 π rdr + lim ∫ ( + 2 πrdr 4dl .
2 0 0 ' 2πσ 2 * R →∞ ' 2πr *
σ 3
The upper limit of the second integral is problematic for a single vortex. However, the airliner
will have two wing-tip vortices of equal and opposite sign so evaluation of the limit is not
necessary at this point.
€
Performing the two integrations yields:
1 Γ 1 2 ) σ
3
R
1 , Γ2 ) 1 σ 4 R,
dKE = ρ + 4 ∫ r dr + lim ∫ dr.dl = ρ + + lim ( ln r )σ.
dl
2 2π *σ 0 R →∞
σ r - 4 π *σ 4 4 R →∞ -
.
Γ2 )1 ,
=ρ + lim (ln R) − ln σ .dl
4 π +* 4 R →∞ -
Use the final part of this equality, divide by the time increment dt, and recognize dl/dt = U = the
aircraft's velocity, to find:
dKE Γ 2U ) 1 ,
€ =ρ + + lim (ln R) − ln σ ..
dt 4 π * 4 R →∞ -
For an aircraft that produces two counter rotating tip vortices, we can write the following
approximate equation
dKE Γ 2U ( 1 +
€ ≅ 2ρ * + ln Rc − ln σ - ≅ UDi
dt 4π ) 4 ,
where Rc is the distance perpendicular to the aircraft's flight path that is large enough for the
induced flow from the aircraft's tip vortices to have effectively cancelled out. The final equality
here follows because the power necessary to overcome the aircraft's induced drag is that
necessary to produce € the aircraft's tip-vortex velocity field.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Thus, for comparable flight speed, air density, and aircraft weight, an estimate of the
induced drag ratio can be obtained from:
$1 '
& + ln Rc − ln σ )
[Di ] modified % 4 (modified 0.25 + ln(90) − ln(1) 4.75
= ≈ = = 0.87 ,
[ Di ] original $ 1 + ln R − ln σ ' 0.25 + ln(90) − ln(0.5) 5.44
&% 4 c )(
original

where Rc has been estimated as three times the wing-span of 30 m. Thus, a ~10% (or more)
reduction in the induced drag is estimated for a doubling of the tip-vortex core radius from 0.5 m
to 1.0 m.
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 14.20. Determine a formula for the range, R, of a long-haul jet-engine aircraft in steady
level flight at speed U in terms of: MF = the initial mass of usable fuel; MA = the mass of the
airframe, crew, passengers, cargo, and reserve fuel; CL/CD = the aircraft's lift-to-drag ratio; g =
the acceleration of gravity; and η = the aircraft's propulsion system thrust-specific fuel
consumption (with units of time/length) defined by: dMF/dt = –ηD, where D = the aircraft's
aerodynamic drag. For simplicity, assume that U, the ratio CL/CD, and η are constants. [Hints. If
M(t) is the instantaneous mass of the flying aircraft, then L = Lift = Mg, M = MF + MA, and dM/dt
= dMF/dt. The final formula is known as the Breguet range equation.]

Solution 14.20. For steady level flight, an aircraft's engine thrust T will equal its drag D. Thus,
the beginning equation is:
dM dM f
= = −η D .
dt dt
Use the two ends of this extended equality, and divide on the left by M and on the right by L/g.
This leads to:
1 dM g 1
= −η D = − gη .
M dt L CL CD
Separate the differential on the left and recognize that Udt = ds = an element of path length along
the flight:
MA
1 1 gη t 1 gη R
∫ dM = − ∫ U dt = − ∫ ds ,
M F +M A M CL CD U 0 CL CD U 0
where t is the time of steady level flight and R is the range of the flight. Perform the integrations
to find:
! MA $ 1 gη
ln # &=− R.
" MA + MF % CL CD U
Solve for R to reach the final form:
U (CL CD ) ! M F $
R= ln #1+ &.
gη " MA %
Interestingly, air density does not explicitly enter this formula. Plus, the apparent proportionality
between R and U may be somewhat misleading. The aircraft's lift must balance its weight so
higher U must be paired with lower CL, and lower U must be paired with higher CL. Plus, CD
tends to increase with increasing CL, too (see Figure 14.28), so maximizing the range of a jet
engine aircraft means achieving a high lift-to-drag ratio at high speed.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.1. Use (15.4), (15.5), and (15.6) to derive (15.7) when the body force is spatially
uniform and the effects of viscosity are negligible.

Solution 15.1. The goal is develop an equation with an inhomogeneous-medium convected-wave

operator acting on p on the left side with source terms involving q, fi, and ∂ui/∂xj on the right.
Dp Dρ
First, use (15.6), = c2 , to eliminate Dρ/Dt from (15.4):
Dt Dt
1 Dρ ∂ui 1 Dp ∂ui 1 Dp ∂u
+ =q= 2 + , or 2
= q− i ,
ρ Dt ∂xi ρ c Dt ∂xi ρ c Dt ∂xi
where the final equation is just a rearrangement of the first. Apply D/Dt to the second equation to
find:
D ! 1 Dp $ Dq D ∂u j
# &= − . (a)
Dt " ρ c 2 Dt % Dt Dt ∂x j
Here, the summed-over index in the final term has been switched from 'i' to 'j'. Next, apply –∂/∂xj
to (15.5) to produce:
∂ Du j ∂ # 1 ∂p & ∂g ∂ # 1 ∂τ ij & ∂fi
− − %% (( = − j − % (− .
∂x j Dt ∂x j $ ρ ∂x j ' ∂x j ∂x j $ ρ ∂xi ' ∂x j
Note that ∂gj/∂xj = 0 because the body force is spatially uniform, drop the viscous stress term,
and move the remaining term that explicitly involves uj to the right side:
∂ # 1 ∂p & ∂f ∂ Du j
− %% (( = − i + . (b)
∂x j $ ρ ∂x j ' ∂x j ∂x j Dt
Add equations (a) and (b):
D ! 1 Dp $ ∂ ! 1 ∂p $ Dq ∂fi ! ∂ D D ∂ $
# &− # &= − +# − &uj , (c)
Dt " ρ c 2 Dt % ∂x j #" ρ ∂x j &% Dt ∂x j #" ∂x j Dt Dt ∂x j &%
The remaining steps involve simplifying the final term. The first operator inside the final large
parentheses may be rewritten:
∂ D ∂ "∂ ∂ % ∂2 ∂u ∂ ∂2 ∂u ∂ D ∂
= $ + ui '= + i + ui = i + .
∂x j Dt ∂x j # ∂t ∂xi & ∂t∂x j ∂x j ∂xi ∂xi∂x j ∂x j ∂xi Dt ∂x j
# ∂ D D ∂ & ∂u ∂u
Thus: %% − (( u j = i j , so (c) reduces to (15.7)
$ ∂x j Dt Dt ∂x j ' ∂x j ∂xi
D ! 1 Dp $ ∂ ! 1 ∂p $ Dq ∂f j ∂ui ∂u j
# &− # &= − + .
Dt " ρ c 2 Dt % ∂x j #" ρ ∂x j &% Dt ∂x j ∂x j ∂xi
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.2. Derive (15.12) through the following substitution and linearization steps. Set q
and fi to zero in (15.7) and insert the decompositions (15.9). Treat Ui, p0, ρ0 and T0 as time-
invariant and spatially uniform, and drop quadratic and higher order terms involving the
fluctuations ui! , p´, ρ´, and T´.

Solution 15.2. Start with (15.7) and set q and fi to zero. This leaves:
D ! 1 Dp $ ∂ ! 1 ∂p $ ∂ui ∂u j
# &− # &= .
Dt " ρ c 2 Dt % ∂x j #" ρ ∂x j &% ∂x j ∂xi
The decomposition equations are:
ui = Ui + ui! , p = p0 + p´ , ρ = ρ0 + ρ´ , and T = T0 + T´ ,
and all derivatives of Ui, p0, ρ0, and T0 are zero. Consider the first term:
Dp # ∂ ∂ & ∂p" ∂p" ∂p"
= % + (Ui + ui") ( ( p0 + p") = +Ui + ui" .
Dt $ ∂t ∂xi ' ∂t ∂xi ∂xi
Here the final term is quadratic in the small fluctuations, so
Dp # ∂ ∂ &
≅ % +Ui ( p) ,
Dt $ ∂t ∂xi '
and this term is linear in the fluctuation quantity p´. Thus, only the leading-order portion of the
operator coefficient of Dp/Dt need be retained because any extra first-order contributions from
this coefficient will wind up being second order when multiplied with Dp/Dt. Therefore, the first
term of (15.7) reduces to:
2
D ! 1 Dp $ ! ∂ ∂ $ 1 !∂ ∂ $ !∂ ∂ $
# & ≅ # +Ui & # +Ui & p) = # +Ui & p) .
Dt " ρ c 2 Dt % " ∂t ∂xi % ρo c 2 " ∂t ∂xi % " ∂t ∂xi %
where c is the speed of sound at pressure p0 and temperature T0. The second term simplifies in a
similar manner:
∂p ∂ ∂p"
= ( p0 + p") = .
∂x j ∂x j ∂x j
Again this term is linear in the fluctuation quantity p´, so its coefficient cannot contain any first-
order contributions from fluctuation quantities
∂ " 1 ∂p % ∂ " 1 ∂p) % 1 ∂2 p)
$ '≅ $ '= .
∂x j $# ρ ∂x j '& ∂x j $# ρ0 ∂x j '& ρo ∂x 2j
The lone remaining right-side term is quadratic in the fluctuation velocity,
∂ui ∂u j ∂ (Ui + ui") ∂ (U j + u"j ) ∂ui" ∂u"j
= = ,
∂x j ∂xi ∂x j ∂xi ∂x j ∂xi
and therefore should be dropped. So, term-by-term replacements in (15.7) using the results above
lead to:
2
1 !∂ ∂ $ ∂ 2 p'
# +U i & p' − ≅0,
c2 " ∂t ∂ xi % ∂ xi∂ xi
which is (15.2)
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.3. The field equation for acoustic pressure fluctuations in an ideal compressible
fluid is (15.13). Consider one-dimensional solutions where p = p(x,t) and x = x1.
a) Drop the x2 and x3 dependence in (15.13), and change the independent variables x and t to
∂ 2 p#
ξ = x − ct and ζ = x + ct to simplify (15.13) to = 0.
∂ξ∂ζ
b) Use the simplified equation in part a) to find the general solution to the original field equation:
p"(x,t) = f (x − ct) + g(x + ct) where f and g are undetermined functions.
€ c) When
€ the initial conditions are: p´ = F(x) and ∂p´/∂t = G(x) at t = 0, show that:
1$ 1 x
€ ' 1# 1 x &
f (x) = &F(x) − ∫ 0 G(x )dx) , and g(x) = %F(x) + ∫ 0 G(x )dx(,
2% c ( 2$ c '
€ where x is just an integration variable.

Solution 15.3. a) The starting point is (15.13) simplified for no x2 and x3 dependence. This is the
€ one-dimensional acoustic wave equation,
classical €
1 ∂ 2 p# ∂ 2 p#
− = 0.
c 2 ∂t 2 ∂x 2
Use ξ = x − ct and ζ = x + ct , and convert the partial derivatives with respect to x and t into
partial derivatives with respect to ξ and ζ.
∂ ∂ξ ∂ ∂ζ ∂ ∂ (x − ct) ∂ ∂ (x + ct) ∂ ∂ ∂
= +€ = + = + , and
€ ∂ x ∂ x ∂ξ ∂ x ∂ζ ∂ x ∂ξ ∂ x ∂ζ ∂ξ ∂ζ
€ ∂ ∂ξ ∂ ∂ζ ∂ ∂ (x − ct) ∂ ∂ (x + ct) ∂ ∂ ∂
= + = + = −c +c .
∂t ∂t ∂ξ ∂t ∂ζ ∂t ∂ξ ∂t ∂ζ ∂ξ ∂ζ
∂ 2 %∂ ∂ (% ∂ ∂ ( ∂ 2
∂ 2
∂ 2
Thus: € 2
= ' + *' + * = 2 + 2 + 2 , and
∂x & ∂ξ ∂ζ )& ∂ξ ∂ζ ) ∂ξ ∂ξ∂ζ ∂ζ
€ ∂ 2 &
2 ∂ ∂ )& ∂ ∂ ) 2& ∂ 2
∂2 ∂2 )
= c ( − +( − + = c ( 2 − 2 + +
∂t 2 ' ∂ξ ∂ζ *' ∂ξ ∂ζ * ' ∂ξ ∂ξ∂ζ ∂ζ 2 *
∂ 2 p# 1€∂ 2 p# ∂ 2 p$ ∂ 2 p$ ∂ 2 p$ ∂ 2 p$ 1 2 ) ∂ 2 p$ ∂ 2 p$ ∂ 2 p$ , 1 ∂ 2 p$
So, = → = + 2 + = ⋅ c + 2 − 2 + .= .
∂x 2 c 2 ∂t 2 ∂x 2 ∂ξ 2 ∂ξ∂ζ ∂ζ 2 c 2 * ∂ξ ∂ξ∂ζ ∂ζ 2 - c 2 ∂t 2
Cancel common € terms across the middle equality to find:
∂ 2 p# ∂ 2 p# ∂ 2 p#
2 = −2 which implies = 0.
€ € ∂ξ∂ζ ∂ξ∂ζ ∂ξ∂ζ
∂p#
b) Use the result of part a) and integrate with respect to ξ to find: = C (ζ ) where C(ζ) is a
∂ζ
function of integration
€ that cannot depend on ξ. Now € integrate this result with respect to ζ to
find: p" = ∫ C (ζ )dζ + f (ξ ) where f(ξ) is a second function of integration that cannot depend on
ζ. Since C is undetermined at this point, a new function€ g can be defined to make the last
equation look a little nicer: g(ζ ) = ∫ C (ζ )dζ . Thus, the general solution of the field equation for
€ p"(x,t) is:
p"(x,t) = f (ξ ) + g(ζ ) = f (x − ct) + g(x + ct) .
c) Use the result of part b) and the condition p´ = F(x) at t = 0 to find:
€
p"(x,0) = f (x) + g(x) = F(x) . (1)
€
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Differentiate the result of part b) with respect to time and use the ∂p´/∂t = G(x) at t = 0 to find:
$∂p#' $ df ∂ (x − ct) df ∂ (x + ct) ' + df (x) dg(x) .
&% ∂t )( = & d(x − ct) + ) = c-, − + 0 = G(x) .
t= 0 % ∂t d(x + ct) ∂t (t= 0 dx dx /
Use the last equality, divide by c, and integrate in x to find:
1 x
− f (x) + g(x) = ∫ 0 G(x )dx + D , (2)
c
€ where D is a constant. Adding equations (1) and (2) together produces:
1# 1x &
g(x) = %F(x) + ∫ G(x )dx + D( ,
€ 2$ c 0 '
while subtracting (2) from (1) produces:
1$ 1x '
f (x) = &F(x) − ∫ G(x )dx − D) .
€ 2% c 0 (
Given that D drops out of the general solution,
1% 1 x−ct 1 x +ct (
p"(x,t) = f (x − ct) + g(x + ct) = 'F(x − ct) + F(x + ct) − ∫ G(x )dx + ∫ G(x )dx − D + D*
€ 2& c 0 c 0 )
1% 1 x +ct (
= 'F(x − ct) + F(x + ct) + ∫ G(x )dx*,
2& c x−ct )
it may be ignored without loss of generality, a situation that is also true for the lower limit of the
x-integral. Therefore:
1$ 1x ' 1# 1x &
€ f (x) = &F(x) − ∫ G(x )dx) and g(x) = %F(x) + ∫ G(x )dx( ,
2% c 0 ( 2$ c 0 '
where the integration lower limit, 0, is chosen for convenience.

€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.4. Starting from (15.15) use (15.14) to prove (15.16)

Solution 15.4. Equation (15.15) is

p"(x,t) = f (x − ct) + g(x + ct) ,
and (15.14) is the linearized integrated Euler equation without the body force:
1 ∂p"
u1"(x,t) = − ∫ dt .
ρ o ∂x
Substitute (15.15) into€(15.14), define the variables ξ = x − ct and ζ = x + ct , and note that
∂ξ ∂x = ∂ζ ∂x = 1 :
1 ∂ 1 ) df ∂ξ dg ∂ζ , 1 ) df dg ,
u1"(x,t) = − ∫ ( f (x − € ct) + g(x + ct)) dt = − ∫ + + .dt = − ∫ + + . dt .
ρ o ∂x € ρ o * dξ ∂x €dζ ∂x - ρ o * dξ dζ -
€ Separate the two terms of the integration, change integration variables to ξ and ζ, and integrate to
reach (15.16).
1 df 1 dg 1 df ( dξ + 1 dg ( dζ + 1 1
€ u1"(x,t) = − ∫ dt − ∫ dt = − ∫ * - − ∫ * -= f (ξ ) − g(ζ )
ρ o dξ ρ o dζ ρ o dξ ) −c , ρ o dζ ) c , ρ oc ρ oc
1
= [ f (x − ct) − g(x + ct)]
ρ oc
From a strictly mathematical point of view, the integrations involved in these final steps should
produce constants of integration. However, the average of an acoustic fluctuation – such as u´– is
zero, so the constants of integration are zero.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.5. Consider two approaches to determining the upper Mach number limit for
incompressible flow.
a) First consider pressure errors in the simplest-possible steady flow Bernoulli equation. Expand
(15.29) for small Mach number to determine the next term in the expansion: p0 = p + 12 ρu 2 + ...
and determine the Mach number at which this next term is 5% of p when γ = 1.4.
b) Second consider changes to the density. Expand (15.30) for small Mach number and
determine the Mach number at which the density ratio ρ0/ρ differs from unity by 5% when γ =
€
1.4.
c) Which criterion is correct? Explain why the criteria for incompressibility determined in a) and
b) differ, and reconcile them if you can.

Solution 15.5. a) Start with (15.29),

$ γ −1 2 'γ (γ −1)
p0 = p&1+ M ) ,
% 2 (
and expand using the Taylor series (1+ ε) β = 1+ βε + 12 β (β −1)ε 2 + ... for ε << 1:
* $ γ '$ γ −1 ' 1 $ γ '$ γ '$ γ −1 2 ' 2 -
2
p0 = p,1+ & )& M )+ & )& −1)& M ) + .../.
+ €% γ −1(% 2 ( 2 % γ −1(% γ −1 (% 2 ( .
€
Keep the first three terms inside the brackets and simplify using M = u /c and c2 = γRT:
2 2 2

$ γ γ ' $ γ u2 γ 4 ' 1* p - γ
p0 ≅ p&1+ M 2 + M 4 ) = p&1+ + M ) = p + , /u 2 + pM 4 .
€ % 2 8 ( % 2 γRT 8 ( 2 + RT . 8
Use the perfect gas law p = ρRT to further simplify the approximate equation:
1 γ
p0 ≅ p + ρu 2 + M 4 p . ($)
€ 2 8
The final term will be 5% of p when
14
γM 4 8 = 0.05 , or M = (8(0.05) 1.4 ) = 0.73 .
b) Start with (15.30) and expand as in part a),
€1 (γ −1)
ρ 0 % γ −1 2 ( % 1 γ −1 2 1 + 1 .+ 1 .+ γ −1 2 . 2 (
= '1+ M * = '1+ M + - 0- −10- M 0 + ...*.
ρ &€ 2 ) &€ γ −1 2 2 , γ −1/, γ −1 /, 2 / )
Keep the first three terms inside the brackets and simplify.
ρ0 M2 2 −γ 4
≅ 1+ + M
€ ρ 2 8
This time the third term is not needed. The ratio ρ0/ρ will differ from unity by 5% when
M2
= 0.05 , or M = 2(0.05) = 0.32 .
€ 2
c) The two Mach number criteria differ by more than a factor of two, and the second one is
correct. The main reason that the approximate Bernoulli equation ($) retains its accuracy to
higher Mach number is that the second € term on the right in ($) contains the gas density, ρ, which
is not constant, and €
this term retains some Mach number dependence that is not explicitly shown.
When ($) is expanded using ρ0, it becomes: p0 − p ≅ 12 ρ o u 2 (1− M 2 2) , which implies a 5% error
in the pressure difference p0 – p when M = 0.32, the same as the part b) result.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.6. The critical area A of a duct flow was defined in §4. Show that the relation
∗

between A and the actual area A at a section, where the Mach number equals M, is that given by
∗

(15.31). This relation was not proved in the text. [Hint: Write
A ρ *c * ρ * ρ 0 c * c ρ * ρ 0 T * T0 1
= = = .
A* ρu ρ 0 ρ c u ρ 0 ρ T0 T M
Then use the other relations given in Section 15.4.]

Solution 15.6. Start with the hint and continue the equality, noting that M* = 1 and using (15.28)
and (15.30) €
A ρ * ρ 0 T * T0 1
=
A* ρ 0 ρ T0 T M
−1 1 1 1
% γ −1 *2 (γ −1 % γ −1 2 (γ −1 % γ −1 *2 (− 2 % γ −1 2 ( 2 1
= '1+ M * '1+ M * '1+ M * '1+ M *
& 2 ) & 2 ) & 2 ) & 2 ) M
−1 1 1 1
% γ −1(γ −1 % γ −1 2 (γ −1 % γ −1(− 2 % γ −1 2 ( 2 1
= '1+ 1+ M * '1+ 1+ M *
& 2 *) '& 2 ) & 2 *) '& 2 ) M
−1 1 1 1
%γ + 1(γ −1 % γ −1 2 (γ −1 %γ + 1(− 2 % γ −1 2 ( 2 1
=' 1+ M * ' 1+ M *
& 2 *) '& 2 ) & 2 *) '& 2 ) M
+ 1 1. + 1 1. 1 + γ +1 .
− + 0 +- + 0 + - 0
%γ + 1( -, γ −1 2/ % γ −1 2 ( , γ −1 2 / 1 % 2 + γ −1 2 .( 2 , γ −1 /
='
& 2 *) '&1+ 2 M *) =' -
M &γ + 1,
1+
2
M 0*
/)
.

€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.7. A perfect gas is stored in a large tank at the conditions specified by p0, T0.
Calculate the maximum mass flow rate that can exhaust through a duct of cross-sectional area A.
Assume that A is small enough that during the time of interest p0 and T0 do not change
significantly and that the flow is isentropic.

Solution 15.7. The mass flux through the tube will be:
−1 (γ −1) −1 2
ρ U c " γ −1 2 % " γ −1 2 %
m! = ρUA = ρ0 c0 A = $1+ M ' M $1+ M ' ρ0 γ RT0 A ,
ρ0 c c0 # 2 & # 2 &
where the subscript '0' indicates reservoir (or stagnation conditions) and (15.28) and (15.30) have
been used to introduce the factors involving the Mach number M. This equation can be
simplified by collecting like factors, and by using p0 = ρ0RT0 to eliminate ρ0.
M p0 γ
m! = 1 (γ +1)
A.
" γ −1 2 %2 (γ −1) RT0
$#1+ 2 M '&
To determine the maximum flow rate differentiate with respect to M, set the derivative equal to
zero, and solve for M, the mach number where the mass flow rate will be maximum.
dm! 1 p0 γ 1 (γ +1) M ( γ −1 +p γ
= 1 (γ +1)
A− 1 (γ +1) *
2M - 0 A=0 .
dM " γ −1 RT 2 (γ −1) +1 ) 2 , RT
2 % 2 (γ −1) " γ −1 2 %2 (γ −1)
0 0
$#1+ 2 M '& $#1+ 2 M '&
Divide out common factors and simplify:
γ −1 2 1 (γ +1) γ −1 2 γ +1 2
1+ M − M (γ −1) M = 1+ M + M = 1− M 2 = 0 .
2 2 (γ −1) 2 2
The two roots are M = ± 1, corresponding to the maximum mass flux out-of or into the tank.
Here we presume outflow is of interest M = +1, so:
1 (γ +1)
1 p0 γ " 2 %2 (γ −1) γ p0 p
m! max = 1 (γ +1)
A =$ ' A = 0.685 0 A ,
" γ −1%2 (γ −1) RT0 #γ +1& RT0 RT0
$#1+ 2 '&
where the final numerical value applies when γ = 1.40.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.8. The entropy change across a normal shock is given by (15.43). Show that this
reduces to expressions (15.44) for weak shocks. [Hint: Let M12 −1 << 1. Write the terms within
the two sets of brackets in equation (15.43) in the form [1 + ε1] [1 + ε2]γ, where ε1 and ε2 are
small quantities. Then use the series expansion ln(1 + ε) = ε – ε2/2 + ε3/3 + … . This gives
equation (15.44) times a function of M1 which can be evaluated at M1 = 1.]

Solution 15.8. From (15.43)

(*" %" (γ −1)M12 + 2 %
γ ,*
S2 − S1 2γ
cv
= ln )$1+ ( M12 −1)'$
&# (γ +1)M1 &
2 '
-.
+*# γ +1 .*
The term in the first set of [,]-brackets is already in an easily managed form:
2γ
1+
γ +1
( M12 −1) ≡ 1+ ε1.
The term in the second set of [,]-brackets can be rearranged as follows:
(γ −1)M12 + 2 (γ + 1)M12 − 2M12 + 2 2(M12 −1)
2
= 2
= 1− 2
≡ 1− ε2 .
(γ + 1)M € 1 ( γ + 1)M 1 ( γ + 1)M 1
Together these two results produce:
S2 − S1 γ

cv
{ }
= ln [1+ ε1 ] [1+ ε 2 ] = ln (1+ ε1 ) + γ ln (1− ε 2 ) .
€
Now expand the natural logarithm functions:
S2 − S1 ε12 ε13 " ε 22 ε 23 %
= ε1 − + +... + γ $ −ε 2 − − +... '
cv 2 3 # 2 3 &.
1 1
≅ (ε1 − γε 2 ) − (ε12 + γε 22 ) + (ε13 − γε 23 )
2 3
Evaluate the terms.
2γ 2(M12 −1) 2γ (M12 −1) % 1 ( 2γ (M12 −1) 2
ε1 − γε2 =
γ +1
( 1 ) (γ + 1)M 2
M 2
−1 − γ = '1− *=
γ + 1 & M12 ) (γ + 1)M12
1

1 2 1 % 4γ 2 2 4(M12 −1) 2 ( 2γ 2+ 1 .
2
(ε1 + γε 2)
2
= '
2 & (γ + 1) 2 (
M 1
2
−1 ) + γ 2 4*
=
(γ + 1) M1 ) (γ + 1) 2 (
M12 −1) -γ + 4 0
, M1 /
€ 1 3 1 % 8γ 3
3 8(M1 −1) (
2 3
8γ 3+ 1 .
3
(ε1 − γε23 ) = '
3 & (γ + 1) 3 (
M12 −1) − γ 3 6*
=
(γ + 1) M1 ) 3(γ + 1) 3 (
M12 −1) -γ 2 − 6 0
, M1 /
€ Reconstruct the entropy difference and simplify.
S2 − S1 2γ (M12 −1)2 2γ (M12 −1)2 # 1 & 8γ (M12 −1)3 # 2 1 &
≅ − %γ + (+ %γ − 6 (
€ cv (γ +1)M12 (γ +1)2 $ M14 ' 3(γ +1)3 $ M1 '
2γ (M12 −1)2 # 1 γ 1 4(M12 −1) 2 4(M12 −1) &
= % − − + γ − (
(γ +1) $ M12 γ +1 (γ +1)M14 3(γ +1)2 3(γ +1)2 M16 '
2γ (M12 −1)2 # 3(γ +1)2 M12 − 3γ (γ +1)M14 − 3(γ +1) + 4(M12 −1)γ 2 M14 − 4(M12 −1)M1−2 &
= % (
(γ +1) $ 3(γ +1)2 M14 '
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

# −2 %
S2 − S1 2γ (M12 −1)2 ' 3(γ +1)#$(−γ M1 +1)(M1 −1)%& + 4(M1 −1)γ M1 − 4(M1 −1)M1 (
2 2 2 2 4 2

≅
cv (γ +1) '$ 3(γ +1)2 M14 (&

2γ (M12 −1)3 # 3(γ +1)(−γ M12 +1) + 4γ 2 M14 − 4M1−2 %

= ' (
(γ +1) $ 3(γ +1)2 M14 &
Given that the first term is cubic in the small quantity, the contents of the final [,]-brackets can be
evaluated at M1 = 1. This produces:
S2 − S1 2γ (M12 −1)3 # 3(γ +1)(−γ +1) + 4γ 2 − 4 & 2γ (M12 −1)3 # 3(−γ +1) + 4γ − 4 &
≅ % (= % (
cv (γ +1) $ 3(γ +1)2 ' (γ +1) $ 3(γ +1) '
2γ (M12 −1)3 # γ −1 & 2γ (γ −1) 2
= % (= (M1 −1)3,
(γ +1) $ 3(γ +1) ' 3(γ +1)2
which is (15.44a).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.9. Show that the maximum velocity generated from a reservoir in which the
stagnation temperature equals T0 is umax = 2cpTo . What are the corresponding values of T and
M?

Solution 15.9. If the flow starts from zero velocity within a reservoir is T0, then the energy
equation is:
1
h0 = h + u 2 .
2
Thus, the maximum flow speed will occur when h = cpT = 0, and this implies:
u = 2cpT0 , T → 0 , and M → ∞.
Such a flow (nearly) occurs in practice
€ when a compressed gas expands into the low-temperature
vacuum of outer space.
€ €
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.10. In an adiabatic flow of air through a duct, the conditions at two points are u1 =
250 m/s, T1 = 300 K, p1 = 200 kPa, u2 = 300 m/s, p2 = 150 kPa. Show that the loss of stagnation
pressure is nearly 34.2 kPa. What is the entropy increase?

Solution 15.10. First, compute the speed of sound at the upstream location:
c1 = γRT1 = (1.4)(287)(300) = 347.2ms−1 .
Therefore, the mach number is: M1 = u1/c1 = 250/347.2 = 0.72. From table 15.1,
p1 p01 = 0.7080 , and T1 T01 = 0.9061, so
p01 = p1 0.7080 = 200 0.7080 = 282.5kPa , and T01 = T1 0.9061 = 300 0.9061 = 331.1K .
€
The stagnation enthalpy at location '1' is:
1 2
€ h01 = cpT1 + 2 u = (1004)(300)
€ + 0.5(250)2 = 332.5kJkg−1 .
€ In adiabatic flow, h01 = h02, therefore:
h01 = h2 + 12 u22 , or h2 = h01 − 12 u22 = 332.5kJkg−1 − 0.5(300) 2 = 287.5kJkg−1.
The temperature at location '2' is:
T2 = h2 cp = 287.5 1.004 = 286.4K
so€the speed of sound€at location '2' is:
c 2 = γRT2 = (1.4)(287)(287.4) = 339.2ms−1 .
Thus, the mach number at location '2' is: M2 = u2/c2 = 300/339.2 = 0.884. From table 15.1, using
the entry for M = 0.88.
p2 p02 = 0.6041 , so p02 = p2 0.6041 = 150 0.6041 = 248.3kPa .
€
The loss of stagnation pressure and the entropy gain are:
p01 − p02 = 282.5 − 248.3 = 34.2kPa , and
s2 − s1 = cp ln (T02 T01 ) − R ln ( p02 p01 ) = 0 − R ln ( p02 p01 )
€ €
= −(287)ln ( 248.3 282.5) = 37.03m 2 s −2 K −1
€
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.11. A shock wave generated by an explosion propagates through a still atmosphere.
If the pressure downstream of the shock wave is 700 kPa, estimate the shock speed and the flow
velocity downstream of the shock.

Solution 15.11. The analysis can be done in a shock-fixed frame of reference, and p1 = 101.3 kPa
and T1 = 295K, are taken as the nominal atmospheric conditions. The shock's pressure ratio is:
p2 700kPa
= = 6.91.
p1 101.3kPa
Locating the closest value in Table 15.2 produces:
M1 = 2.46, M2 = 0.517, and T2 T1 = 2.098 .
Therefore:
€
u1 = shock speed = M1c1 = 2.46 1.4(287)295 = 847ms−1 , and
T2 = 2.098T1 = 2.098(295K) = 619K.
Thus, €
u2 = downstream flow speed = M 2c 2 = (0.517) 1.4(287)619 = 258ms−1 .
€
So, in a frame of reference where the upstream air is still, the velocity downstream of the shock
will be:
u1 – u2 = 847 – 258 = 589 m/s
€
in the direction of shock propagation.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.12. Prove the following formulae for the jump the conditions across a stationary
normal shock wave:
p2 − p1 2γ u2 − u1 2 " 1 % υ 2 − υ1 2 " 1 %
= ( 1 ) c
M 2
−1 , = − $ M 1 − ' , and = − $1− ',
p1 γ +1 1 γ +1 # M1 & υ1 γ +1 # M12 &
where υ = 1/ρ, and the subscripts '1' and '2' imply upstream and downstream conditions,
respectively.

Solution 15.12. The pressure ratio condition (15.39) is:

p2 2γ " 2 $
= 1+ # M1 −1% .
p1 γ +1
Subtract '1' from both sides, and rearrange on the left to find:
p2 p −p 2γ " 2 $
−1 = 2 1 = # M1 −1% .
p1 p1 γ +1
The velocity ratio condition (15.41) is:
u1 (γ +1)M12
= .
u2 (γ −1)M12 + 2
Flip this over, subtract '1' from both sides, multiply by u1/c1, and simplify to find:
" u2 % u1 u2 − u1 " (γ −1)M12 + 2 % u1 " (γ −1)M12 + 2 − (γ +1)M12 %
$ −1' = =$ 2
−1' = $ ' M1
# u1 & c1 c1 # (γ +1)M1 & c1 # (γ +1)M12 &
" 2 − 2M12 % 2 " 1 %
=$ 2'
M1 = − $ M1 − '.
# (γ +1)M1 & γ +1 # M1 &
The specific volume condition can be obtained from the density ratio condition (15.41):
ρ2 υ1 (γ +1)M12
= = ,
ρ1 υ 2 (γ −1)M12 + 2
Flip this over, subtract '1' from both sides, and simplify to find:
" υ 2 % υ 2 − υ1 (γ −1)M12 + 2 (γ −1)M12 + 2 − (γ +1)M12
$ −1' = = −1 =
# υ1 & υ1 (γ +1)M12 (γ +1)M12
2 − 2M12 2 " 1 %
= = − $1− '.
(γ +1)M12 γ +1 # M12 &
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.13. Using (15.1i), and (15.43), determine formulae for p02/p01 and ρ02/ρ01 for a
normal shock wave in terms of M1 and γ. Is there anything notable about the results?

Solution 15.13. Equation (15.1i) applied to stagnation conditions implies:

s2 − s1 "T % R " p %
= ln $ 02 ' − ln $ 02 '
cp # T01 & cp # p01 &
Flow through the shock wave is adiabatic so T02 = T01, and
. +" (γ −1)M12 + 2 % 2
γ
s2 − s1 R " p02 % cv 0( 2γ 0
cp
= − ln $ ' = ln /*1+
cp # p01 & cp 0
( M1 −1)-$
2
2
,# (γ +1)M1 & 0
' 3.
1) γ +1 4
A little algebraic rearrangement using (15.1d) and (15.1g) leads to:
−1 γ
p02 " 2γ %γ −1 ( (γ +1)M12 +γ −1
= $1+
p01 # γ +1
( 1 )'& *) (γ −1)M 2 + 2 -, .
M 2
−1
1

The formula for the stagnation density can be found similarly. Equation (15.1i) applied to
stagnation conditions implies:
s2 − s1 "T % R " ρ %
= ln $ 02 ' − ln $ 02 '
cv # T01 & cv # ρ01 &
Flow through the shock wave is adiabatic so T02 = T01, and
. +" (γ −1)M12 + 2 % 2
γ
s2 − s1 R " ρ02 % 0( 2γ 0
cv
= − ln $ ' = ln /*1+
cv # ρ01 &
( M1 −1)-$
2
2
,# (γ +1)M1 & 4
' 3.
0) γ +1
1 0
A little algebraic rearrangement using (15.1d) and (15.1g) leads to:
−1 γ
ρ02 " 2γ %γ −1 ( (γ +1)M12 +γ −1
= $1+
ρ01 # γ +1
( M1 −1)'& *) (γ −1)M 2 + 2 -, .
2

Yes, the results are 'notable' because they are exactly the same.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.14. Using dimensional analysis, G. I. Taylor deduced that the radius r(t) of the blast
wave from a large explosion would be proportional to (E/ρ1)1/5t2/5 where E is the explosive
energy, ρ1 is the quiescent air density ahead of the blast wave, and t is the time since the blast
(see Example 1.10). The goal of this problem is to (approximately) determine the constant of
proportionality assuming perfect-gas thermodynamics.
a) For the strong shock limit where M12 >> 1 , show:
<Begin Equation>
ρ2 γ +1 T2 γ −1 p2 γ +1 p2
≅ , ≅ , and u1 = M1c1 ≅ .
ρ1 γ −1 T1 γ +1 p1 2 ρ1
</End Equation>
b) For a perfect gas with internal energy per unit mass e, the internal energy per unit volume is
ρe. For a hemispherical blast wave, the volume inside the blast wave will be 23 π r 3 . Thus, set
ρ2 e2 = E 23 π r 3 , determine p2, set u1 = dr/dt, and integrate the resulting first-order differential
equation to show that r(t) = K(E/ρ1)1/5t2/5 when r(0) = 0 and K is a constant that depends on γ.
c) Evaluate K for γ = 1.4. A full similarity solution of the non-linear gas-dynamic equations in
spherical coordinates produces K = 1.033 for γ = 1.4 (see Thompson 1972, p. 501). What is the
percentage error in this exercise's approximate analysis?

Solution 15.14. Start from the normal shock jump conditions (15.39) - (15.42):
p2 2γ $ γ −1 2 ' $ 2 γ −1'
p1
= 1+
γ +1
( M12 −1) , M 22 = &1+
% 2
M1 ) & γM1 −
( % 2 (
),

ρ2 (γ + 1)M12 T2 2(γ −1) $ γM12 + 1' 2

= , and = 1+ & )( M1 −1) .
ρ1 (γ −1)M12 + 2 T1 (γ + 1) 2 % M12 (
Simplify€these four equations for€M12 >> 1 ,
p2 2γ γ −1 2 γ −1
≅ M12 , M 22 ≅ M1 γM12 = ,
€ p1 γ +€1 2 2γ
ρ 2 (γ +€ 1)M12 γ + 1 T2 2(γ −1) % γM12 ( 2 2γ (γ −1) 2
≅ = , and ≅ ' * M1 = M1 .
ρ1 (γ −1)M12 γ −1 T1 (γ + 1) 2 & M12 ) (γ + 1) 2
The third equation€ is in the correct € form so it does not need further manipulation. Use the first
and fourth equation to eliminate M12 from the expression for the temperature ratio:
€ T2 2γ (γ −1) 2 2γ (γ −1) γ + 1 p2 γ −1 p2
≅ € M1 = = ,
T1 (γ + 1) 2 (γ + 1) 2 2γ p1 γ + 1 p1
which is another result€in the correct form. To reach the final part a) result, invert the strong-
shock pressure relationship to find and expression for M1:
12 12 12 12
$ γ + 1 p2 ' " γ +1 p2 % " γ +1 p2 % p1 " γ +1 p2 %
€
M1 ≅ & ) , so that u1 = M1c1 ≅ $ ' γ RT1 = $ ' γ =$ ' ,
% 2γ p1 ( # 2γ p1 & # 2γ p1 & ρ1 # 2 ρ1 &
where R is the gas constant, and the final equality completes the effort for part a).
b) As stated in the question, set
E p p E
€ ρ2 e2 = 2 3 = ρ2 cvT2 = cv 2 = 2 , and solve for p2 = (γ −1) 2 3 .
3
πr R γ −1 3
πr
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Put this into the final result of part a) and set u1 = dr/dt to find:
12 12 12
dr ! γ +1 p2 $ ! γ +1 E $ " 2 3E %
3 2 dr
u1 = = #
dt " 2 ρ1 %
& = ##
2
(γ −1) 2
ρ π r
& , which implies r
3& (
dt #
)
= $ γ −1
4 ρ π
' .
" 3 1 % 1 &

This non-linear first-order differential equation integrates to:

12
(5" 12 25 15
r5 2 " 2 3E % 3 % + "E% 25
52 #
(
= $ γ −1 ) 4 ρ1π &
' t + const , or r(t) = * $ (γ −1) ' - $ ' t ,
*) 2 #
2

4π & -, # ρ1 &
where the initial condition, r(0) = 0, has been used to determine const = 0.
+5 % 12 25
3 ( .
c) For γ = 1.40, K = - ' (γ −1) * 0 = 1.07, a difference of less than 4% from a similarity
2

,2 & 4π ) /
solution of the non-linear gas-dynamic equations. Thus, this simpler analysis has certainly been
worthwhile.
€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.15. Starting from the set (15.45) with q = 0, derive (15.47) by letting station (2) be a
differential distance downstream of station (1).

Solution 15.15. The equation set (15.45) with q = 0 and the second location a differential
distance dx downstream of the first location is:
( )
d ( ρu) = 0 , d ( p + ρu 2 ) = − p1df , and d h + 12 u 2 = 0 .
Use thermodynamic relationships to determine h in terms of p and ρ.
p γ p
h = cpT = cp = .
€ € ρ€R γ −1 ρ
Expand all three equations:
γ dp γ p
ρdu + udρ = 0 , dp + ρdu 2 + u 2 dρ = − p1df , and − dρ + 12 du 2 = 0 .
γ −1 ρ γ −1 ρ 2
Solve the first equation for dρ = − ρdu u , and eliminate dρ from the other two equations:
γ dp γ p du 1 2
€ dp + ρdu 2 − ρudu = − p1df , and + + 2 du = 0 .
€ € γ −1 ρ γ −1 ρ u
Simplify these equations and combine terms:
€
γ % dp p du (
€ dp + ρudu = − p 1df , and ' + * + udu = 0 .
€ γ −1 & ρ ρ u )
Solve the first of these for du = −(1 ρu)( p1df + dp) and substitute this into the second one:
γ % dp p 1 ( 1
2 ( 1
€ ' − p df + dp)* − ( p1df + dp) = 0 .
γ −1 & ρ ρ € ρu ) ρ
Multiply through€by ρ, and collect terms:
% γ γ p ( % γ p (
' − 2
−1* dp − ' 2
+ 1* p1df = 0 .
€ & γ −1 γ −1 ρu ) & γ −1 ρu )
2
Separate the terms across the equals sign, use γp/ρ = c , where c = sound speed, and introduce
the Mach number M = u/c:
$ γ 1€ c 2 ' $ 1 c2 ' $ 1 ' $ 1 '
& − 2
−1) dp = & 2
+ 1) p1df , or &γ − 2 − γ + 1)dp = & 2 + γ −1) p1df , so
% γ −1 γ −1 u ( % γ −1 u ( % M ( %M (
dp 1+ (γ −1)M 2
( M 2 −1)dp = ((γ −1)M 2 + 1) p1df , which implies: − p1
=
1− M 2
df .

€ To reach the second equation of (15.47), multiply

€ by p1 subtract p1df from both sides and divide
by ρu to reconstruct du = −(1 ρu)( p1df + dp) on the left:
€ 1 1 % 1+ (γ −1)M 2 ( 1 % γM 2 (
du = (−dp − p1df ) = ' € −1* p1df = ' * p1df .
ρu ρu & 1− M 2 ) ρu &1− M 2 )
Divide by u€and rearrange the right side:
du p % γM 2 ( p1 1 % M 2 ( p1 1 p1
= 2' 2 * df = 2' 2*
df = df .
€ u ρu & 1− M ) p M & 1− M ) p 1− M 2 p
Thus, when M < 1, positive friction (df > 0) causes the flow to accelerate.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.16. Starting from the set (15.45) with f = 0, derive (15.48) by letting station (2) be a
differential distance downstream of station (1).

Solution 15.16. The equation set (15.45) with f = 0 and the second location a differential
distance dx downstream of the first location is:
( )
d ( ρu) = 0 , d ( p + ρu 2 ) = 0 , and d h + 12 u 2 = h1dq .
Use h = cpT, expand all three equations, and use the perfect gas law (p = ρRT) to find:
ρdu + udρ = 0 , dp + ρdu 2 + u 2 dρ = 0 , cp dT + 12 du2 = h1dq , and dp = ρRdT + RTdρ .
€ dρ = − ρdu u , and€eliminate dρ from the second and last equations to
Solve the first€equation for
reach:
€ dp dT du
€ dp + ρdu 2 − ρudu = dp + ρudu = 0 , and€ = − ,
p T u
€
where the final form of the last equation is obtained by dividing by p or ρRT. Solve the first of
these for dp = –ρudu, and use this to eliminate dp from the second:
€ ρudu γ dT du
− = − 2 udu€= − .
p c T u
2
where γp/ρ = c , where c = sound speed, has been used for the first equality. Solve for udu and
use M = u/c:
γ du du dT u 2 dT
− 2 udu +€ = −(γM 2 −1) = , or udu = 12 du 2 = − .
c u u T T (γM 2 −1)
Substitute this into the differential energy equation (the one that involves cp) and use h1 = cpT1,
u 2 dT
cp dT − = h1dq = cpT1dq .
€ T (γ M 2€−1)
Divide this equation by γR, recognize the factor of M2, and use cp γ R = 1 (γ −1) :
$ 1 M2 '
cp u 2 dT cpT1
& )dT =
1
dT − = dq , or − T1dq .
&%γ −1 (γM −1) )(
2
γR γ RT (γ M −1) γ R
2
γ −1
Solve for dT/T1.
$ (γ −1)M 2 ' dT $ −1+ M 2 ' dT dT 1− γM 2
&1− ) = & ) = dq , or = dq .
&% (γM −1) )( T1 % γM
2
€ 2 −1 ( T1 T1 1− M 2
The final equation is the first part of (15.48) which shows that heat addition leads to cooling of
the gas when 1 γ < M < 1. To reach the second part of (15.48), multiply the result for dT/T1 by
T1 and substitute
€ this into the differential energy equation:€
" 1− γ M 2 %
cp dT + udu = cpT1 $ 2
dq ' + udu = h1dq .
€ # 1− M &
Recognize h1 = cpT1, and collect terms:
udu $ 1− γM 2 ' (γ −1)M 2
= &1− ) dq = dq .
h1 % 1− M 2 ( 1− M 2
This is the second equation of (15.48).

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.17. For flow of a perfect gas entering a constant area duct at Mach number M1,
calculate the maximum admissible values of f and q for the same mass flow rate. Case (a) f = 0;
case (b) q = 0.

Solution 15.17. From Section 15.6,

12
1+ γM 22 $1+ ((γ −1) 2) M1 + q '
2
M2
= & ) .
M1 1+ γM12 − f % 1+ ((γ −1) 2) M 22 (
For maximum values of f or q, the flow is choked at the duct exit so M2 = 1.
a) f = 0; set M2 = 1 to find:
12
1 1+ γ $1+ ((γ −1) 2) M12 + q '
€ = & ) .
M1 1+ γM12 % (γ + 1) 2 (
Solve for q by first clearing the square root.
2 2
# 1+ γM12 & 1+ ((γ −1) 2) M12 + q γ + 1 # 1+ γM12 & * γ −1 2 -
% ( = , so q = % ( − ,1+ M1 / .
$(1+ γ )M1€' (γ + 1) 2 2 $(1+ γ )M1 ' + 2 .
(b) q = 0; set M2 = 1 to find:
12
1 1+ γ $1+ ((γ −1) 2) M12 '
= & ) .
€ M1 1+ γM€12 − f % (γ + 1) 2 (
Solve for f.
12
2
$1+ ((γ −1) 2) M12 ' 2 $ γ −1 2 '1 2
1+ γM1 − f = (1+ γ )M1& ) , or f = 1+ γM1 − 2(1+ γ )M1&1+ M1 ) .
€ % (γ + 1) 2 ( % 2 (
Note, f = 0 when M1 = 1.

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.18. Show that the accelerating portion of the piston trajectory (0 ≤ xp(t) ≤ cot1)
shown in Figure 15.18 is:
2 1+γ
" γ +1 % " t %γ +1 2cot t " 2 %1−γ
x p (t) = $ ' cot1 $ ' − for 1 ≤ ≤ $ ' .
# γ −1 & # t1 & γ −1 t1 # γ +1 &

Solution 15.18. As described in Example 15.7, the C+ characteristics emanate from the origin
during the time that the piston is accelerating. And, as described in the solution to this same
example, the C– characteristics originate on the x-axis where u = 0 and c = co, so the I– invariant
implies:
2c(x, t) 2c(x, 0) 2c γ −1
I − = u(x, t) − = u(x, 0) − = − o or c = co + u.
γ −1 γ −1 γ −1 2
Thus, on the C+ characteristics,
! dx $ ! γ −1 $ ! γ +1 $ !x$
# & = (u + c)C+ = # u + co + u & = # co + u& = # &
" dt %C+ " 2 %C+ " 2 %C+ " t %C+
where the final equality ensures that the C+ characteristics pass through the origin. To determine
the piston trajectory xp(t), require the final equality to hold where any C+ characteristic crosses
the piston path. If the time of this crossing is τ, then the last equality implies:
γ +1 x (τ )
co + x! p (τ ) = p .
2 τ
Replace τ with t, and rearrange to find a first-order ordinary differential equation:
2 x p (t) 2c
x! p (t) − =− o .
γ +1 t γ +1
The solution is the sum of homogeneous and a particular solutions,
2
2c t
x p = At − o ,
γ +1

γ −1
where A is an undetermined constant. The initial condition, xp = cot1 at t = t1, allows A to be
determined from:
2
2cot1 " γ +1 % −γ 2+1
γ +1
cot1 = At1 − , or A = cot1 $ ' t1 .
γ −1 # γ −1 &
Thus,
2 1+γ
" γ +1 % " t %γ +1 2cot t " 2 %1−γ
x p (t) = $ ' cot1 $ ' − for 1 ≤ ≤ $ ' ,
# γ −1 & # t1 & γ −1 t1 # γ +1 &
which is the desired result. In dimensionless form this is:
2
x p (t) " γ +1 %" t %γ +1 2 "t%
=$ '$ ' − $ '.
cot1 # γ −1 &# t1 & γ −1 # t1 &
 –––– T = 2
Fluid Mechanics, 6th Ed. –––T=3 Kundu, Cohen, and Dowling
----T=4
P
Exercise 15.19. ––––
For the flow
T=2 conditions of Figure 15.18, plot u/co and p/po as functions of x/cot1
for xp(t) < x < cot–at– t/t
– 1T==2,3 3, and 4 for γ = 1.4, where co and po are the sound speed and pressure
of the quiescent gas
- - - upstream
- T = 4 of any disturbance from the moving piston. Does the progression
of these waveforms indicate expansion wave steepening or spreading as t increases?

Solution 15.19. Start from the results of Example 15.7,

X X
2 "x % 2 γ −1 x
u(x, t) = $ − co ' and c(x, t) = co + ,
γ +1 # t & γ +1 γ +1 t
and rearrange these equations to introduce dimensionless variables U = u/co, C = c/co, X = x/cot1,
and T = t/t1:
u(x, t) 2 " x 1 % 2 "X %
= $$ −1'' or U = $ −1' , (1)
co γ +1 # cot1 (t t1 ) & γ +1 # T &
and
c(x, t) 2 γ −1 x 1 2 γ −1 X
= + or C = + .
co γ +1 γ +1 cot1 (t t1 ) γ +1 γ +1 T
The final equation can be switched to the dimensionless pressure P = p/po via (15.52):
2γ
" 2 γ −1 X %γ −1
P =$ + ' . (2)
# γ +1 γ +1 T &
For the given parameters, the plots of (1) and (2) look like:

0.5% 1.5"

1"

!0.5%

0.5"

!1.5% 0"
!1% 0% 1% 2% 3% 4% &1" 0" 1" 2" 3" 4"

Examination of both plots suggests that the expansion wave spreads out as time increases.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.20. Consider the field properties in Figure 15.19 before the formation of the shock
wave.
a) Using the piston trajectory from Exercise 15.18, show that the time at which the piston reaches
(1+γ ) (1−γ )
speed co is −t1 ((γ +1) 2 ) = –0.3349t1 for γ = 1.4.
b) Plot u/co and p/po as functions of x/cot1 for xp(t) < x < cot1 at: t/t1 = –1/3, –1/6, and –1/25 for γ
= 1.4, where co and po are the sound speed and pressure of the quiescent gas upstream of any
disturbance from the moving piston. Does the progression of these waveforms indicate
compression wave steepening or spreading as t → 0 ?

Solution 15.20. a) For the situation shown in Figure 15.19, the initial piston location is –cot1, so
the sign of xp must be changed in the formula given in Exercise 15.18, and the piston starts
moving at t = –t1, where t1 is presumed to be positive. Thus, the starting point for this Exercise
is:
2
" γ +1 % " t %γ +1 2cot1 " t %
x p (t) = − $ ' cot1 $ ' + $ '
# γ −1 & # −t1 & γ −1 # −t1 &
where the extra divisor factors of –t1 have been introduced to facilitate the evaluation of xp when
t < 0. Time differentiate this formula,
2
−1
dx p (t) " γ +1 % " 2 %" t %γ +1 " 1 % 2co
= −$ ' cot1 $ '$ ' $ '− .
dt # γ −1 & # γ +1 &# −t1 & # −t1 & γ −1
Simplify and set dxp/dt = co:
2
−1 ( 1−γ ,
dx p (t) " 2 % " t %γ +1 2co 2co * " %
t 1+γ *
= co = $ ' co $ ' − = )$ ' −1- .
dt # γ −1 & # −t1 & γ −1 γ −1 *# −t1 & *
+ .
Divide out the common factor of co and solve for t:
1+γ
" γ +1 %1−γ
t = −t1 $ ' = –0.3349t1,
# 2 &
where the numerical value applies when γ = 1.4.
b) The first or lowest C+ characteristic shown on Figure 15.19 has a slope unity. To the right of
this characteristic, u = 0 and p = po.
Using the part a) result, and equation for the piston's trajectory, the location where the
piston speed reaches co can be found:
!x t$
location = # p , & = (−0.7368, −0.3349 ) .
" cot1 t1 %
The line from this location to the origin is the last or uppermost C+ characteristic shown on
Figure 15.19. To the left of this characteristic, the flow state will be uniform with u = co and p =
constant.
The region in between the two characteristics must provide the appropriate transition
between the left- and right-side flow states. To specify this transition, recognize that x and t are
both negative, and start from the results of Example 15.7:
 Fluid Mechanics, 6th Ed.
P Kundu, Cohen, and Dowling

2 "x % 2 γ −1 x
–––– T = 2 u(x, t) = $ − co '––––andT c(x,
= 2 t) = co + .
–––T=3 γ +1 #t &
–––T=3 γ +1 γ +1 t
Rearrange
----T=4 and introduce dimensionless variables
- - - - TU==4 u/co, C = c/co, X = x/cot1, and T = t/t1.
u(x, t) 2 " x 1 % 2 "X %
= $$ −1'' or U = $ −1' (1)
co γ +1 # cot1 (t t1 ) & γ +1 # T &
X X
c(x, t) 2 γ −1 x 1 2 γ −1 X
= + or C = + .
co γ +1 γ +1 cot1 (t t1 ) γ +1 γ +1 T
The final equation can be switched to the dimensionless pressure P = p/po via (15.52):
2γ
" 2 γ −1 X %γ −1
P =$ + ' . (2)
# γ +1 γ +1 T &
2γ
p ( 2 γ −1 " −0.7368 %+γ −1
The peak pressure then is: P = =* + $ '- = 3.583, when γ = 1.40.
po ) γ +1 γ +1 # −0.3349 &,
For the given parameters, the plots of (1) and (2) look like:

1.25& 4"

1& 3.5"
3"
0.75&
2.5"
0.5& 2"
1.5"
0.25&
1"
0&
0.5"
!0.25& 0"
!1& !0.8& !0.6& !0.4& !0.2& 0& )1" )0.8" )0.6" )0.4" )0.2" 0"

Examination of both plots suggests that the compression wave steepens as time increases.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.21. For the flow conditions of Figure 15.19, assume the flow speed downstream of
the shock wave is co and determine the shock Mach number, its x-t location, and the pressure,
temperature and density ratios across the shock. Are these results well matched to the isentropic
compression that occurred for t < 0? What additional adjustment is needed?

Solution 15.21. The shock wave first forms at the origin in Figure 15.19, and the velocity
difference across the shock wave is co. This is enough to determine the shock Mach number.
Start with the normal shock velocity ratio condition (15.41) for a stationary normal shock wave:
u1 (γ +1)M12
= . (15.41)
u2 (γ −1)M12 + 2
When the situation in Figure 15.19 is subject to a Galilean transformation that creates a shock-
fixed coordinate system, u1 = –M1co and u2 = (1 – M1)co. Thus, (15.41) becomes:
−M1co (γ +1)M12
= 2
or −(γ −1)M12 − 2 = (γ +1)M1 (1− M1 ) ,
(1− M1 )co (γ −1)M1 + 2
which reduces to the quadratic equation:
! 2 $
γ +1 1 # γ +1 ! γ +1 $
& + 4 &.
2
M1 − M1 −1 = 0 , so M1 = ± #
2 2 #" 2 " 2 % &
%
The physically meaningful root comes from the '+' sign and is M1 = 1.766 for γ = 1.4.
Thus, the x-t location of the shock is:
xshock = 1.7662cot,
and the pressure, temperature, and density ratios are:
p2 2γ " 2 $
= 1+ # M1 −1% = 3.473 , (3.583)
p1 γ +1
ρ2 (γ +1)M12
= = 2.305 , and (2.488)
ρ1 (γ −1)M12 + 2
T2 2(γ −1) γ M12 +1 2
T1
= 1+
(γ +1)2 M12
( M1 −1) = 1.506 . (1.440)

Interestingly, these ratios are all slightly different than the ratios produced by the
isentropic compression that occurs before the shock wave forms at t = 0 (provided in parenthesis
at the right). [These numbers are obtained from the solution of Exercise 15.20]. Thus, there will
be an adjustment region near the origin in Fig. 15.19 that will allow the pressure behind the
shock to equilibrate with the adiabatically compressed gas that is between the piston and the
shock wave. The net effect of this adjustment will be to shift the shock wave location slightly
farther ahead – in the positive x-direction – compared to what is calculated above because the
adiabatic compression reaches a higher pressure.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.22. Write momentum conservation for the volume of the small rectangular control
volume shown in Figure 4.20 where the interface is a shock with flow from side 1 to side 2. Let
the two end faces approach each other as the shock thickness → 0 and assume viscous stresses
may be neglected on these end faces (outside the structure). Show that the n component of
momentum conservation yields (15.36) and the t component gives u⋅t is conserved or v is
continuous across the shock.
u2!
(2)! +n!
us!
dA!
l!

(1)!
u1! –n! dA!

Solution 15.22. For a stationary control volume V containing only fluid particles bounded by a
surface A, conservation of mass and momentum may be written:
d d
∫
dt V
ρdV = − ∫ ρu j n j dA , and ∫ ρui dV = − ∫ [ρui u j + pδij − σ ij ]n j dA + ∫ ρgi dV .
dt V
A A V
When applied to a small rectangular control volume like that shown above (a reproduction of
Fig. 4.20) in a steady flow, these simplify to:
0 = − ∫ ρu j n j dA , and 0 = − ∫ [ ρui u j + pδij − σ ij ]n j dA + ∫ ρgi dV .
€ A
€ A V
For the current situation, let the interface define the location of a shockwave, and let the
thickness dimension l → 0 while requiring the upper rectangular surface to remain in region 2
and the lower rectangular surface to remain in region 1. This means V → 0 so the body force
€ be dropped, and that
term may € the flux through the sides of the CV may be ignored because the
curved-side surface area goes to zero as l → 0 . For this specialized limit, the equations reduce
€
to:
€
(ρu j )1 n j = (ρu j )2 n j , and [ρuiu j + pδij − σ ij ]1 n j = [ρuiu j + pδij − σ ij ]2 n j ,
where nj are the components of n, € the primarily-upward-pointing unit vector shown in the figure.
Here, the two larger CV surfaces are chosen to lie outside the shock structure (which is presumed
to be very thin), so that σij = 0. [Inside the shock wave, the viscous stress may be large]. Thus,
the€momentum equation becomes: €
(ui )1 ( ρu j )1 n j + p1n i = (ui ) 2 ( ρu j ) 2 n j + p2 n i . (†)
For the shock-normal component, take the dot product of (†) with n (which has
components ni). The resulting equation is:
€( i i )1 1 ( i i )2
ρ (u n )2 + p = ρ (u n )2 + p2 , or p1 − p2 = −ρ1u12 + ρ2 u22 ,
where (uini)1 = u1 and (uini)2 = u2, and the second equation is (15.36).
For the shock tangent component, take the dot product (†) with either tangent unit vector,
generically represented here by t (which has components ti):
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

t i (ui )1 ( ρu j )1 n j + p1t i n i = t i (ui ) 2 ( ρu j ) 2 n j + p2 t i n i , or t i (ui )1 ( ρu j )1 n j = t i (ui ) 2 ( ρu j ) 2 n j ,

where by definition: t ⋅ n = t i n i = 0 . Now use the final form of the mass conservation equation to
divide out the terms in larger parenthesis that contain ρ; this leaves:
t i (ui )1 = t i (ui ) 2 , or in vector notation: t ⋅ u1 = t ⋅ u2 .
€ €
Thus, the tangential components of u are unchanged across a shockwave since t represents either
tangent unit€vector.

€ €
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.23. A wedge has a half-angle of 50°. Moving through air, can it ever have an
attached shock? What if the half-angle were 40°? [Hint: The argument is based entirely on
Figure 15.22.]

Solution 15.23. From Figure 15.22, no attached shock can produce a deflection angle of 50°.
However, an attached shock with a deflection angle of 40° is possible.
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.24. Air at standard atmospheric conditions is flowing over a surface at a Mach
number of M1 = 2. At a downstream location, the surface takes a sharp inward turn by an angle of
20°. Find the wave angle σ and the downstream Mach number. Repeat the calculation by using
the weak shock assumption and determine its accuracy by comparison with the first method.

Solution 15.24. Using the geometry shown, and assuming M2 > 1, the shock angle σ is
approximately 53° based on Figure 15.22. Therefore:
Mn1 = M1sinσ = 2sin(53°) = 1.6, and
Mn2 = 0.668 (from Table 15.2).
From the geometry:
Mn2 = M2sin(σ – δ), so M 2 = M n 2 sin(σ − δ ) = 0.668 sin(53° − 20°) = 1.227 .
To develop equivalent results from weak shock theory, start from (15.53) and simplify
for δ << 1 and σ → µ = sin−1 (1 M1 ) . Thus,
[ M12 (γ + cos2σ ) + 2] tanδ = 2cot σ ( M12 sin2 σ −1)
€
becomes:
€ [ M12 (γ + 1− 2sin2 σ ) + 2]δ ≅ 2 M12 −1( M12 sin2 σ −1) ,
where tan δ ≅ δ€and cot σ ≅ cot µ = M12 −1 . Substituting sin 2 σ ≅ M1−2 on the right side produces
M12 (γ + 1)
δ ≅ M12 sin 2 σ −1, or
€ 2 M −12
1
€ € 12 €
& ) & - π 0)
12
−1 1 γ +1 −1 1 2.4
σ = sin ( 2
+ +
δ = sin ( + 20/ 2+ = 44.5° .
(' M1 2 M12 −1 +* ' 4 2 4 −1 . 180 1*
So, €
Mn1 = M1sinσ = 2sin(44.5°) = 1.40, and
Mn2 = 0.740 (from Table 15.2).
From€ the geometry:
Mn2 = M2sin(σ – δ), so M 2 = M n 2 sin(σ − δ ) = 0.740 sin(44.5° − 20°) = 1.784 ,
so the error = (1.784 – 1.227)/1.227 = 0.45 or 45%.

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.25. A flat plate is inclined at 10° to an airstream moving at M = 2. If the chord
∞

length is b = 3 m, find the lift and wave drag per unit span.

Solution 15.25. Considering figure 15.24, and using equations (15.58):

L = 2αγM∞2 p∞ b M∞2 −1 = 2(10π 180)(1.4)2 2 (101.3kPa)(3m) 2 2 −1 = 342kNm−1
and
D = αL = (10π 180) = 59.7kNm −1 .
€

€
 Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.26. Using thin airfoil theory calculate the lift and drag on the airfoil shape given by
yu = t sin(πx/c) for the upper surface and y1 = 0 for the lower surface. Assume a supersonic
stream parallel to the x-axis. The thickness ratio t/c << 1.

Solution 15.26. From supersonic thin airfoil theory

p − p∞ γM∞2δ
= ,
p∞ M∞2 −1
where δ is the slope of the streamline on the surface of the foil. On the foil's upper surface:
dy u tπ # πx &
= cos% ( = tan δ ≅ δ for δ << 1.
dx
€ c $c '
On the foil's lower surface dyl/dx = 0. Thus the surface pressures are:
pu − p∞ γM∞2 tπ & πx )
= cos( + , and pl = p∞.
€ p∞ M∞2 −1 c 'c *
The lift, L, per unit span will be:
c
L= ∫ ( pl − pu )e y ⋅ nds = ∫ ( pl − pu )dx
surface 0
€
dx
where e y ⋅ nds = cosδ ⋅ = dx . Thus,
cos δ
2 c
c
γM
€ ∞ p∞ tπ
& πx ) γM∞2 p∞ - & πx )0 γM∞2 p∞
L = −∫ cos( +dx = − t /sin ( +2 = − t(0 − 0) = 0 .
0 M∞2 −1 c 'c * M∞2 −1 . ' c *10 M∞2 −1
€ The drag, D, per unit span will be:
c
D= ∫ ( pu − p∞ )e x ⋅ nds = ∫ ( pu − p∞ )δdx
surface 0
€
dx
where e x ⋅ nds = sin δ ⋅ = tan δdx ≅ δdx . Thus,
cosδ
c 2
& )2 & ) 2
& )2
€ D = ∫ γM∞ p∞ ( tπ + cos 2 ( πx +dx = γM∞ p∞ ( tπ + ⋅ c .
0 M∞2 −1 ' c * 'c * M∞2 −1 ' c * 2
€ The reason for these results is that p > p∞ on the upper foil surface from 0 < x < c/2, and p < p∞
on the upper foil surface from c/2 < x < c with an average value of p∞. Thus, no lift results.
However, with a higher pressure on the upstream half and a lower pressure on the downstream
half, a drag €
is produced.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

Exercise 15.27. Consider a thin airfoil with chord length l at a small angle of attack in a
horizontal supersonic flow at speed M∞. The foil's upper and lower surface contours, yu(x) and
yl(x), respectively, are defined by:
yu(x) = t(x)/2 + yc(x) – αx , and yl(x) = –t(x)/2 + yc(x) – αx ,
where: t(x) = the foil's thickness distribution, α = the foil's angle of attack, and yc(x) = the foil's
camber line. Use these definitions to show that the foil's coefficients of lift and drag are:
) # &2 # 2 ,
4α 4 1 dt dy &
CL = , and CD = + % ( +% c
( +α2..
M ∞2 −1 M ∞2 −1 +* 4 $ dx ' $ dx ' .-

Solution 15.27. Use x-y coordinates and assume the foil extends x = –l/2 to x = +l/2. For
geometrical consistency, the upper and lower foil-surface contours must match at the foil's
leading and trailing edges: yu(–l/2) = yl(–l/2), and yu(+l/2) = yl(+l/2), so t(–l/2) = t(+l/2) = 0.
And, by definition, yc(–l/2) = yc(+l/2) = 0.
To compute the lift coefficient, CL, first determine the pressure coefficient starting from
its definition, then eliminate the pressure difference using (15.68):
p − p∞ γ M ∞2δ p∞ γ M ∞2δ 2 2δ
Cp = 2
= 2
= 2
= .
(1 / 2)ρ∞U∞ M ∞2 −1 (1 / 2)ρ∞U∞ M ∞2 −1 γ M ∞ M ∞2 −1
For small flow deflection angles: δ ≈ sinδ ≈ tanδ = dy/dx and cosδ ≈ 1. Thus,
1 +l/2 1 +l/2 %' −2 % dyl ( 2 % dyu ((
CL = ∫ (C p,l − C p,u ) cos δdx ≅ ∫ ' *− ' ***dx .
l −l/2 l −l/2 '& M ∞2 −1 & dx ) 2
M ∞ −1 & dx ))
where the 'l' and 'u' subscripts indicate the lower and upper foil surfaces, and δ is measured from
the horizontal and is positive when it leads to flow compression. Here, the integration merely
undoes the differentiation of the upper and lower foil contours:
1 2 +l/2
CL ≅ − [ yl (x) + yu (x)]−l/2
l M ∞2 −1
−2 $ αl αl αl αl '
= &%0 + 0 − + 0 + 0 − − 0 − 0 − − 0 − 0 − )( .
2
l M −1
∞
2 2 2 2
4α
=
M ∞2 −1
So, only the angle of attack matters for the production of lift. Neither the thickness nor the
chamber is involved.
To determine the coefficient of drag, the local component of the pressure force in the direction
of flow must be considered:
1 +l/2 %' −2 % dyu ( (*
2 2
1 +l/2 2 % dyl (
CD = ∫ (C p,u − C p,l ) sin δdx ≅ ∫ ' * − ' * dx
l −l/2 l −l/2 '& M ∞2 −1 & dx ) M ∞2 −1 & dx ) *)
+l/2 %
% dyl ( % dyu ( (
2 2
2
= ∫ ''' * + ' * **dx.
l M ∞2 −1 −l/2 && dx ) & dx ) )
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling

2 2 2
dy 1 dt dyc ! dy $ 1 ! dt $ ! dy $ dt dyc dt dy
Here, u = + + α , so # u & = # & + # c & + α 2 + + α + 2 c α , and
dx 2 dx dx " dx % 4 " dx % " dx % dx dx dx dx
2 2 2
dyl 1 dt dyc ! dy $ 1 ! dt $ ! dy $ dt dyc dt dy
=− + + α , so # l & = # & + # c & + α 2 − − α + 2 c α , so
dx 2 dx dx " dx % 4 " dx % " dx % dx dx dx dx
2 2 2 2
! dyl $ ! dyu $ 1 ! dt $ ! dy $ 2 dy
# & +# & = # & + 2 # c & + 2α + 4 c α .
" dx % " dx % 2 " dx % " dx % dx
The chord-length average of the final term is zero, so
+l/2 # 2 2 &
2 1 # dt & # dyc &
CD = ∫ %% % ( + 2 % ( + 2α 2 ((dx
l M ∞2 −1 −l/2 $ 2 $ dx ' $ dx ' '
# # &2 # 2 &
2 % 1 dt dyc & 2(
= % ( + 2% ( + 2α (
M ∞2 −1 %$ 2 $ dx ' $ dx ' '
# # &2 # 2 &
=
4 % 1 % dt ( + % dyc &( + α 2 (.
M ∞2 −1 %$ 4 $ dx ' $ dx ' (
'
This result suggests that airfoil thickness and camber only lead to drag in supersonic flow. Thus,
the wings and control surfaces of supersonic aircraft and missiles are thin and nearly flat.
Chapter 16 Page 1 of 11
 Chapter 16 Page 2 of 11

Problem 2.

∆p = f (L, a, ρ, µ, ω, U )

∆p, L, a, ρ, µ, ω, U ; n = 7, dimensional parameters.

Primary dimensions: M, L, t; r = 3, primary dimensions. Now,

[∆p] = M L−1 t−2 , [L] = L, [a] = L, [ρ] = M L−3 , [µ] = M L−1 t−1 , [ω] = t−1 , [U ] = Lt−1 .

Selecting repeating parameters: a, ρ, U , m = r = 3 repeating parameters.

Then, n − m = 4 dimensionless groups will result.

Setting up the dimensional equations, we have:

Π1 ≡ a C 1 ρ C 2 U C 3 L C 4
≡ LC1 (M L−3 )C2 (Lt−1 )C3 LC4

Therefore,

L : C1 − 3C2 + C3 + C4 = 0
M: C2 =0
t: −3C3 =0

Therefore, C2 = 0 = C3 and C1 = −C4 . Thus,

 C 4
L
Π1 ≡ . (1)
a

Next,

Π 2 ≡ aC 5 ρ C 6 U C 7 µ C 8

C8
≡ LC5 (M L−3 )C6 (Lt−1 )C7 M L−1 t−1

Therefore,

L : C5 − 3C6 + C7 − C8 = 0
M: C6 + C8 =0
t: −C7 − C8 =0

Therefore, C6 = −C8 , C7 = −C8 , and C5 = −C8 . Thus,

 C 8
µ
Π2 ≡ . (2)
aρU

2
 Chapter 16 Page 3 of 11

Next,

Π3 ≡ aC9 ρC10 U C11 ω C8

C12
≡ LC9 (M L−3 )C10 (Lt−1 )C11 t−1

Therefore,

L : C9 − 3C10 + C11 = 0
M: C10 =0
t: −C11 − C12 =0

Therefore, C10 = 0, C11 = −C12 , and C9 = −C12 . Thus,

 aω C9
Π3 ≡ . (3)
U
Next,

Π4 ≡ aC13 ρC14 U C15 ∆p1

1
≡ LC13 (M L−3 )C14 (Lt−1 )C15 M L−1 t−2

Therefore,

L : C13 − 3C14 + C15 − 1 = 0

M: C14 + 1 =0
t: −C15 − 2 =0

Therefore, C15 = −2, C14 = −1, and C13 = 0. Thus,

 
∆p
Π4 ≡ . (4)
ρU 2

From 1-4, we get

 C4  C 8  
∆p L µ aω C9
≡C . (5)
ρU 2 a aρU U
where, C, C4 , C8 , C9 are constants.

Therefore, we may write

 C2
∆p L
= C1 (Re)C3 (St)C4 . (6)
ρU 2 a
where, C1 , C2 , C3 , C4 are constants.

3
 Chapter 16 Page 4 of 11

Problem 3.

Morgan and Young invoke the following assumptions: (1). A collar-like stenosis may be
approximated by a smooth, rigid, axisymmetric constriction in a long straight tube, (2).
The effect of the stenosis geometry is dominant and any influence of wall distensibility is
negligible, (3). Blood can be treated as a Newtonian fluid at the flow rates encountered
in the large arteries where stenosis commonly occur, (4). Blood flow is laminar, and (5).
Steady flow assumption is acceptable although the arterial blood flow is pulsatile.
The stenosis is shown in the following figure: The dimensional variables are designated by

Figure 1: Axisymmetric stenosis.

ˆ. The axial coordinate and velocity are ẑ and û. Û is the centerline velocity. The radial
coordinate is r̂ and the corresponding velocity component is v̂. The dimensionless variables
2
are: r = r̂/R0 , z = ẑ/R0 , R = R̂/R0 , u = û/Ū0 , v = v̂/Ū0 , U = Û /Ū0 , and p = p̂/ρŪ0 ,
where, Ū0 is the average velocity in the unobstructed tube, and p̂ is the pressure and ρ is
the density.
The usual continuity, axial momentum, and radial momentum equations for axisymmetric
flow in a tube are considered. The dimensionless momentum equations involve the Reynolds
number, Re = 2R0 Ū0 ρ/µ, where µ is the fluid viscosity. By integrating the axial momentum
equation over the cross section of the tube and by invoking the no slip condition, u = v = 0 at
the wall of the vessel, the integral momentum equation is developed. Next, by multiplying
the axial momentum equation by ru, and integrating over the cross section, the integral
energy equation is developed. It is now noted that in these two equations, the terms due to
the viscous component of the normal stress in the axial direction (∂ 2 u/∂z 2 ) are negligible.
Next, an important observation is made in regard to the pressure gradient ∂p/∂z. It is
noted that the pressure gradient ∂p/∂z is independent of the radial coordinate r for flow in
a straight tube, whereas for flow in a constricted tube the pressure gradient will, in general,
vary over the cross section. However, if the integral energy equation is multiplied by R2 the
resulting integral involving the pressure gradient is equal to the corresponding integral in
the integral momentum equation for two special cases: (i) ∂p/∂z is independent of r; (ii) the
velocity u is independent of r. Since these two conditions are approached in a constriction,
i.e. in the gradually varying initial portion of the constriction the pressure gradient is nearly
constant and in the rapidly converging portion the velocity profile tends to become flattened,
we may let, Z R Z R
∂p 2 ∂p
r dr ≈ R ru dr. (1)
0 ∂z 0 ∂z

4
 Chapter 16 Page 5 of 11

This approximation would enable the elimination of the pressure gradient term in both the
integral momentum and integral energy equations. Then it is possible to combine the integral
momentum and energy equations into a single equation in terms of axial velocity:
Z R Z R " Z 2   #
R 
1 2∂ ∂ 2 ∂u ∂u
R ru3 dr − ru2 dr = − R2 r dr + R . (2)
2 ∂z 0 ∂z 0 Re 0 ∂r ∂r R

The equation (2) is subject to (i) u = U at r = 0; (ii) u = 0 at r = R; (iii) ∂u/∂r = 0 at

r = 0; (iv) ∂ 3 u/∂r3 = 0 at r = 0; and, (v) the condition that the net flow through any cross
section must be the same for any incompressible fluid which may be expressed by:
Z R
1
r u dr = . (3)
0 2

In the above, the condition (iii) is derived from a consideration of the forces on a cylindrical
element having its axis along the tube centerline. If the pressure and the inertial forces
are to be finite as the radius of the element approaches zero, the viscous force, which is
proportional to ∂u/∂r, must approach zero. The condition (iv) is developed by eliminating
the pressure between the axial momentum and radial momentum equations and considering
the resulting equation as r approaches zero.
Next, it is noted that at high Reynolds numbers, the profile must allow a thin region of
high shear near the wall in the converging section with a relatively flat profile in the core.
To accommodate these requirements, Morgan and Young construct a polynomial fit which
permits the shear near the wall to become large while maintaining a flat core flow. This fit
is given by,
r  r 2  r 4 r
u = U for 0 ≤ ≤ λ, and, u = a + b +c for λ ≤ ≤ 1, (4)
R R R R
where a, b, and c are unknown coefficients and λ is the value of (r/R) at the juncture sepa-
rating the flat and polynomial parts of the profile. The unknown coefficients are determined
from the no slip condition along with two compatibility conditions u = U and ∂u/∂r = 0
at (r/R) = λ. The constraint (v) enables expressing λ in terms of R and U . Thus the
polynomial fit profile for u is entirely in terms of U , r, and R. The profile is now introduced
into the equation (2). The resulting first order, non-linear ordinary differential equation is
numerically solved by assuming that Poiseuille flow prevails far upstream of the stenosis.
The solution provides the desired velocity profiles. These are plotted by Morgan and Young.
The wall shear stress is evaluated from
    2 
∂ û
τ̂w = µ 1 + R̂0 , (5)
∂r̂ R̂

where R̂0 is the slope dR̂/dẑ of the wall. The results are included in the paper by Morgan
and Young.

5
 Chapter 16 Page 6 of 11

Problem 4.

In a pure pressure-gravity flow, the effects of friction are negligible compared with the effects
due to changes of the external pressure and the elevation in the gravity field. Lengthwise al-
terations of speed, pressure, area, etc., are brought about by changes in pe and z. Changes in
pe may be brought about by : (i) active muscle tone; (ii) elastic constrictions, or sphincters,
as where veins pass from the abdominal cavity to the thoracic cavity, and in the pulmonary
system; (iii) weights, (iv) pressurizing cuffs, and (v) clamps etc.,. Changes in z are important
because (i) in the vertical position of a human being, the hydrostatic pressure exceeds the
venous and arterial pressure levels;(ii) during aircraft maneuvers; and, (iiI) during certain
phases of space flight when effective g may be greatly increased.
Retaining only the terms in d(pe + ρgz) in the table of influence coefficients for the tube law,
3
−P ≈ α−n − 1, and, n = , (1)
2
the governing equations of a pure pressure-gravity flow are written as,

1 dα 2 3 dΠ
1 − S2 = − α2 , (2)
α dx 3 dx
and,

1 dS 2 1 3 dΠ
1 − S2 = + α2 , (3)
S 2 dx 3 dx
where,
(pe + ρgz)
Π= . (4)
Kp
Dividing equation (3) by equation (2) and integrating, for a pure pressure-gravity flow we
have,
α A  u −1
= = = S −4 , (5)
α∗ A∗ u∗
where, ∗ denotes the value of the quantity at S = 1. For the tube law given by equation (1),
the wave speed is known to be,
1
nKp α−n 2

3
c= , n= . (6)
ρ 2
From equations(5) and (6),
c  α − 43
= = S 3. (7)
c∗ α∗
Also, from Bernoulli’s theorem,
(p − p∗ ) 8 g (z ∗ − z)
1 = 1 − S + . (8)
2
ρc∗ 2 c∗ 2
With equation (8), and by the elimination of α from equation (3) using equation (5), Shapiro
develops,
1 ∗ 23
α dΠ = S 4 (1 − S 2 )dS 2 . (9)
3

6
 Chapter 16 Page 7 of 11

Integration of equation (9) between limits Π and Π∗ corresponding to S and 1, gives

1 ∗ 32 1 6
1 8
α (Π − Π∗ ) =


S −1 − S −1 . (10)
3 3 4
Shapiro provides graphs of equation(10). The graphs show that increasing values of Π drive
S towards unity and decreasing values of Π drive S away from unity. From these, it is
concluded that: (i) Choking occurs when the value of Π continually increases with either
S < 1 or S > 1; (ii) Continuous transition through S = 1 may be achieved by means of an
increase in Π until S = 1 is reached, with dΠ/dx becoming zero exactly at S = 1 followed
then by a decrease of Π.

7
 Chapter 16 Page 8 of 11

Problem 5.

To determine the volume flux for the flow with the Power-law model, consider an annu-
lar volume element of length L and thickness dr in the flow, as shown in the figure below:

a
dr
r

Figure 2: Representative volume element.

Let there be a constant pressure gradient, −∆p/L, across the element of length L.
∆p
Force due to pressure gradient on the annular element = + (2πrL) dr. (1)
L
This must be balanced by shear force which is given by,
d
Shear force = (2πdr) L (rτ ). (2)
dr
Therefore,
d ∆p
(rτ ) = r (3)
dr L
∆p r c
τ = + (4)
L 2 2
where c is a constant. The stress is finite at axis r = 0. Therefore c = 0 and hence, from 4,
∆p r
τ= . (5)
L 2
For a power-law fluid,
τ = µγ̇ n . (6)
 1
∆p r n
Therefore γ̇ = . (7)
L 2µ
In this problem, since velocity is a function of r only,
du
γ̇ = − . (8)
dr

8
 Chapter 16 Page 9 of 11

where u(r) is velocity component in r direction.

From equations 7 and 8,
 1
du ∆p r n
=− . (9)
dr L 2µ
The no-slip condition at the wall requires that

u = 0 at r = a. (10)

By integrating equation 9 and using equation 10, one obtains

 1/n
∆p n  n+1 n+1

u= a n −r n . (11)
2µL n+1

Now, the flux for flow, Q, is given by

Z R
Q= 2πrudr. (12)
0

With integration by parts and invoking the no-slip conditions at r = a,

Z a  
2 du
Q=π r − dr. (13)
0 dr

With equation 11,

Z a  1/n
2 ∆p r
Q = π r dr (14)
0 L 2µ
 1/n
∆p nπ 3n+1
= a n (15)
2µL 3n + 1

When n = 1,
 
∆p π 4
Q = a (16)
2µL 4
 4
πa ∆p 3n+1
= a n
8µ L

which agrees with equations (17.17) and is the Poiseuille formula.

9
 Chapter 16 Page 10 of 11

Problem 6.

To determine the volume flux for the flow with Herschel-Bulkley model, we first note that
τ = µγ̇ n + τ0 , τ ≥ τ0
and γ̇ = 0 , τ < τ0 (1)
There is yield stress and as a consequence the flow region includes plug flow in the core as
shown in figure Let the radius of the plug flow region be rp .

rp

Core region

Velocity profile

Figure 3: Representative volume element.

For r ≤ rp , τ (r) = τ0 = constant (2)

Therefore in the core, γ̇ = 0. (3)
du
Therefore in the core, = 0. (4)
dr
Therefore,
u = constant (5)
= up (say) (6)
(7)
For a constant pressure gradient, − (∆p/L), across an element of length L in the core of
region, from a force balance,
∆p
2πrp Lτ0 = πrp 2 L (8)
L
2τ0
Thus, rp = (9)
(∆p/L)
∆p rp
τ0 = (10)
L 2
Outside the core region, with u = u(r), just as in problem 5,
 1
du ∆p n 1
=− (r − rp ) n . (11)
dr 2µL

10
 Chapter 16 Page 11 of 11

By integration,
 1/n
∆p n n+1
u=− (r − rp ) n + C. (12)
2µL n+1
where C is a constant of integration.
Now, at r = a, u = 0 (no-slip condition). Therefore,
 1/n
∆p n  n+1 n+1

u= (a − rp ) n − (r − rp ) n . (13)
2µL n+1

We can also calculate the plug velocity by setting r = rp in equation 12. Thus,
 1/n
∆p n n+1
up = (a − rp ) n . (14)
2µL n+1

The volume flux may be calculated from,

Z a
2
Q = πrp up + 2πrudr. (15)
rp

With equations 13 and 14, we can evaluate equation 15.

Let ξ = rp /a. (16)

Then, after integration and considerable algebra,

 1/n
∆p nπ 3n+1 h 2 n+1 2n+1
Q = a n ξ (1 − ξ) n + (1 + ξ)(1 − ξ) n
2µL n+1

2n 3n+1 2n 2n+1
− (1 − ξ) n − ξ(1 − ξ) n (17)
3n + 1 2n + 1

When τ0 = 0, the Herschel-Bulkley model reduces to the Power-law model. Therefore, with
τ0 = 0 and ξ = 0, equation 17 reduces to the result of problem 5.

11
Chapter 16 Page 1 of 11
 Chapter 16 Page 2 of 11

Problem 2.

∆p = f (L, a, ρ, µ, ω, U )

∆p, L, a, ρ, µ, ω, U ; n = 7, dimensional parameters.

Primary dimensions: M, L, t; r = 3, primary dimensions. Now,

[∆p] = M L−1 t−2 , [L] = L, [a] = L, [ρ] = M L−3 , [µ] = M L−1 t−1 , [ω] = t−1 , [U ] = Lt−1 .

Selecting repeating parameters: a, ρ, U , m = r = 3 repeating parameters.

Then, n − m = 4 dimensionless groups will result.

Setting up the dimensional equations, we have:

Π1 ≡ a C 1 ρ C 2 U C 3 L C 4
≡ LC1 (M L−3 )C2 (Lt−1 )C3 LC4

Therefore,

L : C1 − 3C2 + C3 + C4 = 0
M: C2 =0
t: −3C3 =0

Therefore, C2 = 0 = C3 and C1 = −C4 . Thus,

 C 4
L
Π1 ≡ . (1)
a

Next,

Π 2 ≡ aC 5 ρ C 6 U C 7 µ C 8

C8
≡ LC5 (M L−3 )C6 (Lt−1 )C7 M L−1 t−1

Therefore,

L : C5 − 3C6 + C7 − C8 = 0
M: C6 + C8 =0
t: −C7 − C8 =0

Therefore, C6 = −C8 , C7 = −C8 , and C5 = −C8 . Thus,

 C 8
µ
Π2 ≡ . (2)
aρU

2
 Chapter 16 Page 3 of 11

Next,

Π3 ≡ aC9 ρC10 U C11 ω C8

C12
≡ LC9 (M L−3 )C10 (Lt−1 )C11 t−1

Therefore,

L : C9 − 3C10 + C11 = 0
M: C10 =0
t: −C11 − C12 =0

Therefore, C10 = 0, C11 = −C12 , and C9 = −C12 . Thus,

 aω C9
Π3 ≡ . (3)
U
Next,

Π4 ≡ aC13 ρC14 U C15 ∆p1

1
≡ LC13 (M L−3 )C14 (Lt−1 )C15 M L−1 t−2

Therefore,

L : C13 − 3C14 + C15 − 1 = 0

M: C14 + 1 =0
t: −C15 − 2 =0

Therefore, C15 = −2, C14 = −1, and C13 = 0. Thus,

 
∆p
Π4 ≡ . (4)
ρU 2

From 1-4, we get

 C4  C 8  
∆p L µ aω C9
≡C . (5)
ρU 2 a aρU U
where, C, C4 , C8 , C9 are constants.

Therefore, we may write

 C2
∆p L
= C1 (Re)C3 (St)C4 . (6)
ρU 2 a
where, C1 , C2 , C3 , C4 are constants.

3
 Chapter 16 Page 4 of 11

Problem 3.

Morgan and Young invoke the following assumptions: (1). A collar-like stenosis may be
approximated by a smooth, rigid, axisymmetric constriction in a long straight tube, (2).
The effect of the stenosis geometry is dominant and any influence of wall distensibility is
negligible, (3). Blood can be treated as a Newtonian fluid at the flow rates encountered
in the large arteries where stenosis commonly occur, (4). Blood flow is laminar, and (5).
Steady flow assumption is acceptable although the arterial blood flow is pulsatile.
The stenosis is shown in the following figure: The dimensional variables are designated by

Figure 1: Axisymmetric stenosis.

ˆ. The axial coordinate and velocity are ẑ and û. Û is the centerline velocity. The radial
coordinate is r̂ and the corresponding velocity component is v̂. The dimensionless variables
2
are: r = r̂/R0 , z = ẑ/R0 , R = R̂/R0 , u = û/Ū0 , v = v̂/Ū0 , U = Û /Ū0 , and p = p̂/ρŪ0 ,
where, Ū0 is the average velocity in the unobstructed tube, and p̂ is the pressure and ρ is
the density.
The usual continuity, axial momentum, and radial momentum equations for axisymmetric
flow in a tube are considered. The dimensionless momentum equations involve the Reynolds
number, Re = 2R0 Ū0 ρ/µ, where µ is the fluid viscosity. By integrating the axial momentum
equation over the cross section of the tube and by invoking the no slip condition, u = v = 0 at
the wall of the vessel, the integral momentum equation is developed. Next, by multiplying
the axial momentum equation by ru, and integrating over the cross section, the integral
energy equation is developed. It is now noted that in these two equations, the terms due to
the viscous component of the normal stress in the axial direction (∂ 2 u/∂z 2 ) are negligible.
Next, an important observation is made in regard to the pressure gradient ∂p/∂z. It is
noted that the pressure gradient ∂p/∂z is independent of the radial coordinate r for flow in
a straight tube, whereas for flow in a constricted tube the pressure gradient will, in general,
vary over the cross section. However, if the integral energy equation is multiplied by R2 the
resulting integral involving the pressure gradient is equal to the corresponding integral in
the integral momentum equation for two special cases: (i) ∂p/∂z is independent of r; (ii) the
velocity u is independent of r. Since these two conditions are approached in a constriction,
i.e. in the gradually varying initial portion of the constriction the pressure gradient is nearly
constant and in the rapidly converging portion the velocity profile tends to become flattened,
we may let, Z R Z R
∂p 2 ∂p
r dr ≈ R ru dr. (1)
0 ∂z 0 ∂z

4
 Chapter 16 Page 5 of 11

This approximation would enable the elimination of the pressure gradient term in both the
integral momentum and integral energy equations. Then it is possible to combine the integral
momentum and energy equations into a single equation in terms of axial velocity:
Z R Z R " Z 2   #
R 
1 2∂ ∂ 2 ∂u ∂u
R ru3 dr − ru2 dr = − R2 r dr + R . (2)
2 ∂z 0 ∂z 0 Re 0 ∂r ∂r R

The equation (2) is subject to (i) u = U at r = 0; (ii) u = 0 at r = R; (iii) ∂u/∂r = 0 at

r = 0; (iv) ∂ 3 u/∂r3 = 0 at r = 0; and, (v) the condition that the net flow through any cross
section must be the same for any incompressible fluid which may be expressed by:
Z R
1
r u dr = . (3)
0 2

In the above, the condition (iii) is derived from a consideration of the forces on a cylindrical
element having its axis along the tube centerline. If the pressure and the inertial forces
are to be finite as the radius of the element approaches zero, the viscous force, which is
proportional to ∂u/∂r, must approach zero. The condition (iv) is developed by eliminating
the pressure between the axial momentum and radial momentum equations and considering
the resulting equation as r approaches zero.
Next, it is noted that at high Reynolds numbers, the profile must allow a thin region of
high shear near the wall in the converging section with a relatively flat profile in the core.
To accommodate these requirements, Morgan and Young construct a polynomial fit which
permits the shear near the wall to become large while maintaining a flat core flow. This fit
is given by,
r  r 2  r 4 r
u = U for 0 ≤ ≤ λ, and, u = a + b +c for λ ≤ ≤ 1, (4)
R R R R
where a, b, and c are unknown coefficients and λ is the value of (r/R) at the juncture sepa-
rating the flat and polynomial parts of the profile. The unknown coefficients are determined
from the no slip condition along with two compatibility conditions u = U and ∂u/∂r = 0
at (r/R) = λ. The constraint (v) enables expressing λ in terms of R and U . Thus the
polynomial fit profile for u is entirely in terms of U , r, and R. The profile is now introduced
into the equation (2). The resulting first order, non-linear ordinary differential equation is
numerically solved by assuming that Poiseuille flow prevails far upstream of the stenosis.
The solution provides the desired velocity profiles. These are plotted by Morgan and Young.
The wall shear stress is evaluated from
    2 
∂ û
τ̂w = µ 1 + R̂0 , (5)
∂r̂ R̂

where R̂0 is the slope dR̂/dẑ of the wall. The results are included in the paper by Morgan
and Young.

5
 Chapter 16 Page 6 of 11

Problem 4.

In a pure pressure-gravity flow, the effects of friction are negligible compared with the effects
due to changes of the external pressure and the elevation in the gravity field. Lengthwise al-
terations of speed, pressure, area, etc., are brought about by changes in pe and z. Changes in
pe may be brought about by : (i) active muscle tone; (ii) elastic constrictions, or sphincters,
as where veins pass from the abdominal cavity to the thoracic cavity, and in the pulmonary
system; (iii) weights, (iv) pressurizing cuffs, and (v) clamps etc.,. Changes in z are important
because (i) in the vertical position of a human being, the hydrostatic pressure exceeds the
venous and arterial pressure levels;(ii) during aircraft maneuvers; and, (iiI) during certain
phases of space flight when effective g may be greatly increased.
Retaining only the terms in d(pe + ρgz) in the table of influence coefficients for the tube law,
3
−P ≈ α−n − 1, and, n = , (1)
2
the governing equations of a pure pressure-gravity flow are written as,

1 dα 2 3 dΠ
1 − S2 = − α2 , (2)
α dx 3 dx
and,

1 dS 2 1 3 dΠ
1 − S2 = + α2 , (3)
S 2 dx 3 dx
where,
(pe + ρgz)
Π= . (4)
Kp
Dividing equation (3) by equation (2) and integrating, for a pure pressure-gravity flow we
have,
α A  u −1
= = = S −4 , (5)
α∗ A∗ u∗
where, ∗ denotes the value of the quantity at S = 1. For the tube law given by equation (1),
the wave speed is known to be,
1
nKp α−n 2

3
c= , n= . (6)
ρ 2
From equations(5) and (6),
c  α − 43
= = S 3. (7)
c∗ α∗
Also, from Bernoulli’s theorem,
(p − p∗ ) 8 g (z ∗ − z)
1 = 1 − S + . (8)
2
ρc∗ 2 c∗ 2
With equation (8), and by the elimination of α from equation (3) using equation (5), Shapiro
develops,
1 ∗ 23
α dΠ = S 4 (1 − S 2 )dS 2 . (9)
3

6
 Chapter 16 Page 7 of 11

Integration of equation (9) between limits Π and Π∗ corresponding to S and 1, gives

1 ∗ 32 1 6
1 8
α (Π − Π∗ ) =


S −1 − S −1 . (10)
3 3 4
Shapiro provides graphs of equation(10). The graphs show that increasing values of Π drive
S towards unity and decreasing values of Π drive S away from unity. From these, it is
concluded that: (i) Choking occurs when the value of Π continually increases with either
S < 1 or S > 1; (ii) Continuous transition through S = 1 may be achieved by means of an
increase in Π until S = 1 is reached, with dΠ/dx becoming zero exactly at S = 1 followed
then by a decrease of Π.

7
 Chapter 16 Page 8 of 11

Problem 5.

To determine the volume flux for the flow with the Power-law model, consider an annu-
lar volume element of length L and thickness dr in the flow, as shown in the figure below:

a
dr
r

Figure 2: Representative volume element.

Let there be a constant pressure gradient, −∆p/L, across the element of length L.
∆p
Force due to pressure gradient on the annular element = + (2πrL) dr. (1)
L
This must be balanced by shear force which is given by,
d
Shear force = (2πdr) L (rτ ). (2)
dr
Therefore,
d ∆p
(rτ ) = r (3)
dr L
∆p r c
τ = + (4)
L 2 2
where c is a constant. The stress is finite at axis r = 0. Therefore c = 0 and hence, from 4,
∆p r
τ= . (5)
L 2
For a power-law fluid,
τ = µγ̇ n . (6)
 1
∆p r n
Therefore γ̇ = . (7)
L 2µ
In this problem, since velocity is a function of r only,
du
γ̇ = − . (8)
dr

8
 Chapter 16 Page 9 of 11

where u(r) is velocity component in r direction.

From equations 7 and 8,
 1
du ∆p r n
=− . (9)
dr L 2µ
The no-slip condition at the wall requires that

u = 0 at r = a. (10)

By integrating equation 9 and using equation 10, one obtains

 1/n
∆p n  n+1 n+1

u= a n −r n . (11)
2µL n+1

Now, the flux for flow, Q, is given by

Z R
Q= 2πrudr. (12)
0

With integration by parts and invoking the no-slip conditions at r = a,

Z a  
2 du
Q=π r − dr. (13)
0 dr

With equation 11,

Z a  1/n
2 ∆p r
Q = π r dr (14)
0 L 2µ
 1/n
∆p nπ 3n+1
= a n (15)
2µL 3n + 1

When n = 1,
 
∆p π 4
Q = a (16)
2µL 4
 4
πa ∆p 3n+1
= a n
8µ L

which agrees with equations (17.17) and is the Poiseuille formula.

9
 Chapter 16 Page 10 of 11

Problem 6.

To determine the volume flux for the flow with Herschel-Bulkley model, we first note that
τ = µγ̇ n + τ0 , τ ≥ τ0
and γ̇ = 0 , τ < τ0 (1)
There is yield stress and as a consequence the flow region includes plug flow in the core as
shown in figure Let the radius of the plug flow region be rp .

rp

Core region

Velocity profile

Figure 3: Representative volume element.

For r ≤ rp , τ (r) = τ0 = constant (2)

Therefore in the core, γ̇ = 0. (3)
du
Therefore in the core, = 0. (4)
dr
Therefore,
u = constant (5)
= up (say) (6)
(7)
For a constant pressure gradient, − (∆p/L), across an element of length L in the core of
region, from a force balance,
∆p
2πrp Lτ0 = πrp 2 L (8)
L
2τ0
Thus, rp = (9)
(∆p/L)
∆p rp
τ0 = (10)
L 2
Outside the core region, with u = u(r), just as in problem 5,
 1
du ∆p n 1
=− (r − rp ) n . (11)
dr 2µL

10
 Chapter 16 Page 11 of 11

By integration,
 1/n
∆p n n+1
u=− (r − rp ) n + C. (12)
2µL n+1
where C is a constant of integration.
Now, at r = a, u = 0 (no-slip condition). Therefore,
 1/n
∆p n  n+1 n+1

u= (a − rp ) n − (r − rp ) n . (13)
2µL n+1

We can also calculate the plug velocity by setting r = rp in equation 12. Thus,
 1/n
∆p n n+1
up = (a − rp ) n . (14)
2µL n+1

The volume flux may be calculated from,

Z a
2
Q = πrp up + 2πrudr. (15)
rp

With equations 13 and 14, we can evaluate equation 15.

Let ξ = rp /a. (16)

Then, after integration and considerable algebra,

 1/n
∆p nπ 3n+1 h 2 n+1 2n+1
Q = a n ξ (1 − ξ) n + (1 + ξ)(1 − ξ) n
2µL n+1

2n 3n+1 2n 2n+1
− (1 − ξ) n − ξ(1 − ξ) n (17)
3n + 1 2n + 1

When τ0 = 0, the Herschel-Bulkley model reduces to the Power-law model. Therefore, with
τ0 = 0 and ξ = 0, equation 17 reduces to the result of problem 5.

11