Problem 7.

3
DΔpA § ρVD ·
Verify the left-hand side of =φ¨ ¸ is dimensionless using the MLT system.
ρV © μ ¹
2

Solution 7.3
DΔpA § ρVD · M M L
=φ¨ ¸, where D  L , ΔpA  2 2 , ρ  3 , and V 
ρV © μ ¹ T
2
LT L
Thus,
M
L
DΔpA 2 2
 L T = M 0 L0T 0
ρV 2 § M L2 ·
¨¨ 3 2 ¸¸
©L T ¹
DΔpA
That is is dimensionless.
ρV 2
Problem 7.4
The Reynolds number, ρVD / μ , is a very important parameter in fluid mechanics. Verify
that the Reynolds number is dimensionless, using both the FLT system and the MLT
system for basic dimensions, and determine its value for ethyl alcohol flowing at a velocity
of 3 m/s through a 2-in.- diameter pipe.

Solution 7.4

Reynolds number = =
(
ρVD FL T LT ( L )
−4 2 −1
)(
= F 0 L0T 0
)
μ −2
FL T

=
( ML−3 )( LT −1 ) ( L )
= M 0 L0T 0
ML−1T −1
N ⋅S kg
For ethyl alcohol, μ = 1.19 × 10 −3 3
and ρ = 789 .
m m3
§ kg · § m ·§ 2 · § m·
ρVD ©¨ 789 3 ¸ ¨ 3 ¸¨ ft ¸ ¨ 0.3048 ¸
Thus, = m ¹ © s ¹© 12 ¹ © ft ¹
= 1.01× 105
μ N ⋅ S
1.19 ×10 −3 2
m
Problem 7.11
Assume that the flowrate, Q, of a gas from a smokestack is a function of the density of the
ambient air, ρa , the density of the gas, ρg , within the stack, the acceleration of gravity, g ,
and the height and diameter of the stack, h and d , respectively. Use ρa , d , and g as
repeating variables to develop a set of pi terms that could be used to describe this problem.

Solution 7.11

(
Q = f ρa , ρg , g , h, d )
Q  L3T −1 ρa  ML−3 ρg  ML−3 g  LT −2 hL d L
From the pi theorem, 6 − 3 = 3 pi terms required. Use ρa , d , and g as repeating variables.
Thus,
Π1 = Qρaa d b g c
and

( L3T −1 )(ML−3 ) ( L )b ( LT −2 )
a c
= M 0 L0T 0

so that
a=0 (for M )
3 − 3a + b + c = 0 (for L )
−1 − 2c = 0 (for T )

5 1
It follows that a = 0, b = − , c = − , and therefore
2 2
Q
Π1 = 5 1
d 2g2
Check dimensions using FLT system:
Q L3T −1
5 1
= 1
= F 0 L0T 0 ∴ ok
( )
5
d 2g2 (L) 2 LT −2 2

For Π 2:

Π 2 = ρg ρaa d b g c

(ML− )(ML− ) 3 a
( L )b ( LT −2 )
3 c
= M 0 L0T 0
1+ a = 0 (for M )
3 − 3a + b + c = 0 (for L )
−2c = 0 (for T )
It follows that a = −1, b = 0, c = 0, and therefore
ρg
Π2 =
ρa
which is obviously dimensionless.
For Π3:

Π3 = hρaad b gc
( L ) ( ML−3 ) ( L )b ( LT −2 )
a c
= M 0 L0T 0
a=0 (for M )
1 − 3a + b + c = 0 (for L )
−2c = 0 (for T )

It follows that a = 0, b = −1, c = 0, and therefore

h
Π3 =
d
which is obviously dimensionless.
Thus,
Q § ρg h ·
=φ¨ , ¸
© ρa d ¹
5 1
d 2g2
Problem 7.20
The weir shown in the figure below is used to measure the volume flowrate Q. The height H
is a measure of this flowrate. The weir has length L (perpendicular to the paper). Select and
include relevant fluid properties and find the appropriate dimensionless parameters.

Q g
P

Density = ρ
Absolute viscosity = μ

Solution 7.20
Considering head above the weir H as the dependent variable and flowrate Q as an
independent variable.
H = f (Q , P , L , g , ρ , μ ) .
The dimensions of the parameters are
H L LL
L3 L
Q g 2
T T
PL M
ρ 3
L
M
μ
LT
So n = 7, k = 3, and n − k = 4. Choose Q, P, and ρ as repeating variables. The first two
parameters are obvious
H and L
Π1 = Π2 =
P P

The third parameter is The fourth parameter is

Π3 = gQa P b ρ c Π 4 = μQa P b ρ c
a c a c
§ L ·§ L · § M ·§ L ·
3 3
b§M · b§M ·
[Π3 ] = ¨ 2 ¸ ¨¨ ¸¸ ( L ) ¨ 3 ¸ [ Π 4 ] = ¨ ¸ ¨¨ ¸¸ ( L ) ¨ 3 ¸
© T ¹© T ¹ ©L ¹ © LT ¹ © T ¹ ©L ¹
M: c=0 M: 1+ c = 0
L: 1 + 3a + b − 3c = 0 L: −1 + 3a + b − 3c = 0
T: −2 − a = 0 T: −1 − a − 3c = 0
a = −2, b = 5, c = 0. a = −1, b = 1, c = −1.

gP 5 μP
Π3 = Π4 =
Q 2 ρQ
Problem 7.22
The input power, W , to a large industrial fan depends on the fan impeller diameter D, fluid
viscosity μ , fluid density ρ , volumetric flow Q, and blade rotational speed ω . What are the
appropriate dimensionless parameters?

Solution 7.22
Write the dimensional relation considering W as the dependent variable
W = f (Q, D, μ , ρ , ω ) .
Determine the maximum number of dimensionless parameters. There are six parameters
but η is already dimensionless so n = 5. Using the [M , L,T ] system
FL ML2
W   3
T T
L3
Q
T
DL
FT M
μ 2 
L LT
M
ρ 3
L
1
ω
T
There are three fundamental dimensions ( M , L, T ) so we expect k = 3; D, μ , and ρ cannot
form a dimensions group so k = 3 and the maximum number of dimensionless parameters is
( 6 − 3 ) = 3.
Select D, ρ , and ω as repeating parameters. Combining D, ρ , and ω with W gives
 a ρ bω c.
Π1 = WD
The dimensional equation is
a b c
ª¬W º¼ [ D ] [ ρ ] [ω ] = [M ] [ L ] [T ]
0 0 0

or
b c
ª ML2 º a ªM º ª 1 º
« 3 » [ L ] « 3 » « » = [M ] [ L ] [T ]
0 0 0

«¬ T »¼ ¬ L ¼ ¬T ¼
Equating power of each dimension gives
M: 1+ b = 0
L: 2 + a − 3b = 0
T: −3 − c = 0
Solving we obtain
b = −1, c = −3, a = −5.
W
The first dimensionless parameters is Π1 =
ρω 3D5
Combining D, ρ , and ω with Q gives
Π2 = QDa ρ bω c
The dimensional equation is
a b c
Q [ D ] [ ρ ] [ω ] = [M ] [ L ] [T ]
0 0 0

or
b c
ª L3 º a ªM º ª 1 º
« » [ L ] «¬ L3 »¼ «¬ T »¼ = [M ] [ L ] [T ]
0 0 0
T
¬« ¼»
Equating power of each dimension gives
M: b=0
L: 3 + a − 3b = 0
T: −1 − c = 0
Solving we obtain
b = 0, c = −1, a = −3.

Q
The second dimensionless group is Π 2 =
ωD3
Combining D, ρ , and ω with μ gives
Π3 = μDa ρ bωc
The dimensional equation is
[ μ ][D]a [ ρ ]b [ω ]c = [M ]0 [ L ]0 [T ]0
or
b c
ª M º a ªM º ª 1 º
«¬ LT »¼ [ L ] «¬ L3 »¼ «¬ T »¼ = [M ] [ L ] [T ]
0 0 0

Equating power of each dimension gives

M: 1+ b = 0
L: −1 + a − 3b = 0
T: −1 − c = 0
Solving we obtain
b = −1, c = −1, a = −2.
μ
The second dimensionless group is Π3 =
ρωD2
Problem 7.45
A student is interested in the aerodynamic drag on spheres. She conducts a series of wind
tunnel tests on a 10-cm- diameter sphere. The air in the wind tunnel is at 50 DC and 101.3kPa.
She presents the results of her tests in the form of a correlation,
= 0.0015V 1.92,
where is the drag force in Newton and V is the velocity in m/s. Another student realizes
that the results would have been more effectively presented in terms of dimensionless
parameters, say,
Π1 = f ( Π 2 ) = C Π a2,
where Π1 and Π 2 are dimensionless parameters and C and a are constants. What are the
most appropriate dimensionless parameters (Π1 and Π 2) and the corresponding values of C
and a ? What would be the drag on a 2-cm-diameter sphere placed in a 1-m/s, 20 DC water
stream?

Solution 7.45
Assuming Weber and Froude number effects are not relevant,
CD = f ( Re, α )
Since a sphere has no angle of attack,
CD = f ( Re )
or
D § ρVD ·
f¨ ¸
1
ρ D 2V 2 © μ ¹
2
for a sphere of diameter D . The data suggest
CD = C Rea
so Π1 = C D and Π 2 = Re .

This gives
a
§ ρVD ·
=C ¨ ¸
ρD V © μ ¹
2 2

or
a
ρ D2V 2 § ρVD · C ρ1+a ( DV )2+a
= C¨ ¸ = (2)
2 © μ ¹ 2μ a
Comparing Eqs. (1) and (2) gives 2 + a = 1.92 or a = −0.08 and
C ρ1+a D2 +a
= 0.0015 giving
2μ a
0.0030 μ a 0.0030
C= =
ρ1+a D2+a ρ 0.92 D1.92 μ 0.08
Using data for 50 °C air,

§ ·
¨ ¸
¨ ¸§ ·
¨ § · ¸ ¨© ¸
¹
¨¨ ¨ ¸ ¸¸
© © ¹ ¹
C
§ · § ·
¨ ¸ ( ) ¨
−
¸
© ¹ © ¹
C C

Then CD = 0.549 Re−0.08 .

m
For the D = 0.02 m sphere in 20 DC water with V = 1
s

§ m·
VD ¨© s ¸¹
1 (0.02 m)
Re = = = 2.0 ×10 4 ,
ν § −6 m ·
2
¨¨1.0 ×10 ¸
© s ¸¹
CD = 0.549(2.0 ×10 4 )−0.08 = 0.249,
2
§ kg · 2§ m·
¨ 998 3 ¸ ( 0.249 )( 0.02 m ) ¨1 ¸
ρCD D V
2 2
= =© m ¹ © s ¹ ,
2 § kg ⋅ m ·
2¨ ¸
© N ⋅ s2 ¹
= 0.0497 N
Problem 7.48
A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard
sea-level air, while the prototype will be operated in seawater. Determine the speed of the
prototype to ensure Reynolds number similarity.

Solution 7.48

Let ( )m and ( ) p denote the model and prototype, respectively.

Thus, Re m = Re p, or

Vm A m Vp A p 1
= , where A m Ap
νm νp 15
Hence,
§ν ·§ A p · §ν m ·
Vm = ¨ m ¸¨ ¸V p = 15 ¨¨ ¸V p
¨ν p ¸© A m ¸
© ¹ ¹ ©ν p ¹
Also,
ft 2
ν m = 1.57 ×10 −4
s
ft 2
and ν p = 1.26 × 10−5 so that
s
§ −4 ft
2 ·
¨ 1.57 × 10 ¸
Vm = 15 ¨ s ¸V p = 187V p
¨ −5 ft
2
¸
¨ 1.26 × 10 ¸
© s ¹
Thus,
ft
180
Vm s = 0.963 ft
Vp = =
187 187 s
Problem 7.51
The fluid dynamic characteristics of an airplane flying 240 mph at 10,000ft are to be
investigated with the aid of a 1: 20 scale model. If the model tests are to be performed in a
wind tunnel using standard air, what is the required air velocity in the wind tunnel? Is this a
realistic velocity?

Solution 7.51
For dynamic similarity, Reynolds number must be the same for model and prototype.
Thus,

ρ mVm A m ρV A
=
μm μ
So that
μm ρ A
Vm = V (1)
μ ρm A m
Since,
lb ⋅ s slugs
μ = 3.534 × 10 −7 2
; ρ = 1.756 × 10 −3
ft ft 3

lb ⋅ s slugs
μ m = 3.74 × 10 −7 2
; ρ m = 2.38 × 10 −3 (see the table below)
ft ft 3

A
and = 20 it follows from Eq. (1) That
Am

§ −7 lb ⋅ s · § slugs ·
¨ 3.74 ×10 2 ¸ ¨
1.756 × 10 −3 3 ¸
Vm = © ft ¹ © ft ¹ 20 240 mph = 3750 mph
( )( )
§ −7 lb ⋅ s · § −3 slugs ·
¨ 3.534 × 10 ¸ ¨ 2.38 × 10 ¸
© ft 2 ¹ © ft3 ¹

No, it is not a realistic velocity—much too high.

Problem 7.70
In low-speed external flow over a bluff object, vortices are shed from the object (as shown
in the figure below). The frequency of vortex shedding, f ≐ 1/T, depends on ρ, μ, V, and d.
Make a dimensional analysis of this phenomenon. Transform the dimensionless parameters
that you developed into “standard” parameters. What are their names? Two identically
shaped objects with a size ratio 2:1 are tested in air flow. What velocity ratio is necessary
for dynamic similarity? What is the ratio of vortex-shedding frequencies?

Solution 7.70

The dimensional parameters have been identified. Their dimensions are

f 1
T
ρ M
L3
μ F ⋅T M
=
L 2
( L ⋅T )
V L
T
d L
The number of dimensionless parameters will be n – k = 5 – 3 = 2. Note that k = 3 (the
number of fundamental dimensions) since ρ, V, and d cannot form a dimensionless group.

Select the repeating parameters as ρ, V, and d. We now combine ρ, V, and D with f to get
[ f ][ ρ ]a [V ]b [ d ]c = M 0 L0T 0 ,
·§ M · § L ·b
a
§1 c 0 0 0
¨ ¸¸ ¨ ¸ ( L ) = M L T ,
¸ ¨¨ 3
©T ¹© L ¹© T ¹
M: a=0 ½ a=0
°
L : −3a + b + c = 0 ¾ b = −1
T: − 1 − b = 0 °¿ c = 3a − b = 1
The first dimensionless parameter is
fd
Π1 = f ρ 0V −1d or Π1 =
V
Next combine ρ V, and D with μ to get
[ μ ][ ρ ]a [V ]b [ d ]c = M 0 L0T 0 ,
a b
§M ·§ M · § L · c 0 0 0
¨ ¸ ¨ 3 ¸ ¨ ¸ (L) = M L T ,
© LT ¹© L ¹ © ¹T
M: 1 + a = 0 ½ a = −1
°
L: −1 − 3a + b + c = 0 ¾ b = −1
T: − 1 − b = 0 °¿ c = 1 + 3a − b = −1
The second dimensionless parameter is
μ
Π2 = μρ −1V −1d −1 or Π2 =
ρVd
The standard dimensionless parameters are

Π1 = St (Strouhal Number)
Π 2−1 = Re (Reynolds Number)

Equality of the Reynolds number gives

§ ρVd · § ρVd ·
¨ ¸ =¨ ¸
© μ ¹1 © μ ¹2

For (assumed) equal air temperature and pressure, ρ1 = ρ2 and μ1 = μ2 so

V1 d2 V1 1 → smaller d
= =2 or = 2,
V2 d1 V2 2 → larger d

Equality of the Strouhal number gives

§ fd · § fd · f1 § d2 ·§ V1 ·
¨ ¸ =¨ ¸ , = ¨ ¸¨ ¸ ,
© V ¹1 © V ¹2 f2 © d1 ¹ © V2 ¹
f1 § 2 · f1 1 → smaller d
= ¨ ¸ (2), or = 4,
f2 © 1 ¹ f2 2 → larger d
Problem 7.71
In a nutritional products plant, a plate heat exchanger cools infant formula by
approximately 25 °C from an inlet temperature of 68 °C as it flows through the plates as
shown in the figure below. The process engineer would like to be able to cool the infant
formula by as much as 35 °C as it passes through the plates. The plate heat exchanger is
1
currently in use, so he would like to build a -scale model for testing purposes. In the
2
model, water will be used instead of infant formula. What flowrate is needed in the model?
D = 0.05 m

Infant formula
ρ = 1036.6 kg/m3
μ = 1.7 × 10−3 N·s/m2
Q = 16.8 m3/hr
at average formula
1.4 m temperature

30% glycol solution

Solution 7.71

The Reynolds number must be the same for the model (water) and the prototype (formula).
§ ρV A · § ρV A ·
¨ ¸ =¨ ¸
© μ ¹m © μ ¹ p
π
Since Q = D2V
4
ª § 4Q · A º ª § 4Q · A º
«ρ ¨ 2 ¸μ»
= «ρ ¨ 2¸ »
¬ © π D ¹ ¼m ¬ © π D ¹ μ ¼ p
or
2
§ A p · § Dm · § ρ p · § μm ·
Qm = Q p ¨ ¸¨ ¸ ¨ ¸¨ ¸.
© A m ¹ ¨© Dp ¸¹ © ρ m ¹ ¨© μ p ¸¹

Using an average temperature of 50 °C water,

kg N ⋅s
ρ m = 988 3
and μ m = 5.47 × 10 −4 . Also Dm = 0.5Dp and m = 0.5 p. Then
m m2
§ kg ·§ −4 N ⋅ s ·
§ 2 1036.6 3
m3 · § 2 · § 1 · ¨ m ¸ ¨ 5.47 × 10 m 2 ¸
Qm = ¨16.8 ¸¸ ¨ ¸ ¨ ¸ ¨ ¸¨ ¸
¨
© hr ¹ © ¹ © ¹ ¨¨ 988
1 2 kg ¸¸ ¨¨ 1.7 ×10 −3 N ⋅ s ¸¸
© m3 ¹© m2 ¹
m3
Qm = 2.84
hr
Problem 7.77
As winds blow past buildings, complex flow patterns can develop due to various factors
such as flow separation and interactions between adjacent buildings. Assume that the local
gage pressure, p, at a particular location on a building is a function of the air density, ρ; the
wind speed, V; some characteristic length, ; and all other pertinent lengths, i, needed to
characterize the geometry of the building or building complex. (a) Determine a suitable set
of dimensionless parameters that can be used to study the pressure distribution. (b) An
eight-story building that is 100 ft tall is to be modeled in a wind tunnel. If a length scale of
1:300 is to be used, how tall should the model building be? (c) How will a measured
pressure in the model be related to the corresponding prototype pressure? Assume the same
air density in model and prototype. Based on the assumed variables, does the model wind
speed have to be equal to the prototype wind speed? Explain.

Solution 7.77

(a) p = f ( ρ ,V , A, A i )
−2
p = FL ρi  FL−4T 2 V  LT −1 A  L A i  L

From the pi theorem, 5 − 3 = 2 pi terms required, and a dimensional analysis yields

p § A ·
=φ¨ ¸
ρV © Ai ¹
2

(b) For geometric similarity

Am A
=
A im A i
so that
A m A im
=
A Ai
Am
and it follows that all pertinent lengths are scaled with the length scale . Thus, with
A
Am 1 100 ft
= model height = = 0.333 ft.
A 300 300
(c) With geometric similarity satisfied it follows that
p pm
=
ρV 2
ρ mVm 2
Thus, with ρm = ρ
2
§V ·
p=¨ ¸ pm
© Vm ¹
Vm
, so the
With the set of given variables there is no requirement for the velocity scale,
V
model wind speed does not have to be equal to the prototype wind speed. No.
Problem 7.82
A breakwater is a wall built around a harbor so the incoming waves dissipate their energy
against it. The significant dimensionless parameters are the Froude number F and the
Reynolds number Re. A particular breakwater measuring 450 m long and 20 m deep is hit
1
by waves 5 m high and velocities up to 30 m/s. In a -scale model of the breakwater, the
100
wave height and velocity can be controlled. Can complete similarity be obtained using
water for the model test?

Solution 7.82

The Reynolds and Froude number requirements are

Vm A m Vp A p Vm Vp
= and =
vm vp gA m gA p
Assuming that vm = vp, the Reynolds number requirement is
Vm A p 1
= =
Vp A m S
while the Froude number requirement is
Vm Am
= = S
Vp Ap
Obviously, these cannot both be true unless S = 1.

It is not possible to match both

Re and Fr in a scale model test.