Chapter 4.

Inviscid flow (differential approach)

Euler’s equation of inviscid motion

• Inviscid (shear free) flow

⇒ Pascal’s principle applies!

• Mass conservation: Recall that

D
(ρ δVm) = 0
Dt
Dρ
⇒ + ρ∇ · V = 0.
Dt

* Alternative form:
∂ρ
+ ∇ · (ρV) = 0.
∂t

* This is the continuity equation.

Fluid Mechanics, Prof. T.-S. Yang 4-1

 • Newton’s law:
D
(ρ δVmV) = −∇p δV + |ρG{z
{z m}
δVm} .
Dt | {z } |
(1) (2) (3)
1) Momentum of fluid particle with volume
δVm; note that since ρ δVm = const. (mass),
D DV
(ρ δVmV) = ρ δVm .
Dt Dt
2) Net surface (pressure) force
3) Body force; G = body force per unit mass

• We then obtain the Euler’s equation for

inviscid flow:
DV
ρ = −∇p + ρG.
Dt

§ e.g., x-component in Cartesian coordinates:

!
∂u ∂u ∂u ∂u ∂p
ρ +u +v +w =− + ρGx.
∂t ∂x ∂y ∂z ∂x
* The underlined are the nonlinear terms.
* Can you write out the other components?

Fluid Mechanics, Prof. T.-S. Yang 4-2

 § Example: Solid-body rotation
DV
V = rω eθ ⇒ = −rω 2er .
Dt
* So, if Gr = 0, we will have ∂p/∂r = rω 2, as
derived before.

Velocity boundary conditions at solid walls

1) The normal component of fluid velocity

relative to a impermeable solid boundary
Vn = 0.
2) But allow for an arbitrary slip velocity Vt
(artificial; due to the absence of viscosity).

Fluid Mechanics, Prof. T.-S. Yang 4-3

 • Example: Steady uni-directional flow

* Suppose that |dA/dx| ≪ 1, so that the flow

mainly is in the x-direction:

V ≈ u(x) i.
† Cross-sectional variation of u is neglected;
appropriate under certain conditions.
† May think of u(x) as area-averaged velocity
for now.

* Suppose also that body force is negligible,

G = 0.
* The equation of motion then reads
!
∂u ∂p ∂ u2 ∂p
ρu =− or ρ =− .
∂x ∂x ∂x 2 ∂x

Fluid Mechanics, Prof. T.-S. Yang 4-4

 * If ρ = constant, we then have
1 2
p + ρu = const. with x.
2

§ This is a special case of the Bernoulli’s

equation.
§ We will derive a more general form of the
Bernoulli’s equation shortly.
• Example:

* Assume that ρ = const., find p(x, t).

* Mass conservation: u(x, t) = up(t)
† Independent of x indeed.
* Equation of motion:
dup ∂p dup
ρ =− ⇒ p = −ρ · x + const.
dt }
| {z ∂x dt
fn(t)

Fluid Mechanics, Prof. T.-S. Yang 4-5

 * With the boundary condition p = pe at x =
xe for ∀t, we find
dup
p − pe = ρ (xe − x).
dt
† Shown schematically below (left) is the case
with dup/dt > 0.

§ Alternative viewpoint:
* Consider the CV shown below (right).
* Apply ‘F = ma’ on the CV gives the same
result.

Fluid Mechanics, Prof. T.-S. Yang 4-6

Euler’s equation for steady flow expressed in
streamline coordinates

• Assume that:
1) Steady flow
2) ρ = const.
3) G = −g k = −∇(gz)
† Gravity points in the −z direction.
• Recall that
DV ∂V V2
a = = (V · ∇)V = V es − en .
Dt | ∂s
{z } | R{z }
(1) (2)

Fluid Mechanics, Prof. T.-S. Yang 4-7

1) Convective accel. in the streamline dir.
2) Centripetal accel. normal to streamline
* Also,
∂ ∂ ∂
∇ = es + en + el .
∂s ∂n ∂l

• Euler equation: ρ(V · ∇)V = −∇(p + ρgz)

1) s-direction:
 
∂ 1 2
p + ρV + ρgz = 0.
∂s 2
* This is the Bernoulli’s equation; F = ma
along a streamline.
2) n-direction:
∂ ρV 2
(p + ρgz ) = .
∂n R
* F = ma normal to a streamline.
3) l-direction:
∂
(p + ρgz ) = 0.
∂l
* Hydrostatic pressure distribution in the bi-
normal direction.

Fluid Mechanics, Prof. T.-S. Yang 4-8

 * Recall that to derive the above three eqns,
we have made three assumptions:
(1) steady flow;
(2) ρ = const.;
(3) G = −g k = −∇(gz).
* And the Bernoulli’s equation says that along
a streamline p + ρV 2/2 + ρgz = constant.

• Physical interpretation: energy viewpoint

† Read also §3.4, Munson et al.

* Consider the ‘stream tube’ (or ‘stream

filament’ when its cross-sectional area is
infinitesimal) shown above.

Fluid Mechanics, Prof. T.-S. Yang 4-9

 * A section of the stream tube, cut by
infinitesimal A1,2 (that are separated by an
infinitesimal distance δs) is our CV (of fixed
volume in steady flow).

* The mass flowrate through A1,2 is constant

due to mass conservation:

δ ṁ = ρA1V1 = ρA2V2 = const.

† V1,2 essentially are constant over A1,2.

* Flow work: work done by pressure force on

fluid crossing CV boundary.
§ On A1, rate of flow work is

p1A1V1 = δ ṁ · (p1/ρ).
§ Similarly, rate of flow work on A2 is
( ! )
p p ∂ p
−δ ṁ· 2 = −δ ṁ 1 + δs + H.O.T. .
ρ ρ ∂s ρ

Fluid Mechanics, Prof. T.-S. Yang 4-10

 * Rate of mechanical energy flow:
§ At A1, unit fluid mass has a KE of V12/2,
and a PE of gz1. So mechanical energy
enters our CV there at the rate of
!
V12
δ ṁ + gz1 .
2
§ Similarly, rate of mechanical energy exiting
our CV at A2 is
( ! )
V12 ∂ V2
δ ṁ + gz1 + + gz δs + H.O.T. .
2 ∂s 2

* Balance of mechanical energy in steady flow

implies that
! !
∂ V2 ∂ p
+ gz =− ,
∂s 2 ∂s ρ
or
 
∂ 1 2
p + ρV + ρgz = 0.
∂s 2

* So, the Bernoulli’s equation is a statement

of the conservation of mechanical energy.

Fluid Mechanics, Prof. T.-S. Yang 4-11

 § But where did thermal energy go?
* Including thermal energy (i.e., internal
energy; e per unit mass) in energy balance
yields
!
∂ p 1 2
δ ṁ · + V + gz + e δs = δ Q̇,
∂s ρ 2
δ Q̇ being the rate of heat transfer to CV.
* Now, since the fluid is inviscid, the process
is ‘frictionless’ and hence reversible.
* So, δ Q̇ = T δ S˙e , where δ S˙e is the rate of
entropy increase of the CV, and T the
temperature at which heat is transferred.
* But δ S˙e = δ ṁ(se2 − se1) = δ ṁ(∂se/∂s)δs,
where se is the specific entropy (entropy
per unit mass).
* Moreover, by the Gibb’s equation, T dse =
de + p d(1/ρ) = de (since here ρ = const.),
we have
∂se ∂e ∂e
T = ⇒ δ Q̇ = δ ṁ ds.
∂s ∂s ∂s
* So, the thermal energy and heat transfer
terms in the total energy balance cancel
out, resulting in the Bernoulli’s equation.

Fluid Mechanics, Prof. T.-S. Yang 4-12

Pressure–velocity relation in streamline dir.
• Along a streamline (in a steady flow),

p + ρV 2/2 + ρgz = constant.

• Example:

* Bernoulli: p1 + ρV12/2 = p2 + ρV22/2

* Mass: V1A1 = V2A2
* Hence
! !
1 2 2
A1 1 2 A2
p2 = p1 − ρV1 − 1 = p1− ρV2 1− 2 .
2 A2
2 2 A2
1

* As A ↓, V ↑ (because of incompressibility),
and so p ↓.

Fluid Mechanics, Prof. T.-S. Yang 4-13

 § Suppose that at the exit p2 = p0 (ambient
pressure).
* Then at an internal station where the cross-
sectional area is A(x), the pressure there is
related to p0 by
!
1 2 A22
p(x) − p0 = ρV2 1− 2 .
2 A
* The horizontal component of the resultant
force on the wall is calculated as follows:
Z A
2
FH = [p(x) − p0](−dA)
A1
!
A2
Z A
1 2
= − ρV22 1− 2 dA
2 A1 A2
( !)
1 2 2 1 1
= ρV2 (A1 − A2) − A2 −
2 A2 A1
ρV22
= (A1 − A2)2 > 0.
2A1
† Explain −dA in the first line.

* Note that FH > 0 always, no matter which

of A1,2 is greater. Explain.

Fluid Mechanics, Prof. T.-S. Yang 4-14

 • Example:

* Due to suction, V2 < V1, and so p2 > p1.

• Example:

* But here the Bernoulli’s equation does not

apply!
§ Turbulent viscous mixing is important in
the channel.
§ Hence the assumption of inviscid flow is not
justified.
* Can be analyzed by use of CV theorems
(next chapter).

Fluid Mechanics, Prof. T.-S. Yang 4-15

Relation between pressure variation normal to
streamlines and streamline curvature
• Recall that
∂ ρV 2
(p + ρgz) = .
∂n |R
{z }
>0
† n points away from center of curvature.
† R: radius of curvature

• So, if gravity is neglected, we have

∂p
> 0.
∂n

• Example:

* p∞, V∞: free-stream pressure and velocity,

respectively.

Fluid Mechanics, Prof. T.-S. Yang 4-16

 • Example:
* The cause of lift is best explained by
curvature.
§ The top surface of an airfoil has a greater
curvature than that of its bottom surface.
§ Hence the pressure variation in the fluid
above the airfoil is greater than that
under the airfoil.
§ But the free-stream pressure p∞ is the same
for both.
§ The pressure on the top surface therefore
is lower than that on the bottom surface.
† Both lower than p∞.
§ A lift then is produced.

Fluid Mechanics, Prof. T.-S. Yang 4-17

 † Video 3.1: Balancing ball

• Stagnation pressure:

* Suppose that the free-stream velocity and

pressure are V∞ and p∞, respectively.
* At the ‘stagnation point’ V = 0; hence the
‘stagnation pressure’ there is
2
p ≈ p∞ + ρV∞ /2 > p∞.

Fluid Mechanics, Prof. T.-S. Yang 4-18

 * So there is an upward force to support the
weight of the ball.
† Due to streamwise pressure variation

§ The video clip shows that the equilibrium

is stable.
* To see why, suppose that the ball is
displaced to the left.
* On its right side, the air speed then would
increase, resulting in decreased pressure.
* The net pressure force on the ball therefore
is to the right, and pushes the ball back to
its equilibrium position.

Fluid Mechanics, Prof. T.-S. Yang 4-19

 • Coanda effect:

* Note that the jet is bent around the upper

surface of the ball.
* Streamlines therefore are curved.
* And the pressure on the upper surface of
the ball is lower than the atmospheric
pressure at the outside boundary.
* This produces a net pressure force (the
‘Coanda force’) acting on the ball toward
the jet.

† Video 3.2: Free vortex

Fluid Mechanics, Prof. T.-S. Yang 4-20

 • Fluids in the news: Incorrect raindrop shape
* The incorrect representation that raindrops
are teardrop shaped is found nearly every-
where — from children’s books, to weather
maps on the Weather Channel.
* About the only time raindrops possess the
typical teardrop shape is when they run
down a windowpane.
* The actual shape of a falling raindrop is a
function of the size of the drop and
results from a balance between surface
tension forces and the air pressure exerted
on the falling drop.

Fluid Mechanics, Prof. T.-S. Yang 4-21

 § Small drops with a radius less than about
0.5 mm are spherical shaped.
* Because the surface tension effect (which is
inversely proportional to drop size R) wins
over the increased pressure, ρV∞2 /2, caused
by the motion of the drop and exerted on
its bottom.
§ With increasing size, the drops fall faster
and the increased pressure causes the drops
to flatten.
† Drop weight ∝ R3; drag ∝ R2.
* A 2-mm drop, for example, is flattened into
a hamburger bun shape.
* Slightly larger drops are actually concave
on the bottom.
§ When the radius is greater than about
4 mm, the depression of the bottom
increases and the drop takes on the form
of an inverted bag with an annular ring of
water around its base.
* This ring finally breaks up into smaller
drops.
† See Problem 3.28.

Fluid Mechanics, Prof. T.-S. Yang 4-22

 • Fluids in the news:
Armed with a water jet for hunting

* Archerfish, known for their ability to shoot

down insects resting on foliage, are like sub-
marine water pistols.
* With their snout sticking out of the water,
they eject a high-speed water jet at their
prey, knocking it onto the water surface
where they snare it for meal.
* The barrel of their water pistol is formed
by placing their tongue against a groove in
the roof of their mouth to form a tube.

Fluid Mechanics, Prof. T.-S. Yang 4-23

 * By snapping shut their gills, water is forced
through the tube and directed with the tip
of their tongue.
* The archerfish can produce a pressure head
within their gills large enough so that the
jet can reach 2 to 3 m.
* However, it is accurate to only about 1 m.
* Recent research has shown that archerfish
are very adept at calculating where their
prey will fall.
* Within 100 ms (a reaction twice as fats
as a human’s), the fish has extracted all
the information needed to predict the point
where the prey will hit the water.
* Without further visual cues it charges
directly to that point.

† See Problem 3.41.

Fluid Mechanics, Prof. T.-S. Yang 4-24

 • Fluids in the news: Pressurized eyes

* Our eyes need a certain amount of internal

pressure in order to work properly, with the
normal range being between 10–20 mm Hg.
* The pressure is determined by a balance
between the fluid entering and leaving the
eye.
* If the pressure is above the normal level,
damage may occur to the optic nerve,
leading to a loss of visual field termed
glaucoma.

Fluid Mechanics, Prof. T.-S. Yang 4-25

 * Measurement of the pressure within the eye
can be done by several different noninvasive
types of instruments, all of which measure
the slight deformation of the eyeball when
a force is put on it.
* Some methods use a physical probe that
makes contact with the front of the eye,
applies a known force, and measures the
deformation.
* One noncontact method uses a calibrated
‘puff’ of air that is blown against the eye.
* The stagnation pressure resulting from the
air blowing against the eyeball causes a
slight deformation, the magnitude of which
is correlated with the pressure within the
eyeball.
† See Problem 3.29.

Fluid Mechanics, Prof. T.-S. Yang 4-26

Vena contracta effect

• Radius of curvature R = 0 at sharp corner.

• An infinite (and thus impossible!) pressure
gradient across the streamlines is required
to turn the sharp corner.
• So the fluid jet diameter is reduced, with
contraction coefficient
Cc = Aj /Ah = (dj /dh)2.

Fluid Mechanics, Prof. T.-S. Yang 4-27

 • Fluids in the news:
Cotton candy, glass wool, and steel wool
* Although cotton candy and glass wool
insulation are made of entirely different
materials and have entirely different uses,
they are made by similar processes.
* Cotton candy, invented 1897, consists of
sugar fibers.
* Glass wool, invented in 1938, consists of
glass fibers.
* In a cotton candy machine, sugar is melted
and then forced by centrifugal action to
flow through numerous tiny orifices in a
spinning ‘bowl’.

Fluid Mechanics, Prof. T.-S. Yang 4-28

 * Upon emerging, the thin streams of liquid
sugar cool very quickly and become solid
threads that are collected on a stick or
cone.
* Making glass wool insulation is somewhat
more complex, but the basic process is
similar.

Fluid Mechanics, Prof. T.-S. Yang 4-29

 * Liquid glass is forced through tiny orifices
and emerges as very fine glass streams that
quickly solidify.
* The resulting intertwined flexible fibers,
glass wool, form an effective insulation
material because the tiny air ‘cavities’
between the fibers inhibit air motion.
* Although steel wool looks similar to
cotton candy or glass wool, it is made by
an entirely different process.
* Solid steel wires are drawn over special
cutting blades which have grooves cut into
them so that long, thin threads of steel are
peeled off to form the matted steel wool.

Fluid Mechanics, Prof. T.-S. Yang 4-30

Boundary conditions for pressure
1) Exit of jet:

• Near the exit of a jet, the streamlines must

be straight.
† If the streamlines were curved as shown
below:

† Then p < pa as implied by the Bernoulli’s

equation, since V ↓.
† But p > pa because of the streamline
curvature.
† A contradiction!

Fluid Mechanics, Prof. T.-S. Yang 4-31

2) Static pressure tap:

• Across straight streamlines,

p2 ≈ p1 + ρg(z2 − z1).
* As if the fluid were static.
* The static or thermodynamic pressure p1
therefore is measured by a smooth hole on
the wall with no burrs or imperfections.

§ Incorrect and correct designs of static

pressure taps:

Fluid Mechanics, Prof. T.-S. Yang 4-32

Static, stagnation, dynamic, and total pressure
• Bernoulli’s equation: Along a streamline
2
p
|{z}
+ ρV /2 + ρgz
|{z}
= constant.
| {z }
(1) (2) (3)

1) Static pressure (as was just discussed; ∼

thermodynamic pressure)
2) Dynamic pressure
3) Hydrostatic pressure

• Total pressure:

pT = p + ρV 2/2 + ρgz
* The Bernoulli’s equation states pT = const.
along a streamline (under certain precise
conditions).

• Stagnation pressure:
— pressure at stagnation point

Fluid Mechanics, Prof. T.-S. Yang 4-33

 § Example:

* The tube in the sketch above produces a

stagnation point (2), where V2 = 0 and

p2 = p1 + ρV12/2 > p1.

* p2 is the stagnation pressure.
* And is the sum of the static and dynamic
pressures at the upstream point (1).

† Video 3.3: Stagnation point flow

Fluid Mechanics, Prof. T.-S. Yang 4-34

Pitot-static tube

• Neglect hydrostatic pressure variation.

• Then, one can measure the stagnation
pressure

p3 = p2 = p + ρV 2/2.
• And the static pressure

p4 = p1 = p.
• The fluid velocity then can be calculated:
q
2
p3 − p4 = ρV /2 ⇒ V = 2(p3 − p4)/ρ.

Fluid Mechanics, Prof. T.-S. Yang 4-35

 • Some practical considerations:
1) The pressure along the surface of an object
varies from the stagnation pressure at its
stagnation point to values that may be less
than the free-stream static pressure.
* A typical pressure variation for a Pitot-
static tube is shown below.
* Clearly, it is important that the pressure
taps be properly located to ensure that the
pressure measured is actually the static
pressure.

Fluid Mechanics, Prof. T.-S. Yang 4-36

2) In practice, it is often difficult to align the
Pitot-static tube directly into the flow
direction. Any misalignment will produce
an asymmetrical flow field that may
introduce errors.
* Typically, yaw angles up to 12 to 20◦
(depending on the particular probe design)
give results that are less than 1% in error
from the perfectly aligned results.
* Generally, it is more difficult to measure
static pressure than stagnation pressure.
* A direction-finding Pitot tube is illustrated
below. Three pressure taps are drilled into
a small cylinder, fitted with small tubes,
and connected to 3 pressure transducers.
* The two side holes are located at β = 29.5◦
so that they measure the static pressure.

Fluid Mechanics, Prof. T.-S. Yang 4-37

 † Video 3.4: airspeed indicator

Fluid Mechanics, Prof. T.-S. Yang 4-38

 • Fluids in the news:
Bugged and plugged Pitot tubes

* Although a Pitot tube is a simple device for

measuring aircraft airspeed, many airplane
accidents have been caused by inaccurate
Pitot tube readings.

* Most of these accidents are the results of

having one or more of the holes blocked
and, therefore, not indicating the correct
pressure (speed).

* Usually this is discovered during takeoff

when time to resolve the issue is short.

* The two most common causes for such

a blockage are either the pilot (or ground
crew) has forgotten to remove the
protective Pitot tube cover, or that insects
have built their nest within the tube where
the standard visual check cannot detect it.

Fluid Mechanics, Prof. T.-S. Yang 4-39

 * The most serious accident (in terms of
number of fatalities) caused by a blocked
Pitot tube involved a Boeing 757 and
occurred shortly after takeoff from Puerto
Plata in the Dominican Republic.

* The incorrect airspeed data was auto-

matically fed to the computer causing the
autopilot to change the angle of attack and
the engine power.

* The flight crew became confused by the

false indications, the airplane stalled, and
then plunged into the Caribbean Sea killing
all aboard.

† See Problem 3.30.

Fluid Mechanics, Prof. T.-S. Yang 4-40

Flowrate measurement

• Bernoulli: p1 + ρV12/2 = p2 + ρV22/2

• Mass conservation: Q = A1V1 = A2V2
• Then obtain p1−p2 = (ρV22/2)·[1−(A2/A1)2],
q
and hence V2 = 2(p1 − p2)/ρ[1 − (A2/A1)2];
v
u
u 2(p1 − p2)
⇒ Q = A2 t .
ρ[1 − (A2/A1)2]

Fluid Mechanics, Prof. T.-S. Yang 4-41

 • But
Qactual = C · Q,
where the geometry-dependent correction
factor C < 1.

† Video 3.6: Venturi channel

• Fluids in the news: Hi-tech inhaler

* The term inhaler often brings to mind a
treatment for asthma or bronchitis.
* Work is underway to develop a family of
inhalation devices that can do more than
treat respiratory ailments.
* They will be able to deliver medication for
diabetes and other conditions by spraying
it to reach the bloodstream through the
lungs.

Fluid Mechanics, Prof. T.-S. Yang 4-42

 * The concept is to make the spray droplets
fine enough to penetrate to the lungs’ tiny
sacs, the alveoli, where exchanges between
blood and the outside world take place.
* This is accomplished by use of a laser-
machined nozzle containing an array of very
fine holes that cause the liquid to divide
into a mist of µm-scale droplets.
* The device fits the hand and accepts a
disposable strip that contains the medicine
solution sealed inside a blister of laminated
plastic and the nozzle.
* An electrically actuated piston drives the
liquid from its reservoir through the nozzle
array and into the respiratory system.
* To take the medicine, the patient breathes
through the device and a differential
pressure transducer in the inhaler senses
when the patient’s breathing has reached
the best condition for receiving the
medication.
* At that point, the piston is automatically
triggered.

Fluid Mechanics, Prof. T.-S. Yang 4-43

Examples of use of the Bernoulli equation
— Read also §3.6, Munson et al.

• Example:

* Top: Higher pressure near stagnation point

lifts the sheet of paper.
* Bottom: Once lifted, the pressure in the
accelerated flow decreases, and becomes
insufficient to support the weight of the
sheet.
* This makes the sheet flutter!

Fluid Mechanics, Prof. T.-S. Yang 4-44

 • Example: Nozzle

* Suppose that Ar ≫ Ae.

* Hence V2 = Ve ≫ V1.
* Also neglect gravity.
* Bernoulli’s equation then gives

0 = p + ρVe2/2.
p0 + |{z}
†

† Negligible compared with ρVe2/2.

* So we obtain
q
Ve = 2(p0 − p)/ρ.

† Read Example 3.8.

Fluid Mechanics, Prof. T.-S. Yang 4-45

 • Example: Horizontal flow from a tank

* Assume that the half-width (or radius) of

the exit channel is much smaller than the
radius of curvature of the jet: a ≪ R.
* Near the exit, in the water stream,
∂ ρVe2
(p + ρgz) ≈ ≈ const.
∂n R
* We then integrate across the water stream,
and find
ρVe2
(p + ρgz)2 − (p + ρgz)1 ≈ · 2a.
R
* But p1 = pa = p2, and z2 − z1 ≈ 2a; hence

R ≈ Ve2/g.

Fluid Mechanics, Prof. T.-S. Yang 4-46

 * Note that R ↑ as Ve ↑.

* Also, R → ∞ as g → 0.
† Video 3.5: Flow from a tank
† Examples with gravity will be discussed later.

• Example: Blow off the card?

Fluid Mechanics, Prof. T.-S. Yang 4-47

 * Mass conservation: 2πrV (r) = 2πR2Ve
* So V (r) > Ve, and hence p(r) < pe = pa.
* There therefore is a net upward pressure
force on the bottom sheet of paper.
2 = V (r) ·
* Volumetric flowrate: Q = V↓ · πR1
2πrh; hence
V↓ r h
=2· · ≪1
V (r) R1 R1
provided that r (or R2) is not too large.

Fluid Mechanics, Prof. T.-S. Yang 4-48

 * Now apply the Bernoulli’s equation:
p(r) + ρ[V (r)]2/2 = pa + ∆p = pa + ρVe2/2.
* So we find that, for 0 ≤ r ≤ R1,
!2
1 2 1 Q
ρVe = ρ
p − pa = ∆p =
2 2 2πR2h
and that, for R1 ≤ r ≤ R2,
" 2 #
R2
p − pa = −∆p −1 .
r

* Can then calculate the net upward force:

Z R
2
Fup = (pa − p) · 2πr dr
0
( )
2 2 R2 1 2 2
= −∆p · πR1 + 2π∆p R2 ln − (R2 −R1 )
R1 2
!
2 R2
= ∆p · πR2 2 ln −1 .
R1

Fluid Mechanics, Prof. T.-S. Yang 4-49

With gravity
• Example:

* Assume that the exit pipe radius a ≪ h

(h being the water depth), so that it is
appropriate to assume uniform Ve.
* Bernoulli:
pa + ρVe2/2 + 0 = pa + 0 + ρgh
q
⇒ Ve2 = 2gh or Ve = 2gh.
* Pressure variation along the streamline:

† Acquired PE ⇒ KE
† Smooth and gradual velocity and pressure
variations in reality

Fluid Mechanics, Prof. T.-S. Yang 4-50

 • Example: Sink

√
* The exit velocity in case 2, V2 = 2gh2, is
√
smaller than that in case 1, V1 = 2gh1.
* Hence the volumetric flowrates Q2 < Q1.
* Also, in case 2, the stream accelerates, and
so narrows down in order to conserve mass.
* Pressure variations along streamlines:

† In case 2, the tendency of pressure increase

due to gravity is balanced by the tendency
of pressure decrease due to acceleration.

Fluid Mechanics, Prof. T.-S. Yang 4-51

 • Example: Chimney

* 3Ts of combustion: (retention) time,

(treatment) temperature, and turbulence.
* The flow generally is not inviscid in an
incinerator, and so Bernoulli’s equation
does not apply there.
* But we still can use it elsewhere.
* Assume that after combustion the pressure
remains the same, p = pa, but the gas (air)
density decreases, ρH < ρa.
* Consider a streamline connecting the base
and exit of the chimney. Bernoulli’s eqn
then gives

pa + 0 + 0 = pe + ρH Ve2/2 + ρH gh.

Fluid Mechanics, Prof. T.-S. Yang 4-52

 † The fluid velocity at the base is much
smaller than that in the chimney.

* The exit pressure is determined by hydro-

statics, giving

pe = pa − ρagh.
† The air pressure variation with height
outside the chimney no longer is negligible,
especially because ρH < ρa.

* Hence we find

ρH Ve2/2 = (ρa − ρH )gh

or
q
Ve = 2(ρa − ρH )gh/ρH .

† Buoyancy accelerates the fluid.

Fluid Mechanics, Prof. T.-S. Yang 4-53

 • Example: Hovercraft

* The streamlines near the exit have a radius

of curvature r0 ≈ h, and span a width of w.
* Using the force law across streamline,
∂p/∂n = ρVe2/r0 ≈ const.,
we then have
ρVe2 w ρVe2w
Z p Z
b
dp = dn ⇒ pb − pa = .
pa r0 0 r0
* Bernoulli’s equation also gives
∆p = ρVe2/2.
† Given ∆p, one can determine Ve, and then
pb.
* Lift:
Z R
L = (pb − pa ) 2πr dr = (pb − pa) · πR2
0
= 2∆p · πR2 · (w/r0).

Fluid Mechanics, Prof. T.-S. Yang 4-54

 • Example: Channel bend

* Simplest case: w ≪ R
* Along a streamline, the total pressure pT
remains constant.
* Consider an arbitrary point at the entrance
of the bend, with z = z0, p = p0, and

pT = p0 + ρV02/2 + ρgz0.
* Suppose the uniform water depth there is
h0. Then p0 = pa + ρg(h0 − z0), and so

pT = pa + ρV02/2 + ρgh0.

* Note that pT is independent of z0, and

therefore is the same for all streamlines.

Fluid Mechanics, Prof. T.-S. Yang 4-55

 * Consider now a certain section in the bend,
with which two streamlines intersect at
points (1) and (2) on the bottom.
* Since both z1,2 = 0, we have V22 − V12 =
2(p1 − p2)/ρ = 2g(h1 − h2).
* If h2 − h1 is ‘small’ (say, ≪ V02/g; to be
justified later), then V2 − V1 ≪ V0.
* The velocity in that section therefore
approximately is uniform, and is equal to
V0 (to conserve mass).
* Across streamlines on the bottom, we find
ρV02
p2 − p1 = ρg(h2 − h1) ≈ · w.
R
* Hence, for w ≪ R,
V02 w V02
h2 − h1 ≈ · ≪ .
g R g
† The previous assumption then is justified.
* Moreover, since the fluid velocity ≈ V0, we
also have h1 + h2 = 2h0.
† One may proceed to calculate h1,2.
§ Read also the examples in §3.6.

Fluid Mechanics, Prof. T.-S. Yang 4-56

Bernoulli’s integral for some steady isentropic
(compressible) flows
* Inviscid (frictionless) ∼ reversible
* Negligible heat transfer ∼ adiabatic
⇒ Isentropic
I. Steady isentropic flow of a perfect gas:

* Here we shall neglect gravity (which can be

easily included though; see §3.8.1, Munson
et al.).
• Steady: ∂(·)/∂t = 0
• Mass conservation:
∂
∇·V =0 ⇒ (ρV ) = 0.
∂s

Fluid Mechanics, Prof. T.-S. Yang 4-57

 • F = ma: (s-direction)
∂V 1 ∂p
V =− ,
∂s ρ ∂s
or, assuming barotropic fluid, ρ = ρ(p),
!
V2
Z p
∂ 1 ∂p ∂ dp
=− =− .
∂s 2 ρ ∂s ∂s ρ
* Consider two points, 1 & 2, on a stream-
line. We then have the Bernoulli’s equation
V22 V12 (2) dp
Z
− =− .
2 2 (1) ρ

• Ideal gas: p = ρRT

γ
• Isentropic process: p/ργ = p0/ρ0 = const.
* γ = cp/cv ; p0 and ρ0 correspond to a certain
reference state.
• Now let point 1 be in that reference state,
and point 2 be any point on the same
streamline as point 1 (or 0). Then,
Z (2)
dp
V22 − V02 = −2 .
(0) ρ

Fluid Mechanics, Prof. T.-S. Yang 4-58

 * Evaluate the integral on the RHS:
Z (2) Z (2) !1/γ
dp 1 p0
= dp
(0) ρ (0) ρ0 p
 !(γ−1)/γ 
p0 γ  p2
= · − 1 .
ρ0 γ − 1 p0
* Dropping the subscript ‘2’, we then have
 !(γ−1)/γ 
2 2γ p0  p
V − V02 = · 1− .
γ − 1 ρ0 p0

• Suppose that V0 = 0. Then state 0 is

obtained when a flow is decelerated
isentropically from V to rest, and is referred
to as the stagnation condition.
* The state variables are related by
 !(γ−1)/γ 
2γ p p
V2 = · 0 1 − .
γ − 1 ρ0 p0

† Given a particular state with fluid velocity

γ
V , one has p, ρ, and hence p0/ρ0 = p/ργ .
Together with the above expression, p0 and
ρ0 can then be determined.

Fluid Mechanics, Prof. T.-S. Yang 4-59

 • Example:

* The fluid velocity in the pressurized tank

V ≈ 0.
* Given p0 and ρ0, the exit flow velocity can
then be determined from
 !(γ−1)/γ 
2 2γ p0  p
V = · 1− .
γ − 1 ρ0 p0
* Limiting cases:
1) p0 = p + ∆p, with ∆p ≪ p0, p: Then
!(γ−1)/γ !(γ−1)/γ
p ∆p
= 1−
p0 p0
γ − 1 ∆p
= 1− · + H.O.T.,
γ p0
and hence V 2 ≈ 2∆p/ρ0.

Fluid Mechanics, Prof. T.-S. Yang 4-60

 § This is the same as the result from the
Bernoulli’s equation for constant ρ.
§ So, this is the incompressible flow limit.
§ Physically, small ∆p produces small density
variation in the flow, and so the flow is
essentially incompressible.

2) p/p0 → 0: Then
2γ p 2
V2 → · 0= · γRT 0} .
γ − 1 ρ0 γ−1 | {z
2 =c0
√
§ c0 = γRT0 is the speed of sound at the
temperature T0.
§ So the maximum achievable exit flow
velocity is
s
2
Vmax = c0.
γ−1
§ For air, γ ≈ 1.4, and so Vmax > c0.

* But it is not that simple to accelerate gas

to a supersonic speed.

Fluid Mechanics, Prof. T.-S. Yang 4-61

 • Example: 1-D isentropic compressible flow
* Steady flow; consider variation in cross-
sectional area: A = A(x).
1) Mass conservation: d(ρV A)/dx = 0, or
simply
dρ dV dA
d(ρV A) = 0 ⇒ + + = 0.
ρ V A
2) Momentum: ρV (dV /dx) = −(dp/dx), or
simply dp + ρV dV = 0.
* Local sound speed c is given by c2 = (∂p/∂ρ)S .
* So, in an isentropic process,
dp ρV dρ V
dρ = 2 = − 2 dV ⇒ = − 2 dV.
c c
| {z } ρ c
using (2)
* The continuity equation (1) then gives
!
V2 dV dA
1− 2 + = 0;
c V A
dV dA
⇒ (1 − M 2)
+ = 0,
V A
where M = V /c is the local Mach number.

Fluid Mechanics, Prof. T.-S. Yang 4-62

 * It transpires that
dV 1 dA
= 2 · .
V M −1 A

0) If M 2 ≪ 1, then dV /V = −dA/A; the

result for incompressible flow is recovered.
1) M < 1: dA < 0 ⇒ dV > 0
2) M > 1: dA > 0 ⇒ dV > 0

* So, to accelerate gas flow to a supersonic

speed, one needs a convergent–divergent
nozzle.

Fluid Mechanics, Prof. T.-S. Yang 4-63

II. Liquid with constant compressibility:
* Assume isentropic flow as before.
• Euler equation: (for inviscid flow)
DV
ρ = −∇p + ρG
Dt
* In streamline coordinates (steady flow)
!
∂ V2 ∂p
ρ =− .
∂s 2 ∂s
• For a ‘simple compressible substance’ ρ =
ρ(p, se) in general.
* Assuming se = constant (isentropic flow)
as before, we then have simply ρ =
ρ(p) (barotropic fluid), which allows for
integrating the Euler equation to give
V22 V12 (2) dp Z
− =−
2 2 (1) ρ
along a streamline.
• Consider now a liquid with constant
isentropic compressibility
!
1 ∂ρ
β= .
ρ ∂p S

Fluid Mechanics, Prof. T.-S. Yang 4-64

 * We can then write dρ/ρ = βdp ⇒ ρ/ρ0 =
exp[β(p−p0)], where the subscript ‘0’ refers
to a certain reference state on the same
streamline.
* Accordingly,
Z Z
dρ 1 1 −β(p−p0)
= e−β(p−p0)dp = − e .
ρ ρ0 ρ0 β
• So we have
V 2 V02 1 n −β(p−p0) o
− = e −1 .
2 2 ρ0 β

• Take V0 = 0 (if there is a stagnation point

on the streamline); then
2 n −β(p−p0) o
V2 = e −1 .
ρ0 β
† p0 is the stagnation pressure.
• Discussion:
1) p0 = p + ∆p with ∆p ≪ p0, p: Then V 2 ≈
2∆p/ρ0 in this incompressible flow limit.
2) p/p0 → 0: Then V 2 → (2/ρ0β) · (eβp0 − 1).

Fluid Mechanics, Prof. T.-S. Yang 4-65

Criteria for ‘incompressible’ flow
• When the pressure changes isentropically
by ∆p, the corresponding density change is
!
∂ρ
∆ρ = ∆p = ∆p/c2.
∂p S
* Recall that the sound speed c is given by
c2 = (∂p/∂ρ)S .
• Bernoulli’s equation suggests that ∆p ∼
ρV 2. So,
V2
∆ρ ∼ ρ 2 .
c
• The flow essentially is incompressible if
∆ρ/ρ ≪ 1, or if the Mach number

M = V /c ≪ 1.
* As a rule of thumb, compressibility effects
may be negligible in practice if M < 0.3.

• But this criterion is sufficient only in steady

flow.

Fluid Mechanics, Prof. T.-S. Yang 4-66

 • In unsteady flow, an additional condition
must be satisfied.
• Let τ and l be the time and length scales,
respectively, of the flow.
* τ is the time duration in which the flow
condition varies appreciably.
• In the Euler equation, the local acceleration
∂V V
∼ ,
∂t τ
V being the velocity scale of the flow.
* The pressure gradient

|∇p/ρ| ∼ ∆p/ρl.
• Balancing the two terms gives the scale of
pressure variation in the flow:

∆p ∼ ρV l/τ.
* The scale of density variation therefore is

∆ρ ∼ ρV l/τ c2.
• Interpretation: Unsteadiness produces
pressure and density variations.

Fluid Mechanics, Prof. T.-S. Yang 4-67

 • Next consider the continuity equation:
1 Dρ
+ ∇ · V = 0.
ρ Dt
* (Dρ/Dt)/ρ ∼ ∆ρ/ρτ ∼ V l/(τ c)2; ∇ · V ∼
V /l.
• The fluid compressibility is negligible if
∆ρ Vl V
∼ ≪ ⇒ τ ≫ l/c.
ρτ (τ c)2 l

• Interpretation:
* l/c is the time taken for a infinitesimal
perturbation (say, pressure and density
variations produced by unsteady flow
condition) to travel a distance l.
* So, if τ ≫ l/c, the flow readily adapts to
the pressure and density variations caused
by unsteady flow condition.

Fluid Mechanics, Prof. T.-S. Yang 4-68

General form of Bernoulli integral
* for unsteady compressible (barotropic fluid)
flow
• Assumptions:
1) Inviscid fluid
2) Constant gravity: G = −g k = −∇(gz)
† gz is the gravitational (force) potential.
• We then write the Euler equation as
∂V 1
+ (V · ∇)V = − ∇p − ∇(gz).
∂t ρ
• Vector identity:
 
1
(V · ∇)V = ∇ V · V − V × (∇ × V)
2
* Demo:
i j k
!
∂ ∂ ∂ ∂w ∂v
∇×V = = − i
∂x ∂y ∂z ∂y ∂z
u v w
  !
∂u ∂w ∂v ∂u
+ − j+ − k;
∂z ∂x ∂x ∂y

Fluid Mechanics, Prof. T.-S. Yang 4-69

 V × (∇ × V)

i j k
u v w
=
∂w ∂v ∂u ∂w ∂v ∂u
− − −
∂y ∂z ∂z ∂x ∂x ∂y
( ! !
∂ v2 ∂u ∂u ∂ w2
= i −v −w +
∂x 2 ∂y ∂z ∂x 2
! )
∂ u2 ∂u
+ −u + j{· · ·} + k{· · ·}
∂x 2 ∂x
= ∇(V 2/2) − (V · ∇)V.

• Hence [V = |V|; ω = ∇ × V = vorticity]

!
∂V 1 V2
+∇ + gz + ∇p = V × ω.
∂t 2 ρ
• Taking scalar product of the above
equation and an arbitrary dr yields
! !
V2
Z
∂V dp
· dr + d + gz + d
∂t 2 ρ(p)
= (V × ω) · dr,
assuming in addition that the fluid is
barotropic(3), i.e., ρ = ρ(p).

Fluid Mechanics, Prof. T.-S. Yang 4-70

 • The RHS of the above equation vanishes
when
4) drkV (along a streamline), OR
4′) ω = 0 everywhere (irrotational flow).
Then,
!
V2
Z 2 Z
∂V dp
· dr + + gz +
1 ∂t 2 ρ 2
!
V2
Z
dp
= + gz + .
2 ρ 1
• In an irrotational flow, points 1 and 2 need
not be on the same streamline.
• Special case: steady flow, ∂/∂t = 0

V2
Z
dp
+ gz + =C
2 ρ
along each streamline.
* The basic Bernoulli’s eqn is recovered.
* The ‘Bernoulli constant’ C generally varies
from streamline to streamline, but is the
same for all streamlines in an irrotational
flow.

Fluid Mechanics, Prof. T.-S. Yang 4-71

 • If ρ = constant, then along a streamline in
an unsteady flow we have
Z 2
∂V
ρ · dr +p2 + ρV22/2 + ρgz2
∂t
| 1 {z }
transient
= p1 + ρV12/2 + ρgz1.
* The basic-version Bernoulli’s equation is
appended by an extra term that accounts
for transient effects.
• Example:

* Assume that ρ = constant, and that the

flow is ‘quasi-one-dimensional’.
* Mass conservation then implies that V A =
constant with x = Q(t).
* Accordingly,
Q(t) ∂V Q̇(t)
V (x, t) = , = .
A(x) ∂t A(x)

Fluid Mechanics, Prof. T.-S. Yang 4-72

 * Evaluate the transient term as follows:
Z 2 Z 2
∂V dx
ρ ds = ρQ̇(t) .
1 ∂t 1 A(x)
* Neglecting gravity, the Bernoulli’s equation
then gives
Z 2 " #2
dx ρ Q(t)
ρQ̇(t) + p2 +
1 A(x) 2 A2
" #2
ρ Q(t)
= p1 + .
2 A1
* So, we can write the pressure difference as

p1 − p2
! " #2
dx
Z 2
ρ Q(t) 2
A2
= ρQ̇(t) + 1− 2 .
|
1 A(x)
{z }
2 A2 A1
| {z }
transient term quasi-steady term

§ Criterion for ‘quasi-steady flow’:

!
[Q(t)]2 A2
Z 2
dx
Q̇(t) ≪ 1− 2
1 A(x) 2A2
2 A2
1

Fluid Mechanics, Prof. T.-S. Yang 4-73

 • Example: Start-up of flow from reservoir

* Bernoulli’s equation:
Z 2
∂V
ρ ds + p2 + ρV22/2 + ρgz2
∂t 1
= p1 + ρV12/2 + ρgz1
* Three parts of the transient term:
Z 2 Z a Z b Z 2
∂V ∂V ∂V ∂V
ds = ds + ds + ds
1 ∂t ∂t
| 1 {z } | a ∂t
{z } ∂t
| b {z }
(1) (2) (3)
1) Since V ≈ 0 in the reservoir, neglect this
term.
2) ‘Entrance region’ having size of O(R), so
this term has O((dV /dt) · R) [≪ (dV /dt) · l].
3) V = V (t) in the pipe, so this term has the
value of (dV /dt) · l.
* Assuming R ≪ l, only keep the 3rd term.

Fluid Mechanics, Prof. T.-S. Yang 4-74

 * Hence we have (approximately, but with
reasonable accuracy)
dV
ρl+ pa + ρV 2/2 + 0 = pa + 0 + ρgh,
dt
or the separable differential equation (DE)
dV
ρl + ρV 2/2 = ρgh.
dt
§ Solution: Rewrite the DE as
dV dt
2
= ;
2gh − V 2l
√
dV dV 2gh
⇒ √ +√ = dt
2gh − V 2gh + V l
√ √
2gh + V 2gh t
⇒ ln √ = ·t≡ .
2gh − V l τ
† We have used the initial condition V (0) =
0, and defined the characteristic time τ =
√
l/ 2gh.
* It is then straightforward to show that
V (t) et/τ − 1
√ = t/τ .
2gh e +1

Fluid Mechanics, Prof. T.-S. Yang 4-75

 § Asymptotes of V (t):
√
1) t ≫ τ : Then V ≈ 2gh (the steady-state
flow velocity).
2) t ≪ τ : Then
V (t) t gh
√ ≈ ⇒ V (t) ≈ · t.
2gh 2τ l
√
* Note also that for t ≈ 0, V ≪ 2gh, and so
dynamic pressure is negligible.
* We can then derive the same asymptotic
result by simply applying F = ma:
dV 2 gh
ρgh (πR ) = ρlA ·
|{z} dt
⇒ V (t) ≈ · t.
|{z} | {z }
m |{z} l
∆p A a
| {z }
F

Fluid Mechanics, Prof. T.-S. Yang 4-76

 § Pressure distribution in the pipe:

* Consider a streamline connecting point (1)

at position x and the pipe exit (2).
* Accounting for transient effects, we have
dV ρV 2 ρV 2
ρ (l−x)+pa + +0 = p(x, t)+ +0.
dt 2 2

Fluid Mechanics, Prof. T.-S. Yang 4-77

 * Hence the pressure difference
dV
p(x, t) − pa = ρ(l − x)
dt
2gh − V 2
= ρ(l − x)
2l
!
V 2 x

= ρgh 1 − 1− .
2gh l

* The pressure and velocity variations in the

pipe are sketched schematically on the last
page.
* With the present simplified analysis, the
pressure distribution at t = 0− switches
infinitely fast to that at t = 0+ .
† Due to incompressibility

Fluid Mechanics, Prof. T.-S. Yang 4-78

 • Example:

* Mass conservation:
d
(πr2h) + V · 2πrh = 0;
dt
r
⇒ πr2Vc = 2πrhV ⇒ V (r, t) = Vc.
2h(t)
* Unsteady Bernoulli’s equation:
Z 2
∂V
ρ ds + pa + ρV22/2 = p1 + ρV12/2
∂t 1
* Assuming that Vc = const., we have
∂V rVc dh rVc2
=− 2· = 2
. [dh/dt = −Vc]
∂t 2h dt 2h

Fluid Mechanics, Prof. T.-S. Yang 4-79

 * Hence
Vc2 2
Z 2
∂V
ds =
|{z} (R − r2),
1 ∂t dr 4h
and so the Bernoulli’s equation gives
ρVc2 2 2 ρR2Vc2
2
(R − r ) +pa +
4[h(t)] {z
| }
8[h(t)]2
transient
ρr2Vc2
= p(x, t) + 2
.
8[h(t)]
† Note that the transient and steady parts
always have the same order of magnitude.
† Hence the flow is never quasi-steady.

* We then find h(t) = h0 − Vct, and

3 2 2
2 R −r
p(x, t) − pa = ρVc · 2
.
8 [h(t)]
* May proceed to calculate the upward force:
R4
Z R
3π 2
Fup = (p−pa)·2πr dr = · ρVc · 2
.
0 16 [h(t)]

Fluid Mechanics, Prof. T.-S. Yang 4-80

Introductory comments on potential flow
• The velocity potential:
* If the flow is irrotational, i.e., vorticity-free,
then
ω = ∇ × V ≡ 0.
* Since ∇ × ∇φ = 0 for any scalar function φ
(that is twice differentiable), the condition
of irrotationality will be satisfied identically
by choosing
V = ∇φ.
* φ is called the velocity potential.
* Such flow is called potential flow.
• Incompressible flow: (in addition)

∇·V=0 ⇒ ∇2φ = 0.
* Such flow is called potential flow, and is
governed by Laplace’s equation.
• Boundary condition: On solid boundary,
∂φ
Vn = 0 ⇒ = 0.
∂n

Fluid Mechanics, Prof. T.-S. Yang 4-81

 • Equation for pressure:
* Inviscid fluid:
∂V 1
+ (V · ∇)V = − ∇p − ∇G.
∂t ρ
† Conservative body force written as −∇G,
G being the force potential.
* Vector identity:
!
V2
(V · ∇)V = ∇ − V × (∇ × V)
2 | {z }
=ω≡0
* So, assuming that ρ = constant, we have
!
∂φ 1 2 p
∇ + |∇φ| + + G = 0.
∂t 2 ρ
* Hence
∂φ 1 2 p
+ |∇φ| + + G = F (t) ≡ 0.
∂t 2 ρ
† The Bernoulli constant can be arbitrarily
set to zero (which amounts to redefining
φ) without any physical consequences.
R
† Try φ = φ′ + F (t) dt.

Fluid Mechanics, Prof. T.-S. Yang 4-82

 • Example:
Uniform flow past a circular cylinder

!
a2
φ=U r+ cos θ
r
* Verify:
1 ∂ 2φ
 
2 1∂ ∂φ
∇ φ= r + 2 2 = 0.
r ∂r ∂r r ∂θ
* Can also be obtained by solving the Laplace’s
equation with the boundary conditions
∂φ
Vr = = 0 (r = a),
∂r
∂φ ∂φ 1 ∂φ
u = = cos θ − sin θ → U (r → ∞).
∂x ∂r r ∂θ

Fluid Mechanics, Prof. T.-S. Yang 4-83

 * Note that
!
∂φ a2
Vr = = U 1 − 2 cos θ,
∂r r
!
1 ∂φ a2
Vθ = = −U 1 + 2 sin θ.
r ∂θ r
* Hence
!
a4 a2
V 2 = Vr2 + Vθ2 = U 2 1 + 4 − 2 2 cos 2θ .
r r
* Pressure distribution:
1 2
p∞ + ρU
2
!
1 2 a 4 a 2
= p + ρU 1 + 4 − 2 2 cos 2θ ;
2 r r
hence
!
1 2 a2 a4
p − p∞ = ρU 2 2 cos 2θ − 4 .
2 r r
§ On the surface of the cylinder,
1 2
ps − p∞ = ρU (2 cos 2θ − 1)
2
1 2 2

= ρU 1 − 4 sin θ .
2

Fluid Mechanics, Prof. T.-S. Yang 4-84

 * Surface pressure coefficient:
ps − p∞ 2
Cp,s = 2
= 1 − 4 sin θ
ρU /2
§ Only the upstream part agrees with results
of experiments.
† ps = p0 at θ ≈ 29.5◦ (β in above figure;
direction-seeking Pitot tube)
§ Flow separation in reality
† Viscous ‘boundary layer’

Fluid Mechanics, Prof. T.-S. Yang 4-85