incompressible inviscid flow - fcu.edu.twmyweb.fcu.edu.tw/~pwhwang/fluidmechanics2/chap6-1.pdf ·...

as shown. (The stem of the total head tube is placed downstream from the mea-surement location to minimize disturbance of the local flow.)

Two probes often are combined, as in the pitot-static tube shown in Fig. 6.4b. Theinner tube is used to measure the stagnation pressure at point B, while the staticpressure at C is sensed using the small holes in the outer tube. In flow fields where thestatic pressure variation in the streamwise direction is small, the pitot-static tube maybe used to infer the speed at point B in the flow by assuming pB5 pC and usingEq. 6.12. (Note that when pB 6¼ pC, this procedure will give erroneous results.)

Remember that the Bernoulli equation applies only for incompressible flow (Machnumber M # 0.3). The definition and calculation of the stagnation pressure forcompressible flow will be discussed in Section 12.3.

Flow FlowTotalheadtubeA

p p0

B

C

Staticpressure

holes

p0

p

(b) Pitot-static tube(a) Total head tube used with wall static tap

Fig. 6.4 Simultaneous measurement of stagnation and static pressures.

Example 6.2 PITOT TUBE

A pitot tube is inserted in an air flow (at STP) to measure the flow speed. The tube is inserted so that it pointsupstream into the flow and the pressure sensed by the tube is the stagnation pressure. The static pressure is measuredat the same location in the flow, using a wall pressure tap. If the pressure difference is 30 mm of mercury, determinethe flow speed.

Given: A pitot tube inserted in a flow as shown. The flowing fluid is air and the manometer liquid is mercury.

Find: The flow speed.

Solution:

Governing equation:p

ρ1

V2

21 gz 5 constant

Assumptions: (1) Steady flow.(2) Incompressible flow.(3) Flow along a streamline.(4) Frictionless deceleration along stagnation streamline.

Writing Bernoulli’s equation along the stagnation streamline (with Δz5 0) yields

p0ρ

5p

ρ1

V2

2

p0 is the stagnation pressure at the tube opening where the speed has been reduced, without friction, to zero. Solvingfor V gives

V 5

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ðp0 2 pÞ

ρair

s

Air flow

Mercury

30 mm

246 Chapter 6 Incompressible Inviscid Flow

Applications

The Bernoulli equation can be applied between any two points on a streamlineprovided that the other three restrictions are satisfied. The result is

p1ρ

1V2

1

21 gz1 5

p2ρ

1V2

2

21 gz2 ð6:13Þ

where subscripts 1 and 2 represent any two points on a streamline. Applications ofEqs. 6.8 and 6.13 to typical flow problems are illustrated in Examples 6.3 through 6.5.

In some situations, the flow appears unsteady from one reference frame, but steadyfrom another, which translates with the flow. Since the Bernoulli equation was derivedby integrating Newton’s second law for a fluid particle, it can be applied in any inertialreference frame (see the discussion of translating frames in Section 4.4). The proce-dure is illustrated in Example 6.6.

From the diagram,

p0 2 p 5 ρHggh 5 ρH2OghSGHg

and

V 5

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ρH2O

ghSGHg

ρair

s

5

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi23 1000

kg

m33 9:81

m

s23 30 mm3 13:63

m3

1:23 kg3

1 m

1000 mm

vuutV 5 80:8 m=s ß

V

At T5 20�C, the speed of sound in air is 343 m/s. Hence, M5 0.236 and the assumption of incompressibleflow is valid.

This problem illustrates use of a pitottube to determine flow speed. Pitot (orpitot-static) tubes are often placed onthe exterior of aircraft to indicate airspeed relative to the aircraft, andhence aircraft speed relative to the air.

Example 6.3 NOZZLE FLOW

Air flows steadily at low speed through a horizontal nozzle (by definition a device for accelerating a flow), dis-charging to atmosphere. The area at the nozzle inlet is 0.1 m2. At the nozzle exit, the area is 0.02 m2. Determine thegage pressure required at the nozzle inlet to produce an outlet speed of 50 m/s.

Given: Flow through a nozzle, as shown.

Find: p12 patm.

Solution:

Governing equations:

p1ρ

1V2

1

21 gz1 5

p2ρ

1V2

2

21 gz2 ð6:13Þ

Continuity for incompressible and uniform flow: XCS

~V � ~A 5 0 ð4:13bÞ

1

2

CV

Streamline

p2 = patmV2 = 50 m/sA2 = 0.02 m2

A1 = 0.1 m2

6.3 Bernoulli Equation: Integration of Euler’s Equation Along a Streamline for Steady Flow 247

Assumptions: (1) Steady flow.(2) Incompressible flow.(3) Frictionless flow.(4) Flow along a streamline.(5) z15 z2.(6) Uniform flow at sections 1 and 2 .

The maximum speed of 50 m/s is well below 100 m/s, which corresponds to Mach number M � 0.3 in standard air.Hence, the flow may be treated as incompressible.

Apply the Bernoulli equation along a streamline between points 1 and 2 to evaluate p1. Then

p1 2 patm 5 p1 2 p2 5ρ2ðV2

2 2V21Þ

Apply the continuity equation to determine V1,

ð2ρV1A1Þ1 ðρV2A2Þ 5 0 or V1A1 5 V2A2

so that

V1 5 V2A2

A15 50

m

s3

0:02 m2

0:1 m25 10 m=s

For air at standard conditions, ρ5 1.23 kg/m3. Then

p1 2 patm 5ρ2ðV2

2 2V21Þ

51

23 1:23

kg

m3

�ð50Þ2 m

2

s22 ð10Þ2 m

2

s2

�N � s2kg �m

p1 2 patm 5 1:48 kPa ßp1 2 patm

Example 6.4 FLOW THROUGH A SIPHON

AU-tube acts as a water siphon. The bend in the tube is 1 m above the water surface; the tube outlet is 7 m below thewater surface. The water issues from the bottom of the siphon as a free jet at atmospheric pressure. Determine (afterlisting the necessary assumptions) the speed of the free jet and the minimum absolute pressure of the water in thebend.

Given: Water flowing through a siphon as shown.

Find: (a) Speed of water leaving as a free jet.(b) Pressure at point A (the minimum pressure point) in the flow.

Solution:

Governing equation:p

ρ1

V2

21 gz 5 constant

Assumptions: (1) Neglect friction.(2) Steady flow.(3) Incompressible flow.(4) Flow along a streamline.(5) Reservoir is large compared with pipe.

Apply the Bernoulli equation between points 1 and 2 .

1

2

A

8 m

z = 0

1 mz

Notes:

ü This problem illustrates a typicalapplication of the Bernoulliequation.ü The streamlines must be straight

at the inlet and exit in order tohave uniform pressures at thoselocations.


p1ρ

1V2

1

21 gz1 5

p2ρ

1V2

2

21 gz2

Since areareservoir c areapipe, then V1 � 0. Also p15 p25 patm, so

gz1 5V2

2

21 gz2 and V2

2 5 2gðz1 2 z2Þ

V2 5ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2gðz1 2 z2Þ

p5

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi23 9:81

m

s23 7 m

s

5 11:7 m=s ßV2

To determine the pressure at location A , we write the Bernoulliequation between 1 and A .

p1ρ

1V2

1

21 gz1 5

pAρ

1V2

A

21 gzA

Again V1 � 0 and from conservation of mass VA5V2. Hence

pAρ

5p1ρ

1 gz1 2V2

2

22 gzA 5

p1ρ

1 gðz1 2 zAÞ2V2

2

2

pA 5 p1 1 ρgðz1 2 zAÞ2 ρV2

2

2

5 1:013 105N

m21 999

kg

m33 9:81

m

s23 ð21mÞ N � s2

kg �m

21

23 999

kg

m33 ð11:7Þ2 m

2

s23

N � s2kg �m

pA5 22.8 kPa (abs) or 278.5 kPa (gage) ßpA

Notes:ü This problem illustrates an applica-tion of the Bernoulli equation thatincludes elevation changes.

ü It is interesting to note that whenthe Bernoulli equation appliesbetween a reservoir and a free jetthat it feeds at a location h belowthe reservoir surface, the jet speedwill be V 5

ffiffiffiffiffiffiffiffiffi2 gh

p; this is the same

velocity a droplet (or stone) fallingwithout friction from the reservoirlevel would attain if it fell a distanceh. Can you explain why?ü Always take care when neglectingfriction in any internal flow. In thisproblem, neglecting friction is rea-sonable if the pipe is smooth-surfaced and is relatively short. InChapter 8 we will study frictionaleffects in internal flows.

Example 6.5 FLOW UNDER A SLUICE GATE

Water flows under a sluice gate on a horizontal bed at the inlet to a flume. Upstream from the gate, the water depth is1.5 ft and the speed is negligible. At the vena contracta downstream from the gate, the flow streamlines are straightand the depth is 2 in. Determine the flow speed downstream from the gate and the discharge in cubic feet per secondper foot of width.

Given: Flow of water under a sluice gate.

Find: (a) V2.(b) Q in ft3/s/ft of width.

Solution:Under the assumptions listed below, the flow satisfies allconditions necessary to apply the Bernoulli equation. Thequestion is, what streamline do we use?

21

z D1 = 1.5 ft

Sluice gate

Vena contracta

V2D2 = 2 in.

g

V1 – 0~


Governing equation:p1ρ

1V2

1

21 gz1 5

p2ρ

1V2

2

21 gz2

Assumptions: (1) Steady flow.(2) Incompressible flow.(3) Frictionless flow.(4) Flow along a streamline.(5) Uniform flow at each section.(6) Hydrostatic pressure distribution (at each location, pressure increases linearly with depth).

If we consider the streamline that runs along the bottom of the channel (z5 0), because of assumption 6 thepressures at 1 and 2 are

p1 5 patm 1 ρgD1 and p2 5 patm 1 ρgD2

so that the Bernoulli equation for this streamline is

ðpatm 1 ρgD1Þρ

1V2

1

25

ðpatm 1 ρgD2Þρ

1V2

2

2

or

V21

21 gD1 5

V22

21 gD2 ð1Þ

On the other hand, consider the streamline that runs along the free surface on both sides and down the inner surfaceof the gate. For this streamline

patmρ

1V2

1

21 gD1 5

patmρ

1V2

2

21 gD2

or

V21

21 gD1 5

V22

21 gD2 ð1Þ

We have arrived at the same equation (Eq. 1) for the streamline at the bottom and the streamline at the free surface,implying the Bernoulli constant is the same for both streamlines. We will see in Section 6.6 that this flow is one of afamily of flows for which this is the case. Solving for V2 yields

V2 5ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2gðD1 2D2Þ1V2

1

qBut V2

1 � 0, so

V2 5ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2gðD1 2D2Þ

p5

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi23 32:2

ft

s23 1:5 ft2 2 in:3

ft

12 in:

0@

1A

vuuutV2 5 9:27 ft=s ß

V2

For uniform flow, Q5VA5VDw, or

Q

w5 VD 5 V2D2 5 9:27

ft

s1 2 in:3

ft

12 in:5 1:55 ft2=s

Q

w5 1:55 ft3=s=foot of width ß

Q

w


Example 6.6 BERNOULLI EQUATION IN TRANSLATING REFERENCE FRAME

A light plane flies at 150 km/hr in standard air at an altitude of 1000 m. Determine the stagnation pressure at theleading edge of the wing. At a certain point close to the wing, the air speed relative to the wing is 60 m/s. Computethe pressure at this point.

Given: Aircraft in flight at 150 km/hr at 1000 m altitude in standard air.

Find: Stagnation pressure, p0A, at point A and static pressure, pB, at point B.

Solution:Flow is unsteady when observed from a fixed frame, that is, by an observer on the ground. However, an observer onthe wing sees the following steady flow:

At z5 1000 m in standard air, the temperature is 281 K and the speed of sound is 336 m/s. Hence at point B,MB5VB/c5 0.178. This is less than 0.3, so the flow may be treated as incompressible. Thus the Bernoulli equationcan be applied along a streamline in the moving observer’s inertial reference frame.

Governing equation:pairρ

1V2

air

21 gzair 5

pAρ

1V2

A

21 gzA 5

pBρ

1V2

B

21 gzB

Assumptions: (1) Steady flow.(2) Incompressible flow (V, 100 m/s).(3) Frictionless flow.(4) Flow along a streamline.(5) Neglect Δz.

Values for pressure and density may be found from Table A.3. Thus, at 1000 m, p/pSL5 0.8870 and ρ/ρSL5 0.9075.Consequently,

p 5 0:8870pSL 5 0:88703 1:013 105N

m25 8:963 104 N=m2

and

ρ 5 0:9075ρSL 5 0:90753 1:23kg

m35 1:12 kg=m3

Since the speed is VA5 0 at the stagnation point,

p0A 5 pair 11

2ρV2

air

5 8:963 104N

m21

1

23 1:12

kg

m3

�150

km

hr3 1000

m

km3

hr

3600 s

�2

3N � s2kg �m

p0A 5 90:6 kPaðabsÞ ßp0A

Vair = Vw = 150 km/hrpair @ 1000 m

AB VB = 60 m/s

Observer

Vair = 0pair @ 1000 m

Vw = 150 km/hrA B

VB = 60 m/s(relative to wing)

Observer


Cautions on Use of the Bernoulli Equation

In Examples 6.3 through 6.6, we have seen several situations where the Bernoulliequation may be applied because the restrictions on its use led to a reasonable flowmodel. However, in some situations you might be tempted to apply the Bernoulliequation where the restrictions are not satisfied. Some subtle cases that violate therestrictions are discussed briefly in this section.

Example 6.3 examined flow in a nozzle. In a subsonic nozzle (a converging section)the pressure drops, accelerating a flow. Because the pressure drops and the wallsof the nozzle converge, there is no flow separation from the walls and the boundarylayer remains thin. In addition, a nozzle is usually relatively short so frictional effectsare not significant. All of this leads to the conclusion that the Bernoulli equation issuitable for use for subsonic nozzles.

Sometimes we need to decelerate a flow. This can be accomplished using a subsonicdiffuser (a diverging section), or by using a sudden expansion (e.g., from a pipe into areservoir). In these devices the flow decelerates because of an adverse pressure gra-dient. As we discussed in Section 2.6, an adverse pressure gradient tends to lead torapid growth of the boundary layer and its separation. Hence, we should be careful inapplying the Bernoulli equation in such devices—at best, it will be an approximation.Because of area blockage caused by boundary-layer growth, pressure rise in actualdiffusers always is less than that predicted for inviscid one-dimensional flow.

The Bernoulli equation was a reasonable model for the siphon of Example 6.4because the entrance was well rounded, the bends were gentle, and the overall lengthwas short. Flow separation, which can occur at inlets with sharp corners and in abruptbends, causes the flow to depart from that predicted by a one-dimensionalmodel and theBernoulli equation. Frictional effects would not be negligible if the tube were long.

Example 6.5 presented an open-channel flow analogous to that in a nozzle, forwhich the Bernoulli equation is a good flow model. The hydraulic jump is an exampleof an open-channel flow with adverse pressure gradient. Flow through a hydraulicjump is mixed violently, making it impossible to identify streamlines. Thus the Ber-noulli equation cannot be used to model flow through a hydraulic jump. We will see amore detailed presentation of open channel flows in Chapter 11.

The Bernoulli equation cannot be applied through a machine such as a propeller,pump, turbine, or windmill. The equation was derived by integrating along a stream

Solving for the static pressure at B, we obtain

pB 5 pair 11

2ρðV2

air 2V2BÞ

pB 5 8:963 104N

m21

1

23 1:12

kg

m3

��150

km

hr3 1000

m

km3

hr

3600 s

�2

2 ð60Þ2 m2

s2

�N � s2kg �m

pB 5 88:6 kPaðabsÞ ßpB

This problem gives a hint as to how awing generates lift. The incoming air hasa velocityVair 5 150 km=hr 5 41:7 m=sand accelerates to 60 m/s on the uppersurface. This leads, through theBernoulli equation, to a pressure drop of1 kPa (from89.6kPa to88.6kPa). It turnsout that the flow decelerates on thelower surface, leading to a pressure riseof about 1 kPa. Hence, the wing experi-ences a net upward pressure differenceof about 2 kPa, a significant effect.

CLASSIC VIDEO

Flow Visualization.


It looks likeweneeded restriction (7) to finally transform the energy equation into theBernoulli equation. In fact, we didn’t! It turns out that for an incompressible and fric-tionless flow [restriction (6), and the fact we are looking only at flows with no shearforces], restriction (7) is automatically satisfied, as we will demonstrate in Example 6.7.

For the steady, frictionless, and incompressible flow considered in this section, it istrue that the first law of thermodynamics reduces to the Bernoulli equation. Each termin Eq. 6.15 has dimensions of energy per unit mass (we sometimes refer to the threeterms in the equation as the “pressure” energy, kinetic energy, and potential energy perunit mass of the fluid). It is not surprising that Eq. 6.15 contains energy terms—after all,we used the first law of thermodynamics in deriving it. How did we end up withthe same energy-like terms in the Bernoulli equation, which we derived from themomentum equation? The answer is because we integrated the momentum equation(which involves force terms) along a streamline (which involves distance), and by doingso ended up with work or energy terms (work being defined as force times distance):The work of gravity and pressure forces leads to a kinetic energy change (which camefrom integrating momentum over distance). In this context, we can think of the Ber-noulli equation as a mechanical energy balance—the mechanical energy (“pressure”plus potential plus kinetic) will be constant. We must always bear in mind that for theBernoulli equation to be valid along a streamline requires an incompressible inviscidflow, in addition to steady flow. It’s interesting that these two properties of the flow—itscompressibility and friction—are what “link” thermodynamic and mechanical energies.If a fluid is compressible, any flow-induced pressure changes will compress or expandthe fluid, thereby doing work and changing the particle thermal energy; and friction, aswe know from everyday experience, always converts mechanical to thermal energy.Their absence, therefore, breaks the link between the mechanical and thermal energies,and they are independent—it’s as if they’re in parallel universes!

Example 6.7 INTERNAL ENERGY AND HEAT TRANSFER IN FRICTIONLESS INCOMPRESSIBLE FLOW

Consider frictionless, incompressible flow with heat transfer. Show that

u2 2 u1 5δQdm

Given: Frictionless, incompressible flow with heat transfer.

Show: u2 2 u1 5δQdm

.

Solution:In general, internal energy can be expressed as u5 u(T, v). For incompressible flow, v5 constant, and u5 u(T). Thusthe thermodynamic state of the fluid is determined by the single thermodynamic property, T. For any process, theinternal energy change, u22 u1, depends only on the temperatures at the end states.

From the Gibbs equation, Tds5 du1 ρ dv, valid for a pure substance undergoing any process, we obtain

Tds 5 du

for incompressible flow, since dv5 0. Since the internal energy change, du, between specified end states, is inde-pendent of the process, we take a reversible process, for which Tds5 d(δQ/dm)5 du. Therefore,

u2 2 u1 5δQdm

ß

6.4 The Bernoulli Equation Interpreted as an Energy Equation 255

In summary, when the conditions are satisfied for the Bernoulli equation to bevalid, we can consider separately the mechanical energy and the internal thermalenergy of a fluid particle (this is illustrated in Example 6.8); when they are notsatisfied, there will be an interaction between these energies, the Bernoulli equationbecomes invalid, and we must use the full first law of thermodynamics.

Example 6.8 FRICTIONLESS FLOW WITH HEAT TRANSFER

Water flows steadily from a large open reservoir through a short length of pipe and a nozzle with cross-sectional areaA5 0.864 in.2 A well-insulated 10 kW heater surrounds the pipe. Find the temperature rise of the water.

Given: Water flows from a large reservoir through the system shown anddischarges to atmospheric pressure. The heater is 10 kW; A45 0.864 in.2

Find: The temperature rise of the water between points 1 and 2 .

Solution:

Governing equations:p

ρ1

V2

21 gz 5 constant ð6:8ÞX

CS~V � ~A 5 0 ð4:13bÞ

� gz) V � dAV 2

2Q � Ws � Wshear �

CV

e dV �

CS

(u � pv �

� 0(4) � 0(4) � 0(1)

� �� tð4:56Þ

Assumptions: (1) Steady flow.(2) Frictionless flow.(3) Incompressible flow.(4) No shaft work, no shear work.(5) Flow along a streamline.(6) Uniform flow at each section [a consequence of assumption (2)].

Under the assumptions listed, the first law of thermodynamics for the CV shown becomes

_Q 5

ZCS

�u1 pv1

V2

21 gz

�ρ~V � d~A

5

ZA1

�u1 pv1

V2

21 gz

�ρ~V � d~A1

ZA2

�u1 pv1

V2

21 gz

�ρ~V � d~A

For uniform properties at 1 and 2

_Q 52ðρV1A1Þ u1 1 p1v1V2

1

21 gz1

� �1 ðρV2A2Þ u2 1 p2v1

V22

21 gz2

� �

From conservation of mass, ρV1A1 5 ρV2A2 5 �m ; so

_Q 5 �m u2 2 u1 1p2ρ

1V2

2

21 gz2

� �2

p1ρ

1V2

1

21 gz1

� ��

13

2

410 ft

HeaterCV


6.5Energy Grade Line and Hydraulic Grade LineWe have learned that for a steady, incompressible, frictionless flow, we may use theBernoulli equation (Eq. 6.8), derived from the momentum equation, and also Eq.6.15, derived from the energy equation:

p

ρ1

V2

21 gz 5 constant ð6:15Þ

We also interpreted the three terms comprised of “pressure,” kinetic, and potentialenergies to make up the total mechanical energy, per unit mass, of the fluid. If wedivide Eq. 6.15 by g, we obtain another form,

For frictionless, incompressible, steady flow, along a streamline,

p

ρ1

V2

21 gz 5 constant

Therefore,

_Q 5 �mðu2 2 u1ÞSince, for an incompressible fluid, u22 u15 c(T22T1), then

T2 2T1 5_Q�mc

From continuity,

�m 5 ρV4A4

To find V4, write the Bernoulli equation between the free surface at 3 and point 4 .

p3ρ

1V2

3

21 gz3 5

p4ρ

1V2

4

21 gz4

Since p35 p4 and V3 � 0, then

V4 5ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2gðz3 2 z4Þ

p5

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi23 32:2

ft

s23 10 ft

r5 25:4 ft=s

and

�m 5 ρV4A4 5 1:94slug

ft33 25:4

ft

s3 0:864 in:2 3

ft2

144 in:25 0:296 slug=s

Assuming no heat loss to the surroundings, we obtain

T2 2T1 5_Q�mc

5 10 kW3 3413Btu

kW � hr 3hr

3600 s

3s

0:296 slug3

slug

32:2 lbm3

lbm �� R1 Btu

T2 2T1 5 0:995 �R ßT2 2T1

This problem illustrates that:ü In general, the first law of thermo-dynamics and the Bernoulli equationare independent equations.

ü For an incompressible, inviscid flowthe internal thermal energy is onlychanged by a heat transfer process,and is independent of the fluidmechanics.

6.5 Energy Grade Line and Hydraulic Grade Line 257

and

Vr 52@φ@r

and Vθ 521

r

@φ@θ

ð6:33Þ

In Section 5.2 we showed that the stream function ψ is constant along anystreamline. For ψ5 constant, dψ5 0 and

dψ 5@ψ@x

dx1@ψ@y

dy 5 0

The slope of a streamline—a line of constant ψ—is given by

dy

dx

�ψ52

@ψ=dx@x=@y

522 vu

5vu

ð6:34Þ

Along a line of constant φ, dφ5 0 and

dφ 5@φ@x

dx1@φ@y

dy 5 0

Consequently, the slope of a potential line — a line of constant φ — is given by

dy

dx

�φ52

@φ=@x@φ=@y

52u

vð6:35Þ

(The last equality of Eq. 6.35 follows from use of Eq. 6.29.)Comparing Eqs. 6.34 and 6.35, we see that the slope of a constant ψ line at any

point is the negative reciprocal of the slope of the constant φ line at that point; thismeans that lines of constant ψ and constant φ are orthogonal. This property ofpotential lines and streamlines is useful in graphical analyses of flow fields.

Table 6.1Definitions of ψ and φ, and Conditions Necessary for Satisfying Laplace’s Equation

Definition Always satisfies . . .

Satisfies Laplace equation . . .

@2ðÞ@x2

1@2ðÞ@y2

5 r2ðÞ 5 0

Stream function ψ

u 5@ψ@y

v 52@ψ@x

. . . incompressibility:

@u

@x1

@v@y

5@2ψ@x@y

2@2ψ@y@x

� 0

. . . only if irrotational:

@v@x

2@u

@y52

@2ψ@x@x

2@2ψ@y@y

5 0

Velocity potential φ

u 52@φ@x

v 52@φ@y

. . . irrotationality:

@v@x

2@u

@y52

@2φ@x@y

2@2φ@y@x

� 0

. . . only if incompressible:

@u

@x1

@v@y

52@2φ@x@x

2@2φ@y@y

5 0

Example 6.10 VELOCITY POTENTIAL

Consider the flow field given by ψ5 ax22 ay2, where a5 3 s21. Show that the flow is irrotational. Determine thevelocity potential for this flow.

Given: Incompressible flow field with ψ5 ax22 ay2, where a5 3 s21.

Find: (a) Whether or not the flow is irrotational.(b) The velocity potential for this flow.

6.7 Irrotational Flow 263

Elementary Plane Flows

The ψ and φ functions for five elementary two-dimensional flows—a uniform flow, asource, a sink, a vortex, and a doublet—are summarized in Table 6.2. The ψ and φfunctions can be obtained from the velocity field for each elementary flow. (We saw inExample 6.10 that we can obtain φ from u and v.)

Solution: If the flow is irrotational, r2ψ 5 0. Checking for the given flow,

r2ψ 5@2

@x2ðax2 2 ay2Þ1 @2

@y2ðax2 2 ay2Þ 5 2a2 2a 5 0

so that the flow is irrotational. As an alternative proof, we can compute the fluid particle rotation (in the xy plane,the only component of rotation is ωz):

2ωz 5@v@x

2@u

@yand u 5

@ψ@y

v 52@ψ@x

then

u 5@

@yðax2 2 ay2Þ 522ay and v 52

@

@xðax2 2 ay2Þ 522ax

so

2ωz 5@v@x

2@u

@y5

@

@xð22axÞ2 @

@yð22ayÞ 522a1 2a 5 0 ß

2ωz

Once again, we conclude that the flow is irrotational. Because it is irrotational, φ must exist, and

u 52@φ@x

and v 52@φ@y

Consequently, u 52@φ@x

522ay and@φ@x

5 2ay. Integrating with respect to x gives φ5 2axy1 f(y), where f(y) is an

arbitrary function of y. Then

v 522ax 52@φ@y

52@

@x½2axy1 f ðyÞ�

Therefore, 2 2ax 522ax2@f ðyÞ@y

522ax2df

dy; so

df

dy5 0 and f5 constant. Thus

φ 5 2axy1 constantß φ

We also can show that lines of constant ψ and constant φ are orthogonal.

ψ 5 ax2 2 ay2 and φ 5 2axy

For ψ5 constant, dψ 5 0 5 2axdx2 2aydy; hencedy

dx

�ψ 5 c

5x

y

For φ5 constant, dφ 5 0 5 2aydx1 2axdy; hencedy

dx

�φ 5 c

52y

x

The slopes of lines of constant φ and constant ψ are negative reciprocals.Therefore lines of constant φ are orthogonal to lines of constant ψ.

This problem illustrates the relationsamong the stream function, velocitypotential, and velocity field.The stream function ψ andvelocity potential φ are shown inthe Excel workbook. By entering theequations for ψ and φ, other fields canbe plotted.


Much of this analytical work was done centuries ago, when it was called “hydro-dynamics” instead of potential theory. A list of famous contributors includes Ber-noulli, Lagrange, d’Alembert, Cauchy, Rankine, and Euler [11]. As we discussed inSection 2.6, the theory immediately ran into difficulties: In an ideal fluid flow no bodyexperiences drag—the d’Alembert paradox of 1752—a result completely counterto experience. Prandtl, in 1904, resolved this discrepancy by describing how real flowsmay be essentially inviscid almost everywhere, but there is always a “boundary layer”adjacent to the body. In this layer significant viscous effects occur, and the no-slipcondition is satisfied (in potential flow theory the no-slip condition is not satisfied).Development of this concept, and the Wright brothers’ historic first human flight, ledto rapid developments in aeronautics starting in the 1900s. We will study boundarylayers in detail in Chapter 9, where we will see that their existence leads to drag onbodies, and also affects the lift of bodies.

An alternative superposition approach is the inverse method in which distributionsof objects such as sources, sinks, and vortices are used to model a body [12]. It is calledinverse because the body shape is deduced based on a desired pressure distribution.Both the direct and inverse methods, including three-dimensional space, are todaymostly analyzed using computer applications such as Fluent [13] and STAR-CD [14].

Example 6.11 FLOW OVER A CYLINDER: SUPERPOSITION OF DOUBLET AND UNIFORM FLOW

For two-dimensional, incompressible, irrotational flow, the superposition of a doublet and a uniform flow representsflow around a circular cylinder. Obtain the stream function and velocity potential for this flow pattern. Find thevelocity field, locate the stagnation points and the cylinder surface, and obtain the surface pressure distribution.Integrate the pressure distribution to obtain the drag and lift forces on the circular cylinder.

Given: Two-dimensional, incompressible, irrotational flow formed from superposition of a doublet anda uniform flow.

Find: (a) Stream function and velocity potential.(b) Velocity field.(c) Stagnation points.(d) Cylinder surface.(e) Surface pressure distribution.(f) Drag force on the circular cylinder.(g) Lift force on the circular cylinder.

Solution: Stream functions may be added because the flow field is incompressible and irrotational. Thus from Table6.2, the stream function for the combination is

ψ 5 ψd 1ψuf 52Λ sin θ

r1Ur sin θ ß

ψ

The velocity potential is

φ 5 φd 1φuf 52Λ cos θ

r2Ur cos θ ß

φ

The corresponding velocity components are obtained using Eqs. 6.30 as

Vr 52@φ@r

52Λ cos θ

r21U cos θ

Vθ 521

r

@φ@θ

52Λ sin θr2

2U sin θ


The velocity field is

~V 5 Vrer 1Vθeθ 5 2Λ cos θ

r21U cos θ

� �er 1 2

Λ sin θr2

2U sin θ� �

eθ ß

~V

Stagnation points are where ~V 5 Vrer 1Vθeθ 5 0

Vr 52Λ cos θ

r21U cos θ 5 cos θ U2

Λr2

� �

Thus Vr 5 0 when r 5

ffiffiffiffiffiΛU

r5 a: Also,

Vθ 52Λ sin θr2

2U sin θ 52sin θ U1Λr2

� �

Thus Vθ5 0 when θ5 0, π.Stagnation points are ðr; θÞ 5 ða; 0Þ; ða;πÞ: ß

Stagnation points

Note that Vr5 0 along r5 a, so this represents flow around a circular cylinder, as shown in Table 6.3. Flow isirrotational, so the Bernoulli equation may be applied between any two points. Applying the equation between apoint far upstream and a point on the surface of the cylinder (neglecting elevation differences), we obtain

pNρ

1U2

25

p

ρ1

V2

2

Thus,

p2 pN 51

2ρðU2 2V2Þ

Along the surface, r5 a, and

V2 5 V2θ 5 2

Λa2

2U

� �2

sin2θ 5 4U2 sin2θ

since Λ5Ua2. Substituting yields

p2 pN 51

2ρðU2 2 4U2 sin2θÞ 5 1

2ρU2ð12 4 sin2θÞ

or

p2 pN1

2ρU2

5 12 4 sin2 θ ß

Pressuredistribution

Drag is the force component parallel to the freestream flow direction.The drag force is given by

FD 5

ZA

2p dA cos θ 5

Z 2π

0

2pa dθ b cos θ

since dA5 a dθ b, where b is the length of the cylinder normal to the diagram.

Substituting p 5 pN 11

2ρU2ð12 4 sin2 θÞ,

a

p dA

U

p�

θ


FD 5

Z 2π

0

2 pNab cos θ dθ1Z 2π

0

21

2ρU2ð12 4 sin2 θÞab cos θ dθ

52pN ab sin θ�2π0

21

2ρU2ab sin θ

�2π0

11

2pU2ab

4

3sin3 θ

�2π0

FD 5 0ßFD

Lift is the force component normal to the freestream flow direction. (By convention, positive lift is an upward force.)The lift force is given by

FL 5

ZA

p dAð2sin θÞ 52

Z 2π

0

pa dθ b sin θ

Substituting for p gives

FL 52

Z 2π

0

pNab sin θ dθ2Z 2π

0

1

2ρU2ð12 4 sin2 θÞab sin θ dθ

5 pNa b cos θ�2π0

11

2ρU2ab cos θ

�2π0

11

2ρU2ab

�4 cos3 θ

32 4 cos θ

�2π0

FL 5 0 ßFL

This problem illustrates:ü How elementary plane flows can becombined to generate interestingand useful flow patterns.ü d’Alembert’s paradox, that potentialflows over a body do not generatedrag.

The stream function and pres-sure distribution are plotted in

the Excel workbook.

Example 6.12 FLOW OVER A CYLINDER WITH CIRCULATION: SUPERPOSITION OF DOUBLET,UNIFORM FLOW, AND CLOCKWISE FREE VORTEX

For two-dimensional, incompressible, irrotational flow, the superposition of a doublet, a uniform flow, and a freevortex represents the flow around a circular cylinder with circulation. Obtain the stream function and velocitypotential for this flow pattern, using a clockwise free vortex. Find the velocity field, locate the stagnation points andthe cylinder surface, and obtain the surface pressure distribution. Integrate the pressure distribution to obtain thedrag and lift forces on the circular cylinder. Relate the lift force on the cylinder to the circulation of the free vortex.

Given: Two-dimensional, incompressible, irrotational flow formed from superposition of a doublet, a uniform flow,and a clockwise free vortex.

Find: (a) Stream function and velocity potential.(b) Velocity field.(c) Stagnation points.(d) Cylinder surface.(e) Surface pressure distribution.(f) Drag force on the circular cylinder.(g) Lift force on the circular cylinder.(h) Lift force in terms of circulation of the free vortex.

Solution:Stream functions may be added because the flow field is incompressible and irrotational. From Table 6.2, the streamfunction and velocity potential for a clockwise free vortex are


ψfv 5K

2πln r φfv 5

K

2πθ

Using the results of Example 6.11, the stream function for the combination is

ψ 5 ψd 1ψuf 1ψfv

ψ 52Λ sin θ

r1Ur sin θ1

K

2πln r ß

ψ

The velocity potential for the combination is

φ 5 φd 1φuf 1φfv

φ 52Λ cos θ

r2Ur cos θ1

K

2πθ ß

φ

The corresponding velocity components are obtained using Eqs. 6.30 as

Vr 52@φ@r

52Λ cos θ

r21U cos θ ð1Þ

Vθ 521

r

@φ@θ

52Λ sin θr2

2U sin θ2K

2πrð2Þ

The velocity field is

~V 5 Vr er 1Vθ eθ

~V 5 2Λ cos θ

r21U cos θ

� �er 1 2

Λ sin θr

2U sin θ2K

2πr

� �eθ ß

~V

Stagnation points are located where ~V 5 Vr er 1Vθ eθ 5 0. From Eq. 1,

Vr 52Λ cos θ

r21U cos θ 5 cos θ U2

Λr2

� �

Thus Vr5 0 when r 5ffiffiffiffiffiffiffiffiffiffiffiΛ=U

p5 a ß

Cylinder surface

The stagnation points are located on r5 a. Substituting into Eq. 2 with r5 a,

Vθ 52Λ sin θa2

2U sin θ2K

2πa

52Λ sin θΛ=U

2U sin θ2K

2πa

Vθ 522U sin θ2K

2πa

Thus Vθ5 0 along r5 a when

sin θ 52K

4πUaor θ 5 sin21 2K

4πUa

� �

Stagnation points: r5 a θ 5 sin2 1 2K

4πUa

� �ß

Stagnation points

As in Example 6.11, Vr5 0 along r5 a, so this flow field once again represents flow around a circular cylinder, asshown in Table 6.3. For K5 0 the solution is identical to that of Example 6.11.


The presence of the free vortex (K. 0) moves the stagnation points below the center of the cylinder. Thus thefree vortex alters the vertical symmetry of the flow field. The flow field has two stagnation points for a range of vortexstrengths between K5 0 and K5 4πUa.

A single stagnation point is located at θ52π/2 when K5 4πUa.Even with the free vortex present, the flow field is irrotational, so the Bernoulli equation may be applied between

any two points. Applying the equation between a point far upstream and a point on the surface of the cylinder weobtain

pNρ

1U2

21 gz 5

p

ρ1

V2

21 gz

Thus, neglecting elevation differences,

p2 pN 51

2ρ ðU2 2V2Þ 5 1

2ρU2 12

U

V

� �2" #

Along the surface r5 a and Vr5 0, so

V2 5 V2θ 5 22U sin θ2

K

2πa

� �2

and

V

U

� �2

5 4 sin2 θ 12K

πUasin θ1

K2

4π2U2a2

Thus

p 5 pN 11

2ρU2 12 4 sin2θ2

2K

πUasinθ2

K2

4π2U2a2

� �ß

pðθÞ

Drag is the force component parallel to the freestream flow direction. As in Example 6.11, the drag force is given by

FD 5

ZA

2p dA cos θ 5

Z 2π

0

2pa dθb cos θ

since dA5 a dθ b, where b is the length of the cylinder normal to the diagram.Comparing pressure distributions, the free vortex contributes only to the terms containing the factor K. The

contribution of these terms to the drag force is

FDfv

1

2ρU2

5

Z 2π

0

2K

πUasin θ1

K2

4π2U2a2

� �ab cos θ dθ ð3Þ

FDfv

1

2ρU2

52K

πUaab

sin2 θ2

3752π

0

1K2

4π2U2a2ab sin θ

�2π0

5 0 ß

FD

Lift is the force component normal to the freestream flow direction. (Upward force is defined as positive lift.) Thelift force is given by

FL 5

ZA

2p dA sin θ 5

Z 2π

0

2pa dθ bsin θ

p�

V

θ

a

p

U


Comparing pressure distributions, the free vortex contributes only to the terms containing the factor K. The con-tribution of these terms to the lift force is

FLfv

1

2ρU2

5

Z 2π

0

2K

πUasin θ1

K2

4π2U2a2

0@

1Aab sin θ dθ

52K

πUa

Z 2π

0

ab sin2θdθ1K2

4π2U2a2

Z 2π

0

ab sin θ dθ

52Kb

πUθ22

sin2 θ4

24

352π

0

2K2b

4π2U2acos θ

�2π0

FLfv

1

2ρU2

52Kb

πU2π2

24

35 5

2Kb

U

Then FLfv5 ρUKb ß

FL

The circulation is defined by Eq. 5.18 as

Γ �I

~V � d~s

On the cylinder surface, r5 a, and ~V 5 Vθ eθ; so

Γ 5

Z 2π

0

22U sin θ2K

2πa

0@

1Aeθ � a dθ eθ

52

Z 2π

0

2Ua sin θ dθ2Z 2π

0

K

2πdθ

Γ 5 2K ßCirculation

Substituting into the expression for lift,

FL 5 ρUKb 5 ρUð2ΓÞb 52ρU Γb

or the lift force per unit length of cylinder is

FLb

52ρUΓ ß

FLb

This problem illustrates:ü Once again d’Alembert’s paradox,that potential flows do not generatedrag on a body.ü That the lift per unit length is2ρUΓ.It turns out that this expression forlift is the same for all bodies in anideal fluid flow, regardless of shape!The stream function and pres-

sure distribution are plotted inthe Excel workbook.

6.8 Summary and Useful EquationsIn this chapter we have:

ü Derived Euler’s equations in vector form and in rectangular, cylindrical, and streamline coordinates.ü Obtained Bernoulli’s equation by integrating Euler’s equation along a steady-flow streamline, and discussed its restrictions. We

have also seen how for a steady, incompressible flow through a stream tube the first law of thermodynamics reduces to theBernoulli equation if certain restrictions apply.


incompressible inviscid flow - fcu.edu.twmyweb.fcu.edu.tw/~pwhwang/fluidmechanics2/chap6-1.pdf ·...

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