Additional Solutions VFF3e

8/10/2019 Additional Solutions VFF3e

1/48

1

VISCOUS FLUID FLOW Third Edition - Frank M. White

SOLUTIONS TOADDITIONAL PROBLEMS NOT IN THE TEXT

Chapter 1. Preliminary Concepts

(Text problems in Chapter 1 are solved on pages 1-16 of the Manual)

1-27 Consider the tilted free surface of a liquid,

as in Fig. P1-27. Show that, if the fluid weight is

taken into account, the tilting causes shear stresses

in the liquid and hence the liquid willflowand

cannot be in a hydrostatic condition.

Solution: Draw a freebody of a wedge of liquidat the surface and show that it cannot be held in

equilibrium by normal stresses only.

As we move down from the surface on the rightside of the wedge, the pressure must rise due to

the liquid weight, as shown, to the value po+ gy.For the hydrostatic assumption, this same pressuremust act vertically on the right point of the bottom side.The vertical forces on the wedge are in balance!

But the horizontal forces are not, there is an extraleft-acting force on the right side, equal to the area

of the gray triangle shown on the right side.

The wedge is out of balance, shear stresses must occur, and the liquid will flow.______________________________________________________________________________

PROPRIETARY MATERIAL. The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be

displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or

used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course

preparation. If you are a student using this Manual, you are using it without permission

Gas,po

Liquid,

density0

o

y

o

o

po+gy

o+g

Fig. P1-27


2/48

2

1-28 Following up on Prob. 1-4, consider laminar flow in a pipe of radiusR, vz=K(R2 r

2),

for 0 r R, whereKis a constant. Determine (a) the vorticity distribution; (b) the strain rates;

and (c) the average velocity V = Q/Apipe, where Q is the volume flow through the pipe.

Solution: These are math exercises using the paraboloid distribution vz(r), with vr= v= 0. The

vorticity and strain rates are defined for cylindrical coordinates in Appendix B:

The average velocity is found by integrating the velocity distribution over the pipe:

______________________________________________________________________________

1-29 A realistic steady flow field, in Cartesian coordinates, has the following strain rates:

whereKis a constant. Deduce formulas for the velocity components (u, v, w) and state

whether your result is unique. Sketch a few streamlines of your velocity distribution.

Solution: Integrate each strain rate separately and compare your results for consistency:

).(;0

).(0;)2(;0

2

2

bAnsr

v

aAnsrKr

v

Kr

Kr

zrzrzzzrr

zz

r

=

======

===

==

).(2

42

22)(

11

2][ 0

422

2

2

0

2

2 cAns

rrR

R

KdrrrRK

RdAv

AA

QV

KRRR

pipe

z =====

0;0;; ====== zxyzxyzzyyxx KK

constantarehfdx

dh

dz

dfw

z

u

constantarefhdz

df

y

h

z

g

y

h

z

v

y

w

zffgy

f

x

g

y

u

x

vyxhwz

w

zxgKyvy

vKzyfKxu

x

uK

zx

yz

xyzz

yyxx

&,)(2

1)(

2

10

&,)(2

1)(

2

1)(

2

10

)(,)(2

1)(2

10;),(,0

),(,;),(,

11

1

+=

+

==

+

=

+

=

+

==

==+

=

+

===

==

+=

==+=

==


3/48

3

The analysis is similar to the traditional concept of separation of variables, e.g., a function of x

can only equal a function of yif both are equal to a constant. The velocities are unique:

u = Kx + constant ; v = -Ky + constant ; w = constant Ans.

If the constants are zero, this is (irrotational)stagnation flow, Fig. 3-24 of the text.

______________________________________________________________________________

1-30 Consider a wide film of liquid flowing steadily

down an inclined plane, as shown in Fig. P1-30.

The density and viscosity of both liquid and gas

are constant. The film depth h is constant.

List some simplifying assumptions which would

help us to analyze this problem and find the film

velocity distribution without the use of a parallel

set of supercomputers. Pay especial attention to boundary conditions.

Do not solve the problem, but give a nice discussion.

Solution: First, it would simplify the analysis if the flow were laminar, not turbulent.

(This would require a low Reynolds number, say, umaxh/< 1000 or so.)

Second, we require no-slip at the bottom, u(y=0) = 0. (A slipping micro-film? No thanks.)Third, a velocity vnormal to the wall would only happen if the film were overturningas it flows

down the incline. That only happens if there is a thermal or saline instability in the film. We

assume that is not the case, so we let v0 and solve only for a udistribution.

Fourth, at the surface,y = h, gas viscosity and density are typically much lessthan liquid values.

Thus the gas has little interaction with the film and we assume negligible gas shear stress on the

surface. Neglecting air shear means that liquid=(u/y) 0 at y = h. The film velocity will

reach its maximum at the surface.Finally, fifth, we assume constant atmospheric pressure on the surface: p/x|y=h= 0.We can then

find an analytic solution using the equations of motion of a newtonian fluid in Chapter 2.

___________________________________________________________________________

, u

, v

hGAS

LIQUID

Fig. P1-30


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4

1-31 Consider a flat interface between a gas and a liquid. Letxandydenote coordinates

parallel and normal to the interface, respectively. Properties do not vary in the z direction. Gas

shear on the interface is negligible. Suppose that the surface tension coefficient varies due to

temperature or concentration gradients, T = T (x). By summing forces in thexdirection along an

elemental thin piece of interface, derive a boundary condition that relates liquid velocity uto T (x).

Solution: Make a freebody of an elemental strip

of interface dxlong, as shown. Sum forces in the

xdirection, noting that pressure forces (not shown)

are in theydirection. Surface tension force

difference must balance with liquid shear force.

Let the interface be of width binto the paper. Then

Thus a gradient in surface tension coefficient acts as a driving force for flow of the fluid. This

type of flow is calledMarangoni convection[see, e.g., Sasmal and Hochstein (1994)].

___________________________________________________________________________

1-32 Data for the viscosity of ethyl alcohol (C2H6O) are given as follows:

T, C -40 -20 0 20 40 60 80

, kg/m-s 4.81E-3 2.83E-3 1.77E-3 1.20E-3 8.34E-4 5.92E-4 4.30E-4

Fit this data to a formula similar to Eq. (1-56), plot it, and comment upon the accuracy.

Solution: Rearrange the data, in a spreadsheet, taking the form of ln(/o) versus To/T and fit it

to a second-order parabola. The results for least squares fitting are accurate, even for a linear fit,

and excellent for a quadratic fit:

dx

Tb (T+dT)b

u

.|

:or,0)(

Ansdx

d

y

u

dxby

ubdb

liquid

liq

T

TTT

+++


5/48

5

As seen from the plot below, the accuracy is very good (quadratic fit shown, linear fit OK also).

-1.5

-1.0

-0.5

0.0

0.5

1.0

0.7 0.8 0.9 1.0 1.1 1.2

ln(

/ o)

To/T

ETHYL ALCOHOL

Quadratic Fit

_____________________________________________________________________________

1-33 Consider the cone-plate viscometer in Fig. P1-33. The fluid to be tested fills the gap and

one measures the momentMrequired to steadily rotate the cone at angular velocity . Suppose

thatR= 8 cm and = 4. Assuming a locally linear velocity profile and laminar flow, estimate

the fluid viscosity if a moment of 0.9 N-m is required to rotate the cone at 650 rpm.

%)1.1Quadratic,()(424.1)(789.8349.7

2.6%)Linear,()(025.6034.6ln

2

)(

+

+

T

T

T

T

T

T

oo

o

o

Heavy fluid,

Fig. P1-33


6/48

6

Solution: For any local radius r, the gap in the fluid is h= rtan. The moment about the axis is

From the given data we estimate the viscosity of the fluid. Note = 650 r/min = 68.1 rad/s:

This is the nominal viscosity of SAE 50 oil, but that may be a coincidence.____________________________________________________________________________

1-34 Analyze the interface (x) between a liquid and a gas near a plane wall, as in Fig. P1-34.

Assume constant surface tension T and small curvature, 1/Rx= d2/dx2(see Eq. 1-107 of the

text). The contact angle is atx= 0. (a) What are the appropriate boundary conditions? (b)

Find a formula for (x) and the interface heightHat the wall.

Solution: (a) Two boundary conditions are needed:

(b) In Eq. (1-107), the pressure difference is due to the weight of the fluid above the interface:

3

3

0

2

2

sin3Thus

sin3

2

sin

2;)

cos

2)(

tan

(

R

M

RdrrMr

drr

r

rrdAdM

R

ww

=

=

.)08.0)(/1.68(2

)4sin()9.0(33

Ansmsrad

mN

sm

kg

=

0.86

o

=HFig. P1-34

).(0)(;cot)0( aAnsxxdx

d===


7/48

7

The boundary condition at infinity requires thatA = 0. Atx = 0, we require a slope

d/dx= - cot= -B . Thus B= cot/. But alsoB= the wall height,H. Thus

Since we assumed small slope and curvature, should lie between about 70and 110._____________________________________________________________________________

Chapter 2. Fundamental Equations of Compressible Viscous Flow


2-22 A simple demonstration of how the Grashof number arises in free convection is to

consider a parcel of lightfluid,< o, rising a distancex due to buoyancy. Equate the decrease

in potential energy of the parcel to its increase in kinetic energy to form a characteristic

velocity of the parcel. Use this velocity-scale to form a Reynolds number Rexand then

interpret its square, Rex2.

Solution: Let the parcel have volume d and lie, as shown, within the

boundary layer. The average density difference in the boundary layer

would be about (1/2)(w). At heightx, equate itslost potential energy with its gained kinetic energy:

d

V?

xgV

Vdxgd

w

w

)/1(:or

]])[( 2[2

1

2

1

0

T

T

/,)exp()exp(:Solution

2

2

gxBxA

dx

dgppa

=+=

=

).()/exp(/

cotbAnsgx

gT

T

=


8/48

8

We could leave it this way, or we could note from the thermal expansion approximation, Eq.

(2-86), that (1 -w/) = (Tw T). Then the local Reynolds number of the parcel is

Thus the square of the Reynolds number is equivalent to the Grashof number Grx, as can be seen

by comparing this expression to Eq. (2-87) of the text.

___________________________________________________________________________

2-23 In slip flow of a gas near a fixed wall, R. Raju and S. Roy, Hydrodynamic Model for

Microscale Flows in a Channel with Two 90 Bends,J. Fluids Engineering, vol. 126, May 2004,

pp. 489-492, propose the following model for slip velocity at the wall:

wherexandyare parallel and normal to the wall, respectively, Tis absolute gas temperature, and

is a tangential momentum accommodation coefficient whose value is approximately unity.

(a) Note the tangential temperature gradient term. What might this represent? (b) Using typical

reference properties to define dimensionless variables, rewrite this boundary condition in

dimensionless form and discuss any nondimensional parameters which arise.

Solution: (a) The streamwise temperature-gradient term is usually neglected in slip-flow

analyses. Walls are often nearly isothermal and have negligible temperature gradient. This term

is often called thermal creepor thermal transpirationand can be important in high-Knudsen-

number orfree moleculeflow. See, e.g., Chap. VIII of Kennard (1938).

(b) Let the dimensionless variables be u* = u/Uo, x* =x/L, y* =y/L, and T* = T/To, where Uo,

To, and L are reference properties, such as inlet velocity and temperature and body width.

Substitute these variables into the above equation, clean up, and rewrite as follows:

.)(

Re:or,])([

Re2

32

2/1

AnsTTxgxgxTTVx w

xw

x

=

==

wwwx

T

Ty

u

RTu )(

4

3)(

2

25

16

+

=


9/48

9

where ReL=UoL/is the basic Reynolds number of the flow, and Uo/(RTo)1/2is almost the

Mach number, except that 1/2is missing. The coefficient is already dimensionless.

_____________________________________________________________________________

2-24 For one-dimensional flow through a porous medium of variable thickness h(x), the

differential equation for thickness relates to the pressure gradients:

wherepis the fluid pressure andK(x) is the variablepermeabilityof the medium (usually

measured in m2). Using typical reference properties to define dimensionless variables, rewrite

this equation in dimensionless form and discuss any nondimensional parameters that arise.

Solution: The writer thinks this is a bit tricky. There is no characteristic velocity U, and density

does not appear, so usingU2to nondimensionalize pressure is not appropriate. Flow in porous

media is generally creeping motion; Reynolds number is not important. The writer therefore

chooses entrance valuesKoand ho, system lengthL, and ambient pressure poas constant

parameters. The dimensionless variables are defined as follows:

Substitute these variables into the above differential equation, clean up, and we obtain

).()*

*(

Re

1

*4

3)

*

*(

2

Re

1

25

16*bAns

x

T

Ty

u

RT

Uu w

Lw

o

o

Lw

+

=

)]()[(1

2

2

x

pKh

x

p

x

Kh

x

hK

t

h

+

+

=

/*;/*;/*;/* oooo pttpppKKKhhh ====

)]*

*(**

*

*)

*

**

*

**[()(

*

*2

2

2 x

pKh

x

p

x

Kh

x

hK

L

K

t

h o

+

+

=


10/48

10

Viscosity, initial height, and ambient pressure vanish with these variables, leaving only the single

dimensionless porous-media parameter Ko/L2 . Ans.

____________________________________________________________________________

Chapter 3. Solutions of the Newtonian Viscous Flow Equations


3-59 In Section 3-5, for Stokes 1stproblem of the suddenly accelerated plate, the solution for

u(y, t) in Eq. (3-107) is given subject to the initialcondition u(y, 0) = 0 in Eq. (3-106). In

Stokes 2nd

problem, the oscillating plate, the solution of Eq. (3-111) is found withouta listed

initial condition. Is this an unfortunate omission or not? Please explain.

Solution: The omission was intentional. In Stokes 1stproblem, the problem beginsat t= 0 and

the entire flow history is shown in Figure 3-18. In Stokes 2ndproblem, the solution given, Eq.

(3-111), is thefinal steady oscillationwhich occurs after the initial transient dies out. This steady

oscillation is independent of any initial condition. Varying the initial condition would cause a

different transient, which develops after that into the same steady oscillation, Eq. (3-111).

____________________________________________________________________________

3-60 The solution to Stokes 1stproblem, Eq. (3-107), was given without any ceremony. Let

/z0 in Eq. (3-105). Show that the similarity variable u/U0=f(), where =y/[2(t)],

reduces Eq. (3-105) to an ordinarydifferential equation whose solution is an error function.






11/48

11

Solution: Lets drop the prime and just call the velocity u. Substitute into the basic equation:

The constants (C,D) are found from f(0) = 1 and f() = 0, which leads to f = erfc():

____________________________________________________________________________

3-61 A belt moves upward at velocity V, dragging a film

of viscous liquid of thickness h, as in Fig. P3-61. Near

the belt, the film moves upward due to no-slip. At its

outer edge, the film moves downward due to gravity.

Assuming that the only non-zero velocity is v(x), with

zero shear stress at the outer film edge, derive a formula

For (a) v(x); (b) the average velocity Vavgin the film;

and (c) the velocity Vo for which there is no net flow

either up or down. (d) Sketch v(x) for case (c).





V

h const

,

, u

,v

Fig. P3-61

BELT

Air

g

+===+

=

=

=

=

=

=

DdCfCf

ff

tfU

y

fU

y

u

tfU

t

fU

t

fU

t

uooooo

)exp(:againIntegrate;)exp(':onceIntegrate

0'2'':upClean

)4

1('')

2('

22

2

2

2

2

.)(erfc)exp(2

1)(

0

2 AnsdxxfU

u

o

===


12/48

12

_____________________________________________________________________________________

3-62 Consider the following proposed incompressible flow field in cylindrical coordinates:

whereBand Care constants. (a) Determine if this distribution satisfies the continuityequation.

Does a stream function exist? (b) Does it satisfy theNavier-Stokes equationif gravity is

neglected? If so, find the pressure distribution p(r, , z).

Solution: (a) From Eq. (B-3) of Appendix B, the continuity equation is

Yes, continuity is satisfied by this distribution. But a simple stream function does notexist

because there are three, rather than two, velocity components. Ans.(a)

(b) Substitute these velocities into the steady (r, , z) momentum relations, Eqs. (B-6,7,8):

3)(

3)(

:)()(2)(:

2

2

2

ghVc

ghVVb

ccaseofsketchAdxg

xgh

VvaSolution

o

avg

=

=

+=

-0.50

-0.25

0.00

0.25

0.50

0.75

1.00

0.0 0.2 0.4 0.6 0.8 1.0x/h

v/ V

zBvrCvrBv zr 2/ ===

022)2()(1

])[(1

0)()(1

)(1

+=

+

+

==

+

+

BBBzzr

C

rBrr

rrv

zv

rvr

rr zr


13/48

13

After substitution of (vr, v,vz), these equations yield the three pressure gradients:

The Navier-Stokes equations are satisfied, and integration yields the pressure distribution:

______________________________________________________________________________

3-63 Consider low-speed laminar flow between two walls,

caused by natural convection. The walls are at (y = +h) and

(y = -h) respectively, as shown in the figure. The boundary

conditions are no-slip at each wall, T(x, -h) = To, and

T(x, +h) = To. Assume u= u(y), v= 0, T= T(y) andneglect viscous dissipation. Include buoyancy effects.

Find the velocity and temperature distributions.

Solution: For these simplifications, the momentum and energy equations reduce to

The energy equation integrates to T= a + by = 0 + Toy/h. Ans. Momentum becomes

)(1)(:)8(Eq.,momentum

)2

(1

)(:)7(Eq.,momentum

)2

(11

)(:)6(Eq.,momentum

2

22

2

22

22

zz

rr

rrr

vz

pvBz

r

vv

rv

p

rr

vvvB

v

rr

vv

r

pv

rvBr

+=

++

=+

+

=

V

V

V

zBz

pprB

r

p 22 402

=

=

=

).(constant)2( 222 bAnszrBp ++=

, u

, v

Fig. P3-63

g

h

, ,

( )

oo

p

meanmean

ThTThThu

dydu

dyTdk

dtdTc

Tdy

udTTg

dt

d

===

+=

=+=

)()(0)(:conditionsBoundary

termlasttheneglectweand)(0

casefor this0where,0

22

2

2

2


14/48

14

_____________________________________________________________________________

3-64 The height of the equilateral triangular duct in Fig. 3-9

is h = (31/2

/2)a 0.866a. In terms of heightinstead

of side length, investigate the following velocity

distribution for fully developed laminar flow:

where C is a constant.

(a) Determine if this distribution satisfies the no-slip condition in the duct. (b) If your answer to

part (a) is Yes, find the value of Cwhich satisfies the fully-developed-flowx-momentum

equation. (c) Show that the volume flow rate is proportional to h4and the pressure gradient.

Solution: (a) The formula requires that u= 0 at z = hand at y= z/3, which denotes the

walls of the duct. So the answer is Yes, no-slip on the duct walls.

(b) Substitute the formula for uinto the Navier-Stokes equation:

(c) The flow rate is found by integrating u. Do one half of the triangle and double it:

This is identical to Eq. (3-49) of the text, which is written in terms of a4(the side length).

___________________________________________________________________________

.obtaintoconditionsslip-notwotheApply

6:twiceIntegrate;

3

2

2

Ans

BAyy

h

Tgu

h

yTg

dy

ud oo

)yy(hh6Tgu 32o =

++==

0)3()(

22

=== wvz

yhzCu

ah

Fig. 3-9

).(4

3:

3

41)22()2

3

2(

2

2

2

2

bAnsdx

dp

hCor

hC

dx

dphzCz

hC

y

u

x

u

=

==+=

+

).(180

3

2039

4)

45(

39

4)

39

2()(2

33

)(2)

3

()()2(

45

0

453

0

3/0

2

0

32

0

3/

0

2

|

|)(

cAnsdx

dphhChzzCdzzhzC

dzyzy

hzCdzdyz

yhzCsidesdAuQ

hh

zhh z

=

=

=

=

====


15/48

15

3-65 A dashpot can be modeled as a cylinder moving

through a tube filled with oil, as in Fig. P1-33. The

clearance is small, V. Find an expression for the forceF

required to push the piston through the tube at velocity V,

if the resistance is due to pressure drop in the clearance.

HINT: Use the one-dimensional continuity relation as part of the solution.

Solution: The dashpot motion sweeps out a volume of fluid

which must reverse through the annulus at average velocity Vc:

If, as stated,R>> , then Vc>> V. Now simplify Eq. (3-51), laminar annular flow, for


16/48

16

Solution: Plug these numbers into one-dimensional continuity and laminar annular flow:

The simplified formula in Prob. 3-65,F 6LR 3V/ 3, gives F 3490 N.

____________________________________________________________________________

3-67 For some similarity-theory practice, try this problem, modified from a suggestion of

Clement Kleinstreuer,Engineering Fluid Dynamics, Cambridge University Press, 1997. Figure

P3-67 below shows a simplified model of laminar incompressible boundary layer convection.

The boundary layer thickness is constant and the velocity profile is simplified to be linear.

Boundary values U, Tw, and Teare constant. Dissipation is negligible. Balancex-directed

convection with conduction and find a suitable similarity variable. Solve as best you can and

plot the temperature profile and calculate the wall heat transfer. [HINT: The similarity variable

is proportional to y/x1/3

.]

.)08.0()169800(Finally,

800,169forSolve

])08.0/084.0ln()08.0084.0(08.0084.0)[

4.0(

8]

}/)ln{(}){()[(

8

/solve;])08.0()084.0[(])[()08.0()2.0(

22

22244

22244

222222

AnsRpF

Pap

pRR

RRRRLpQ

smVVRRVRVQ ccc

N3410

1.95

===

=

=+++

==+===

Tw

TeU

u(y)

T(x,y)

0

= constant

Fig. P3-67


17/48

17

Solution: The appropriate energy equation is

Its hard work, but its good practice to find similarity variables. The writer finds this:

This is a second-order linear differential equation. Integrating twice gives

To make () = 0, the constantD= 0. To make (0) = 1, the constant C = 1/[31/3(4/3)] =

0.7765. The final solution, then, is

The Nusselt number is Nu = h/k, where h = qw/T:

A plot of the temperature distribution is given below.

Uc

kB

y

TB

x

Ty

y

Uuy

T

kx

T

uc

p

p

=

=

=

=

where,:or

where,

2

2

2

2

1)0(,0)(areconditionsboundaryThe

0d

toleading,)(,)3(

2

2

2

3/1

==

=

+

==

d

d

dTT

TT

Bx

y

ew

e

+==

DdCC )3/exp(and)3/exp(' 33

.)3

exp(7765.03

AnsdTT

TT

ew

e

=

=

.)Re(Pr12.1)3(7765.0

)3

(7765.0)3(

7765.0

)(

)/(

)(

3/13/1

3/1

3/1

Ansxx

U

k

c

Uc

xk

BxkTT

yTk

kTT

qNu

p

pew

wall

ew

w

==

==

=

=


18/48

18

0.0

0.2

0.4

0.6

0.8

1.0

0 0.5 1 1.5 2

The boundary layer must be thick enough that > 2 aty = .

____________________________________________________________________________

3-68 SAE 30 oil at 20C (See Table A-3) flows through a smooth pipe of diameter 2 cm.

The pressure drop is 123 kPa/m. If the flow is fully developed, according to the writer, the

resulting flow rate will be 6 m3/h. If the pressure drop and fluid remain the same, determine the

size of a smooth equilateral triangle which would pass the same flow rate. Is the flow laminar?

Solution: For SAE 30 oil at 20C,= 891 kg/m3and= 0.29 kg/m-s. Check the Reynolds

number of the writers pipe-flow solution. Check the writers flow rate, also.

The flow rate checks and it is indeed laminar flow, ReD < 2000. The equilateral triangle is given

by Eq. (3-49), with athe side length of the triangle:

The equivalent triangle side length is slightly longer than the pipe diameter. The triangle area,

however, is 0.000369 m2, only 18% larger than the pipe area of 0.000314 m2. The triangle

hydraulic diameter is a/3 = 0.0168 m, and its Reynolds number is 234, also laminar. Ans.

_____________________________________________________________________________

)laminar(29.0

)02.0)(3.5(891Re;30.5;)02.0(

4001666.0

)OK(001666.0)000,123()/29.0(8

)01.0(

8

23

3344

326

6.0

======

==

==

VD

s

mVVm

s

mQ

h

m

s

m

m

Pa

smkg

m

L

prQ

D

opipe

.solve,)000,123()29.0(320

3

320

3

001666.0

443

Ansma

a

L

pa

s

m

Q triangle 0.0292==

==


19/48

19

Chapter 4. Laminar Boundary Layers

(Text problems in Chapter 4 are solved on pp. 88-137 of the Manual.)

4-56 For laminarboundary layer separation, a purely algebraic alternative to Thwaites

method was given by Stratford (1954). Define the local freestream pressure coefficient by

Then Stratfords inner-outer velocity-matching scheme predicts separation when

where xB is the position of the freestream velocity maximum. The formula is applied in the

decelerating region downstream ofxB. Test this criterion for the Howarth (1938) flow,

Eq. (4-114), and compare with the Thwaites separation estimate.

Solution: The Howarth flow is U= 1 x. Thus the pressure coefficient is

The position xB= 0, where the velocity is maximum, U= 1. Thus Stratfords formula yields

Compared to the exact value xsep = 0.120, Stratford is +1.7% high, not bad!

___________________________________________________________________________

PROPRIETARY MATERIAL. The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may bedisplayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or



equation)sBernoulli'from(1)( 2max

2

2max2/1

min

U

U

U

ppCp =

=

(1954)Stratford0104.0)()( 22 dx

dCCxx p

pB

xdx

dCxxxC

pp 22and2)1(1

22 ===

.:ror whateveerror&or trialEESoriterationafterSolution,

0104.0)22()2()0()()( 22222

Ansx

xxxxdx

dCCxx

sep

ppB

0.122=

==


20/48


21/48

21

So the duct only protrudes up about halfway into the boundary layer. We cannot just calculate

* and subtract it from h, because * represents the wholeboundary layer, not just half. No, we

need thestream function , which gives the exact volume flow up to any positiony:

This is only 42% of the volume flow that would enter if there were no boundary layer!

______________________________________________________________________________

4-59 As a variation of Prob. 4-58 above, if the geometry and air velocity are the same, find the

spacing hfor which the volume flow into the duct will be 0.017 m3/s per meter of plate width.

Solution: Problem 4-58 showed that the key to the problem is to find the Blasius stream

functions and f() which match this flow rate. Thus we calculate

The duct roof should be placed just below the boundary layer thickness of 0.0056 m._____________________________________________________________________________





.)804.0()]5.0)(6)(55.1(2[)(2Then

804.0)(Read:1-4Table;897.1)5.0)(55.1(2

/6)003.0(

2

2/1 AnsEfUx

fE

smm

x

Uy

m/sm0.007633 ===

=

==

.forsolveFinally,

)5.0)(55.1(20.6

20.3Read:1-4Table.792.1)(forSolve

/017.0)5.0)(/6)(/55.1(2)(2)( 32

Ans

Eh

xUyf

msmmsmsmEfUxf

m0.0047

===

===

h


22/48

22

4-60 Consider two-dimensional steady laminar boundary layer flow with zero pressure gradient

and constant boundary thickness. The velocity profile does not change shape, that is, u/U=

fcn(y/). Show that such a flow would be related to a uniformly permeable wallaty= 0.

Solution: Write out the steady-flow momentum integral relation, Eq. (4-121):

For zero pressure gradient, dU/dx= 0, and for an unchanging velocity profile shape, /is

constant. Therefore, since is given to be constant, is also constant, or d/dx= 0. The

momentum integral relation reduces to Cf /2 = - vw/U. Since Uis constant and the profile

shape, including its slope at the wall, does not change, it follows that wis constant also. We

then conclude that Cf = 2w/(U 2) is constant, thus it must be that vw = constant < 0. Our

final conclusion, therefore, is that this is flow with uniform wall suction aty= 0. We cannot

determine more details from the integral relations. The analytic solution of Navier-Stokes was

given as Prob. 2-16 of the text, and the velocity profile was found to be

___________________________________________________________________________

4-61 In Table 4-5, the laminar-flow test cases of Tani (1949) are missing a sixth power case.

Use the method of Thwaites to predict the separation point for U= Uo(1 x6).

Solution: We have to integrate U5and then calculate Thwaites parameter :

The integral is rather messy, too much algebra for this writer, but no trouble to evaluate

numerically. A plot of the computed (x) is shown below.

U

v

dx

dU

UH

dx

dC wf ++=

)2(2

0constantfor,)/

1(


23/48

23

-0.10

-0.08

-0.06

-0.04

-0.02

0.00

0 0.1 0.2 0.3 0.4 0.5 0.6x

Separation

According to Thwaites method, separation, = -0.09, occurs at xsep 0.553. Ans.

This is reasonable, lying between the 4th-power and 8th-power results in Table 4-5 of the text.

______________________________________________________________________________

4-62 Consider laminar flat plate flow with the following approximate velocity profile:

which satisfies the conditions u = 0 at y= 0 and u= 0.993U at y = . (a) Use this profile in

the two-dimensional momentum integral relation to evaluate the approximate boundary layer

thickness variation (x). Assume zero pressure gradient. (b) Now explain why your result in

part (a) is deplorably inaccurate compared to the exact Blasius solution.

Solution: (a) First evaluate the momentum thickness of this profile:

Now evaluate the approximate wall shear stress:

Introduce these two approximations into the momentum integral relation:

)]/5exp(1[ yUu

1.00987.0105

)11)(1()()1(105

51

0

51

0

=+=+==

eedee

yd

U

u

U

u

UUC

U

y

u wfyw

102;

5)(

20 =

= =


24/48

24

These are more than twice as highas they should bethe 14.2 should be about 5, and the 1.41

should be about 0.664. The discrepancy occurs because the guessed velocity profile does not

satisfy the important flat-plate boundary condition 2u/y2 = 0 at y= 0. The exponential

profile above has a highly negativecurvature at the wall, 2u/y2 = -25U 2/ 2. In other words,

theshapeof the exponential profile is not representative of flat plate flow.

_____________________________________________________________________________

4-63 A proposed approximate velocity profile for a boundary layer is a 4thorder polynomial:

(a) Does this profile satisfy the same three conditions as the 2nd

order polynomial of Eq. (4-11)

of the text? (b) What additionalcondition is satisfied by the 4thorder polynomial aty=

which is notsatisfied by the 2nd

order form? (c) What pressure gradient dp/dxis implied by this

profile? (d) Evaluate the displacement and momentum thicknesses as compared to . (e) Use

the 4thorder profile to evaluate the momentum integral relation for flat-plate flow, Cf=

2(d /dx), and show that the boundary layer thickness is approximated by

Solution: (a) Yes, this polynomial, first proposed by Pohlhausen (1921) in his pioneeringintegral theory, doessatisfy u(0) = 0, u() = U, and u/y() = 0. Ans.(a)

(b) This profile satisfies the additional smoothness condition2u/y

2() = 0. Ans.(b)

(c) The curvature at the wall is zero, that is, 2u/y2(0) = 0. From the boundary layer

x-momentum equation (4-35b), this means that U(dU/dx) = -dp/dx= 0. Ans.(c)

y

U

u=+ where,22 43

U

x

84.5

.Re

41.1and

Re

2.14:or,

7.202:Integrate

4.101:or,)0987.0(22

10

2Ans

xxU

x

dxU

ddx

d

dx

d

UC

xx

f

=

====


25/48

25

(d) Use the definitions of displacement and momentum thickness:

Now evaluate the wall shear stress and substitute into the momentum integral relation:

_____________________________________________________________________________

4-64 In Chapter 3 of the text we used the simple quadratic laminar flat plate profile, Eq. (4-11),

to make an integral analysis that had about a 10% error for thickness and friction:

Now use this approximate profile to estimate the velocity normalto the wall, v(x,y), assuming

two-dimensional steady laminar flow. Plot this (approximate) normal velocity profile and

compare its value aty = with the Blasius result, v(x, ) = 0.86/(Rex)1/2. HINT: You will need

to use the thickness estimate (x) from Eq. (4-14) to obtain v.

Solution: This is done by using the continuity equation to integrate for v:

).()()221()()1(*

).()()221)(22()()1(

431

0

1

0

431

0

431

0

dAnsdydU

u

dAnsdy

dU

u

U

u

10

3

315

37

=+==

=++==

).(;37

1260:Integrate;

37

630

:or),315

37(22

42;

2)(

2

20

eAnsU

xdx

Ud

dx

d

dx

d

UUC

U

y

u wfyw

U

x 5.84==

===

= =

2

2

2

yy

U

u

.)3

2(

30

2

1:integrateandCombine

30

2

1,

3014),-(4Eq.From

)22

()2(

3

3

2

2

0

3

2

22

2

Ansyy

Uxdy

x

uv

Uxdx

d

U

x

yy

dx

dyy

xx

u

y

v

y

=

=

+=

=


26/48

26

Aty= d, we obtain the approximation v(x, ) 0.91/(Rex)1/2

, or about 6% higher than Blasius.

The graph is as follows. The shapes are similar, the quadratic formula is a bit higher.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

Polynomial

Blasius

y/

v/URex1/2

_____________________________________________________________________________

4-65 Consider laminar boundary layer flow of airwith the Howarth (1938) linear freestream

velocity distribution, U/Uo= 1 x/L. As a text example of Thwaites method, we found

separation atx/L0.123. For this same adverse pressure gradient, use a similar integral method

to calculate the local heat transferfor a constant wall temperature, TwTe. Note what happens

at the separation point and comment.

Solution: The appropriate integral technique is the Smith-Spalding method of section 4-8.1.

For air, take Pr= 0.72. The method leads to the following relation for h(x) = qw/(Tw Te):

For air,Pr= 0.72, take a-1/2

= 0.296 and b= 2.88. The denominator is readily evaluated:

The final solution is as follows and is plotted below. Heat transfer does not separate. The

Howarth prediction is only slightly below flat-plate flow. Note the different shape of Cf(x).

x

b

o

b

o

o

L

dx

U

U

U

U

a

LU

khL

0

2/11

2/1

][ )()(/

/

]1)/1[(88.2

1)()( 88.2

0

1 = LxLdx

U

U

U

U x b

o

b

o


27/48

27

0.0

0.5

1.0

1.5

2.0

2.5

3.0

0.00 0.05 0.10 0.15 0.20 0.25 0.30

Howarth

Flat Plate

Cf unscaled

x/L

NuL/ReL

1/2

_____________________________________________________________________________





2/188.2

2/1

]1)/1[(

)88.2(296.0

/

/

LxLU

khL

o


28/48

28

Chapter 5. The Stability of Laminar Flows


5-32 For transition in accelerating pipe flow, fit the data of Fig. 5-37bto a power law and then

discuss how transition time varies with pipe diameter, acceleration, and kinematic viscosity.

Solution: The data in Fig. 5-37bare slightly scattered and wiggly, but they fit a power-law

Collecting terms, we find the following proportionalities for the transition time:

Transition time varies moderately with diameter and acceleration but only very slightly with the

kinematic viscosity.

_____________________________________________________________________________

5-33 Consider water at 20C filling a horizontal pipe of diameter 7 cm. If the water

accelerates from rest uniformly, with a constant acceleration of 2 gs, estimate (a) the time of

transition to turbulence, and (b) the Reynolds number at this time. Comment if appropriate.

Solution: For water at 20C take 1.0E-6 m2/s. Given a= 2g= 19.62 m/s2. The only data we

have are in Fig. 5-37b. The dimensionless diameter is

From Fig. 5-37b, the dimensionless transition time is approximately 52020. Then calculate

4.03/123/12 ])/([6.24)/( aDattr

.15/815/15/2 AnsaDttr

1890)/60.1(

/62.19)07.0()/(*

3/1

22

23/12 ][ =

==

smE

smmaDD


29/48

29

At this time, the average velocity would be U= a tt r = (19.62 m/s2)(0.72 s) = 14.0 m/s. Then

the diameter Reynolds number of the flow at this instant would be

This is much larger than the traditional pipe flow transition value ReD2000 and occurs

because the accelerating flow is very flat and far from a Poiseuille parabola distribution.

___________________________________________________________________________





).(hence,520727]/60.1

)/62.19([][ 3/1

2

223/1

2

aAnsttsmE

smt

at trtrtrtr s0.020.72=

=

).(/60.1

)07.0)(/0.14(Re

2 bAns

smE

msmUDD 30,000980,000

==


30/48

30

Chapter 6. Incompressible Turbulent Mean Flow


6-48 Let us revisit Prob. 6-45, which modifies Prob. 6-43 to be a cone.

The problem was: Consider a conical flat-walled

diffuser as in Fig. P6-8. Assume incompressible

flow with a one-dimensional freestream velocity

U(x) and entrance velocityUoatx=0. The entrance

diameter is W. Use two-dimensionaltheory.

Assume turbulent flow atx=0, with momentum thickness

o/W= 0.02,H(0) = 1.3, and UoW/= 105. Using the method of Head, Sect. 6-8.1.2, numerically

estimate the angle for which separation occurs at x = 1.5 W. [The answer was = 4.6.]

Now: Make this a parameter study. Experiments with conical diffusers show that, for a given

length (L/W= 1.5 here), the separation angle increases if inlet boundary layer thickness decreases

[see, e.g., White (2003), pp. 399-404]. Can Heads theory predict this effect? Calculate and plot

the separation angle sep for inlet momentum thicknesses in the range 0.0002 o/W0.05.

Solution: The theory doespredict that sepdecreases as o/Wdecreases, but it is not a large

effect. For small o/W, the given initial conditionH(0) = 1.3 is unrealistic and should be larger.

But we stuck with the given conditions, and our resulting plot is as follows:

4.0

4.5

5.0

5.5

6.0

6.5

7.0

0 0.01 0.02 0.03 0.04 0.05/W|o

sep

degrees

Inlet boundary-layer thickness vs. separation angle

for a conical diffuser of lengthL /W= 1.5

_____________________________________________________________________________

Uo

Fi . P6-48

W


31/48

31

6-49 For turbulent boundary layer methods that cannot calculate Cf0, some other criterion

is needed for predicting separation. The most common strategy is to assumeH3 at separation.

In a paper published after this 5thedition was put to bed, L. Castillo, X. Wang, and W. K.

George, Separation Criteria for Turbulent Boundary Layers Via Similarity Analysis,J. FluidsEngineering, vol. 126, May 2004, pp. 297-304, show, with both theory and experiment, that

turbulent separation occurs at a dimensionless pressure gradient parameter whose value is

where is the momentum thickness and U is the freestream velocity. Using the turbulent

two-dimensional steady momentum integral relation, show that the above criterion is equivalent

to a shape factor Hsep 2.76 0.23. Which criterion has less uncertainty?

Solution: The two-dimensional momentum integral relation, Eq. (6-28), contains a grouping

similar to , and we may rewrite it as follows:

Collect terms and note, at the separation point, that Cf = 0:

The uncertainty inHsep is 8%, while the uncertainty in is only 5%.

_____________________________________________________________________________





01.021.0/

/, =

dxd

dxdU

Usep

)()2(2

)2(dx

dH

dx

dC

dx

d

UH

dx

d f ++==++

.201.021.0

12

1

:or,separationat02

)]2(1[

AnsH

CH

dx

d

sep

f

0.232.76

=

=

==+


32/48

32

6-50 Show that, if the turbulent logarithmic law-of-the-wall is valid, then the two-dimensional

stream function (x,y), when nondimensionalized by kinematic viscosity, can be written entirely

in terms of wall-law variables u+andy

+. Find an analytic function for this stream function.

Solution: The streamwise velocity u= /y, so we may integrate at constantxto find :

We cant evaluate this at the wall,y+= 0, because of the logarithmic singularity. For actual flow

rate numbers, then, combine a linear sublayer, u+= y

+, up toy

+ = 10.8, after which the log-law

applies. Then we can apply limits to the integration, withB= 5.0 and = 0.41:

This formula should give the volume flow rate in a log-law turbulent boundary layer.

____________________________________________________________________________

6-51 Let us revisit the boundary layer displacement

problem when a flow enters a duct near a wall,

similar to Prob. 4-58. The flow is two-dimensional,

with U= 40 m/s of air at 20C and 1 atm. The duct

entrance is 1.5 m downstream of a flat plate tip.

If the duct height is h= 13 mm, estimate the

volume flow of air into the duct per unit

width into the paper. Assume the flow is

turbulent from the leading edge onward.

.constant)1

()ln(1

])ln(1

[ }{ AnsyByydyBydyudyu ++=+=== +++++++

.)8.10(32)1

()ln(1

:or

66.2768.62)1

()ln(1

32.58])ln(1

[

8.10

8.10

0

AnsyyByy

yByydyBydyy

y

>+

++=++=

++++

+++++++

+

= 0

U = 40 m/s

= 1.5 m

DUCT

(x)

h

Fig. P6-5


33/48

33

Solution: First check the Reynolds number at the duct entrance, taking air at 20C and 1 atm to

have a kinematic viscosity = 1.5E-5 m2/s and a density = 1.2 kg/m3:

This is about 5 times thicker than the laminar boundary layer in Prob. 4-58. To find the flow rate

up to a height y= 13 mm, you can integrate the log-law numerically, or you can use the nice

formula derived in Prob. 6-50 for a linear sublayer combined with a logarithmic overlap:

From Eq. (6-88), Cf 0.455/ln2(Rex) = 0.455/ln

2(4E6) = 0.00197, whence w=CfU

2/2 =

0.00197(1.2kg/m3)(40m/s)2/2 = 1.89 Pa, and then v* = (w/)1/2 = (1.89/1.2)1/2 = 1.26 m/s.

With v* known, the roof of the duct is at a dimensionless wall-law heighty+=yv*/=

(0.013m)(1.26m/s)/(1.5E-5 m2/s) = 1090. The nice formula above then predict a stream function

If there were no boundary layer, the flow into the duct would be Uh= (40)(0.013) =

0.52 m3/s-m, or about 62% more.

____________________________________________________________________________


displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, orused beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course


mmE

x

turbulentsmE

msmUxmx

x

x

0274.0)5.1()64(

16.0

Re

16.0Calculate

)flow(000,000,4/55.1

)5.1)(/40(Re,5.1At

7/17/1

2

==

====

32)1

()ln(1

+ +++ yByy

.)55.1)(21350(isducttheintorateflowThe

2135032)1090)(41.010.5()1090ln()1090(

41.01

AnsE m/sm0.323 ==

+=


34/48

34

6-52 As a variation of Prob. 6-51 above, if the geometry and air velocity are the same, find the

spacing hfor which the volume flow into the duct will be 0.5m3/s per meter of plate width.

Solution: This time we are looking for a value of = Q/ = (0.5 m3/s-m)/(1.5E-5 m2/s) =

33,333 (dimensionless). From Prob. 6-51, we calculated the local friction velocity as v* = 1.26

m/s. Thus we solve fory+in our nice formula from Prob. 6-50:

This is still well within the boundary layer, which is 0.0274 m thick from Prob. 6-51.

_____________________________________________________________________________

6.53 Water, at 1 atm and 20C, flows through a 30-cm

smooth square duct at 0.4 m3/s as in Fig. P6-53.

Fifty thin flat plates of chord-length 3 cm are stretched

across the duct at random positions. Their wakes do not

interfere with each other. (a) How much additional

pressure drop do these plates contribute to the duct flow loss?

(b) By what per cent do the plates increase the pressure drop over the 1-meter length?

Solution: For water at 20C and 1 atm,= 998 kg/m3and= 0.0015 kg/(m-s). First find the

average velocity through the duct from one-dimensional continuity:

Check the Reynolds number: ReDh = (998)(4.44)(0.3)/(0.001) 1,330,000. Thus the duct

flow is turbulent, and we will assume (why not?) that each tiny plate sees this average approach

./26.1

/55.1(1620

*1620Then

*1620forSolve;333,3332)

1()ln(

1

2

Anssm

smE

vh

hvyyByy

m0.0193

==

==+ ++++

Fig. P6-53

50 plates

s

m

m

sm

A

QV 44.4

)3.0(

/4.02

3

===


35/48

35

velocity V = 4.44 m/s. Theplate lengthReynolds number is only ReL = 133,000, low enough

that the flow over the plates is laminarand nottripped by the stream turbulence.

Calculate the drag of one plate from the Blasius formula:

(a) The total drag of 50 plates is thus approximately Ftotal = (50)(0.645 N) = 32 N. A freebody

of the plates shows that this force can only be balanced by an overall pressure drop across the

array of plates:

(b) Without the plates, the pressure drop would be solely due to the Moody wall-friction loss:

The plates increase the pressure drop by a factor of ten, or 1000% more! Ans.(b).

______________________________________________________________________________

6-54 A model two-dimensional hydrofoil has the following theoretical potential-flow surface

velocities on its upper surface at a small angle of attack:

x/C 0.0 0.025 0.05 0.1 0.2 0.3 0.4 0.6 0.8 1.0

V/U 0.0 0.97 1.23 1.28 1.29 1.29 1.24 1.14 0.99 0.82

The stream velocity U is varied to set the Reynolds number. The chord length Cis 40 cm.

The fluid is water at 20C and 1 atm. Assume that x is a good approximation to the arc length

along the upper surface. Using any turbulent-boundary-layer method of your choice, find the

predicted separation point, if any, forReC = 6E6. For simplicity, assume turbulent flow from

the leading edge onward.

NmmsidesbLVCF

C

Dplate

L

DL

645.0)2)(03.0)(3.0()44.4)(2

998(00364.0)2(

2

00364.0133000

328.1

Re

328.1,000,133Re

221 ===

====

).()3.0(

322

aAnsm

N

A

Fp

duct

totalplates Pa358==

Pam

mV

D

LpMoody

DhDh

36==

==

22 )44.4)(2

998)(

3.0

1)(00111.0(

2

00111.0Solve;000,330,1Re,)51.2

Relog(0.2

1


36/48


37/48

37

Separation is predicted by Heads method at approximately x/C 0.65. Ans.

NOTE: For a laminar boundary, Prob. 4-50, separation was predicted atx/C 0.45. Since the

adverse gradient does not start untilx/C= 0.3, the turbulent boundary layer is nearly three times

as resistant to separation as laminar flow.

______________________________________________________________________--

6-55 Air at 20C issues from a 2-cm-diameter nozzle at a near-uniform velocity of 20 m/s. At

sectionx, a considerable distance downstream, measurements of the turbulent jet velocity

distribution have been fit to the following formula:

(a) Evaluate the jet momentum at sectionxand compare with the nozzle momentum. (b)

Determine the volume flow of air entrained into the jet, if any, between the nozzle and sectionx.

(c) Estimate the distance that sectionxis downstream of the nozzle.

Solution: The jet Mach number is less than 0.06, thus the flow is incompressible. So we can

skip density and evaluate the momentum in the form J/= u2 dA. At the inlet and the exit,

The factor of 106occurs because the radius rwas given in millimeters. We see that the jet

momentum remains nearly constant, as predicted for all free jets in Eq. (6-144). Ans.(a)

(b) Now evaluate the volume flow at the nozzle and in the jet:

mminwith,)00104.0exp()/1.9( 2 rrsmu

24

6

222

0

2

242222

/)10)(00104.0(2

)1.9(2)]00104.0(exp1.9[|/

/)01.0()/20(|/

smdrrrdAuJ

smmsmRUJ

jet

nozzle

0.125

0.126

====

===

).(4.3/0063.00275.0flowvolumeentrainedDifference

/0275.0)10(00104.0

1.92)00104.0exp(1.9

/0063.0)01.0()/20(

3

3

60

2

322

bAnsQsm

smdrrrdAuQ

smmsmRUQ

nozzle

jet

nozzle

===

====

===

0.0212


38/48

38

The entrainment is a dominant part of the flow. (c) The Grtler jet theory, Eq. (6-152), predicts

x= 7.4(J/)1/2/Umax= 7.4(0.125)1/2

/9.1 287 mm. Fitting Grtlers profile formula (6-152) to

the data yields x 313 mm. Fitting the sech2profile formula (6-153) to the data yields x

333 mm. These are empirical theories, not perfect analyses. The best we can say is that section

xis approximately 300 mmdownstream of the nozzle, or about 15 nozzle diameters. Ans.(c)

_____________________________________________________________________________

6-56 Theodore von Krmn, in NACA Technical Memorandum No. 611, 1931, used a

similarity approach to derive the following model for turbulent eddy viscosity:

(a) If wnear the wall, show that this model leads to the logarithmic overlap law (6-38a).

(b) In fully developed pipe flow, the shear stress is linear across the duct:

whereRis the pipe radius andyis measured from the wall (y = 0) to the centerline (y = R).Krmn (1931) applied his model above to pipe flow and integrated to the following result:

Plot this formula versus y/Rfor the special case Umax/v* = 25.0 and compare it with the

logarithmic law for the same value of Umax/v*. Comment if appropriate. (c) Discuss some

possible reasons why Krmns model and his pipe-flow result have not become popular.

Solution: (a) Assuming constant near-wall shear stress makes the model easily integrable:

41.0and)/(

)/(where,

222

32 ==

dyud

dyud

dy

udtt

)/1( Rywflowpipe =

RyZZZv

Uu /1where,)]1ln([*

max =++=

).()ln(*

:againIntegrate;*

:onceIntegrate;)(*

:or,)/(

)/( 22

2

222

42

aAnsconstyv

uy

v

dy

ud

dy

ud

vdy

ud

dyud

dyudw

+==

==


39/48

39

(b) If Umax/v* = 25.0, thenR+= exp[0.41(25-5.0)] = 3641, so we have to formy/R=y

+/3641 to

make the plot of the log-law and compare with Krmns model as shown below. The two

velocity profiles are quite similar, so both formulas can be considered a success.(c) Krmns model is too complex algebraically. Applying 4

thand 2

ndpowers of derivatives to

general turbulent shear flows is too involved. The pipe-flow result is difficult to place in

Reynolds number or friction factor form. The present writer cannot even integrate the pipe-flow

formula to find the average velocity. Better models are available and popular.

0

5

10

15

20

25

0 0.2 0.4 0.6 0.8 1

Karman

log law

Plot made for Umax+

= 25.

/R

u+

______________________________________________________________________________

6-57 Consider fully developed turbulent flow in a pipe of radiusR. Recall that the shear stress

varies linearly from zero at the centerline to a maximum, w, at the wall. Assume that the log-

law holds all the way across the pipe and (a) find an expression for the variation of eddy

viscosity tacross the pipe, in terms of the coordinateymeasured from the wall inward. (b) Also

find the position of maximum eddy viscosity.


40/48

40

Solution: From the log-law, evaluate the velocity gradient and then set equal to the shear stress:

This simple theory indicates that the eddy viscosity in a pipe is a parabola going from zero at the

wall to zero at the centerline. Differentiate to find the maximum eddy viscosity:

______________________________________________________________________________

6-58 Water at 20C flows through a smooth pipe of diameter 2 cm. The pressure drop is 123

kPa/m. If the flow is fully developed, according to the writer, the resulting flow rate will be 21.3

m3/h. If the pressure drop and fluid remain the same, determine the size of a smooth equilateral

triangle that would pass the same flow rate. Is the flow turbulent?

Solution: For water at 20C,= 998 kg/m3and= 0.0010 kg/m-s. Check the Reynolds

number of the writers pipe-flow solution. Check the writers flow rate, also, for asmoothpipe:

The flow rate checks and it is indeed turbulent flow, ReD>> 4000. The proposed equivalent

equilateral triangle has a side length a. Its hydraulic diameter isDh= 4A/P= a/3.

)(.)(*Solve;*

)1(*

*;)

*ln(

1*/

22 aAns

R

yyv

y

v

dy

du

R

yv

y

v

dy

duB

yvvu

ttt ====

=+=

RvbAnsR

yatR

yv

dy

dt

t *4

|;).(2

0)2

1(* max

====

)checksallitYes,()8.18)(2

998(

02.0

0139.0

2

0139.0;)51.2

376000log(0.2

1;000,376

001.0

)02.0)(8.18)(998(Re

8.18,)02.0(4

00592.03600

3.21

22

23

m

PaV

DL

pThen

SolveVD

s

mVthenVm

s

mQIf

D

123,000==

=

=

=

===

====


41/48

41

The equivalent triangle side length is slightly longer than the pipe diameter. The triangle area,

however, is 0.000356 m2, only 13% larger than the pipe area of 0.000314 m2. The triangle

hydraulic diameter is a/3 = 0.0166 m, and its Reynolds number is 275,000, also turbulent. Ans.

_____________________________________________________________________________





.solveIterate,1230002

)51.2

Relog(2

1;Re;

33

4;

30cos5.0

/00593.0

2

2

3

Ansmam

PaVDL

p

DVa

a

AD

a

sm

A

QV

triangleh

DhhDhh

0.0287==

=

=

=====

o


42/48

42

Chapter 7. Compressible Boundary Layer Flow


7-28 Analyze steady compressibleCouette flow between an upper moving plate and a lower

fixed plate, as shown in Fig. P7-28. There is no pressure gradient, and the variables (, u, T)

vary withyonly. Assume constant cp,Pr, andg. Viscosity and thermal conductivity may vary

with temperature, but their ratio /kis assumed constant. Neglect . (a) Set up continuity,

momentum, and energy and carry out the solution as far as you can a complete solution is not

possible because the variation(T) is not known. (b) Find an expression for the heat transfer at

the lower wall. (c) Examine the special case of an adiabatic lower wall.

Solution: For the assumed parallel flow, u = u(y), v = w = 0. Continuity reduces to the

triviality that /x(u) = 0. (a) Thex-momentum equation becomes

As stated, we need to know the variation of (T) before we could complete the integration.

(a, Part 2) The energy equation reduces to

H

T = Te, u = U

T = Tw, u = 0

Fig. P7-28

).(:Integrate

constant:or,0)(0

0

aAnsdy

u

dy

du

yy

u

yx

p

Dt

Du

y

ew

=

====

+=

+

==


43/48

43

Divide byand integrate again from the lower wall upward, assuming constant /k:

This gives Tin terms of u, and in principle we know ufromAns.(a) above.

(b) ApplyAns.(a) above at the upper wall, u= Ue, noting that k/ cp/Pr. Solve for qw:

(c) If the lower wall is adiabatic, qw= 0, and the only way this can happen is if

This a duct-flow example of the recovery temperaturethat was studied for boundary layers in

Chapter 7 of the text.

____________________________________________________________________________

PROPRIETARY MATERIAL. The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may bedisplayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or



w

p

qudy

dTk

dy

du

dy

dTk

dy

d

y

u

y

Tk

yDt

DTc

==+

+=

+

==

constant:Integrate

)()()(0 2

).(2

)(2

aAnsuq

yqu

TTk ww

w

==+

).()2

(2

bAnsc

UPrTTUPrcq

p

eew

e

pw =

).(2

2

cAnsc

UPrTTT

p

eeaww +==


44/48

44

7-29 Air,Pr= 0.71, flows at supersonic speed on a flat plate and encounters a double-wedge

configuration as in Fig. P7-29. The boundary layers are laminar. Compare the adiabatic wall

temperature in region 3 with the corresponding value in region 1 and comment cogently.

Solution: Both wedges cause oblique shock waves, shown above as heavy dashed lines. We

have to work our way through to region 3 using oblique shock theory [see, for example, White

(2003), Chap. 9]. Express the temperatures in absolute units, e.g., T1= 273 K. The recovery

factor for laminar constant-pressure flow is rPr1/2= 0.843. The calculated results are:

The adiabatic wall temperature slowly rises through the shock waves, due to additional

dissipationinside the shocks. The maximum increased dissipation would be caused by a normal

shock, where Tawdownstream would be 942 K for the present data.

_____________________________________________________________________________





15

18

Fig. P7-29

Ma1= 3.5

T1= 0C

(2)

(3)

.;7.566;819.1

19.29;18:theoryShock:3Region857;5.399;605.2

13.39;15:theoryShock:2Region

.])5.3)(2.0(843.01)[273()2

11(:1Region

333

222

22111

AnsTKTMa

anglewaveangleDeflectionKTKTMa

anglewaveangleDeflection

AnsKMarTT

aw

aw

aw

K883

K837

===

=====

==

=+=

+=

oo

oo


45/48


46/48

46

_____________________________________________________________________________

7-31 Supersonic turbulentflow of air,Pr= 0.71,

passes through a Prandtl-Meyer expansion fan, as (1)

in Fig. P7-31. Approach conditions are Ma1= 3.0,

p1= 150 kPa, and T1= 120C. Assume steady isentropicsupersonic flow through the fan [see, e.g., White (2003), Chap. 9].

(a) Estimate the adiabatic wall temperaturein region (2).

(b) If the actual wall temperature in region 2 is Tw2 = 100C, estimate the wall heat transferat a

point where the local Reynolds number is 4E7. [NOTE: If the algebra in part (b) is tiring, make

your estimate from the graphs in the text in Chap. 7.]

Solution: (a) This is a straightforward application of Prandtl-Meyer theory, plus a turbulent

recovery factor. The Prandtl-Meyer angle is a function of Mach number:

Follow through with these calculated results:

).()]/ln(2/[

|:2to1fromIntegrate

02)/(

2

21

22

21

2

bAnsppDLRT

pp

A

m

p

dp

D

dx

RTAm

dpp

flowisothermal +

=

=+

&

&

20

(2)

Fan

Fig. P7-31

1

1where,])1[(tan])

1[(tan 2/1212/1

212/1

+

=

=

KMa

K

MaK

).(])318.4)(2.0(892.01)[7.232()2

11(

892.0)71.0(Pr

7.232])318.4(2.01/[)1100()2

1

1/(

1100])0.3(2.01)[393()2

11(

318.4,76.6920,76.49,0.3

22122

3/13/1

22

222

222

111

21211

aAnsKMarTT

r

KKMaTT

TKKMaTT

MaMa

aw

turb

o

oo

K1007=+=

+=

===+=

+=

==+=

+=

==+===

ooo


47/48


48/48

Solution: Actually, increasing the wedge half-angle increasesthe heat transfer, for two reasons:

(1) the adiabatic wall temperature increases, hence Tincreases; and (2) the density increases

sharply inside the shock but velocity and Stanton number do not change much. Recall the data

(see Prob. 7-14 on page 230 of the Solutions Manual): approach conditions Ma1= 3.5,p1= 30

kPa, T1= 20C, and1 = 0.357 kg/m3. The wedge wall temperature is 300K. For a flat plate of

length 19.32 cm, ReL= 4.6E6, CH= 0.000363, U2= U1= 1201 m/s, since there is no shock. The

laminar adiabatic wall temperature is 898K. Then the mean heat transfer to the plate is:

This is 43% less than the heat transfer at a half-angle of 15. Here is a plot for various angles:

0

50000

100000

150000

200000

250000

0 5 10 15 20 25 30 35

Wedge half-angle, degrees

Heattransfer,W/m^2

Problem 7-14 Extended

163,000 at 15 degrees

The heat transfer increases steadily and levels off near the detachment angle of 36.5.

___________________________________________________________________________




.)300898)(1005)(1201)(357.0)(00363.0()(2,

Ansm

WTTcUCq wawpeeeHmeanw 93,000===

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