Viscous Fluid Flow Solution Manual (2022) [4 ed.] 1260597806, 9781260597806

Table of contents :
White_VFF_4e_Chap001_ISM
White_VFF_4e_Chap002_ISM
White_VFF_4e_Chap003_ISM
White_VFF_4e_Chap004_ISM
White_VFF_4e_Chap005_ISM
White_VFF_4e_Chap006_ISM

Citation preview

PROBLEM SOLUTIONS CHAPTER 1. PRELIMINARY CONCEPTS

1-1 A 1.4-cm diameter sphere placed in a freestream at 18 m/s at 20°C and 1 atm. Compute the diameter Reynolds number for 3 cases: (a) Air: Table A-2 - at 20°C,  = 1.205 kg/m3 ,  = 1.81 E-5 Pa-s. Then

Re D = VD/ =

(1.205)(18 )( 0.014 ) = 16, 800

(Ans.)

1.81E-5

(b) Water: Table A-1 - at 20°C,  = 998 kg/m3 ,  = 1.002 mPa-s:

ReD = ( 998)(18)( 0.014) / ( 0.001002) = 251,000

(Ans.)

(c) Hydrogen: Table A-3, M = 2.016, then R = 8313/M = 4124 m2 /s 2 -K. Thus estimate  = p/RT = (101350 ) / ( 4124 )( 293) = 0.0838 kg/m3. From Table 1-2 for hydrogen,

  o ( T/To ) = (8.411E-6)( 293/273) n

068

= 8.83 E-6 Pa-s

Then ReD = ( 0.0838)(18)( 0.014) / (8.83 E-6) = 2,400

(Ans.)

At what wind velocity will an 8-mm-diameter wire “sing” at middle C (256 Hz)? For air at 20°C, assume v  1.5E-5 m 2/s. From Fig. 1-8 guess a vortex-shedding Strouhal number of 0.2 [check the Reynolds number afterward]. Then fD/U  0.2 = ( 256 )( 0.008) /U, or U  10.24 m/s. At this speed the Reynolds number is 1-2

ReD = UD/v = (10.24)( 0.008) / 1.5E-5 = 5400. This is nicely in the range where fD/U = 0.2. Perhaps we could iterate just a little more closely to obtain

fD/U  0.205, Re = UD/v  5300, or U = 10.0 m/s

(Ans.)

1-3 If U = 12 m/s in Prob. 1-2 above, what is the wire drag in N/m? For air assume  = 1.205 kg/m3 and v = 1.5E-5 m2/s. The Reynolds number is

ReD = UD/v = (12)( 0.008) / ( l.5E-5) = 6400

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From Fig. 1-9 at this Reynolds number, estimate a drag coefficient of 1.1. Then 2 1 Fdrag = CD   V2 ( DL ) = 1.1( 0.5)(1.205)(12 ) ( 0.008)(1.0 ) = 0.76 N /m 2

(Ans.)

1-4 Given, without proof, the Poiseuille-paraboloid laminar-pipe-flow formula from Chap. 3, u = (C/)(R 2 − r 2 ), find the wall shear stress if u max = 30 m/s, D = 1 cm, and  = 0.3 kg/(m-s). [The exact analysis will be given in Sect. 3-3.1.] Examining the formula, we see that the maximum velocity occurs on the centerline:

(

u mx = u ( r = 0 ) = CR 2 / = 30 m/s = C ( 0.005 ) / ( 0.3) , or: C = 3.6E5 N/ m 2 -s 2 2

)

With C thus known for this data, we may evaluate wall shear stress by differentiation:

wall = 

u r

r =0

 2RC  =   = 2RC = 2 ( 0.005 )( 3.6E5 ) = 3600 Pa   

(Ans.)

We should check the Reynolds number Re D but we don’t know the density. But “oil” is usually in the range   900 kg/m3. Then ReD = u max D/ = ( 900 )(30 )( 0.01) / ( 0.3)  900, which is well within the laminar-flow range.

1-5

Glycerin at 20 C is confined between two large parallel plates. One plate is fixed and the other moves parallel at 17 mm/s . The distance between the plates is 3 mm . Assuming no-slip, estimate the shear stress in the glycerin, in Pa.

Solution: Glycerin at 20 C is confined between two large parallel plates. One plate is fixed and the other moves parallel at V = 17 mm/s . The distance h between the plates is 3 mm. u=V Moving plate h Glycerin u=0 Fixed plate

For glycerin at 20 C , the viscosity  = 1.5 kg/m  s . Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -2-

Assuming no-slip, the shear stress  =

1-6

V h

=

(1.5 kg/m  s ) (17  10−3 m/s )

( 3 10

−3

m/s

)

= 8.5 Pa .

(Ans.)

Given a plane unsteady viscous flow in polar coordinates:

v r = 0; v =

 r 2  C 1 − exp   −   r   4vt  

Compute the vorticity and sketch some profiles of vorticity and velocity. From Appendix B, the vorticity is

 r2  1 C z = ( rv ) = exp  −  r r 2vt  4vt  The instantaneous velocity and vorticity profiles are plotted at top. At t = 0, the flow is a “line” vortex, irrotational everywhere except at the origin (  =  ) .

1-7

Given the two-dimensional unsteady flow u = x/ (1+t ) , v = y/ (1+2t ) , find the equation

for the streamlines which pass through the point (x 0 , y0 ) at time ( t = 0 ) . From the geometric requirement for two-dimensional streamlines at any instant,

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dy v y (1 + t ) = = dx streamline u x (1 + 2t )

Holding time t constant, we may separate the variables and integrate to obtain y = C x n , where

n = (1 + t ) / (1 + 2t ) . To satisfy the initial condition, we must have C = y0 / ( x0 ) . The final result for the (unsteady) streamlines is n

n

y  x  1+ t =  , n = y0  x 0  1 + 2t

(Ans.)

1-8 For the inviscid streamline approaching the forward stagnation point of the cylinder in Fig. 1-5, evaluate the strain rates and the time to go from ( 2R, ) to ( R, )

(

)

(

)

From Eqs. (1-2), v r = U  1 − R 2 /r 2 cosθ, v = − U  1 + R 2 /r 2 sinθ Then, from Appendix B, evaluate the normal and shear-strain rates along the line  = :

vr 2U R 2 cos  2U R 2 rr = = |= = − r r3 r3 

2U R 2 cos  2U R 2 1 v vr = + =− |= = + r  r r3 r3

r =

1 vr v v 4U R 2 sin  + − = |= = 0 r  r r r3

(Ans. a)

The particle moving toward the stagnation point gets shorter in the “r” direction and fatter by the same amount in the “  ” direction, thus maintaining constant volume for this incompressible flow. The shear strain rate is zero because we are on a line of symmetry. For part (b), by definition, the radial velocity along the stagnation line (  =  ) is

vr =

(

dr = − U 1 − R 2 /r 2 dt

)

We may separate the variables and integrate to find the time of travel between ( 2R ) and ( R ) : R

− U t =

R

 R  r − R   r 2 − R 2 = r + 2 ln  r + R  = − 2R 2R r 2dr

(Ans. b)

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It takes infinite time to actually reach the stagnation point, where V = 0 .

1-9

Use the approximate equation of state for water,

p/po  ( A + 1) ( /o ) − A n

with A  3000, n = 7

to compute the following quantities for water, po = 1 atm, o = 998 kg/m3 : (a) the pressure required to double the density of water:

p/po = ( 3000 + 1)( 2.0) − 3000 = 381128, or: p = 381,000 atm 7

(Ans. a)

(b) the bulk modulus K of water at 1 atm. By definition,

K =

dp nn −1 |T = po ( A + 1) n = npo ( A + 1) = 21007 po at 1 atm. d o

Thus the bulk modulus is K  21,007 atm  2.13 E9 Pa

(Ans. b)

(c) the speed of sound at 1 atm: a 1 atm = ( K /o )

1/2

1/2

 2.13 E9 Pa  =  3   998 kg/m 

= 1460 m /s

(Ans. c)

These are accurate estimates of the measured compressibility and sound speed of water.

1-10 As shown, a plate slides down an incline on a film of oil of viscosity  = 5E-4 slug/ft-s

(a) Estimate the terminal sliding velocity V*: Acceleration is zero, so

W sinθ = A = 

V A y

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where A is the plate area touching the oil film. Assuming a linear velocity profile, V = V* and y = the film thickness, hence

 V*  V W sinθ = 40sin ( 30 ) =  A = ( 0.0005 )  9  0.005/12  ( ) y   in English units.

Hence solve for V* ( terminal ) = 1.85 ft /s

(Ans. a)

(b) Estimate the time for the plate to accelerate from rest to 99% terminal velocity: If x is down the incline, then a dynamic force balance gives

V

 Fx = W sinθ −  y A = or:

W dV , g dt

dV  gA  +  V = g sinθ dt  W y 

The solution to this first-order linear ordinary differential equation is

  gA   4.605W y V = V* 1 − exp  − t   = 0.99V* if t* = gA  Wy    For our data, then, the time to reach 99% of terminal velocity is t* =

1-11

4.605 ( 40 )( 0.005/12 ) = 0.53 sec ( 32.2)( 0.0005)( 9.0)

(Ans. b)

Estimate the viscosity of nitrogen at 86 MPa and 49°C. From Appendix A-3, for N 2 , read

Tc = 226R = 126K, pc = 33.5 atm, c = 18.0 E-6 Pa-s. At this high pressure, we cannot use “low density” formulas but rather must use Fig. 1-17. Compute ratios: T 49 + 273 p 86E6  = = 2.55; = = 25.3; Read  2.5  0.1 Tc 126 pc 33.5 (101350 ) c Then our estimate is  = 2.5 c = 2.5 (18.0) = 45 ± 2 μPa-s

(Ans.)

The agreement with the measured value (also 45 Pa-s ) is excellent. 1-12 Estimate the thermal conductivity of helium at 420°C and 1 atm. This is truly “lowdensity”, since p  pc and T  Tc . A power-law estimate would be based on 0°C:

k  k o ( T/To )

n

 420 + 273   ( 0.142W/m-K )    273 

072

 0.278 W /m-K

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Alternately, we could use the kinetic theory formula, Eq. (1-41). From Appendix A-5 for helium,  = 2.551 Å, T = 10.22K, and M = 4.003. First use Eq. (1-34) to compute

 420 + 273  v  1.147    10.22 

−0.145

.20

 420 + 273  + + 0.5   10.22 

= 0.6226

[check with Table 1-1]

Our estimate from (1-41) then is

k=

0.0833 T 2  v

M

=

( 693) = 0.27 W /m-K ( 2.551)2 ( 0.6226 ) 4.003 0.0833

The agreement with the experimental value of 0.28 W/m-K is good for both estimates.

1-13

According to Table C-5 and Fig. 1-15, at what pressure is the viscosity of CO2 equal to approximately 30 10−5 Pa  s when the temperature is 800°R ?

Solution: T = 800°R = 444.4444 K , Tc = 548°R = 304.4444 K , Tr =

 = 30 10−5 Pa  s , c = 3.43  10−5 Pa  s ,  r =

 c

=

T 444.4444 K = = 1.46 Tc 304.4444 K

30 10−5 Pa  s = 8.75 34.3 10−6 Pa  s

Thus, pr = 25 (from Fig. 1-15). Then, pressure p = pc pr = ( 72.9 atm)  25 = 1822.5 atm .

(Ans.)

1-14 Fit the given viscosity-vs-temperature data for ammonia gas to power-law and Sutherlandlaw formulas. (a) The power-law is an excellent fit to this data. Taking To = 300K, we obtain, by least-squares to a log (  ) vs. log ( T ) plot,

1.051

 T   o  To 

 0.3% for To = 300K

(Ans. a)

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(b) The Sutherland-law is not an especially good fit to the data, which has  rising with T at an increasing rate. It may be fit to least squares by minimizing the functional

( ) (

 * 3/2 T 1 + S* i  *  i − T* + S* i =1  i  6

)

2

  * T * S  *  where  =  , T = T ,S = T o o o  

for the given six data points. The minimum is found by differentiating the functional with respect to S* , v with the result S* = 1.91, or:

Sbest fit  573°K

(Ans. b)

The error is 2.4%, or eight rimes more than the power-law fit.

1-15

Experimental data for the viscosity of helium at low pressure are as follows: T, °C

0

100

200

300

400

500

µ, Pa·s 1.87 × 10–5 2.32 × 10–5 2.73 × 10–5 3.12 × 10–5 3.48 × 10–5 3.48 × 10–5 Fit these values to a suitable formula. Solution: Experimental data for the viscosity of helium in Kelvin scale at low pressure are as follows: T, K

273

373

473

573

673

773

µ, Pa·s 1.87 × 10–5 2.32 × 10–5 2.73 × 10–5 3.12 × 10–5 3.48 × 10–5 3.48 × 10–5 n

 T  =  . Using power law curve,  0  T0 

For 0 = 1.87  10−5 Pa  s and T0 = 273 K T 373 K 473 K 573 K 673 K 773 K n 0.691 Therefore, the mean value

 n=

0.688 5

n

i =1 i

5

0.69

0.688

0.597

= 0.671 ( 4 % accuracy for 250 K  T  1000 K ).

(Ans.)

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1-16 Analyze newtonian flow between parallel plates (Fig. 1-15) with a finite slip velocity u = l ( du/dy ) at both walls. The velocity profile is still linear, but with slip at both walls the slope is less, as shown in the sketch:

du/dy = ( V − 2u ) /h

Introducing u from the slip relation, we obtain

du V V = or:  w = at both walls dy h + 2l h + 2l

(Ans.)

1-17 Derive Eq. (1-106) from a balance of forces on the differential surface-area alement shown in the problem. Since the sliver of area is negligibly thin, it has no weight. The pressures act on a projected surface area dSx dSy . The surface tension forces are slanted slightly upward, at angles

(dx /2) and (dy /2), respectively. The force balance is

(

)

2T dSysin ( dx /2 ) + 2T dSxsin dy /2 + ( p − pa ) dSx dSy = 0 For differentially small angles, sin ( d) = d. Clean up this equation and rearrange:

 d dy  p = pa − T  x +  = pa − T  dSx dSy   

 1 1  +    Rx Ry   

(Ans.)

since, by definition, d/dS = 1/R, where R is the radius of curvature.

1-18 Two bubbles of radii R1 and R 2 coalesce isothermally into a single bubble R 3. Find the radius of the new (single) bubble. Because of surface tension, the pressure inside a bubble (which has two surfaces) is higher than ambient, p = po + 4T /R. Assuming that no interior-bubble air mass escapes during the coalescence, m1 + m2 = m3 , or, for an ideal isothermal gas of temperature T,

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po + 4I /R1 4 3 po + 4I /R2 4 3 po + 4I /R3 4 3 R1 + R2 = R3 T 3 T 3 T 3 where  is the gas constant. Canceling common terms and cleaning up, we have

(

)

(

po R33 + 4I R32 = po R13 + R23 + 4I R12 + R22

)

(Ans.)

This must be solved numerically or algebraically for the new radius R3 .

1-19 In Prob. 1-1, if the temperature, sphere size, and velocity remain the same for air flow, at what air pressure will the Reynolds number Re D be equal to 10,000?

Solution: From Prob. 1 −1 , T = 293 K, D = 0.014 m, V = 18 m/s, and  = 1.81E-5 Pa-s. Use the specified Reynolds number to compute the required air density:

Re D =

VD  (18 m/s )( 0.014 m ) = = 10, 000 Solve  = 0.718 kg /m3  1.81E -5 kg /m-s

Ideal gas:  = 0.718 =

p p = , Solve for p = 60400 Pa  60 kPa RT ( 287 )( 293)

(Ans.)

1-20 A solid cylinder of mass m, radius R, and length L falls concentrically through a vertical tube of radius R + R, where R  R. The tube is filled with gas of viscosity  and mean free path . Neglect fluid forces on the front and back faces of the cylinder and consider only shear stress in the annular region, assuming a linear velocity profile. Find an analytic expression for the terminal velocity of fall, V, of the cylinder (a) for no-slip; (b) with slip, Eq. (1-91). Solution: (a) For no-slip, the shear stress in the thin annular region between cylinders is

 =

u V  V  = , then W = mg = Fshear =  Awall =    ( 2 RL ) y R  R  Solve for Vno-slip =

mg R 2 RL

Ans. (a)

(b) For slip, modify the shear stress (see Prob. 1.16 for another example):

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u =

du V -2 u du V V = , or: = , = dy R dy R + 2 R + 2

As above in part (a), mg =  Aw , Vslip =

1-21

mg ( R + 2 2 RL

)

Ans. (b)

Solve P1-20 for the terminal fall velocity for no-slip if the cylinder is aluminum, with diameter 4 cm and length 10 cm . The tube has a diameter of 4.02 cm and is filled with argon gas at 20 C .

Solution: For no-slip, the shear stress 𝜏 in the thin annular region is  = 

u V = . y R

V  Then, W = mg = Fshear =  Awall =    ( 2 RL ) .  R 

 Al = 2710 kg/m 3 , R = 0.02 m , L = 0.1 m , g = 9.81 m/s2 ; then, mg =  Al R 2 L = 3.3408 N .

R = ( R + R ) − R = 0.0201 m − 0.02 m = 0.0001 m  Ar = 2.24 10−5 Pa  s

Therefore, Vno − slip =

mg R

2 RL Ar

=

( 3.3408 N )( 0.0001 m )

( 0.01256 m )( 2.24 10 2

−5

Pa  s )

= 1187.4431 m/s

(Ans.)

1-22 In Fig. Pl-22 a disk rotates steadily inside a disk-shaped container filled with oil of viscosity . Assume linear velocity profiles with no-slip and neglect stress on the outer edges of the disk. Find a formula for the torque M required to drive the disk.

Fig. Pl-22 Solution: The disk tangential velocity varies with radius, V = r, hence the local shear stress is  =   r /h on the top and bottom of the disk. The torque on a circular strip dr wide is Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -11-

 r  dM = ( dA) r ( 2 sides ) = 2r    2 r dr  h  or: M = 4

 h

R

 r dr = 3

0

  R 4 h

Ans.

1-23 Show, from Eq. (1-86), that the coefficient of thermal expansion of a perfect gas is given by  = 1/T . Use this approximation to estimate  of ammonia gas ( NH3 ) at 20°C and 1 atm and compare with the accepted value from a data reference. Solution: Introduce the ideal-gas law into the definition of  : 1    1   p  1  −p  1    1  =−   =− = =   =−    T  p  T  RT  p   RT 2    T  T

Ans.

It doesn’t matter what gas we are considering, ammonia or carbon dioxide or whatever, the ideal gas approximation predicts  = 1/T = 1/293K = 0.00341 K -1. Ans. This estimate is very close to estimates for ammonia in the literature, e.g., White (1988).

1-24 The rotating-cylinder viscometer in Fig. P1-24 shears the fluid in a narrow clearance r, as shown. Assuming a linear velocity distribution in the gaps, if the driving torque M is measured, find an expression for μ by (a) neglecting, and (b) including the bottom friction.

Fig. Pl-24 Solution: (a) Analyze the annular region only. The shear stress equals  ( du/dy )   ( R /R ) . The shear force on the cylinder side is normal to the radius, and the driving moment must be

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M =  RdF =  R ( dAw ) =

2

 0

R L   R R RLd = 2 R  R  3

Solve for  =

M R

Ans. (a)

2  R3 L

(b) On the bottom, the shear stress varies linearly with radius: R

M bottom

R

2   3 2   R 4  r =  r dAw =  r   r dr =  2 rdr = R  R 0 4R 0 

Thus M total =

2   R3 L 2   R 4 M R + , Solve  = R 4R 2  R3 ( L + R /4 )

Ans. (b)

1-25 Consider 1 m3 of a fluid at 20°C and 1 atm. For an isothermal process, calculate the final density and the energy, in joules, required to compress the fluid until the pressure is 10 atm, for (a) air; and (b) water. Discuss the difference in results. Solution: (a) The work done is −  pd , where  is the volume. From the ideal-gas law,

p = mRT . Thus

2

W1−2 = −  pd = −  1

mRT



  p  d = −mRT ln  2  = p11 ln  2   1   p1 

( )

 10  = (101350Pa ) 1m3 ln   = 233,000 J 1

Ans. (a)

(b) For water, we could use the compressed-liquid tables, but we can estimate the (very small) result from the bulk modulus K =  ( dp /d  )T = 2.23E9 Pa for water, Eq. (1-84). The change in volume of the water is very small when the change in pressure is only 9 atm:

  −

p K

1m3 ) ( 9 )(101350 Pa )  ( =−  −0.0041 m3 2.23E 9 Pa

A slightly more accurate estimate from Prob. 1-9, or from the compressed-liquid tables, gives

  −0.00042 m3. Then the work required to compress water from 1 atm to 10 atm is

(

)

W1−2 = −  pd  − pavg  = − ( 5.5 )(101350 Pa )  −0.00042 m3  230 J

Ans. (b)

This is 1000 times less than Ans.(a) for air above, since water is nearly incompressible. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -13-

1-26 Equal layers of two immiscible fluids are being sheared between a moving and a fixed plate, as in Fig. P1-26. Assuming linear velocity profiles, find an expression for the interface velocity U as a function of V , 1, and 2 .

Fig. P1-26

Solution: The shear stress is the same in each layer:

1 = 1

V −U U =  2 = 2 , h /2 h /2

solve for

U=

1 V 1 + 2

Ans.

1-27 Utilize the inviscid-flow solution of flow past a cylinder, Eqs. (1-3), to (a) find the location and value of the maximum fluid acceleration amax along the cylinder surface. Is your result valid for gases and liquids? (b) Apply your formula for amax to air flow at 10 m/s past a cylinder of diameter 1 cm and express your result as a ratio compared to the acceleration of gravity. Discuss what your result implies about the ability of fluids to withstand acceleration. Solution: Along the cylinder surface, r = R, and Eqs. (1-3) reduce to vr = 0 and v = −2U  sin  . Thus, along the surface, the absolute velocity is V = 2U sin ( s /R ) , where s is the arc length along the surface, measured from the front stagnation point. There is a convective acceleration given by

a =V

dV  s   2U s =  2U  sin   cos  ds  R  R R

(a) The acceleration is a maximum at  = 135, or s/R =  /4. Thus amax = 2U 2 /R. Ans. (a) This result is valid for all fluids, gases or liquids, in the inviscid approximation. (b) For the given data, R = 0.005 m, U  = 10 m/s, compute, independent of fluid properties,

amax = 2 (10 m/s ) / ( 0.005 m) = 40,000 m/s2  4080 g’s 2

Ans. (b)

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The lesson is that fluids have no fear of huge accelerations that would defeat a human being.

1-28

The coefficient of thermal expansion is defined as  =−

1   T

p

Determine  for an ideal gas with p =  RT . Show your work in detail. Solution: Given, the coefficient for thermal expansion is  = −

1   T

p

Using p =  RT for ideal gas,  =−

1-29

1   p  1 p  1   RT  1 = − − =   = − − 2   T  RT  p   RT    RT 2  T

(Ans.)

2  a, 3 and Newton’s expression of the wall shear stress as a function of the velocity gradient,  u   u   w =    , express Maxwell’s slip velocity, uw =   ,  y  w  y w Starting with Maxwell’s low-density approximation of the viscosity, namely,  

(a) as a function of the shear stress, density, and speed of sound a ; (b) as a function of the Mach number, the mean-flow velocity U , and the skin friction coefficient, 2 C f = w2 . U Solution: Given, Maxwell’s low-density approximation of viscosity is   density,

= mean free path, and a = speed of sound.

2  a , where  = 3

 u  Newton’s wall shear stress 𝜏𝑤 as a function of velocity gradient is  w =    .  y w

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Slip velocity 𝑢𝑤 is uw =  w , and constant  =



.

   w  3 w  u  w Therefore, uw    = . =  = 2   y w   a  2 a 3  (Ans.) Dividing by mean-flow velocity U and arranging gives Mach number is Ma = Therefore, uw =

1-30

uw 3 U 2 w = . U 4 a U 2

U 2 , and skin friction coefficient is C f = w2 . U a

3 Ma U  C f . 4

(Ans.)

Consider a hydraulic lift with a 50 cm diameter shaft sliding inside a housing with an inside diameter of 50.02 cm . If the shaft travels at 0.25 m/s , calculate the shaft resistance to motion per unit length. You may use water as the working fluid.

V  Solution: Shaft resistance F to motion is F =  Awall =    ( 2 RL ) .  R  R = 0.25 m , L = 1 m , V = 0.25 m/s

R = ( R + R ) − R = 0.251 m − 0.25 m = 0.001 m  water = 1.02  10 −5 Pa  s (at 20 C and 1 atm ) Funit -length

1-31

(1.02 10 =

−5

Pa  s ) ( 0.25 m/s ) 2 ( 0.25 m )(1 m )

( 0.001 m )

= 4.0055 10−3 N

(Ans.)

Consider a thin air gap of 1 mm that is formed between two parallel surfaces that are maintained at 20 C and 40 C , respectively. In the case of a quiescent medium (say still air), calculate the heat transfer rate across the gap per unit area.

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Solution: The rate of heat transfer per unit area (in one-dimensional space) is T ( x + x ) − T ( x ) q  −k . x The parallel plates are at 20 C and 40 C . Then, k needs to be determined for 30 C . n

Using power law curve, kair

T   303 K  = k0   = ( 0.0241 W/m  K )    273 K   T0 

0.81

313 K − 293 K  2 Therefore, q  − ( 0.0262 W/m  K )   = 524 W/m . 0.001 m  

1-32

= 0.0262 W/m  K .

(Ans.)

In the presence of viscosity, the pressure drop associated with a fully developed laminar motion in a horizontal tube of length L and diameter D may be evaluated analytically. One finds:

p = p1 − p2 =

128 LQ  D4

1 where  stands for the dynamic viscosity and Q =  D 2V denotes the volumetric flow rate. 4 Show that the corresponding head loss may be written as hL =

p1 − p2 L V2 = f lam g D 2g

What value of f lam do you obtain? Solution: Consider fully developed flow and apply steady-flow energy equation between section 1 and section 2.

 p   p  V2 V2 +  + z = +  + z  + hS − hL    2g 2g  g 2   g 1 Use z1 = z2 (horizontal), V1 = V2 (constant cross-section), 1 =  2 (same velocity profile), hS = 0 p − p2 p = (no pump); to simplify as hL = 1 … (1) g g

1

2

Q

D

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L

Empirical data on viscous losses in straight sections of pipe are correlated by the dimensionless Darcy friction factor f =

p

D … (2) 1 2 L V 2

From equation (2), p = f

1 L V 2 … (3) 2 D

Combining equations (1) and (3) gives hL = f lam

L V2 (where f = f lam , for laminar flow). D 2g

(Ans.) In the presence of viscosity, the pressure drop associated with a fully developed laminar motion in 128 LQ a horizontal tube of length L and diameter D is p = p1 − p2 = … (4)  D4

1 Dynamic viscosity is  , and volumetric flow rate is Q =  D 2V … (5) 4 For fully developed laminar flow, using equations (4) and (5) in equation (2) gives

f lam

1-33

1  128 L   D 2V  4   2 D = 64 = 64  4 D V 2 L VD  Re D

(Ans.)

A time-dependent, two-dimensional motion has three velocity components that are given by

u=

x 1 + at

v=

y 1 + bt

w=0

where a and b are pure constants. The objective of this problem is to compare and contrast the streamlines in this flow with the pathlines of the fluid particles. (a) Find the equations governing the streamline that passes through the point (1,1) at time t .

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(b) Calculate the path of a particle that starts at r0 = ( x0 , y0 ) = (1,1) at t = 0 . Determine the location of a particle at t = 1 , denoted as r1 . (c) Use the results of part (a) to determine the condition under which the streamlines and pathlines coincide. Solution: The geometric requirement for two-dimensional streamlines is as follows: v dy y (1 + at ) = = u dx x (1 + bt )

By separating the variables,

dy  1 + at  dx =  . y  1 + bt  x

1 + at  By integrating, ln ( y ) =   ln ( x ) + ln ( C ) .  1 + bt   1+ at   

Solving this gives y = Cx 1+bt  (where C is the integration constant). The condition of the streamline passing through the point (1, 1) at time t is C = 1 must be satisfied.

Therefore, the governing equation is y = x

 1+ at     1+bt 

… (1)

(Ans.)

The rate of change of x component of particle velocity with respect to time is

dx x = . dt 1 + at

1

By separating the variables and integrating, x = C1 (1 + at ) a (where C1 is the integration constant). The rate of change of y component of particle velocity with respect to time is

dy y = . dt 1 + bt

1

By separating the variables and integrating, y = C2 (1 + bt ) b (where C2 is the integration constant). At t = 0 , x = x0 = 1 = C1 and y = y0 = 1 = C2 . 1

Therefore, the path of the particle is

(1 + bt ) b y=

1

(1 + at ) a

x … (2)

(Ans.)

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1

1

Therefore, at t = 1 , x = x1 = (1 + a ) a and y = y1 = (1 + b ) b . 1 1 Then, r1 = ( x1 , y1 ) =  (1 + a ) a , (1 + b ) b  .  

(Ans.)

Comparing equations (1) and (2), we can say that the condition under which the streamlines coincide with pathlines is a = b = 0 . (Ans.)

1-34

A tornado may be simulated as a two-part circulating flow in cylindrical coordinates, with vr = vz = 0 ,

 r ( r  R )  v =   R 2 (r  R)   r (a) Calculate the divergence of the velocity. Is the flow compressible or incompressible? (b) Determine the vorticity. Is the flow rotational or irrotational? (c) Determine the strain rates in each segment of the flow. What is the sum of the three normal strain rates? Solution:

(1)

vϴ

(2)

R

Divergence

of

velocity

is

v =

r

1  1   ( rvr ) + ( v ) + ( vz ) = 0 r r r  z

(incompressible).

(Ans.)

 1  ( rv ) 1 vr   1 vz v   vr vz  − − − Vorticity is  =  v =   ez .  e +   er +  r    z r   r  z   r r Only nonzero component of vorticity is  z =

1  ( rv ) . r r

For segment (1),  z = 2 (nonzero; thus rotational).

(Ans.) (Ans.)

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For segment (2),  z = 0 (irrotational).

(Ans.)

1 v v Only nonzero component of tangential strain rate is  r =   −   . 2  r r 

(Ans.)

For segment (1),  r = 0 .

(Ans.)

For segment (2),  r = −

 R2

.

r2

(Ans.)

The sum of normal strain rate components is  rr +  +  zz =

vr  vr 1 v + + r  r r 

 vz =0. +  z

(Ans.)

1-35

In modeling the motion of an 8-meter diameter tornado rotating at an angular speed of  at the point of maximum swirl, it is possible to use the Maicke–Majdalani profile (Maicke and Majdalani 2009) as a piecewise approximation for which vr = vz = 0 and the tangential velocity is given by

16 r 1 − ln ( r 2 )    v ( r ) = 16   r

0  r  1 (inner, forced vortex segment) r  1 (outer, free vortex segment)

(a) State whether the flow is 1D, 2D, or 3D; steady or unsteady; and specify v ( r ) as r →  . (b) Calculate the divergence of the velocity. Is the flow compressible or incompressible? (c) Determine the vorticity. Is the flow rotational or irrotational? (d) Determine the strain rates and the shear stresses in the inner and outer flow segments. (e) What is the limit of v ( r ) as r → 0 ? Hint: In taking the limit, it is helpful to remember that u' ( ln u ) = and that, in the inner segment, the tangential velocity can be rewritten as u '

1 − ln ( r 2 )   v = 16  −1 r Solution:

vϴ (1) 0

(2) 1

r

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The velocity varies with respect to the radial distance r from the centerline and is independent of the axial distance z or of the angular position  . This represents a typical onedimensional flow. (Ans.) Since

vr v vz = = = 0 , the flow is time invariant (steady). t t t

(Ans.)

As r →  , v ( r ) → 0 .

(Ans.)

Divergence of velocity is   v =

1  1   ( rvr ) + ( v ) + ( vz ) = 0 (incompressible). r r r  z

 1 vz v − Vorticity is  =  v =   r  z

  vr vz −  er +   z r 

Only nonzero component of vorticity is  z =

 1  ( rv ) 1 vr  −  e +  r  r r   

(Ans.)

  ez . 

1  ( rv ) . r r

1 For segment (1),  z = 32 ln  2  (rotational). r 

(Ans.)

For segment (2),  z = 0 (irrotational).

(Ans.)

1  v v Only nonzero component of tangential strain rate is  r =   −  2  r r

 . 

For segment (1),  r = −16 . For segment (2),  r = −

(Ans.)

16 . r2

(Ans.) Then, shear stress for segment (1) is  r = −32 and segment (2) is  r = −

32 . r2

(Ans.)

As r → 0 , v ( r ) → 0 . (Ans.)

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1-36

The Taylor profile, which has been used to describe the bulk gaseous motion in planar, slab rocket chambers (Maicke and Majdalani 2008), corresponds to a self-similar profile in porous channels that bears symmetry with respect to the chamber’s midsection plane. Using normalized Cartesian coordinates, the streamfunction may be written as 1   = x sin   y  ; 0  y  1 , and 0  x  l , where l represents the aspect ratio of the 2   chamber (i.e., the length of the porous surface normalized by the chamber half height). In this problem, the velocity vector, normalized by the wall injection speed, may be expressed as V( x, y) = ui + vj .

(a) Determine the axial and normal velocity profiles using d d and v=− dy dx (b) Evaluate the velocity divergence and determine whether the motion is compressible or incompressible. u=

(c) Evaluate the vorticity and determine whether the motion is rotational or irrotational.

Solution: The axial and normal velocity profiles of the stream function are given as follows: u=

d d  x 1  1  = − sin   y  = cos   y  , and v = − dx dy 2 2  2 

Divergence of velocity is

  V ( x, y ) =

   2x 1 ( u ) + ( v ) = − sin   y  y x 4 2 

(Ans.)

(compressible)

(Ans.) Only nonzero vorticity component is

 v u   2 x  1   z =   Vz ( x, y ) =  −  = sin   y  4 2   x y  (Ans.)

(rotational)

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1-37

In the Taylor flow problem described above, determine the following:

(a) The strain rates. (b) The Lagrangian time-dependent coordinates x j ( t ) and y j ( t ) for a particle j that enters the porous chamber at the sidewall where y j = 1 and x j = X j at t = 0 . Recall that dx j dt

=u

dy j

and

dt

=v

(c) The pathlines of a particle j entering the chamber at X j = 1, 2, 3, 4, 5 . Display your results in the ( x, y ) plane. Solution: Only nonzero components of strain rate are as follows:

 xx =

 2x  1  u   1 1   sin   y  = cos   y  ,  yy = − cos   y  , and  xy = − 8 x 2 2 2  2 2  

The rate of change of x j ( t ) is u =

1 1  =  x j cos   y j  . dt 2 2 

dx j

By separating variables and integrating, x j = C1e The rate of change of y j ( t ) is v =

(Ans.)

1 1  t  cos  y j  2 2 

1  is the integration constant).  C1 

, ( ln 

1  = − sin   y j  . dt 2 

dy j

 −t    1  By separating variables and integrating, y j = tan  C2e 2  , ( ln   is the integration   C2    constant).

4

−1

Given, at t = 0 , ( x j , y j ) = ( X j ,1) . Then, C1 = X j , and C2 = 1 . Therefore, the time-dependent coordinates for the particle are 1 1  t  cos  y j  2 2 

x j = X je

and y j =

 −t   tan −1  e 2  .    4

(Ans.)

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 The equation of the pathline is x j = X j    (Ans.)

  1  tan   y j   4  

1  cos  y j  2 

1

.

Pathlines of the particle j entering the chamber at X j = 1, 2, 3, 4, 5 are plotted on the ( x, y ) plane as follows:

Xj = 1 Xj = 2 Xj = 3 Xj = 4

y

Xj = 5

x

1-38

(Ans.)

The Taylor–Culick profile, which describes the bulk gaseous motion in solid rocket motors (Culick 1966), corresponds to an axisymmetric, self-similar profile in porous tubes. Using normalized cylindrical coordinates, the streamfunction may be written as 1   = z sin   r 2  ; 0  r  1, and 0  z  l , where l represents the aspect ratio of the 2  motor (i.e., the length of the porous surface normalized by the chamber radius). In this problem, the velocity vector, normalized by the wall injection speed, may be expressed as V(r , z ) = vr er + vz e z .

(a) Determine the axial and radial velocity profiles using Stokes’ definition:

vz =

1  r r

and

vr = −

1  r z

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(b) Evaluate the velocity divergence and determine whether the motion is compressible or incompressible. (c) Evaluate the vorticity and determine whether the motion is rotational or irrotational.

Solution: The axial and normal velocity profiles of the stream function are given as follows: vz =

1  1  1 1 1   =  z cos   r 2  and vr = − = − sin   r 2  r r r z r 2 2  

Divergence of velocity is  V ( r , z ) =

(Ans.)

1  1   ( rvr ) + ( v ) + ( vz ) = 0 (incompressible). r r r  z

(Ans.)

 1  ( rv ) 1 vr  1 vz v   vr vz  − − − Vorticity is given by  =  V ( r , z ) =   e +   er +  r   z r   r  z   r r

  ez . 

Only nonzero vorticity component is

 =

1-39

vr vz 1  − =  2 zr sin   r 2  (rotational) z r 2 

(Ans.)

In the Taylor–Culick flow problem described above, determine the following:

(a) The strain rates. (b) The Lagrangian time-dependent coordinates rj ( t ) and z j ( t ) for a particle j that enters the cylindrical rocket chamber at the sidewall where rj = 1 and z j = Z j at t = 0 . Recall that drj dt

= vr

and

dz j dt

= vz

(c) The pathlines of a particle j entering the chamber at Z j = 1, 2, 3, 4, 5 . Display your results in the ( r , z ) plane. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -26-

Solution: Only nonzero components of strain rate are as follows:

 rr =

vr 1  1 1  = − cos   r 2  + 2 sin   r 2  , r 2  r 2  1  v v v 1   zz = z =  cos   r 2  , and  zr =  r + z 2  z r z 2  (Ans.)

The rate of change of rj ( t ) is vr =

drj dt

=−

1 1  sin   rj 2  . rj 2 

By separating variables and integrating, rj = The rate of change of z j ( t ) is vz =

vr 1 v 1 1  + = − 2 sin   r 2  , r r  r 2  1 2  1 2  = −  zr sin   r  2 2  

 =

1 tan −1 ( C1e− t ) , ( ln   is integration constant).   C1  4

1  =  z j cos   r 2  . dt 2 

dz j

1 2

 

 t cos  r 2 

By separating variables and integrating, z j = C2e

 1   is integration constant).  C2 

, ( ln 

Given, at t = 0 , ( rj , z j ) = (1, Z j ) . Then, C1 = 1 , and C2 = Z j . Therefore, the time-dependent coordinates for the particle are

rj =

4



tan

−1

( e ) and z

1 2

 

 t cos  r 2 

− t

j

= Z je

 The equation of the pathline is z j = Z j    (Ans.)

(Ans.)

  1  tan   r j 2   4   1

1  cos  r j 2  2 

.

Pathlines of the particle j entering the chamber at Z j = 1, 2, 3, 4, 5 are plotted on the ( r , z ) plane as follows:

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Zj = 1 Zj = 2 Zj = 3 Zj = 4

rj

Zj = 5

zj

1-40

(Ans.)

The Vyas–Majdalani profile, which describes the helical motion of a cyclonic chamber (Vyas and Majdalani 2006), consists of a three-component velocity profile, V ( r , z ) = vr e r + v e + vz e z , where

vr = −U

 r2  z a  r2  a sin   2  , v = U , and vz = 2U cos   2  ; a r r  a   a 

0  r  a  0  z  L

Here a denotes the radius of the cyclonic chamber, U stands for the average tangential velocity at r = a , and  represents a dimensionless offset swirl parameter that gauges the relative importance of axial and tangential speeds. (a) Is the flow one-dimensional, two-dimensional, or three-dimensional? (b) Is the flow steady or unsteady? (c) Calculate the divergence of the velocity. Is the flow compressible or incompressible? (d) Determine the vorticity. Is the flow rotational or irrotational? (e) Determine the strain rates. What is the sum of the three normal strain rates? (f) Assuming a circular opening of radius r = b = a

2 at z = L , calculate the volumetric flow

rate by integrating the axial velocity from r = 0 to r = a

2.

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Solution: The velocity varies with respect to the radial distance r from the centerline and axial distance z but it is independent of the angular position  ; which corresponds to a typical two-dimensional flow. (Ans.) Since

vr v vz = = = 0 , the flow is time invariant (steady). t t t

Divergence of velocity is  V ( r , z ) =

(Ans.)

1  1   ( rvr ) + ( v ) + ( vz ) . r r r  z

 r2   r2  2 2  V ( r , z ) = − U cos   2  + 0 + U cos   2  = 0 (incompressible) a a  a   a   1  ( rv ) 1 vr  1 vz v   vr vz  − − − Vorticity is  =  v =   e +   er +  r   z r   r  z   r r Only nonzero component of vorticity is  = −

(Ans.)

  ez 

 r2  vz 4 = − 3  2Uzr sin   2  (rotational). r a  a 

(Ans.) Only nonzero components of tangential strain rate are 1  1 v

v

v 

1

a

a



a

r  r =  +  −   = 0− 2 U − 2 U  = − 2 U 2  r  r r  2 r r r 

(Ans.)

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1  v

v 

1

z



r2 

z



r2 

 zr =  r + z  =  0 − 4 2U 3 r sin   2   = −2 2U 3 r sin   2  2  z r  2  a a  a   a  (Ans.) The sum of the normal strain rate components is vr  vr 1 v + + r  r r 

 rr +  +  zz =

 vz =0 +  z

(Ans.) The volumetric flow rate is given by a

Q=



2

r =0

1-41

a

vz dA =



2

r =0

 2

 r2  z 2U cos   2  d ( r 2 ) = 2 aUL  cos  d = 2 aUL a  a   =0

(Ans.)

The Maicke–Majdalani profile, which may be used to describe the motion of an unbounded tornado (Maicke and Majdalani 2009), can be expressed as a simple piecewise approximation for which vr = vz = 0 and

 r   r 2  1 − ln  2  2 2   ; r  XR X   X R  v ( r ) =   R 2 r  XR  r ; where R denotes the radius at which the wind’s angular speed  may be measured and X represents the fraction of the radius R where the free vortex behavior ceases. (a) Is the flow one-dimensional, two-dimensional, or three-dimensional? (b) Is the flow steady or unsteady? (c) Calculate the divergence of the velocity. Is the flow compressible or incompressible? (d) Determine the vorticity. Is the flow rotational or irrotational? (e) Determine the strain rates and the shear stresses in each segment of the flow. (f) What is the limiting value of v ( r ) as r → 0 ? Solution: The velocity varies with respect to the radial distance r from the centerline and is independent of the axial distance z or of the angular position  . This represents a

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typical one-dimensional flow. (Ans.) Since

vr v vz = = = 0 , the flow is time invariant (steady). t t t

Divergence of velocity is   v =

1  1   ( rvr ) + ( v ) + ( vz ) = 0 (incompressible). r r r  z

 1 vz v − Vorticity is  =  v =   r  z

  vr vz −  er +   z r 

Only nonzero component of vorticity is  z = For segment ( r  XR ),  z = −

(Ans.)

 1  ( rv ) 1 vr  −  e +  r    r r

(Ans.)

  ez . 

1  ( rv ) . r r

2  r 2  ln   (rotational). X 2  X 2 R2 

(Ans.)

For segment ( r  XR ),  z = 0 (irrotational). (Ans.) 1  v v  Only nonzero component of tangential strain rate is  r =   −   . 2  r r 

For segment ( r  XR ),  r = − For segment ( r  XR ),  r = −

 X2

.

R 2 . r2

(Ans.)

(Ans.)

Then, shear stress is given as follows: For segment ( r  XR ),  r = −

2 . X2

(Ans.)

For segment ( r  XR ),  r = −

2  R 2 . r2

(Ans.)

As r → 0 , v ( r ) → 0 . (Ans.)

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1-42

Assuming  to be a scalar and V a vector, evaluate the following quantities:

(a)  (b) (  )  (  ) (c) (  )  (  ) (d)  (  ) (e)  (  V ) Solution:  = 0

(Ans.)

        i+ j+ k z x x y   y z

(  )  (   ) = 2 

2

     (  )  (  ) =   +   +    x   y   z  2

(Ans.)

2

(Ans.)

 2  2  2   (  ) = 2 + 2 + 2 x y z

(Ans.)

 (  V ) = 0

(Ans.)

1-43

Assuming U , V , and W are Cartesian vectors, prove the following identities:

1 (a) ( V  ) V =  ( V  V ) − V  (  V ) 2

(Lamb’s vector identity)

(b)  (  V ) =  ( V ) − 2 V (c)  ( U  V ) = V  (  U ) − U  ( V ) (d) ( U  V )  W = U  ( V  W )

(mixed vector scalar product)

(e) U  ( V  W ) = ( U  W ) V − ( U  V ) W

(vector triple product)

(f) ( U  V )  ( U  V ) = ( U  U )( V  V ) − ( U  V )( U  V )

(Lagrange’s identity)

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(g) ( U − V )  ( U + V ) = 2U  V (h)  ( V ) =  V + V  (i)  ( V ) =  V + (  )  V (j)  ( UV ) = (  U ) V + U  (V ) Solution:

1  ( V  V ) − V  (  V ) = V V − V V − ( V  ) V  = ( V  ) V 2

 (   V ) −  2 V =  (   V ) −  (   V ) −   (   V )  =   (  V )  (U  V) =

(Ans.) (Ans.)

 (VzU y − VyU z ) + y (VxU z − VzU x ) + z (VyU x − VxU y ) x

  U z U y   U y U x     Vz Vy   Vy Vx    U x U z   Vx Vz  = Vx  − − − − − −  + Vy    − U x   +U y    + Vz   +Uz  z  x  y     y z  x  y    z  z  x  x   y = V  (  U ) − U  (  V ) (Ans.)

( U  V )  W = (VzU y − VyU z )Wx + (VxU z − VzU x )Wy + (VyU x − VxU y )Wz = (VzWy − VyWz )U x + (VxWz − VzWx )U y + (VyWx − VxWy )U z = U (V  W)

(Ans.)

U (V  W)

= (U x i + U y j + U z k )  (VzWy − VyWz ) i + (VxWz − VzWx ) j + (VyWx − VxWy ) k 

= U z (VxWz − VzWx ) − U y (VyWx − VxWy )  i + U x (VyWx − VxWy ) − U z (VzWy − VyWz )  j + U y (VzWy − VyWz ) − U x (VxWz − VzWx )  k = (U xWx + U yWy + U zWz )(Vx i + Vy j + Vz k ) − (U xVx + U yVy + U zVz )(Wxi + Wy j + Wz k ) = ( U  W) V − (U  V) W

(Ans.)

( U  V) (U  V)

= (U zVy − U yVz ) i + (U xVz − U zVx ) j + (U yVx − U xVy ) k   (U zVy − U yVz ) i + (U xVz − U zVx ) j + (U yVx − U xV y ) k  = (U zVy − U yVz ) + (U xVz − U zVx ) + (U yVx − U xVy ) 2

2

2

= (U zVy ) + (U yVz ) + (U xVz ) + (U zVx ) + (U yVx ) + (U xVy ) − 2 U zVyU yVz + U xVzU zVx + U yVxU xVy  2

2

2

2

2

= (U x 2 + U y 2 + U z 2 ) (Vx 2 + Vy 2 + Vz 2 ) − (U xVx + U yVy + U zVz )

2

2

= ( U  U )( V  V ) − ( U  V )( U  V )

(Ans.)

( U − V) ( U + V) = (U x − Vx ) i + (U y − Vy ) j + (U z − Vz ) k   (U x + Vx ) i + (U y + Vy ) j + (U z + Vz ) k  = 2 (U zVy − U yVz ) i + 2 (U xVz − U zVx ) j + 2 (U yVx − U xVy ) k = 2U  V Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -33-

(Ans.)

  ( V )      = i+ j + k   (Vx i + Vy j + Vz k ) z   x y Vx Vy Vz = + + x y z  V Vy Vz       = x + + + Vy + Vz  +  Vx  y z   x y z   x =   V + V    ( V )

(Ans.)

     = i+ j + k   (Vxi + Vy j + Vz k ) z   x y  Vy Vz   Vz Vx   Vx Vy   =  − − − i +   k  j+  y   x z   y x    z  Vy Vz   Vx Vz   Vy Vx        =   − − − (Vx − Vz ) j + (Vy − Vx ) k  i +   k  +  (Vz − Vy ) i +  j+  y   z x   x y    x y z   z =   V + (  )  V

(Ans.)

(   U ) V + U  ( V )  U x U y U z    V  Vx V =  + + i + y j+ z  (Vx i + Vy j + Vz k )  + (U x i + U y j + U z k )   y z  y z  x  x   U xVx U yVy U zVz = + + x y z      = i+ j + k   (U xVxi + U yVy j + U zVz k ) z   x y =   ( UV )

 k  

(Ans.)

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PROBLEM SOLUTIONS CHAPTER 2. FUNDAMENTAL EQUATIONS OF COMPRESSIBLE VISCOUS FLOW

2-1

Derive conservation of mass using the elemental control volume shown at right.

C.V. mass loss equals flow-out − flow-in: (a) flow through front and back sides:

   −vdzdr + v + ( v ) d dzdr    (b) flow through left and right sides:

   −vr rddz + vr r + ( vr r ) dr  ddz r   (c) flow through bottom and top sides:

   −vz rddr + vz + ( vz ) dz  rddr z   (d) mass loss within the clement: −  (  r d dr dz ) t

The sum of (a,b,c) must equal (d) to conserve mass. Since ( r,θ,z ) are independent of time, we may cancel the differentials ( dr d dz ) and rewrite the final result: 1 1    ( rvr ) + ( v ) + ( vz ) + = 0 r r r  z t

(Ans.)

This is the (compressible) Equation of Continuity in cylindrical polar coordinates.

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2-2 Simplify the result of (2-1) above to the case of steady plane compressible flow in polar coordinates and derive a suitable stream function for this case. For plane flow,

 /z = 0 and vz = 0; for steady flow,  /t = 0. The continuity equation above in

Prob. 2-1 reduces to the following relation, with its stream-function equivalence:

      ( rvr ) + (v ) = 0 =   +  −  r  r      r  By comparing terms, we see that the velocities are related to stream function as follows:

vr =

1  1  ; v = − r   r

(Ans.)

This is the steady plane polar-coordinate compressible stream function.

2-3 Simplify the result of (2-1) above to the case of steady plane compressible flow in axisymmetric coordinates and derive a suitable stream function for this case. For axisymmetric flow,  / = 0 and v = 0; for steady flow,  /t = 0. The continuity equation above in Prob. 2-1 reduces to the following relation, with its stream-function equivalence:

      ( rvr ) + ( rvz ) = 0 =  −  +   r z r  z  z  r  By comparing terms, we see that velocities are related to stream function as follows:

vr = −

1  1  ; vz = r z r r

(Ans.)

This is one form of the steady plane axisymmetric compressible stream function.

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2-4

Use the Navier-Stokes equation to derive the “weak” form of the Bernoulli relation.

For steady, incompressible flow with zero viscosity, the Navier-Stokes equation reduces to

(

)

 ( V  ) V = −p + g, where ( V  ) V =  V 2 /2 +  V where the latter identity follows from Eq. (1-10) of the text. Now “dot” this entire equation with a differential arc length dr, giving the intermediate result

(

)

1  2   p +  V /2 − g   dr = ( V  )  dr  

For any direction dr, the left hand side consists of three exact differentials. The right hand side vanishes for the particular direction dr along a streamline of the flow, since a streamline is everywhere perpendicular to the cross-product of velocity and vorticity. For this case, then, the dot product above becomes

( )

2 dp d V + + g dz = 0,  2

or:

p/ + V2 /2 + g z = constant

(Ans.)

This is the “weak” form of the steady, incompressible Bernoulli equation [that is, it does not account for heat transfer or shaft work] and assumes that the coordinate z is “up”. The “constant” may vary in value from streamline to streamline.

2-5 Show that the incompressible energy equation (2-40) reduces, for zero  and k, to the “strong” form of Bernoulli’s equation. If we simply set  and k equal to zero in Eq. (40), we obtain the oversimplified result De/Dt = 0, or e = constant as a particle moves through the flow. This is a “weak” isothermaltype result, implying no interaction between work and heat transfer in a flow. If, instead, we combine energy and momentum, using (2-19) to substitute for the stress tensor in (2-40), we obtain the result ij

u i Dp DV = −p div ( V ) − − V +  V  g + viscous terms x j Dt Dt

Substituting into Eq. (2-40) and setting , k, and div ( V ) equal to zero, we obtain Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -3-



(

)

D e + p/ + V 2 /2 − g  r = 0, Dt

or:

e + p/ + V2/2 + g z = constant along a streamline

(Ans.)

where, again, z is “up”. This is the “strong” form of Bernoulli’s equation and is related to the steady-flow-energy-equation of thermodynamics. It holds even for compressible flow as long as viscous and heat-conduction effects are negligible.

2-6

Consider the proposed incompressible axisymmetric flow field vz = C ( R 2 − r 2 ), vr = 0

in the region 0  z  L,0  r  R, where C is a constant. Neglect gravity. (a) Determine if this is an exact solution to the Navier-Stokes equation. (b) What might it represent? (c) If an axisymmetric stream function  ( r, z ) exists for this flow, find its form.

Solution: Substitute the given flow field into the radial and axial momentum Eqs. (B-6, 8): r – momentum:

0=−

p +0 r

z – momentum:

0=−

p      2 2 +  r C R − r z r r  r

(

)  +  z 2 C ( R 2 − r 2 ) 2

The first of these tells us that p is independent of r and only varies axially. The second relation yields a constant axial pressure gradient, p/z = −4C. (a) The given flow field is indeed an exact solution. (b) It represents Poiseuille flow in a tube, Sect. 3-3 of the text. (c) The solution is exact and a stream function  ( r, z ) exists, from Appendix B or Prob. 2-3:

(

)

  = 0; = rvz = C R 2 r − r 3 . z r

 R2r  r 4  Integrate:  = C  −  + const  2 4  

Ans. (c)

2-7 Investigate the stream function  = Cxy, with C  0, to determine if it is a realistic solution for (a) inviscid; or (b) viscous, incompressible flow. The stream function, by definition, satisfies continuity exactly, with velocity components

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u=

 = Cx; y

v=−

 = −Cy x

(1)

There are no shear strains, but the normal strain rates are finite: xx = C,  yy = −C, (hence there are finite viscous normal stresses). The sum of these strain rates is zero, hence the flow is incompressible. Note that the vorticity is also zero:

z =

v u − =0 x y

Thus the flow is irrotational, and a velocity potential also exists. We conclude that the stream function  = Cxy represent a valid potential-flow field. When plotted in the x-y plane for various values of C, the result is the pattern of streamlines shown below. We could interpret this as flow (a) between two intersecting

streams; (b) of a stream against a plane wall; or (c) of flow around a 90° corner. Patterns (b) and (b) would not be realistic for viscous flow, because the “walls” are not no-slip lines of zero velocity. Since the flow is irrotational, the viscous momentum term (  V) vanishes identically [see Eq. (2-107) in the text] and, if we neglect gravity for convenience, the Navier-Stokes equations yield the pressure gradients 2

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p = − C2 x; x

p = − C2 y y

Integration of these simple relations yields the exact pressure distribution in the fluid:

(

)

1 p = −  C2 x 2 + y2 + constant 2

(Ans.)

But this is precisely Bernoulli’s equation, which is what we expect when an irrotational flow is tested directly in the Navier-Stokes equation.

2-8 Investigate the incompressible stream function y = C(x y − y /3), where C  0. Follow the same procedure as Prob. 2-7 above. The velocity components are 2

u=

(

)

 = C x 2 − y2 ; y

v=−

3

 = −2Cxy x

from which we may compute that the vorticity, z = v/x − u/y, is identically zero and the flow is irrotational. The streamlines of this exact potential flow are plotted below.

Only the (symmetric) upper half plane is shown for convenience. We could interpret this flow as (a) flow caused by three intersecting streams; (b) flow against a 120° corner; or (c) flow around a

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60° corner. Patterns (b) and (b) would not be realistic for viscous flow, because the “walls” are not no-slip lines of zero velocity. Since the flow is irrotational, the viscous momentum term (  V) vanishes identically [see Eq. (2-107) in the text] and, if we neglect gravity for convenience, the Navier-Stokes equations yield the pressure gradients 2

(

)

p = −2C2 x 3 + xy 2 ; x

(

p = −2C2 y3 + x 2 y y

)

Integration of these simple relations yields the exact pressure distribution in the fluid:

(

)

(

)

1 1 p = −  C2 x 4 + 2x 2 y 2 + y 4 + constant = −  u 2 + v 2 + constant 2 2

(Ans.)

But this is precisely Bernoulli’s equation, which is what we expect when an irrotational flow is tested directly in the Navier-Stokes equation.

2-9

Analyze the following plane polar unsteady incompressible flow:

vr = 0

 r 2  C v = 1 − exp  −  4t   r    

with C and v constant and gravity neglected. This flow satisfies continuity exactly. If we substitute these velocities into the θ-momentum equation in Appendix B, assuming radial symmetry or p/ = 0, we find that it is also satisfied exactly. [The pressure distribution p ( r, t ) could then be found from the radial momentum equation but is not shown here.] Thus the given distribution is indeed an exact solution to the Navier-Stokes equations. This problem was discussed earlier, without proof of its exactness, in Prob. 1-6 on page 3. The instantaneous vorticity profiles,

z =

 r2  1 C ( rv ) = exp  −  r r 2 t  4 t 

and velocity profiles are plotted for various times on page 3. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -7-

Using  from Eq. (2-46), prove the inequalities in Eq. (2-47).

2-10

We are to find the conditions under which the dissipation function,

(

)

 =   2 a 2 + b2 + c2 + f 2 + g 2 + h 2  +  ( a + b + c )  

2

is positive, where we adopt the simpler notation ( a, b, c, f , g, h ) for the strain rates

( xx , yy , zz , xy , yz , zx ). The first thing to notice is that only the coefficient  is concerned with the shear strain rates ( f ,g, h ) . It is easy to visualize pure shear flow with zero normal strain, e.g. Couette flow or fully-developed duct flow. For this case, we require positive

(

)

 =  f 2 + g 2 + h 2 , with no contribution from . It follows that  cannot be positive in such flows unless 0

(Ans. a)

This is the first condition to be proved. With  thus guaranteed positive, it follows that the normal-strain-rate terms will also have to stay positive to keep  positive in cases where there is no shear strain (such as flow through normal shock waves). Thus we require

(

)

 ( a + b + c ) + 2 a 2 + b 2 + c 2  0, 2

or:

 / + R  0,

where

R=

(

2 a 2 + b2 + c2

)

( a + b + c )2

This is equivalent to requiring that  /  ( −R min ) . To minimize R, we simply set

R/a = R/b = R/c = 0 and solve for a = b = c. Then R min = 2/3, and thus   −2/3

(Ans. b)

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This was the second condition to be proved.

2-11

Non-dimensionalize the differential equation of compressible irrotational flow:  2 t 2

+

(

 u 2 + v2 t

) + u2 − a 2 2 + v2 − a 2 2 + 2uv 2 = 0 ( ) x2 ( ) y2 xy

where a is the speed of sound of the gas. We only need a velocity scale U and a length scale L, since mass units are not present. Appropriate nondimensional variables are: * =

 UL

u*, v*, a* =

u, v, a U

x*, y* =

x, y L

t* =

Ut L

Substituting these variables into the basic differential equation and dividing out the leading coefficient, we obtain the following nondimensional differential equation:

(

)

2 2 2 2  2*  u* + v*  2* 2 2  * 2 2  * + + u* − a* + v* − a* + 2u*v* =0 t* x*y* t*2 x*2 y*2

(

)

(

)

Examining this relation, we see that no dimensionless parameters appear (Ans.) [Actually, correlating a* with temperature and velocity would in fact lead to two important parameters, the Mach number and the specific heat ratio.]

2-12

Repeat the nondimensionalization of the Navier-Stokes equation for slow viscous flow by

noting that the pressure should scale as ( U/L ) . What happens if Re  1 ?

To proceed, it is only necessary to modify dimensionless pressure in Eq. (2-83) to

p* = ( p − po ) L/ ( U ) . Let us neglect gravity and buoyancy and consider only the Navier-Stokes equation with constant  and : 

DV = −p +  2 V Dt

Introduce the variables from Eq. (2-83), including the new p*, and clean up by dividing out the coefficient of the viscous term. The result is the following dimensionless equation:

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Re

DV* = −*p* + *2V*, Dt*

where

Re = UL/

The only parameter is the Reynolds number, which occurs only in the “inertia” or acceleration term. If the Reynolds number is very small, Re  1 , then inertia is negligible, and the balance of fluid forces is between pressure and viscosity:

p = 2V

Re  1

if

(Ans.)

This is the creeping-flow or Stokes-flow approximation. It linearizes the Navier-Stokes equation and enables many slow viscous flows to be solved analytically [see Section 3.9].

2-13 Nondimensionalize the equations of motion for free convection near a hot vertical plate, using dimensionless variables which have no direct velocity scale (“U”):

u* =

uL 

v* =

vL 

where L is a length scale and



x* =

x L

y* =

u L

T* =

T − T1 To − T1

is kinematic viscosity.

The equations of motion are given in the statement of Prob. 2-13. Introducing these variables into the continuity equation gives u*/x* +v*/y* = 0, with no parameters appearing. Substitute these variables into the momentum equation and divide by the leading coefficient. The result is the dimensionless momentum equation: u*

u* u*  2 u*  2 u* + v* = GrL T* + + , x* y* x*2 y*2

where

GrL =

g ( To − T1 ) L3 2

The single parameter, which occurs only in the buoyancy term, is the Grashof Number. Substitute the same variables into the energy equation and divide by the leading coefficient. The result is the dimensionless energy equation:

u*

T* T* 1   2T*  2T*  + v* =  + , x* y* Pr  x*2 y*2 

where

Pr =

cp k

Here a second parameter arises, associated with the conduction term: the Prandtl Number.

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By merely nondimensionalizing the equations of motion, we conclude, correctly, that free convection problems are dependent upon both the Grashof and Prandtl numbers.

2-14 Make a control-volume analysis of laminar flow in the pipe entrance as shown and find a formula for the friction drag on the pipe walls.

(

)

2 2 Let the exit velocity be parabolic: u = C ro − r .

The proper control volume includes the entrance and exit and passes just inside the tube walls, exposing the wall-friction drag force acting to the left on the fluid. First relate “C” to the inlet by using the integral continuity relation:

(

 V dA = 0 = −Uoro +   CV

or:

)

(

)

C ro2 − r 2 2r dr = Uo  −ro2 + Cro4 /2 ,

2

exit

C = 2Uo /ro2

With C known, the drag is computed by applying the integral x-momentum equation:

 Fx = ( po − px ) ro2 − Drag =



u ( V dA ) = −Uo2ro2 +

CV

 exit

(

)

2

 C ro2 − r 2  2r dr  

The value of the last integral on the right is

C2ro6/3 = Uo2ro2/3

since

C = 2Uo2 ro2

The force equation above may thus be solved for the desired drag force:

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1   Drag = ro2  po − px −  Uo2  3  

2-15

(Ans.)

Illustrate “boundary-layer” behavior with Prandtl’s model differential equation: 

d2u dy

2

+

du + u = 0, with u ( 0 ) = 0 and u (  ) bounded dy

assuming that   1. For finite , we assume a solution u = exp ( my ) , leading to m2 + m + 1 = 0,

or:

m1,2 =

1  −1  2 

Then the general solution has the form

(1 − 4 ) 

u = A exp(m1y) + B exp(m2 y). To satisfy the initial

condition u(0) = 0, we must have B = −A. Then our desired “boundary layer” type solution is given by

u = C e m1y − e m 2 y    For   0, both m1,2 are negative and thus the solution is bounded at large y. For very small , we may approximate m1 = −1 and m2 = −1/. If we neglect the second derivative term entirely (  = 0, corresponding to “inviscid” or “slip” conditions), the basic differential equation reduces to

du + y = 0, dy

with solution u = Ce− y

This is equivalent to the first term of the boundary-layer solution. It is impossible in this “inviscid” case to satisfy the “no-slip” initial condition, u ( 0) = 0.

These two solutions, inviscid and boundary-layer, are compared in the figure on p. 28. The boundary-layer solutions satisfy the no-slip condition and, for small ε, merge into the

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inviscid solution at larger y. The “boundary-layer” is very thin. This mathematical behavior can be generalized into the asymptotic expansion methods of Section 4-11.

2-16

Consider the plane, incompressible, cartesian stream function in the region 0  y   b c

 = ax + by + e−cy

where ( a, b, c ) are positive constants. (a) Determine if this is an exact solution to the continuity and Navier-Stokes equations if gravity and pressure gradient are neglected. (b) What are the

dimensions of ( a, b, c ) ? (c) If y = 0 represents a wall, does the no-slip condition hold there? (d) Is there any vorticity in the flow field? If so, what is its form? Solution: (a) First use the stream function  ( x, y ) to determine the velocity components:

u=

 = b − be−cy; y

Check continuity:

v=−

 = −a x

u v + = 0+0 x y

OK, satisfied

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The Navier-Stokes y-momentum equation simply gives 0  0, while x-momentum yields

  2u  2u   u u  − cy   u + v  =  0 − acbe =   2 + 2  =  −c 2be −cy  x y  y   x  

(

)

Thus Navier-Stokes is satisfied if

(

)

a = c /

Ans. (a)

(b) The units of a and b are velocity, m/s. The units of c are inverse length, (c) At y = 0, the “wall”, u = 0 (no-slip) and

v = −a

m −1.

Ans. (b)

(wall suction).

Ans. (c)

(d) This is a boundary-layer flow with suction, and there is vorticity, highest at the wall:

z =

2-17

v u − = 0 − cbe −cy x y

Ans. (d)

Use Eq. (2-86) for density in the gravity term of the Navier-Stokes equation (2-29b) and

  0 in the acceleration term and let  = constant. For free convection, with U replaced by  / (  0 L ) and the hydrostatic assumption, p = 0 g = − 0g k, nondimensionalize Navier-Stokes

let

and show that the Grashof number appears. Solution: With these assumptions, the Navier-Stokes equation for free-convection becomes 0

DV  0  (T − T0 ) gk +  2 V Dt

Now introduce the dimensionless variables from Eqs. (2-81), with the single change

V* = 0 LV / , and collect terms, divide out the leading coefficient, and rearrange: 2 3 DV* 0 g  (Tw − T0 ) L  T *k + *2 V* = GrT * k + *2 V* 2 Dt* 

The Grashof number appears by itself, as the coefficient of the T * term. If we nondimensionalized the energy equation also, the Prandtl number would appear there.

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2-18 Flow through a well designed contraction or nozzle is nearly frictionless, as shown for example in White (2003), Sec. 6.12. Suppose that water at 20°C flows through a horizontal nozzle at a weight flow of 50 N/s. If entrance and exit diameters are 8 cm and 3 cm, respectively, and the exit pressure is 1 atm, estimate the entrance pressure from Bernoulli’s equation. 3 Solution: For water at 20°C,  = 998 kg/m . First find the inlet velocity from the weight flow:

w = 50

N  kg  m  m  2 =  g D12V1 =  998 3  9.81 2  ( 0.08m ) V1, solve for V1 = 1.02 s 4 s m  s 4 

Now use one-dimensional continuity to estimate V2 in the exit and substitute into Bernoulli:

 2  D1 V1 = D22V2 , 4 4 p1 +

 2

V12 = p1 +

2

or:

2  D1  m m  0.08m   V2 =   V1 =   1.02  = 7.23 s s  0.03m    D2 

998  998 (1.02 )2 = p2 + V22 = 101350Pa + ( 7.23)2 , solve p1 = 127, 000 Pa Ans. 2 2 2

2-19 Show, using Gauss’ theorem [Kreyzig (1999), Sect. 9.8] that the control volume mass relation, Eq. (2-111), leads directly to the partial differential equation of continuity, Eq. (2-6). Solution: Gauss’ theorem relates an area integral to a volume integral over the same region:

For any vector Q,

   Qd ( vol ) =  Q  dA vol

area

For continuity, Eq. (2-111), the relevant vector Q =  V. Also, for a fixed control volume, we can slip the time derivative inside the volume integral of Eq. (2-111) and then use the theorem:

d   d ( vol ) =  d ( vol )  dt CV  t CV

Hence Eq. (2-111) becomes

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    d ( vol ) +   V  dA =   +    V  d ( vol ) = 0 t t  CV CS CV 



Since the elemental volume d ( vol ) is arbitrary, it follows that the terms inside the braces [] are identically equal to zero. These terms are in the fact the equation of continuity, Eq. (2-6). NOTE: A similar derivation for a moving control volume is given by Panton (1995), p. 88.

2-20 In discussing incompressible flow with constant , Eq. (2-30), we cavalierly said, “many terms vanish” from Eq. (2-29a). Be less cavalier and show that the many viscous terms in Eqs. 2 (2-29a) do indeed reduce to the single vector term  V, in three dimensions.

Solution: The writer goes ahead and laboriously works out each viscous term for constant viscosity :

  u     u v      u w     v     u v    2  +    +   +    +  +  2  +    +   x  x  y   y x   z   z x   y  y  x   y x   +

   w v     w     u w      w v   +   +  2   +  +   +  =   + z   y z   z  z  x   z x   y   y z  



  2u  2 v  2 w    u v w  + + +   2 + 2 + 2  + . . . exactly similar terms in v and w   x  x y z  y z   x

The original terms split into two groups, the first of which contains  V = 0 and other is  2 multiplied by  2 ( u, v, w ) . Add up these latter three terms and you get  V.

Ans.

2-21 In deriving the basic equations of motion in this chapter, we skipped over the partial differential equation of angular momentum. Did we forget? Do some reading, perhaps in Lai et al. (1995) or Malvern (1997), and explain the significance of the angular momentum differential equation.

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Solution: In heavy-duty continuum mechanics, differential elements are allowed to have concentrated couples or moments per unit volume. Let us denote such a vector moment by T .

Take Fig. 2-1 of the text and specialize to moments about the z axis only, as shown in the sketch at below.

Summation of moments about the central z axis gives

( yx dx dz ) dy/2 + yx + ( yx /y ) dy  ( dx dz ) dy/2 − ( xy dy dz ) dx/2 −  xy + ( xy /x ) dx  ( dy dz ) dx/2 = Tz dx dy dz   Neglecting 4th-order differentials, we obtain ( yx − xy ) dx dy dz = Tz dx dy dz, or:

( yx − xy ) = Tz . Similarly, for the other two axes, ( zy − yz ) = Tx and ( xz − zx ) = Ty . So the concentrated couple causes shear-stress differences. If T = 0, ij = ij , the stress tensor is symmetric.

2-22

Normalize the Navier–Stokes equation with constant properties, specifically,

V p + ( V  ) V = − +  2 V + g t  (a) by setting:

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x L

y* =

y L

z* =

z L

* = L

t * = t

V* =

V U

p* =

p P

g = − gk

x* =

(b) Define the dimensionless parameters that appear in the following nondimensional form:

V* 1 1 St + ( V* * ) V* = − Eu* p* + *2 V* − k t Re Fr Solution: Rearrange the dimensional variables x = x* L

y = y* L

1

V = V*U

t=



t*

z = z*L p = p* P

=

1 *  L

g = − gk

Use the rearranged variables to normalize the Navier–Stokes equation. *  P  * *    *2 *  gL    L  V * * *  p +  2  V −  2  k   * + ( V  ) V = −  2   U  t  UL  U   U 

The dimensionless parameters are Strouhal number St =  L , Euler number Eu = U

Reynolds number Re =

(Ans.)

P , U 2

UL2 U2 , and Froude number Fr = .  gL

(Ans.)

2-23 In order to better analyze the boundary-layer structure over a flat plate with a characteristic length L , the traditional variables are rescaled using

x* =

x y u  p , y* = , u* = ,  * = , p* = L L U U U 2

 2u  2u where   in the axial 1 and  denotes the boundary-layer thickness. Show that x 2 y 2 L momentum equation written for the boundary-layer region. For steady-state motion with constant fluid properties, you may start with the axial momentum equation given by: 

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u

  2u  2u  u u p +  =− + 2 + 2  x y x y   x

Solution: Rearrange the variables and use them in the axial momentum equation

 u*U 2 u*  *U 2 u* U 2 p* U  2u * U  2u * + = − +  +  L x* L y* L x* L2 x*2  2 L2 y*2 2 u * * u * p* 1   2u *  L   2u *  u * +  * = − * +  *2 +   *2  x y x Re  x    y 

where, Re =

*

UL 

For steady-state motion with constant fluid properties on a flat plate parallel to the direction of flow, there is no axial pressure gradient. Therefore, the equation reduces to 2 u * * u * 1   2u *  L   2u *  u * + = +    x y* Re  x*2    y*2  *

Since it is assumed that the boundary layer is thin, i.e., 

L

2

 L  will be large.   

1 then 

Since each of the derivatives is assumed to be order one through proper scaling, then the second term  2u  2u in parentheses must be much larger than the first term, 2 (Ans.) x y 2

2-24

Given the dimensionally scaled expressions of the basic fluid dynamical forces:

3 Gravitational force: Fg = mg   L g

Viscous force: F =  A = ( dV dy ) A  UL Steady inertial force: Fi = mVdV dx  U L

2 2

Unsteady inertial force: Fu = mdV dt   L U  3

2 Pressure force: Fp = pA  pL 2 2 Compressibility force: Fc =  ( dp d  ) A  c L

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Surface tension force:

F = D  L

where the velocity gradient is approximated using dV dy = U L , verify the following characteristic ratios: Inertia and viscosity:

Fi UL = : Osborne Reynolds (British) F 

Inertia and compressibility:

Inertia and gravity:

Fi U 2 = = Fr : William Froude (British) Fg gL

Fi U 2 L = = We : Moritz Gustav Weber (German) F 

Inertia and surface tension:

Pressure and inertia:

Fi U 2 = = Ma 2 : Ernst Mach (German) Fc c 2

Fp Fi



Unsteadiness and inertia:

p = Eu : Leonhard Euler (Swiss) U 2

Fu  L = = St : Vincenc Strouhal (Czech physicist) Fi U

Unsteadiness and viscosity:

Fu L2 = = Sk 2 : Sir George Gabriel Stokes (Irish) F 

Viscosity and surface tension:

F F

=

U = Ca : Capillary Number 

Solution: Inertia and viscosity: Fi mVdV dx U 2 L2 UL : Osborne Reynolds (British) =  = F (  dV dy ) A UL 

(Ans.)

Inertia and compressibility: Fi mVdV dx U 2 L2 U 2 =  = 2 = Ma 2 : Ernst Mach (German) Fc  ( dp d  ) A  c 2 L2 c

(Ans.)

Inertia and gravity: Fi mVdV dx U 2 L2 U 2 =  = = Fr : William Froude (British) Fg mg  L3 g gL

(Ans.)

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Inertia and surface tension: Fi mVdV dx U 2 L2 U 2 L =  = = We : Moritz Gustav Weber (German) F D L 

(Ans.)

Pressure and inertia: Fp Fi

=

pA pL2 p  = = Eu : Leonhard Euler (Swiss) 2 2 mVdV dx U L U 2

(Ans.)

Unsteadiness and inertia: Fu mdV dt  L3U   L =  = = St : Vincenc Strouhal (Czech physicist) Fi mVdV dx U 2 L2 U

(Ans.)

Unsteadiness and viscosity: Fu mdV dt  L3U  L2 =  = = Sk 2 : Sir George Gabriel Stokes (Irish) F (  dV dy ) A UL 

(Ans.)

Viscosity and surface tension: F F

=

2-25

(  dV

dy ) A UL U  = = Ca : Capillary Number D L 

(Ans.)

In 1851, Stokes introduced an expression for the drag force, FD = 3VD , in the context of

1 2

 

2 laminar flow over a smooth sphere of diameter D . Express the drag coefficient CD = FD  V A 

as a function of the Reynolds number. Recall that A should be the projected area of the sphere.

 1 2  D 2  24 24 = Solution: CD = 3VD  V     =  2   ( VD  ) Re 2

(Ans.)

2-26 In normalizing the basic equations of motion in Sec. 2-9.1, we have assumed a characteristic time constant that scales with the unsteady period, i.e., the reciprocal of the frequency,  = 1  , when

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we set t * = t . Show that by taking a different t = tU L , for which  = L U represents the time for a fluid particle to travel a distance L at the characteristic mean-flow velocity, *

(a) the continuity and momentum equations reduce to

 * * * * Continuity: * +   (  V ) = 0 t * Navier–Stokes: 





T DV* 1 1 = −  *k − * p* + *   * *V* + ( *V* )  *   Dt Fr Re

(b) the energy equation reduces to

DT* Dp* 1 Ec = Ec * + *  ( k **T* ) + * Energy:  * Dt Dt Re Pr Re *

Solution: From Sec. 2-9.1, the continuity equation is given by St

 * + *  (  * V * ) = 0 . * t

Here, Strouhal number St =  L . From the question,  = 1  . Then the equation becomes U

L  * + *  (  * V * ) = 0 *  U t If a new  = L U

 * * * * is chosen, then the continuity equation reduces to * +   (  V ) = 0 . (Ans.) t

From Sec. 2-9.1, the Navier–Stokes equation is given by 

 *  St 





T  V * 1 * 1 * + ( V* * ) V*  = −  k − * p* +    * *V* + ( *V* )  *   t Fr Re 

Here, Strouhal number St =  L . From the question,  = L U . Then the equation reduces to U





* T  V * 1 * 1 * * * * * DV + V  V =  =−  k − * p* +    * *V* + ( *V* )  ( )  * *   Dt Fr Re  t 

* 

(Ans.)

From Sec. 2-9.1, the energy equation is given by 

 *  St 

T* Dp* 1 Ec * * * * + V  T = Ec + *  ( k **T* ) +  ( )  * * t Dt Re Pr Re 

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Here, Strouhal number St =  L . From the question,  = L U . Then the equation reduces to U

*  T* Dp* 1 Ec * * * * * DT + V  T =  = Ec + *  ( k **T* ) +  ( )  * * * Dt Dt Re Pr Re  t 

* 

(Ans.)

2-27 Renormalize the basic equations of motion in Sec. 2-9.1 assuming an incompressible fluid with negligible dissipation function  as well as constant viscosity and thermal conductivity,

 = 0

and k = k0 . The normalization will correspond to

xi* =

xi L

t * = t * =

 0

V* = T* =

V U

p* =

T − T0 Tw − T0

p p0

=

=0

cp c

V = 0

* = L

(a) Show that the continuity and momentum equations can be simplified into Continuity: St

 * + *  (  * V * ) = 0 * t

 V*  1 * 1 *2 * Navier–Stokes:  *  St * + V* *V*  = −  k − Eu* p* +  V Fr Re  t 

(b) Show that the energy equation reduces to  T*   − 1  p*  1 Energy:  *  St * + ( V* * ) T*  = St * + ( V* * ) p*  + *2T*   t   t Re Pr    

(c) Under what circumstances does the Euler number become important? (d) What happens to this set of equations if the density

 = 0

is assumed to be constant?

Solution: Continuity equation is given by  +   (  V ) = 0 . t

* * Using  = 0  , V = UV ,  =

 * U  0 * * t*   (  * V* ) = 0 . , and t = gives 0 * + t L L 

 * U 0 Divide both sides by and using St =  L gives St * +   t L U *

( V ) = 0. *

*

(Ans.)

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Navier–Stokes equation for incompressible flow 

DV  V  =   + ( V  ) V  =  g − p + 2 V Dt  t 

Using normalized variables in the above equation gives 

0  * U 

 p U V * U 2 + V* * ) V*  = − 0 gk0  * − 0 * p* + 02 *2 V* ( * t L L L 

Dividing by

 0U 2   L V *  gk L p 0 gives  *  + ( V* * ) V*  = − 02  * − 0 2 * p* + *2 V* * L U 0U 0UL  U t 

 V *  1 * 1 *2 * Or,  *  St * + ( V* * ) V*  = −  − Eu* p* +  V Fr Re  t 

(Ans.)

The energy equation for incompressible flow with negligible dissipation function  is cp

DT Dp = T +   ( k T ) Dt Dt

 T   p  + ( V  ) T  =  T  + ( V  ) p  +   ( kT )  t   t 

Or,  c p 

Using normalized variables in the above equation gives

 T − T   T * U  0  *c p  w 0   * + ( V* * ) T *  = L   T0   t *  T −T   p U  k  T −T    w 0  T * p0  * + ( V* * ) p*  + 02  w 0  *2T * L  t  L  T0   T0   T −T U Dividing by 0c p  w 0  gives  T0  L   L T *   T * p0 L   L p* k0 * * * + V  T = + ( V* * ) p*  + *2T * ( )   * * U  t  c U U  t  c UL     0 p 0 p

* 

Using the expression for coefficient of thermal expansion  and specific-heat ratio  

 *  St 

 T *  − 1  p* 1 * * * + V  T = St * + ( V* * ) p*  + *2T * ( )   * t   t Re Pr   

(Ans.)

For such boundary conditions when the inlet and outlet flows are approximately uniform and parallel, the viscous normal stresses become locally negligible and the only important boundary

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work is due to pressure forces at the inlet and outlet. Then, due to the presence of significant pressure difference, the Euler number becomes important. (Ans.) If density

 = 0 is assumed to be constant, the  2 terms in the set of equations become zero. (Ans.)

2-28 In deep water, the average speed V of a free-surface wave can be expressed as a function of the density,  , gravity, g , depth, D , and spatial wavelength,  . Obtain a dimensionless representation of the free-surface wave speed as a function of these controlling parameters. Solution: Apply the Buckingham  procedure: Number of parameters n = 5: Primary dimensions r = 3:

 V D   Where  L M  L L3 t

g L t2

Repeat parameters m = 3:

V

M

D



g



L t

     L  D



g

Then n − m = 2 dimensionless groups will set up one dimensionless equation. a b V L M c L Summing exponents, 1 = g a  b DcV =  2   3  ( L )   = M 0 L0t 0 . Hence, 1 = . gD t   L  t

 L M c M Summing exponents, 2 = g a  b Dc =  2   3  ( L )  2  = M 0 L0t 0 . Hence,  2 = . g  D2 t   L  t  a

By checking using F The relation between

L

b

t as primary dimensions, give 1 = 1 and 2 = 1 .

1 and  2

is

   1 = f ( 2 ) or, V = gD f  2   gD 

(Ans.)

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2-29 A model sphere of diameter D is to be immersed in a water tunnel running at a speed of V in order to obtain data that can be used to predict the drag force on a weather balloon that is traveling in air at 1.5 m/s . For a model diameter of 50 mm , the measurement of the drag force on the small sphere yields 3.78 N . Assuming similarity conditions have been established between the spherical lab-scale model and the air balloon, what should be the water tunnel speed if the diameter of the balloon is 3 m ? Estimate the drag force on the full-scale balloon. Solution: For water at 20o C temperature, take For water at 20o C temperature, take

 = 998 kg/m3 and

 = 0.001 kg/m  s .

 = 1.2255 kg/m3 and  = 1.78 10−5 kg/m  s .

The balloon velocity follows dynamic similarity, which requires identical Reynolds numbers Remodel =

998 (1.5)( 0.05) 1.2255 (Vballoon )( 3) VD = = 74850 = Reprototype =  model 0.001 1.78 10−5 −1

Therefore, Vballoon  3.624  10 m/s . (Ans.) Then, the two spheres will have identical drag coefficients.

CD ,model =

F V 2 D 2

= model

3.78 998 (1.5) ( 0.05) 2

Therefore, Fballoon  9.754 10

−1

2

= 0.6733 = CD ,prototype =

N.

Fballoon

1.2255 ( 3.624 10−1 ) ( 3) 2

2

(Ans.)

2-30 Surface tension is responsible for causing capillary waves to develop on a liquid-free surface with an average speed, V , that is dependent on the fluid density,  , surface tension,  , and spatial wavelength,  . Obtain the nondimensional relations linking the speed of the capillary waves to these three independent quantities. Solution: Number of parameters n = 4 : Primary dimensions r = 3:

 V  Where  L  t

 M t2

M

V







L t

    M  L L3 



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Repeat parameters m = 3:







Then n − m = 1 dimensionless group will set up one dimensionless equation. a c  L  M  bM   Summing exponents, 1 = V   V =   2  ( L )  3  = M 0 L0t 0 . Hence, 1 = V .   t  t  L  a

By checking using F

b

c

t as primary dimensions, give 1 = 1 .

L

 =C. 

The functional relationship is 1 = V

Then, the nondimensional speed of capillary waves is V = C





.

(Ans.)

2-31 The lifting force F on an air-launched rocket depends on its length L , velocity V , diameter D , angle of attack  , density  , dynamic viscosity  , and speed of sound c . Using the Buckingham Pi theorem, determine the dimensionless groups or Pi parameters needed to characterize the functional relation between the lifting force and the quantities that affect it. Solution: Apply the Buckingham  procedure: Number of parameters n = 8:

F

Primary dimensions r = 3:

 F L  Where   ML  2 L  t

L

V

M

L t



V

D



L t

L

1

M L3

Repeat parameters m = 3:



V

 M Lt

D







c

 c   L  t 

D

Then n − m = 5 dimensionless groups will set up one dimensionless equation. a

b

F c  ML  M  L 1 =  V D F =  3    ( L )  2  = M 0 L0t 0 . Hence, 1 = V 2 D 2 L  t  t  a

b

c

a

(Ans.)

b

c L M   L 2 =  aV b Dc L =  3    ( L ) ( L ) = M 0 L0t 0 . Hence,  2 = D L  t

(Ans.)

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 cM  M   L 3 =  aV b Dc  =  3    ( L )   = M 0 L0t 0 . Hence,  3 = VD L  t  Lt  a

b

a

b

a

b

(Ans.)

c L cV M   L 4 =  V D c =  3    ( L )   = M 0 L0t 0 . Hence, 4 = 2 D L  t t (Ans.) a

b

c

c M   L 5 =  V D  =  3    ( L ) (1) = M 0 L0t 0 . Hence, L  t a

b

c

5 = 

(Ans.)

2-32 A solid propellant rocket chamber is often modeled as a porous tube with sidewall injection (Majdalani 1999). The oscillatory Stokes boundary layer that develops inside a porous tube with sidewall injection (due to pressure oscillations) is affected by variations in several parameters. These include the injection velocity U , the frequency of oscillations  , the kinematic viscosity  , the hydraulic radius of the chamber R , the chamber length L , and the axial location within the chamber X . The dependence of the Stokes boundary-layer thickness  on these parameters may be expressed as  = f (U , ,, R, L, X ) . Using scaling principles, show that the penetration depth of the unsteady rotational wave, y p =  R , may be related to the following dimensionless parameters

yp =

 R X  = F  Mj , St , ,  R L L 

where Mj =

U3 L and St = 2  R U

Note that the square, square-root, or reciprocal of a dimensionless parameter as well as the product of two dimensionless parameters, such as X R and R L , can also produce a nondimensional group. Solution: Number of parameters n = 7 : Primary dimensions r = 3:

M

   Where   L 

 L X    L L  

U





R

L t

1 t

L2 t

L

Repeat parameters m = 2 :

U

 U

  R L X

L t

R

Then n − m = 5 dimensionless groups will set up one dimensionless equation.

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The dimensionless groups in this case can be easily identified through inspection (check using F , L , t as primary dimensions is not really needed here). 3

X  = U  R , 4 , 5 =  L 1 = ,  2 = , 3 = 2  R U L R L The relation between

1 ,  2 ,  3 ,  4 , and  5 is

3 1 = f ( 2 , 3 , 4 , 5 ) or, y p =  = f  R , X , U2 ,  L  = f  Mj, St , R , X 

 L L  R U 

R

2-33



L L

(Ans.)

The fluid hammer or hydraulic shock can occur when a valve is closed rapidly in a fluid

supply line. Given a circular tube, the pressure impulse p = p1 − p2 , also known as the strength of the fluid hammer, depends on the fluid density,  , the speed of sound in the fluid, c , the diameter of the tube, D , the dynamic viscosity,  , and the mean-flow velocity, V . Using the Buckingham Pi theorem, identify the dimensionless groups that will minimize the number of test cases needed to fully characterize the pressure impulse sensitivity to these five independent parameters. Note that the dimensionless pressure impulse is traditionally referenced to the speed of sound and that the Reynolds number based on the speed of sound is known as the acoustic Reynolds number. Solution: Number of parameters n = 6 : Primary dimensions r = 3:

 p  Where  M  2  Lt

 M L3

M

c L t

Repeat parameters m = 3:

D





c

D



V

L t

 M Lt

L

p

D

 V   L   t  V

Then n − m = 3 dimensionless groups.

The dimensionless groups are as follows: a b p c M  M   L 1 =  aV b Dc p =  3    ( L )  2  = M 0 L0t 0 . Hence, 1 = V 2 L  t  Lt 

(Ans.)

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a

b

c L c M   L 2 =  aV b Dc c =  3    ( L )   = M 0 L0t 0 . Hence,  2 = V L t t      

(Ans.)

 cM  M   L 3 =  V D  =  3    ( L )   = M 0 L0t 0 . Hence,  3 = VD L  t  Lt 

(Ans.)

a

a

b

b

c

2-34 The drag force acting on a submarine can be taken to be a function of the water density,  , speed, V , viscosity,  , and displaced volume,  . Using a 1: 50 scaled-down model of the submarine in a laboratory-size water tunnel, start by specifying the dimensionless parameters needed to guide your drag force measurements. Assuming a leveling of the drag force coefficient at increasing speeds, provide a best guess of the drag force on a full-scale prototype traveling at 27 knots if a value of 13 N is obtained in a water tunnel running at 10 knots . Solution: Number of parameters n = 5: Primary dimensions r = 3:

 F  Where   ML  2  t



V

M L3

L t

Repeat parameters m = 3:

M

 M Lt 

L

F



V





t

     L3   V



Then n − m = 2 dimensionless parameters.

The dimensionless parameters are as follows: a

b

F c  ML  M   L 1 =  aV b c F =  3    ( L3 )  2  = M 0 L0t 0 . Hence, 1 = V 2 2 3 L  t  t  (Ans.)

 cM  M   L 2 =  V   =  3    ( L3 )   = M 0 L0t 0 . Hence,  2 = V 1 3 L  t  Lt  a

a

b

b

c

(Ans.)

From second dimensionless parameter,

p m = 13  pVp p mVmm1 3

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13

13 13  p   p   Vm  m   10 knots  1   10 knots  1  = Therefore,     = (1)    =     m  m   Vp   27 knots   50   27 knots  50  p  

From first dimensionless parameter, Fp

 pVp  p 2

23

=

where p and

Fm mVm 2m 2 3 m

subscripts in these equations represent the prototype and the model, respectively.

Therefore,

   V  Fp = Fm  p  p    m  Vm 

2

 p     m 

23

13

2

 10 knots  1   27 knots   50  = 13 N         27 knots  50   10 knots   1 

23

= 129.3095 knots (Ans.)

2-35

Experimental measurements of the pressure drop across a sudden contraction in a circular

p = p1 − p2 can be a function of the density,  , dynamic viscosity,  , meanflow velocity, V , and both large and small diameters, D1 and D2 . Identify the dimensionless groups tube suggest that

that will minimize the number of test cases needed to fully characterize the pressure dependence on these five independent parameters.

Solution: Number of parameters Primary dimensions r = 3:

M

n = 6 : p  D1 D2  V L t

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 p  Where  M  2  Lt

 M L3

Repeat parameters

D1

D2

L

 M Lt

L

 V   L   t 

m = 3:  D1 V

Then n − m = 3 dimensionless groups.

The dimensionless groups are as follows: a b p c M  M  L 1 =  aV b D1c p =  3    ( L )  2  = M 0 L0t 0 . Hence, 1 = V 2 L  t  Lt  a

(Ans.)

b

D2 c M   L 2 =  V D D2 =  3    ( L ) ( L ) = M 0 L0t 0 . Hence, 2 = D1 L  t

(Ans.)

 cM  M   L 3 =  aV b D1c  =  3    ( L )   = M 0 L0t 0 . Hence,  3 = VD L  t  Lt 

(Ans.)

a

b

c 1

a

b

2-36 Water at 15 C flows through a 0.1 m internal diameter galvanized iron pipe (  = 0.00015 m ) at a mass flow rate, m , of 15 kg/s . We are told that: The total pipe length, which is L = 150 m , runs in a horizontal plane. The pipe contains one check valve ( Le D = 150 ) and four 90 elbows ( Le The properties of water are:

D = 30 each).

 = 1000 kg/m3 ,  = 0.00114 N  s/m2 , and g = 9.81 m/s2 .

(a) Find the Reynolds number. Is the flow turbulent? (b) Find the Darcy friction factor f . (c) Find the major head loss,

hM .

(d) Find the minor head loss,

hm .

(e) What percentage of the total head loss is due to minor losses? (f) Find the pressure drop across this pipe. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -32-

Solution: We know that m = VA . Thus, V =

m = A

15 kg/s = 1.9099 m/s . 2 3  2 (1000 kg/m )  4 ( 0.1) m 

VD (1000 kg/m ) (1.9099 m/s )( 0.05 m ) Re = =  83767  ( 0.00114 N  s/m2 ) 3

Darcy friction factor is calculated from

(

3500 ) . (Turbulent)

 d 1 2.51  −2.0log  +  3.7 Re f f 

  ; f  0.0241 . 

(150 m ) (1.9099 m/s ) = 6.7209 m L V2 = ( 0.0241) Major head loss hM = f . D 2g ( 0.1 m ) ( 2  9.81 m/s2 )

(Ans.)

(Ans.)

2

Sum of loss coefficients

(Ans.)

 K = 0.0241 (150 + 4  30) = 6.507 .

(1.9099 m/s ) = 1.3146 m V2 = ( 6.507 ) Minor head loss hm = (  K ) . 2g ( 2  9.81 m/s2 ) 2

(Ans.)

Total head loss hL = hM + hm = 6.7209 m + 1.3146 m = 8.0355 m . Therefore, % contribution of

hm to hL is %contribution =

hm 1.3146 m 100 = 100 = 16.36% . (Ans.) hL 8.0355 m

Pressure drop p = 1 V 2  K = 1 (1000 kg/m3 ) (1.9099 m/s )2 ( 6.507 ) = 11.8679 kPa . 2

2

(Ans.)

Consider a water pipe of length 100 m , diameter of 0.1541 m ( 6 inches ), roughness of 4.6 10−5 m (DN 150 Schedule 40 Steel), viscosity of 1.08 10−5 m2 /s , specific weight of 8630 N/m3 , and an inlet pressure of 120 kPa at Point 1. In the presence of minor losses, determine:

2-37

(a) The maximum permissible volumetric flow rate (in m3 /s ) and average velocity at Point 2 that will ensure that the pressure drop in the pipe does not exceed 60 kPa . (b) The total head loss in meters. (c) The percentage of the total head loss that corresponds to minor losses. We are told that the pipe runs in a horizontal plane and that it contains an open butterfly valve (

Le D = 45 ) and two standard 90° elbows ( Le D = 30 each). Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -33-

Solution: Pressure drop across the pipe is p = 60 kPa . Cross-sectional area A = Density  =

 D2 4

=

 ( 0.1541 m )

2

4

= 0.01865 m 2 .

 8630 N/m3 = = 879.7146 kg/m3 . g 9.81 m/s 2

Dynamic viscosity  =  = ( 879.7146 kg/m3 )(1.08 10−5 m 2 /s ) = 0.0095 N  s/m 2 . Velocity in terms of friction factor and pressure drop is given by

V=

2 Dp = f

2  ( 0.1541 m )( 60000 Pa )

(100 m ) (879.7146 kg/m ) ( f ) 3

=

Initial guess for friction factor f = 0.02 gives V =

0.210205 f 0.210205 0.210205 = = 3.242 m/s . f 0.02

3 VD (879.7146 kg/m ) ( 3.242 m/s )( 0.1541 m ) =  46262 . For f = 0.02 , Re =  ( 0.0095 N  s/m2 )

Using Haaland formula, friction factor can be estimated from 1.11  6.9  6.9   D 1.11   4.6 10−5 m 0.1541 m   1 = −1.8log  + +     = −1.8log  3.7 f  46262   Re  3.7    

Thus, f = 0.021946 . An improved guess of f = 0.022168 is estimated through an iterative approach for convergence. The respective velocity is V = 3.0793 m/s . Then, volumetric flow rate is Q = VA = ( 3.0793 m/s )  ( 0.01865 m2 ) = 0.05743 m3 /s . Sum of loss coefficients

(Ans.)

 K = 0.022168  ( 45 + 2  30) = 2.32764 .

Total head loss hL = hM + hm = f

Therefore, % contribution of

L V2 V2 + ( K ) = 8.0772 m . D 2g 2g

hm to hL is %contribution =

(Ans.)

hm 1.1249 m 100 = 100 = 13.9269% . hL 8.0772 m

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2-38

A 200 m long cast iron pipe has a roughness of  = 0.00025 m and a diameter of 0.05 m . 3

It carries water with a density of 1000 kg/m and a kinematic viscosity of 10−6 m2 /s . The gauge pressure in the water main is 1500 kPa . The supply line will require the installation of eight elbows ( Le D = 30 each), two standard tees ( Le D = 60 each), and one square-edged entrance ( K = 0.5 ), in the total length of 200 m . The gage pressure required at the discharge point is 500 kPa . Find the discharge flow rate Q . Solution: Pressure drop across the pipe is p = p1 − p2 = 1500 kPa − 500 kPa = 1000 kPa . Cross-sectional area A =

 D2 4

=

 ( 0.05 m ) 4

2

= 0.0019635 m 2 .

Dynamic viscosity  =  = (1000 kg/m3 )(10−6 m2 /s ) = 0.001 N  s/m 2 . Velocity in terms of friction factor and pressure drop is given by

V=

2  ( 0.05 m )(1000000 Pa ) 2 Dp 0.5 = = 3 f f ( 200 m ) (1000 kg/m ) ( f )

Initial guess for friction factor f = 0.02 gives V =

0.5 0.5 = = 5 m/s . f 0.02

3 VD (1000 kg/m ) ( 5 m/s )( 0.05 m ) = = 250000 . For f = 0.02 , Re =  ( 0.001 N  s/m2 )

Using Haaland formula, friction factor can be estimated from 1.11  6.9   D 1.11   6.9 1  0.00025 m 0.05 m   = −1.8log  +  +   = −1.8log    3.7 f    Re  3.7    250000 

Thus, f = 0.030775 . An improved guess of f = 0.03086 is estimated through an iterative approach for convergence. The respective velocity is V = 4.0252 m/s . Then, volumetric flow rate is Q = VA = ( 4.0252 m/s )  ( 0.0019635 m2 ) = 0.0079 m3 /s .

(Ans.)

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2-39 Water at 15 C flows through a galvanized iron pipe with a roughness of  = 0.00015 m , at a mass flow rate, m , of 20 kg/s . We are told that: The total pipe length, which is L = 200 m , runs in a horizontal plane. The pipe contains one check valve ( Le The properties of water are:

D = 55 ), and four

90 elbows ( Le

D = 30 each).

 = 1000 kg/m3 ,  = 0.00114 N  s/m2 , and g = 9.81 m/s2 .

The total pressure drop across the pipe is 169 kPa . What is the diameter of the pipe? Solution: We know that Q =

m 20 kg/s = = 0.02 m3 /s . 3  1000 kg/m

Assume a pipe diameter of D = 0.25 m .

4 (1000 kg/m3 )( 0.02 m3 /s ) VD 4 Q = =  89350 . Reynolds number in terms of Q is Re =   D  ( 0.00114 N  s/m2 ) ( 0.25 m ) Using Haaland formula, friction factor can be estimated from

 6.9   D 1.11   6.9  0.00015 m 0.25 m 1.11  1 = −1.8log  +  +   = −1.8log    3.7 f    Re  3.7    89350  Thus, f = 0.020758 . Pressure drop in terms of Q is p = f

L V 2 8 fL  Q 2 = 2 5 . D 2  D 1

2 5 3 3  8 fL  Q 2   8 ( 0.020758)( 200 m ) (1000 kg/m )( 0.02 m /s )  Then, D =  2 .  =   2 (169000 Pa )   p    1 5

An improved guess of D = 0.02614 m is estimated through iteration for convergence.

(Ans.)

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2-40 Water with a density of  = 1000 kg/m negotiates a 180 turn through an elbow as shown below. As the water enters the elbow, its gage pressure is 200 kPa . The pressure at the outlet section 3

is

atmospheric.

Assuming

uniform

properties

everywhere,

we

have

A1 = 0.002 m 2 ,

A2 = 0.0006 m 2 , and U1 = 2 m / s . Determine the horizontal force that is needed to prevent the elbow from separating. Are the bolts in tension or compression?

Solution: Momentum flux in

x

direction for the elbow Fx = FSx + FBx =

  u  dV +CS u VdA t CV

assuming steady, uniform, and incompressible flow with atmospheric pressure at exit. Also assume that the horizontal force Rx to hold the elbow at its place is from left to right. Therefore, Rx = − p1g A1 −  (U12 A1 + U 22 A2 ) . From continuity,

U1 A1 = U 2 A2 .

0.002 m2 ) ( A1 = ( 2 m/s ) = 6.6667 m/s . Then, U 2 = U1 A2 ( 0.0006 m2 ) Then,

(

Rx = − ( 200000 kg/m  s 2 )( 0.002 m 2 ) − (1000 kg/m3 ) ( 2 m/s ) ( 0.002 m 2 ) + ( 6.6667 m/s ) ( 0.0006 m 2 )

Rx = −434.6669 N

2

2

(Ans.)

The negative sign indicates that the horizontal force needed to prevent the elbow from getting separated is acting from right to left. Therefore, the bolts are in compression. (Ans.)

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)

2-41

Consider a two-dimensional planar channel with one inlet and two outlets that include a

reducing bend. The velocity distribution at the inlet varies linearly from zero to U1 . The velocity at the two outlets is uniform and the motion may be taken to be incompressible and steady. Determine (a) the maximum velocity at Section 1 using the integral, control-volume approach and (b) the reaction forces needed to support the channel.

Solution: The control-volume around sections 1, 2, and 3 yields Q1 = Q2 + Q3 ; in other words,

U1 A1 = U 2 A2 + U 3 A3 

Thus, U1 =

U 2 A2 + U 3 A3 = A1

2    + ( 6 m/s )  ( 0.1 m )   4  = 0.6667 m/s 2   ( 0.6 m )  4 

( 2 m/s )  ( 0.3 m ) 4

Considering water having a density of

2

 = 998 kg/m3 flowing through the channel at

(Ans.)

20 C ,

the horizontal reaction force is Rx = m2u2 + m3u3 − m1u1 = ( Q2 )U 2 + ( Q3 )U3 cos 45 − ( Q1 )U1 . Therefore,

Rx = 356.2828 N .

The vertical reaction force is Ry = m3u3 = ( Q3 )U3 sin 45 = 199.5299 N .

(Ans.) (Ans.)

2-42 Consider the steady incompressible motion of a fluid across an axisymmetric tube of length L and radius R = 3 cm . (a) Determine the uniform velocity at the inlet, U , if the velocity profile at the outlet can be   r2  approximated using u ( r ) = U max cos  2  .  2R  Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -38-

(b) Evaluate your result for U max = 10 m/s and water with a density of

 = 1000 kg/m3 .

(c) Determine the difference in the fluid momentum flux between the inlet and outlet sections of the tube.

Solution: For the given steady incompressible fluid motion inside the tube

−min + mout = 0

Then, with  = constant , we have −Qin + Qout = 0 . R   r2  −U  R 2 +  U max cos  2  2 rdr = 0  2R  0 R   r2    r2  2 2 −U  R + 2 R U max  cos  2  d  2  = 0  2R   2R  0 R

   r 2  −U  R + 2 R U max sin  2   = 0   2R  0 2

U=

2



2

(Ans.)

U max

For U max = 10 m/s , the inlet velocity is U = 3.1416 m/s .

(Ans.)

(

)

Difference in fluid momentum flux is mout u ( r ) − minU =  Qout u ( r ) − QinU .   r2    r2  d − U 2 R 2 2   2   2R   2R 

R

2  ( Qout u ( r ) − QinU ) = 2 U max R 2  cos 2  0

= 2 U

2 max

R

1   r   r  4 2 R  sin  2  + 2  − U max R2  4  R  4R 0  2

2

2

 2  2 = 2 U max R2  −  4 

(Ans.)

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2-43 Consider a contraction between two pipes in a hydraulic system, which is known as a reducer. With the pressure and velocity measurements available across the reducer, and given the empty volume and mass of the reducer (r , mr ) , determine the reaction forces that must be supplied by the surrounding system to hold the reducer. You may assume that gravity is downward.

Solution: From conservation of mass, we have U1 A1 = U 2 A2 . Therefore, U 2 = U1

A1 D2 = U1 12 . A2 D2

From conservation of momentum in x direction, we have

F = R + F x

surface, x

+ Fbody, x = −mU 1 1 + m2U 2

Since there is no body force (weight) in the x direction,

Rx + Fsurface, x = Rx + p1 A1 − p2 A2 = U1 A1 (U 2 − U1 )    2 2  D12  2 2 Or, Rx = U1 A1 (U 2 − U1 ) − p1 A1 + p2 A2 =  U1 D1  2 − 1 − p1D1 + p2 D2  4  D2  

(Ans.)

From conservation of momentum in y direction, we have

F = R

y

+ Fsurface, y + Fbody , y = 0

Since there is no surface force (pressure) in the y direction,

Ry + Fbody , y = Ry − mr g − r g = 0

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Or, Ry = g ( mr + r ) .

2-44

(Ans.)

A two-dimensional velocity field is given by

u=−

Ay Ax v= 2 2 2 , x +y x +y 2

where A is a constant. Does the field satisfy continuity? Solution: The continuity equation should satisfy

Here,

u v + =0. x y

u v   Ay    Ax  2 xyA 2 xyA + = − 2 +  2 = − =0. 2 2  2  x y x  x + y  y  x + y  ( x 2 + y 2 ) ( x 2 + y 2 )2

Therefore, the field satisfies continuity.

(Ans.)

2-45 For a certain incompressible, three-dimensional flow field, the velocity components in the x and y directions are given by u = x 2 + y 2 + z 2  v = xy + yz + z

(a) Determine the simplest velocity component rate is zero.

w

in the z direction so that the volume dilatation

(b) Determine the components of the vorticity vector. Is this an irrotational flow field? Solution: The volume dilatation rate is given by   V =

Stated that

u v w + + . x y z

w u v w u v = − − = − ( 3x + z ) . + + = 0 . Therefore, z x y x y z

Then, the velocity component in the

z

 z2  direction is w = −  3xz +  . 2   (Ans.)

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w=

1  w v   u w   v u   1 − i +  −   k  ==  − ( y + 1) i + 5 z j − y k   0  j+  − 2  y z   z x   x y   2

Thus, the flow is rotational.

(Ans.)

2-46 For a two-dimensional flow field, the velocity is defined by V = ( x 2 − y 2 ) i − 2 xyj . Determine whether this motion is compressible or not. Solution: Given V = ( x 2 − y 2 ) i − 2 xyj

(Eq-1)

Generally, the two-dimensional fluid flow is given by V = ui + vj

(Eq-2)

where u and v are velocity components. Comparing (Eq-1) and (Eq-2), we get

u = x2 − y 2 v = −2xy

Partially differentiating the above equations with respect to x and y, respectively, u = 2x x

u = −2 x y For the flow to be incompressible,

u u + =0 x y

2x − 2x = 0 Hence, the flow is incompressible.

2-47 For an incompressible, two-dimensional flow field, the velocity component in the y direction is given by

v = 3xy + x 2 y Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -42-

Determine the velocity component u in the x direction so that the volume dilatation rate is zero. Solution: Given:

v = 3xy + x 2 y

The flow is incompressible, and for zero volumetric rate in a 2D flow,

u v + =0 x y

(Eq-1)

Since

v = 3x − x 2 y From (Eq-1), we have u = −3x + x 2 x

(Eq-2)

(Eq-2) can be integrated with respect to x as follows:

 du = − 3xdx +  x dx + f ( y) 2

u=

−3 2 x3 x + + f ( y) 2 3

(Ans)

where f(y) is the undetermined function of y.

2-48 The three components of the velocity in a flow field are given by  u = x 2 + y 2 + z 2  2 v = xy + yz + z  1  w = −3xz − z 2 + 4  2

Determine the divergence (volumetric dilatation rate) and interpret the results. Determine an expression for the vorticity vector. Is this an irrotational flow field? Solution: The three components of the velocity in a flow field are given by

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 u = x 2 + y 2 + z 2  2 v = xy + yz + z  1  w = −3xz − z 2 + 4  2

For 3D flow, we find w. w u v = −3z = 2x , =y , x x x

v u w = x+z , = 2y , =0 y y y w v u = −3x − z = y + 2z , = 2z , z z z

Rotation component along x direction is Wx =

1  w v  1 y  −  = ( − y − 2z ) = −  + z   2  y z  2 2 

Rotation component along the y direction is

1  u w  1 5z Wy =  −  = ( 2 z − ( −3z ) ) = 2  z x  2 2 Rotation component along the z direction is 1  v u  1 −y Wz =  −  = ( y − 2 y ) = 2  x y  2 2

Rotation vector is given by

 y   5z   y  W = −  + z  iˆ +   ˆj −   kˆ 2   2  2 Since no component of the rotation vector is zero, this is rotational flow field. Volumetric dilation rate is given by

v =   v =

u v w + + x y z

 2z  = 2 x + x + z + ( −3x ) −   = 0  2 Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -44-

There is no change in the volume of the fluid element as it moves from one location to the other.

2-49 The velocity vector in a certain flow field is prescribed by V( x, y, t ) = 2xti − 2 ytj

where (x, y) have units of feet and t has units of seconds. Determine expressions for the local (unsteady) and convective (spatial) acceleration terms. Evaluate the magnitude and direction of the velocity and acceleration at the point x = y = 2 ft at time t = 0 s. Solution: Given, velocity vector is V( x, y, t ) = 2xti − 2 ytj

Comparing the above equation with v = uiˆ + vjˆ + wkˆ ,

u = 2 xt , v = −2 yt , w = 0 Acceleration is given by a = axiˆ + a y ˆj + az kˆ . ax =

du u u u u =u +v +w + dt x y z t Convective part

Local part

dv v v v v =u +v +w + dt x y z t dw w w w w az = =u +v +w + dt x y z t

ay =

(Eq-1)

Local acceleration point can be calculated as follows:

u = 2 xt

u = 2x t

v = 2 yt

v = 2y t

w=0

w =0 t

Local acceleration = 2 xˆi + 2 yˆj Convective part can be calculated as follows:

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u = 0 , u = 0 u = 2 xt , u = 2t , x

v = 2 yt ,

y

z

v v = 2t , v = 0 =0 , y x z

So, combining the local and convective parts of acceleration in (Eq-1),

ax =

u u u u +u +v +w t x y z

= 2 x + ( 2 xt ) 2t

ax = 2 x + 4 xt 2 ay =

v v v v +u +v +w t x y z

= 2 y + ( 2 yt )( 2t ) ay = 2 y + 4 yt 2 2 2 And convective acceleration = 4 xt ˆi + 4 yt ˆj .

Net acceleration = ax ˆi + a y ˆj = ( 2 x + 4 xt 2 ) ˆi + ( 2 y + 4 yt 2 ) ˆj . Velocity x = y = 2ft, t = 0.

V ( x, y, t ) = 2 xtˆi − 2 ytˆj Substituting, x = y = 2 and t = 0, V=0 Acceleration at x = 2, y = 2, and t = 0 is a = ( 2 x + 4 xt 2 ) ˆi + ( 2 y + 4 yt 2 ) ˆj

a = ( 4 + 0 ) ˆi + ( 4 + 0 ) ˆj=4iˆ + 4jˆ The magnitude of the acceleration = a = 42 + 42 = 4 2 ft/s 2 (Ans)

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The direction of the acceleration is tan  =

ay ax

=

4 =1 4

 = tan −1 (1) = 45

(Ans)

2-50 For steady flow, show that the curl of the inviscid momentum equation reduces to  (V  ω) = 0 , where  =  V . Start with the steady expression of Euler’s equation, namely, (V )V =

1 (V  V) − V  (  V) = −p /  + g 2

Solution: Steady expression of Euler’s equation is (V )V =

1 (V  V) − V  (  V) = −p /  + g 2

where  =  V  V2  −p  ( V  ) V =   +g  − V  =   2  2

2 where V = V = V  V

By taking the curl of Euler’s equation, we get the velocity equation.

  (Euler’s equation)  V2     ( V  ) V =     −   (V  )  2 

p  = −    + g   

(Eq-1)

Applying the identity  scalar = 0 , the pressure term vanishes, provided that the density is uniform.  p  V2      + g   = 0 and       2  

 =0 

Therefore, (Eq-1) remains as Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -47-

  ( V  ) V = − ( V   ) For inviscid fluid, V = 0

  ( V   ) = 0

(Ans)

2-51 For planar flow in the xy plane, show that the vorticity has only one nonvanishing component in the z direction, specifically, ω = z k . Show that the vorticity, when written in terms of the incompressible stream function, becomes   2  2  + 2  2 y   x

z = − 

, where u =

 y

v=−

 x

Solution: For 2D planar consideration,

u v + =0 x y   2u  2u  u u u 1 p +u +v = gx − +  2 + 2  t x y  x y   x

And

  2u  2u  v v v 1 p +u +v = gy − +  2 + 2  t x y  y y   x

The pressure and gravity can be eliminated by cross-differentiation, that is, by taking the curl of 2D vectorial momentum equation. The result is 

  2z  2z  z  z + u z + =  +  2 t x y y 2   x

 v u  where  z =  −   x y 

(Eq-1)

(Eq-2)

So the vorticity has only one nonvanishing component in the z direction, specifically

ω = z k .

For incompressible stream functions,

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u=

 y

v=−

 (given) x

Therefore, (Eq-2) becomes   2  2  − 2  2 y   x

z =  −

  2  2   z = −  2 + 2  y   x

(Ans)

2-52 For steady, inviscid, planar flow in the xy plane, (a) show that  (V  ω) = 0 reduces to

  z  z   −   = 0 , where u = y y x   x y (b) Show that any solution of the form

v=−

 x

z = f ( ) will satisfy this condition, including the linear

relation  z = C  , where C is a constant. 2

Solution: (a) For steady, inviscid, planar flow in the xy plane,  (V  ) = 0

Problem 2-50 shows that steady Euler equation reduces to  (V  ) = 0 for inviscid fluid, and problem 2-51 shows that for 2D planar consideration, it can be written as   2z  2z  z  z + u z + =  +  2 t x y y 2   x

(Eq-1)

 v u  where  z =  −   x y    2  2  Also  z = −  2 + 2  y   x

2 = − 2

(Eq-2)

Substituting (Eq-2) in (Eq-1), we get

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

 2   2   2   )+   )−   ) =  4 ( ( ( t y x x y

(Eq-3)

Considering the given assumptions,

  2 ) = 0 (steady) ( t v 4 = 0 (inviscid)

(eq-4)

Substituting (Eq-4) in (Eq-3), we get



  2   2   )− ( (  ) = 0 y x x y

   z    z − =0 y x   x y

(Ans)

(b) Eq-2 shows that 2 = −  . 2

Due to symmetry of all the above equations, it can be easily shown that the resulting vorticity transport equation is fully satisfied by a rotational flow so long as its vorticity can be expressed as

z = f ( ) including the linear relation  z = C  . 2

2-53 For planar flow in the xy plane, any solution of the form  z = C  can be shown to satisfy the steady, inviscid vorticity transport equation. When this relation is inserted into the vorticity expression, one gets 2

  2  2 + 2 + C 2 = 0 , where u = 2 y x y

v=−

 x

(a) Apply the classification test given by Eq. (2-68) to determine the character of the equation. (b) How many boundary conditions will be needed in order to solve this equation? Solution: Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -50-

 z = C 2

(Eq-1)

When (Eq-1) is inserted into the vorticity expression, we get

 2  2 + 2 + C 2 = 0 2 x y

(Eq-2)

(a) Applying classification test given by (Eq 2-68), to find the character of the equation. Compare (Eq-2) with respect to (Eq 2-67), then we get A = 0, C = 0, B = 0  B2 − 4 AC = 0

Equation is parabolic. (b) Equation is parabolic, boundary conditions must be closed in one direction. If the boundary condiations remain open at the other end of the other direction it becomes a mixed initial and boundary-value problem.

2-54 The Taylor profile, which has been used to describe the bulk gaseous motion in twodimensional rocket chambers, corresponds to a planar profile in porous channels that bears symmetry with respect to the chamber’s midsection plane at y = 0. Assuming unit wall injection and unit chamber half height, the domain may be defined over 0  x  and 0  y  1 , as shown in Fig. 2-54. Using the stream-function vorticity approach, one may take z = C 2 to secure the vorticity transport equation and then eliminate the velocities in favor of the stream function everywhere. The resulting vorticity equation in terms of the stream function becomes  2  2 + 2 + C 2 = 0 2 x y

where u =

 y

=−

 x

(a) Using the method of separation of variables with  ( x, y) = X ( x)Y ( y) , find the general form of the solution. (b) Express the problem’s four fundamental boundary conditions in terms of the stream function knowing that:

u ( x,1) = 0  ( x,1) = −1    ( x, 0) = 0 u (0, y ) = 0

no axial speed at the wall as the fluid enters only normally normal speed matches the unit injection value at the wall symmetry about the midsection plane prevents flow crossing no axial inflow at the chamber headwall

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(c) Find the solution that satisfies these boundary conditions. (d) Determine the velocities associated with this solution and verify that they satisfy the four boundary conditions stated in Part (b).

FIGURE P2-54

Solution:

(a) By inserting the product solution back into the vorticity equation, dividing through by  ( x, y) = X ( x)Y ( y) , and rearranging, we get

Y  X  + C2 = − =  2 Y X where  is the separation constant. We take  = 0 because any other finite value can be shown to be unphysical, as it would lead to either trigonometric or exponential variations in the axial direction. By solving each of the separated ODEs individually, we obtain the general expression

 ( x, y ) = ( C1 x + C2 ) C3 cos ( Cy ) + C4 sin ( Cy ) 

Ans. (a)

(b) Next, we use the velocity-stream function definitions to relate the velocities to  : u ( x, y ) =

  and  ( x, y ) = − y x

We then convey the 4 boundary conditions to the stream function derivatives:  ( x,1)    ( x,1) u ( x,1) = y = 0  y = 0    ( x,1)    ( x,1)  ( x,1) = − x = −1  x = 1 or    ( x, 0 ) = −  ( x, 0) = 0   ( x, 0) = 0   x x    (0, y )  (0, y ) u ( 0, y ) =  =0 =0 y   y

Ans. (b)

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(c) Given the general form of the stream function, we can first evaluate    y = C ( C1 x + C2 )  −C3 sin ( Cy ) + C4 cos ( Cy )     = C C cos ( Cy ) + C sin ( Cy )  1 3 4   x

Direct substitution into the 4 boundary conditions yields:

( C1 x + C2 ) C C4 cos ( C ) − C3 sin ( C )  = 0  C1 C3 cos ( C ) + C4 sin ( C )  = 1  C1C3 = 0  CC2 C4 cos ( Cy ) − C3 sin ( Cy )  = 0 This set can now be solved simultaneously. The last boundary condition returns C2 = 0 especially that any other choice leads to an unphysical outcome. The third boundary condition leads to C3 = 0 since the choice of C1 = 0 is no longer viable as it eliminates the axial dependence altogether. At this juncture, the first boundary condition reduces to the problem’s solvability condition, namely,

CC1C4 x cos C  0; x To avoid a trivial outcome, it may be readily shown that the constants C , C1 , and C4 cannot be permitted to vanish. This leaves us with cos C = 0 or C = (2n +1) / 2; n = 0,1,2,

Realizing that the cases of n = 1,2, lead to an increasing number of flow reversal regions, we select n = 0 or C =  / 2 for a unidirectional motion. The stream function becomes:

 ( x, y ) = C1C4 x sin ( 12  y ) This enables us to set C4 = 1 with no loss of generality (i.e., whenever the product of two undetermined constants appears jointly as yet another undetermined constant, one of the original constants can be set equal to unity; alternatively, their product can be defined as a new undetermined constant that supersedes the original two constants). Lastly, the second boundary C =1 condition enables us to retrieve C1C4 sin ( / 2) = 1 or 1 . We thus arrive at the celebrated profile due to Taylor (1956): Taylor Profile:

 ( x, y ) = x sin ( 12  y )

Ans. (c)

As a testament to his ingenuity, G. I. Taylor was able to guess this solution with very little analysis. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -53-

(d) To verify that the above solution can indeed satisfy our boundary conditions, we first derive the velocities directly from the stream function. We get u ( x, y ) = 12  x cos ( 12  y )

and  ( x, y ) = − sin ( 12  y )

These velocities can be evaluated at the appropriate locations at the wall, midsection plane, and headwall. We find

u ( x,1) = 12  x cos  12  (1)  = 0   ( x,1) = − sin  12  (1)  = −1  1  ( x, 0 ) = − sin  2  ( 0 )  = 0  1 1 u ( 0, y ) = 2  ( 0 ) cos ( 2  y ) = 0

Ans. (d)

The solution clearly reproduces the 4 boundary conditions that are prescribed for this problem.

2-55 For axisymmetric flow in the rz plane, show that the vorticity has only one nonvanishing component in the θ direction, specifically, ω =  e . Show that the vorticity, when written in terms of the incompressible stream function, becomes

1  2 1  1  2  = − + − , where vr = − 1  r z 2 r 2 r r r 2 r z

vz =

1  r r

Solution: The vorticity equation can be equally simplified for axis symmetric flows. The radial and axial velocities can be related to the incompressible Stoke’s steam function using the given relations vr = −

1  and vz = 1  r r r z

Since the flow is confined to the rz plane, the vorticity reduces to a single component in the  direction.

 v v  ω =  e =  r − z  e  z r   ω =  e =

(Eq-1)

−1   2 1   2  − + 2  e  r  r 2 r r z 

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Therefore,  =

−1  2 1  1  2 + − . r r 2 r 2 r r 2 z 2

(Ans)

2-56 For steady, inviscid, axisymmetric flow in the rz plane, (a) show that the vorticity transport equation  (V  ω) = 0 reduces to

 ( / r )   ( / r )  − = 0 , where vr = − 1  r z z r r z (b) Show that any solution of the form

 = rg ( )

vz =

1  r r

satisfies this condition, including the linear

relation  = C r , where C is a constant. 2

Solution: (a) We know that the vorticity transport equation, i.e.,  (V  ω) = 0 , can be obtained by taking curl as D = ( − )V + v2 Dt

(Eq-1)

As we know that  z = −  in 2D planar flow, we stumble upon the stokes operator D 2 , which is a minor variant of the Laplacian in cylindrical coordinates. Hence 2

 1   2 −1 2 + 2  = D  with D = 2 − r r r z r 2

(Eq-2)

As for vorticity transport equation, since (  ) = 0 for all axis-symmetric flows bearing only one vorticity component, (Eq-1) becomes 

     −  t z r  r

2 =

2        2 1  2 1  D   ; + =   −  = − v  −          r2  r2  r     r z  r 

2 1  2 + + r 2 r r z 2

(Eq-3)

Expanding the expressions, (Eq-3) can be written as 

  D 2  t  r

    D 2     D 2  +  2 −  2 =  r z  r  z r  r 

 1   4  4  4  2   3  3  3  2 3   = v  4 + 4 +2 2 2 − 2  3 + −  + z r z  r  r r z 2  r 3 r 2 r 4 r   r  r

(Eq-4)

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For limiting our attention with steady flow and inviscid flow (i.e., v = 0) with Re>>1, we get    D 2     D 2   −  =0 r z  r 2  z r  r 2 



 ( / r )   ( / r )  − =0 r z z z (b) Any rotational motion with equation identically.

(Ans)

 = rf ( ) can be shown to satisfy the vorticity transport

By using  = C r , several solutions have been advanced and shown viable for Reynolds number as low as 200. 2

2-57 For axisymmetric flow in the rz plane, any solution of the form  = C r can be shown to satisfy the steady, inviscid vorticity transport equation. When this relation is inserted into the vorticity expression, one gets 2

 2 1   2 − + 2 + C 2 r 2 = 0 where vr = − 1  2 r r r z r z

vz =

1  r r

(a) Apply the classification test given by Eq. (2-68) to determine the character of the equation. (b) How many boundary conditions will be needed to solve this equation? (c) How should you define B(ψ) and H(ψ) in the Bragg–Hawthorne Eq. (2-30e) to reproduce the same partial differential equation considered here? Solution: (a) We have  = C r 2

(Eq-1)

When inserted into the vorticity expression, we have

 2 1   2 − + 2 + C 2 r 2 = 0 2 r 2 r z

(Eq-2)

Comparing this with (Eq 2-67), we have B2 − 4 AC = 0

Therefore, the equation is parabolic (i.e., no real characteristic solution).

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(b) Since the equation is parabolic, boundary conditions must be closed in one direction but remain open at one end of the other direction. So it becomes a mixed initial- and boundary-value problem. (c) Here the considered equation is

 2 1   2  2 − + 2 + C 2 r 2 = 0 r r r z And Bragg–Hawthorne equation is (Eq 2-30e)  2 1   2 r 2 dH ( ) dB( )  2 − + 2 = −B r r r z d d So, we can get B(ψ) and H(ψ) to reproduce the same partial differential equation as r 2 dH ( ) BdB ( ) − = −C 2 r 2 d d

(Ans)

2-58 The Taylor–Culick profile, which has been extensively used to describe the bulk gaseous motion in axisymmetric rocket chambers (Culick 1966), corresponds to a self-similar axisymmetric profile in porous tubes. Assuming a unit wall injection and a unit chamber radius, the domain may be defined over 0  r  1 and 0  z  , as shown in Fig. 2-58. Using the streamfunction vorticity approach, one may take  = rC 2 to secure the vorticity transport equation and then eliminate the velocities in favor of the Stokes stream function everywhere. The resulting vorticity equation in terms of the Stokes stream function becomes  2 1   2 − + 2 + C 2 r 2 = 0 2 r r z r

where

r = −

1  r z

z =

1  r r

(a) Using the method of separation of variables with  (r, z) = R(r)Z ( z) , find the general form of the solution. (b) Express the problem’s four fundamental boundary conditions in terms of the stream function knowing that: no axial speed at the wall as the fluid enters only radially  z (1, z ) = 0  (1, z ) = −1 radial speed matches the unit injection value at the wall  r  symmetry about the centerline prevents flow crossing r (0, z ) = 0  z (r , 0) = 0 no axial inflow at the chamber headwall (c) Find the solution that satisfies these boundary conditions.

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(d) Determine the velocities associated with this solution and verify that they satisfy the four boundary conditions stated in Part (b).

FIGURE P2-58 Solution: (a) By inserting the product solution back into the vorticity equation, dividing through by  (r, z) = R(r)Z ( z) , and rearranging, we get: R 1 R Z  − + C 2 r 2 = − =  2 R r R Z where  is the separation constant. We take  = 0 because any other finite value can be shown to be unphysical, as it would lead to either trigonometric or exponential variations in the axial direction. By solving each of the remaining ODEs individually, we obtain  ( r , z ) = ( C1 z + C2 ) C3 cos Cr 2 + C4 sin Cr 2  Ans. (a) (b) Next, we use the Stokes stream function definitions to relate the velocities to  : 1  1  r ( r , z ) = − and  z (r , z ) = r r r z We then convey the 4 boundary conditions to the stream function derivatives, thus leading to a well-posed problem in which both the governing equation and its boundary conditions are cast under the same umbrella in terms of  : 1  (1, z )    (1, z ) =0  z (1, z ) = 1 r =0  r    (1, z ) = − 1  (1, z ) = −1   (1, z ) = 1 r  1 z  z  or    (r , z ) lim 1  (r , z ) = 0   z =0  r →0 r r (0, z ) = lim z r →0 r   (r , 0)    1  (r , 0) =0 =0  r  z (r , 0) = r r It should be noted that the third boundary condition will be valid if, and only if, the limit is zero despite the singularity in the denominator as r → 0 . This can occur if the ratio itself is indeterminate to begin with, implying that the numerator should also vanish. Then, by applying L’Hôpital’s rule, we must have  (r , z )  2 (r , z ) =0 lim = 0 and lim r →0 r →0 zr z

(

)

(

)

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So finally, the fundamental constraints on the stream function become:

  (1, z )  r = 0    (1, z ) = 1  z  2 lim  (r , z ) = 0 and lim   (r , z ) = 0  r →0 z r →0 zr   (r , 0)  =0  r (c) Given the general form of the Stokes stream function, we can write

Ans. (b)

  2 2    r = 2C ( C1 z + C2 ) r  −C3 sin Cr + C4 cos Cr     = C1 C3 cos Cr 2 + C4 sin Cr 2      z  2   = 2rC1C  −C3 sin Cr 2 + C4 cos Cr 2      zr Direct substitution into the main boundary conditions in Part (b) yields: 2C ( C1 z + C2 )  −C3 sin ( C ) + C4 cos ( C )  = 0 C ( C1 z + C2 ) ( C4 cos C − C3 sin C ) = 0   C1 C3 cos ( C ) + C4 sin ( C )  = 1 C1 ( C3 cos C + C4 sin C ) = 1  or  C C = 0 C1 C3 cos Cr 2 + C4 sin Cr 2  = 0 lim  1 3   r →0  CC r C cos Cr 2 − C sin Cr 2  = 0 2 2 3  2  4  2CC2 r  −C3 sin Cr + C4 cos Cr  = 0 Note that there is no need to impose the auxiliary constraint that accompanies the third boundary condition, lim  2 / zr = lim 2rC1C −C3 sin Cr 2 + C4 cos Cr 2  = 0   r →0 r →0 The latter proves to be self-satisfied due to the algebraic nature of the general solution obtained in Part (a).

(

)

(

(

)

(

(

)

(

(

)

(

)

)

(

)

(

)

)



( )

(

)

(

)

)

(

)

( )

The last boundary condition returns C2 = 0 especially that any other choice leads to an unphysical outcome. The third boundary condition leads to C3 = 0 since the choice of C1 = 0 is no longer viable (it eliminates the axial dependence altogether). At this juncture, the first boundary condition reduces to the problem’s solvability condition, namely,

CC1C4 z cos C  0; z

To avoid a trivial outcome, it may be readily shown that the constants C , C1 , and C4 cannot be allowed to vanish. This leaves us with cos(C) = 0 or C = (2n +1) / 2; n = 0,1, 2,

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Realizing that the cases of n = 1, 2, lead to an increasing number of flow reversal regions, we select n = 0 or C =  / 2 for which the flow remains unidirectional. The stream function becomes:  ( r , z ) = C1C4 z sin ( 12  r 2 ) This enables us to set C4 = 1 with no loss of generality (i.e., whenever the product of two undetermined constants appears jointly as yet another undetermined constant, one of the original constants can be set equal to unity; alternatively, their product can be defined as a new undetermined constant that supersedes the original two constants). Lastly, the second boundary condition enables us to retrieve C1C4 sin ( / 2) = 1 or C1 = 1 . We thus arrive at the celebrated

profile due to Taylor (1956) and Culick (1966), known colloquially as the Taylor–Culick profile: Taylor–Culick Profile: Ans. (c)  ( r , z ) = z sin ( 12  r 2 ) As a testament to his ingenuity, G. I Taylor was able to guess this solution with very little analysis. Conversely, F.E.C. Culick derived it 10 years later with the intent of achieving a rotational mean flow model of the bulk gaseous motion for an internal burning solid rocket motor with a circular grain perforation. (d) To verify that the above solution can indeed satisfy our imposed boundary conditions, we first derive the velocities directly from the stream function: r ( r , z ) = − sin ( 12  r 2 ) and  z ( r , z ) =  z cos ( 12  r 2 ) 1 r

Next, we can evaluate these velocities at the appropriate locations at the wall, centerline, and headwall. We find  (1, z ) =  z cos  1  (1)2  = 0 2   z  1 2 r (1, z ) = − sin  12  (1)  = −1   1   Ans. (d)    r cos 12  r 2   ( 0, z ) = lim  − 1 sin 1  r 2  = 0 = lim  − =0 2  0 r →0  r →0   r 1   r     r , 0 = 0  cos 1  r 2 = 0 2   z( ) ( ) Clearly, the analytical solution that we obtain is observant of the 4 principal boundary conditions that are prescribed for this problem.

(

(

)

(

)

)

2-59 In order to derive an exact solution for the velocity field in a wall-bounded cyclonic flow evolving in a cylindrical chamber (Fig. P2-58), Vyas and Majdalani (2006) assumed that the flow is steady, incompressible, inviscid, and axisymmetric with respect to the centerline. Because of the presence of swirl, a three-component velocity profile is used, namely,

V(r , z ) = r er +  eθ +  z ez . In this problem, a and b(= a / 2) denote the outer and outlet radii of the cyclonic chamber, U stands for the circumferential tangential velocity at r = a, Ai = Qi / U denotes the injection area for a volume flow rate of Qi , and

 = Ai / (2 aL) represents a

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dimensionless offset swirl parameter that gauges the relative importance of the axial-to-tangential speed ratio. The boundary conditions for this problem can be chosen to be   (r = a) = U  ( z = 0) = 0 z   r (r = 0) = 0  (r = a) = 0  r Q = 2 b  (r , L) rdr 0 z   i

tangential speed matches imposed injection value no axial flow penetration at chamber headwall no flow crossing at centerline no radial flow penetration at chamber sidewall conservation of volume flowrate, Qi = UAi at z = L

(a) Guided by experimental data and numerical simulations, we may assume

 =  (r ) to be an

axially invariant function. Under this assumption, the azimuthal momentum equation in cylindrical coordinates, namely,

r

      1 p + + z  +  r = − r r  z r  r 

decouples from its axial and radial counterparts. Recalling that the motion is axisymmetric (i.e., there are no variations in  ), show that the solution to the azimuthal momentum equation given above reduces to

   + r  r

r 

 =0 

or

1 (r ) =0 r r

(b) Realizing that the solution to Part (a) constitutes a statement of conservation of angular momentum, r = B = const, solve for B and

 (r ) using the first boundary condition,

 (a) = U . (c) Express the remaining four boundary conditions as functions of the Stokes stream function using 1  1  and  z = r = − r z

r r

(d) Because the vorticity transport equation  ( V  ω ) = 0 reduces to   ( / r )   ( / r )   −  =0 r z z r  

use partial differentiation to prove that any solution of the form

 = rg ( ) satisfies this

2 condition, including the linear relation  = C r , where C is a constant.

(e) Substitute the Stokes stream function into the tangential vorticity to show that

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 =

r  z 1  1  2 1  2 − = − − z r r 2 r r r 2 r z 2

2 (f) Substitute  = C r , which satisfies the vorticity transport equation, into the vorticity

equation to obtain

 2 1   2 − + 2 + C 2 r 2 = 0 2 r r z r (g) Using the method of separation of variables with  (r, z) = R(r)Z ( z) , find the general form of the solution to the resulting partial differential equation. (h) Find the particular solution that satisfies the four boundary conditions formulated in Part (c). (i) Show that the velocities associated with this profile are: a



r2 

a

z



r2 

0  r  a

r = −U   sin   2  ;  =  U ;  z = 2U   cos   2  ;  r  a  r a  a  0  z  L (j) Verify that the velocities satisfy the problem’s boundary conditions. (k) Show that the same governing equation may be restored from the Bragg–Hawthorne Eq. (22 2 30e) where B = const and H ( ) = H0 − 12 C  may be readily substituted. The form of the

latter may be easily deduced from a pressure analysis along a streamline.

FIGURE P2-59 Solution: (a) Let us consider the azimuthal momentum equation:

r

    + + z  r r  z axisymmetry

axial invariance

+

r r

=−

1 p  r  axisymmetry

Since  =  (r ) , the tangential velocity has no dependence on  (axisymmetry) or z (axial invariance). Therefore, the second and third terms on the left-hand side vanish. Moreover, the overarching condition of axisymmetry allows no variations in the azimuthal direction of any property, including the pressure. As a result, the right-hand side vanishes; we are left with

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    + =0 r   r

r 

or

The latter can be grouped conveniently using

 ( r ) r

=r

 +  r

and so

The azimuthal momentum equation becomes

   + r  r

r 

 =0 

or

 1 +  = 0 r r

1  ( r )   = + r r r r

  1  ( r ) + = =0 r r r r

Ans. (a)

(b) A single integration of Part (a) in the radial direction leads to r = B , where B must be determined from one of the boundary conditions. We use the first condition,  (a ) = U , to deduce that B = aU . Ans. (b) (c) Next, we may express the remaining 4 boundary conditions as a function of the Stokes stream function. We get   (r , 0) 1  (r , 0)  =0 =0  r  z (r , 0) = r r     2   (0) = lim  − 1   = 0 lim = 0 and lim =0     r r →0  r →0 zr r z   r →0 z  or  Ans. (c)   (a, z ) 1  (a, z )  r (a ) = − = 0 =0  z a z   b  (r , L)  b Q  UA  1   (r , L)    dr = i  = i = UaL   2  rdr = Q  i   r 2  2    0 0 r  r It should be noted that the second boundary condition will be valid if, and only if, the limit is zero despite the singularity in the denominator as r → 0 . This can occur if the ratio itself is indeterminate to begin with, implying that the numerator must also vanish. Then, by applying L’Hôpital’s rule, we are able to identify a total of two physical requirements:

  =0 lim = 0 and lim r → 0 z r r →0 z 2

Moreover, the last boundary condition can be further simplified into: b   (r , L) dr =  (b, L) − (0, L) =  (b, L)  0 r

(d) Due to axisymmetry, and since the vorticity vector reduces to a single component ω =  e , the vorticity transport equation collapses into    1  1       ( V  ω ) =  e r + e z    − er + e z   ( e )  = 0 z   r z r r   r  By evaluating the second cross product on the right-hand side, we get

         ez −  er = 0  er + e z    − z   r z r r   r

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Then, by evaluating the first cross product on the left-hand side, we collect a single azimuthal vector component              r  r z  − z  r r   e = 0      To make further headway, we may expand the derivatives and write                       + −    −  =0  r r  z  z  r  r r z  r    r  r  z

Due to the cancellation of the mixed partial derivatives, we are left with

 ( / r )  r

z

−

 ( / r )  z

r

=0

or

g  g  − =0 r z z r

Ans. (d)

where g ( )   / r . At this stage, it can be easily shown that any g ( ) will satisfy the resulting vorticity transport equation (VTE). For example, one may use the chain rule to express the partial derivatives of g ( ) by putting

g dg  = r d r

and

g dg  = z d z

Direct substitution into the VTE yields

dg   dg   − =0 d r z d z r

Clearly, the above relation is true irrespective of g ( ) , i.e., as long as the ratio

Ans. (d)

 / r can be

2 written in terms of the stream function. This includes any linear relation of the form  / r = C  . When the latter is substituted into the VTE, one recovers

 2   2    C − C =0  r z z r 

Ans. (d)

which satisfies the VTE identically. (e) By combining the azimuthal vorticity definition and the Stokes stream function definitions, we obtain the vorticity equation, which relates the vorticity directly to the stream function:     1     1   1  1  2 1  2  = r − z =  − − − Ans. (e) −  = z r z  r z  r  r r  r 2 r r r 2 r z 2 2 (f) By substituting the linear relation  = C r into the vorticity equation, we arrive at

1  1  2 1  2 C r = 2 − − r r r r 2 r z 2 2

Multiplying by

r

and rearranging, the vorticity equation becomes

 2 1   2 − + 2 + C 2 r 2 = 0 2 r r z r

Ans. (f)

(g) By inserting the product solution back into the resulting partial differential equation, dividing through by  (r, z) = R(r)Z ( z) , and rearranging, we get

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R 1 R Z  − + C 2r 2 = − =  2 R r R Z

where  is the separation constant. We take  = 0 because any other finite value can be shown to be unphysical, as it would lead to either trigonometric or exponential variations in the axial direction. By solving each of the remaining ODEs individually, and consolidating the results, we arrive at

 ( r , z ) = ( C1 z + C2 ) C3 cos ( Cr 2 ) + C4 sin ( Cr 2 ) 

Ans. (g)

(h) Before applying the boundary conditions obtained in Part (c), it is useful to express   2 2    r = 2C ( C1 z + C2 ) r  −C3 sin Cr + C4 cos Cr     = C1 C3 cos Cr 2 + C4 sin Cr 2      z   2 = 2rC1C  −C3 sin Cr 2 + C4 cos Cr 2      zr Then the 4 boundary conditions become 2CC r  −C sin Cr 2 + C cos Cr 2  = 0 2  3 4   C1C3 = 0  C C cos Ca 2 + C sin Ca 2  = 0 4   1 3  b Q   2C ( C1L + C2 ) r  −C3 sin Cr 2 + C4 cos Cr 2  dr = i 2 0 Note that the auxiliary constraint is not included because it is satisfied identically, namely,

(

(

)

)

(

)

)

(

(

(

)

)

(

(

(

(

)

)

)

)

(



)

(

)

( )

 2 = lim 2rC1C  −C3 sin Cr 2 + C4 cos Cr 2  = 0   r →0 zr r →0

lim

( )

The first boundary condition returns C2 = 0 since any other choice leads to an unphysical outcome. The second boundary condition leads to C3 = 0 since the choice of C1 = 0 is no longer viable as it eliminates the axial dependence altogether. At this juncture, the third boundary condition reduces to the problem’s solvability condition, namely,

(

)

C1C4 sin Ca 2 = 0

To avoid a trivial outcome, it may be readily shown that the constants C , C1 , and C4 cannot be permitted to vanish. This leaves us with

sin(Ca 2 ) = 0 or Ca 2 = (2n + 2) / 2; n = 0,1, 2, We select n = 0 or C =  / a2 for which only one flow reversal will occur, as expected of the bidirectional motion of a cyclonic flow field. Cases of n = 1, 2, are not considered here as they give rise to an increasing number of flow reversal regions. For n = 0 , the stream function becomes:  ( r , z ) = C1 z sin ( r 2 / a 2 ) where C4 = 1 is set with no loss of generality (i.e., whenever the product of two undetermined constants appears jointly as yet another undetermined constant, one of the original constants can Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -65-

be set equal to unity). Lastly, the mass conservation condition may be used to retrieve C1 . We obtain b

2   r 2  2 r Q   Qi C1L cos   2  2 dr = i , C1L sin   b 2  = Qi and so C1 = 2 2 L sin ( b 2 / a 2 ) 0  a  a  a  2 Alternatively, one may use the equivalent form given in Part (c) to express, more straightforwardly, b Qi   (r , L) dr =  (b, L) = C L sin  b2 / a 2 = Qi and so  C1 = 1 0 r 2 2 L sin ( b 2 / a 2 )

(

)

At the outset, the stream function becomes:

( (

2 2 Qi sin  r / a  ( r, z ) = z 2 L sin  b2 / a 2

) )

Ans. (h)

Note that for the special case of b = a / 2 , we get sin ( b2 / a 2 ) = 1 ; and since Ai = 2 aL and

Qi = UAi , the resulting expression can be further simplified into  ( r, z ) =

 r2   r2  2 aLU z sin   2  =  aUz sin   2  2 L  a   a 

Ans. (h)

The resulting profile was first presented by Vyas and Majdalani (2006); it is sometimes referred to as the quasi complex-lamellar model of a wall-bounded cyclonic motion. (i) The radial and axial velocities may be deduced directly from the stream function; we get  r2   r2  1  1  a z r = − = 2U cos   2  Ans. (i) = − U sin   2  and  z = r r r z r a  a   a  As for  , we may recall the angular momentum conservation principle leading to r = B = aU from Part (b). This implies  = U

a r

Ans. (i)

(j) It is always a good habit to verify that the solution satisfies the originally prescribed boundary conditions. To do so, we check one-by-one:

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  (r = a ) = Ua = U a   2  z ( z = 0) = 2U  0  cos   r  = 0 2  a  a     r2    2  r cos   2     1  r 2  0   a  = 0 − aU sin   2   = = lim  − aU r (r = 0) = lim   r →0 r 1  a   0 r →0             (a) 2  − aU sin  2  = 0 r (r = a) = a  a    2 2 b Qi = 2 2U  L  cos   r  rdr = 2 aLU sin   b  = 2 aLU   2 2 0  a  a   a 

Ans. (j)

Note that the continuation of the limit in the third boundary condition is carried out using L’Hôpital’s rule. This confirms that the velocity profile fully satisfies the problem’s boundary conditions. (k) The Bragg–Hawthorne equation is defined as dH ( ) dB ( )  2 1   2 − + 2 = r2 −B 2 r r z d d r Since B = Ua is constant, the last term on the right-hand side vanishes. Additionally, since we are given a prescribed variation of the form H ( ) = H0 − 12 C 2 2 , the first term on the right-hand side becomes dH ( ) / d = −C 2 . Forthwith, the Bragg–Hawthorne equation reduces to  2 1   2 − + 2 + C 2 r 2 = 0 2 r r z r The resulting expression is identical to the vorticity equation obtained in Part (f).

Ans. (k)

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CHAPTER 3. SOLUTIONS OF THE NEWTONIAN VISCOUS FLOW EQUATIONS

3-1

Solve for constant-pressure Couette flow between parallel plates, as shown at right, for a

non-newtonian fluid,  = K ( du/dy ) , n  1. Compare with the newtonian solution. n

Since the boundary conditions are independent of x, we assume that velocity varies only with y. The non-newtonian version of Eq. (3-4) is 0=−

dp   + = 0+ dx y y

since the pressure is constant. Thus we integrate to  = K ( du / dy ) = constant, or n

du = constant, or: u = a + b y dy

Introducing no-slip boundary conditions at each wall, u ( −h ) = 0, u ( +h ) = V, we obtain

u=

U y 1 +  2  h

(Ans.)

This is exactly the same linear velocity distribution as Eq. (3-6), Fig. 3-2a, for newtonian flow. The result is independent of the particular behavior of the non-newtonian fluid.

3-2

Solve for u ( r ) in annular Couette flow, Fig. 3-3, when both cylinders are moving. With

u = u ( r ) only and pressure constant, the analysis follows Section 3-2.2: 1 d  du   r  = 0, or: u = C1 ln ( r ) + C2 r dr  dr 

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The no-slip boundary conditions are u ( ro ) = Uo and u ( r1 ) = U1, which enable us to find the two constants: C1 =

U1 − Uo , ln ( r1 /ro )

C2 = Uo − C1 ln ( ro )

Substituting these constants into the solution above yields the desired velocity distribution: u = Uo

ln ( r/ro ) ln ( r1 /r ) + U1 ln ( r1 /ro ) ln ( r1 /ro )

(Ans.)

This is exactly the sum of the separate solutions for moving inner or outer cylinder, Eqs. (3-18) and (3-19). This superposition is possible because the Navier-Stokes equations are linear for this particular flow. We are asked to plot u ( r ) for (a) U1 = Uo ; (b) U1 = − Uo ; and (c) U1 = 2Uo . The results are shown below for the particular geometry r1 = 2ro . Also shown, for comparison, is the case where the upper cylinder is not moving.

3-3 For axial Couette flow with only the inner cylinder moving. Eq. (3-18), find the temperature distribution T ( r ) if the cylinder walls are at To and T1 , respectively. If T = T(r) only and there are no radial or circumferential velocities, the energy equation from Appendix B, Eq. (B-9), reduces to 0=

2 ln ( r1 /r ) k d  dT   du  r +  from Eq. (3-18)     , where u = U o r dr  dr  ln ( r1 /ro )  dr 

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Introducing u(r), separating the variables, and integrating once, we obtain

ln ( r ) C Uo2 dT =− + , where C = constant 2 dr r k ln ( r1 /ro ) r Integrating again, noting that  ln ( r ) /r = ln 2 ( r ) /2, we obtain 2 U o2 ln ( r/ro ) T=− + C ln ( r ) + D 2 k ln 2 ( rr /ro )

Evaluating C and D from the boundary conditions T ( ro ) = To and T ( r1 ) = T1, we obtain the desired result, after rearranging and cleaning it up a bit: 2  U o2  ln ( r/ro ) U o2 ln ( r/ro ) T = To + T1 − To + −  2 k  ln ( r1 /ro ) 2 k ln 2 ( r1 /ro ) 

(Ans.)

When plotted, these look very similar to the “channel” temperature profiles in Fig. 3-2b.

3-4

A long thin rod of radius R is pulled axially at speed U through fluid with constant ( ,  ) .

Find the velocity distribution u z = u ( r ) . The analysis is similar to that of Section 3-2.2:

1 d  du   r  = 0, or: u = C1 ln ( r ) + C2 r dr  dr  Application of the boundary conditions immediately leads to a paradox:

u ( ro ) = U = C1 ln ( R ) + C2 ; u ( ) = 0 = C1 ln (  ) + C2

(?)

It is impossible to find steady-flow constants if the rod moves through an infinite expanse of fluid. Physically, it would require the finite-diameter rod to deliver an infinite amount of kinetic energy to the fluid with only a finite wall shear stress. 3-5 A cylinder of radius R rotates at angular rate  in an infinite expanse of fluid. Find the velocity and pressure distributions and compare with an inviscid “potential” vortex. The velocity follows from the -momentum equation in Section 3-2.3: d2u  dr

2

+

d  u  = 0, dr  r 

or : u  = C1r +

C2 r

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Introducing the boundary conditions, we obtain

u  ( R ) = R = C1R +

C2 C R 2 ; u  (  ) = C1 (  ) + 2 , or: u  = R r ()

(Ans.)

This is the solution given by Eq. (3-25). It does indeed correspond to a potential vortex. The pressure is found from the r-momentum equation: dp u 2 2 R 4 2 R 4 = = , or: p = − +C dr r r3 2r 2

The constant C = po + 2R 2 / 2 is found from the single condition p = po at r = R. The final expression for the pressure distribution is thus

(

1 p = po + 2R 2 1 − R 2 /r 2 2

)

(Ans.)

This is exactly what one would find by using Bernoulli’s equation for a potential vortex.

3-6

Find p ( r ) between rotating cylinders using the velocity distribution from Eq. (3-22).

With velocity known, the pressure is found from the r-momentum equation (3-20b): r r (  − 1 ) r  /r − r  /r dp u 2 b = , where u  = ar + , and a = 1 1 o o o 1 , b = o 1 o dr r r r1 /ro − ro /r1 r1 /ro − ro /r1

(1)

Introducing u 2 and integrating, we find the pressure: 1 b2  p =   a 2 r 2 + 2ab ln ( r ) − 2  + C 2r   2

The constant C = po − [a 2 ro2 /2 + 2ab ln ( ro ) − b 2 / 2ro2 ] is found from the single condition p = po at r = ro . The final result may be rearranged in the form

 1 1  1  p = po +  a 2 r 2 − ro2 + 4ab ln ( r/ro ) + b 2  2 − 2   r  2   o r  

(

)

(Ans.)

This is not at all what one would obtain by using the (invalid) Bernoulli equation.

3-7 Using one-dimensional integral analysis, derive the differential equation for the oscillation amplitude X ( t ) of the U-tube given in the figure.

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For one-dimensional incompressible flow at constant area, it follows from the continuity relation, AV = constant, that the fluid velocity

V=

dX = fcn ( t ) only dt

The fluid thus moves as a single “slug” of mass L long, with forces applied due to gravity (the excess weight when the fluid is displaced) and wall friction. Newton’s law yields

F = or:

−gro2

( 2X ) − w 2ro L

d2X

2w 2g + X=0 ro L

dt

2

+

2 2 d X = ro L 2 ,

dt

The problem is closed if we relate wall shear to velocity through the friction factor:

d 2X dt

2

+

2 w Cf dX dX 2g | | + X = 0, where Cf = 2 ro dt dt L  ( dX/dt )

(Ans.)

where the absolute value is needed to make the friction force reverse when the velocity reverses. Given an initial displacement X ( 0) , we can solve for a damped oscillation X ( t ) . For laminar flow, we could approximate C f by the pipe-flow value 16/ReD , as in Eq. (3-40). The differential equation then becomes d2X dt 2

+

r2 2 dX 2g + X = 0, where t* = o t* dt L 4v

(1)

The quantity t* is a characteristic damping time of a laminar U-tube oscillation. If we displace the fluid to a fixed initial point X o and release it, the solution of (1) above is X = Xoe−t/t* cos ( n t ) , where n = ( 2g/L )

1/2

(Ans.)

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The natural frequency is the same as that of a (frictionless) pendulum of length (L/2), and the oscillation damps to 1/2 amplitude in a time t = 0.693 t*. For turbulent flow, the basic differential equation is nonlinear and a numerical solution is required.

3-8 Air at 20°C and 1 atm is driven between parallel plates 1 cm apart by a pressure gradient and by the upper plate’s moving at 20 cm/s. (a) Find the volume flow per unit width if the pressure gradient is –0.3 Pa/m. The velocity distribution is given by Eq. (3-42): u=

U  y  h 2  dp   y 2  1 +  +  −  1 −  2  h  2  dx   h 2 

The volume flow per unit depth is found by integrating this across the fluid between plates: +h

Q=



−h

u dy = U h +

2h 3  dp  −  3  dx 

(1)

We are given h = 1/2 cm, U = 20 cm/s, and dp/dx = −30 Pa/m. For air at 20C,  = 1.81E-5 Pa-s. Equation (1) above may thus be evaluated for this case:

2 ( 0.005 m ) ( 0.3 Pa/m ) Q = ( 0.2 m/s )( 0.005 m ) + = 0.00100 + 0.00138 3 (1.81E-5 Pa-s ) 3

= 0.00238 m2 / 2 = 2380 cm3 /s/m

(Ans. a)

(b) Find the pressure gradient for which the shear stress at the lower plate is zero. This is the ‘separation criterion’ given by Eq. (3-43) – [see Fig. 3-8, dashed line]: dp 2U 2 (1.81E-5 Pa -s )( 0.2 m /s ) = = = + 0.0724 Pa/m dx ( 2h )2 ( 0.01 m )2

(Ans. b)

It takes a positive pressure gradient to cause flow separation.

3-9

Derive the solution u ( y, z ) for flow through an elliptical duct, Fig. 3-9, by solving

Eq. (3-30). Begin with an guessed quadratic solution, u = A + By 2 + Cz 2 , and work your way through to the exact solution.

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Solution: The guessed solution seems OK since 2u = constant, as required by Eq. (3-30) for fully-developed duct flow. Since u must be zero on the ellipse walls of semi-axes a and b, as seen in Fig. 3-9, the constants ( A, B, C ) must be related to the ellipse formula. That is,  y2 z2  u = K 1 − 2 − 2   a b  

where K = constant

Substitute this u, which has no-slip (u = 0) at the walls, into the momentum equation (3-30):

 2u

 2u

1  dp  a 2b2  −2 −2  1 dp + =K 2 + 2 = , solve K = −  2  dx  a 2 + b2 y 2 z 2 b   dx a 1  dp  a 2b 2  y2 z2  −  Thus, finally, for an ellipse, u = 1 − −  2   dx  a 2 + b 2  a 2 b 2 

Ans.

3-10 Air at 20°C and 1 atm flows at an average velocity of 1.7 m/s through a rectangular 1-cm by 4-cm duct. Estimate the pressure drop in Pa/m by (a) an exact calculation; and by (b) the hydraulic diameter approximation. Solution: First establish the hydraulic diameter: Dh = 4A/P = 4[(0.01m)(0.04m)/[2(0.01m+ 0.04m)] = 0.016 m

For air at 20°C and 1 atm,  = 1.205 kg/m3 and  = 1.81E-5 kg /( m-s ) . Then calculate

ReDh = VDh /  = (1.205)(1.7 )( 0.016) / (1.81E-5) = 1810  2000, thus the flow is laminar. (a) Exact analysis: Eq. (3-48) should apply to this rectangular section. Evaluate the flow:

Q=

4ba 3  dp   192 a  tanh ( ib/2a )     −  1 − 3  dx   5 b 1,3,5,... i5 

3 4 ( 0.005 )( 0.02 )  dp   192 ( 0.02 )  0.3737 0.8269 0.9614  = − 1 − + + + ...     3 (1.81E-5 )  dx   5 ( 0.005 )  15 35 55   = 0.0001554 ( −dp/dx )

But Q/A = V, which is known. Dividing Q by the duct area gives the desired result: Q/A = V = 1.7 m/s =

0.0001554 ( −dp/dx )  dp  = 0.3885  −  ( 0.01)( 0.04 )  dx 

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or:

dp = − 4.38 Pa/m dx

(Ans. a)

This answer could also be obtained by using Cf Re = 18.2 from Figure 3-13. (b) Approximate: Simply assume that Cf  16/Re as if the duct were circular: Cf 

16 1 2 = 0.00884, or: avg = Cf V 2 = ( 0.00884 )( 0.5 )(1.205 )(1.7 ) 1810 2 = 0.0154 Pa

Then, for fully developed flow, the relation between wall shear and pressure drop is  dp  4avg 4 ( 0.0154 Pa ) = = 3.85 Pa/m −  = Dh 0.016 m  dx 

(Ans. b)

For this case, the hydraulic-diameter approximation predicts about 12% low.

3-11

Consider swirling motion superimposed upon axial flow and find the velocities.

We assume, for developed flow in a circular pipe, that

vr = 0

v = v ( r, t )

vz = vz ( r )

Substitute these into the complete incompressible equations of motion in cylindrical coordinates, Appendix B. The continuity equation is satisfied identically, and the z-momentum equation becomes

   vz  p r = r r  r  z This is independent of the circumferential motion v and thus may be solved directly, leading simply to Poiseuille flow in a duct, Section 3-3.1. The -momentum equation reduces to

v v   v  = r t r r  r  This is uncoupled from axial motion v z and may be solved independently, subject to the swirlflow boundary conditions. Finally, with v known, the pressure may be found from the r-momentum equation, p/r =  v2 /r. Further discussion of superposition of swirl flows is given in the text by Langlois (1964).

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3-12 If light oil viscosity (0.02 to 0.1 Pa-s) is measured by passing 1 m3 /h of fluid through an annulus 30 cm long, with inner and outer radii of 9 mm and 10 mm, find p . For an annulus, the hydraulic diameter is Dh = 2 ( a − b ) = 2 ( 0.010 − 0.009) = 0.002 m. The annular area is (a 2 − b2 ) = 5.97E-5 m2 . Then the average velocity is V = Q/A  4.65 m/s. Estimating the density of light oils to be about 900 kg/m3 , the Reynolds numbers are ReDh =

VDh ( 900 )( 4.65)( 0.002 ) = = 80 to 400  0.02 to 0.10

Thus the flow is laminar, and Eq. (3-51) applies:

(

)

 2 2 2 a − b  p  4 4  Q= a −b −  8 L ln ( a/b )      Introducing the given numerical values, we obtain the pressure-drop estimate:



 2 2 0.01) − ( 0.009 ) (  1   p 4 4 m3 /h = ( 0.01) − ( 0.009 ) − 3600 8 ( 0.02 to 0.10 )(.3)  ln ( 0.01/0.009 )  or:

p  335,000 to 1,680,000 Pa  3 to 17 atm

  2

 

(Ans.)

For such a large pressure drop, a mechanical (e.g. Bourdon tube) gage might be recommended. 3-13 Lubricating oil at 20°C flows at 2 m/s into a tube, D = 3 cm, L = 2 m, with a wall temperature of 10°C. Estimate (a) the heat transfer rate at x = 10 cm; and (b) the exit temperature of the oil. The oil properties are  = 890 kg/m3  = 0.8 Pa-s k = 0.15

W m-K

c p = 1800

J kg-K

The Reynolds number is Re = VD/ = (890)( 2.0)( 0.03) / ( 0.8) = 67, hence the flow is laminar. The Prandtl number is cp /k = ( 0.8)(1800 ) / ( 0.15 ) = 9600 ( oils ) . (a) At x = 10 cm, the local Graetz number is x* =

x 0.1 = = 5.20E-6 D Re Pr ( 0.03)( 67 )( 9600 )

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At this very small value, we estimate the local Nusselt number from Eq. (3-83):

Nu x  1.076 ( 5.20E-6 )

−1/3

− 1.064  66.1 =

qw D k T

This is close to the entrance, so we assume that T still equals 10°C. Then we calculate q w |x =10cm = 66.1

( 0.15)(10) = 3060 W/m2 k T = 66.1 D ( 0.03)

(Ans. a)

(b) At the exit, L* = L/ ( D Re Pr ) = 2.0/ ( 0.03)( 67 )( 9600 )  = 0.000104. This is too small to make the chart in Fig. 3-15. Therefore we use Eq. (3-93) as an estimate: Nu m  3.66 +

0.075/ ( 0.000104 ) 1 + 0.05/ ( 0.000104 )

2/3

= 34.2 = −

1

* ln Tm ( L )  4L *

where the latter equality follows from Eq. (3-90). We may solve for * Tm ( L ) = 0.986 =

Tw − Tm 10 − Tm = , or: Tm ( L ) = 19.86C Tw − To 10 − 20

(Ans. b)

The fluid has hardly cooled down at all over a two-meter length! This poor convection rate is characteristic of oil flow under laminar conditions.

3-14

For plane flow with circular streamlines, solve the vorticity equation,   2 1    = v 2 +   r t r r  

for the decay of a line vortex of strength  o initially concentrated at the origin. The boundary conditions are symmetry at the origin and no motion at infinity:

 ( 0, t ) =  ( , t ) = 0 r This is a classic problem in linear PDE theory and has an exact analogy in heat conduction (a line heat pulse at the origin). The solution is given by Carslaw and Jaeger (1959):  r2  1  o = exp  − = rv  4t  r r (  ) 4t  

where  o =  V  ds is the initial circulation about any closed curve enclosing the origin.

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The vorticity may be integrated to yield the instantaneous circumferential velocity:

v =

r  r 2  o  1  r dr = 1 − exp   −   r 0 2  r   4t  

These distributions were discussed earlier, without derivation, in Prob. 1-6, and both the velocity and vorticity distributions are plotted on page 3.

3-15 A wide liquid film of thickness h flows due to gravity down the inclined plane shown below. Find the (laminar) velocity distribution.

Since the boundary conditions are independent of x, we may assume u = u ( y ) , v = w = 0, p/x = 0, and the momentum equation reduces to



d2u dy2

= − g sin  = constant, or: u =

−g sin  2 y + Ay + B 2

One boundary condition is no-slip at the wall:

u ( 0) = 0 = B The second condition is hinted at in the problem statement: negligible shear at the free surface due to weak interaction with the constant-pressure, low density atmosphere:  xy ( y = h ) = 0 = 

or:

u −g sin  |y = h = ( 2h ) + A, y 2

A = gh sin /

Substituting for A and B, we find the complete solution for laminar developed film flow:

u=

ρgsinθ y ( 2h − y ) 2

(Ans.)

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This may be integrated to find the volume flow per unit width of film: h

Q =  u dy = 0

ρg h 3sinθ 3μ

(Ans.)

The flow rate varies as h3, so the rate of draining is highly dependent upon the film thickness. Note the similarity of this (parabolic) solution with pressure-driven Poiseuille flow between parallel plates (2h) apart, Eq. (3-44).

3-16 A laminar film drains at constant thickness down a vertical rod, as shown. Find formulas for the velocity distribution and the flow rate.

Assume, due to the constant-pressure atmosphere outside the film, that p/z = 0, and, for a fully developed film, u z = u ( r ) only. The z-momentum equation, Appendix B, reduces to

0 = g +

 d  du  r  r dr  dr 

This second-order ordinary differential equation may readily be integrated twice: u=−

g r 2 + A ln ( r ) + B 4

The boundary conditions are no-slip at the wall and no shear stress (weak atmospheric interaction - see the top of page 44): No-slip at wall:

g a 2 u ( a ) = 0 gives B = − A ln ( a ) 4

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No surface shear:

du |y=b = 0 gives B = ρgb2 /2μ dr

With A and B known, the final expression for the film velocity profile is

g b2 u= 4

2 2  r   r  a   2ln   −   +     a   b   b   

(Ans.)

Considerable integration skill (not shown) is needed to evaluate the film flow rate: b

Q =  u ( r ) 2 r dr = a

g a 4  4 b 4 ln (  ) + 42 − 1 − 34  ,  =  1   8 a

(Ans.)

For thin films approximating a flat wall, 1.0  b/a  1.2, Q increases (approximately) as the cube of the film thickness (b-a) - see the solution to Prob. 3-15 above. For b/a  1.2, Q increases even faster than the cube of the thickness, as film area increases with radius.

3-17 By extension of Prob. 3-15, consider a double layer of immiscible fluids 1 and 2, flowing steadily down an inclined plane, as in Fig. P3-17. The atmosphere exerts no shear stress on the surface and is at constant pressure. Find the laminar velocity distribution in the two layers.

Fig. P3-17 Solution: We solve for the velocity distribution in each layer and patch them together at y = h1. Assuming u1,2 = fcn( y ) only, the layers satisfy similar momentum equations:

x − momentum:

0 = 1 g sin  + i

y − momentum:

0=−

d 2ui dy 2

pi + i g cos  , y

, i = 1, 2

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The pressure gradient pi /x = 0 because the surface pressure is constant. Thus the pressure is hydrostatic in i g cos  and is uncoupled from velocity. Forget pressure, find velocity:

u1 = −

1g sin  2  g sin  2 y + C1 y + D1 ; u2 = − 2 y + C2 y + D2 21 2 2

(1a, b)

By inspection, D1 = 0 to satisfy the no-slip condition at the bottom, y = 0. The other three boundary conditions are negligible shear at the surface (weakly interacting atmosphere) and matching velocities and shears in each layer at the interface, y = h1. Surface shear: τ 2  0 = 2 Interface velocity: u1 = −

  g sin   du2 | y =h1+h2 = 2  − 2  ( h + h ) + 2C2 dy 2  1 2 

1g sin  2  g sin  2 h1 + C1h1 = u2 = − 2 h1 + C2h1 + D2 21 2 2

  g sin  Interface shear: 1 = 1  − 1 1 

  2 g sin    h1 + 1C1 =  2 = 2  −  h1 + 2C1    2 

Solve simultaneously for the three constants:

  g sin   C2 =  2  ( h1 + h2 ) ;  2 

C1 =

g sin 

1

( 1h1 + 2h2 )

h  h  h  h  D2 = h1g sin   1 1 + 2 2 − 2 1 − 2 2  1 2 2 2   21 Substitute these constants into Eqs. (1a, b ) above. (It’s too much typing to repeat here. Au) To illustrate this flow, an example is shown below. The two layers are of equal thickness, h1 = h2 = 0.5, and the densities are the same. Only the effect of upper-layer viscosity is shown. If 2 = 1, as in the center curve, the velocity profile is a continuous parabola, as calculated earlier in Prob. 3-15. If 2 = 51, the lower curve, the outer velocity is less and two parabolas meet with different slopes at the interface, y1 = 0.5. Similarly, if 2 = 0.21, the upper layer goes faster, as shown, and again the slopes are different at y1 = 0.5.

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Two-layer flow down an inclined plane, for the special case  2 = 1 and h 2 = h1 .

3-18 The film surface in Prob. 3-15 is known to become unstable at a critical Reynolds number Re* = hu max /v = Q (10) , which depends on , g, , , and surface tension T. . Make a dimensional analysis of the correlation Re* = fcn ( , g, , , T ) , noting the dimensions of each in the Mass-Length-Time system:

 Re* and  = 0; g  = [L/T 2 ]; [] = [M/L3 ];  = M/LT ; T  = [M/T 2 ] We expect two dimensionless groups (“Pi’s”). One of them obviously is  itself. The second one, assumed proportional to , could be written as a power-group:

2 =  ga bT c = M0L0T0 , from which we find a = −1, b = −4, and c = +3. Dimensional analysis thus predicts that two Pi groups influence the Reynolds number:  T 3  Re* = fcn  ,  g  4   

(Ans.)

The last parameter, which depends only upon gravity and fluid parameters, is a form of the Weber number for capillary motions. In his review of plane film motion, Fulford (1964) points out that Weber number has only a small effect on film instability, so that the simple correlation Re *  12 csc ( ) is a reasonable approximation for a draining plane film.

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3-19

Derive Eq. (3-96) for start-up of flow in a circular pipe.

This is a good exercise for the student. The first difficulty is that the basic differential equation (3-94) contains the inhomogeneity ( −dp/dx ) , which must be removed before separation of variables can work. To do this, define velocity V such that

(

)

dp r 2 V = u − u max 1 − r *2 , where r = r/ro and u max =  −  o  dx  4

We thus work with the difference between u and the final Poiseuille-flow profile at t = . Substitution into Eq. (3-94) gives the following homogeneous PDE for V:

V  2 V 1 V = + , where r* = r/ro and t* = vt/ro2 2 t* r* r* r*

V (1,t*) = 0

subject to:

(no-slip)

(1)

(2)

2 V ( r*, 0 ) = u max 1 − r*   

(3)

The solutions to (1) subject to condition (2) are all of the form V = J o ( λ n r*) exp(−  n 2 t*), where  n are the roots of the Bessel function J o (see values in Table 3-3). The general solution of the linear PDE (1) is thus a Fourier-Bessel series of such terms: 

(

V =  Cn J o (  n r *) exp − n2 t * n =1

)

(4)

To evaluate C n , we use condition (3) and the Bessel function orthogonality condition: 1

 xJo ( i x ) Jo (  jx ) dx

=

0

and

= 0

1 J1 ( i ) 2

if i = j if i  j

We thus multiply Eq. (3) by  r* J o (  j r*)  and integrate the whole equation from 0 to 1. The orthogonality condition enables us to pick off the constant one by one, with the result Cn = −

8 u max

 3n J1 (  n )

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With V thus known, we return to the original variable u and the desired solution:

u u max



= 1 − r*2 − 

8 J o (  n r*)

3 n =1  n J1

( n )

(

exp − 2n t*

)

Some typical profiles of u ( r*, t *) are shown in the text in Fig. 3-16. The profiles for very early

t* are shown at right (as requested in the problem) and are seen to consist of a linearly accelerating core with thin growing boundary layers. 3-20 Consider air at 20°C and 1 atm initially at rest between fixed parallel plates h = 2 cm apart. If the lower plate moves at speed Uo = 30 cm/s at time t = 0, evaluate u at the center. For air at 20°C and 1 atm, v  1.50E-5 m2 /s. The Reynolds number of the flow is Re = Uo h/v = (0.30)(0.02)/ (1.50E-5) = 400, hence the flow is laminar. The analysis of laminar start-up between suddenly moved plates is given in Section 3-5.2, with velocity profiles plotted in Fig. 3-20. If we insert the center point, y = h/2, into this formula, we obtain the velocity prediction

u/Uo = 0.5 −

2  1 vt  nπ  exp −n 22 t* sin   , t* = 2   n =1 n  2  h

(

)

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(a) Compute the midpoint velocity after 2 seconds. The corresponding dimensionless time is

t* = (1.50E-5)( 2.0) / ( 0.02) = 0.075. The formula above predicts 2

u midpoint Uo

 0.1966, or: u midpoint = ( 0.1966 )( 30 ) = 5.9 cm/s

(Ans. a)

(b) At what time will the midpoint velocity be 14 cm/s? This corresponds to u/U o = 14/30 = 0.4667. If we evaluate the formula above, searching for this ratio, we will find that

( 0.299)( 0.02) = 8.0 s u = 0.4667 occurs at t* = 0.299, or t = Uo (1.50 E-5) 2

(Ans. b)

One may check these results (approximately) by reading the midpoint values in Fig. 3-20.

3-21 Derive simplified formulas for the volume flow and pumping power of a Couette pump (see figure in problem statement). Illustrate for SAE 30 oil.

For small clearance,

( a − b )  a,

we assume a linear

velocity distribution, as shown at right. Let the annular region have depth “L” into the paper. Then the flow area is A = ( a − b ) L, with average velocity ( V/2) . Thus a simplified expression for volume flow rate is

1 Q = Vavg A =  b ( a − b ) L 2

(Ans. a)

As shown in the figure, the wall shear stress is approximately  = V/ ( a − b ) , so the total torque imparted to the fluid by the inner cylinder is T =  ( 2 bL ) b. Thus the power needed to drive this flow is

P = T =  2b2 L =

2 2 b3L (a − b)

(Ans. b)

As a numerical example, take SAE-30 oil at 20°C, with a = 10 cm, b = 9 cm, L = 1 m, and a rotation rate of 900 rpm. From Appendix A, Fig. A-2, read  = 0.29 kg/m-s for SAE-30 oil. The

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angular velocity is w = ( 900 rev/min )( 2π rad/rev ) / ( 60 s/min ) = 62.83 rad/s. From formula (a) above, we compute a volume flow rate

1 Q  b ( a − b ) L = ( 0.5)( 62.83)( 0.09 )( 0.10 − 0.09 )(1.0 ) 2

= 0.0283 m3 /s

(100 m3 /hr )

(Ans.)

From formula (b) above, we compute the power required:

2 ( 0.29 )( 62.83) ( 0.09 ) (1.0 ) = 520 W ( 0.10 − 0.09 ) 2

P=

3

(Ans.)

Although it’s a lot of power for a small flow rate, the values seem reasonable for heavy oil.

3-22

Determine if the “Taylor vortex’’ is a solution to the Navier-Stokes equations: v = ( const )

 r2  exp  −  t 2  t  r

The vorticity is computed as in the “Oseen vortex” result, Prob. 3-14, page 34: =

 r2  1   v  2 ( const )  r 2  r = 1 − exp    −  r r  r  t 2  t   t 

By direct substitution into the -momentum equation (Appendix B) and the vorticity-decay equation (top of page 43), we may verify after considerable algebra that the Taylor vortex is an exact solution. The velocity and vorticity profile shapes are shown on the next page. By comparison to the Oseen vortex profiles on page 3, the Taylor profiles are much flatter, less peaked, and decay by dropping quickly near the axis with a reversal in vorticity away from the axis.

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3-23

Consider purely radial outflow between parallel disks, as shown below. Assume u r = fcn ( r, z ) with constant ( ,  ) and p = p ( r ) only. Solve as far as possible.

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The continuity equation (Appendix Eq.B-3) reduces to

 ( ru r ) = 0, or: ru r = f ( z ) r Substitution of u r = f /r into the r-momentum relation (Appendix Eq.B-6) gives u r

or:

 1   u r u r p = − + r r r  r r  r −

f 2 r3

=−

2  ur  ur  − + ,  2 z 2   r

dp  d 2f + dr r dz 2

(1)

to be solved for f ( z ) . (The first two terms in the brackets [] on the right cancel for the particular distribution u r = f /r.) At any given section r, Eq. (1) is a nonlinear ordinary differential equation which can be solved for f ( z ) , subject to two boundary conditions:

f = 0 at z =  L;

No-slip:

Symmetry:

df = 0 at z = 0 dz

The parameters involved in Eq. (1) are made clearer if we integrate the equation in the radial direction between r = r1 and r = r2 [see page 114 of Bird et al. (1960)]. The result is  r2  d 2f f 2  1 1  − =  p +  ln     2 2  r22 r12   r1  dz

(2)

where p = p1 − p2 . For “creeping” flow, the inertia term on the left hand side is negligible, and the solution can be found immediately [Bird et al. (1960), page 114]: Very low Reynolds number: f = ru f 

p L2  z 2  1 −  2 ln ( r2 /r1 )  L2 

(3)

This parabolic distribution is the exact analogy to Poiseuille flow in a channel. At higher Reynolds numbers, Eq. (2) is nonlinear, and we may nondimensionalize it using Eq. (3) as a guide:

d 2 dz *2

= −K 2 − 2, where  =

and

K=

(

pL2f z , z* = , 2 ln ( r2 /r1 ) L

)

 r12 − r22 p L4 4 2 ln 2 ( r2 /r1 )

(5)

The single parameter K is proportional to the Reynolds number of the flow. The boundary conditions become  ( 1) = 0 and d/dz*( 0 ) = 0. Equation (5) may be solved numerically by,

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e.g., the Runge-Kutta method. For a given K, the problem is to find the proper value of f ( 0) which causes f ( l ) to be zero. The following solutions were found by the writer:

K=

f ( 0) =

0

0.2

0.4

0.6

0.7

1.0

1.0861

1.2135

1.4575

1.8569

Interesting, no solutions could be found for K  0.75 (approximately) - the reader is invited to investigate this difficulty and explain it to the writer. The five tabulated solutions for velocity profile (z*) are plotted below - all of them look approximately parabolic.

3-24

A porous cylinder of radius R exudes fluid radially at velocity U o into an unbounded

fluid of constant ( ,  ) . Find the velocity and pressure distributions if p = po at r = R. Let u r = u ( r ) only, with u  = u z = 0. The continuity equation is easily solved:

1d R ( r u ) = 0, or: r u = constant, or: u = Uo r dr r

(Ans. a)

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This is the desired velocity distribution. The pressure follows from the r-momentum equation (Appendix Eq. B-6) if we neglect gravity:

u

du dp d 1 d = − +  ( ru ) dr dr dr  r dr 

Introducing u ( r ) from solution (a) above, we find that the viscous term vanishes identically (a consequence of u being proportional to 1/r and therefore a “potential” type of distribution) and we obtain the result

Uo2R 2 dp Uo2 R 2 = , or: p = − + constant dr r3 2 r2 The constant is evaluated from the condition p = po at r = R. The final result is

  R 2  1 p = po +  Uo2 1 −    2   r  

(Ans. b)

For this particular case, the pressure relation is equivalent to Bernoulli’s equation.

3-25

A cylinder of radius ro rotates at surface vorticity o with a wall suction velocity of

( −vw ) at the surface. Find the vorticity distribution assuming that  / = 0 and vz = 0.

With pressure and velocities dependent only upon r, the continuity equation (B-3) becomes

1 ( rvr ) = 0, r r

or: vr = −

ro v w r

since vr |r =ro = −vw

With the radial velocity thus known, the θ-momentum equation, vr

dv v r v v  d  dv  v  + =  r − dr r r  dr  dr  r 

can be solved for tangential velocity. The general solution is

v = A r −1 + Br1−Re ,

where Re = v w ro / is the wall Reynolds number

The first term is the ‘potential’ vortex of a solid cylinder. The second term is a combined effect of viscosity and wall suction and is unbounded if Re  1, which is unrealistic. Further, the flow circulation,  = 2 r v , is proportional to r 2-Re and is unbounded if Re  2, also unrealistic. We conclude that, for physical realism.

B=0

if

Re  2

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The vorticity in the flow is obtained by differentiation:

=

1d ( r v ) = B ( 2 − Re ) r − Re r dr

Utilizing the boundary condition for surface vorticity,  = o at r = ro , we obtain the constant B = o roRe / ( 2 − Re ) if Re  2. The final desired solution for vorticity is thus

r   = o  o  r

Re

(Ans.)

With B known, the constant A follows from the condition v = roo at r = ro . The final solution for velocity is in two parts, depending upon the wall Reynolds number: v =

o ro2  Re-2  1 − Re + ( ro /r )    r ( 2 − Re )

=

3-26

o ro2 /r

if

Re  2

if

Re  2

For laminar flow in a porous tube, similar to the channel example in Section 3-6.3, use

the similarity variable  = ( r/ro )

2

and find the proper stream function and its differential

equation and boundary conditions. For uniform wall normal velocity, the stream function must have the form  = ( a + bx ) g (  ) , and the constants are found by volume-flow balance between 0 and x:

1   =  ro2 u ( 0 ) − v w ro x  g (  ) 2 

(Ans.)

The velocity components are obtained by differentiation: vr = − vx =

1  v w g = r x 

2v x  dg 1   = u ( 0) − w  r r  ro  d

If these velocities are substituted into the r- and z-momentum relations, Appendix Eqns. (B-6) and (B-8), it is found that the pressure satisfies the relation  2p =0 x

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exactly analogous to the porous channel in Section 3-6.3. Cross-differentiation to eliminate the pressure thus gives the following differential equation for the stream function:

2  g + 4g+ Re ( gg − g g) , where Re = vw ro / The boundary conditions at r = 0 are vr = vz /r = 0, which translates to As  − 0:

g = 0 and g  = 0 

The boundary conditions to the wall, r = ro , are v r = v w and v z = 0, or:

g (l) = 1

and

g ( l ) = 0

This problem was first developed by Yuan and Finkelstein (1956), and a complete array of numerical solutions - including vexing non-uniqueness and non-existence difficulties - is presented by White (1962). 3-27 Modify the Ekman spiral solution for turbulent flow by using an “eddy viscosity” correlation suggested by Clauser (1956):

veddy  0.04 D ( o /)

1/2

From Section 3-7.2, use the Ekman solution for surface velocity and penetration depth: Vo =

o /

( 2v sin)

(1)

D =   v/ ( ω sin )

(2)

For our particular case, Vwind = 6 m/s, air = 1.205 kg/m3 , and latitude 41°N, and  = 2/

(86400 s ) = 7.27E-5 s−1. We have the surface-wind-shear estimate, Eq. (3-141): 2 o  0.002 air Vwind = ( 0.002)(1.205)( 6.0) = 0.0868 Pa 2

Dividing this by seawater density gives the so-called “friction velocity”, v*; v* =

( o /) = ( 0.0868/1025)1/2 = 0.00920 m/s

Substituting Clauser’s v into Eq. (2) above gives the penetration depth directly:

D = 2

( 0.04)( 0.00920) = 76 m 0.04 v* = 2 ω sin ( 7.27E-5)(sin41°)

(Ans. a)

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We may then compute the eddy viscosity as v  0.028 m 2 /s from Clauser’s formula. The surface seawater velocity may finally be computed from Eq. (1) above: 2 0.0092 ) ( Vo = 1/2 ( 2 )( 7.27E-5 ) sin ( 41 )( 0.028 ) 

= 0.052 m/s

(Ans. b)

These are realistic results: surface velocity a few centimeters per second and penetration depth of the order of 100 meters (varying with the wind velocity).

3-28

Repeat the Ekman-flow analysis of Section 3-7.2 for shallow water of depth h.

We are to solve the same differential equation with one changed boundary condition:

v w − 2i  sin  w = 0, subject to:

w ( −h ) = 0,

w = u + iv, i =

( −1)

w ( 0) = i K, K = o /

The (linear) differential equation has a hyperbolic-function general solution:

w = Acosh ( bz ) + Bsinh ( bz ) , b = 2isin /v

1/2

= (1 + i ) sin /v

1/2

From the boundary conditions, we find that A = Btanh ( bh ) , B = iK/b. The desired solution for (general) velocity and surface velocity is thus

w=

iK iK [tanh(bh) cosh(bz) + sinh(bz)], and w(0) = tanh(bh) = u o + i vo b b

We use the identity tanh(x + iy) = [sinh(2x) + isin(2y)]/[cosh(2x) + cos(2y)] to untangle real and imaginary parts of the surface velocity:

uo =

1 1 K sinh 2h − sin 2h  , vo = [sinh2βh + sin2βh], Q = Q Q 2 cosh 2h + cos 2h 

where  = /D. The ratio of these two is the tangent of the desired surface flow angle:

tan(θ) = u o /vo =

sinh(2πh/D) − sin(2πh/D) sinh(2πh/D) + sin( 2πh/D)

(Ans.)

A plot of  versus h/D is shown on the next page. The angle rises slowly from zero through an interesting overshoot of 47° at h/D = 0.63 to level off at 45° in “deep” water. From the figure (or the formula) we may estimate that  = 20 at h/D = 0.24 (Ans.).

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3-29 Air at 20°C and 1 atm flows at U  = 1 m/s across a cylinder of diameter 2 cm with wall temperature of 40°C. Estimate (a) ; (b)  w ; and (c) q w at the stagnation point. Use the properties of air at an average temperature of 30°C. From Appendix A, the kinematic viscosity is approximately  = 1.60E-5 m2 /s. We can estimate B from inviscid theory, Eq. (1-3): at the surface, U = 2Usin ( x/R ) , hence dU/dx at x = 0 equals B = 2U /R = 2 (1 m/s ) /

( 0.01 m ) = 200 s−1. From the analysis of Section 3-8.1.1. (a) the shear layer thickness at the stagnation point is 1/2

1.60E-5 m2s   = 2.4 (  /B) = 2.4   −1  200s 

= 0.00068 m = 0.68 mm

(Ans. a)

(b) From Eq. (3-159), wall shear is proportional to x, hence at the stagnation point, x = 0,

τw = 0

(Ans. b)

(c) for the heat transfer, estimate, at 30°C, from Appendix A, that k  0.0263 W/ ( m-K ) . The Prandtl number is 0.71. From Section 3-8.1.4, the heat transfer rate is q w  −k ( T − Tw ) G ( Pr )

( B/ )

= − ( 0.0263)( 20 − 40 ) [0.570 ( 0.71)

0.4

1/2

 200  ] 1.60E-5 

 920 W/m2

(Ans. c)

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3-30

Solve Bodewadt’s “rotating disk” problem, with a fixed disk and a rotating stream.

This problem is very similar to the Kármán rotating-disk problem, with the same variables ( F,G, H, P ) as in Section 3-8.2 as functions of z*. There is one important modification (what some students might call a “trick”): in order for the radial momentum equation to balance at z* =  with a rotating stream, there must be a positive nonzero radial pressure gradient. In dimensionless form, this radial equation becomes

F = − G2 + F2 + FH + 1 This guarantees that, in the farfield, where G (  ) = 1, a rotating stream, the derivatives F' and

F will vanish and the radial and axial motion will decay to zero. The other three differential equations are just the same as in Section 3-8.2: H = −2F; G = 2FG + HG; P = 2FH − 2F

The boundary conditions change on G only, to make the wall fixed and the stream rotating:

F ( 0) = P ( 0) = H ( 0) = G ( 0) = 0;

F (  ) = 0; G (  ) = 1

This problem was first solved by hand by U. T. Bodewadt (Z. angew. Math. Mech., vol. 20, 1940, p. 241). Accurate numerical solutions were later given by Rogers and Lance (1960), who also considered cases where both the disk and the freestream rotated (they concluded that no-slip solutions were possible only if both rotated in the same direction). Similar to Section 3-8.2, one must find the proper initial conditions F ( 0) and G ( 0 ) which make the radial velocity F vanish and the tangential velocity G approach unity at infinity. After considerable searching, we find the correct conditions:

F ( 0) = −0.94197

G ( 0) = +0.77289

(Ans.)

The velocity and pressure profiles are shown plotted below - they oscillate slowly before decaying to their freestream values. [We used the Runge-Kutta routine in Appendix C.]

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3-31 Evaluate Kármán’s problem for a numerical case: one side of a disk rotating at 1200 rpm in air at 20°C and 1 atm. Estimate (a) flow rate; (b) torque; and (c) power. For air at 20°C and 1 atm,  = 1.205 kg/m3 ,  = 1.81E-5 kg/ ( m-s ) , and  = 1.50E-5 m2 /s. The angular velocity is  = (1200)( 2/60) = 40 = 125.7 rad/s. (a) One way (the hard way) to evaluate flow rate is to integrate the radial velocity at the edge of the disk, r = R: 

Q =  v r ( r = R ) 2 R dz = 2 R



2

( v)  Fdz*

0

0

However, the last integral, which equals 0.441, is not shown or tabulated in the text. More straightforward, the flow rate should also equal the fluid ‘pumped’ axially toward the disk:

Q = R 2 | vz (  ) |, where | vz (  ) | = 0.883 or:

(

)(

( ),

)

1/2

2 Q =  ( 0.25 m ) ( 0.883)  1.5E-5 m2 /s 125.7 s−1   

= 0.0075 m3 /s (Ans. a)

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(b) The torque on one side of the disk is given by Eq. (3-190):

M= =

  R 4Go 2

( 3 )

1/2  (1.205)( 0.25)4 ( 0.61592) (1.50E-5)(125.7 )3  = 0.025 N-m 2

(Ans. b)

(c) the power required to drive one side of the disk is

P = M = ( 0.025)(125.7 ) = 3.1 W

(Ans. c)

It should be pointed out that these numbers slightly exceed the expectations for laminar flow, that is, the Reynolds number Re = R 2 /v is about 524,000.

3-32

Solve Jeffery-Hamel wedge flow for creeping flow, Re = 0 but   0 .

Equation (3-195) in Section 3-8.3 reduces for Re = 0 to: f  + 4 2 f  = 0. This is a third-order linear differential equation which has the general solution

f = A1 sin ( 2) + A2 cos ( 2) + A3 Using the boundary conditions, f ( 0 ) = 1 and f ( l ) = 0, we find the constants:

A1 = 0; A2 = 1/ (1 − cos2 ) ; A3 = 1 − A2 Thus the desired solution for creeping Jeffery-Hamel wedge flow is f=

cos2αη − cos2α 1     = 1 + csc2α sin  − 2  − 1 1 − cos2α 2    2

(Ans.)

Some representative velocity profiles are plotted below. The case (  = 0) is the Poiseuille parabola for channel flow, and the case ( = 90) is the separation point. For   90, separation or backflow must occur in a diverging flow even at zero Reynolds number.

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3-33 Consider the following exact incompressible Navier-Stokes stream function in spherical polar coordinates. Plot some streamlines and velocity profiles.

2v r sin 2θ  ( r,  ) = , 1 + a − cosθ

a = constant

(1)

The radial velocity is computed from the stream function:

2v  2cosθ (1 + a − cosθ ) − sin 2θ   ur = 2 =  2 r sinθ r (1 + a − cosθ )  /

(2)

This flow represents a laminar round jet issuing from the origin. The streamlines (Eq. 1) and jet velocity profiles (Eq. 2 for constant r and unit centerline magnitude) are shown below for a = 0.01. The jet becomes progressively thinner as a decreases [not shown].

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The jet width  is defined as the point where u r = 0.01 u max . From Eq. 2 we compute that the maximum velocity (  = 0 ) equals 2v/ ( ra ) . Then the jet velocity ratio is

ur u max

a  2 cosθ (1 + a − cosθ ) − s in 2θ   = 0.01 at  =   =  ( ) 2 2 (1 + a − cosθ )

That is, the jet “thickness” is a constant value of , or  varies linearly with r. Some computed values of the “1%-velocity” angle corresponding to jet thickness are as follows: a

= 0.001 0.003 0.006 0.010

 ( ) =

7.5°

12.8°

17.6° 21.8°

The jet mass flow is found by integration over the jet profile at a given value of r: (  )

m=

 u r dA =  jet

0

f ( , a )  2 r ( r d ) = ( const )( r ) r

(  )

 f ( , a ) d = ( const ) r 0

Thus the (laminar) mass flow increases linearly with r along the jet axis due to “entrainment” from the ambient fluid outside the jet. [See Eq. (4-204) for details.]

3-34 A sphere of specific gravity 7.8 is dropped in oil of specific gravity 0.88 and viscosity  = 0.15 Pa-s. Estimate its terminal velocity for D = (a) 0.1 mm. (b) 1.0 mm, and (c) 10.0 mm. At terminal velocity, the sphere’s net weight in oil equals its drag:

   Wnet = sph −  g D3 = Drag = CD V 2 D2 6 2 4

(

)

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or:

(

) 

 4 D g sph / − 1 V= 3CD  

1/2

 

, where CD = fcn ( Re ) , Re =

VD 

For creeping (Stokes) motion, we have CD = 24/Re, or V = W/ (3D ) , which could serve as a first estimate. At higher Reynolds numbers, we could use Eq. (3-225): C D ( sphere ) =

24 6 + + 0.4 Re 1 + Re

(a) D = 0.1 mm: The smallest sphere is most likely to be Stokes flow. Then estimate

( 7.8 − 0.88)( 998)  ( 9.81)( /6 )( 0.0001) W V= = 3 D  3 ( 0.0001)( 0.15 )

3

= 0.00025 m /s

(Ans. a)

This answer is accurate, as we may verify by checking the Reynolds number: Re =

VD ( 0.88)( 998) ( 0.00025)( 0.0001) = = 0.00015  1  0.15

(Stokes flow)

(b) D = 1.0 mm: Use the Stokes-flow estimate V = W/ ( 3D ) to get a first guess of

V  0.025 m/s. Cheek the Reynolds number, Re = VD/ = 0.15, which is  1: V  0.25 m/s

(Ans. b)

(c) D = 10 mm: Here the Stokes estimate is V = 2.51 m/s, with Re = 147, which is not small. Therefore re-compute CD = 1.02 and re-estimate V from our formula above:

(

) 

 4Dg sph / − 1 V 3 CD  

1/2

 

 4 ( 0.01)( 9.81)( 7.8/0.88 − 1)  =  3 (1.02 )  

1/2

= 1.00 m/s

This is a big change, so we need to iterate the formulas until they converge to Re = 46, CD = 1.70,

3-35

V = 0.78 m /s

(Ans. c)

Verify Eq. (3-230), for the Stokes drag of a droplet, with a complete analysis.

We are to solve the Stokes stream function differential equation, 2

  2 1  2 cot     2 + 2 2 − 2   = 0   r r  r  

(1)

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for  ( r,  ) subject to zero radial velocity, equal tangential velocity, and equal shear stress at the droplet surface, r = a. By analogy with the sphere solution in Section 3-9.2, the stream function yields to separation of variables,  = Usin 2  f (r), and substitution into Eq. (1) above gives a fourth-order linear “equidimensional” equation: f  −

4 r

2

f  +

8 r

3

f−

8 r4

f =0 ,

with the general solution

f=

A + Br + Cr 2 + Dr 4 r

(2)

This result is valid for both inside and outside flows. Let subscripts “i” and “o” denote the inside and outside flows, respective. The eight constants ( A, B, C, D )i,o are found by laborious systematic application of the boundary conditions. First, at large r, the outside flow must approach the uniform stream,  o = (1/2 ) Ur 2 sin 2 , which requires that Co = 1/2

Do = 0

Second, the inside flow must have a finite velocity at the origin, r = 0, which requires that Ai = Bi = 0

(

)

2 Third, u r = (  / ) / r sinθ must equal zero at r = a on both sides. This requires

Outside:

1 Ao = −Boa 2 − a 3 ; 2

Inside: Ci = −Dia 2

This leaves only two free constants, B o and Di , which are found by (a) equating the tangential velocities, ( −/r ) / ( r sinθ ) , at the interface:

3 u o ( r = a ) = u i ( r = a ) , or: Bo = Dia 3 + a 4 Finally, equate the shear stresses at the interface:

o

u o u a ( 2o + 3i ) ( r = a ) = i i ( r − a ) , or: Bo = r r 4 ( o + i )

(Ans.)

This is the desired solution for the inner and outer stream functions. The problem is completed by evaluating the drag from the integrals in Eq. (2-219), with the result

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F = 8o U Bo = 6 a o U

1 + 2o /3i 1 + o /i

which was the liquid-droplet Stokes drag result to be proved.

3-36

Repeat Prob. 3-23, for radial outflow between parallel disks, for creeping flow.

Referring back to Prob. 3-23, continuity requires that ru r = fcn ( z ) only, and, if inertia or convective acceleration is neglected (creeping flow), the r-momentum equation reduces to

dp u d 2 constant = r u = ( ) r dr r dz 2 r where the latter equality follows from separation of variables. This may be integrated once with respect to r and twice with respect to z, giving the velocity as a function of pressure:

ur =

L2 ( p1 − p 2 ) 1 − z 2 /L2   2 r ln ( r2 /r1 ) 

(Ans.)

for any two points “1” and “2” between the disks. Thus, for creeping motion in this ‘slowly varying’ duct, by analogy with lubrication theory, the local velocity profile is everywhere parabolic, and the velocity level varies inversely with r to satisfy continuity. The above result may be rewritten to show the logarithmic variation of pressure:

p = p1 −

2 2

L

r   r1 

( ru max )1 ln 

(Ans.)

The pressure drops logarithmically in the radial direction, at a rate proportional to the “level” of flow ( ru max ) , i.e., the flow rate entering from the center of the disks.

3-37

Analyze creeping flow between parallel disks. The lower disk ( z = 0) is fixed, and the

upper disk ( z = L ) rotates at angular rate . Find the velocity and pressure distributions. This is, in principle, a three-dimensional flow but with radial symmetry, i.e.  / = 0. The boundary conditions on velocity are At z = 0: u r = u  = u z = 0;

at z = L: u r = u z = 0, u  = r

With inertia (acceleration) negligible in creeping flow, the -momentum equation (Eq. B-7 of Appendix B) reduces, for p/ = 0 and zero gravity, to

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 2 v =

1   v   2 v v r + = 2 r r  r   2 r z

The hint u  = r f (z) is a good one and cracks the puzzle, reducing this equation to f  = 0, or f (z) = A + Bz. The no-slip boundary conditions above yield A = 0 and B = /L. The desired solution for creeping circumferential flow between the disks is thus

v = r 

z L

(Ans.)

The flow simply rotates in linear fashion, like Couette flow between plates. Since this particular tangential velocity does not influence continuity, r-momentum, or z-momentum, the remainder of the flow reduces, with no-slip between disks, to the trivial results vr = vz = 0;

p = constant

(Ans.)

For creeping flow, no secondary motion is generated. As Reynolds number increases, acceleration causes nonzero v r and v z to arise, like the Kármán rotating-disk solution.

3-38

Set-up the method of separation of variables for finding u ( y, z ) for a rectangular duct,

Fig. 3-9 and Eq. (3-48), by analyzing Eq. (3-30). First note that the separation will not work until one defines a new variable U = u − F , where  2 F = (1/ )( dp /dx ) , so that 2U = 0. Then separate U into z and y parts and find the form of each part. Show how an infinite series would be required to satisfy the boundary conditions, but do not determine the series coefficients. Solution: First find the function F. From the figure at right,

(

)

2 2 a suitable guess is F = K y − a , whence 2 F = 2 K =

(1/ )( dp/dx ) . If U = u − F , then 2U = 0. Assume that U can be written as a separation of variables:

U ( y, z ) = Y ( y ) Z ( z ) , whence 2U = Y Z + Y Z  = 0, or: Y  Z  = − = constant =   0 Y Z We chose   0 so that Y would be trigonometric and Z would be exponential. Thus

U ( y, z ) =  A cos ( y) + B sin( y)C cosh( z) + D sinh( z)

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The boundary conditions are no-slip, u = U + F = 0 at the duct walls, and symmetry, zero spatial slopes along the center lines of the rectangle: U ( a, z ) = 0;

(

)

U U ( 0, z ) = 0; U ( y, b ) = − F ( y, b ) = − K y 2 − a 2 ; ( y, 0 ) = 0 y z

The second of these conditions requires that B  0, the fourth requires that D  0. Now absorb C into A and we are down to

U = A cos ( y ) cosh ( z ) . Satisfying the first boundary condition above requires that cos ( a ) = 0, or:

a=

 2

,

3 5 , , ... 2 2

or

=

i , 2a

i = 1,3,5, 7, ......

The third (and final) boundary condition requires that U equal a quadratic function along the upper walls of the rectangle:

(

)

U ( y, b ) = − K y 2 − a 2 = A cos ( y ) cosh ( b )

?

This looks impossible, but since Laplace’s equation is linear, we may take U to be the sum of an infinite series of cosine terms:

U ( y, z ) =



 i Ai cos   2a i =1,3,5...



  i  y  cosh  z    2a 

Then we can find the Fourier coefficients Ai such that the above boundary condition for

U ( y, b ) is satisfied. We asked you not to carry this part out. The complete solution procedure is

given in the classic text by Rosenhead (1963), Laminar Boundary Layers, and we put the final solution for u ( y, z ) in the text as Eq. (3-48).

3-39 Find the pressure distribution for lubrication flow [Section 3-9.8.1] in a parabolically varying gap:  x2  h ( x ) = hL + ( h0 − hL ) 1 − 2   L   

The relevant relation to be solved is Reynolds’ equation (3-23g). Using dimensionless variables

( )

x* = x/L, h* = h/h 0 , and p* = ph 02 / ( UL ) , Eq. (3-239) becomes

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d  3 dp*  dh * , h* =6 dx *  dx *  dx *

with h* =

hL  hL  2 + 1 −  (1 − x *) h0  h0 

If we let “ p  ” be zero for convenience, the boundary conditions on pressure are:

p* = 0 at x* = 0 and at x* = 1 Because h* is cubed and thus leads to the sixth power of x*, it is very laborious to solve for p * analytically. We therefore program the differential equation above for a computer, using Subroutine RUNGE from Appendix C. Some numerical values of slope are: h L /h 0 =

dp*/dx *(0) =

0.9 0.7 0.5 0.4 0.3 –0.417 –1.357 –2.459 –3.078 –3.743

The pressure profiles from these results are plotted below

For this type of gap - with a steep change initially, decreasing to a shallow change at the end the pressure drops in the gap and there is a strong possibility of cavitation for this direction of flow. 3-40 Set up Stokes’ first problem of the suddenly moved wall (Fig. 3-18a) on a digital computer by the explicit finite-difference method of Eq. (3-247).

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Let us use dimensionless variables, y* = y/L, u* = u/U, and t* = vt/L2 , so the basic differential equation and its explicit model are

(

)

t * u*  2 u* j j j+1 j u =  u + u + 1 − 2  u ,  = = , with model: ( ) n n n + 1 n − 1 t* y*2 ( y*)2

(1)

The initial conditions at t = 0 are u1 = 1, u n 1 = 0 everywhere. The boundary conditions are u1 = 1 and u n = = 0. For stability we must have   0.5. For convenience take y* = 1.0 and let t* = 0.25, that is,  = 0.25. Some numerical results are shown below: t* 0.000 0.250 0.500 0.750 1.000 1.250 1.500 1.750 2.000 2.250 2.500 2.750 3.000 3.250 3.500 3.750 4.000

u1 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

u2 0.000 0.250 0.375 0.453 0.508 0.549 0.581 0.607 0.629 0.648 0.664 0.678 0.690 0.701 0.711 0.720 0.728

u3 0.000 0.000 0.063 0.125 0.180 0.227 0.267 0.302 0.332 0.359 0.383 0.405 0.424 0.442 0.458 0.473 0.487

u4 0.000 0.000 0.000 0.016 0.039 0.065 0.092 0.118 0.143 0.167 0.189 0.210 0.230 0.248 0.265 0.281 0.296

u5 0.000 0.000 0.000 0.000 0.004 0.012 0.022 0.035 0.049 0.064 0.078 0.093 0.108 0.122 0.136 0.150 0.163

u6 0.000 0.000 0.000 0.000 0.000 0.001 0.003 0.007 0.013 0.019 0.027 0.035 0.043 0.052 0.061 0.071 0.080

u7 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.002 0.004 0.007 0.011 0.015 0.019 0.024 0.029 0.035

u8 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.001 0.003 0.004 0.006 0.008 0.011 0.014

u9 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.002 0.002 0.003 0.005

u10 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.001 0.001

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When plotted versus  = y*/ t*, we get a nearly unique error-function curve (Eq.3-108):

3-41 Repeat Prob. 3-40 using the implicit method, Eq. (3-249). We follow the ideas of Prob. 3-40 except that we use the implicit model:

−u nj++11 + (1 + 2) u nj+1 −  u nj+−11 = u nj ,  = t*/ ( y*)

2

There is no limitation on , but now we have a set of tri-diagonal simultaneous equations to solve. Again let y* = 1.0 for convenience. We may use any time step without instability. Let us choose t* = 1.0 (four times as large as Prob. 3-40), or  = 1.0. The computed results are tabulated below: t* 0.000 1.000 2.000 3.000 4.000

u1 1.000 1.000 1.000 1.000 1.000

u2 0.000 0.382 0.553 0.642 0.696

u3 0.000 0.146 0.276 0.374 0.445

u4 0.000 0.056 0.130 0.203 0.267

u5 0.000 0.021 0.059 0.105 0.151

u6 0.000 0.008 0.026 0.052 0.082

u7 0.000 0.003 0.011 0.025 0.043

u8 0.000 0.001 0.005 0.012 0.022

u9 0.000 0.000 0.002 0.005 0.011

u10 0.000 0.000 0.001 0.002 0.005

Compared to the exact (error-function) solution, the accuracy is about 4%. We obtained 1.5% accuracy in Prob. 3-40 (see table on page 71) but used a time step 4 times smaller.

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3-42 Set up Stokes’ second problem, the steadily oscillating wall, Fig. 3-18b. Use the explicit finite-difference model. We can follow the same procedure as Prob. 3-40, with variables y* = y/L, u* = u/U, and t* = vt/L2 . For the wall velocity oscillation, take

u* ( 0, t*) = cos ( ω*t*) , where * = L2 /v

(1)

For simplicity, take y* = 0.5 (to get the needed spatial definition) and * = 1.0. To keep  less than 0.5, use the value  = 0.25 from Prob. 3-40, which requires that t* = 0.0625. Run the same explicit model as in Prob. 3-40:

(

)

u nj+1 =  u nj +1 + u nj −1 + (1 − 2 ) u nj ,  = 0.25, u1 = cos ( t*) From the print-out (not shown), we have plotted some velocity profiles on the next page (p. 60). Initial transient effects do not die out until t*  10, so we have only plotted ‘steady-state’ profiles for every unit value of t* = 10 to 20. Note the striking similarity with the exact profiles in Fig. 3-18b. The numerical accuracy is about ±1% for these particular step sizes y* = 0.5 and t* = 0.0625.

3-43

Repeat Prob. 3-42 (Stokes’ 2nd problem) using an implicit method.

By analogy with Prob. 3-41, by using the implicit model (see top of page •••) we can choose, e.g., a step size four times larger, that is, t* = 0.25, with y* = 0.5 and  = 1.0, and of course let u*( 0,t*) = cos ( t*) . We get good accuracy (about ±4%) over the entire cycle of varying wall motion. The plot of instantaneous profiles would look almost exactly as above, so we will not repeat those results.

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3-44

The solutions f n ( r*) to Eq. (3-76) are called Graetz functions, but they are not tabulated.

Set up a numerical solution of Eq. (3-76), perhaps using the Runge-Kutta subroutine of Appendix C, and solve iteratively for the first three functions f1-2-3 and their eigenvalues 1-2-3. Compare with Table 3-1, but no fair using that table for your initial guesses for eigenvalues. Solution: The basic differential equation (3-76) should be solved for the highest derivative:

f n = −

(

)

1  f n − n2 1 − r 2 f n r

With initial conditions fn ( 0) = 1 and fn (1) = 0. Because r = 1 is the centerline, we also know from symmetry that f n ( 0) = 0, and, further, the value f n ( r ) /r |r =0 = 0 also. Thus we can start at r = 0 and integrate numerically outward, for a small r, varying  n until f n (1) = 0. We can use Subroutine RUNGE or any effective numerical method. The writer used an Excel spreadsheet. Only certain discrete eigenvalues  n work, and we won’t cheat by selecting them from Table 3-1 of the text! For example, for the first eigenvalue, we can guess  0 = 2.0, 2.5 and 3.0 and compute f1 (1) = 0.368, 0.104, − 0.141. The correct value of λ0 , for which f1 (1)  0, lies between 2.5 and 3.0, and we can interpolate to find guessed λ0  2.7 or, even more closely, 2.704, just like Table 3-1. In a similar manner, we can home in on λ1  6.68 and λ 2  10.67. The three eigenfunctions are shown in the chart below. They look like Bessel functions, but no, they are Graetz functions.

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3-45 Extend Prob. 1-21, where we only found the Knudsen number, Kn  0.17, by using all the data given there. (a) Find the Reynolds number and see if it is less than 2000 (laminar flow). (b) Estimate the required pressure gradient in Pa/m. (c) Estimate the flow rate in mm3 /s. Solution: From Prob. 1-21, for oxygen at 1200 Pa and 293 K,  = 0.0158 kg/m3 and  = 2.0E-5 kg/m-s. The average velocity is 10 cm/s, and D = 35 μm. Thus the Reynolds number is small and definitely laminar:

(

)

3 VD 0.0158 kg /m ( 0.1m/s )( 0.000053m ) = = 0.0028 (a) Re D =  2.0E -5 kg / ( m  s )

Ans. (a)

With velocity and diameter known, the volume flow rate follows immediately:

Q = AV =

 4

( 0.000035m )2 ( 0.1m/s ) = 9.6E − 11m3 /s = 0.096

mm3 s

Ans. (c)

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With Kn = 0.17  0.01 from Prob. 1-21, we know slip flow is important, so Eq. (3-40) applies:

 ( 0.0000175m )  dp  m3  ro4  dp  Q = 9.6 E -11 =  −  (1 + 8Kn ) =  −  1 + 8 ( 0.17 )  s 8  dx  8 ( 0.00002 kg /m  s )  dx   4

 dp  Solve for  −  = 22,000 Pa /m  dx 

3-46

Ans. (b)

Starting from the axial momentum equation, derive Eqs. (3-40) for slip flow in tubes.

Solution: From Sect. 3-3.1 for the no-slip laminar pipe flow, the general solution of the fullydeveloped-flow equation,  2u = K = (1/ )( dp /dx ) , is u = Kr 2 /4 + C1 ln ( r ) + C2 . As before, we must have C1 = 0 to enforce centerline symmetry and avoid a logarithmic singularity. Then we find C 2 from the slip condition, noting that K is negative: 2 − ( dp /dx ) du K 2  Kr  Kr |w =  − o  = o + C2 , or: C2 = − ro + 2ro , K = dr 4 4   2  − ( dp /dx ) 2 2 ro − r + 2ro Thus the final solution is: u = Ans. 4

(

uw = |

(

)

)

3-47 For the geometry of Fig. 3-1 assume a constant pressure gradient with both walls fixed. Solve continuity and x-momentum for laminar slip flow between the plates. Find the velocity distribution and the volume flow rate per unit depth. Does the Knudsen number appear?

Solution: The appropriate fully-developed x-momentum equation is Eq. (3-41):

d 2u dy Integrate twice:

2

=

1 dp =K 0  dx

du K = Ky + C1; u = y 2 + C1 y + C2 dy 2

For symmetry, du /dy = 0 at y = 0, hence C1 = 0. To find C2 , apply the slip condition at y − h:

u y −h =

 du  K 2 h + C2 =  −  = 2  dy  y −h

( − Kh ) ,

or: C2 = −

Kh (h + 2 2

)

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The final slip-flow velocity distribution is thus u=

( −dp /dx ) 2

( h2 + 2 h − y 2 )

Ans.

The volume flow rate, with slip, does indeed contain the Knudsen number. +h

 udy =

Q=

−h

2 ( −dp /dx ) 3 h (1 + 3Kn ) , Kn = Knudsen number = 3 h

Ans.

3-48 Hadjiconstantinou (2003) has updated a second-order slip theory by Cercignani (2000) to give new numerical coefficients for the wall-slip velocity, to be compared with Eq. (1-91):

uw  1.11

u |w − 0.61 y

2

 2u y 2

|w

Repeat Prob. 3-52 with this formulation to solve continuity and x-momentum for laminar slip flow between the plates. Find the velocity distribution and the volume flow rate per unit depth. Does the Knudsen number appear? Solution: From Prob. 3-52 we can get this far without no-slip:

d 2u dy Integrate twice:

2

=

1 dp =K 0  dx

du = Ky + C1 ; dy

u=

K 2 y + C1 y + C2 2

Symmetry, du /dy = 0 at y = 0, hence C1 = 0. To find C2 , apply the new slip condition at y = h:

u y =h

 du  K = h 2 + C2 = 1.11  −  − 0.61 2  dy  y =h Solve for

2d

u  2   dy  y =h 2

 h2 C2 = − K  + 1.11 h + 0.61  2 

2

 

The final modified slip-velocity distribution is:

u=

( −dp /dx ) 2

( h2 + 2.22 h + 1.22 2 − y 2 )

Ans.

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The volume flow rate contains the Knudsen number twice: +h

Q=



u dy =

−h

2 ( −dp /dx ) 3 h 1 + 3.33Kn + 1.83 kn 2 , Kn = Knudsen number = 3 h

(

)

Ans.

3-49 Glycerin, with a density of 1260 kg/m3 and viscosity of 1.5 kg/m-s, flows at 1 m/s past a small smooth sphere. (a) What sphere diameter will cause the Reynolds number to be exactly unity? For the sphere size in part (a), according to Stokes-flow theory, what is the gage pressure (relative to freestream pressure) (b) at the front stagnation point; (c) at the rear stagnation point? (d) What is the surface shear stress at these two points? Solution: Numerical problems are a refreshing relief from heavy theory. Calculate Re D :

(

)

3 VD 1260 kg /m (1.0m /s ) D Re D = 1.0 = = , solve for D = 0.00119m  1.5 kg /m − s

Ans. (a)

(b, c) From Eq. (3-215), with p = 0 pascals gage, the fluid gage pressure at the surface is

3 (1.5kg /m − s )(1.0 m/s ) 3U cos  = − cos  = −3780cos  2a  0.00119m  2a 2  2   = −3780cos180 = +3780Pa Ans. (b); prear = −3780 cos 0 = −3780 Pa Ans. (c)

p |r =a = −

p front

3 aU 2

cos  = −

(d) From Eq. (3-216), the surface shear  rθ is zero at both the front (180 ) and back ( 0 ) .

3-50 Hill and Power (1956) proved that the creeping-flow (Stokes) drag of a solid object is greater than the drag of any inscribed shape but less than the drag of any circumscribed shape. Verify this result for the spheroid of Fig. 3-36 by comparing it to inscribed and circumscribed spheres. Do the relative drag forces differ markedly or only by a few per cent?

Solution: The spheroid, shown as a heavy line above, has a length 2a and a diameter 2b. The drag estimates for the spheroid are given by Eqs. (3-221): Parallel to 2a:

C1  6 ( 4 + a/b ) /5; Normal to 2a: Cn  6 (3 + 2a/b ) /5

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For simplicity, let a  b, the results would simply be reciprocal for a  b. The force ratios may be calculated as follows:

Ft −circumscribed 6 U a 5a /b = = Ft −actual 6 ( 4 + a /b ) /5 Ub 4 + a /b Ft −inscribed 6 U b 5 = = Ft −actual 6 ( 4 + a /b ) /5 Ub 4 + a /b Fn−circumscribed 6 U a 5a /b = = Fn−actual 6 ( 3 + 2a /b ) /5 Ub 3 + 2a /b

Fn−inscribed 6 U b 5 = = Fn−actual 6 ( 3 + 2a /b ) /5 Ub 3 + 2a /b When plotted below in the curve-fit range, 1  a/b  5, we see that both “circumscribed” forces are greater and both “inscribed” forces are less than the actual forces. The ratios are significantly different from unity and vary by a factor of 2 to 3.

3-51 The elemental creeping-flow solution,  = r (ln r )sin  , is called an Oseenlet. Is this a solution for plane flow, Eq. (3-206), or axisymmetric flow, Eq. (3-211), or both? How might Oseenlets be used in analysis of more complex creeping flows? Solution: Well, let’s first try axisymmetric creeping flow, Eq. (3-211):

( D )  = 0?, where D 2 2

2

=

2 r

2

+

1 2 r  2

2

−

cot   r 2 

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With much labor, work out D 2 ( r ln r sin  ) =

sin  ln r sin  cos2  ln r − − r r sin  r

With even more labor, D 2  D 2 ( r ln r sin  )  = Six terms, nowhere near zero!   No, this particular Oseenlet is not a solution for axisymmetric creeping flow. Then let us try plane creeping flow, Eq. (3-206):

(

)

 4 =  2  2 = 0? where  2 =

1    r r r  r

2  1  +   r 2  2

1 1 Work out 2 ( r ln r sin  ) = sin  , then 2  sin   = 0 r r 

Yes, this particular Oseenlet is indeed a solution for plane creeping flow. Since the creeping-flow differential equation is linear, it follows that more complicated creeping flows could be constructed as a superposition of Oseenlets.

3-52 Using a numerical method such as Subroutine RUNGE of Appendix C, solve the axisymmetric stagnation flow Eq. (3-165) for the function F ( ) . Use an iterative scheme to determine the proper value of F  ( 0 ) . Solution: From Eqs. (3-165) and (3-152), the axisymmetric equation and conditions are: F  + 2 F F  + 1 − F 2 = 0 F ( 0) = F  (0) = 0 ; F  (  ) = 1

The challenge is to find the initial value F  ( 0) which achieves F  (  ) → 1. We could estimate (guess) that this initial value is of order unity. The solutions are not as well behaved as the Blasius equation – selecting F  ( 0) too low causes F  (  ) to keep falling far below unity, and selecting F  ( 0) too high causes F  (  ) to keep rising far above unity. Anyway, we can make a few guesses for F  ( 0) and select  = 4.0 as “infinity”. The equation above was solved numerically, and here are some results:

F  ( 0)

1.00

1.25

1.50

F  ( 4.0 ) –0.8036 0.6686 1.9645

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So we see the correct solution  F  ( 4.0 )  1.0  is somewhere between F  ( 0 ) = 1.25 and 1.50, at about 1.31. We could interpolate parabolically (giving F  ( 0)  1.311 ) or we could just run off two more solutions to narrowly bracket that region:

F  ( 0)

1.30

1.32

F  ( 4.0 ) 0.9369 1.0427 Now we can interpolate linearly to F  ( 0)  1.3119, the same as Table 3-4 of the text. Ans. And we can recompute the desired axisymmetric stagnation flow as in Fig. 3-25 of the text.

3-53 The rotating disk of Fig. 3-28 acts rather like a centrifugal pump. Consider a disk 60 cm in diameter, rotating at 120 rev/min in air at 20°C and 1 atm. Is the flow at the disk tip laminar, transitional, or turbulent? Using Kármán’s theory of Sect. 3-8.2, estimate the volume flow of air, in cm3 /s, pumped outward by one side of the disk. Solution: For air at 20°C and 1 atm, take  = 1.205 kg/m3 and  = 1.81E-5 kg/ ( m-s ) . Convert

 = (120 rev/min )( 2π/60) = 12.57 rad/s. Then the disk Reynolds number, as defined by Eq. (3-191), is

 ro2 (1.205kg /m3 ) (12.57rad /s )( 0.3m ) = = 75300  300, 000 ( transition )  1.8E − 5 kg /m − s 2

Re =

The Reynolds number is low, the flow is laminar.

Ans.

There are two ways to find the volume flow (l) integrate the radial outward flow proportional to F ( z*) ; or ( 2) calculate the inward axial flow proportional to H ( z*) . The latter is easier, using the approach velocity from Eq. (3-188):

vz (  ) = −0.8838 v = −0.8838 (12.57 )(1.81E − 5 /1.205) = −0.01214 m/s The negative sign means flow toward the disk. The total volume flow is thus 2 Q = vz (  ) Adisk = ( 0.01214m/s )  ( 0.3m )  = 0.00343 m3 /s = 3430cm 3 /s  

Ans.

There may be edge effects, since Kármán’s solution is for an infinite disk. But they cannot be large, because the boundary layer thickness is only  = 5.4 ( v / )

1/2

 5.9 mm.

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3-54 Consider the fully developed flow in a long axisymmetric annulus, which is driven by a pressure gradient. (a) Using an appropriate coordinate system, make the necessary assumptions that will simplify the incompressible Navier–Stokes equations. (b) Write the differential equation of motion for this problem. (c) Specify a judicious assortment of boundary conditions. (d) Solve for the steady-state velocity profile u(r) for b ≤ r ≤ a.

Solution: Consider a strip of width dx and radius dr. The flow is steady, incompressible, and fully developed. The forces on the free body in the direction of flow must be zero. Using the momentum and conservation of mass equation, we have  P    T      P ( 2 r ) dx −  P + x dx  2 rdx  + T ( 2 r ) dx −  T + r dr  2 ( r + dr ) dx  = 0          P    T  T    − x dx  2 rdr −  r dr  2 rdx  − T ( 2 r ) dxdr + r dr ( 2 dx ) dr  = 0       

Dividing the above equation throughout by the volume of the element 2 rdrdx and neglecting the last term that is a higher-order differential, we obtain

P 1 T + T+ =0 x r r P 1  + (Tr ) = 0 x r r

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P 1  =− (Tr ) x r r  P (Tr ) = −r r x Integrating the above equation, we obtain r 2 P Tr = − + C1 2 x

T =−

r P C1 + 2 x r

We know that T = − 

V . r

Therefore, we have

−

V r P C1 =− + r 2 x r

V r P C1 = − r 2  x  r

V=

r 2 P C1 − ln r + C2 4 x 

Applying boundary conditions at r = Ri and r = Ro, we obtain

  1 P  2 2 Ri2 − Ri2  Ri V =− Ri − r − ln  4 x   Ri   r ln      Ro 

      

(Ans.)

3-55 A fluid of constant thickness h flows steadily down a flat surface that is inclined at an angle θ, as shown below. (a) Make the necessary assumptions to reduce the Cartesian Navier–Stokes equations. (b) Write the differential equation of motion for this problem. (c) Specify a judicious assortment of boundary conditions. (d) Solve for the steady-state velocity profile u(y) for 0 ≤ y ≤ h. (e) Calculate the volumetric flow rate per unit width.

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Solution: τ is expanded in a Taylor series about the center of the differential control volume shown in the figure.

t =  +

d  dy    dy  2 

and

b =  +

d  dy  −  dy  2 

The boundary conditions are as follows: For no slip at y = 0,

u=0

For no shear stress at y = h,

du =0 dy

Applying the x component of the momentum equation to the differential control volume, we obtain

FSx + FBx =  updV +  upV  dA Assuming steady and fully developed flow (i.e., u = u(y) and τ = τ(y)) and that there is no variation of pressure in the x direction, we obtain   d  dy   d  dy   FSx + FBx = 0 =  +    dx dz +  +  −  dx dz +  g sin  dx dy dz dy  2   dy  2    

Thus,

d = −  g sin  . dy

Integrating the above equation, we obtain

 = − (  g sin  ) y + C1 We know that τ = 0 at y = h Therefore,

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C1 = (  g sin  ) h

and

du  g sin  = hy dy 

Integrating again, we obtain

u=

 g sin   y2  hy −   + C2   2 

Also, u = 0 at y = 0 Therefore, C2 = 0

and

 g sin   y2  u=  hy −    2 

(Ans. d)

The volumetric flow rate per unit width can be determined as follows: h

 g sin  h  y2   g sin   hy 2 y 3  Q = udy = hy − dy = −    0   2   6 0  2 0 h

Thus, Q = h3  g sin  / 3 .

(Ans. e)

3-56 Consider the flow configuration of two different fluid bodies that are separated by a rigid interfacial surface that can be assumed to be infinitesimally thin. The heights and dynamic viscosities of the two bodies are specified on the sketch. The upper body is bounded by a moving surface that translates in the positive x direction at a uniform speed V. What should be the axial speed of the interfacial surface in the absence of a pressure gradient in the streamwise direction?

Solution: At the interface,

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Z1 = Z2 and V1 = V2 Therefore, V=U

1V1 h1

=

2V2 h2



1U

F = Z1 A + Z 2 A =

U=

D1

=

2U D2

1 0 2V h1

+

h2



1 D1

= 0+

=

2 D2

2U D2

FD2

2

(Ans.)

3-57 Consider fully developed laminar flow in the annulus between two concentric pipes as shown below. The outer pipe is stationary, and the inner pipe moves in the x direction at a constant speed V. Assume the axial pressure gradient is zero (dp/dx = 0). From the differential form of the conservation laws, obtain a general expression for the velocity profile, u(r), in terms of two integration constants, C1 and C2. Apply the no-slip boundary conditions to solve for C1 and C2. Give the final expression for u(r).

Solution: Applying the x component of momentum equation for the given annulus, we obtain

FSx + FBx =  updV +  upV  dA and

 rx = 

du = dr

The axial pressure gradient is zero. For a fully developed laminar flow, we have

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d dr   dr  d dr   dr    FSx =  +  2  r +  dx −  −  2  r −  dx = 0 dr 2   2 dr 2   2   Neglecting the products of differentials, the above equation reduces to

 +r

d =0 dr

or

d ( r ) dr

=0

Therefore, r = C1 or  =

C1 . r

We know that

 =

du dr

Therefore, du C1 = dr  r

and u=

C1



ln r + C2

Applying no-slip boundary conditions to the above equation gives us the following: At u r = r = Vo ,

Vo =

At u r = r = 0 ,

0=

i

o

C1



C1



ln ri + C2

ln ro + C2

C1

or

C2 = −

C2 = −

Vo ln ro ln ( ri / ro )



ln ro

Solving for C1 and C2, we obtain

C1 =

Vo

ln ( ri / ro )

and

(Ans.)

3-58 Consider a fully developed laminar flow in the annular space formed by the two concentric cylinders shown in Prob. 3.57, but with a stationary inner cylinder and a nonzero pressure gradient (dp/dx ≠ 0). Let ro = R and ri = kR, where 0 < k < 1 is a constant. (a) Using the Navier–Stokes equations and proper boundary conditions, show that the velocity profile and volume flow rate can be written as

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2 R 2 dp   r  1− k 2  r  u (r ) = − ln    1 −   + 4 dx   R  ln (1/ k )  R  

(1 − k )   R 4 dp  Q=− 1− k 4 − 8 dx  ln (1/ k )  

and



2

2





(b) Obtain an expression for the location of the maximum velocity as a function of k. (c) Obtain an expression for the average velocity in the annulus, u . (d) By taking the limiting case of k → 0, compare your expressions to those corresponding to the Poiseuille flow in a circular pipe.

Solution:

The differential control volume for nonzero dp/dx is

u=

r 2 dp C1 + ln r + C2 4 dx 

Applying boundary conditions to the above equation, we obtain the following:

R 2 dp C1 + ln R + C2 = 0 . At u r =r , i 4 dx  At u r =r , o

k 2 R 2 dp C1 + ln kR + C2 = 0 . 4 dx 

Solving for C1 and C2, we obtain

(

)

2 R 2 dp 1 − k C1 = − 4 dx ln (1/ k )

and

(

)

2 R 2 dp R 2 dp 1 − k C2 = − + ln ( R ) 4 dx 4 dx ln (1/ k )

Substituting C1 and C2 in the expression of u, we obtain

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(

)

(

)

2 2 r 2 dp R 2 dp 1 − k R 2 dp R 2 dp 1 − k u= − ln r − + ln ( R ) 4 dx 4 dx ln (1/ k ) 4 dx 4 dx ln (1/ k )

Further simplification of this expression gives

u (r ) = −

2 R 2 dp   r  1− k 2  r  1 − + ln       4 dx   R  ln (1/ k )  R  

(Ans. a)

The volume flow rate is given by R

R

kR

kR

Q =  udR = u 2 r dr = 2 ur dr Using the previously obtained expression for u(r), we obtain  R 2 dp  R  r3 1− k 2  r  Q = 2  − r − + r ln    dr  2 ln (1/ k )  R    4 dx  kR  R

 R 4 dp 1  r   r  1 − k 2  r   r   r  − +       ln    d   2 dx k  R   R  ln (1/ k )  R   R    R  3

Q=−

1

 R 4 dp  1  r  1  r  1 − k 2  r   1  r  1   Q=− ln   −     −   +   2 dx  2  R  4  R  ln (1/ k )  R   2  R  4   k 2

4

2

Further simplification of this expression gives

(1 − k )   R 4 dp  Q=− 1− k 4 − 8 dx  ln (1/ k )  



2

2



(Ans. a)



(b) The condition for maximum velocity is

 =

du =0 dr

Using the expression of u from part (a), we obtain 0=−

R 2 dp  2r 1 − k 2  1  − 2 +   4 dx  R ln (1/ k )  r  

The maximum velocity is located at α = r / R. Therefore,

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1/2

1  2 R 1 − k 2  1   1− k 2   or,  = 0=− 2 +      R ln (1/ k )   R   2   ln (1/ k )  

(Ans. b)

(c) The average velocity in the annulus is given by

u=

Q A

The area is

r r A =  2 r dr = 2 R 2    d   =  R 2 1 − k 2 R R k 1

(

)

Therefore,

(1 − k )   R 4 dp  − 1− k 4 − 8 dx  ln (1/ k )  

Q u= = A

2

  R2 1 − k 2

(

)

2



(

)

2 2    = − R dp 1 + k 2 − 1 − k  8 dx  ln (1/ k )   

(Ans. c)

(d) Considering the limiting case of k → 0, we have

Q=u =−

 R 4 dp 8 dx

Thus, the results are congruent with the Poiseuille flow in a circular pipe.

(Ans. d)

3-59 Consider the problem associated with the spinning porous tube depicted in Fig. 3-28 of Sec. 3-6.4. By substituting the similarity transformations given by Eq. (3-182) into Eq. (3-181), derive Eq. (3-183). Subsequently, recognizing that υθ = g(r) is a function of a single variable, derive Eq. 1 (3-184). Finally, introduce  = r 2 into Eq. (3-184) and differentiate once to obtain Eq. (32 185). Verify that the four boundary conditions for this problem translate into

1 1. f   = 1 from ur(1, z) = −1 (uniform radial injection at the wall) 2 1 2. f    = 0 = 0 from uz(1, z) = 0 (zero axial speed at the injecting wall) 2 3. f(0) = 0 from ur(0, z) = 0 (no radial crossflow at the centerline) 4.

2 f  ( 0 ) = 0 from ∂uz(0, z)/∂r = 0 (symmetric axial velocity at the centerline).

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The spinning angular rate is ωs, and the tangential speed at the wall is vs. For a constant radial injection speed at the porous wall, we have

vr ( r = ro ) = −v  0 ro = Radius of the tube r=

r* z* v* ; z = ; vr = ro ro v

v =

v* v* P* ; vz = z ; P = 2 v v v

For injection, we have

Re =

v ro 0 v

The steady, incompressible, and normalized Navier–Stokes equation simplifies to 1  ( rvr ) vz + =0 r r z

vr

vr v v 2 P 1   2vr 1 vr  2vr vr  + vz r −  = − + + + −   r z r r Re  r 2 r z r 2 r 2 

vr

v v v v 1   2v 1 v  2v v  + vz  + r  = + + −   r z r Re  r 2 r r z 2 r 2 

vz vz P 1   2vz 1 vz  2vz  vr + vz =− + + +   r z z Re  r 2 r r z 2     The axisymmetric conditions  = 0  have been enforced.    A stream function that varies linearly in the flow direction is introduced.

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 ( r, z ) = zf ( r ) vr = − vz =

f r

zf  r

The pressure can be eliminated by cross-differentiation.

z

 1  1 1  1    2 v 2  2  ff  − ff  − f  +  rf  − f  + f     = v r  r  r Re  r   r z

Also, v = g ( r ) .

ff  −

1 1  1  2 ff  − f 2 +  rf  − f  + f   = r C r Re  r 

(1)

We know that

=

r2 2

Differentiating equation (1) with respect to η, we obtain

 d 3 f df d 2 f  2  d 4 f d3 f  − +2 3 =0 f +  3 d d 2  Re  d 4 d   d f(η) is subject to the following boundary conditions:

1 1. f   = 1 from ur(1, z) = −1 (uniform radial injection at the wall) 2 1 2. f    = 0 = 0 from uz(1, z) = 0 (zero axial speed at the injecting wall) 2 3. f(0) = 0 from ur(0, z) = 0 (no radial crossflow at the centerline) 4.

2 f  ( 0 ) = 0 from ∂uz(0, z)/∂r = 0 (symmetric axial velocity at the centerline).

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CHAPTER 4. LAMINAR BOUNDARY LAYERS

Repeat the momentum-integral analysis of the flat plate in Sec. 4-1 for Pohlhausen’s

4-1

3

u 3 y  1 y  =   −   , where δ(x) is the boundary-layer or disturbance U 2  2  thickness as pictured in Fig. P4-1. Is this profile any more (or less) realistic than the approximation of Eq. (4-11)? For the above profile, compute (a) ( / x ) Re x , (b) ( * / x ) Re x

cubic polynomial profile

, (c) ( / x ) Re x , (d ) ( C f / x ) Re x , and (e) ( CD / x ) Re x .

Solution: We know that momentum thickness is approximated by 

1

(

)(

)

u u 39 y 3 3 ,  = 1 −  dy    1.5 − 0.5 1 − 1.5 + 0.5 d = U U 280  0 0

=

In a similar manner, we can find that * =   (1 − u/U ) dy  3/8. [Note H =  */  = 2.69 ] Substitution into the integral relation (4-10) gives Cf =

or:

2w U

2

=

2 (1.5 U/ ) U

2

=2

d d  39  =2  , dx dx  280 

 d = 140  dx/ (13  U )

Integrating, assuming  = 0 at x = 0, yields

2 

280  x , 13  U

or

 4.64  x Re x

(Ans. c)

The other two thicknesses follow from  = .1393  and * = 0.375 :  Re x = 0.646 x

* Re x  1.740 x

(Ans. a, b)

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The errors, compared to the Blasius solution, are (–2.7%) and (+1.1%), respectively. The skin friction follows by substituting  ( x ) into the approximation w = 1.5U/. Thus

Cf Re x  0.646

CD ReL  1.293

(Ans. d, e)

The error is (−2.7%) for each. This is a better accuracy than the 10% errors of the parabolic profile (see the Table on page 221), probably because the cubic profile is more realistic. 4.2 Repeat the integral flat-plate heat-transfer analysis of Sect. 4-1.7 by using the parabolic velocity from Eq. (4-11) and the quartic temperature profile

 2y 2y3 y 4  T − Te = ( Tw − Te ) 1 − + 3 − 4   T T  T  This is a realistic temperature profile, satisfying the five boundary conditions T ( 0) = Tw ,  2T/y 2 ( 0 ) = 0, T (  ) = Te , T/y (  ) = 0, and  2T/y 2 (  ) = 0. Substitution into the integral

energy equation, Eq. (4-21), gives   2k ( Tw − Te ) d  r   cp U(Tw − Te )(2 −  22 )(1 − 2 + 23 − 4 )T d qw = =  dx  0    2 2  3   d   = −  , where  = T  1 cp U ( Tw − Te )    dx    15 42  

Now assume that  = constant (flat plate with heating starting at the leading edge) and substitute

 ( d/dx )  15v/U from page 220 (the integral momentum analysis). After cancellation of T and U, we obtain 3 −

5 4 1  = , 28 Pr

or:

=

T = Pr −1/3 

(Ans.)

where we neglect the quartic term in the last approximation. This is the same as Eq. (4-26), which uses a parabolic temperature profile. The heat transfer estimate is Nu x =

qw x x  2kT  2 2 = Re1/ Pr1/ 3   x k ( Tw − Te ) kT     4.64

(Ans.)

This is similar to Eq. (4-27), except the constant ( 2/4.64) = 0.431 is about 30% higher than the accurate estimate of 0.332 of Pohlhausen (1921). This is probably because we have mixed a crude (parabolic) velocity with an accurate (quartic) temperature profile. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -2-

4-3

Repeat the momentum analysis of C f from Sect. 4-1.4 with the better profile

 y  u = U sin    2 

(1)

Whereas the parabolic profile from Eq. (4-11) only satisfies three conditions, u ( 0) = 0, u (  ) = U, and u/y (  ) = 0, the sine-wave profile (1) above also satisfies the zero-pressure-gradient condition  2 u/y 2 ( 0 ) = 0 appropriate for flat-plate flow - as was also true for the cubic profile in Prob. 4-1 on p. 86 of this Manual. The wall shear estimate is

w = 

 U u | y =0  y 2

Similarly, the momentum thickness estimate is 1

4−        =  sin    1 − sin     d =  = 0.1366  2 2 2        0 Then the momentum integral relation yields

Cf =

2 ( U/2 ) U 2

d d =2 = 2 ( 0.1366 ) , dx dx

1/ 2

or:

    Re x =   x  0.1366 

= 4.80

The accuracy is good. Putting  ( x ) back into the C f estimate gives Cf Re x =

/2 = 0.3276 4.80

(Ans.)

The error is only (–1.4%), but we can’t usually expect such accuracy in integral theories.

4-4 Air at 20°C and 1 atm flows past a smooth flat plate as in Fig. P4-4. A pitot stagnation tube, placed 2 mm from the wall, develops a water manometer head h = 21 mm. Use this information with the Blasius solution, Table 4-1, to estimate the position x of the pitot tube. Check to see if the flow is laminar.

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Fig. P4-4 Solution: For air at 20°C and 1 atm, take  = 1.205 kg/m3 and  = 1.81E-5 kg/(m-s). Estimate the velocity u at y = 2 mm from the pitot stagnation tube formula:

p = ( water − air ) gh = (998 − 1.205 kg/m3 )(9.81 m/s 2 )(0.021 m) = 205 Pa u = 2p/air = 2(205 Pa)/(1.205 kg/m3 ) = 18.46 m/s Since u  U = 20 m/s, the pitot tube is inside the boundary layer. The velocity ratio is

u /U = 18.46/20 = 0.923 = f  ( ) We can interpolate in Table 4-1 of the text to find that this ratio corresponds approximately to   2.55. Everything is known in the definition of  except the position x:

 = 2.55 = y

(1.205)( 20 ) U = ( 0.002m ) 2 x 2 (1.81E − 5) x Solve for x  0.41m = 41cm

Ans.

Check to see if the flow is laminar, so that the Blasius flat-plate theory holds:

Re x =

Ux (1.205)( 20 )( 0.41) =  545, 000  1.81E − 5

For a smooth plate and a quiet freestream, transition to turbulence occurs for Re x  106. Thus we conclude that the flow at x = 41 cm is indeed laminar.

4-5

Repeat the unheated-starting-length analysis on p. 224 of the text by keeping the term

(  /5 ) . Solve numerically in the range 1  x  5 for Pr = 1 and x 3

o

= 1.

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The relevant differential equation is on p.

and, for Pr = 1 and 2 = 30vx/U, becomes

 2 6 3  d  3  3  12 = 0.8, with  = 0 at x = x o  4 −  +   −  = 5  dx  5  15  This is first-order and nonlinear and easily solved numerically by, e.g., the Runge-Kutta routines of Appendix C. If we neglect the second terms in each parentheses, we obtain the simple text −1/3

3/4 solution  = 1 − ( x o /x )  . The numerical and approximate solutions are shown below and   are seen to be in good agreement - which would be better if Pr > 1.

4-6 Develop an iterative solution of the Blasius equation (4-45) which will start with a reasonable guess, say, f  ( 0) = 0.3, and converge to the exact solution, f  ( 0) = 0.4696.

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Solution: The Blasius equation is beautifully behaved, always yielding a constant value of f *( ) at large  , so interpolation is quite easy. Note, for example, the following results:

f  ( 0) =

0.30

0.35

0.40

0.45

0.50

f  ( ) =

0.74176

0.82204

0.85857

0.97198

1.042705

One can almost interpolate linearly for the proper value f  ( 0) = 0.4696 to give f  (  ) = 1.0.

4-7 Consider a long flat plate emerging from a wall at velocity U, as in Fig. P4-7. There is no freestream. Show that the Blasius Eq. (4-45) holds for this ease, with f ( 0 ) = 0,

f  ( 0) = 1, and f  (  ) = 0. Solve the equation numerically and show that Cf  0.444/Re1/x 2 . Also

evaluate v (  ) and discuss. [Hint: Note that f  ( 0 ) is negative.]

Fig. P4-7 Solution: This is a moving flat plate, with zero pressure gradient and a boundary layer that grows starting at x = 0. Therefore the Blasius equation is valid with boundary conditions reversed on the streamwise ( u ) velocities:

f  + f f  = 0

with

f ( 0) = 0, f  ( 0) = 1, f  ( ) = 0

The correct initial condition which matches these conditions is f  ( 0)  −0.6276. The profiles of

f ( ) , f  ( ) , and f  ( ) are calculated (from Subroutine RUNGE or other routine) and are plotted on the next page. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -6-

Stream function ( f ), velocity ( f ) and shear stress ( f ) near a moving flat plate. The wall shear stress is similar in form to the fixed-plate, moving-stream relation, Eq. (4-52):

w = 

Uf  ( 0 ) u | y =0 = ; y 2vx /U

Cf =

2 |w |

U

2

=

0.6276/ 2 0.4438 = Ux / Re x

Ans.

The vertical velocity at the outer edge of the boundary layer is

v () =

vU vU vU ( f  − f ) | → = ( −1.142 ) = −0.808 2x 2x x

Ans.

The negative sign means that the outer (still) fluid is entraining fluid into the boundary layer.

4-8 Develop a numerical solution to iterate the Falkner-Skan equation (4-71) for any value of   0. Start with a reasonable but incorrect value of f  ( 0 ) . Solution: Not as well behaved as Blasius, but still OK. We illustrate for  = +0.3, “” = 5:

f  ( 0) =

0.6

0.7

0.8

0.9

1.0

f  ( ) =

0.27076

0.70748

1.09402

1.44830

1.77985

We may interpolate to f  ( 0) = 0.77476 for this value  = 0.3, as given in Table 4-2 of the text.

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4-9

The Blasius Eq. (4-45) must be iterated to find the value of f ( 0) which cause f  (  ) to

equal 1.0. Panton (1996) suggests the following scheme to avoid iteration: Define

f ( ) =  F ( ) ,

where  is a constant

(a) Show that the function F also satisfies the Blasius equation. (b) If we arbitrarily set an initial condition F ( 0) = 1.0, explain how  can immediately be found without iteration. (c) Even with

 found, explain why the solution for F is still awkward for filling out Table 4-1. (d) If an integration (not required of you) then yields F  (  ) = 1.6552, what is the proper value of  ? (e) Show that the value of  from part (d) leads to the result f  ( 0) = 0.4696. Solution: (a) Introducing f  =  2 F  ( ) etc., we find indeed that F  + FF  = 0.

Ans. (a)

(b) From the transformation, f  (  ) =  2 F  (  ) = 1, compute  = 1/F  (  ) 

Ans. (b)

1/2

.

(c) The numerical solution, in even increments of , will not be spaced nicely in .

Ans. (c)

(d) Given the writer’s result F  (  ) = 1.6552, compute  = 1/1.6552

Ans. (d)

1/2

= 0.7773.

(e) With  known, f  ( 0) =  3 F  ( 0) = ( 0.7773) (1.0) = 0.4696 3

4-10

Ans. (e)

Show that Clauser’s parameter, ( */w ) ( dp/dx ) , is a constant for the Falkner-Skan flows.

What is its value for the separating-flow condition? Knowing that, for Falkner-Skan flow,  = y ( m + 1) U/2vx 

1/2

, and f  ( ) = u/U, we may

compute 

1/2

 2 x  u  * =  1 −  dy =   U  ( m + 1) U  0

1

* =  (1 − f  ) d

*,

0

 ( m + 1) U  u |y = 0 =  U f  ( 0 )   y  2 x 

1/2

w = 

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Then, since U = Kx m , the pressure gradient is

dp dU mU2 2 2m−1 = − U = − m K x =− dx dx x

[Bernoulli relation]

Combining these three, we obtain the Clauser parameter for Falkner-Skan flow: Clauser Parameter =

* dp 2m * =− w dx ( m + 1) f  ( 0 )

(Ans.)

These values are independent of x, hence the Clauser parameter is a constant for a given wedge flow. We may take , etc., from Table 4-2, page 243. For separating flow,  = −0.19884, m = −0.09043, since f  ( 0 ) = 0, the Clauser parameter = +.

(Ans.)

When the (different) similarity variable  = y ( IUI/vx ) equation reduces to

1/2

4-11

f  +

(

)

m +1 f f + m f 2 − 1 = 0 2

is used, the Falkner-Skan

(1)

subject to the boundary conditions of Eq. (4-72): f ( 0) = f  ( 0) = 0 and f  (  ) = 1. Find the solution for the special case flow U = −K/x = ( −K ) x −1. This case is for m = −1, so the differential equation becomes 1/2

f − f  + 1 = 0, 2

 K/x  with  = y    x 

=

y x

( K/ )

But this is exactly Eq. (4-88) for point-sink flow, and the solution is given by Eq. (4-89):

f=

u    = 3 tanh 2  + 1.146  − 2 U  2 

(Ans.)

Other details of this solution are given on p. 250 of the text.

4-12 The thin equilateral triangle below is subjected to airflow at 20°C and 1 atm. Estimate the (laminar) plate drag in newtons. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -9-

Assume that the local wall shear stress varies only with distance x from the leading edge, from laminar flat-plate theory. Then the strip dA in the figure would have a uniform shear stress given by Eq. (4-52), with f  ( 0) = 0.4696: 1/2

   w = 0.332    x 

U3/2

For a triangle of side length a, the strip has an area dA = Ldx (1 − x/L ) , where L = a sin60° and a = 2m. Then the total shear force on the plate is the integral of local strip force:

F =  w dA ( 2 sides ) = 2 ( 0.332 )(  )

1/2

L

U3/2 =  0

(

= 0.664 U3 = 0.885 (  )

)

1/2 

1/2

1/2

 2Lx 

−

L−x x

dx

2 3/2  L x  |0 3 

( UL)3/2

(Ans.)

For our particular case, U = 12 m/s, L = 2sin 60 = 1.732 m, and for air  = 1.2 kg/m3 and  = 1.8E-5 kg/m  s. The numerical value of the laminar drag is thus F = 0.885 (1.2 )(1.8E-5 ) 

1/2

(12 )(1.732 ) 

3/2

= 0.39 N

(Ans.)

We should check the Reynolds number to see if the flow is truly laminar: ReL = UL/ =

(1.2)(12)(1.732) / (1.8E-5)  462,000 (OK ). Transition should occur at

ReL  1E6.

4-13 Model a flow straightener as an array of thin-walled square ducts as in the figure below. Estimate the pressure drop across an N  N bundle of such ducts, using laminar flat-plate boundary-layer theory. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -10-

Each square cell has four surfaces, and each ( L  a ) surface experiences a shear force given by laminar boundary-layer theory, Eq. (4-53):

CD =

1.328 , ReL

or

F = 0.664 ( L )

1/2

for one ( L  a ) surface

U3/2a

An N  N array of boxes has N 2 cells of four surfaces each, or a total shear force

Ftotal array = 2.656 N2 ( L )

1/2

U3/2a

A freebody around the entire array shows that this shear force has to be balanced by a pressure drop across the array, multiplied by the array frontal area: parray =

Ftotal

( Na )

2

=

2.656 ( L )1/2 U3/2 a

(Ans.)

The result is independent of the array size N. This pressure drop is modest unless the cell width a is very small. For example, for air flow at U = 10 m/s and a cell size a = 1 cm and L = 10 cm, we may estimate p  12 Pa.

4-14 Develop a numerical solution for laminar flat-plate flow with wall suction or blowing and compare your results with Fig. 4-15.

We are to solve the Blasius Eq. (4-45), f  + f f  = 0, subject to the blowing/suction conditions of Eqs. (4-84,85):

f  ( 0) = 0

f () = 1

f ( 0 ) = − 2v*w  0

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The required FORTRAN statements are the same as for the Blasius problem (bottom of p. 233 of the text). The numerical problem is to find the unknown initial condition f  ( 0) = Y1 ( x = 0) which causes the farfield velocity profile to approach f  (  ) = 1. For a particular example, let us take

blowing, v*w = 0.4, or f ( 0 ) = −0.4 2 = −0.5657. From Fig. 4-15(a) we see that “infinity” extends out to about “ = 8”. The initial slope f  ( 0 ) is clearly much smaller than the Blasius (no blowing) value of 0.4696. Let us guess an array of initial values and tabulate f  (  ) :

f  ( 0) =

0.05

0.10

0.15

0.20

Y2 (8) = f  (  ) =

0.7618

0.9531

1.1077

1.2423

The step size was  = 0.05, adequate for 5-digit accuracy. The solutions for f  (  ) are well behaved and nearly equally spaced. Linear interpolation indicates f ( 0) = 0.1152 will cause

f  (  ) = 1.0. A slightly more accurate value is

v*w = ( v w /U ) Re x = 0.4 :

Blowing:

f  ( 0 ) = 0.1143

(Ans.)

With this initial value, the velocity profile f  ( ) is exactly as shown in Fig. 4-15(a) for v*w = 0.4. To check this, note that, for Pr = 1, from Eq. (4-60), the local heat transfer should be

Nu x / Re x = f  ( 0 ) / 2  0.0810. This checks well with Fig. 4-15(b).

4-15 Derive a relation for local skin friction versus Reynolds number for point-sink flow. Fig. 4-16, p. 250 of the text. Compare with the Falkner-Skan results if possible. This is a favorable gradient, U = −K/x, and the velocity profile solution is Eq. (4-89):

   u = f  ( ) = 3 tanh 2  + 1.146  − 2 U  2  Differentiating this, we find that the wall shear is given by

W = 

u y

y =0

=

U K f  ( 0) , x v

where f  ( 0 ) =

6 tanh (1.146 ) sech 2 (1.146 ) 2

= 2/ 3 = 1.1547 Then the skin friction coefficient is

Cf =

2w U

2

= 2.3094 ( v/K )

1/ 2

=

2.3094 Re x

(Ans.)

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where the last relation follows from the substitution K = |U|x. It is difficult to compare this result with the Falkner-Skan solutions because we cannot easily take the limit, as m approaches (–1), of, say, the listed values in Table 4-2, p. 243. The result does compare well with a comparably strong favorable gradient, stagnation flow, Eq. (3-160), p. 158.

4-16 Develop a numerical solution for a laminar mixing layer between parallel streams and compare with Fig. 4-17 for any value of k = ( 22 ) / ( 11 ) . We must simultaneously solve two Blasius profiles and match them at  = 0:

  0: f1 + f1f1 = 0;

  0: f 2 + f 2f 2 = 0

At  = 0:

f1 = f 2 = 0;

f1 = f 2  0;

At  = :

f1 = 1.0;

At  = −: f 2 = 0

f1 = f 2 k (Assume)

For simplicity, we take the lower fluid to be at rest far below the interface ( U2 = 0 ) . As a sample case, we take k = 10, which is plotted in Fig. 4-17(c). The numerical solution of this problem is an order of magnitude more difficult than the standard Blasius profile. We must match the (unknown) interface velocity and the (unknown) interface shear stress, at the same time ensuring that both the upper and lower velocity profiles approach their asymptotic values of 1.0 and 0.0, respectively. And we must integrate one profile to ( + ) and the other to ( − ) . For k = 10, Fig. 4-17(c) indicates that the interface velocity f  ( 0)  0.35, which we may take as a first guess. Now guess a value of f 2 ( 0 ) and, using f1 ( 0) = f 2 ( 0) 10, integrate the upper profile and see if the final velocity f1 (  ) = 1.0? If not, try another value of f 2 ( 0 ) . When the upper velocity finally does approach 1.0, then integrate the lower profile to see

if it approaches f1 ( − ) = 0.0? It probably will not because our initial value for f  ( 0 ) was only a

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guess. Change this guess and try again. Take “infinity” to correspond to a value  = 10, which is well off the chart in Fig. 4-17(c). Tabulate some results as follows: f  ( 0 ) = 0.35: f1 ( +10 ) = 1.0 at f 2 ( 0 ) = 0.12503, but

f 2 ( −10 ) = +0.0199  0

f  ( 0 ) = 0.34: f1 ( +10 ) = 1.0 at f 2 ( 0 ) = 0.12623, but

f 2 ( −10 ) = −0.0067  0

Our first guess for f  ( 0 ) was very good, so we didn’t have to iterate much. Clearly we are close to the correct answer, and we may interpolate linearly to obtain k = 10: f  ( 0)  0.3425,

f 2 ( 0) = 0.12593,

f1 (0 ) = 0.39823

(Ans.)

The velocity profiles are in close agreement with those plotted in Fig. 4-17(c).

4-17

Air at 20°C and 1 atm issues from a slot and forms a plane laminar jet. At x = 50 cm,

u max = 20 cm/s. Estimate the jet (a) width; (b) mass flow; and (c) Reynolds number.

For air take  = 1.2 kg/m3 and  = 1.8E-5 kg/m  s. Knowledge of u max and its position allows us to compute the jet momentum J from Eq. (4-104): 1/3

u max

 J2  = 0.2 m/s = 0.4543   x   

1/3

  J2 = 0.4543    (1.2 ) j (1.8E-5 )( 0.5 ) 

,

J = 0.00096 kg/s 2

or:

Then the jet width estimate follows from Eq. (4-106): 1/3

 x 2 2  b = 21.8   J    

1/3

 ( 0.5)2 (1.8E-5 )2   = 21.8   ( 0.00096 )(1.2 ) 

= 0.090 m

(Ans. a)

= 0.0072 kg/s/m

(Ans. b)

The jet mass flow per unit depth is given by Eq. (4-107): m = ( 36Jx )

1/3

= 36 ( 0.00096 )(1.2 )(1.8E-5 )( 0.5 ) 

1/3

The jet Reynolds number could be expressed in three essentially equivalent ways: Re jet

m = = 400; 

1/3

 Jx  or:  2   

= 120;

or:

u max b = 1200 

(Ans. c)

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4-18

Air at 20°C and 1 atm flows at 1 m/s past a slender two-dimensional body of length L = 30 cm and CD = 0.05 based on ‘plan’ area (bL). At 3 m downstream of the trailing edge, estimate (a) wake velocity defect; (b) wake thickness; and (c) Reynolds number. For air take  = 1.2 kg/m3 and  = 1.8E-5 kg/m  s. The body-length Reynolds number is

ReL = UL/ = (1.2)(1.0)( 0.3) / (1.8E-5) = 20,000. For x = 3 m, the centerline wake defect velocity is computed by applying Eq. (4-112) at y = 0: 1/2 1/2 1/2 1/2 u1 ( y = 0 )  ReL   L   20000   0.3  = CD  = 0.05 ( )     = 0.315,    U  16   3.0   16   x 

or:

u1 ( y = 0) = u max  0.32 m/s

(Ans. a)

[This is perhaps pushing wake theory a bit, since u is not truly small compared to U.] By analogy with jet theory, the wake half-thickness could be defined as the point where the defect velocity drops to 1% of its maximum velocity, from the gaussian profile:  Uy 2  exp  − = 0.01,  4xv   

or:

 4.605 ( 4xv )  =  U  

1/2

or:

y1% = 0.0144 m,

y1%

 4.605 ( 4 )( 3.0 )(1.8E-5/1.2 )  =  1.0  

Width b = 2y1%  0.029 m

1/2

(Ans. b)

Finally, the wake Reynolds number could be expressed in terms of wake width and the maximum defect velocity:

Re wake 

 u max b (1.2 )( 0.32 )( 0.029 ) = = 600  1.8E-5

(Ans. c)

There are alternate Reynolds numbers, such as m/, similar to jet flow, Eq. (4-107).

4-19 Derive the steady-flow two-dimensional momentum-integral relation, Eq. (4-120 or 121), including wall suction or blowing, by applying conservation laws to the control volume shown.

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First apply conservation of mass, starting at the bottom left and working clockwise: 





0

0

0

  ( V n )dA = 0 = − ubdy − Ubd +  ubdy + d  ubdy − vw b dx CS

Rearranging this gives a net continuity relation to be used later as a substitution: 

U

d d = u dy − v w dx dx 0

In a similar manner, the x-momentum relation, again starting at the bottom left point, is 

dp    Fx = pb +  p + 2  b d − ( p + dp )(  + d ) − w b dx = −  u 2b dy − U 2b d 0 



+  u b dy + d  u 2 b dy − v w ( 0 ) b dx 2

0

0

where the last (vanishing) term is included only to show that there is zero x-momentum coming in at the wall because u = 0 there. Clean up and drop second order terms dpd, etc: 

dp d d w = − −  u 2dy +  U 2 dx dx 0 dx

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Now eliminate ( d/dx ) using the continuity relation above: 

w = −



d d dp u 2dy + U  u dy − Uv w −   dx 0 dx 0 dx

This is one common form of the momentum-integral relation [see, e.g., Kays and Crawford (1980), p. 48.] To put this equation in the form of Eq. (4-121), we use the Bernoulli relation and the definitions of momentum and displacement thicknesses: 

dp dU = −U ; dx dx

 u dy = U (  − *) ;



2 2  u dy = U  u dy − U 



0

0

0

After these three substitutions, we obtain Eq. (4-121) for steady flow (  /t = 0):

w U

4-20

2

=

d 1 dU v w + ( 2 + *) − dx U dx U

(Ans.)

Solve Eq. (4-121) for flat-plate flow with nonzero blowing, v w  0, with u/U given by

Eq. (4-11) and both  and 1/v w proportional to x1/2 . Compare to Fig.4-15(b). The velocity profile is a simple (and unrealistic for large blowing) parabola

u y y2 =2 − 2 , U  

with



2  15

and

w =

2U 

For flat-plate flow with dU/dx = 0, substituting these formulas into Eq. (4-121) gives 2v 2 d v w = − U 15 dx U

Now assume (correctly) a square-root relationship, to which the blowing parameter v*w in Fig. 4-15(b) can be related: Let

 = B x1/ 2

and

v w = K x −1/ 2 ,

whence v*w =

vw U

Re x =

K

( vU )

(1)

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Substitution into the momentum integral relation above gives a quadratic equation in B: B2 −

15K 30v B− =0 U U

Note that the no-blowing ( K = 0) solution is Bo2 = 30/U, or o  ( 30x/U ) , as in Eq. (4-14). Then the solution to the quadratic equation above can be written in the form 1/2

(

 B = Bo  1 + 1.875 v*2w 

)

1/2

+ (1.875)

1/2

 v*w  , 

where we have substituted for K = v*w

Bo = ( 30 v/U )

1/2

where

( U )

|noblowing

from Eq. (1) above. Since w = 2U/ and

 = Bx1/2 , the skin friction on the plate may be written in the following final form: Cf =

Cfo

(

)

 1 + 1.875v*2 + (1.875 )1/2 v*  w w 

Using the Reynolds analogy for

Pr = 1,

,

Cfo =

0.73 Re x

(Ans.)

this may also be written in the form

Nu x / Re x = Cf /2 = 0.365/  , where   denotes the bracketed factor above. The agreement with Fig. 4.15(c) for suction is excellent, as shown by the following calculated values: v*w =

Nu x /Re x =

0.0

–1.0

–2.0

–3.0

–4.0

–5.0

0.365

1.12

2.06

3.04

4.03

5.02

Note the good agreement with the suction values in Fig. 4-15. The reason is that the u ( y ) profiles remain approximately parabolic when suction is applied [see Fig. 4-15(a)]. For blowing, however, the agreement of the approximate theory is more qualitative: v*w = Nu x /Re x =

0.0

0.1

0.2

0.4

0.6

0.8

0.365

0.318

0.278

0.216

0.173

0.142

Blow-off is not predicted, and the friction/heat-transfer estimates are too high. This is because the blowing profiles become S-shaped and the parabolic formula is not accurate.

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4-21

Improve Prob. 4-20 by developing a parametric polynomial profile u ( y ) which accounts

for blowing and suction similar to Fig. 4-15(a). Do not solve or integrate for Cf . We need profiles which develop the S-shape when there is wall blowing [we did OK with the simple parabola in Prob. 4-20 for suction cases]. At a minimum, we add one more boundary condition at the wall, obtained from the x-momentum boundary-layer equation: non-zero curvature when there is blowing or suction: At y = 0, u = 0, but v w  0, thus: v w

u  2u |wall =  2 |wall y y

(1)

We therefore try a cubic polynomial for the velocity profile: u = a  + b 2 + c 3 , U

 = y/

This already satisfies the no-slip condition. We further require that it merge smoothly with the freestream, plus condition (1) above: At y = :

u = U, or:

a + b + c = 1

At y = :

u/y = 0, or:

a + 2b + 3c = 0

At y = 0, Condition (1) above:

Re*b = 2c,

Re* = v w /

The parameter Re* is directly related to the “blowing parameter” v*w from Fig. 4-15:

Re* =

vw  vw   Const = Re x = ( Const ) v*w if =  U x x Re x

Solving, we obtain a = 6/ ( 4 + Re*) , b = 3Re*/ ( 4 + Re*) , and c = −2 (1 + Re*) / ( 4 + Re*) . The desired polynomial profile is thus listed and plotted as follows: u 6 3Re* 2 2 + 2Re* 3  +  −  U 4 + Re* 4 + Re* 4 + Re*

(Ans.)

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We see below that the profiles are realistic for Re*  −2.0, but they do not “blow off”.

4-22

Modify Prob. 4-20 by using the same parabolic profile, Eq. (4-11), but let the wall velocity v w  0 (suction) be constant. Solve for Cf ( x ) and  ( x ) and compare with Fig. 4.21 and Eq. (2-123) of the text. This problem begins the same way as Prob. 4-20 on p. 79 of this Manual. The parabolic velocity profile leads to a differential equation for  ( x ) : 2 2 d v w = − U 15 dx U

The difference here is that we assume constant suction velocity, v w  0, not a square-root variation. We may solve for  ( x ) by separating the variables: 

x

 d 15  ( 2 /U ) + ( vw /U )  =  2 dx 0 0 where we have assumed that  = 0 at x = 0. Both integrals are easy to evaluate, and the result may be rearranged as follows:

−

2  v w   15xv w ln 1 + = v w  2  2U

If we rewrite this in terms of the variables plotted in Fig. 4-21, we have a better result: Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -20-

( −vw )  − ln 1 − ( −vw )   = 15 ,  

2v

2v

 

4

where

=

( −vw ) U

(Ans.)

Re x

Meanwhile, the skin-friction estimate comes from differentiation of the parabolic profile:

w =

2U 4 4 , hence Cf  = CQ ,  U / ( − v w )  /

CQ =

−vw U

(Ans.)

Thus, with ( −vw ) / known as a function of  from the first answer, C f may be computed and is directly related to the “suction coefficient” ( −v w ) /U. Some numerical values from these integral estimates may be tabulated as follows:

( −vw ) / =

0.1

0.3

0.5

1.0

1.5

1.8

1.9

2.0

=

0.027

0.083

0.143

0.318

0.570

0.854

1.052



Cf (U/ − v w ) =

40.0

13.33

8.00

4.00

2.67

2.22

2.11

2.0

The thicknesses are a little small, e.g., at

  0.6, ( −v w ) /  1.54 whereas Fig. 4.21 indicates

a value of about 2.0. At z = , ( −vw ) / = 2.0, whereas Eq. (2-232) for ‘asymptotic’ (exponential profile) suction indicates a value of 4.6. The skin friction estimate is realistic and, as expected, smaller than the CD values plotted in Fig. 4-21. For example, ReL = 106 and (− v w )/U = 0.001 are equivalent  = 1.0, for which C f from the tabulation above would equal about 0.0022, whereas CD from Fig. 4-21 is about 0.0025, which is quite reasonable agreement.

4-23 Apply the method of Thwaites (Sect. 4-6.6) to laminar boundary flow on a circular cylinder. Compute local wall friction and compare with Fig. 4-24(b). We select the inviscid-flow theory for freestream velocity: U(x) = 2Uo sin(x/a), where a is cylinder radius and x starts at the front stagnation point. [See Fig. 4.24(a) for the geometry.]

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The momentum thickness from Thwaites’ method is given by Eq. (4-138):

 = 2

x

0.45  U

U = 2Uo sin ( x/a )

 U dx, 5

6

0

Thwaites’ factor  equals (2 /) ( dU/dx ) , where dU/dx = ( 2Uo /a ) cos ( x/a ) . After substitution into the momentum-thickness relation above, we obtain, denoting  = x/a,

=

0.45 cos  sin  6



 sin

5

 d, where

0

 = 15 8 − cos  (3sin 1

4

)

 + 4sin 2  + 8  

Then, with  known, the wall shear stress is given by Eq. (4-139):

w 

Ux) S() , (x)

Finally, we are asked to plot ( Cf /2)

where S = (  + 0.09 )

0.62

( )

( Ux/ ) ,

where Cf /2 =  w / U 2 . Putting all this together

from the above formulas results in the following desired expression: 

1/2

  sin 5   1 Cf Re x = S (  )   2  0.45 F 

,

where

F =  sin 5  d

(Ans.)

0

where  = ( x/a ) is the local angle in radians. Some numerical values from this formula may be listed as follows for various positions along the cylinder: =

0°

30°

45°

60°

75°

90°

95°

100°

103.1°

=

0.075

0.072

0.068

0.059

0.041

0.0

–0.025

–0.060

–0.090

0.5Cf Re =

1.195

1.148

1.084

0.984

0.830

0.575

0.444

0.256

0.0

The skin friction values are very close to Terrill’s digital computer results plotted in Figure 4-24(b). Thwaites predicts separation at  = 103.1, quite close to Terrill’s value, 104.5°. (b) We could also calculate using the measured velocity distribution of Eq. (4-144): 3 5 U  U  1.814 ( x /a ) − 0.271( x /a ) − 0.0471( x /a )    2 4 dU /dx  (U  /a ) 1.814 − 0.813 ( x /a ) − 0.2355 ( x /a )   

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Carrying out Thwaites’ integral (4-138) for ( 2 / ), we can find Thwaites’ parameter  ( x ) :

 ( x) =

 2 dU v dx

=

0.45

(U /U  )

6

d (U /U  ) d ( x /a )

x /a

5  (U /U  ) d ( x /a ) 0

The results are shown in the following plot. Separation is at x /a = 1.372 radians = 78.6°.

4-24 Apply Thwaites’ method to the test cases in Table 4-5 of the text. Have each student take a different case. Compute and plot Cf Re x and compute the separation point. The cases U = 1 − x and U = sin ( x ) have already been covered, but Table 4-5 still contains 9 other cases to try, with (presumably accurate) separation points listed by two different methods. We illustrate here with only one case: Tani (1949 ) : U = 1 − x 2 ;

U5 = 1 − 5x 2 + 10x 4 − 10x 6 + 5x 8 − x10

We have to integrate this for Thwaites method, Eq. (4-138), to obtain  and : 1/2 0.45vF) ( =

U3

and  = −

0.9xF U6

7 5x 3 5x 9 x11 5 10x where F = x − + 2x − + − 3 7 9 11

The separation point,  = −0.09, occurs at x = 0.268. The local skin friction is given by

Cf =

2 w U

2

,

where  w 

(Ans.)

[see Table 4-5]

U 0.62 S (  ) , and S = (  + 0.09 ) 

For the present case, this may be rearranged and evaluated as follows: Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -23-

Cf Re x =

(

2S 1 − x 2

)

3/2

( 0.45 G )1/2

,

where G = 1 −

5x 2 10x 6 5x 8 x10 F + 2x 4 − + − = 3 7 9 11 x

This wall friction distribution is plotted below and is very close to the exact (digital computer) solution for this problem.

4-25 Consider a two-dimensional flat-walled diffuser as in Fig. P4-25. Assume incompressible flow with a one-dimensional freestream velocity U ( x ) and entrance velocity U o ( x ) . The entrance height is W and the constant depth into the paper is b.

Fig. P4-25 Using Thwaites’ method, find an expression for the angle  at which separation will occur at x = L. What is the value of  if L = 1.5 W ? Solution: By one-dimensional continuity, the average velocity U ( x ) is given by Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -24-

U obU = U b (W + 2 x tan  ) ,

or:

U=

Uo 1 + ( 2 x tan  ) /W

Thwaites’ method requires us to compute the position where  ( x ) = −0.09, where x dU  2 dU  0.45 5  = = U dx   dx v dx  U 6 0 

with U ( x ) given by the expression above.

As it happens, this velocity U ( x ) is the same as Görtler’s third case in Table 4-5, (1 + x ) (“x” here is not the same quantity). Thus we can save ourselves a laborious calculation and use Görtler’s Thwaites-method result: −1

 = −0.09 ( separation)

if

2 x tan  0.079W  0.158, or xsep = Lsep  W tan  sep

If L = 1.5W , then tan sep = 0.079/1.5 = 0.0527,

or

sep  3.0.

Ans

Laminar flow has much less resistance to separation than turbulent flow.

4-26 Apply the explicit model, Sect. 4-7.1, to laminar boundary layer flow over a circular cylinder, with U ( x ) given by either Eq. (4-143) or (4-144). Compute and plot Cf . In the explicit model, we step forward directly to u m +1,n using Eq. (4-146), then compute

v m +1,n from Eq. (4-148). Step sizes are limited by Eqs. (4-147): x  u min y 2 / ( 2 )

y  2 /v max |

We can use either the inviscid or the actual freestream. U ( x ) - see p. of the text. Actual numbers for u, v are immaterial, and C f will scale, when completed, with Re x . For example, we could take Uo = 1, D = 1, and v = 1E-5, in arbitrary units, simulating a Reynolds number U o D/v = 1E5. The computed results will, if stability limits are satisfied, be very close to those plotted in Fig. 4-24(b), p. 270, for either potential flow or actual flow. The computed separation point will be somewhat inaccurate unless very small step sizes x are used as we approach separation.

4-27

Repeat Prob. 4-26 using the implicit finite-difference model of Section 4-7.2.

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Here we use Eq. (4-149) instead of (4-146) and solve simultaneously for u m +1,n either by iteration or the TDMA [Sect. 4-7.2.1], after which v m +1,n are again computed by Eq. (4-148). There are no stability limitations, but for numerical accuracy the step size y should be no larger than about 5% of the local boundary layer thickness, and the marching step x should be less than about 1° of arc, say, x* = 0.01. Computed skin friction will be in good agreement with Fig. 4-24(b) except possibly near separation, where smaller x* should be used as we approach this singular point.

4-28 Investigate the Crank-Nicolson (1947) finite-difference method as implemented by, e.g., Blottner (1970). What are its advantages and disadvantages in boundary layers? The Crank-Nicolson method splits the difference between an implicit and an explicit model for the (second-order) viscous term in the x-momentum equation:  2u y2

=

1 u m,n +1 − 2u m,n + u m,n −1 1 u m+1,n +1 − 2u m+1,n + u m+1,n −1 + 2 2 y2 y2

where the last term is implicit, involving velocities at the next station ( m + 1) . We thus have to solve simultaneously for these new values. However, since the scheme is only fifty-per-cent implicit, so to speak, the iteration for the new values of u m +1,n will be faster and the TDMA will not usually be necessary. The Crank-Nicolson model is unconditionally stable for linear systems, e.g. one- or two-dimensional linear diffusion. Eqs. (3-246) or (3-250), and it is generally quite stable for weakly nonlinear systems such as the laminar boundary layer equations. See Blottner (1970) for further details.

4-29 Apply the explicit finite-difference method of Sect. 4-7.1 to one of the laminar-flow test cases in Table 4-5. Compute CfRex and compare with Table 4-5 at separation. The explicit model works well for all these cases in Table 4-5, as long as the stability limits of Eqs. (4-147) are satisfied. As an example, we show computations for the same case used in Prob. 4-24 on page 85 of this Manual: U = 1 − x 2 (Tani’s first case). The computed skin friction is compared below with the Thwaites’ results from Prob. 4-24. The step size x should be smaller near separation to match the “exact” xsep = 0.271.

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4-30

Sherman (1990) gives a CFD solution for laminar flow due to a freestream U o approaching

a parabolic cylinder, as in Fig. P4-30. The cylinder surface is defined by y /R = ( 2 x /R )

1/ 2

(

R is the cylinder nose radius. The arc length s along the surface is defined by ds = R 1 +  2

(

where  = y /R. From potential theory, the surface velocity is U = U o / 1 +  2

)

1/2

, where

)

1/ 2

d ,

. (a) Show that

the surface velocity approaches the stream velocity U o as one moves up the surface. (b) Using

Thwaites’ method, estimate the distance s /R along the surface where  ( U ) is within 10% of the flat plate value of 0.22.

Fig. P4-30

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Solution: First plot the surface velocity. It is easiest to use  = y /R as the abscissa, as follows:

We see that U  U o , or zero-pressure-gradient flat-plate behavior, at about y /R  6. Ans. (a)

(

(b) Apply Thwaites’ method by first evaluating s /R =  1 +  2

(

 2 /v = 0.45/U 6

)

1/2

)  U 5ds and thence  = ( 2 /v ) ( dU /ds ) and plot 

d  , after which we compute w /U

We see that  w /U is within 10% of 0.22 at about s /R  9, or y /R  4.

 (  + 0.09)

0.62

.

Ans. (b)

4-31 Apply the Smith-Spalding thermal-integral method of Eq. (4-161) for Pr = 1 to the Howarth distribution, U = Uo (1 − x/L ) , for constant wall temperature. Compute and plot local Nusselt number and also the Reynolds analogy factor, Cf /2C h .

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The skin friction for Thwaites’ method can be computed from the results of the sample Howarth analysis on pp. 268–269 of the text, including Eq. (4-142): 1/ 2

 ( x ) = −0.075 (1 − x/L ) 

−6

− 1 ; 

Cf

  x/L   Re x = 2S (  )  − 5    0.075 (1 − x/L ) − (1 − x/L )  

where S = (  + 0.09) . The computed values of Cf Re x vary from 0.664 at x = 0 to zero at separation, x/L = 0.123, as shown in Fig. 4-26, p. ••• of the text. Meanwhile, the heat transfer is estimated from the Smith-Spalding formula, Eq. (4-161), for Pr = 1: 0.62

0.330 ( 2.95 ) Nu L 0.332 = = 1/2 1/2 x ReL  (1 − x/L )−2.95 − 1 2.95 1.95 dx   (1 − x/L )  (1 − x/L ) L    0  1/2

(Ans.)

Since Nu = Ch RePr, we may convert this, for Pr = 1, to Ch Re x by the substitution

Ch

1/ 2

Nu L  Uo x  Re x = = Re x ReL  U L  Nu x

where

U/Uo = 1 −

x L

(Ans.)

The values of Cf Re x are plotted below and decrease slightly, from 0.332 at x = 0 to 0.311 at

x/L = 0.123 (separation). The Reynolds analogy factor ( Cf /2Ch ) is also shown below.

4-32

Modify Prob. 4-31 by using an explicit or implicit finite-difference method.

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Prob. 4-31, a Howarth velocity distribution, U = Uo (1 − x/L ) with an isothermal wall, was programmed by the implicit finite-difference model of Eq. (4-165), p. 282 of the text. The following numerical values were used, for simplicity: Uo = 1, L = 1, Tw − T = 1, x = 0.001,

(

)

y = 0.00003, and  = 1E-5. Computation was carried out until x/L  0.12, where separation occurred. Two hundred y-steps were used, and, after the velocities and temperatures were computed, the wall heat flux q w = −k ( T/y )w was evaluated and non-dimensionalized in the

form ( hx/k ) /

( Ux/ ).

The results are plotted below for Pr = 1.

We see that the Nusselt numbers predicted by the simple Smith/Spalding quadrature method from Prob. 4-31 are somewhat high. The finite-difference results are more realistic - the value of 0.22 at separation agrees with the Falkner-Skan result on p. 248.

4-33 Apply the Smith-Spalding method, Eq. (4-161) with Pr = 0.72, to flow past an isothermal circular cylinder, Eq. (4-167), and compare results with Fig. 4-29. By substituting U ( x ) from Eq. (4-167) into the Smith-Spalding method, Eq. (4-1671), and using the constants a and b from Table 4-6 for Pr = 0.72, we obtain the following estimate for local Nusselt number around the cylinder:

 Nu R  0.296  1.82z − 0.4z3 ReR 

(

)

−2.88 z

 (1.82z − 0.4z 0

)

3 1.88

 dz  

−1/2

, z = x/R

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No closed form integral is known to this writer, but the integration is readily performed by, e.g., the Runge-Kutta Subroutine of Appendix C. However, the values plotted in Fig. 4-29 are based on diameter D as the length scale. Hence we need to multiply the above results by 2. That is,

Nu D = ReD

( 2 ) NuReR

R

The results are nearly identical to the dotted line plotted as “Smith-Spalding” in Fig. 4-29. Some particular values may be tabulated as follows (for air, Pr = 0.72 ): = Nu/ Re =

0°

10°

20°

30°

40°

50°

60°

70°

80°

0.958

0.953

0.936

0.907

0.865

0.811

0.743

0.659

0.558

4-34 In Eq. (4-144) for Rea = 9500, U /U   1.814 x /a near the front of the cylinder. For air at 20°C and 1 atm, this corresponds approximately to a = 5 cm and U  = 2.85 m/s. Using the

Falkner-Skan theory, Tables 4-2 and 4-3, and a temperature difference (Tw − T ) = 12C, estimate (a) the momentum thickness in mm; and (b) the heat transfer rate in W/m2 at the front of this cylinder. Solution: For air at 20°C and 1 atm, take  = 1.205 kg/m3 ,  = 1.81E-5 kg/( m-s ) , Pr = 0.71, and

k = 0.026 W/( m-K ) . Calculate v =  / = 1.50E-5 m2 /s. (a) Use Table 4-2 for m =  = 1 (stagnation flow) to estimate the momentum thickness.

 * = 0.29235 = 

( m + 1)U =  (1 + 1)(1.814U  /a ) 2v x

2v x

va = 0.29235 Solve for   0.29235 1.814U 

(1.50E − 5m2 /s ) ( 0.05m ) = 0.00011m = 0.11mm 1.814 ( 2.85 m /s )

An alternate computation by Thwaites’ method would yield   0.10 mm.

Ans. (a)

(b) For the heat transfer, use Table 4-3 or Eq. (4-79) for m =  = 1 (stagnation flow). Both formulations give the same numerical result: Nu x =

hx 0.4  1.814U  x  x = 0.57 Pr 0.4 Re1/2 x = 0.57 ( 0.71)   , or: k a  v

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h = ( 0.497 ) k

Then

1.814 ( 2.85m /s )

1.814U  W   = ( 0.497 )  0.026  av m−K  

( 0.05m ) (1.5E − 5m

2

/s

)

= 34

W  W  qw = h (Tw − T ) =  34 2  (12 K )  410 2 m  m K

W m2 K

Ans. (b)

Note that h is proportional to a −1/2 , that is, small cylinders have higher heat transfer.

4-35

(

)

3 3 For flat-plate flow with a wall temperature difference Tw − To = To 1 − x /L , find the

point x at which the local wall heat transfer changes sign. This is a short exercise in understanding of the superposition technique of Sect. 4-8.4. If T is a polynomial, the local heat transfer is given by Eqs. (4-172) and (4-173). In this particular case, all polynomial coefficients are zero except a o and a 3 . Thus, from Eq. (4-173), we obtain

(

)

q w = ( const ) To 1 − 2.2091 x 3 /L3 , or:

4-36

qw = 0

at

x/L = 2.2091−1/3 = 0.768

(Ans.)

Modify Prob. 4-35 by using an explicit or implicit finite-difference method.

We may verify the integral-theory method of Prob. 4-35 (taken from Sect. 4-8.4) by programming the problem of Blasius flow with a cubic wall temperature difference. This writer used the implicit method of Eq. (4-165) with the following conditions: U = 1, L = 1, Te = 0,

Tw = 1 − x3 , v = 1E-5, x = 0.005, and y = 0.0001. The computed results are plotted below in the form of local Nu x / Re x , which would equal 0.332 everywhere if the wall temperature were constant. We see that the integral theory is very accurate.

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4-37

Analyze laminar flow in the entrance region between parallel plates ( 2H ) apart, if the

entrance velocity is uniform, u = Uo . Approximate the flow as a potential ‘core’, where

u = U ( x ) , plus parabolic-shaped boundary layers on each wall [Sparrow (1955)].

The geometry is shown on the previous page. Boundary layers grow out from each wall: u ( x, y ) y y2 =2 − 2 , U(x)  

 = (x)

where y is measured from the wall. The boundary layer merges at y =  with a flat inviscid core of velocity U ( x ) . At any x, the volume flow must match the entrance flow: Continuity:

2  2HUo = 2 ( H −  ) U + 2   U  , 3 

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or:

 = 3H (1 − Uo /U )

(1)

Meanwhile, the x-momentum equation is given by boundary layer theory, Eq. (4-120): 



d dU w / =  u ( U − u ) dy + ( U − u ) dy dx 0 dx 0 Substituting for u ( x, y ) from the parabolic approximation above, we obtain 2U 2 2 d 9 dU = U + U  15 dx 15 dx

Now we eliminate  using Eq. (1) above and rewrite our differential equation for U ( x ) in terms of the dimensionless variables  = U/Uo and x* = x/H:

 16 7  d 10  = = constant 9 − + 2  2  dx*  3U H   o The initial condition (at the entrance) is  = 1.0 at x* = 0. The solution is

 x/ ( 2H )  U  Uo 3  U = 9 − 16 ln  − 7 − 2   U o ( 2H ) / 40  U o U  Uo  

(Ans.)

Note that the left hand side matches the grouping ( x/D) / ReD plotted for “exact” entrance flow in Fig. 4-33, p. 288 of the text. This (approximate) equation of Sparrow (1955) is plotted on the next page of this Manual and may be compared with Fig. 4-33. The predicted “entrance length”, x = Xe , occurs when U = 1.5 Uo :

Xe /2H  0.02594 2HUo /

(Ans.)

Figure 4-33 indicates that this estimate may be low - at least, based on the pressure drop parameter K in Fig. 4-33, fully-developed flow seems to be reached at a somewhat higher value,

( Xe /2H) /Re2H = 0.04.

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4-38 For potential freestream flow past a sphere, use the Rott-Crabtree method, Eq. (4-191), to predict the separation point for a laminar boundary layer. The Rott-Crabtree method consists of the quadrature (4-191) for a given U ( x ) and ro ( x ) :

2 =

0.45  ro2 U6

x

 ro U dx, 2

5

x U ( x ) = 1.5Uo sin   , a

for

0

x ro = a sin   a

after which the shape factor  = ( /)(dU/dx). Combining all the above relations, we obtain the laminar shape factor for potential flow past a sphere: 2

=

0.45cos  sin  8



 sin

7

 d, where

0

 = 35 16 − cos (5sin 1

6

)

 + 6sin 4  + 8sin 2  + 16  

We may evaluate this relation for various angles  and tabulate the results as follows: =

30°

60°

75°

90°

95°

100°

103°

103.6°

=

0.0545

0.0459

0.0330

0.0

–0.0220

–0.0553

–0.0836

–0.0900

Thus, separation is predicted at  = 103.6.

(Ans.)

If these values of  are used to compute wall shear stress from Eq. (4-139), the result is the solid curve for ‘potential flow’ in Fig. 4-35, page 295 of the text.

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4-39 In the spirit of Eq. (4-146) for two-dimensional flow, develop an explicit finite difference model for the thick axisymmetric-flow momentum relation, Eq. (4-195b). Use the same mesh as shown in Fig. 4-25, with yn being the radial coordinate, but do not analyze the axisymmetric continuity Eq. (4-195a). Do the parameters  and  still appear? Note that there is no pressure gradient, U m+1 = U m . Solution: We are to model the momentum equation (4-195b): u

u u v   u  +v = ( a + y )  , where  x y a + y y  y 

a

is the body radius,

After substituting in the finite-difference approximations for the derivatives and solving laboriously for the explicit value of um +1,n , the writer found the following result:   2a + yn + yn+1     4a + 2 yn + yn+1 + yn−1   um+1,n     −   um,n+1 + 1 −     um , n 2 a + 2 yn     2 a + 2 yn      2a + yn + yn−1   +    +   um,n−1   2 a + 2 yn  

(

)

2 Here  = vx / um,n y and  = vm,n x / ( 2um,n y ) have the same meanings as in Eq. (4-146).

Compare the above formulation with Eq. (4-146) for two-dimensional momentum, which does not contain the complicated parenthesis terms which multiply  in the above equation.

4-40 Air at 20°C and 1 atm issues from a circular hole to form a round laminar jet. At 20 cm downstream of the hole the maximum jet velocity is 35 cm/s. Estimate (a) the “1%” jet thickness; (b) the jet mass flow; and (c) an appropriate Reynolds number. 3 2 For air at 20°C take  = 1.2 kg/m ,  = 1.8E-6 kg/m-s, and v = 1.5E-5 m /s. With u max and known, we may compute jet momentum J at this section from Eq. (4-207):

u max =

3J , or 8x

J=

x

8  kg  m kg-m  1.8E-5  ( 0.2 m )  0.35  = 1.06 E-5 2 3 m-s  s  s

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With J known, we evaluate the (dimensionless) constant C from Eq. (4-206): 1/2

 3J  C= 2  16v 

1/2

 3 (1.06 E-5)   = 16 (1.2 )(1.5E-5)2 

= 48.3

The “1%” jet thickness occurs when the profile shape function in Eq. (4-207) equals 0.01:  C22  1 +  4  

−2

= 0.01,

C = 3, 2

or

or:

=

r 6 = 1% = 0.124 48.3 x

Jet thickness b = 2r1% = 2 ( 0.124)( 0.2 m) = 0.050 m

(Ans. a)

The jet thus spreads outward at a half-angle of tan −1 ( 0.124 )  7.1. The jet mass flow at this station is given by Eq. (4-208):

m = 8x = 8 (1.8E-5)( 0.2) = 9.05 E-5 kg/s

(Ans. b)

Finally the local jet Reynolds number could be expressed in at least two ways: Re =

J v2

= 39100;

or:

Re =

u max b ( 0.35)( 0.050 ) = = 1160 v 1.5E-5

(Ans. c)

4-41 Air at 20°C and 1 atm flows at 1 m/s past a slender body of revolution, L = 15 cm, whose drag coefficient is 0.008 based on area L2 . Assuming laminar flow at 3 m downstream of the trailing edge, estimate (a) the maximum wake velocity defect; (b) the “l%” wake thickness; and (c) the wake-thickness Reynolds number. For air at 20°C take  = 1.2 kg/m ,  = 1.8E-6 kg/m-s, and  = 1.5E-5 m2 /s. With x, L, and CD known, the maximum wake defect velocity follows from Eq. (4-211): 3

U2 L2 (1.0) ( 0.15) = 0.159 m/s = 16 cm/s = CD o = ( 0.008) 8x 8 (1.5E-5)( 3.0 ) 2

u max

2

(Ans. a)

The 1% thickness occurs when the Gaussian profile in Eq. (4-210) equals 0.01:

 U r2  exp  − o  = 0.01, or:  4x    or:

Wake thickness

 4.605 ( 4 )( 3.0 )(1.5E-5 )  =  (1.0 )  

1/ 2

r1%

b = 2r1% = 0.0576 m = 58 cm

= 0.0288 m, (Ans. b)

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Finally, an appropriate wake Reynolds number is

Re =

u max b ( 0.159 )( 0.0576 ) = = 610  1.5E-5

(Ans. c)

4-42 Given a streamlined airfoil as pictured in Fig. P4-42a, axial velocity profiles may be measured at both the upstream and downstream sections labeled 1 and 2 using a streamlinebounded control volume (i.e., one that forms a streamtube). Assuming equal pressure around the control volume and symmetry with respect to the midsection plane, determine (a) the half height of the upstream station, H , and (b) the drag coefficient on this airfoil if the vertical dimension at the downstream station is Y = φc , where c is the cord. The downstream velocity is given by  1 1  u = U 1 − cos   y / Y   . Hint: You may take advantage of symmetry by selecting a control 2   2 volume as pictured in Fig. P4-42b.

Solution:  u  * =  1 − dy 0  U 

1 2

 y   dy  2 

 * =  1 − cos  0

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



0

0

 * =  dy − 

 y  1 cos   dy 2  2 

 1 2   y    = −  sin   2 0   2   *

     * =  −   sin − 0   2   

   * =  −   sin  2  * = −

  1  ( − 1)  =  1 −  =    

* =

( −)  

(Ans.)

4-43 Consider the steady, incompressible, and two-dimensional flow past a circular cylinder of diameter d, as pictured in Fig. P4-43. The upstream velocity U may be taken to be uniform while the downstream velocity may be set to be

  y  ; +0  y  + u ( y) sin  =   2   U 1; +  y  +  where both U and d are known. Calculate (a) δ and (b) the drag coefficient CD.

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Consider the conservation of mass in integral form:

( )

d   dv +CS  v  n dA = 0 dt CV For steady-state conditions,

d   dv = 0 dt CV

   (v  n)dA = 0 CS

a.

  (v  n)dA +   (v  n)dA +   (v  n)dA +   (v  n)dA = 0

AB

BC

CD

(eq-1)

AD

  (v  n)dA =   (v  n)dA = 0

BC

AD

Since streamlines are the upper boundary and lower boundary of the control volume, along surfaces BC and AD, the velocity vector v and normal vector to the surface are perpendicular such that v  n =| v || n | cos90 = 0 .

  (v  n )dA = − U (d 1) AB

Here v and n are in opposite directions.

v  n =| v | +n∣ cos180 = −v = −U

  (v  n )dA =  

CD

 0

  y  u ( y )(dy 1) =   U sin   dy 0 2



  y   − cos  2       = U        2   0

2   ( v  n )dA = U   ( − ( 0 − 1) )

CD

=

2  U



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Therefore, (Eq-1) becomes − Ud +

Ud =

 =

2  U



=0

2 U

 2

 d

(Ans) b. Linear momentum equation in integral form is

F =

d   vdv + CS  v(v  n)dA dt CV

Under steady-state conditions,

d  vdv = 0. dt CV

 ΣF =   v (v  n )dA CS

Assuming only drag force is the external force acting, D =   v(v  n )dA CS

D=

  (v  n)dA +   (v  n)dA +   (v  n)dA +   (v  n)dA

AB

BC

CD

AD

For similar reasoning made during conservation of mass equation simplification

  (v  n)dA =   (v  n)dA = 0

BC

AD

  v (v  n )dA = − U ( d 1) 2

AB

= − U 2 d

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 v (v  n)dA =  

 0

(u( y)) 2 (dy 1)

CD

 y  =  0 U 2 sin 2   dy  2  1   y  = U 2 0 1 − cos 2    dy 2  2   =



   y  U 2 0 1 − cos    dy 2     

  y  sin        = U2 y − 2            0

 U 2 = 2

 U 2  D = − U d + 2 2

= − U 2 d +

  2



2

dU 2

  = U 2 d  −1 +  4   

2 Drag on the body = U d 1 −



 4

  2 U 2 d 1 −  D  4 = Drag coefficient = 1 U 2 d U 2 ( d 1) 2 4 = 4 − d

(Ans)

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4-44

Plot the rotating-disk solution, Table 3-5, in the form of a hodograph of ‘streamwise’ flow

( r− v ) versus ‘transverse’ flow vr . Compare with Fig. 4-42(c).

The rotating disk is a fully three-dimensional (laminar) boundary layer. As seen in Fig. 3-28, p. 156 of the text, the ‘secondary’ flow is radially outward, and the ‘main’ flow is circumferential. To view this flow relative to the wall, we subtract out the moving-wall velocity ( r ) . Thus we

plot dimensionless streamwise velocity, 1 − G ( z*)  , versus secondary or transverse velocity,

F ( z*) . We could do this from the values listed in Table 3-5, p. 159. Or, if your PC, like

mine, has a plotting program, you could just re-run the Runge-Kutta solution, Eqs. (3-186) of the text, and make a running plot of Y ( 3) versus 1 − Y ( 5 )  . In this latter manner the hodograph below was prepared. It clearly resembles ordinary unidirectional skewing, as in Figs. 4-42(a,c) on p. 311 of the text.

4-45 Verify that the free-convection similarity variables of Eqs. (4-255) do indeed lead to the coupled ordinary differential equations of Eqs. (3-256). This is a standard exercise in variable-substitution, using the chain rule. It is very good. I think, for students to perform such an exercise at least once in their graduate training, since similarity solutions do crop up in many areas of applied mechanics.

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4-46 Develop a numerical solution to the vertical-plate free-convection relations, Eqs. (4-256) and (4-257), for a particular Prandtl number not listed in Table 4-10. This is a challenging numerical-integration exercise, since there are two unknown initial conditions, f  ( 0 ) and Θ ( 0 ) . The table is a big help, as otherwise we would have to make very crude guesses of these two values. As an example, let us take the case Pr = 8. The table indicates that the proper values of

f ( 0) and Θ ( 0) are about +0.44 and –1.09, respectively. Equations (4-256) may be modelled by the Runge-Kutta Subroutine of Appendix C, letting Y1 = f , Y2 = f , Y3 = f , and Y5 = Θ. The proper FORTRAN relations are then

F (1)

=

−3* Y ( 3) * Y (1) + 2* Y ( 2 ) * Y ( 2 ) − Y ( 5 )

F( 2)

=

Y (1)

F ( 3)

=

Y ( 2)

F( 4)

=

−3* Pr* Y ( 3) * Y ( 4 )

F ( 5)

=

Y ( 4)

Y4 = Θ,

with known initial conditions Y ( 2) = Y ( 3) = 0 and Y ( 5) = 1. Let us guess that f  ( 0 ) is somewhere between 0.43 and 0.45, and that  ( 0 ) is between –1.08 and –1.10. From Fig. 4-50, page 326 of the text, we see that “infinity” is at about  = 6. Based on these four guesses, we run off four sets of solutions and check to see whether the values of f (  ) or Y2 ( 6 )  and  (  ) or Y5 ( 6 )  are near zero. We list the following results:

f ( 0)

 ( 0 )

f  ( 6)

 ( 6)

0.43

–1.08

–0.0466

–0.0055

0.43

–1.10

+0.0756

–0.0203

0.45

–1.08

–0.1373

+0.0327

0.45

–1.10

–0.0103

+0.0174

We may interpolate linearly to find that f  ( 6)   ( 6 )  0 and f  ( 0 ) = 0.438 and  ( 0) = −1.094. We then repeat the procedure using a narrower range of guesses. The process converges this second time to an accurate estimate of the solution: Pr = 8.0:

f  ( 0 ) = 0.4386

 ( 0 ) = −1.0957

Nu/Gr1/ 4 = 0.775

(Ans.)

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The solutions for velocity ( f  ) and temperature ( ) are plotted on the next page.

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4-47 A vertical isothermal plate 40 cm high and 30 cm wide is immersed in air at 20°C and 1 atm. Each side of the plate is to dissipate 100 W of heat to the air. Calculate (a) the wall temperature (in °C ) and (b) the Grashof number. Is the flow laminar? Hint: Assuming an average wall-fluid (film) temperature of 100°C, evaluate the necessary thermophysical properties, and repeat until the properties no longer change. With T unknown, we cannot compute the Grashof number, but let us assume the flow is laminar and proceed (checking later for transition). Make a guess that the average fluid temperature is 50°C—from Table A-2, k  0.0276 W/m-K and  = 1.8E − 5m 2 /s . For air, Pr  0.71, the laminar flow relation (4-260) in the text predicts Nu x = 0.354 Grx1/4 , or Nu L,mean =

4 Nu x = 0.472 GrL1/4 3

For air, an ideal gas,  = 1/T ( K ) . So we know everything except T, and the above relation becomes, for qw = 100 W ,

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Nu L =

qw L (100 W )( 0.4 m ) = A w kT ( 0.4 )( 0.3) m 2  ( 0.0276 W/m-K ) T   1/4

 g  = 0.472  2 L3T  v 

(

)(

 2 −1  9.81 m/s 1/323 K = 0.472  2 1.8E-5 m 2 /s  

(

)

)

1/4

 3 ( 0.4 m )   

Solve this for T5/4  517, or T  148C. A better estimate of average fluid temperature is thus (1/2) ( T + Tw )  94C (367 K ) , and we may correct the fluid properties to k  0.0306 W/m-K,

v  2.23E-5 m2 /s,  = 1/367 K−1. Repeating the calculation gives little change, and we may take the desired result to be T  150C,

or:

Tw = 170 °C

(Ans.)

3 2 9 Finally, check the Grashof number, gL T/ = 5.2E8, which is less than 10 ; hence the flow is presumed to be laminar and our estimate is accurate.

4-48 A horizontal pipe of diameter 5 cm is immersed in air at 20°C and 1 atm. If the cylinder surface is at 300°C, estimate the heat loss in watts per meter of pipe length. Estimate the fluid properties at the average temperature between the wall and the ambient air: T ( avg ) = (1/2)( 20 + 300 )  160 C = 433 K. From Table A-2, p. 575 of the text, −1 k  0.0350 W/m-K and   2.96E-5 m2 /s. Also,  = 1/T = 1/433 K . The Rayleigh number of the cylinder is thus Ra D =

 ( 9.81)(1/433)  3 3   ( 0.05 ) ( 300 − 20 )( 0.71) = 643000 D  T Pr = 2 2   ( 2.96E-5 ) 

g

The relevant heat-transfer correlation for free convection to horizontal cylinders is Eq. (4-265), p. of the text: 0.387 ( 643000 )

1/6

Nu1/2 L

or:

= 0.60 +

9/16 8/27

1 + ( 0.559/0.71) 

= 3.58,

or:

Nu L = 12.85 =



hD , k

h = 12.85 ( 0.035 W/m-K ) / ( 0.05 m ) = 8.99 W/m 2 -K

The total heat loss from cylinder to air is thus estimated as Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -47-

(

)

q w = h ( DL ) T = 8.99 W/m 2 -K  ( 0.05 m )(1.0 )( 300 − 20 C ) = 400 W / m

(Ans.)

4-49 A model two-dimensional airfoil 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 C is 30 cm. The fluid is air at 20°C and l atm. Assume that x is a good approximation to the arc length along the upper surface. Using any laminar-boundary-layer method of your choice, find the predicted separation point, if any, for (a) ReC = 1E6; and (b) ReC = 4E5. Solution: For air at 20°C and 1 atm,  = 1.205 kg/m and  = 1.81E-5 kg/(m-s). We need to integrate V 5 versus x, but the data are sparse and highly variable, as shown below: 3

We either need a terrific curve-fit, or we can just plow along. Thwaites’ method requires

2 v



0.45 V6

x

5  V dx and then find where  = 0

 2 dV = −0.09 (separation) v dx

The writer plowed along and produced the reasonably accurate plot of  versus x on the next page.

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Separation is predicted by Thwaites’ method at approximately x/C  0.45. Ans. (a, b) The problem called for calculated separation “for (a) ReC = 1E6; and (b) ReC = 4E5. ” However, for a laminar boundary layer, the separation point is independent of ReC ! The reader is invited to prove this as Problem 4-51.

4-50 Air, at about 1 atm and 20°C, flows through a 12 cm square duct at 0.4 m3 /s as in Fig. P4-50. Two hundred thin flat plates of chord-length 1 cm are stretched across the duct at random positions. Their wakes do not interfere with each other. How much additional pressure drop do these plates contribute to the duct flow loss?

Fig. P4-50 Solution: For air at 20°C and 1 atm,  = 1.205 kg/m and  = 1.81E-5 kg/(m-s). First find the average velocity through the duct from one-dimensional continuity: 3

V=

Q 0.4 m3 /s m = = 27.8 A ( 0.12m )2 s

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Check the Reynolds number: ReDh = (1.205)( 27.8)( 0.12)(1.81E-5)  220,000. Thus the duct flow is turbulent, and we will assume (why not?) that each tiny plate sees this average approach velocity V = 27.8 m/s. The plate length Reynolds number is only Re L = 18, 500, low enough that the flow over the plates is laminar and not tripped by the stream turbulence. Calculate the drag of one plate from the Blasius formula: Re L = 18500, CD =

F1 plate = CD

1.328 1.328 = = 0.00977 Re L 18500

 2 2  1.205  V bL ( 2sides ) = 0.00977   ( 27.8) ( 0.12m )( 0.01m ) (2) = 0.0109 N 2  2 

The total drag of 200 plates is thus approximately Ftotal = ( 200)( 0.0109N ) = 2.18 N. A freebody of the plates shows that this force can only be balanced by an overall pressure drop across the array of plates: p plates =

4-51

Ftotal 2.18 N =  150 Pa Aduct ( 0.12m )2

Ans.

Nondimensionalize Thwaites’ method, using U o and L as reference values, for laminar

boundary layer flow with an external potential flow distribution U ( x ) . Show that the predicted

separation point is independent of the Reynolds number U o L /v. Apply this conclusion to discuss the expected results for Prob. 4-49. Solution: For typing ease, let V = U /U o and let  = x /L. Then Thwaites’ method calls for the calculation of the Thwaites parameter  ( x ) , with separation as follows:  0.45 x

 =  2 /v  ( dU /dx ) = 

 U

6

 dU 5 U dx = −0.09   dx  0 

(separation)

Now introduce the dimensionless variables and rewrite:  0.45  5 5  d (U oV )  0.45  5  dV  =  6 6  U oV Ld  = −0.09 =  6  V d  U oV 0  d ( L )  V 0  d

The final expression depends only on the dimensionless velocity distribution and is entirely independent of U o , L, and v, that is, it is independent of Reynolds Number! See Problem 4-49 for a numerical example. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -50-

4-52

A conical diffuser of initial radius R expands at a uniform angle  , as in Fig. P4-52.

The flow enters at uniform velocity U o . Assuming a one-dimensional freestream, use any laminar boundary layer method of your choosing to estimate the angle  for which flow separation occurs at x = 2 R.

Fig. P4-52 Solution: The writer is going to choose the Rott-Crabtree axisymmetric modification, Equation (4191), of Thwaites’ integral method. The cone radius is ro ( x ) , and we estimate the velocity U ( x ) from one-dimensional continuity and compute momentum thickness: x

 2 0.45 2 5 R2 2 tan   2 6  ro U dx, U = U o 2 , ro = R (1 + Kx ) , K = v ro U 0 R ro Substitute for U and ro in the  2 equation and integrate, with the result

 2 0.45 (1 + Kx )  −7 = 1 − (1 + Kx )  ,   v 7U o K 10

where K = 2 tan  R

Now evaluate the velocity gradient: −2 K U o dU d  −2 = U o (1 + Kx )  =  (1 + Kx )3 dx dx 

Then the Thwaites’ parameter  ( x ) is given by

 2 dU 0.9  2 tan  7 = =− 1 + Kx ) − 1 , K = (  v dx 7  R

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Separation occurs at  = −0.09, or (1 + Kx )7 = 1.7, or Kx = 0.0788. If, as specified for this problem, x = 2R, then Kx = 4 tan  = 0.0788, tan  = 0.0197, θ = 1.13°. Ans. This is smaller than the (laminar) result for the flat diffuser in Prob. 4-25, because the conical geometry causes U ( x ) to drop about twice as fast.

4-53 Show that the point-sink boundary-layer solution of Eqs. (4-86) and (4-89) may be interpreted through Thwaites’ method as a constant value of Thwaites’ parameter,  = 9/80. (a) How does this value of  compare to that for plane stagnation flow? (b) Use Eq. (4-89) to derive an expression for skin friction C f as a function of Rex. Solution: From Fig. 4-16, sink flow is a strong acceleration, U = ( − K /x ) . Evaluate the Thwaites’ parameter  ( x ) . We have to integrate from x to



to avoid the x = 0 singularity:

 0.45  5  dU  0.45x6  − K 5  d = = U dx  =  ( − K /x ) 6  5 v dx  U 6 0  dx  K x x   dx

 2 dU

 0.45 x6 K 5   K  0.45 9 = = = = 0.1125 = constant 6 4  2  4 80  K 4 x   x 

Ans.

The comparable Thwaites estimate for plane stagnation flow is stag  0.075, or 33% less.

4-54 Show that the exact Falkner-Skan solutions for two-dimensional stagnation flow are equivalent to Thwaites-method parameters of  = 0.0855, S = 0.360, and H = 2.216. How do these values compare with the use of Thwaites’ method for the freestream U = Bx ? Solution: For stagnation flow,  = m = 1, the dimensionless wall shear, displacement thickness, and momentum thickness are given in Table 4-2: fo = 1.23259, * = 0.64790, and  * = 0.29235. We calculate the desired Thwaites-type parameters if U = Bx, dU /dx = B:

* = 

m +1 U v = 0.29235, U = Bx, m = 1,  = 0.29235 , 2 vx B

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=

Thus

w = 

 2 dU 2 = ( 0.29235) = 0.0855 v dx

Ans.

  u m +1 U |w = U fo , S = w = (1.23259 )( 0.29235 ) = 0.360 Ans. y 2 vx U H=

 * * 0.64790 = = = 2.216   * 0.29235

Ans.

Use Thwaites’ method, Eq. (4-138), to estimate  for stagnation flow, U = Bx, dU /dx = B: x

x

 2 0.45 5 0.45 0.45  2 dU 0.45  6  U dx = 6 6  B5 x5 dx = , = = = 0.075 v 6B v dx 6 U 0 B x 0

Ans.

By comparison, from Thwaites’ correlation, Table 4-4, for   0.075 (12% low from FalknerSkan), S  0.327 (9% low), and H  2.36 (6% high).

4-55 When a jet is formed normal to a wall, it is called a wall jet, as shown in Fig. P4-55. The jet origin can be thought of as a source of momentum, and further down the wall the velocity profiles are similar. Glauert (1956) showed that both plane and axisymmetric wall jets satisfy the same ordinary differential equation for laminar flow: 2

 df  +f + 2  = 0 3 2 d d  d 

d3 f

d2 f

Fig. P4-55

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where   y, f  , f   u, and f    . The boundary layer thickness  is proportional to x3/4 .

(a) Verify Glauert’s choice of boundary conditions: f ( 0) = f  ( 0) = f  (  ) = 0. (b) To enforce the normalized condition f (  ) = 1, Glauert showed that f  ( 0) = 2/9. For this value, using any convenient method, numerically integrate the similarity equation out to  = 8 and plot both

f ( ) and f  ( ).

Solution: Read Glauert’s paper to see how this interesting similarity solution was deduced. There is no freestream, the jet discharges into still air, so f   u has to be zero at both the wall and at infinity. Letting f ( 0) = 0 simply sets the wall stream function to zero. A numerical solution for

f  ( 0) = 2/9 (Glauert’s choice to make f (  ) = 1) is plotted as follows:

4-56 Consider the steady two-dimensional flow past a circular cylinder of diameter d, as pictured in Fig. P4-56. The upstream velocity U may be taken to be uniform while the downstream velocity is allowed to vary from U⧸2 to U according to the following symmetric profile:

1  1    y   1  ; 0  y   u ( y )  1 + sin  u (  )  1 + sin     ; 0    1   = 2  = 2  or , where  2    2  U U 1;    y 1     1; both U and d are known. Calculate (a) δ as a function of d and (b) the drag coefficient CD. Here, ξ = y⧸δ.

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Solution:

 

*  a.  =  0 1 −

u( y)   dy U 

1   y  = 0 dy − 0 1 + sin    dy 2  2    1  y   = [ y ]0 −  0 dy + 0 sin   dy   2    2

  2 = − +  2 

    y    − cos  2       0  

  2 = − + 2 

     0   − cos 2 −  − cos 2     

  2  =  −  + [0 + 1] 2     2  = − +  2     + 4   +4 =  =     2   2 

 +4 2 d  =    = 2   +2   b. cD =

FD  = 1  Av 2 (2 +  ) 2

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4-57 Repeat the previous problem using the following polynomial profile at the downstream section of the control volume:

−  y  − 1; −    −1 1;  2  u( y)  1  y  u ( )  1 =  1 + 2  ; −  y  + or =  (1 +  2 ) ; −1    +1 U U 2    2 +1    + 1; 1; +  y  + 

Solution:

1; −   y  −   u ( y )  1  y2  =  1 + 2  ; −   y  + Given U 2     1; +   y  + 

From RTT,

dN dt

= system

   ndv + CS  n(v ˆ )dA  t CV

a. Using conservation of mass, N = m, n = 1

0 = 0 + 2  0 udyw − 2  vd = 0

 d =



0

udy

U

=

U 2U





0

 y2  1 + dy  2    

 1   1  y3   = [ y]0 + 2    2   3  0  

1   1  4 =  +  =  2 3  2 3

 2 =  3

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=

3d 2

(Ans)

b. Using conservation of linear momentum,

N = mv, n = v 

 Fx = 0 + 2   u 2 dy − 2 U 2 d  0

  u 2  y 2 2  Fx = 2    1 + 2  dy − u 2 d    0 4    

  u2   y2 y4  = 2    1 + 2 + 4  dy − u 2 d   2 0     4  u2   3 5  = 2     + 2 2 + 4  − u 2 d  3 5  4  1  2 1   = 2  u 2   1 + +  − d  4  3 5  

3 1 FD = − u 2 wd = cD u 2 wd 5 2

cD = 1.2

(Ans)

4-58 Consider the steady, incompressible, and two-dimensional flow past an ellipsoidal cylinder with unit depth W in the z direction. The upstream velocity, U, may be taken to be uniform while U the downstream velocity varies from until it reaches U according to the following profile: 2

−  y  −Y 1;  u( y)  1  y 4  =  1 + 4  ; −Y  y  +Y U 2  Y  1; +Y  y  + 

−    −1 1;  u ( )  1 =  (1 +  4 ) ; −1    +1 or U 2 +1    + 1;

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where U is known and Y defines the edge of the boundary-layer region. Calculate (a) Y in terms of H and (b) the drag coefficient CD using the major axis of the ellipse as the characteristic length. You may use half of the domain shown below for simplicity.

Solution:

1; −   y  −   u ( y )  1  y4  =   1 + 4  ; −   y  + Given: U 2     1; +   y  + 

* a.  =

 u( y)  1− dy 0  u  





  1 1 y4   =  1 −  + dy 4  0  2 2   

4   1 y  1  =  d y −   dy +  dy  4 0 0 2   0 2 

 1 1 1 =  + [ − 0] +  4   5  0 2 2 5

= −



5   = − + 4 2 10 2 10 +

* = H =

10 − 5 +  3 = 10 5

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 =

5H 3

(Ans) b. cD =

2 FD 20 H  cD = 2  u w 276

(Ans)

4-59 The Blasius equation can be expressed as

f  + f =0 f  where the primes denote differentiation with respect to  = y / U / (2 x) . Given that the Blasius constant is

 = f  (0) , show that an integrodifferential form of the solution consists of  f  ( ) =  exp − f (t )dt   0 

Solution: The Blasius solution is given by

f "' + f =0 f"

(eq-1)

The initial condition is given by,

f "( 0) =  and  =

y u 2vx

From equation (1),

f "' =−f f" Integrating both sides from 0 to 





0

 f  dt = 0 − f (t )dt f 

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



0

0

 ln f " ( t )  = −  f ( t )dt 

 ln f " ( ) − ln f " ( 0 ) = −  f ( t )dt 0

  f  ( )  ln   = −  f ( t )dt  0  f (0) 

" But f ( 0) = 

 f " ( )    ln   = − 0 f ( t )dt    Taking exponents on both sides,

(

 f " ( )     = exp − 0 f ( t )dt   

(



f " ( ) =   exp − f ( t )dt 0

) )

(Ans)

4-60 The Blasius solution arises in the context of laminar, incompressible, constant property, boundary-layer flow over a flat plate, where the freestream velocity above the plate is U. As usual, the boundary-layer equations for flow over a flat plate reduce to

 u v  x + y = 0   2 u u + v u =   u  x y y 2 

with

u ( x, 0) = 0, u ( x,  ) = U ,

and

u y

=0 y =

An equally common form of the Blasius similarity variable does not carry the arbitrary “2” factor under the radical. More specifically, the Blasius equation can be rederived using the following transformations:

=

U y,  =  xU f , x

and

u ( x, y ) df ( ) = U d

Show that the corresponding Blasius equation becomes f  +

1  ff = 0 with the same boundary 2

conditions as before. Given: Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -60-

Boundary layer equations:

u

du dv d2y + v = 2 dx dy dy

u v + =0 x y

(eq-1)  =

 

(eq-2)

Boundary conditions are as follows: At y = 0 , u = v = 0 At y =  , u = v The local free stream velocity u(x) at section x is scale factor for u, because the dimensionless u(x) varies in x direction and is unity with all values of y at all sections. The scale factor for y, denoted by g(x), is proportional to local boundary layer thickness so that “u” itself varies between 0 and unity. Now, if Blausius flow, it is possible to identify g(x) with boundary layer thickness  , we know that

=

 L

1 Re L

In terms of “x,”

 x

=

1  = v x u

v u

    u y  That is, = F = f ( )  v  u     u 

Where  = y and  

=

vx v

u v y  y =  x  dy = x  d u u ux u

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The stream function can be obtained in terms of velocity component

 =  udy =  u F ( )

vx d u

= u vx  F ( ) d  = uvx  f ( ) + D

Also,

where D = constant

(Ans)

 F ( ) d = F ( ) , and the constant of integration is zero if the stem function at solid

surface is equal to zero. Now the velocity component and their derivates

u=

du d d = u f −1 ( ) = , dy d  dy

      −d 1 1 1 y  −1   v= = − v  v  f  + x f ( ) − 2 x ( )  2n v   dx x     u        du d 1 " " Therefore, = u f ( ) = u f ( )  −  2 dx dx  

  y 1 vx x   u 

    dy d 1  = u f " ( ) = u f " ( )   v  du dy x   u    u du = v  f  (n) dy vx

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−

2

d v = u dy 2

    v 1     f ( )  vx  x    u 

 2u v2 =  f  ( n) 2 y vx

(eq-3)

Substitute (eq-3) in (eq-1)



u2 u2   u2   f ( ) f  ( ) + 0  f  ( x) − f ( )   f  (n) =  f  ( ) 2 x 2x x

u2  1 u2  =− f ( ) f ( ) =  f ( x) 2 x x

 2 f  ( ) + f (n)  f  ( ) = 0 Where f ( ) =

4-61

 f ( )d + C u  u d + C

(Ans)

and  =

y vx u

Use the Runge–Kutta method with shooting to solve the Blasius equation f  + ff  = 0

and thus capture the flat plate boundary layer using  = y U / (2 x). Use the far-field value of

 = 6 and a step size of  = 0.02. (a) Select f (0) = {0.3,0.5} as the first two guesses for the unknown initial condition. Report the number of additional guesses required to converge on f ()  1 within an absolute tolerance of ò = 10−5. (b) Plot f , f , and f  versus . Tabulate f , f , and f  for increments of  = 0.2 (i.e., every tenth data point). Feel free to check your results against Table 4-3. (c) Use the 99% definition of the boundary-layer thickness to estimate the value of  =  that corresponds to the “edge” of the boundary layer. (d) Compute the parameters: *  Rex , Rex , and C f Rex x x Solution: Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -63-

A finite difference formulation of the Blasius equation may be implemented using a Runge– Kutta solver with a shooting method. To begin, we use the following finite-difference discretization scheme, fi +2 = 3 fi +1 − 3 fi + fi −1 − fi ( fi +1 − 2 fi + fi −1 ) with f − 2 fi + fi −1 f − 3 fi +1 + 3 fi − fi −1 f −f fi = i +1 i fi= i +1 and fi= i + 2 2   3 (a) If we take f (0) = {0.3,0.4} as the first two initial guesses, we find that both of these values undershoot our desired convergence on f ()  1. However, if we use a value of f (0) = 0.5 , we overshoot the initial “Blasius” slope needed for convergence. The use of f (0) = 0.5 to bracket the initial guesses is therefore more practical. In fact, it is possible to determine how many additional guesses are needed over an interval [a, b] for the bisection method to converge. For the bisection method, the number of iterations leading to convergence corresponds to: 1 b−a  b − a  ln ( b − a ) − ln(2 ) n = log 2  ln  = = ln 2  2  ln 2  2  To obtain a tolerance of  = 10−5 when converging on f ()  1 , we determine that our guesses for f (0) must actually fall within a tolerance of  = 10−7 . Thus, using a bracketing interval of [0.4,0.5] , the total number of guesses needed to reach convergence may be estimated to be  0.5 − 0.4   = 3 + 18.93  22 guesses Ans. (a) Total Number of Guesses = 3 + n = 3 + log 2   2 10−7    However, this number can be subjective as it depends on the numerical scheme used by each individual.

(

)

(b) To check our results, we may compare the plots and tabulated data to Figure 4-10(a) and Table 4-3. Our results agree with the tabulated values within 2 significant digits, as the solution method employed implements a far coarser discretization than that produced in Table 4-3.

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 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8 6.0

f ( ) 0.00000 0.00675 0.03204 0.07600 0.13843 0.21895 0.31694 0.43151 0.56148 0.70541 0.86165 1.02844 1.20395 1.38646 1.57436 1.76628 1.96105 2.15780 2.35585 2.55472 2.75409 2.95375 3.15358 3.35349 3.55345 3.75343 3.95342 4.15341 4.35341 4.55341 4.75340

f  ( ) 0.00000 0.08434 0.17789 0.27081 0.36231 0.45128 0.53641 0.61629 0.68952 0.75494 0.81170 0.85942 0.89820 0.92861 0.95158 0.96828 0.97995 0.98778 0.99282 0.99594 0.99779 0.99884 0.99941 0.99971 0.99986 0.99993 0.99997 0.99998 0.99999 0.99999 0.99999

f  ( ) 0.00000 0.46843 0.46688 0.46220 0.45281 0.43736 0.41501 0.38552 0.34945 0.30813 0.26358 0.21820 0.17443 0.13443 0.09974 0.07115 0.04876 0.03209 0.02026 0.01227 0.00713 0.00397 0.00212 0.00108 0.00053 0.00025 0.00011 0.00005 0.00002 0.00001 0.00000

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5 f f' f" f (η) at large η

4.5 4 3.5 3 2.5 2 1.5 1 0.5 0

0

1

2

3

η

4

5

6 Ans. (b)

(c) Being limited to a discretization of  = 0.02 , the closest value we are able to determine is   3.46 Ans. (c) This value can be obtained numerically and checked graphically. The actual value to 16 significant digits is  = 3.471886880405967 , which is given in Table 4-3; however, the relative coarseness of the  discretization size used here leads to a value of either 3.46 or 3.48, with 3.46 being closer to the true value of 3.47188688… (d) Based on the present numerical calculations,   U / (2 x) can be determined from the  intercept of a 45˚ line, which is asymptotic to the curve of f ( ) at large  . Using the slopeintercept for the asymptotic line and a data pair ( , f ) , one may deduce   U / (2 x) . For example, using (6.000,4.793) and a slope of m = 1 for a 45˚ line, we can write, for y = mx + b , 6 = 4.793 + b and so b = 1.207 Then using the expression given by Eq. (4-68), we have U U   1.207 or    1.706 2 x x This leaves us with   1.706  Rex  1.706  or Ans. (d) x x Rex

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Using the present discretization scheme, and using f (0) =  U / (2 x)  0.4686 as determined from Part (a) –and consistently with Eq. (4-69),– we can perform a similar manipulation for the momentum thickness. We get



Rex  0.4686 2



Rex  0.662 Ans. (d) x x Finally, using Eq. (4-70), we are able to retrieve  0.662 Cf =  or C f Rex  0.662 Ans. (d) x Rex These results generally agree with those obtained through Eqs. (4-68)–(4-70) and Table 4-1. Their differences from the ‘true values’ serve as an important reminder of the appreciable sensitivity of the solution to the discretization size. The so-called classic values associated with the Blasius problem, given in Table 4-1, were originally obtained using a relatively coarse grid. and so

4-62 Under standard temperature and pressure conditions, consider a laminar, incompressible flow of air over a flat plate, as pictured in Fig. P4-1, where the velocity may be taken to be linear, specifically, u( y) = Uy /  . Using von Kármán’s momentum-integral approach, evaluate (a) the fundamental boundary-layer properties, δ, δ*, θ and τw; (b) the drag force per unit width ( FD / b) as well as the drag coefficient, CD = 2FD / ( U 2 Lb ) . You may use a constant freestream velocity, U = 10 m/s, b = 1 m, and L = 1 m. Solution: a. Given equation of velocity profile is u( y) =

Uy



u( y) y = U 

According to von Kármán momentum-integral equation,

 d = 2 U dx Momentum thickness   u  u  =   1 −  dy 0 U   U    y  y =   1 −  dy 0     

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=



0

 y  y 2   −   dy       

 y 2 y3     = − 2 = − =  2 3  0 2 3 6 =



(Ans)

6

 du   d  Uy   U w =    =     =   dy  y =0  dy     y =0

w d = U 2 dx U  1 d =  u 2 6 dx 6  d = u dx 6

 U dx =   d 6 2  x2 = Ux 2 

2 =

12  x2 ( Re x )

2

12 Re x

x

 x

2

=

=

(Ans)

3.4641 Re x

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6 2 2 x =  ux 2



2 =

12 x 12  8.90 10−4 1 = u 1000 10

 = 1.03344 10−3 m = 1.03344mm =

 6

=

1.03344 = 0.1722 mm 6

Disp. Thickness ( * ) is calculated as follows:   u y  * =  1 − dy =  1 − dy 0 0  U  



 y2      =  y −  =  −  =  2  0  2  2 * =

 2

* =

1.03344 = 0.51672 mm 2

w =

U  10  8.9 10−4 = = 8.612 N / m 2 −3  1.03344 10

b. Drag force per unit width can be calculated as follows: FD =   wb  dx LV FD U   L 10  8.9 10−4 1 =  dx = = = 8.612 N / m b 0   1.03344 10−3

Drag coefficient can be calculated as follows:

cD =

2 FD 2  8.612 = = 0.0001722  v 2 Lb 1000 102 1

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4-63 Under standard temperature and pressure conditions, consider a laminar, incompressible flow over a flat plate where the velocity may be taken to be of the Pohlhausen type, specifically,

u / U = c0 + c1 + c2 2 + c3 3 + c4 4 ;   y /  . Evaluate (a) the fundamental boundary-layer properties, δ, δ*, θ, and τw; (b) the drag force per unit width

( FD / b)

as well as the drag

coefficient, CD = 2 FD / ( U 2 Lb ) . Assume a uniform freestream velocity with no pressure gradient. a. For Karman–Pohlhausen approximation, u y = C0 + c1 + c2 2 + c3 3 + c4 4 ;  = U 

For boundary conditions, i. u(x, 0)=0 ii. u( x, s) = u( x) iii. iv. v.

u ( x, s ) = 0 y  2u u ( x) d (u ) ( x,0) = −  2 y  dx 2 u ( x,  ) = 0 y 2

Karman–Pohlhausen approximation got the following result:

0 = C0 − = 2C2 1 = C0 + C1 + C2 + C3 + C4 0 = C1 + 2C2 + 3C3 + 4C4 0 = 2C2 + 6C3 + 12C4 So, C0 = 0 C1 = 2 + C2 =

 6

− 2

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C3 = −2 + C4 = 1 −

 2

 6

And rewrite the equations as follows: u  3 = 1 − (1 +  )(1 −  ) +  s (1 −  )3 U 6 

u = C5 ( ) + C6 ( ) , where −12  ( x)  12 U 1





* So,  =  1 − 0

3   =  −   10 120 

u d U (from K–P approximation)

 3 2C  =  + 2   10 120  Similarly,  37  2  − −   315 945 9072 

 = 

 37 2C2 4C22  =  + −   315 945 9072 

And   = 

 = 

U  2+   6 

2C2  U 2−  Here, the answer is written in terms of C2.  6 

(Ans)

Here are the results obtained by Pohlhansen approximation.

 0.686 = x Re U = constant,

du =0 dx

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=

37  and 315

And   =

* =

1.75 x Re

1 0.686 U 2 2 Re

So, we have been asked

 = nominal boundary thickness  = 0.99U

(Ans)

Here, U is the flow velocity.  * = displacement boundary thickness

* =

1.75 x Re

(Ans)

Re = Reynolds number

 = momentum boundary thickness

=

0.686 x Re

(Ans)

And stress  w is calculated as follows:

w =

1 0.686 U 2 2 Re

(Ans)

b. Drag force per unit width can be calculated as follows: Let L be length of plate and b be its width. So, FD = CD

So,

FD 1 0.686 = CD U 2 L b 2 Re

CD =

2 FD U 2 Lb

CD =

0.686 Re

1 0.686 U 2  Lb 2 Re

(Ans)

(Ans)

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4-64 (a) Consider steady, incompressible, laminar flow over a flat plate at 20°C, 1 atm, and 0.1 m/s. Does water or air produce a thicker boundary layer at the trailing edge? Show your work and any assumptions that you need to make.

(b) Consider the attached two-dimensional laminar motion over two streamlined airfoils as pictured below.

If the only difference between the two airfoils is that Airfoil # 1 has twice the length of Airfoil # 2, what is the relation between their boundary-layer thicknesses at the trailing edge? You may treat the two airfoils as two-dimensional flat plates. You may also resort to any suitable laminar correlation. Solution: a. Given data is At 20⁰C, 1 atm vx = 0.1m/s

Let x = 1 m, where x is the length of the plate. Given fluid is Air:  air = 1.214 kg/m3

air = 1.812 10−5 N-s/m 2 Water:  water = 998 kg/m3 Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -73-

 water = 1.002 10−3 N-s/m 2 Let the velocity profile be linear, i.e.,

x  v

So,  air = 3.46

 air

u y = . v s

1.812 10−5 1 = 3.46 1.214  0.1

 air = 0.4224 m

(Ans)

Now for water,

 Water = 3.46

1.002 10−3 1 998  0.1

 water = 0.1096 m So,

(Ans)

 ain 0.4224 =  water 0.1096

 air = 3.85401  water  air = 3.86 water

(Ans)

b. First streamlined airfoil length = L Second streamlined airfoil length = 2L We know that

 =c

x  v

i.e.,   x



1 x = 1 2 x2



1 1 = 2 2

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 2 = 21

(Ans)

4-65 For nearly a century, the boundary-layer problem for flow over a flat plate, which was first reduced to a third-order nonlinear differential equation by Blasius (1908), could only be solved numerically or through the use of infinite series. In seeking practical alternatives, several polynomial approximations have been attempted, and these are mainly attributed to Pohlhausen (1921). Because the polynomial profiles are intended to capture the flow behavior inside the boundary layer only, the profiles are expressed as piecewise solutions of the form: 2 3 4  y  y  y  y u c0 + c + c2   + c3   + c4   ; 0  y   =        U    y  1;

To ensure smooth merging between the velocity in the boundary layer ( u ) and the outer, freestream velocity (U ) , several boundary conditions have been suggested by Pohlhausen (1921). These are: (1) No slip condition at the wall: u ( x,0 ) = 0 . (2) Matching the far-field velocity at the boundary-layer edge: u ( x,  ) = U . (3) Smooth merging with the far field, where the shear vanishes: (4) Axial momentum balance at the wall: 

 2u y 2

layer equation, and where the pressure gradient

= y =0

u y

= 0. y −b

dp , which is recovered from the boundarydx

dp dU in conformance with Euler’s far= − U dx dx

field equation. (5) No further variations in the slope of the velocity at the boundary-layer edge (zero curvature):  2u = 0 Note that in the case of a zero pressure gradient, which corresponds to the Blasius y 2 y = problem, condition # 4 reduces to

 2u y 2

= 0 . For this particular case, determine which of the y =0

five boundary conditions (1 to 5) are satisfied by the following Pohlhausen profiles:

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u y y2 =2 − 2 U   u 3y y3 = − 3 (b) Third-order, cubic polynomial: U 2 2

(a) Second-order, quadratic polynomial:

(c) Fourth-order, quartic polynomial:

u y y3 y 4 = 2 −2 3 + 4 U   

. Hint: Although condition #5 is true as y →  , it proves to be inaccurate at the edge of the boundary layer, where the slope is still changing. Fortuitously, this condition only affects the quartic Pohlhausen polynomial, which explains why it tends to be less precise in predicting boundary-layer characteristics than its lower-order counterparts (cf. Table 4-1). Solution: The given profile is 2

3

u y  y  y  y = c0 + c1 + c2   + c3   + c4   U       

4

The boundary conditions are as follows: (i) y = 0, u = 0 (ii) y =  , u = U (iii) y =  ,

u =0 y

 2u (iv) y = 0, 2 = 0 y  2u (v) y =  , 2 = 0 y Conditions (iv) and (v) are valid when the pressure gradient is zero, i.e.,

P = 0. x

The given Pohlhausen profile is u y y2 = 2 − 2 (Second-order, quadratic polynomial) (a) U   Therefore, we have Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -76-

u y  y = c0 + c1 + c2   U   

2

Applying boundary conditions, we have (i) y = 0, u = 0

0 = c0 + c1 ( 0) + c2 ( 0)

c0 = 0

(1)

(ii)

y = 6, u = U 1 = 0 + c1 ( /  ) + c2 ( /  )2

1 = c1 + c2

(2)

(iii)

y =,

u =0 y

 u  1 c1 y = + 2c2 2     y  U 

0=

c1

+ 2c2

0=

c1

+2



y

2 c  0 = 1 + 2c2 2   

c2



c1 + 2c2 = 0

(3)

Using Eq (2) and Eq (3), we obtain

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Therefore, the profile is u y  y = 2 −  U   

2

This profile satisfies the first three boundary conditions. P  0. x

In this case,

(b)

(Ans. a)

u 3y y3 = − 3 (Third-order, cubic polynomial) U 2 2

Therefore, we have 2

u  y  y  y = c0 + c1   + c2   + c3   U      

3

Applying boundary conditions, we have (i) y = 0, u = 0

0 = c0 + c1 ( 0) + c2 ( 0) + c3 ( 0)

c0 = 0 (ii)

(4)

y =  , u = U

1 = 0 + c1 ( /  ) + c2 ( /  )2 + c3 ( /  )3

1 = c1 + c2 + c3

(5)

(iii) y =,

u =0 y

 u  1 c1 y y2  = + 2 c + 3 c   2 3 2 3  y  U   0=

c1



+ 2c2

 2 + 3 c 3 2 3

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0=

c1



+2

c2



+3

c3



c1 + 2c2 + 3c3 = 0

(6)

(iv)

y = 0, 1 U

 2u =0 y 2

  2u  c3 y c2  2  = 0+2 2 +6 3    y 

0=2

c2

2

c2 = 0

+0

(7)

Using equations (4) and (5), we obtain 3 −1 c1 = , c3 = 2 2

Therefore, the profile is u 3 y  1 y  =  −   U 2    2   

3

This profile satisfies the first four boundary conditions. (c)

(Ans. b)

u y y3 y 4 = 2 − 2 3 + 4 (Fourth-order, quartic polynomial) U   

Therefore, we have 2

3

u  y  y  y  y = c0 + c1   + c2   + c3   + c4   U        

4

Applying boundary conditions, we have (i)

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y = 0, u = 0

c0 = 0 (ii)

(8)

y =  , u = U

1 = c1 + c2 + c3 + c4

(9)

(iii) y =,

u =0 y

 u  1 c1 y y2 y3 = 0 + + 2c2 2 + 3c3 3 + 4c4 4        y  U 

0=

c1

+2



c2



+3

c3



+4

c4



c1 + 2c2 + 3c3 + 4c4 = 0

(10)

(iv)

y = 0, 1 U

 2u =0 y 2

  2u  c3 y c2 c4 y 2 = 0 + 2 + 6 + 12  2 2 3 4  y 

0=2

c2

2

c2 = 0

+0+0

(11)

(v)

y =, 1 U

 2u =0 y 2

  2u  c3 y c2 c4 y 2 = 2 + 6 + 12  2 2 3 4  y 

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0=2

2

c2



2

c2

+6

+6

c3

2



2

c3

+ 12

3

+ 12

c4

2

c4 2

4

=0

c2 + 3c3 + 6c4 = 0 Using equation (11), we have

c3 + 2c4 = 0

(12)

Using equations (9), (10), and (12), we obtain

c1 = 1, c3 = −2, c4 = 1 Therefore, the profile is 3

u y  y  y = − 2  +   U     

4

This profile satisfies all five boundary conditions.

(Ans. c)

4-66 Consider the quartic Pohlhausen profile for the laminar, incompressible flow over a flat plate with a norzero pressure gradient: u y = c0 + c1 + c2 2 + c3 3 + c4 4 ; 0    1 where  = U 

(a) Apply Pohlhausen’s five boundary conditions to show that the unknown coefficients must be 1 1 1 1 c0 = 0, c1 = 2 + Λ, c2 = − Λ, c3 = Λ − 2,   and c4 = 1 − Λ . 6 2 6 2

 2 dp  2 dU where Λ = − = U dx  dx Here, Λ represents the Pohlhausen pressure gradient parameter. Simplify your solution into the form: 1 u / U = 2 − 2 3 +  4 + Λ (1 −  )3 6 Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -81-

(b) For what value of Λ will flow separation occur? Recall that separation occurs when u / y = 0 at y = 0 . (c) Plot u / U versus  for Λ = −12, −6,0, +6, +12 .

Solution:

u = c0 + c1 + c2 2 + c3 3 + c4 4 U

 ( x, y ) =

y



The boundary conditions are as follows: (i) u ( x,0 ) = 0 (ii) u ( x,  ) = u ( x ) (iii)

u ( x,  ) = 0 y

(iv)

−u ( x ) du ( x )  2u x, 0 ) = 2 ( y U dx

(v)

 2u ( x,  ) = 0 y 2

The Kármán–Pohlhausen approximation yields the following results:

0 = c1 −Λ = 2c2 1 = c1 + c2 + c3 + c4 + c0 0 = c1 + 2c2 + 3c3 + 4c4 0 = 2c2 + 6c3 + 12c4 c0 = 0 Λ 6 Λ c2 = − 2 Λ c3 = −2 + 2

c1 = 2 +

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c4 = 1 −

Λ 6

The profile is rewritten as follows: u ( y) U u ( y) U

= 1 − (1 +  )(1 −  ) + 3

Λ 3  (1 −  ) 6

(Ans. a)

= f ( ) +ΛG ( )   , where −12  Λ ( x )  12

Therefore,  0

1

 =   1 − *

u U

Λ   3 2c3    3  d =   −  =  +    10 120   10 120 

Similarly, we have  37 2c3  37 4c32  Λ Λ2  − − =  + −     315 945 9072   315 945 9072 

 =  and

w = u

U  Λ U  2c  2+  = u 2−  8 6 8 6

Here are the results obtained by Pohlhausen approximation.

 2 = 0.4 x

ux U  0.686 x

=

Re

U = constant dU =0 dx

=

37  315

* =

1.75 x Re

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 =

1 0.686 U 2 2 Re

δ = nominal boundary thickness = 0.99U δ* = displacement boundary thickness =

 = momentum thickness =

 w = shear stress =

1.75x Re

0.686 x Re

1 0.686 U 2 2 Re

Let L be the length of the plate and b be its width. 1 0.686 U 2 Lb 2 Re FD 1 0.686 = CD U 2 L b 2 Re

FD = CD

0.686  1 2  U 2 Lb  2 2 FD Re   = 0.686 CD = = 2 2 U Lb U Lb Re

4-67

(Ans.)

Consider the quartic Pohlhausen profile for the laminar, incompressible flow over a flat plate at zero incidence and with no pressure gradient. Using dimensionless variables, we have: F ( ) =

y u 2 − 2 3 +  4 ; 0    1 where  = =  U 1; 1  

(a) Show that the dimensionless displacement thickness, momentum thickness, and shape factor lead to:

 * 1 u  = = 1−  0 U *

(b)

 u  *  d = 0.3  = =   0U  1

u  1 −  U

*  and H = = 2.554 d  = 0.1175   

Show that the Kármán–Pohlhausen momentum-integral equation reduces to w d  U dF = U 2 * = s; s  = dimensionless velocity slope at the wall.  dx  d   −0

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(c) After separating variables and integrating from  ( 0 ) = 0 to  ( x ) , show that the boundary-layer thickness can be written as a function of the local Reynolds number, Rex = Ux /  , specifically,

 x

=

a ; a = 5.836 Rex

(d) Recalling that the friction coefficient is C f = 2 w / ( U 2 ) = 2 *d / dx , and that

 = ax1/2 / U /  , show that d ( a / 2 ) = and C f = dx Rex

b ; b = a * = 0.6855 Rex

(e) Recalling that the displacement thickness can be deduced from  / x =   / x , show that *

*

c ; c = a * = 1.751 Rex

=

x

*

(f) Recalling that the momentum thickness can be deduced from  / x =  * / x , show that

 x

=

d ; d = a * = 0.6855 Rex

v  u = −  0y dy , integrate the axial profile to show that the transverse U x U velocity can be written as

(g) Using continuity,

v  2 3 4 4 5  ( a / 2 )  2 3 4 4 5  2.9178 =  −  +   =  −  +   U  2 5  Rex  2 5  Rex

(h) Show that the maximum transverse velocity occurs at the edge of the boundary layer where it can be obtained from v U

= max

v U

= y −

( 3a / 20 ) = 0.8753 Rex

Rex

(i) Verify that the errors in predicting  / x and C f are 17 and 3.2 percent relative to the classic Blasius results. Solution: The given quartic Pohlhausen profile is Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -85-

F ( ) =

u 2 − 2 3 +  4 ; 0    1 = U 1; 1  

where

=

y



0   1 0  

 

 * =  1 − 0

u U

y



1 0  y  

3 4  y     y   y   3 4   d  = 1 − 2  − 2  +  d  = 1 − 2 − 2 +   0  0           dy     

(

)

3 4   y  y   y   2  4 5   * =  dy −  2 dy − 2 3 dy +  4 dy  =  −  − 3 + 4  0 0  0  5   0    2

    3  * =  −  − +  = 2 5  10  * 3 = = 0.3  10

*  = = 0.3  *

 =



 =



0

0

2

 u u u u 1 −  dy =0 −   dy U U U U  2

 u  u dy −    dy 0 U U  

  y 3 y3 y 4  = −   2 − 2 3 + 4  dy 10 0      2

 = 0.1175 

* =

1 u   u =  1 −  d = 0.1175 0  U U

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H=

* *  0.3 = = = 2.554    0.1175

(Ans. a)

(b) According to the von Karman equation, we have

w

=2

 x

1 U 2 2 2   w2 = 2 U dx 

w d u = . where  w = u 2 U dx y

y =0

u 2 y 2 y3 y 4 = − 3 + 4 U     du  1 2 6 y2 3y4  = − 3 + 4     dy  V    du  1 2 du 2U =  =    dy   dy  y =0 V 

w = u

2



=

2U



2uU d =  U2 dx 2u d =  U dx  2udx =   u d

  2  x =   U d

 2  x = U  *

2 2

where

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* =

   =  *  d =  *d  =

U s 

(Ans. b)

(c) For the given profile, we have

w d = 2 U  dx 1  2uU  U 2  

 d  2   =   dx  15 

u dx =  d U 

15

 d =

15u dx U 

Integrating the above equation, we obtain

2 2

=

15ux +C U 

When x = 0,  = 0 , C = 0.

 x

30 5.836 a = = Re x Re x Re x

=

(d) c f =



w 1 U 2 2

=2

(Ans. c)

d dx

2 w d = 2 * 2 U  dx

where

* =

 d d   =  *  =*  dx dx

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   1/2   d ax   w2 =*  U  dx  U     v 

ax1/2 U / v

=

d =

1 −1/2 x d 2 dx  = dx U v

a

Using the value of

cf =

a 2 xU v

d in the above equation, we have dx

w a * b * ( a / 2) =   c = = f 2 U  xU Re x Re x

(Ans. d)

v

(e) Now,

* x

= *

 x

* x

* x

* x

= *

a  a , where = x Re Re x

=

c , where c = a * = 1.751 Re x

=

c 1.751 = Re x Re x

(f)  * =

* =

(Ans. e)

 

 x   a d 0.6855  =* =* = = x x x Re x Re x Re x

(Ans. f)

(g) The given profile is Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -89-

u = 2 − 2 3 +  4 U

=

0   1

y



Using continuity equation, we have

v  u = −  dy U x 0 U y

y y y y v −   y y 3 y 4   −  y y3 y4  = 2 − 2 + dy = 2 dy − 2 dy +      0  3 0  4 dy  U x  0    3  4   x  0   2 4 5 −  2 y 2 y y  = − 3 + 4  x  2 4 5 



It is given that

 x

=

a ax  = Re x Re x

Therefore,

we        v −  y 2 y4 y 5  −  y 2 y4 y5 = − 3 3 + 5 5 = − + 2a x 5a x  x  ax 2a 3 x 3 5a 5 x 5 U x  ax 3/2 5/2  Re  Re3/2 Re5/2  V x  V x  x  V x  x x        u  u   u  

v −  y 2 =  U x  a v  y2 = U  a

V 1 u

V

y 4  V  − 3  x 2a  u 

1 y 4  V  −   u 2 x3/2 2a3  u 

3/2

3/2

1 y 5  V  +   x3/ 2 5a5  u 

3 y 5  V  +   2 x5/2 5a5  u 

5/2

5/2

have         

1   x5/ 2  5 1   2 x7 / 2 

v  1 y 2 ax 1 3 y 4 a3 x3 1 1 y 5 a5 x5 1  = − +  U  2 a  x 2 4 a3  3 x 4 2 a5  5 x6 

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v  2 3 4 4 5  ( a / 2) =  −  +   U  2 5  Re x

(1)

v  2 3 4 4 5  2.9178 =  −  +   U  2 5  Re x

(Ans. g)

(h) Differentiating with respect to y gives 2 3 y 4 4 y5  ( a / 2)  2 y y3 y 4  ( a / 2)  d y = − + = − 6 + 4  2   2  4 5  5  Re x   2 4  5  Re x  dy  

d u  dy  U

To obtain the stationary point, set

(2)

 u   = 0. y  U 

 2y y3 y4  − 6 + 4 =0 y   2 4  5  2

−

2

16 y 3



5

8 y3

3 8 y3

18 y 2

4 −

+

18 y 2



−

9 y2

−

9 y2

2

4

16 y 3

5 =−

=0 2

2

= −1 +1 = 0

3 2 8 3 − 9 2 + 1 = 0

This equation is satisfied when  = 1 . Therefore,

y



=1 y = 

v v   =   U  max  U  y =

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v  y 2 3 y 4 4 y5  ( a / 2)   2 3  4 4  5  ( a / 2)  3 4  ( a / 2) = − + = − + = 1 − +    U   2 2  4 5  5  Re x   2 2  4 5  5  Re x  2 5  Re x v  3a  1 0.8753 =  = U  20  Re x Re x

(Ans. h)

    −      x Quartic Pohlhausen  x Blasius % error in = 100% x      x Blasius =

5.836 − 5 100%=16.61% 17% 5

% error in c f =

(c ) f

Quartic Pohlhausen

(c ) f

=

− (c f )

(Ans. i)

Blasius

100%

Blasius

0.6855 − 0.664 100% = 3.162%  3.2% 0.664

(Ans. i)

4-68 Consider the quartic Majdalani–Xuan profile for the laminar, incompressible flow over a flat plate: 1 4 5 3 y u   − +  ; 0   1 where  = [Majdalani and Xuan (2020)] F ( ) = =  3 3  U  1   1; (a) Show that the dimensionless displacement thickness, momentum thickness, and shape factor lead to:

 * 1 u   1u  u  * *  = = 1 −  d = 0.3500,  = =  1 −  d = 0.1337, and H = = 2.618  0 U   0U  U   *

(b) Show that

a =  Rex / x = 4.993, b = C f Rex =  Rex / x = 0.6676, and c =  * Rex / x = 1.748 . (c) Calculate the total skin friction force and the drag coefficient for a plate of length L. Express ReL = UL /  your result in terms of (d) Using the continuity equation, integrate the axial velocity profile to show that

v  2 3 4 4 5  ( a / 2 )  2 3 4 4 5  2.9178 =  −  +   =  −  +   U  2 5  Rex  2 5  Rex Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -92-

(e) Show that the maximum transverse velocity occurs at the edge of the boundary layer where it can be obtained from

v U

= max

v U

= y −

( 7a / 40 ) = 0.8738 Rex

Rex

(f) Verify that the errors in predicting  / x and C f are 1.7 . and 0.53 percent relative to the classic Blasius results in Table 4-4. (g) Show that the quartic Majdalani–Xuan profile satisfies all of the fundamental boundary conditions except for the vanishing curvature at the edge of the boundary layer. Explain why better predictions are obtained when this condition is relaxed. Hint: Although the zero-curvature boundary condition is true as y →  , it is inaccurate at the edge of the boundary layer, where the velocity slope continues to change. By relaxing this condition, Majdalani and Xuan ( 2020 ) are able to derive a simple approximation that is more accurate than Pohlhausen’s quartic polynomial by one order of magnitude. Consequently, the Majdalani–Xuan polynomial profile may be viewed as a practical equivalent to the Blasius solution in several engineering applications. Solution: The given profile is 1 4 5 3 u   − +  ; 0   1 F ( ) = =  3 3 U  1   1;

where

=

y



0   1 0   u  =  1 − 0  U *



y



1 0  y  

 5  y   y 3 1  y  4     1 4  5 3 d = 0 1 −   −  +   d = 0 1 −    −   +    dy 3    3   3       3     

3   y 5  y  5 2  4 1  y4  5   * =  dy −   dy −  3 dy +  4 dy  =  −  − 3+ 4 0 0  3 0 3 0    6 4 15 

 5    21 − + =  6 4 15  60

* = −

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 * 21 = = 0.3500 =  *  60

 =



 =



0

0

2

 u u u u 1 −  dy =0 −   dy U U U U  2

 u  u dy −    dy 0 U U  



  5 y y3 1 y 4   5 y y3 1 y 4   =  − 3+ dy −  0  3  −  3 + 3  4  dy 3  34  0 2

* =

 = 0.1337 

H=

* *  0.35 = = = 2.618    0.1337

(Ans. a)

(b) For the given profile, we have

w 1 U 2 2

u

du dy

1 U 2 2

=

d dx

=

d dx

5 U 3  = d  11    1 U 2 dx  31250  2

u

31250 u d = 33 U  dx

 d =

31250 u dx 33 U 

Integrating the above equation, we obtain

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2 2

=

31250 ux +C 33 U 

When x = 0,  = 0 , C = 0.

 x

 Re x 1 43.52 = a= = 4.993 x Re x Re x

62500 33

=

4.993 ( 0.1337 ) 0.6676  Re x a * b cf = = = =  b = c f Re x = x Re x Re x Re x Re x Now

* x



*

x

* x

= *

 x

= * =

a Re x

c Re x

where

 *a = c c = 0.35  4.993 c = 1.748 c=

 * Re x x

= 1.748

(Ans. b)

(d) The given profile is u 5 1 =  − 3 +  4 U 3 3

= a=

y





Re x x a = 4.993

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Using continuity equation, we have v   5 y y3 1 y 4 = −  − + U x 0  3   3 3  4 y

  dy 

y y 3 y −  5 y y 1 y4  = dy −  3 dy +  4 dy   x  3 0   30  0

=

−  5 y 2 y4 y5  − +   x  3 2 4 3 15 4 

It is given that

 x

=

a  = Re x

ax Re x

Therefore, we have 2  y 5 Re5/2 v −  5 y Re x y 4 Re3/2 x x = − +   U x  6 ax 4a 3 x 3 15a 5 x5 

v −  5 y 2 =  U x  6 a v  5 y2 = U  6a

V 1 u

y 4  V  − 3  x 4a  u 

V  −1 

3/2

1 y 5  V  +   x3/2 15a 5  u 

5/2

1   x5/2 

y 4  V   −3  y 5  V   − +      u  2 x3/2  4a3  u   2 x5/2  15a5  u 

v  2 3 4 4 5  ( a / 2) =  −  +   U  2 3  Re x

3/2

5/2

 −5    7/2    2 x  

(1)

u  y 2 3 y 4 4 y5  ( a / 2) = − +  U   2 2  4 3  5  Re x

(Ans. d)

Differentiating with respect to y gives d  u  2y y3 y 4  ( a / 2)   =  2 −6 4 +4 5  dy  U       Re x

(2)

7 ( a / 2 )  7a  1 0.8738 v v = =   =  =  Re x  U max  U  y = 20 Re x  40  Re x

(Ans. e)

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    −      x Quartic M − x  x Blasius % error in = 100% x      x Blasius =

1.748 − 1.72 100%=1.7% 1.72

% error in c f =

=

(c ) f

Quartic M − x

(c f

)

(Ans. f) − (c f

)

Blasius

100%

Blasius

0.668 − 0.664 100% = 0.53% 0.664

(Ans. f)

4-69 Consider the continuous Majdalani–Xuan exponential profile for the laminar, incompressible flow over a flat plate: y − s  (1+ 12 s  + 2 ) u = 1 − e ; =   U [Majdalani and Xuan (2020)]  Re  a  x s = =  1.6304, 0      x 2 2  (a) Show that the dimensionless displacement thickness, momentum thickness, and shape factor lead to:  u   u u * =  1 −  d = 0.35068 ,  * =  1 −  d = 0.13559 , and H = 2.5862  0 U 0  U  U (b) Show that  Rex / x = 4.9039 , C f Rex =  Rex / x = 0.66494 , and  * Rex / x = 1.7197 (c) Verify that the errors in predicting  / x , C f , and  * / x are negligible, being, respectively, 0.12%, 0.13%, and 0.06% relative to the computed Blasius solution in Table 4-4. (d) In addition to the fundamental boundary conditions that are satisfied by this profile, verify that the continuous Majdalani–Xuan exponential profile captures the behavior of the Blasius solution very robustly: (1) it exhibits the same slope as the exact Blasius solution at the wall, (2) it equals, as it should, 0.99U at y =  , (3) it matches the slope of the Blasius velocity profile at the edge of the boundary layer with reasonable accuracy, and (4) it approximates the curvature of the exact Blasius solution at y =  , where the Blasius curvature does not vanish. (e) Contrary to the piecewise approximations that are confined to the 0    1 range, this profile remains valid beyond the edge of the boundary layer, specifically as  →  , with u / U → 1 asymptotically, as it should, thus allowing for the smooth and seamless merging of the viscous solution with the far-field freestream. Explain why this simple analytic profile is Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -97-

practically equivalent to the “exact” Blasius solution given the small truncation error entailed in the derivation of the Blasius equation itself. (f) By writing  = y U / (2 x ) = C  , and recalling that  = a  x / U , show that the conversion constant between the Blasius similarity variable and



is simply

C = a / 2  3.47 . (g) Plot this solution for 0      , where  = 1.5 corresponds to   5.2 written in terms of the Blasius variable. Add to your plot the Blasius derivative function df / d , which represents the normalized axial velocity. Solution: (a) We are asked to determine the dimensionless displacement and momentum thicknesses as well as the shape factor for the given velocity profile. Using the defining expressions for  * ,  *, and H , we have:    − s  (1+ 1 s  + 2 ) u   2 d  0.35068  * =  1 −  d = 0 e *  0  U   2.5862 and H = Ans. (a)  *   2 2 1 1 − s  (1+ 2 s  + ) − s  (1+ 2 s  + )  *   =  1− e e d  0.13559  0  The above integrals may be evaluated numerically using an available symbolic program or integrator (e.g., Mathematica, Matlab, Maple, Python, C, Fortran, Basic, etc.). An example of a Mathematica code is provided below.

(

)

s:=1.63040 u[x]:=1-Exp[-s*x*(1+(1/2)*s*x+x^2)] EtaStar:=NIntegrate[1-u[x],{x,0,Infinity}] ThetaStar:=NIntegrate[u[x]*(1-u[x]),{x,0,Infinity}] H:=EtaStar/ThetaStar a:=Sqrt[2*s/ThetaStar] b:=Sqrt[2*ThetaStar*s] c:=a*ThetaStar*H ErrorDeltax := (Abs[(a-4.90999)]/4.90999)*100 ErrorCf:=(Abs[(b-0.66411)]/0.66411)*100 ErrorDeltaxStar:=(Abs[(c-1.72079)]/1.72079)*100 (b) Given the properties in Part (a) and noting that s  1.6304 stands for the slope of F at the wall, we can calculate   2s Rex =  4.9039 a = x *    b = C f Rex = Ans. (b) Rex = 2 * s  0.66494 x   * Rex = a * H  1.7197 c = x  Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -98-

(c) We can compute the relative errors using the robustly computed or ‘exact’ Blasius values located in Table 4-4 as benchmarks. We get   Rex / x − 4.90999   error:  100%  0.124% x 4.90999  C f Rex − 0.66411  C error:  100%  0.125% Ans. (c)  f 0.66411   *  * Rex / x − 1.72079   error:  100%  0.064%  x 1.72079  These coincide with the reported relative errors and show that the relative errors are virtually insignificant. (d) We are asked to verify 4 conditions, namely, 1. F  ( 0 )  s 2. F (1) = 0.99 3. F  (1) matches the slope of the Blasius velocity with reasonable accuracy 4. F  (1) approximates the curvature at y =  First, we may proceed by defining the function and its derivatives using    1 2   F ( ) = 1 − exp  − s  1 + s  +     2      3  2s 2   2 2 −s   F  ( ) = 3 s +  s + s exp  − s − 2     2 2  F  ( ) = − s  9 3 s + 6 2 s 2 +  s s 2 + 6 + 2 s 2 − 3  exp  − 3 s −  s −  s     2   Checking these conditions one-by-one, we find F  ( 0 ) = s identically;

(

)

(

) (

)

  1  F (1) = 1 − exp  − s 1 + s    0.99 ;   2   s2  2 F  (1) = s + 4s exp − − 2s   0.0932 which mimics Blasius’s value of 0.0904 in Table 4 2  2;  −s 2  3 2 F  (1) = − s s + 8s + 15s − 6 exp  − 2s   −0.729 which mimics the -0.709 Blasius value.  2  Ans. (d)

(

)

(

)

(e) The Blasius equation is accompanied by a small truncation error that depends on the size of the flow Reynolds number. This is due to its original derivation requiring the neglect of the Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -99-

second-order diffusion term in the streamwise direction. We also recall that the Blasius solution is derived assuming laminar conditions, thus ensuring that the Reynolds number remains of modest size. As a result, the truncation error associated with the Blasius equation itself may be viewed to be of order Re −1 = O(10−5 ) . As shown in Table 4-4, since the largest L2 error associated with the present analytical solution is O (10 −4 ) , it makes its predictions practically equivalent to the Blasius solution obtained numerically, especially that the latter carries a finite truncation error that depends on the discretization scheme employed by the user. Ans. (e) (f) The following derivation enables us to specify the conversion constant between the Blasius similarity variable  and  , i.e., the fractional distance spanning the viscous boundary layer:

 = C  , y

y U = C , and so 2 x 

C a 4.9039 U =  3.47 Ans. (f) = or C = 2 x a x / U 2 2

Since  = ax / Rex = a x / U from Part (b). (g) The results shown below resemble those of Fig. 4-13 and confirm that the Majdalani–Xuan and Blasius solutions remain imperceptible, not only within the viscous layer, but also in the far field as  → . This may be viewed as being significant due to the simplicity of the Majdalani– Xuan profile. In fact, a feature article on this subject has appeared in the Physics of Fluids where the systematic approach leading to a family of exponentially-decaying formulations by Majdalani and Xuan (2020) is explained.

1

0.8

0.6 u U

Blasius (1908) Majdalani–Xuan (2020)

0.4

0.2

0

0

0.2

0.4

0.6

0.8 y/

1

1.2

1.4

1.6 Ans. (g)

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4-70 Consider the sinusoidal Schlichting profile for the laminar, incompressible flow over a flat plate:

u  y  = sin  ; 0  y   U  2 

[Schlichting (1979)]

(a) Using continuity, v = −   0y u dy , integrate the axial profile to show that the transverse x

U

U

velocity can be written as

 Re x v   y   y  y   a = cos   + sin   − 1 ;a = U   2  2 x  2    Re x (b) Plot u / U and v / U versus y /  and show that the maximum transverse velocity occurs at the edge of the boundary layer where it can be obtained from v U

= max

v U

= y =

 − 2 ( a / 2 ) 0.871264 =  Re x Re x

(c) Evaluate at (v / U )max at x = 0.6 m and  = 0.006 m . Hint: First show that, irrespective of the Reynolds number,

v U

= max

 −2  2 x

Solution: Given that

u  y  = sin  ; 0  y   U  2  (a)

v v  yu     y  2   y  = = −  dy = −  0y sin  − 1  dy =  cos  0 U U x U x x  2  2   y

    v   y 2ax = cos − 1 2ax   Re x U x    Re x  

     

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v    y V  2a ux  =  cos − 1  U x  2ax ux   V     y Re x = − sin   2ax   =

   − y     y Re x     + cos    2x     2ax  

  a  − 1     Re x

y  y    y   a sin   + cos   − 1 2 x  2    2    Re x

 y   y    Re x    y   a =   sin  + cos    − 1   2    Re x   2 x  2   a v  y a  y   y   = sin   + cos   − 1 U  2  2   2    Re x

(b)

(Ans. a)

v  y a  y   y   = sin   + cos   − 1 U  2  2   2    Re x

Differentiating the above equation with respect to y, we obtain 2 d v   y     y        =  − sin    + sin   +   y cos  dy  U    2  2 2  2   2   2 2 d  v       y  a   =   y cos   dy  U   2   2    Re x

To obtain the stationary points, set

 a     Re x

(1)

d v  =0. dy  U 

   2   y  a =0   y cos    2  2   Re       x y = 0, 

Differentiating equation (1) with respect to y, we obtain 3 2 d2  v       y       y  a = − y sin +         cos   2  dy  U    2 y   2   2   2    Re x

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Substituting y =  , we obtain 3  a d2  v      = − y + 0       dy 2  U    2 y    Re x

3

d2  v     a 0 =  y 2  dy  U   2 y   Re x Therefore, v is maximum at y =  . U

v v   =   U  max  U  y =    = cos   2 

    sin  +  2  2

a   a  − 2 ( a / 2)    = 0 + − 1 =  − 1 2   Re x  Re x    Re x 

(Ans. b)

(Ans. b) (c) Given that a =

 Rex x

 −2    −2 v =   =   2 x   2  x  U max At x = 0.6 m and  = 0.006 m,

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v   − 2  0.006 0.18169 =   =  100  U max  2  0.6 4-71

0.182%

(Ans. c)

Consider the cubic Pohlhausen profile for the laminar, incompressible flow over a flat plate:

u 3 y y3 = − ;0 y  U 2 2 3

[Pohlhausen 1921]

(a) Using continuity, integrate the axial profile to show that the transverse velocity can be written as v 3  y2 y4  a 3  y2 y4 = 2 2 − 4  = 2 2 − 4 U 16     Re x 16   

  x

(b) Plot u / U and v / U versus y /  and show that the maximum transverse velocity occurs at the edge of the boundary layer where it can be obtained from v U

= max

 U

= y −

3a 0.8702 = 16 Re x Re x

(c) Evaluate at (v / U )max at x = 0.7 m and  = 0.007 m . Solution: Given that

u 3 y y3 = − ;0 y  U 2 2 3 (a) Using continuity, we have v  yu =− dy U x 0 U

v  y  3y y3    3y2 y4  = − 0  − −  dy = −   U x  2 2 3  x  4 8 3 

Also,

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a=

 Re x

3 =

x

 =

ax = Re x

ax = V x u

u ax V x

a 3 x3/2  V     u 

3/2

3/2 3/2  2 V  4  V  4  V   y 3 y 3y      2 4 v  u −  u   = 1 y 2 V −  u  = 3 2 y − y  a =−    U x  4a x 8a 3 x 3/2  8 x 3/2 u 16a 3 x 5/2 16   2  4  Re x    

v 3  y2 y4   = 2 2 − 4  U 16    x

(Ans. a)

2 4   (b) v = 3  2 y 2 − y 4  a U 16     Re x

Differentiating the above equation with respect to y, we obtain d  v  3  4 y 4 y4  a −    = dy  U  16   2  4  Re x

To obtain the stationary points, set

(1)

d v  =0. dy  U 

3  4 y 4 y4  a − =0   16   2  4  Re x

4 y 4 y4 − =0 2 4 4 y2  y2 1 −  2  2

 =0 

4 y2 =0 y =0 2

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1−

y2 y2 = 0  = 1  y =  2 2

y = 0,  , −

Differentiating equation (1) with respect to y, we obtain d 2  v  3  4 16 y 3  a − 4   =  dy 2  U  16   2   Re x

Substituting y =  , we obtain d 2  v  3  4 16  a −  0  =  dy 2  U  16   2   Re x

Therefore, v is maximum at y =  . U

v v   =   U  max  U  y = =

3  2 4  a 3 = 2 2 − 4  16     Re x 16

a 0.8702 = Re x Re x

(Ans. b)

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(Ans. b) v 3 a (c)   =  U max 16 Re x

a=

 Re x a   = x Re x x

3 v   =  U max 16 x At x = 0.7 m and  = 0.007 m,

3  0.007  v   =   = 0.188%  U max 16  0.7 

(Ans. c)

4-72 Consider the continuous, one-parameter, tangent hyperbolic Yun profile for laminar flow over a flat plate:

( )

u y = tanh s ;  = , s = 1.6304, 0     U 

[Yun (2010)]

(a) Show that the dimensionless displacement thickness, momentum thickness, and shape factor lead to: 

 0

 * = 1 −

u U



u u  *  d = 0.4251,  =  1 − U U  0

  d = 0.1882 and H = 2.2589 

(b) Show that

 Re x / x = 4.1624, C f Re x =  Re x / x = 0.7834, and  * Re x / x = 1.7696

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(c) Verify that the errors in predicting  / x and C f are 15.2 and 18 percent relative to the computed Blasius values in Table 4 − 4 . Solution: The given profile is

( )

u y = tanh s ;  = , s = 1.6304, 0     U 

=

y



0    0 

y



  0  y 

The possible condition is 0  y  

 

 * =  1 − 0

u U

(

( ))

    y  d = 0 1 − tanh s d = 0 1 − tanh  s  dy     

  * 0.9748  * =  − ln cosh s − ln1 =  − 0.9748 − 0  = 1 − = 0.4251  1.6304 s s  =



 =



0

0

2

 u u u u 1 −  d =0 −   dy U U U U  2

  y  y  tanh  s dy −   tanh  s   dy 0      

Integrating and solving gives * =

 = 0.1882 

 *  *  0.4251 H= = = = 2.2587    0.1882

c fx =

w 1 U 2 2



=

(Ans. a)

u y

d d  = 1 dx U 2 dx 2

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 =

   y  U  tanh  s   y  y   y  y     d  =   tanh  s  − tan 2 h  s  dy  1 dx  0      U 2 2

Applying the boundary condition x = 0,  = 0 , we obtain 

Re x = 4.1624 = a

x

cf =

a *  * = Re x b = a  x Re x

cf =

b 0 1882  4 1624 = Re x Re x

c f Re x = 0.7834 Also,

* x

= *

 x

= *

a Re x

 *a = c  c = 0.4251 4.1624 = 1.7696

* x

=

c 1.7696 = Re x Re x

* Re x = 1.7696 x

(Ans. b)

      −    x  Yun  x  Blasius % error in = 100% x      x  Blasius =

4.1624 − 4.90999 100% = 15.22% 4.90999

(Ans. c)

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% error in c f =

=

(c ) f

Yun

− (c f )

(c f )

Blasius

100%

Blasius

0.783 − 0.66411 100% = 18% 0.66411

(Ans. c)

4-73 Consider the continuous, two-parameter tangent hyperbolic Savaş profile for laminar flow over a flat plate: u  y 5/3 3/5 = tanh ( )  ;  = ,  = s = 1.6304, 0     [Savaş (2012)]   U  (a) Show that the dimensionless displacement thickness, momentum thickness, and shape factor lead to:  u   u u * =  1 −  d = 0.3509 ,  * =  1 −  d = 0.1363 , and H = 2.5738 0 0 U U  U (b) Show that  Rex / x = 4.8906 , C f Rex =  Rex / x = 0.6667 , and  * Rex / x = 1.7161 (c) Verify that the errors in predicting  / x and C f are 0.40% and 0.40% relative to the computed Blasius solution in Table 4-4. Solution: (a) We are asked to find the dimensionless displacement and momentum thicknesses as well as the shape factor for the given velocity profile. Using the defining expressions for  * ,  *, and H , we have:

 

 

   u 5/3 3/5     * =  1 −  d  =  1 −  tanh ( )  d  0.3509 0 * 0  U    2.5738 Ans. (a) and H =   * 5/3 3/5 5/3 3/5  *   tanh ( )  1 −  tanh ( )  d   0.1363  =       0 

The above integrals may be evaluated numerically using an available symbolic program or integrator. An example of a Mathematica code is provided below. s:=1.6304 u[x]:=(Tanh[(s*x)^(5/3)])^(3/5) EtaStar:=NIntegrate[1-u[x],{x,0,Infinity}] ThetaStar:=NIntegrate[u[x]*(1-u[x]),{x,0,Infinity}] H:=EtaStar/ThetaStar a:=Sqrt[2*s/ThetaStar] b:=Sqrt[2*ThetaStar*s] c:=a*ThetaStar*H ErrorDeltax := (Abs[(a-4.90999)]/4.90999)*100 Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -110-

ErrorCf:=(Abs[(b-0.66411)]/0.66411)*100 (b) Given the properties in Part (a) and noting that  = s  1.6304 stands for the slope of F at the wall, we can write a =  Re / x = 2s /  *  4.8906 x   * Ans. (b) b = C f Rex =  Rex / x = 2 s  0.6667  * c =  * Rex / x = a H  1.7161  (c) We can compute the relative errors using the robustly computed or ‘exact’ Blasius values located in Table 4-4 as benchmarks. We get   Rex / x − 4.90999   error: 100%  0.40% x 4.90999 Ans. (c)  C Re − 0.66411  f x 100%  0.40% C f error: 0.66411  These coincide with the reported relative errors in Table 4-4.

4-74 Consider the continuous, one-parameter Moeini–Chamani profile for laminar flow over a flat plate: u y = erf ( );  = ,  = 1.59261, 0     [Moeini and Chamani (2017)] U  (a) Show that the dimensionless displacement thickness, momentum thickness, and shape factor lead to:  u   u u * =  1 −  d = 0.3543 ,  * =  1 −  d = 0.1467 , and H = 2.4142 0 0 U U  U (b) Show that  Rex / x = 4.9491 , C f Rex =  Rex / x = 0.7262 , and  * Rex / x = 1.7532 (c) Verify that the errors in predicting  / x and C f are 0.8% and 9.4% relative to the computed Blasius solution in Table 4-4. Solution: (a) We are asked to find the dimensionless displacement and momentum thicknesses as well as the shape factor for the given velocity profile. Using the defining expressions for  * ,  *, and H , we have:

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     u  * =  1 −  d = 0 1 − erf ( )  d  0.3543 0  U       *  u u  = Ans. (a)   1 −  d = 0  erf ( ) 1 − erf ( )  d  0.1467 U U    0   *  2.4142 H = *  The above integrals may be evaluated numerically using an available symbolic program or integrator (e.g., Mathematica, Matlab, Maple, Python, C, Fortran, etc.). An example of a Mathematica code is provided below. s:=1.7971 u[x]:=Erf[1.59621*x]] EtaStar:=NIntegrate[1-u[x],{x,0,Infinity}] ThetaStar:=NIntegrate[u[x]*(1-u[x]),{x,0,Infinity}] H:=EtaStar/ThetaStar a:=Sqrt[2*s/ThetaStar] b:=Sqrt[2*ThetaStar*s] c:=a*ThetaStar*H ErrorDeltax := (Abs[(a-4.90999)]/4.90999)*100 ErrorCf:=(Abs[(b-0.66411)]/0.66411)*100

(b) Given the properties in Part (a), we must find the value of s for this particular profile. The parameter s is defined as dF ( 0 ) s= d In view of the Moeini–Chamani error function, we have 2 dF dF (0) 2 −( 0 )2 d 2 = = e  1.7971  erf ( ) = e− and so s = d d d   This value of s can be used in the remaining analysis. For example, we readily find a =  Re / x = 2s /  *  4.9491 x   * Ans. (b) b = C f Rex =  Rex / x = 2 s  0.7262  * c =  * Rex / x = a H  1.7532  (c) We can compute the relative errors using the robustly computed or ‘exact’ Blasius values located in Table 4-4 as benchmarks. We obtain   Rex / x − 4.90999   error: 100%  0.80% x 4.90999 Ans. (c)  C f Rex − 0.66411  100%  9.4% C f error: 0.66411  These coincide with the reported relative errors. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -112-

4-75 Because of the popularity of approximate profiles in laminar boundary-layer studies, let us consider a quartic mean flow in the case of no pressure gradient:

u y = c0 + s + c2 2 + c3 3 + c4 4 ;  = 0   1 U  where  represents the fractional distance within the laminar boundary layer, and s  c1 = F (0) is the axial velocity slope at the wall, also known as the Blasius constant or connection parameter. F ( ) =

(a) Show that the particular profile that will satisfy Pohlhausen’s four essential conditions described in Sec. 4–1 must be:

F ( ) =  s + ( 4 − 3s )  3 + ( 2s − 3)  4 Hint: These conditions translate into F (0) = F (1) = F (0) = 0 and F (1) = 1 . (b) Show that the normalized boundary-layer properties,  * ,  * , and H , can be written as a direct function of s : 1 * 3   *   = 0 (1 − F ) d  = 20 ( 4 − s )  1  4 13s 19 s 2   *  = F (1 − F ) d  = + −  0  35 210 630   189 ( 4 − s )  * * = = H    * 144 + 78s − 38s 2 

( normalized displacement thickness ) ( normalized momentum thickness ) ( momentum shape factor )

(c) Making use of C f = 2d / dx , show that the wall shear stress and skin friction coefficient can be written as

w =

U s 

and

Cf =

 4 13s 19s 2  d 2 s = 2 + −  U  35 210 630  dx

(d) Use separation of variables to identify

630s  dx 2 72 + 39s − 19s U (e) After integrating from  (0) = 0 to  ( x) , show that the boundary-layer thickness and its

 d =

derivative can be written as functions of the local Reynolds number, Rex = Ux /  , and the Blasius constant s :

 x

=

a 1260s ; a= 72 + 39s − 19s 2 Rex

and

d  ( a / 2) = dx Rex

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(f) Using the definition of the friction coefficient, C f = 2 w / ( U 2 ) , show that

(

b s ; b= 72 + 39 s − 19 s 2 315 Rex

Cf =

)

(g) Recalling that the displacement thickness may be specified as  * =  * , show that

* x

=

c 3a ; c = (4 − s) 20 Rex

(h) Recalling that the momentum thickness may be deduced from  =  * , show that  d = ; d = a * = b x Rex (i) Using the continuity equation  = −

 y udy , show that the transverse velocity component x 0

becomes



3 2 s  a =   2 + ( 4 − 3s )  4 + ( 2s − 3)  5  U 4 8 5  Rex (j) Show that the maximum value of the transverse velocity occurs at the edge of the boundary layer where

     =  U max U

= y =

e 567 s ; e = (1 − 14 s ) 360 + 195s − 95s 2 Rex

(k) Calculate (a, b, c, d , e) for the two values of s = 2 and 5 / 3 , which correspond to the quartic Pohlhausen and Majdalani–Xuan profiles. Blasius values in Table 4-1.

Compare your results to the classic

Solution: (a) In order to determine the 4 unknown polynomial coefficients, we revert back to Pohlhausen’s essential boundary conditions and apply them systematically to the generic quartic polynomial: u F ( ) = = c0 + s + c2 2 + c3 3 + c4 4 U Condition 1: F (0) = 0 (no slip condition at the wall) c0 + s (0) + c2 (0) 2 + c3 (0)3 + c4 (0) 4 = 0

or

c0 = 0

Condition 4: F (0) = 0 (zero pressure gradient in the far field) F ( ) = 2c2 + 6c3 + 12c4 2 ,

2c2 + 6c3 (0) + 12c4 (0) 2 = 0

and so

c2 = 0

This leaves us with only 2 unknowns. Condition 2: F (1) = 1 (matching the freestream velocity at the boundary-layer edge) s + c3 + c4 = 1 and so c4 = 1 − s − c3 Condition 3: F (1) = 0 (negligible shear at the boundary-layer edge)

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F ( ) = s + 2c2 + 3c3 2 + 4c4 3 ,

F (1) = s + 3c3 + 4c4 = 0

and so

c4 = − 14 s − 34 c3

Equating the last two expressions for c4 , we get c4 = − 14 s − 34 c3 = 1 − s − c3 ; this enables us to deduce c3 = 4 − 3s and c4 = 2 s − 3 We thus arrive at

F ( ) =  s + ( 4 − 3s )  3 + ( 2s − 3)  4

Ans. (a)

(b) The normalized displacement thickness can be computed from: 1 1 * * = =  (1 − F )d  =  1 − s − (4 − 3s ) 3 − (2 s − 3) 4  d  0 0  1

 2 4 5  s 3s 2s 3 3 3s =  − s − (4 − 3s ) − (2s − 3)  = 1 − − 1 + − + − 0 = − 2 4 5 0 2 4 5 5 5 20 

3 (4 − s) 20 The normalized momentum thickness can be computed from:

* =

or

*=

Ans. (b)

1  =  F (1 − F )d  0 1

=   s + (4 − 3s) 3 + (2 s − 3) 4  1 − s − (4 − 3s) 3 − (2 s − 3) 4  d 0 1

=  ( s − s 2 2 + 4 3 − 3s 3 − 3 4 − 6 s 4 + 6 s 2 4 + 6 s 5 − 4 s 2 5 − 16 6 + 24 s 6 0

−9s 2 6 + 24 7 − 34s 7 + 12s 2 7 − 9 8 + 12s 8 − 4s 2 8 )d 4 13s 19s 2 + − 35 210 630 The momentum shape factor can be computed from: 3 3 (4 − s) (4 − s) 189 ( 4 − s ) * 20 20 H= = = or H = 2 2 4 13s 19 s 144 + 78s − 38s  144 + 78s − 38s 2 + − 35 210 630 1260 or

*=

Ans. (b)

Ans. (b)

(c) Here we derive two different forms of the skin friction coefficient starting with the fundamental shear stress definition: u   u U U  U F U and so  w =  s Ans. (c) w =  = = y y =0   y    =0       y =0 Therefore, based on the shear stress definition, we have, on the one hand,

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

U

s w 2 s  Cf = 1 =1 = 2 2 U 2 U 2 U

2 s U On the other hand, using Kármán’s momentum-integral equation, we have d ( * )  4 13s 19s 2  d d d Cf = 2 =2 = 2 * = 2 + −  dx dx dx  35 210 630  dx

and so

Cf =

Ans. (c)

(d) Equating the two forms of C f found in Part (c), we get:

s s  4 13s 19s 2  d 2 s U U dx =  d  dx =  d  and so = 2 + − or  2 72 4 + 39 13 s − 19 s 2 s 19 s U 35 210 63 0 dx   + − 35 210 630 630 This can be rearranged into 630s  Ans. (d)  d = dx 2 72 + 39s − 19s U (e) In order to express the boundary layer as a function of the local Reynolds number, the result in Part (d) can be integrated from the leading edge of the plate to any location x . This can be accomplished by taking x  2 630 s x 630s   0  d = 0 72 + 39s −19s2 U dx or 2 = 72 + 39s − 19s 2 U Solving for the displacement thickness, we find  1260s 1260s x 1260s x  = = = 2 2 2 2 x 72 + 39s − 19s Ux 72 + 39s − 19s U 72 + 39s − 19s Ux and so

 x

=

1260s 72 + 39s − 19s 2

1 a 1260s = ;a = 72 + 39s − 19s 2 Rex Rex

As for the spatial gradient of the disturbance thickness, we obtain ( 12 a ) d d d  1260s x  1 1260s  and so = =  =  dx dx  72 + 39s − 19s 2 U  2 72 + 39s − 19s 2 Ux dx Rex

Ans. (e)

Ans. (e)

(f) We start with the outcome of Part (c):  4 13s 19 s 2  d  4 13s 19 s 2   1 1260 s   Cf = 2 + − = 2 −    +   2  35 210 630  dx  35 210 630   2 72 + 39 s − 19 s Ux   72 + 39 s − 19 s 2   1260 s   =    2 630    72 + 39 s − 19 s Ux 

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or

Cf =

s 1 b s 72 + 39s − 19s 2 ) = ;b = 72 + 39s − 19s 2 ) ( ( 315 315 Rex Rex

(g) Starting with  * =  * , we may simply divide both side by x and simplify. We get 3a (4 − s) * c 3a *  *  a  3  20 or = ;c = (4 − s) =  =   ( 4 − s ) =  Re   20 x 20 x x Rex Rex  x  

Ans. (f)

Ans. (g)

(h) Starting with  =  * , we may simply divide both side by x and simplify. We get  a  a *   =   * =   * =  Re  x x Rex x  



Ans. (h)

Specifically, we have   72 + 39 − 19s 2    a   4 13s 19s 2   1   1260s = − =  +      x  Rex   35 210 630   Rex   72 + 39s − 19s 2   630  Collecting terms and simplifying, we retrieve  b d s   1  s = = ;d = 72 + 39s − 19s 2 = 72 + 39s − 19s 2 ) or  ( x  Rex  315 x 3 1 5 Rex Rex Ans. (h)

(

)

(i) To obtain the normal velocity component, the continuity equation may be integrated, starting  y with  = −  udy : x 0 y 3 4    y u      y   y  y   =−  dy = −   s   + (4 − 3s)   + (2s − 3)    dy  x  U x 0 U          0   2 4 5     s   y   4 − 3s   y   2s − 3   y     = −      +  +  3     4   x    2      4      5        2 4 5   d   s   y   d    4 − 3s   y   d    2s − 3   y   = − −  − 3 − 4              dx   4      dx   5        dx   2     4  d    s  2 3 4 5 =     + ( 4 − 3s )  + ( 2s − 3)   4 5  dx   2  

=

(a / 2)  s  2 3 4 4 5  2   + 4 ( 4 − 3s )  + 5 ( 2s − 3)   Rex   

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

3 2 s  a =   2 + ( 4 − 3s )  4 + ( 2 s − 3)  5  U 4 8 5  Rex

Ans. (i)

(j) To show that the maximum transverse velocity occurs at the boundary-layer edge, we set d ( / U ) =0 d Substituting the results of Part (i) and differentiating, we get d ( / U )  s 3  a =   + ( 4 − 3s )  3 + 2 ( 2s − 3)  4  d 2 2  Rex s 3  a =   + ( 4 − 3s )  2 + 2 ( 2s − 3)  3  =0 2 2  Rex

Accordingly, the extrema will correspond to the roots of  [ 12 s + 32 ( 4 − 3s )  2 + 2 ( 2s − 3)  3 ] = 0 . We get  3 11s 2 − 16 s − s (always negative, unphysical) − 2 8 s − 12 ( )  0 (minimum of  where  = 0)   = (maximum of  within 0    1) 1  2  s + 3 11s − 16 s (negative or out of range for 1  s  2)  2 ( 8s − 12 )

Interestingly, for s = 2 , we recover a double root at  = 1, namely, when both the 3rd and 4th roots predict the maximum to occur at the edge of the boundary layer. For s  2 , the initial slope starts gravitating away from the true Blasius slope of 1.63; as such, it starts affecting the physical behavior of the transverse velocity and should not be considered. Having determined the location of the peak transverse velocity, it is now possible to evaluate its magnitude. We have      =  U  max U  =1 Forthwith, the maximum transverse velocity may be calculated at the edge of the boundary layer by taking:  3 9 4 6 1260s 1 1 = s+ − s+ s−  2 U  =1  4 2 8 5 5  72 + 39 s − 19s Rex 1260 s  12  =  (1 − 14 s )  2  40  72 + 39s − 19s

or

    =  U max

  3 2 1 1260s 1 = (1 − 4 s )      10  72 + 39s − 19s 2 Rex 

e 567 s ; e = (1 − 14 s ) 360 + 195s − 95s 2 Rex

 1   Rex 

Ans. (j)

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(k) An evaluation of these parameters and their errors relative to the Blasius values are provided in the table below.

Profile

a=

 x

Re x

b = C f Re x = d =

 x

Re x

c=

* x

Re x

  e=  Re x  U max

5.83558515 16.72%

0.685449684 3.16%

1.750675545 1.80%

0.8753 1.73%

 −  3 + 13  4

4.993399337 0.13%

0.667547918 0.53%

1.747689767 1.63%

0.8738 1.55%

Blasius (1908)

5

0.664

1.72

0.8604

2 −  3 +  4 5 3

Clearly, the Majdalani–Xuan profile outperforms Pohlhausen’s quartic polynomial in predicting the classic Blasius values, especially in its expressions for the boundary-layer thickness and skin friction coefficient. For example, instead of a 17% error in predicting the boundary-layer thickness, the use of the Majdalani–Xuan profile leads to a rather negligible error which, at 0.13%, proves to be two orders of magnitude smaller than that associated with Pohlhausen’s quartic polynomial.

4-76 Because of the popularity of approximate solutions in laminar boundary-layer studies, let us consider a quartic mean-flow profile in the presence of a pressure gradient:

u y = c0 + s + c2 2 + c3 3 + c4 4 ;  = 0   1 U  where  represents the fractional distance within the laminar boundary layer, and s  c1 = F (0) is the axial velocity slope at the wall which can be written as function of the zero-pressure slope s0 and a pressure gradient correction: F ( ) =

s = s0 + s1 (a) Show that the particular profile that will satisfy Pohlhausen’s four essential conditions with a non-vanishing pressure gradient [Sec. 4–1] must be:

F ( ) = s − 12  2 + ( 4 − 3s +  )  3 + ( 2s − 3 − 12  )  4 ;  = −

 2 dp  2 dU = U dx  dx

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Hint: These translate into F (0) = F (1) = 0 , F (1) = 1 and F (0) = − , where  is Pohlhausen’s pressure parameter. (b) Show that flow separation will occur for  = − s0 / s1 . Recall that separation occurs when

u / y = 0 at y = 0. (c) Show that the normalized boundary-layer properties,  * ,  * , and H , can be written as a function of s and  :

1 ( 36 − 9s +  ) ( normalized displacement thickness ) 60 19s 2 (  − 12 )( 24 +  ) s (156 + 17 ) *  =− − + 630 2520 2520 s (156 + 17 ) − (  − 12 )( 24 +  ) − 76 s 2 = ( normalized momentum thickness ) 2520 42 ( 9s − 36 −  ) H= ( momentum shape factor ) 2 76s + (  − 12 )( 24 +  ) − s (156 + 17 ) (d) Making use of C f = 2 * d  / dx + 2(2 + H ) * (U  / U ) , where U  = dU / dx , show that the

* =

wall shear stress and skin friction coefficient can be written as, U  2 w = s and so C f = 1 w 2 = s = 2 * d / dx + 2(2 + H ) * (U  / U ) U    U 2 or

U

 3 − (17 s + 9 )  2 + ( 76 s 2 + 33s − 1044 )  + 1260 s d    2s = − 2 2 + H  = − 4 ( )   dx  U    *  2 + (12 − 17 s )  + 76s 2 − 156s − 288

(e) For the case of Hiemenz flow where the velocity outside the boundary layer is given by U = ax and V = −ay in Sec. 3–8.1, the analytic solution leads to a constant boundary-layer thickness, show that  = ( 2 /  )U  = const. (f) Using Pohlhausen’s quartic profile with s = 2 +  / 6 , determine  and the corresponding skin friction coefficient,  w  / U  / (U ) , for the Hiemenz flow; compare your predictions to those obtained from the numerical solution provided in Table 3-4. Show that:

 Pohlhausen = 7.052

and

 Hiemenz = 5.760 and

w  s = = 1.196 U U  

versus

the

numerical

value

of

w  = 1.233 U U 

The latter appears as “ F (0) ” in Table 3-4.

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(g) Using Majdalani–Xuan’s quartic profile with s = 5 / 3 + 0.2075 , determine  and the corresponding skin friction coefficient,  w  / U  / (U ) , for the Hiemenz flow; compare your predictions to those in Part (f ). Show that:  Majdalani-Xuan = 5.761 matches the Hiemenz value and

w  s = = 1.192 is off by U U  

3.3%. (h) Use Thwaites’ approach to predict the value of  and the corresponding skin friction coefficient  w  / U  / (U ) by applying Eqs. (4-139) and (4-140). Compare the skin friction coefficient predicted by Thwaites’ approach to the one obtained from the numerical solution provided in Table 3-4. Show that:

w  6 = S ( ) = 1.195 U U  0.45 (i) Compute the normalized displacement and momentum thicknesses as well as the shape factor using Pohlhausen’s quartic profile for the Hiemenz flow. Show that: * = 0.241,  * = 0.105,  = 0.075 ,

S ( ) = 0.3272 , and

and H =2.308 . (j) Compute the normalized displacement and momentum thicknesses as well as the shape factor using Majdalani–Xuan’s quartic profile for the Hiemenz flow. Show that: * = 0.267,

 * = 0.115, and H = 2.318 . Hint: The reference Hiemenz solutions obtained by numerical integration are * = 0.270,

 * = 0.122, and H = 2.219. Solution: (a) In order to determine the 4 unknown polynomial coefficients, we revert back to Pohlhausen’s essential boundary conditions and apply them systematically to the generic quartic polynomial: u F ( ) = = c0 + s + c2 2 + c3 3 + c4 4 U Condition 1: F (0) = 0 (no slip condition at the wall) c0 + s (0) + c2 (0) 2 + c3 (0)3 + c4 (0) 4 = 0

or

c0 = 0

Condition 2: F (1) = 1 (matching the freestream velocity at the boundary-layer edge) s + c2 + c3 + c4 = 1 and so c2 = 1 − s − c3 − c4 Condition 3: F (1) = 0 (negligible shear at the boundary-layer edge) F ( ) = s + 2c2 + 3c3 2 + 4c4 3 , F (1) = s + 2c2 + 3c3 + 4c4 = 0 Substituting the expression of c2 found through Condition 2, we get c3 = s − 2 − 2c4 . Condition 4: F (0) = − (pressure gradient in the far field) Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -121-

 2 This value of c2 may be readily substituted into the expression found through Condition 2 and F ( ) = 2c2 + 6c3 + 12c4 2 , F (0) = 2c2 = −

and so

c2 = −

retrieve c3 = 1 − s + 2 − c4 . This reduces the problem to two equations with two unknowns for c3 and c4 . Combining this expression for c3 with the expression found through Condition 3, we have  s − 2 − 2c4 + c4 = 1 − s + or c4 = 2s − 3 − 12  2 This enables us to deduce c3 = s − 2 − 2c4 = s − 2 − 2 ( 2s − 3 − 12  ) and so c3 = 4 − 3s +  We thus arrive at F ( ) = s − 12  2 + (4 − 3s + ) 3 + (2s − 3 − 12 ) 4

Ans. (a)

(b) At separation, the flow detaches. This means that there is no change in velocity between the wall and the flow directly above it (zero shear stress). This condition translates into F (0) = 0 . Starting with F ( ) = s −  + 3(4 − 3s +  ) 2 + 4(2 s − 3 − 12  ) 3 setting F (0) = 0 leaves us with F ( ) = s −  (0) + 3(4 − 3s +  )(0) 2 + 4(2 s − 3 − 12  )(0)3 or s = s0 + s1 = 0 We deduce s Ans. (b) =− 0 s1 (c) The normalized displacement thickness can be computed from: 1 * 1 * = =  (1 − F )d  =  1 − s + 12  2 − (4 − 3s +  ) 3 − (2 s − 3 − 12  ) 4  d 0 0  1

 2 1 3 4 5  1 =  − s +  − (4 − 3s + ) − (2s − 3 − 2 )  2 6 4 5 0  s  3s  2s 3  3 3s  = 1 − + −1 + − − + + − 0 = − + 2 6 4 4 5 5 10 5 20 60 1  * = ( 36 − 9s +  ) 60

or

Ans. (c)

The normalized momentum thickness can be computed from:

 *=

1  =  F (1 − F )d  0

1

=   s − 12  2 + (4 − 3s +  ) 3 + (2 s − 3 − 12  ) 4  1 − s + 12  2 − (4 − 3s +  ) 3 − (2 s − 3 − 12  ) 4  d 0 Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -122-

*=

or

s (156 + 17 ) − (  − 12 )( 24 +  ) − 76 s 2 2520

Ans. (c)

The momentum shape factor can be computed from: 1 ( 36 − 9s +  ) 42 ( 36 − 9s +  ) * 60 H= = = 2 s (156 + 17 ) − (  − 12 )( 24 +  ) − 76s  s (156 + 17 ) − (  − 12 )( 24 +  ) − 76s 2 2520 42 ( 9 s − 36 −  ) or Ans. (c) H= 2 76s − s (156 + 17 ) + (  − 12 )( 24 +  ) (d) Here we derive two different forms of the skin friction coefficient starting with the fundamental shear stress definition: u   u U U  U F U Ans. (d) w =  = = = s y y =0   y    =0       y =0 Based on the shear stress definition, we now have, on the one hand, U  s w 2 s 2 s  and so C f = Ans. (d) Cf = 1 =1 = 2 2 U U 2 U 2 U On the other hand, using Kármán’s momentum-integral equation, we have 2  2s d 2 s  2 d  U  =U  + 2 2 + H and so C f = 2 * + 2 ( 2 + H ) *    = ) U   ( * dx U   dx  U    To make further headway, we recall that from the definition of Pohlhausen’s pressure parameter, we have 2  U  2 d  d  d   2  2 d   = U  and so = =    =    dx  U   dx  U   dx     dx     When these groupings are employed, the momentum-integral equation becomes 2s d    2s d  = U   + 2 ( 2 + H )  or U   = − 2(2 + H )  * dx  U    * dx  U   Next, we may substitute H and  * from Part (c) into the right-hand side to obtain: U

and so

d    == dx  U  

2s s (156 +17  ) −( −12 )( 24 + ) −76 s 2 2520

  42 ( 9s − 36 −  ) − 2 2 +  2  76s − s (156 + 17 ) + (  − 12 )( 24 +  ) 

 3 − (17 s + 9 )  2 + ( 76s 2 + 33s − 1044 )  + 1260s d  U   = −4 dx  U    2 + (12 − 17 s )  + 76s 2 − 156s − 288

Ans. (d)

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(e) For the Heimenz flow, both  and U  = a are constant (because U = ax ). This leaves us with  2 dU 2 = =a = const. Ans. (e)  dx  Since a is constant as well as  , the pressure parameter  must also be invariant. (f) Since  is constant for the Hiemenz flow, and U  = a is independent of x , the left-hand side term in Part (d) becomes: d  U  =0 dx  U   We now have, referring to the right-hand side of Part (d):  3 − (17 s + 9 )  2 + ( 76 s 2 + 33s − 1044 )  + 1260 s −4 =0  2 + (12 − 17 s )  + 76 s 2 − 156 s − 288 or

3 − (17 s + 9 )  2 + ( 76s 2 + 33s − 1044 )  + 1260s = 0

At this stage, we may insert s = 2 +  / 6 for Pohlhausen’s quartic profile to obtain 2  3 − 17 ( 2 + 6 ) + 9   2 + 76 ( 2 + 6 ) + 33 ( 2 + 6 ) − 1044   + 1260 ( 2 + 6 ) = 0   5 3 79 2  +  − 464 + 2520 = 0 or 18 6 Based on the resulting cubic polynomial, three roots may be deduced, namely, 1 = −72.26,  2 = 17.803, and 3 = 7.052 Both 1 and  2 fall outside the range of acceptable  , which is −12    12 . The only acceptable outcome is, therefore,  Pohlhausen = 7.052 Ans. (f) This value entails a 22.43% error compared to the numerical value of  Hiemenz = 5.760 . As for the skin friction coefficient, we get   s 2+ 6 (C f ) Pohlhausen = w = = = 1.1957 Ans. (f) U U    This value entails a 3.00% error compared to the numerical value of (C f ) Hiemenz = 1.233 , which appears as F (0) in Table 3-4. (g) We now repeat the same analysis using s = 5 / 3 + 0.2075 for Majdalani–Xuan’s quartic profile. Substitution into 3 − (17 s + 9 )  2 + ( 76s 2 + 33s − 1044 )  + 1260s = 0 yields 3 − 17 ( 53 + 0.2075 ) + 9  2 2 + 76 ( 53 + 0.2075 ) + 33 ( 53 + 0.2075 ) − 1044   + 1260 ( 53 + 0.2075 ) = 0   3 2 or 0.744775 + 22.08083 − 516.43888 + 2100 = 0 Based on the resulting cubic polynomial, three roots may be deduced, namely,

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1 = −46.04,  2 = 10.63 and  3 = 5.761 Both 1 and  2 fall outside the range of acceptable  , which is −8.03    8.03 . The only acceptable outcome is, therefore, Ans. (g)  Majdalani-Xuan = 5.761 This value entails a 0.017% error compared to the numerical value of  Hiemenz = 5.760 . As for the skin friction coefficient, we get 5 + 0.2075  v s (C f ) Majdalani-Xuan = w = =3 = 1.1924 Ans. (g) U U    This value entails a 3.3% error compared to the numerical value of (C f ) Hiemenz = 1.233 , which appears as F (0) in Table 3-4. Clearly, by using the Majdalani–Xuan profile in lieu of Pohlhausen’s, we gain a staggering 3 orders of magnitude improvement in predicting the pressure parameter for the Hiemenz flow. (h) Based on Thwaites’ method, the momentum thickness may be written as a function of the mean flow using Eq. (4-159), 0.45 x 5 2 = U dx U 6 0 2  2U   2 dU dU . = Then based on Thwaites’ parameter,  = , we can retrieve dx =    dx Substituting back into Thwaites’ integral, we have x

 0.45  5   2  = U  dU  , 6  U 0    2

=

0.45 x 5 U dU U 6 0

and so

=

0.45  U 6  0.45 3 =  = U6  6  6 40

 = 0.075 or As for Thwaites’ shear correlation, which is given by Eq. (4-161), it becomes 0.62 0.62 S ( ) = (  + 0.09 ) = ( 0.075 + 0.09 ) S ( ) = 0.3272

and so

Ans. (h)

Ans. (h)

This enables us to deduce the wall shear stress using Eq. (4-160), namely, U w = S ( )



The skin friction coefficient follows. We get

 U  S ( )        = S ( )  Cf = w = U U  U U  U

However, since  =  2U  /  , , we have  / U  =  2 /  . Inserting this ratio under the radical leaves us with S ( )  2 S ( ) 0.3272 and so C f = 1.195 Ans. (h) Cf = = =    0.075 Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -125-

(i) Before computing the normalized displacement and momentum thicknesses for the Hiemenz flow, we recall that, from Part (c), we have  1  * = ( 36 − 9 s +  ) 60   −76s 2 + s (156 + 17  ) − (  − 12 )( 24 +  ) *  =  2520   42 ( 9 s − 36 −  ) H = 2 76s − s (156 + 17 ) + (  − 12 )( 24 +  )  For Pohlhausen’s profile,  Pohlhausen = 7.052 and s = 2 + 6 . Backward substitution yields: 1  or  * = 0.241  * = 36 − 9 ( 2 + 7.052 Ans. (i) 6 ) + 7.052  60 Next, we calculate 2 7.052 −76 ( 2 + 7.052 6 ) + ( 2 + 6 ) (156 + 177.052 ) − 7.052 ( 7.052 − 12 )( 24 + 7.052 ) *  = 2520  * = 0.105 or Ans. (i) At this point, the shape factor may be readily evaluated. We get 42 9 ( 2 + 7.052 6 ) − 36 − 7.052   H= 7.052 2 7.052 76 ( 2 + 6 ) − ( 2 + 6 ) (156 + 17  7.052 ) + ( 7.052 − 12 )( 24 + 7.052 ) H = 2.308

and so

Ans. (i)

(j) For the Majdalani–Xuan profile,  Majdalani-Xuan = 5.761 and s = + 0.2075 . Backward 5 3

substitution yields:

* =

1 (36 − 9 ( 53 + 0.2075  5.761) +  ) 60

or

 * = 0.267

Ans. (j)

Next, we calculate 2 −76 ( 53 + 0.2075  5.761) + ( 53 + 0.2075  5.761) (156 + 17 ) − (  − 12 )( 24 +  ) *= 2520  * = 0.115 or Ans. (j) Finally, the shape factor may be readily evaluated. We obtain 42 9 ( 53 + 0.2075  5.761) − 36 − 5.761 H= 2 76 ( 53 + 0.2075  5.761) − ( 53 + 0.2075  5.761) (156 + 17  5.761) + ( 5.761 − 12 )( 24 + 5.761) and so

H = 2.318

Ans. (j)

The errors associated with the Majdalani–Xuan profile in predicting the key boundary-layer characteristics may be determined relative to the reference Hiemenz values obtained by numerical integration. These are * = 0.270 and  * = 0.122 . In comparison, the Majdalani– Xuan characteristics entail errors of 1.1% and 5.7%, respectively. These constitute one order of Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -126-

magnitude improvements over the same two properties computed using Pohlhausen’s quartic polynomial and yielding, respectively, relative errors of 11% and 14%.

4-77 Consider a shape-preserving mean-flow profile in the laminar boundary layer over a flat plate with no pressure gradient: u y = = F ( ); 0   1 U  As usual, let s = F (0) denote the axial velocity slope at the wall. For an arbitrary profile F ( ) , the normalized boundary-layer properties, * =  * / ,  * =  /  , and H =  * / *, consist of pure constants that can be determined from their defining integrals once F ( ) is specified. (a) Recalling that C f = 2d / dx , show that the wall shear stress and skin friction coefficient can be reduced to:  2 s (definition)  U  U w = s and C f =   2 * d (momentum-integral relation)  dx  (b) By equating the defining expression for C f and its momentum-integral relation to  , show

that

 d =

s dx U *

(c) By integrating from the plate’s leading edge to any station x , show that  ( x) and its derivative can be expressed in terms of Rex = Ux /  ,  * , and the slope s :

 x

=

a 2s ; a= * Rex

and

d  ( a / 2) = dx Rex

(d) Using the defining expression for the friction coefficient, C f = 2 w / ( U 2 ) , show that Cf =

b ; b = 2 s * Rex

(e) Prove the two identities ab = 2s and b / a =  * . (f) Recalling that the displacement and momentum thicknesses may be specified as  * =  * and

 =  * , show that * x

=

c ; c = a * Rex

and

 x

=

d ; d = a * = b Rex

(g) Using CD , plate = 2 ( L) / L from Eq. (4-10) and  ( x) above, show that Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -127-

2a * ReL (h) Evaluate (a, b, c) for the Majdalani–Xuan quartic profile and compare your results to the CD, plate =

classic Blasius values. Solution: (a) From the definition of the wall shear stress, we have, on the one hand, u   u U U  U F U w =  = = = s y y =0   y    =0       y =0 The corresponding skin friction coefficient returns U  s 2 s w Cf = 1 =1  2= 2 U 2 U 2 U On the other hand, using Kármán’s momentum-integral equation, we have  U  d d C f = 2 * + 2 ( 2 + H ) *    = 2 * dx dx U 

Ans. (a)

Ans. (a)

Ans. (a)

(b) Equating the two expressions of C f from Part (a), we get, in general,

2 *

d 2 s = dx  U

and so

 d =

s dx U *

Ans. (b)

(c) Using the result of Part (b), both sides of the equation may be integrated from the leading edge of the plate to any position x . We get: x 2    s dx ,  =  xs and so  = 2s  x or  = 2s   d  =  0 2 U * x * U  * Ux 0 U * This shows that in general, the disturbance thickness coefficient can be easily predicted using

 x

=

a 2s ;a= * Rex

Ans. (c)

(d) Using the second expression for the skin friction coefficient from Part (a), we have 1 2s d 2s * * C f = 2 * = 2 * 2  = dx Rex Rex and so

Cf =

b ; b = 2s * Rex

Ans. (d)

(e) It is straightforward to show that Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -128-

b 2s * 2s  *  * = = = * a 2s 2s * (f) Using the displacement thickness definition and dividing both sides by x , we have *  a *  * =  * and so = * = x x Rex ab =

2s 2s * = 2s *

and

*

hence

x

=

c ; c = a * Rex

Ans. (e)

Ans. (f)

Similarly, using the momentum thickness definition and dividing both sides by x , we have

 =  * and so 

hence

x

=

  * a * =  = x x Rex

d ; d = a * = b Rex

Ans. (f)

(g) Using CD , plate = 2 ( L) / L from Eq. (4-10) and  ( x) from Part (f) at x = L , we have: 2 2 a *L CD, plate =  ( L) = L L ReL Simplifying, we get 2a * Ans. (g) CD, plate = ReL (h) Before evaluating the key boundary-layer properties for the Majdalani–Xuan profile, we recall from the foregoing analysis that: 2s a= ; b = 2s * ; c = a* ; ab = 2s *



For the Majdalani–Xuan profile, the slope is set at s = 5 / 3 . This means 3 4 13s 19 s 2 − = 0.1337 * = ( 4 − s ) = 0.35 and  * = + 35 210 630 20 By substituting these values into the above expressions, we find: a = 4.993; b = 0.668; c = 1.748

Ans. (h)

For the classic Blasius (1908) solution:

a = 5; b = 0.664; c = 1.72 The boundary-layer predictions based on the Majdalani–Xuan profile deviate from the classic Blasius values for a, b and c by 0.13%, 0.53% and 1.6%, respectively. These percentages depend on the use of double precision in computing all properties directly from their defining expressions. If 10 significant digits are retained before computing the errors in Table 4-1, one would obtain the following results: Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -129-

Table 4-1 Boundary-layer predictions from five piecewise analytic profiles with their errors relative to the classic Blasius values showing 10 significant digits

u = F ( ) U

2 −  2 3 2

 − 12  3

* =

sin ( 12  )

5 3

 −  3 + 13  4

Blasius (1908)

* =

 

H=

* 

 x

Re x

Cf

Re x

* x

Re x

L2 error

0.333333333 0.133333333 2.500000000 5.477225575 0.730296743 1.825741858 0.020 3.1% 0.25% 3.5% 9.5% 10% 6.1% 0.375000000 0.139285714 2.692307692 4.640954808 0.646418705 1.740358053 0.034 9.0%

2 − 2 3 +  4

* 

4.7%

4.0%

7.2%

2.6%

1.2%

0.300000000 0.117460317 2.554054054 5.835585150 0.685449684 1.750675545 0.054 13% 12% 1.4% 17% 3.2% 1.8% 0.363380227 0.136619772 2.659792366 4.795326227 0.655136377 1.742526735 0.021 5.6% 2.7% 2.7% 4.1% 1.3% 1.3% 0.350000000 0.133686067 2.618073878 4.993399337 0.667547918 1.747689767 0.008 1.7%

0.52%

1.1%

0.13%

0.53%

1.6%

0.344

0.133

2.590

5.000

0.664

1.720

n/a

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CHAPTER 5. THE STABILITY OF LAMINAR FLOWS

P5-1

While holding ( g , 1, 2 , J

)

constant, show that the right hand side of Eq. (5-9) has a /2

minimum at the wavenumber  −  g ( 1 − 2 ) /J  . Find experimental data somewhere and estimate this “critical” wavelength and velocity difference for air blowing over gasoline. Solution The equation (5-9) to be studied is repeated here:

(U1 − U 2 )

2

g ( 1 −  2 ) +  2J  ( 1 +  2 )  

(5-9)

1 2

Instead of >, we will use an equality ( = ) to find the minimum value of (U1 − U 2 ) which we will 2

(

)

call Y for short. The equation is of the form Y = a + b 2 / . Differentiating with respect to  , we find that dY /d = 0 when  2 = a/b. Thus our immediate result:

 2 |min velocity =

g ( 1 − 2 ) ( 1 + 2 ) / ( 12 ) = g ( 1 − 2 ) J

( 1 + 2 ) / ( 12 )

Ans.

J

2 Plot (U1 − U 2 ) versus dimensionless wavenumber  J /  g ( 1 −  2 )

1/2

The curve is rather flat, but the minimum is at  =  g ( 1 −  2 ) /J 

1/2

as follows:

, as predicted.

(b) For air blowing over gasoline, 1  680 kg/m3 , 2  1.2 kg/m3 , J  0.022 N/m. We may calculate a minimum at   550 m−1 = 2 / , or a wavelength   0.011 m. This corresponds to a minimum value of (U1 − U 2 )  4.5 m/s.

Ans. (b)

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P5-2 Show that, if the upper and lower velocities in Fig. 5-2 are negligible, and if surface tension is neglected, a disturbance of the interface will propagate at the phase speed

g  ( 1 − 2 ) 2 ( 1 + 2 )

c=

where  is the wavelength of the disturbance. Discuss what might happen if 1  2 . Estimate this propagation speed for an air-water interface when the wavelength is 3 meters. Solution: If surface tension and the velocities are negligible, then we are studying the effect of gravity on an interface between two still fluids of different density. The expression for wave frequency, from Eq. (5-8), reduces to

 −

 g ( 1 − 2 ) ( 1 + 2 )

If an interfacial wave is produced, its propagation speed is

C=

 T

=

2 /T  = 2 / 

Thus, for the present case of an interfacial travelling wave,

C=

g ( 1 − 2 ) g  ( 1 − 2 )  =i =i   ( 1 + 2 ) 2 ( 1 + 2 )

Ans.

This is the speed of a deep-water wave, far from any upper or lower boundaries. For air layered over fresh water, 1  998 kg/m3 , 2  1.2 kg/m3. If  = 3 m, the propagation speed is

C  9.81( 3m )( 998 − 1.2 ) / 2 ( 998 + 1.2 )

1/2

= 2.16 m/s.

Ans.

If 1  2 , then the argument of the square root is positive and the wave is unstable. Presumably the two layers will overturn so that the heavy fluid goes to the bottom. Ans.

P5-3

Derive a linearized disturbance equation for the celebrated van der Pol equation:

d2X dt

2

(

+ X = 0, ) dX dt

+ C X2 − 1

C = constant

(1)

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Assume that X o (t) is a known solution to (1). Substitute the proposed solution X o +  into (1), where  ( t ) is a (small) disturbance. Write out all the resulting terms. Solution:

(

)

(

)

Xo" +  " + C Xo2 − 1 Xo' + C Xo2 − 1 + 2  ' + 2 Xo CXo '  + 2CX o 2 + + CXo ' 2 + C2 ' + Xo +  = 0 where the prime ( ') denotes differentiation with respect to time. The sum of the boldface terms vanishes, because X o is a known solution to (1). What remains is the disturbance equation, which we linearize by neglecting power and products of e and e' . What then remains is the desired linearized disturbance equation:

(

)

 ''+ C X o2 − 1  '+ (1 + 2CX o X o ' )  = 0

(2) (Ans.)

This is a second order linear differential equation with variable (oscillatory) coefficients. The original solution Xo ( t ) is stable if the solutions to (2) do not grow indefinitely with time. It helps

(

)

a little to modify (2) by defining a new variable Z = exp   (1/2 ) C Xo2 − 1 dt  , resulting in a   differential equation without a first-derivative term:

Z" + f ( t ) Z = 0,

where f ( t ) = 1 + CXo Xo ' −

(

)

1 2 2 C Xo − 1 4

(4)

The coefficient f ( t ) is periodic, thus (4) is related to the Mathieu Equation. We will not proceed any further with the analysis. It has been proven in the literature that solutions of the van der Pol equation (1) are stable, that is, bounded, for any positive value of the constant C.

P5-4

Verify the Orr-Sommerfeld equation (5-18) in the text, page 347

Solution: Begin with the linearized two-dimensional (parallel-flow) equations (5-15a,b,c): iu + v ' = 0

(

iu ( U − c ) + U ' v = −ip/ + v u "−  2u

(

iv ( U − c ) = −p ' / + v v"−  2 v

)

)

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Eliminate p by differentiating the second relation, multiplying the third relation by ( i ) , and subtracting. The result is:

(

v ( U − c )  2 + U"− i3v  = −i ( U − c ) u '+ v u '''− 2 2 u '  

)

Now eliminate u by substituting u = ( −v'/i ) , u ' = ( −v"/i ) ,etc., from the first relation. The result is the desired Orr-Sommerfeld equation which was to be proved:

( U − c ) ( v"− 2 v ) − U"v +

P5-5

(

)

iv v ''''− 22 v ''+ 4 v = 0 

(Ans.)

Consider Rayleigh’s inviscid stability equation (5-21). The perturbation amplitude v ( y )

is complex, and, for temporal stability analysis,  is real and c is complex. Show that Rayleigh’s equation may be split into real and imaginary parts, as follows:

 U " (U − cr )  U " ci  vr + vr "−  2 + =0 2 2  (U − cr ) + ci  (U − cr )2 + ci2  U " (U − cr )  U " cr  vi "−  2 + v − =0 i 2 2 2 2  U − c + c U − c + c ( ( r) i  r) i Explain, in words, a possible method for solving this system numerically for a given U ( y ) . Solution: This is a good exercise for students to split a linear differential equation into real and imaginary parts. The numerical scheme would typically begin at large y, where vr and vi are exponential in y, and integrate inwards, varying cr and ci to match the wall conditions.

P5-6

For an approximate quartic polynomial flat-plate velocity distribution,

u/ U = 2− 23 + 4 ,

= y/

solve the inviscid Orr-Sommerfeld equation, find the eigenvalues, and plot ci versus  . Nondimensionalize with respect to  and U. Solution: The Orr-Sommerfeld relation becomes Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -4-

2  d 2 v* 2  d U* U* − c* −  * v* = v*, ( )  2  2 d  d   

where v* =

v c , c* = , * =  U U

and where U* = 2− 23 + 4 , d2 U*/d2 = −12+ 122 This quartic polynomial is a reasonable approximation to U( y ) but is not very accurate for

U"( y ) , since it does not satisfy U"' ( 0 ) = 0 as required for flat plate flow. [One might try, as an alternate profile, a 5th or 6th order polynomial to meet this wall condition.] As suggested in the problem statement, we start the integration out at  = 2, where

v* = exp ( −*) ,

and integrate inward numerically to satisfy the wall conditions

v*( 0) = v*' ( 0) = 0. We guess values of c* = c*r + ic*i . When passing through the ‘critical’ layer, where U* = c*, there will be an unavoidable blip in the integrated curve, which we avoid by not letting the denominator |U* − c*| drop below, say, 0.001. The integrated solution then passes smoothly through this singularity. The (inviscid) eigenvalues and eigenfunctions are found for a given  by varying ( cr ,ci ) until we satisfy v ( 0) = v' ( 0) = 0. Here we show only the solutions for the neutral condition,

ci = 0. The eigenvalues vary with wavenumber as follows: 

=

0.5

1.0

1.5

2.0

2.5

c r /U o

=

0.190

0.332

0.469

0.620

0.762

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The neutral eigenfunctions v* ( y*) for this case are shown in the following graph.

They are not especially accurate compared to solutions using the exact Blasius functions for U ( y ) and U" ( y ) - see, for example, the text by Betchov and Criminale (1967), pp. 46–50.

P5-7 For two-dimensional stagnation flow, U = Kx, estimate the position Re x where instability first occurs. Solution: From Table 5-1, p. 354 of the text, stagnation flow ( = 1.0 ) becomes unstable at Re = 5636. Meanwhile, from the Falkner-Skan solutions in Table 4-2, p. 243, the dimensionless momentum thickness in stagnation flow is θ* = 0.29235. This gives the prediction 1/ 2

 m +1 U    2 vx 

= 0.29235, for m = 1, or:

 0.29235 = x ( Ux/v )1/ 2

Solve this for Re and set it equal to the critical value: Re = 0.29235 Re1/2 x = 5636, or

Re x,crit = 3.72 E8

(Ans.)

This is a large Reynolds number, illustrating the extreme (but not infinite) stability of the favorable-gradient stagnation-flow velocity profile. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -6-

We can check this result against the Wazzan et al. (1981) shape factor correlation in Fig. 5-31, page 381. The shape factor for stagnation flow is, from Table 5-1, H = */ = 12490/5636 = 2.22. Entering Fig. 5-31 at H = 2.22, we can read Re x,crit = 3 E8, in good agreement with our estimate above.

P5-8 For the separating Falkner-Skan wedge flow,  = −0.19884, estimate the position Re x where instability first occurs. Solution: From Table 5-1, p. 354 of the text, separating flow ( = −0.19884) becomes unstable at

Re  17. Meanwhile, from the Falkner-Skan solutions in Table 4-2, p. •••, the dimensionless momentum thickness in separating flow is θ* = 0.58544. This means that 1/2

 m +1 U    2 vx 

= 0.58544 for m = −0.09043, or:

 0.8681 = x Re1/2 x

Solve this for the momentum-thickness Reynolds number and set it equal to the critical value: Re = 0.8681 Re1/2 x = 17, or: Re x,crit  380

(Ans.)

This is a remarkably small value of the critical Reynolds number, illustrating the extreme instability of the S-shaped separating-flow velocity profile. We cannot check this result very well against the correlation of Wazzan et al. (1981) in Fig. 5-31, page 381, because H in that figure only goes up to 3.1. The shape factor for FalknerSkan separating flow is, from Table 4-2, H = 2.35885/0.58544 = 4.0, which is off the bottom right of Fig. 5-31. However, our value of Re x,crit = 380 seems to be a reasonable extrapolation of Fig. 5-31.

P5-9 For the Howarth freestream velocity U = Uo (1 − x/L ) , estimate the point ( x/L ) where boundary layer instability first occurs, if Uo L/v = 1 E6. Solution: This is not a similarity solution, so we need estimates of  ( x ) and H ( x ) along the wall, assuming that instability occurs before boundary-layer separation. We could use Thwaites’ method from Sect, 4-6.7, pp. 265–268 of the text, for which

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2 = −

vL , Uo

  −0.075 (1 − x/L ) 

where

Thus: Re = ( − ReL )

1/2

= 1000 ( − )

1/2

−6

− 1 

since ReL = 106

One computes Re ( x ) and, with  ( x ) known from Table 4-4 or Eq. (4-141) on p. 267, one computes H ( x ) and consults the shape factor correlation, Fig. 5-12 on p. 360 to see if instability has been reached. For example: At x/L = 0.02:  = −0.0097, Re = 98, H = 2.63, Re = 257 Check this against H = 2.63 in Fig. 5-12, for which Re*,crit = 400. Thus we have not yet reached the point of instability. Move a bit further downstream: At x/L = 0.03:  = −0.015,

Re = 123,

H = 2.65, Re* = 325

At x/L = 0.04:  = −0.021,

Re = 144,

H = 2.68, Re* = 386

For H = 2.65, from Fig. 5-12, read Re*,crit  350 - perhaps just a bit shy of instability. For H = 2.68, from Fig. 5-12, read Re*,crit  300 - perhaps just a bit downstream of instability. Our best estimate, then, reading a very small-scale correlation curve, is: Howarth flow:

Uo L/v = 106 :

( x/L )instability = 0.035

(Ans.)

This result is equivalent to a point Rex = (1E6)( 0.035) = 35,000. We can check this estimate from the Wazzan correlation in Fig. 5-31: for H = 2.65, read Re x,crit = 35, 000, which is excellent agreement for this somewhat uncertain prediction.

P5-10 Generalize Prob. 5-9 to calculate and plot the instability point ( x/L )crit as a function of nominal Reynolds number ReL = Uo L/v. Solution: We follow the same procedure as Prob. 5-9, systematically varying the Reynolds number:  = −0.075 (1 − x/L ) 

−6

− 1 , Re = 

( − ReL ) , H = H (  ) , Re = H Re , Rex = ReL ( x/L )

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We then read Fig. 5-12 to see if Re is critical for the local H, and check by reading Fig. 5-31 to see if the local Re x is critical for that value of H. The results may be plotted as follows for a wide range of nominal Reynolds numbers:

The value x/L = 0.035 for ReL = 1E6 is taken from Prob. 5-9 above. We see that the point of instability always occurs before the point of boundary-layer separation ( x/L = 0.120) for a very wide range 1E4  Uo L/v  1E8. The two predictions, Figs. 5-12 and 5-31, give similar results.

P5-11 For potential freestream flow across a circular cylinder, U = 2Uo sin ( x/a ) , with

Re D = 1E6, estimate the position ( x/a )crit where instability first occurs.

Solution: This is a variation of Prob. 5-9, with a different freestream velocity distribution. We need to estimate  ( x ) ,  ( x ) , H ( x ) , and Re ( x ) for cylinder flow using, say, Thwaites’ method. The results of this prediction were given in Prob. 4-23 on page 107 of this Manual: 2 =

v a () 0.45cos   x , where  = 8 − cos  3sin 4  + 4sin 2  + 8  ,  = 6  2Uo cos  a 15sin  

(

)

We compute H (  ) from Table 4-4 or Eq. (4-141) on p. 267, and we also compute

Re = sin  (  ReD / cos ) , Re = H Re , Rex = ReD  sin  1/2

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Then we check Fig. 5-12 to see if Re is critical for this value of H and ReD . A sample calculation is as follows for ReD = 1E6:

 = 60:  = +0.059, Re = 210, H = 2.41, Re = 507, Rex = 9.1E5 For H = 2.41, Fig. 5-12 predicts that Re,crit = 25, 000, so we have not yet reached instability. Also, for H = 2.41, Fig. 5-31 predicts that Re x,crit  8E6, so this correlation also indicates that we are too early to be unstable. Try a position further along:

 = 80:  = +0.031, Re = 420, H = 2.49, Re = 1050, Re x = 1.4 E6 For H = 2.49, we read from Fig. 5-12 that Re,crit = 1200, so we are before the point of instability. For H = 2.49, we read from Fig. 5-31 that Re x,crit = 8 E5, just past the point of instability. This seems to bracket our prediction well enough: Cylinder Potential Flo w:

Uo D/v = 106 :

crit = 80, or:

( x/a )crit = 1.4

(Ans.)

P5-12 Using the guidance of Probs. P5-5 and P5-6 and any numerical method of your choosing, solve the Rayleigh equation (5-21) for a simplified boundary layer flow, U = tanh ( y ) , 0  y  , with a typical value of   0.2 to 0.3. Do you expect any inviscid instability? If time permits, plot some computed values of cr versus  . [HINT: Begin at large y  4 with the exponential solution to Eq. (5-6) and integrate backwards to the wall.] Solution: This assignment is quite similar to Prob. 5-6: a simple function for U ( y ) , numerical integration backward from large y, and a search for neutral eigenvalues ( , cr ) with ci = 0. Here are some neutral pairs computed by the writer, plotted on the next page:

=

0.5

1.0

1.5

cr =

0.30

0.41

0.55

There is no point of inflection in tanh ( y ) , except y = 0, thus there is no inviscid instability.

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P5-13 Using the guidance of Probs. P5-5 and P5-6 and any numerical method of your choosing, solve the Rayleigh equation (5-21) for the Blasius boundary layer flow, which you should generate from Eqs. (4-45) and (4-46). Do you expect any inviscid instability? Select a value of  in the range 0.2 to 0.3. [HINT: Begin at large   5 with the exponential approximation of Eq. (5-6) and integrate backwards to the wall.] Solution: First we have to generate the Blasius solution numerically, using a small enough step size, say,  = 0.05, to enable us to compute an accurate neutral Rayleigh solution. The Rayleigh procedure follows the examples set in Problems 5-6 and 5-12. We need U  f  and U   f  and then solve Eq. (5-21), splitting it up, real and imaginary, as in Prob. 5-5. There is no inviscid instability because the Blasius profile only has a point of inflection at  = 0. The neutral eigenvalues ( ci = 0) found by the writer are as follows:

=

0.3

0.5

1.0

cr =

0.42

0.54

0.83

The eigenfunctions vr ( ) are shown on the next page.

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Neutral inviscid eigenfunctions vr ( ) for the Blasius velocity profile.

P5-14 For stagnation-point flow, U = Kx, estimate the position Re x where transition occurs, using the method of Michel, Eq. (5-38). Why is Granville’s method, Section 5-5.1.1, inappropriate? Solution: Michel’s method requires an estimate of Re ( x ) , which we could take from the Falkner-Skan solution of Table 4-2, p. 243 of the text: 1/2

* = 0.29235,

 vx  or:  = 0.29235   U

= 0.29235 ( v/K )

1/2

The same formula was used in Prob. 5-7, page 143 of this Manual. This relation is equivalent to Re = 0.29235 Re x . Thus Michel’s correlation predicts that 10

 2.9  0.4 Re = 0.29235 Re1/2 x  2.9 Re x , or: Re x,trans =    9.2 E9  0.29235 

(Ans.)

This is a very large transition Reynolds number, showing the extreme stability of stagnation flow, which has a strong favorable pressure gradient. An alternate estimate, always possible whether the flow is “similar” or not, is to use Thwaites’ method, Eq. (4-138), to estimate the momentum thickness for the given U ( x ) :

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x

2 0.45 = v ( Kx )6

 ( Kx ) dx = 0

5

0.075 , K

1/2

or:

v   0.274   K

This estimate for  is about 6% less than the Falkner-Skan result and leads to the result 10

 2.9  Rex,trans. =   = 1.8 E10  0.274 

(Ans.)

[The 6% discrepancy is greatly magnified after taking the tenth power.] Both estimates are acceptable to this writer. The method of Granville, Eq. (5-37), is inappropriate because it is limited to a Thwaites parameter   0.4. From our Thwaites calculation above, stagnation flow has  = 0.075, or too high for Granville’s correlation. Should we choose to extrapolate Granville’s formula to such a high , it would predict Re,tr = 42100, or Re x,tr  2.3 E10, about a factor of 2 too large.

P5-15 For the separating Falkner-Skan flow,  = −0.19884, use any appropriate correlation to estimate the point of transition, Re x,tr . Compare with Fig. 5-32. Solution: We may use Michel’s method, which requires an estimate of Re ( x ) , which we could take from the Falkner-Skan solution of Table 4-2, p. 243 of the text: 1/2

 m +1 U    2 vx 

= 0.58544 for m = −0.09043, or:

 0.8681 = x Re1/2 x

Michel’s method, Eq. (5-38), then predicts 10

 2.9  0.4 Re = 0.8681 Re1/2 x  2.9 Re x , or: Re x,tr    = 1.73 E5  0.8681 

(Ans.)

This estimate is in agreement with Fig. 5-32. Other methods could be used as well, as illustrated in Fig. 5-32.

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P5-16 For the Howarth flow, U = Uo (1 − x/L ) , use Michel’s method, Eq. (5-38), to estimate the transition point if Uo L/v = 4E6. Neglect freestream turbulence. Solution: For Michel’s method we need momentum thickness  ( x ) , which was computed in Prob. 5-9 on page 144 of this Manual:

2 = −

Thus

vL −6 , where  = −0.075 (1 − x/L ) − 1   Uo

Re = U/v = ( U/Uo ) ( − ReL )

1/2

for this particular flow. Noting that

x/L =

until it is ( Ux/v ) / ( ReL ) ( U/Uo ) , we may evaluate Michel’s correlation, Re = 2.9 Re0.4 x satisfied. Some numerical results, for Uo L/v = 2 E6, are as follows:

x/L =

0.02

0.04

0.06

0.08

0.10

0.12

Rex =

39200

76800

112800

147200

180000

211200

Re =

136

196

244

287

327

366

2.9Re0.4 x = 199

261

304

339

367

391 (?)

We see that, at the separation point, ( x/L )  0.120, the momentum-thickness Reynolds number is still not quite high enough for transition. Therefore, for ReL = 2E6, this flow does not undergo transition before separation is reached. If we increase Re L , we do get transition before separation, as listed below:

Re L =

2 E6

3 E6

4 E6

6 E6

8 E6

1 E7

( x/L )tr =

none

0.131

0.119

0.102

0.091

0.081

P5-17 Generalize Prob. 5-16 into a parametric computer plot of ( x/L ) tr versus Re L . Solution: All we do is follow the same procedure of Prob. 5-16 above with different values of Re L . Some results are tabulated above, and we may plot a broad spectrum of results as follows:

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P5-18 For (potential) cylinder flow, U ( x ) = 2Uo sin ( x/a ) , use Michel’s method, Equation (5-38), to estimate the transition point if Uo D/v = 2 E6. Neglect freestream turbulence.

Solution: Let us reproduce the Thwaites’ result for momentum thickness in potential-stream cylinder flow from Prob. 5-11, page 145 of this Manual: 2 =

v a () 0.45cos   x , where  = 8 − cos  3sin 4  + 4sin 2  + 8  ,  = 6   2Uo cos  a 15sin 

(

)

where a = D/2 is the cylinder radius. Thus Re = U/v = sin  (  ReD /cos)

1/2

for this particular

flow. Noting that x/D = ( Ux/v ) / ( ReD ) ( U/U o )  , we may evaluate Michel’s correlation,

Re = 2.9 Re0.4 x , continuously (as  increases) until it is satisfied. Some numerical results for our case, ReD = 2 E6, may be listed as follows: =

50°

60°

70°

80°

90°

100°

110°

120°

2.75E6

3.14E6

3.44E6

3.61E6

3.63E6

Rex =

1.34E6

Re =

345

420

501

589

693

820

990

1237

2.9 Re0.4 x = 818

924

1016

1092

1151

1194

1217

1220

1.81E6 2.30E6

We see that the Michel criterion is satisfied just before 120°, say, tr  119.5° (Ans.). However, the laminar separation point, from Prob. 4-23, page 107 of this Manual, is at   103.1. Therefore our estimate of 119.5° is unrealistic according to the limitations of boundary Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -15-

layer theory. Furthermore, it is known from experiment (see, e.g., Fig. 3-38a) that transition in cylinder flow actually occurs at about ReD  2.5E5, or much less than the value 2E6 studied here. Thus it appears to this writer that Michel’s method may be unrealistic in general for estimating the transition point in strong favorable gradients. Meanwhile, the method of Wazzan et al, Fig. 5-31, indicates that transition occurs at about tr = 92, which is quite reasonable when compared to experiment. Wazzan’s method is used in Prob. 5-19 below.

P5-19 Air at 20°C and 1 atm flows quietly toward a wedge of half-angle 36°, resulting in a powerlaw freestream and a laminar boundary layer along the surface. Use Wazzan’s method, Eq. (5-42), to estimate the transition Reynolds number Rex,tr . Solution: This problem is dimensionless, and it doesn’t matter if the fluid is air or water or whatever. For wedge flow, Fig. 4-10, the half-angle is 36 =  /2, or  = 0.4, where the power exponent in U = Cx m is m = 1 4. The shape factor H is constant, and we can either use Thwaites’ method or simply interpolate in Table 4-4 to estimate H  2.34. (a) Wazzan’s transition correlation, Eq. (5-42), gives log10 ( Rex,tr )  8.03, or Rex,tr  1.1E8. Ans. (a) (b) The Thwaites/Michel approach, Eqs. (4-138) and (5-38), yields

2 v



0.45

( )

1/4 6

K6 x

x

K 0

5

( x1/4 )

5

dx =

0.45 3/4 0.2 x  0.2 x = ; = ; 2.25K U x Re x

Re = 0.2 Re x = 2.9 Re0.4 x ,tr , Solve for Re x ,tr  1.3E8

Ans. (b)

The two estimates are remarkably close but, as seen in Fig. 5-32 of the text, the two theories spread apart as the pressure gradient becomes more favorable,   0.4.

P5-20 Repeat Prob. 5-14 (stagnation flow) for a freestream turbulence level T = 1%. Solution: We must change methods here, since the Michel and Wazzan correlations are valid only for negligible freestream turbulence. Equation (5-47) is quite simple, and predicts Re x,tr  1.4E6 ± 40%, reasonable, but it is only suggested for high turbulence, T  1%.

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Dunham’s correlation, Eq. (5-46) or Fig. 5-34(b), is not valid for   0.04, whereas stagnation flow has   0.075. This leaves only the van Driest & Blumer semi-empirical theory, Eq. (5-45), p. 385 of the text. For stagnation flow, m = 1.0 and  = 2.4 from Table 4-2, page 243. Equation (5-45) then becomes, for T = (1%) = 0.01:

1690 Re1/2 x or:

= 0.312 (1 + 0.11)

−0.578

+ 1.6 ( 2.4 ) Re1/2 x ( 0.01) , 2

2

Z2 + 320 Z − 1.834E6 = 0, Z = Re1/2 x , with solutions Z = 1203 and −1524

We reject the negative solution, Z = −1524, as spurious. Therefore the predicted transition Reynolds number in stagnation flow, for a turbulence level of 1%, is Re1/2 x = 1203,

or:

Re x,tr = 1.45 E6

(Ans.)

This is approximately verified by the plotted value in Fig. (5-34a) at m = 1 and T = 1%. This is a factor of 6000 less than the value predicted for T = 0% in Prob. 5-14.

P5-21 Repeat Prob. 5-15 (separating flow) for a freestream turbulence level T = 1%. Solution: Since   0.04 for this case, both the Van Driest & Blumer (Fig. 5-34a) and Dunham (Fig. 5-34b) correlations are valid. The Michel, Granville, and Wazzan methods are not, since they assume negligible freestream turbulence. For the Van Driest and Blumer formula. Equation (5-45), take

m = −0.09043 for separating flow and read  = 4.9 from Table 4-4. Then, for T = (1%) = 0.01, Eq. (5-45) predicts 1690 Re1/2 x or:

 ( 0.312 )( −0.09043 + 0.11)

−0.528

+ 1.6 ( 4.9 ) Re1/2 x ( 0.01) , 2

2

Z2 + 648 Z − 440000 = 0, z = Re1/2 x , with solutions Z = +414 and Z = −1062

We reject the negative solution, Z = −1062, as spurious. Therefore the predicted transition Reynolds number in separating flow, for a turbulence level of 1%, is Re1/2 x = 414,

or:

Re x,tr = 172, 000

(Ans.)

This value is plotted (approximately) in Fig. 5-34(a).

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For the Dunham correlation, Eq. (5-46), we take T = 0.01 and we need an estimate for the value of  at separation. For Falkner-Skan flow, m = −0.09043, we find from Table 4-4, page 267 of the text, that  2   dU   ( 0.58544 )2 ( 2x )   mU    =   = = −0.0682 for m = −0.09043  v   dx   ( m + 1) U x       

We may then evaluate Dunham’s correlation:

 680 − 80 0.01  Re,tr = 0.27 + 0.73e ( )( )  550 +  = 447   1 + 100 ( 0.01) − 21( −0.0682 )  Meanwhile, from Prob. 5-15, for separating flow, Re = 0.8681 Rex . Therefore 2

 447  Rex,tr =   = 266, 000  0.8681 

(Ans. - Dunham)

Considering the uncertainty of transition-point estimation, this result is quite comparable to the Van Driest & Blumer value of 172,000 predicted above. Note - exasperatingly - that these values for T = 1% are about the same as the ( T = 0% ) Michel prediction in Prob. 5-15! This result casts further doubt upon the accuracy of the Michel correlation.

P5-22 Repeat Prob. 5-16 (Howarth flow) for a freestream turbulence level T = 1%. Solution: Since, strictly speaking, the Van Driest & Blumer method, Eq. (5-45), applies only to similar (Falkner-Skan) flows, we should not apply it to Howarth (nonsimilar) flow. That leaves only the Dunham correlation. Eq. (5-46), which for T = 0.01 becomes

 680  Re,tr = 0.598 550 +  2 − 21  tr   We follow the same procedure as in Prob. 5-16, this time monitoring  ( x ) and using the above relation to compute Re,tr for comparison with the local Re until they are equal. Some example results for Re L = 2 E6 are as follows:

x/L = =

Re =

0.02

0.04

0.06

0.08

0.10

0.12

–0.0097

–0.0208

–0.0337

–0.0487

–0.0661

–0.0865

136

196

244

287

327

366

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Re,tr =

513

487

479

463

449

435

Thus, even at separation ( x/L = 0.12) for a turbulence level of 1%. Dunham’s method does not predict transition for Howarth flow at ReL = 2 E6 - which was also the case with Michel’s method at T = 0% from Prob. 5-16. If we try some higher values of Re L , we do find transition before the separation point, as follows:

Re L =

( x/L )tr

2 E6

3 E6

4 E6

6 E6

8 E6

N/A

0.116

0.096

0.0723

0.0583

1 E7 0.0489

These transition points are earlier than those predicted for T = 0% by Michel’s method in Prob. 5-17 - see the plot at the bottom of page 150 of this Manual.

P5-23 Repeat Prob. 5-18 (cylinder flow) for a freestream turbulence level T = 1%. Solution: As discussed in Prob. 5-22 above, the Van Driest & Blumer method is not strictly valid, since cylinder flow is not ‘similar’. We therefore use the Dunham correlation, Eq. (5-46): For T = (1% ) = 0.01:

Re,tr

 680  0.598 550 +  2 − 21 tr  

We follow the same procedure as in Prob. 5-18, this time monitoring  ( x ) and using the above relation to compute Re,tr for comparison with the local Re until they are equal. Some example results for ReD = 2 E6 are as follows: =

40°

50°

60°

70°

80°

90°

100°

Re =

274

345

420

501

589

693

820

Re,tr =

1081

977

862

743

631

532

453

We see that transition is predicted somewhere between 80° and 90°. If we use smaller increments, we will find the point of equality of Re at   82.1° (Ans.). The effect of Reynolds number for this same turbulence level T = 1% is as follows:

Re L =

1 E6

2 E6

3 E6

4 E6

6 E6

8 E6

1 E7

 tr =

86.4°

82.1°

75.9°

71.5°

65.3°

60.8°

57.3°

These occur (plausibly) earlier than the predictions for T = 0% by Michel’s method in Prob. 5.18, p. 151 of this Manual. They seem to be reasonable predictions. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -19-

P5-24

For a pipe flow started from rest with acceleration a, as in Fig. 5-37b, the momentum

thickness initially grows according to the formula   0.35 ( vt )

1/2

. Apply this relation to Michel’s steady-flow transition correlation, Eq. (5-38), assuming that “V” and “x” are given by constantacceleration formulas. Show that the result is a constant value of the dimensionless transition time

(

ttr a 2 /v

)

1/3

. Why does the diameter D not appear?

Solution: By “constant-acceleration formulas”, we simply mean to take “V” = at and “x” = (1/2 ) at 2 . Substitute into Michel’s transition correlation, Eq. (5-38):

Re

V ( at ) ( 0.35 = = v

vt

   v 

x

v

1/2

 a 2t 3  0.35 Clean up:    v 

 a 2t 3   2.9   2v   

1/3

Solve:

)  2.9 Re0.4 = 2.9  Vx 0.4 = 2.9  ( at ) ( at 2 /2) 

 a2  t tr    v   

0.4

 

  2 1/3  a , or: 0.35 ttr       v    

v

3/2

0.4

 

1/3   2.9   a 2   = 0.4 ttr   2   v    

6/5

10/3

  2.9 =  0.4   2 ( 0.35 ) 

 457

Ans.

This simple estimate, which is not too bad, is given as a straight line in Fig. 5-37b of the text. Here there is no diameter effect on ttr , because the momentum thickness estimate,   0.35 ( vt )

1/2

, is based on unsteady boundary layer theory, with no constraint or outer wall at the outer edge. The data in Fig. 5-37b show that there is indeed an effect of diameter.

P5-25 Air at 20°C and 1 atm flow at U = 12 m/s past a smooth flat plate. A 1-mm diameter wire trip is placed across the plate at x = 1 m. If freestream turbulence is negligible, at what position x tr will transition occur? What wire location x will cause earliest transition? Solution: For air at 20°C and 1 atm, take v = 1.5 E-5 m2 /s. Then, at x = 1 m, Re x = Ux/v = 8 E5, at which point the displacement thickness is, from laminar flat-plate theory,

* =

1.72 x Re1/2 x

=

1.72 (1.0 )

( 800,000 )1/2

 0.00192 m = 1.92 mm

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Thus, at this position, a wire of diameter k = 1 mm has a ratio k/δ* = 1/1.92  0.52, which will certainly cause early transition. From Fig. 5-36(a), read Re(rough)/Re ( smooth )  0.40. We need an estimate for Retr ( smooth ) : from Fig. 5-33, for the data of Schubauer and Skramstad, read, at zero turbulence, Retr ( smooth ) = 3 E6. Then our estimate for the transition point location with the wire trip is Re x,tr = 0.4 ( 3 E6 ) = 1.2 E6 =

Ux (12 ) x = , or: v 1.5 E-5

x tr = 1.5 m

(Ans.)

The wire causes transition at about 50 cm further downstream, whereas without the wire transition would not occur until x = 3.75 m. Since Re ( smooth ) = 3 E6 is a basic parameter in this problem, the earliest transition occurs when the wire causes the lowest possible value of Re ( rough ) . From Fig. 5-36(a), the lowest measured data give Re ( rough ) /Re ( smooth ) = 0.1 at k/* = 0.8. From flat-plate theory, this corresponds to

k 0.1( 3 E6 )  k Re1/2 k x = 0.8 =   * 1.72 x 1.72 x

1/2

, or x = 398 k = 0.4 m

(Ans.)

A 1-mm wire placed at x = 0.4 m would cause transition almost immediately downstream.

P5-26 Repeat Prob. 5-25 for a freestream turbulence level T = 1 %. Solution: This analysis is identical to Prob. 5-25 except that the reference value Re tr,smooth must be modified for T = 1%. Fig. 5-33 does not extend out to T = 1%, therefore use the Van Driest & Blumer correlation, Eq. (5-44), which is based upon the same data:

Re1/2 x,tr =

2 −1 + 1 + 132500 ( 0.01)   

39.2 ( 0.01)

2

 708, or:

Rex,tr ( T = 1% ) = 500, 000

Other parameters follow from Prob. 5-25: at x = 1 m and k = 1 mm, Rex = 8 E5. Thus the wire is too late, transition has already occurred at Rex = 5 E5, or at the position x  0.625 m (Ans.). At this high level of freestream turbulence, a trip wire is not needed.

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P5-27 Find out more about the narrow white band in the chaotic region of Fig. 5-38, for 3.825  r  3.865, by making computer iterations in that region for, say, r  0.0005. Plot the resulting iterates only in this expanded region and comment on the resulting remarkable pattern which appears. Solution: The scale of Fig. 5-38 is so large that it hints little of the beauty of the narrow white area. However, plotting the iterates in the narrow region 3.825  r  3.865 discussed above immediately reveals a beautiful pattern, as shown below [one needs a personal computer with point-plotting ability to do this - this writer uses the PLOT routine that is part of the TRUE BASIC software].

The “chaos” stops at about r = 3.827, and the iterates then regroup into three single lines, each of which bifurcates at about r = 3.842 into a pattern resembling the original “period-doubling” pattern of Fig. 5-38. In fact, each pattern - labelled #1, #2, and #3 in the above figure - is quite similar in shape to the original pattern. A greatly enlarged version of pattern #3, which lies in the region 0.95  x  0.97, is shown below. We see that it closely resembles the original chaotic large-scale pattern of Fig. 5-38. Thus chaotic phenomena often contain - hidden in the chaos - microscopic and submicroscopic images - presumably occurring multiply at smaller and smaller scales - of the

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chaos. There is a very good discussion of this phenomenon in the popular-science book by James Gleich, Chaos - Making a New Science, Viking Press, New York, 1987.

P5-28 Repeat Prob. P5-19 if the freestream has a turbulence level of 4%. Find the estimated transition Reynolds number Rex,tr by two different methods and compare. Solution: Recall that Prob. 5-19 involves flow past a wedge of half-angle 36°, with a power-law freestream and a laminar boundary layer along the surface. We showed there that U = Kx1/2 , with  = 0.4. That problem shows that, by Thwaites’ method,  2 /v  0.2 x /U and   0.05, a bit too high to use Dunham’s freestream-turbulence correlation, Eq. (5-46). (a) If used as an estimate, Dunham’s formula would yield too high a transition value:

Re ,tr  217, Re x,tr  ( Re ,tr ) /0.2 = 234, 000 2

Ans. (a) – Dunham

(b) As a second estimate, for m = 1 4,  = 0.4, Table 4-2 indicates   2.8, and we can estimate from the van Driest and Blumer correlation, Eq. (5-45), that

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1690 Re1/2 x,tr

= 0.312 ( 0.25 + 0.11)

−0.528

+ 1.6 ( 2.8) Re1/2 x,tr ( 0.04 ) , solve Re x,tr  77, 000 2

2

Ans. (b)

(c) The simple estimate of Schmid and Henningson, Eq. (5-47), is quick and reasonable:

T ( Re x,tr )

1/2

 1200, T = 4%,

solve

Re x,tr  90, 000

Ans. (c)

This is the limit of our prediction skills. A reasonable transition estimate is Rex,tr  100,000.

P5-29 The famous neutral curve of Taylor (1923), for Couette flow between rotating cylinders, is in Fig 5-22b. The region above the curve is simply labeled unstable. Some amazingly diverse flow regimes lie in this region, as shown in a wonderful chart by Andereck, Liu, and Swinney (1986). Report to the class on this chart and its many unstable flow patterns. Solution: The writer’s best effort at scanning the chart is shown below. The writer lacks words to discuss the nineteen different unstable flow regimes in the chart. Good luck!

Flow regimes observed by Andereck et al. (1986) between two rotating cylinders. The Reynolds numbers are Reinner = ri2i /v and Reouter = ro2o /v.

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P5-30 For two-dimensional inviscid flow, the vorticity equation (2-116) may be written as    +u + v = 0, t x y

Defining a basic flow

( ,U ,V ,)

where

=

v u − = − 2 x y

and disturbances

( , u, v, ) ,

derive a linearized

disturbance equation for this flow. Then assume normal modes as travelling waves:

( , u, v,  ) = ˆ ( y ) , uˆ ( y ) , vˆ ( y ) , ˆ ( y ) exp i ( x − ct ) Derive the disturbance equations and, if possible, combine them to obtain a single differential equation for a single disturbance amplitude. Solution: The instructions above are pretty clear. Substitution of (basic flow + disturbances) into the vorticity equation yields Ω  ' Ω  '   ' Ω  '   ' + +U +U + u' + u' +V +V + v' + v' =0 t t x x dx x y y y y

Ω + ' = −2 −2 ' The boldface terms are basic flows and cancel themselves. The products of fluctuations are neglected, with an arrow through them. The disturbance differential equations are thus  '  '   '  +U + u' +V + v' 0 t x x y y v' u'  ' = − 2 ' = − x y

These can be solved, as suggested in the problem statement, by assuming travelling waves. After substitution, all terms contain exp i ( x − ct )  , which can be cancelled out. Thus we have the following disturbance equations:

ˆ ( −i c ) + ˆ ( iU ) + uˆ ˆ = − ( i ) ˆ − 2

 dˆ  +V + vˆ =0 x dy y d 2ˆ dy

2

= i vˆ −

duˆ dy

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To make this work, the “basic” coefficients (U , V , /x, /y ) have to vary with y only. In fact, it makes sense to neglect the term uˆ ( /x ) as being unlikely. To boil this down to one single equation in one single variable, clearly we could substitute for  in terms of  . We could also use the definition of stream function to eliminate the two velocities: u=

  , v=− , y x

hence uˆ =

dˆ dy

and

vˆ = −iˆ

Making these substitutions, we obtain the final third-order ordinary differential equation for the amplitude of the stream function:

 d 2ˆ   d 3ˆ    dˆ  i (U − c )  2 −  2ˆ  − V  3 −  2  − iˆ  =0 dy   dy   dy  y 

Ans.

We would have to know, in advance, the (y) variations of the coefficients U, V, and .

P5-31 Use the airfoil surface-velocity data of Prob. P4.49 and Michel’s method, Eq. (5-38), to estimate the position of transition to turbulence if Rec = 2 106. Assume air at 20°C and 1 atm. If you note an ambiguity in the results, please criticize them. Solution: Recall the data from Prob P4.49: A model two-dimensional airfoil has these potentialflow 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 C is 30 cm. The fluid is air at 20°C and 1 atm. Assume that x is a good approximation to the arc length. For air at 20°C and 1 atm,  = 1.205 kg/m3 and  = 1.81E-5 kg/(m-s). Thus, if

Rec = 2 106 , the appropriate velocity is U = Re v /C = ( 2E6)( 0.000015) / ( 0.3) = 100 m/s.

Problem 4-49 shows how Thwaites’ method was used to estimate  ( x ) and  ( x ) . One can then calculate Re ( x ) and compare it with Michel’s transition line, 2.9 Re0.4 x . Here is a graph:

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According to this estimate, transition occurs at Rex  1,080,000, or x /C  0.54. The ambiguity is that this is past the point of separation ( x /C  0.45) predicted in Prob. 4-49!

P5-32 Consider the cubic flat-plate velocity distribution: u 3 1 y =  − 3 = U 2 2  Solve the Orr–Sommerfeld Eq. (5-23) numerically for the inviscid case with  = 0 . Begin at − y and integrate inwardly to satisfy the wall conditions  =   = 0 at y = 2 assuming that   e y =  = 0. Assuming temporal amplification, find some (damped) eigenvalues and plot ci versus  in dimensionless form. Solution: We are asked to numerically solve the Orr-Sommerfeld equation for the inviscid case with  = 0 . Based on Eq. (5-29), we can derive a non-dimensional Rayleigh equation of a form similar to Eq. (5-27). The nondimensional Orr-Sommerfeld equation can be written as i U * − c*   − 2 − U * +   − 22  + 4 = 0  Re with  u c U = U* = c* =  =  Re = U U U  This step aids in the formulation by casting the problem into a nondimensional form. Next, we let  = 0. This can be thought of as letting Re → . The result is the Rayleigh equation that stands at the basis of inviscid-stability theory: U * − c*   − 2 − U * = 0

(

)(

)

(

(

)(

)

)

Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -27-

Next, a computational strategy needs to be chosen given several traditional techniques to solve the stability equation. For example, one could rely on a shooting method as traditionally done in this case. However, the problem is formulated as a boundary-value problem, and the shooting method will require solving an initial value problem multiple times until convergence is achieved. Since our goal is to retrieve the eigenvalues, we can rewrite the above expression as an eigenvalue problem. For more insight, please refer to the Matlab code provided as part of the solution for P535 below. Note that this formulation can be extended from the inviscid Rayleigh equation to the full Orr-Sommerfeld equation rather straightforwardly. To proceed, we may reformulate our continuous differential eigenproblem from L1 = c* L2 to a discrete, generalized eigenproblem of the form M  = c* N  where M and N represent finite-difference operator matrices that only depend on the nondimensional wave number and discretization scheme. The nondimensional Rayleigh stability equation can then be separated into two matrices, namely,   U * −  U *2 + U * = c*   − 2

)

(

( )

(

)

Now that we have our assumed form, we may discretize  . Using a second-order finite difference scheme, we may put d 2  j +1 − 2 j +  j −1 = + O  2 2 2 d  Our continuous eigenvalue problem can thus be discretized. We get B j j −1 + Aj + B j j +1 = c* Bˆ j −1 + Aˆ j + Bˆ j +1

(

(

)

)

where

 A = − 2 2U * − 2U * − U * 2   j *  B j = U  2 2  Aˆ = − 2 +     Bˆ =  2 From this set-up, two tridiagonal matrices can be computed for M and N . To solve the general eigenvalue problem, we can multiply both sides by N −1 and retrieve N −1M  = c* This problem can be simply solved using linear algebra and a code of your choice for the spectrum of eigenvalues.

(

)

As we go through this exercise, it may be useful to recall the properties of the Rayleigh stability equation (for an additional reference, see Drazin, Introduction to Hydrodynamic Stability). In the Rayleigh equation, if a complex eigenvalue exists, so does its complex conjugate pair. Therefore, to each damped stable mode, there exists a corresponding amplified, unstable mode which may be Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -28-

attributed to the time reversibility of the problem. In other words, the only way that an inviscid flow may be stable is if all modes are neutrally stable. Although such behavior may at first seem paradoxical; it should be rather expected for a non-dissipative system. A Hamiltonian system with one degree of freedom has centers for stable equilibrium points and saddles for unstable equilibrium points in the phase plane analysis. For this reason, the spectral plot will contain both positive and negative complex conjugate pairs, if modes are not all neutrally stable. For this profile, all eigenvalues predicted from the Rayleigh equation are found to be neutrally stable. Specifically for  = 0 , we obtain ci*  0 . This should not be surprising because the velocity profile has an inflection point at the outer edge of the viscous boundary-layer region, i.e., not inside the viscous domain. In fact, the absence of unstable modes can be further investigated by solving the complete OrrSommerfeld equation using an artificially large Reynolds number. By way of illustration, a plot of the eigenvalues for a Reynolds number of 1,000,000 is reproduced below for a nondimensional wave number ranging from 1 to 10 and a discretized  coordinate that is based on 20 points.

We see that this Orr-Sommerfeld solution with very small viscosity is stable within the domain given that ci*  0 . Before leaving this problem, it may be useful to remark that the presence of eigenvalues is quite sensitive to the discretization level. In our framework, we are approximating a continuous eigenvalue problem with a discrete formulation. In any such analysis, it is important to ensure that the discretization is sufficient by repeatedly increasing the number of points until further changes in the eigenvalues become negligible.

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P5-33 Consider the sinusoidal flat-plate velocity distribution: u y    = sin   = U   2  Solve the Orr–Sommerfeld Eq. Error! Reference source not found. numerically for the inviscid case with  = 0 . Begin at y = 2 assuming that   e− y and integrate inwardly to satisfy the wall conditions  =   = 0 at y =  = 0. Assuming temporal amplification, find some (damped) eigenvalues and plot ci versus  in dimensionless form. Solution: We are asked to numerically solve the Orr-Sommerfeld equation for the inviscid case with  = 0 . Based on Eq. (5-29), we can derive a non-dimensional Rayleigh equation of a form similar to Eq. (5-27). The nondimensional Orr-Sommerfeld equation can be written as i U * − c*   − 2 − U * +   − 22  + 4 = 0  Re with  u c U = U* = c* =  =  Re = U U U  This step aids in the formulation by casting the problem into a nondimensional form. Next, we let  = 0. This can be thought of as letting Re → . The result is the Rayleigh equation that stands at the basis of inviscid-stability theory: U * − c*   − 2 − U * = 0

(

)(

)

(

(

)(

)

)

Next, a computational strategy needs to be chosen given several traditional techniques to solve the stability equation. For example, one could rely on a shooting method as traditionally done in this case. However, the problem is formulated as a boundary-value problem, and the shooting method will require solving an initial value problem multiple times until convergence is achieved. Since our goal is to retrieve the eigenvalues, we can rewrite the above expression as an eigenvalue problem. For more insight, please refer to the Matlab code provided as part of the solution for P535 below. Note that this formulation can be extended from the inviscid Rayleigh equation to the full Orr-Sommerfeld equation rather straightforwardly. To proceed, we may reformulate our continuous differential eigenproblem from L1 = c* L2 to a discrete, generalized eigenproblem of the form M  = c* N  where M and N represent finite-difference operator matrices that only depend on the nondimensional wave number and discretization scheme. The nondimensional Rayleigh stability equation can then be separated into two matrices, namely,   U * −  U *2 + U * = c*   − 2

( )

(

)

(

)

Now that we have our assumed form, we may discretize  . Using a second-order finite difference scheme, we may put Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -30-

d 2  j +1 − 2 j +  j −1 = + O  2 2 2 d  Our continuous eigenvalue problem can thus be discretized. We get B j j −1 + Aj + B j j +1 = c* Bˆ j −1 + Aˆ j + Bˆ j +1

(

(

)

)

where

 A = − 2 2U * − 2U * − U * 2   j  B j = U *  2 2  Aˆ = − 2 +     Bˆ =  2 From this set-up, two tridiagonal matrices can be computed for M and N . To solve the general eigenvalue problem, we can multiply both sides by N −1 and retrieve N −1M  = c* This problem can be simply solved using linear algebra and a code of your choice for the spectrum of eigenvalues.

(

)

As we go through this exercise, it may be useful to recall the properties of the Rayleigh stability equation (for an additional reference, see Drazin, Introduction to Hydrodynamic Stability). In the Rayleigh equation, if a complex eigenvalue exists, so does its complex conjugate pair. Therefore, to each damped stable mode, there is a corresponding amplified, unstable mode which may be attributed to the time reversibility of the problem. In other words, the only way that an inviscid flow may be stable is if all modes are neutrally stable. Although such behavior may at first seem paradoxical; it should be rather expected for a non-dissipative system. A Hamiltonian system with one degree of freedom has centers for stable equilibrium points and saddles for unstable equilibrium points in the phase plane analysis. For this reason, the spectral plot will contain both positive and negative complex conjugate pairs, if modes are not all neutrally stable. For this profile, all eigenvalues predicted from the Rayleigh equation are found to be neutrally stable. Specifically for  = 0 , we obtain ci*  0 . This should not be surprising because the velocity profile has an inflection point at the outer edge of the viscous boundary-layer region, i.e., not inside the viscous domain. In fact, the absence of unstable modes can be further investigated by solving the complete OrrSommerfeld equation using an artificially large Reynolds number. By way of illustration, a plot of the eigenvalues for a Reynolds number of 1,000,000 is reproduced below for a nondimensional wave number ranging from 1 to 10 and a discretized  coordinate that is based on 20 points.

Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -31-

We see that this Orr-Sommerfeld solution with very small viscosity is unstable at the nondimensional wave numbers of  = 4 and 5 where ci*  0 . Before leaving this problem, it may be useful to remark that the presence of eigenvalues is quite sensitive to the discretization level. In our framework, we are approximating a continuous eigenvalue problem with a discrete formulation. In any such analysis, it is important to ensure that the discretization is sufficient by repeatedly increasing the number of points until further changes in the eigenvalues become negligible.

P5-34 Consider the Majdalani–Xuan polynomial profile, which represents a surprisingly simple and precise quartic velocity approximation to the Blasius boundary-layer problem for flow over a flat plate with no pressure gradient: u 5 1 y =  − 3 +  4 = U 3 3  Solve the Orr–Sommerfeld Eq. Error! Reference source not found. numerically for the inviscid case with  = 0 . Begin at y = 2 assuming that   e− y and integrate inwardly to satisfy the wall conditions  =   = 0 at y =  = 0. Assuming temporal amplification, find some (damped) eigenvalues and plot ci versus  in dimensionless form. Solution: We are asked to numerically solve the Orr-Sommerfeld equation for the inviscid case with  = 0 . Based on Eq. (5-29), we can derive a non-dimensional Rayleigh equation of a form similar to Eq. (5-27). The nondimensional Orr-Sommerfeld equation can be written as i U * − c*   − 2 − U * +   − 22  + 4 = 0  Re

(

)(

)

(

)

Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -32-

with



u c U c* =  =  Re = U U U  This step aids in the formulation by casting the problem into a nondimensional form. Next, we let  = 0. This can be thought of as letting Re → . The result is the Rayleigh equation that stands at the basis of inviscid-stability theory: U * − c*   − 2 − U * = 0

=

U* =

(

)(

)

Next, a computational strategy needs to be chosen given several traditional techniques to solve the stability equation. For example, one could rely on a shooting method as traditionally done in this case. However, the problem is formulated as a boundary-value problem, and the shooting method will require solving an initial value problem multiple times until convergence is achieved. Since our goal is to retrieve the eigenvalues, we can rewrite the above expression as an eigenvalue problem. For more insight, please refer to the Matlab code provided as part of the solution for P535 below. Note that this formulation can be extended from the inviscid Rayleigh equation to the full Orr-Sommerfeld equation rather straightforwardly. To proceed, we may reformulate our continuous differential eigenproblem from L1 = c* L2 to a discrete, generalized eigenproblem of the form M  = c* N  where M and N represent finite-difference operator matrices that only depend on the nondimensional wave number and discretization scheme. The nondimensional Rayleigh stability equation can then be separated into two matrices, namely,   U * −  U *2 + U * = c*   − 2

)

(

( )

(

)

Now that we have our assumed form, we may discretize  . Using a second-order finite difference scheme, we may put d 2  j +1 − 2 j +  j −1 = + O  2 2 2 d  Our continuous eigenvalue problem can thus be discretized. We get B j j −1 + Aj + B j j +1 = c* Bˆ j −1 + Aˆ j + Bˆ j +1

(

(

)

)

where

 A = − 2 2U * − 2U * − U * 2   j *  B j = U  2 2  Aˆ = − 2 +     Bˆ =  2 From this set-up, two tridiagonal matrices can be computed for M and N . To solve the general eigenvalue problem, we can multiply both sides by N −1 and retrieve

(

)

Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -33-

N −1M  = c* This problem can be simply solved using linear algebra and a code of your choice for the spectrum of eigenvalues.

As we go through this exercise, it may be useful to recall the properties of the Rayleigh stability equation (for an additional reference, see Drazin, Introduction to Hydrodynamic Stability). In the Rayleigh equation, if a complex eigenvalue exists, so does its complex conjugate pair. Therefore, to each damped stable mode, there is a corresponding amplified, unstable mode which may be attributed to the time reversibility of the problem. In other words, the only way that an inviscid flow may be stable is if all modes are neutrally stable. Although such behavior may at first seem paradoxical; it should be rather expected for a non-dissipative system. A Hamiltonian system with one degree of freedom has centers for stable equilibrium points and saddles for unstable equilibrium points in the phase plane analysis. For this reason, the spectral plot will contain both positive and negative complex conjugate pairs, if modes are not all neutrally stable. For this profile, all eigenvalues predicted from the Rayleigh equation are found to be neutrally stable. Specifically for  = 0 , we obtain ci*  0 . This should not be surprising because the velocity profile has an inflection point at the outer edge of the viscous boundary-layer region, i.e., not inside the viscous domain. In fact, the absence of unstable modes can be further investigated by solving the complete OrrSommerfeld equation using an artificially large Reynolds number. By way of illustration, a plot of the eigenvalues for a Reynolds number of 1,000,000 is reproduced below for a nondimensional wave number ranging from 1 to 10 and a discretized  coordinate that is based on 50 points.

We see that this Orr-Sommerfeld solution with very small viscosity is stable within the domain given that ci*  0 .

Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -34-

Before leaving this problem, it may be useful to remark that the presence of eigenvalues is quite sensitive to the discretization level. In our framework, we are approximating a continuous eigenvalue problem with a discrete formulation. In any such analysis, it is important to ensure that the discretization is sufficient by repeatedly increasing the number of points until further changes in the eigenvalues become negligible.

P5-35 Consider the Majdalani–Xuan exponential profile, which represents a very accurate velocity approximation to the Blasius boundary-layer problem for flow over a flat plate with no pressure gradient: u y = 1 − exp  − s  1 + 12 s  +  2  =   U  and s = 1.630398038629397 is the Blasius slope at the wall, also known as the Blasius constant or connection parameter. Solve the Orr–Sommerfeld Eq. Error! Reference source not found. numerically for the inviscid case with  = 0 . Begin at y = 2 assuming that   e− y and integrate inwardly to satisfy the wall conditions  =   = 0 at y =  = 0. Assuming temporal amplification, find some (damped) eigenvalues and plot ci versus  in dimensionless form.

(

)

Solution: We are asked to numerically solve the Orr-Sommerfeld equation for the inviscid case with  = 0 . Based on Eq. (5-29), we can derive a non-dimensional Rayleigh equation of a form similar to Eq. (5-27). The nondimensional Orr-Sommerfeld equation can be written as

(U

*

)(

)

− c*   − 2 − U * +

with



(

)

i   − 22  + 4 = 0  Re

u c U c* =  =  Re = U U U  This step aids in the formulation by casting the problem into a nondimensional form. Next, we let  = 0. This can be thought of as letting Re → . The result is the Rayleigh equation that stands at the basis of inviscid-stability theory:

=

U* =

(U

*

)(

)

− c*   − 2 − U * = 0

Next, a computational strategy needs to be chosen given several traditional techniques to solve the stability equation. For example, one could rely on a shooting method as traditionally done in this case. However, the problem is formulated as a boundary-value problem, and the shooting method will require solving an initial value problem multiple times until convergence is achieved. Since our goal is to retrieve the eigenvalues, we can rewrite the above expression as an eigenvalue problem. For more insight, please refer to the attached Matlab code. Note that this formulation can be extended from the inviscid Rayleigh equation to the full Orr-Sommerfeld equation rather straightforwardly. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -35-

To proceed, we may reformulate our continuous differential eigenproblem from L1 = c* L2 to a discrete, generalized eigenproblem of the form

M  = c* N  where M and N represent finite-difference operator matrices that only depend on the nondimensional wave number and discretization scheme. The nondimensional Rayleigh stability equation can then be separated into two matrices, namely,

)

(

  (U * ) −  U *2 + U * = c* (  − 2 )

Now that we have our assumed form, we may discretize  . Using a second-order finite difference scheme, we may put d 2  j +1 − 2 j +  j −1 = + O  2 2 2 d  Our continuous eigenvalue problem can thus be discretized. We get

(

)

(

B j j −1 + Aj + B j j +1 = c* Bˆ j −1 + Aˆ j + Bˆ j +1

)

where

 A = − 2 2U * − 2U * − U * 2   j  B j = U *  2 2  Aˆ = − 2 +     Bˆ =  2

(

)

From this set-up, two tridiagonal matrices can be computed for M and N . To solve the general eigenvalue problem, we can multiply both sides by N −1 and retrieve

N −1M  = c* This problem can be simply solved using linear algebra and any code of your choice for the spectrum of eigenvalues. As we go through this exercise, it may be useful to recall the properties of the Rayleigh stability equation (for an additional reference, see Drazin, Introduction to Hydrodynamic Stability). In the Rayleigh equation, if a complex eigenvalue exists, so does its complex conjugate pair. Therefore, to each damped stable mode, there is a corresponding amplified, unstable mode which may be attributed to the time reversibility of the problem. In other words, the only way that an inviscid flow may be stable is if all modes are neutrally stable. Although such behavior may at first seem paradoxical; it should be rather expected for a non-dissipative system. A Hamiltonian system with one degree of freedom has centers for stable equilibrium points and saddles for unstable equilibrium points in the phase plane analysis. For this reason, the spectral plot will contain both positive and negative complex conjugate pairs, if modes are not all neutrally stable.

Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -36-

The following plot is for 200 points in the  domain and a nondimensional  that is discretized from 1 to 50 with 50 points.

Based on the solution of the Rayleigh equation, both stable and unstable eigenvalue pairs are detected, albeit with very small growth rates. This behavior seems to indicate that the exponential profile is very weakly unstable in the absence of viscosity. Sample Code: clear; clc; %% Orr-Sommerfeld equation solver clear; clc; format long % Mean Flow Profile % % P5-32 u = @(y) (3/2)*y - (1/2)*y.^3; upp = @(y) -3*y; % % P5-33 % u = @(y) sin(pi()*y/2); % upp = @(y) ((-pi()^2)/4)*sin(pi()*y/2); % % P5-34 % u = @(y) (5/3)*y - y.^3 + (1/3)*y.^4; % upp = @(y) 4*y.^2 - 6*y; % % P5-35 % s = 1.630398038629397; % u = @(y) 1 - exp(-s*y*(1+(1/2)*s*y+y.^2)); % upp = @(y) -y*(9*y.^3 *s + 6*y.^2 * s^2 + y*s*(s^2 + 6)+2*(s^2 3))*s*exp(-y.^3 *s-(y.^2 * s^2)/2 - y*s); % Parameters alpha = linspace(1,30,30); Re = 1000000; Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -37-

for k = 1:length(alpha) yCoord = linspace(0,2,50); dy = yCoord(2) - yCoord(1); P = @(y) u(y) - 2*alpha(k)*1i/Re; Q = @(y) (alpha(k)^3)*1i/Re -1*alpha(k)^2 * u(y) - upp(y); A = @(y) 6*1i/(alpha(k)*Re) + dy^2 * (dy^2 * Q(y) - 2*P(y)); B = @(y) -4*1i/(alpha(k)*Re) + dy^2 * P(y); C = 1i/(alpha(k)*Re); Ahat = -1*dy^2 * (2+alpha(k)^2 * dy^2); Bhat = dy^2; SizeMatrix = length(yCoord)-1; M = zeros(SizeMatrix, SizeMatrix); N = zeros(SizeMatrix, SizeMatrix); for i = 2:SizeMatrix-1 N(i,i) = Ahat; N(i,i+1) = Bhat; N(i+1,i) = Bhat; end N(1,1) = Ahat; N(1,2) = Bhat; N(end,end-1) = Bhat; N(end,end) = Ahat; for i = 3:SizeMatrix-2 M(i,i) = A(yCoord(i+1)); M(i,i+1) = B(yCoord(i+1)); M(i+1,i) = B(yCoord(i+1)); M(i,i+2) = C; M(i+2,i) = C; end M(1,1) = C+ A(yCoord(2)); M(1,2) = B(yCoord(2)); M(1,3) = C; M(2,1) = B(yCoord(3)); M(2,2) = A(yCoord(3)); M(2,3) = B(yCoord(3)); M(2,4) = C; M(end,end) = C + A(yCoord(end)); M(end,end-1) = B(yCoord(end)); M(end,end-2) = C; M(end-1,end) = B(yCoord(end-1)); M(end-1,end-1) = A(yCoord(end-1)); M(end-1,end-2) = B(yCoord(end-1)); M(end-1,end-3) = C; EigenValues{k} = eig(N^(-1)*M); EigenValuesNonzero{k} = find(imag(eig(N^(-1)*M))); end alphaPlot = []; for i = 1:length(alpha) if EigenValuesNonzero{i} ~=0 alphaPlot = [alphaPlot; alpha(i)]; EigenP = EigenValues{i}; Index = EigenValuesNonzero{i}; for j = 1:length(Index) Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -38-

EigenPlot(j) = imag(EigenP(Index(j))); figure(1) scatter(alpha(i),EigenPlot(j)) hold on end end end title('Problem 5-34') xlabel('Dimensionless wave number') ylabel('Dimensionless imaginary wave speed') hold off %% Rayleigh equation (inviscid stability) clear; clc; format long % Mean Flow Profile % % P5-32 u = @(y) (3/2)*y - (1/2)*y.^3; upp = @(y) -3*y; % % P5-33 % u = @(y) sin(pi()*y/2); % upp = @(y) ((-pi()^2)/4)*sin(pi()*y/2); % % P5-34 % u = @(y) (5/3)*y - y.^3 + (1/3)*y.^4; % upp = @(y) 4*y.^2 - 6*y; % % P5-35 % s = 1.630398038629397; % u = @(y) 1 - exp(-s*y*(1+(1/2)*s*y+y.^2)); % upp = @(y) -y*(9*y.^3 *s + 6*y.^2 * s^2 + y*s*(s^2 + 6)+2*(s^2 3))*s*exp(-y.^3 *s-(y.^2 * s^2)/2 - y*s); % Parameter alpha = linspace(1,100,500); for k = 1:length(alpha) yCoord = linspace(0,2,500); dy = yCoord(2) - yCoord(1); P = @(y) u(y) ; Q = @(y) -1*alpha(k)^2 * u(y) - upp(y); A = @(y) 1 * (dy^2 * Q(y) - 2*P(y)); B = @(y) P(y); C = 0; Ahat = -1 * (2+alpha(k)^2 * dy^2); Bhat = 1; SizeMatrix = length(yCoord)-1; M = zeros(SizeMatrix, SizeMatrix); N = zeros(SizeMatrix, SizeMatrix); for i = 2:SizeMatrix-1 N(i,i) = Ahat; N(i,i+1) = Bhat; N(i+1,i) = Bhat; end N(1,1) = Ahat; Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -39-

N(1,2) = Bhat; N(end,end-1) = Bhat; N(end,end) = Ahat; for i = 3:SizeMatrix-2 M(i,i) = A(yCoord(i+1)); M(i,i+1) = B(yCoord(i+1)); M(i+1,i) = B(yCoord(i+1)); M(i,i+2) = C; M(i+2,i) = C; end M(1,1) = C+ A(yCoord(2)); M(1,2) = B(yCoord(2)); M(1,3) = C; M(2,1) = B(yCoord(3)); M(2,2) = A(yCoord(3)); M(2,3) = B(yCoord(3)); M(2,4) = C; M(end,end) = C + A(yCoord(end)); M(end,end-1) = B(yCoord(end)); M(end,end-2) = C; M(end-1,end) = B(yCoord(end-1)); M(end-1,end-1) = A(yCoord(end-1)); M(end-1,end-2) = B(yCoord(end-1)); M(end-1,end-3) = C;

% % end

EigenValues{k} = eig(N^(-1)*M); EigenValuesNonzero{k} = find(imag(eig(N^(-1)*M))); Index = EigenValuesNonzero{1}; Values = EigenValues{1};

alphaPlot = []; for i = 1:length(alpha) if EigenValuesNonzero{i} ~=0 alphaPlot = [alphaPlot; alpha(i)]; EigenP = EigenValues{i}; Index = EigenValuesNonzero{i}; for j = 1:length(Index) EigenPlot(j) = imag(EigenP(Index(j))); figure(1) scatter(alpha(i),EigenPlot(j)) hold on end end end title('Problem 5-35') xlabel('Dimensionless wave number') ylabel('Dimensionless imaginary wave speed') hold off

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P5-36 Determine whether the following profiles are prone to instability in the context of an inviscid, incompressible, external fluid domain ( 0 ≤ y ≤ ∞ ):

(a) U = 1 + y 2 ; (b) U = y3 ; (c) U = y 4 ;(d )U = e− y ; (e) U = e− y ;( f )U = e− y cos( y);( g )U = tanh( y) 2

2

Solution: The general stability criteria are i. ii. iii.

Stable: U "  0 Stable: U "  0 Stable: U " = 0 and also U "(U − U s )  0

iv.

Unstable: U " = 0 and also U "(U − U s )  0

U s = U ( ys ) , where ys is a point at which U " = 0 . Assumption: Assuming that the profile is an inviscid, incompressible, external fluid domain ( 0 ≤ y ≤ ∞ ). (a) U = 1 + y

2

U ' = 2y

U"= 2 U"0 It is stable. (b) U = y

(Ans. a)

3

U ' = 3y2 U " = 6y U " y 0  0

It is stable. U " y = 0 = 0 and U "(U − U s ) = 0

Therefore, at y = 0 this profile is prone to instability. (c) U = y

(Ans. b)

4

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U ' = 4 y3

U " = 12 y 2 U " y  0  0 . It is stable. U " y = 0 = 0 and U "(U − U s ) = 0

Therefore, at y = 0 this profile is prone to instability.

(Ans. c)

(d) U = e− y U ' = −e− y

U " = e− y U " y 0  0

It is stable as e0 = 1 . (e) U = e

(Ans. d)

− y2

U ' = −2 y e− y

2

U " = 4 y 2e− y − 2e− y 2

2

U " = e− y ( 4 y 2 − 2 ) 2

U " y = 0  0 . It is stable. U " y  0  0 . It is stable.

(Ans. e)

(f) U = e− y cos ( y ) 2

U ' = −e − y sin ( y ) + cos ( y ) e − y 2

2

( −2 y ) = −e− y ( sin ( y ) + 2 y cos ( y ) ) 2

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U " = e− y ( 4 y 2 cos ( y ) + 4 y sin ( y ) − cos ( y ) ) 2

U " y = 0  0 . It is stable. U " y  0  0 . It is stable.

(Ans. f)

(g) U = tanh ( y )

U ' = sech 2 ( y ) U " = 2sech ( y ) sech ( y ) tanh ( y ) = 2sech 2 ( y ) tanh ( y ) U " y  0  0 . It is stable. U " y = 0 = 0 and U "(U − U s ) = 0

Therefore, at y = 0 this profile is prone to instability.

(Ans. g)

P5-37 Given an inviscid motion U(y), show that if σ represents an eigenvalue of the Rayleigh equation given by Eq. (5-27) then so does its complex conjugate.

 U"  +  2  v = 0 (Eq. 5-27) U −c 

Solution: Rayleigh equation: v "−  Boundary layer: U ( 0) = U (  ) = 0

U "( y; c ) is a non-trivial solution of Rayleigh’s equation. Integrating Eq. (5-27) from 0 to ∞ gives





0

( v'

2

+ 2 v

2

) dy + 



0

U" 2 v dy = 0 (U − c)

c represents an eigenvalue. Therefore * U " 2 U " (U − c ) 2 U " (U − Cr + ici ) 2 v = v = v 2 2 U −c U −c U −c

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where ci is the complex conjugate of c. So we can write the imaginary part as ci 



0

U" U −c

v dy = 0 2

2

For stable flow, ci = 0 or for instability ci  0 . Hence proved.

(Ans)

P5-38 Compare the benefits and deficiencies of the Orr–Sommerfeld equation to the en method in the treatment of boundary-layer instability. Solution: Benefits of the Orr–Sommerfeld equation: i. ii. iii.

It is used for linear stability theory for parallel flow problems. It governs small perturbations in the velocity component normal to the wall in parallel shear flow. Boundary layer responds to external forcing.

Benefits of the en method: i. ii. iii.

It is a useful tool for both linear and nonlinear stability theory. It links the transition location to a certain degree of amplification. Small deviations may accumulate during the integration over a large streamwise distance.

Deficiencies of the Orr–Sommerfeld equation: i. ii.

iii.

It cannot predict where breakdown of laminar boundary layer will occur. Factors not taken into consideration in the equation: a. The normal velocity of the mean flow profile b. Upstream conditions c. Wall and freestream disturbances d. Nonlinear mode of interaction for large amplitude disturbances and mean flow distortion Solutions are lengthy and sophisticated.

Deficiencies of the en method: i. ii. iii.

Boundary layer responding to the external forcing has not been considered. Amplitude A0 is same for all frequencies. Initial disturbance predicted by linear theory will have a factor of en.

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P5-39 Discuss the implications of Squire’s theorem on the stability of two-dimensional planar motions that are subject to three-dimensional disturbances under (a) inviscid and (b) viscous conditions. Solution: (a) Inviscid case: Squire’s theorem implies that to each unstable three-dimensional disturbance there corresponds a more unstable two-dimensional one. It is sufficient to look at 2D perturbations governed by the Rayleigh equation. Proof: If c = f ( ) is the solution of the two-dimensional problem of Eqs. (5-27) and (5-28), then

c = f ( ) is the solution of the equivalent two-dimensional problem. 1/2 By Squire’s transformation, c = f ( 2 +  2 )  is the solution of the three-dimensional   problem, where  is the wave number.

Thus, to each unstable three-dimensional disturbance with growth rate  ci , there corresponds a two-dimensional disturbance with growth rate  ci which is more unstable since    if  = 0 .

(Ans.a)

(b) Viscous case: Squire’s theorem implies that if a 3D mode ei x+i y becomes unstable at a particular value of Re, then there is an equivalent 2D mode which is unstable at a smaller value of Re. It therefore suffices to consider only 2D modes. Proof: Consider the 3D perturbation qˆ = q ( y ) exp i ( x +  y − ct )  and a simplified form of the eigenvalue problem.

i u + v + i = 0

(Eq. 1)

(c + iU )u + vu  = −i p + (c + i u )v = − p +

1 n u −  2u −  2u ) ( Re

1  v −  2v −  2v ) ( Re

(c + i ) = −i p +

1   −  2 −  2 ) ( Re

(Eq. 2)

(Eq. 3)

(Eq. 4)

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Variables u=

u and p can be defined as

( u +  ) and p =  p 



 2 = ( 2 +  2 )

(Eq. 1) becomes

i u + v ' = 0

(Eq. 5)

  (Eq.2) +   (Eq.4) we obtain, (c + i u ) +  vU  = −i 2 p +

R=

1  ( u  −  2v ) Re

 Re c , c=  

(c + iU )u + vU  = −i p + (c + iU )V = − p +

1  ( u −  2u ) R

1  ( v −  2v ) R

(Eq. 6)

(Eq. 7)

Comparing the above equations to the first four equations, we find that for 3D problems, finding c in terms of  and R is mathematically identical to the 2D problem obtained by setting  = 0 . Then c = −i , c = −i If Re is finite, then the 3D mode ( ,  ) will become unstable at some critical Reynolds

 Rc .  By mathematical similarity, the 2D mode ( ,0 ) will go unstable at Re = Rc . Since number Re = Rc , corresponding to R = Rc =

Rc  Rc , an equivalent 2D mode will go unstable at a lower Reynolds number. Thus, 2D modes are the first to go unstable as Re increases, and it is sufficient to consider 2D only.

P5-40 Rayleigh’s criterion for rotational instability states that “an inviscid rotating flow is unstable if the square of its circulation decreases outward.” Using cylindrical polar coordinates (r, θ, z), consider the motion of an inviscid rotating field prescribed by V(r ) = 0,V (r ),0 over a

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finite interval ri ≤ r ≤ ro, where the vorticity retains a single nonzero component in the z direction, namely, ω = ( 0,0, z ) . (a) Show that 1 d z = ( rV ) ; r dr

dz  0; dr

and

d ( r 2V2 ) dr

0

Is the flow stable or unstable? (b) Assuming Couette flow [Eq. (3-22)] between two long cylinders where the subscripts i and o denote the inner and outer surfaces rotating at angular speeds Ωi and Ωo, show that a condition leading to instability corresponds to

i ri 2  o ro2 From this condition, what can you say about the stability of the motion if the outer cylinder is stationary and the inner cylinder is rotating? Conversely, what happens if the inner cylinder is stationary and the outer cylinder is rotating? Finally, what happens anytime the two cylinders rotate in opposite directions? Solution: (a) The vorticity is  =  v . Here v = ( 0, v (r ),0) = v ( r ) and  = ( 0,0, z ) are in cylindrical coordinates. Now, we know that in cylindrical coordinates,  1 vz v   vr vz v =  − −  rˆ +  z   z r  r 

 ˆ 1   ( rv ) vr −  +  r  r  

  zˆ 

Here vr = 0, v = v ( r ) and vz = 0 . Also,

v = 0 (as v is a function of r only). z

Hence only   v =

 z =

zˆ component remains.

1  ( rv ) zˆ r r

1 d ( rv ) r dr

(Ans)

Hence proved.

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dv  dv v 1 z =  v + r   =  +  r dr  dr r d  z d 2 v v = − 2 0 dr dr 2 r

(Ans)

Hence proved. d ( r 2 v2 ) dr

= 2rv2 + r 2 2v

dv 0 dr

This solution for v holds good only for the finite interval ri ≤ r ≤ ro. The flow is stable. (b) (i) When the outer cylinder is stationary and the inner cylinder is rotating, then 0 = 0 . Then the flow is always unstable. (ii) When the outer cylinder is rotating and the inner cylinder is stationary, then the flow is linearly stable. Hence, this is the basis for the modern viscometer, which works by avoiding the vortical structure and obtains an accurate measurement of the viscosity of fluids. (iii) For two cylinders rotating in opposite directions, the region close to the inner cylinder is unstable and the region close to the outer cylinder is stable. According to Taylor, “In the inner region, the square of the circulation decreases outward, so that centrifugal force tends to make the flow unstable. In the outer region the square of the circulation increases so that centrifugal force tends to make the flow stable.”

P5-41 Consider the following time-dependent convective equation for u(x, t) , u u +c = u t x

where c and σ are positive constants. (a) Show that any disturbance of the form u(x, t) = εeσt f (x − ct) represents a valid solution. (b) By specifying disturbances to the zeroth-order solution u0 = 0 using the normal mode form u(x, t) = εeσt exp [iα(x − ct)] , determine the condition for stability. (c) Assuming u(x, t) = εeσt sech2(x − ct) , illustrate the behavior of this disturbance graphically. (d) How does the solution behave as t → ∞ at a fixed point in space? (e) After a very long time, if u becomes unbounded at a fixed point in space, it will signal the presence of an absolute instability. Show that this disturbance becomes absolutely unstable for σ > 2c. Solution: (a) Disturbance of the form Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -48-

u =  e t f ( x − ct ) u =  e t f ( x − ct ) − c e t f  ( x − ct ) t u =  u − c e t f  ( x − ct ) t u =  e t f  ( x − ct ) x

Substituting the partial derivatives in the following time-dependent convection equation u u +c =  u , we get t x

 u − c e t f  ( x − ct ) + c e t f  ( x − ct ) =  u t

Hence, the disturbance in the form u =  e f ( x − ct ) is a valid solution.

(Ans.a)

(b) For obtaining the zeroth-order solution using the normal mode form, we can write the timedependent convection equation as



u u +c = u t x

u =  e t exp[i ( x − ct )] u =  t  e t exp[i ( x − ct )] − i c e exp[i ( x − ct )] t

u = ( − i c ) u t u = i e t exp[i ( x − ct )] x

Hence the time-dependent convection equation becomes

  − i c u + ci cu =  u Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -49-

 u −  i cu + ci cu =  u

 u ( −1) = i cu ( − 1)

 = ic Therefore, the condition for stability is

 = ic .

(Ans.b)

(c) Behavior of the disturbance u ( x, t ) =  e t sech 2 ( x − ct ) is shown in the following graphs.

Fig. (a)

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Fig. (b)

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Fig. (c) (d) The solution behaves asymptotically stable as t →  at fixed point in space.

(Ans.d)

(e) This disturbance becomes absolutely unstable for a  2c as shown in fig. (c). It states that small perturbation grows above a given threshold at a fixed point of the flow. (Ans.e)

P5-42 Consider the following time-dependent convection-diffusion equation for u(x, t), u u  2u +c = + ( −  0 ) u t x x 2

where c, σ, and σ0 are positive constants. (a) Show that the flow will be stable so long as σ < σ0. (b) If, instead, we have σ > σ0, find the range of c that will give rise to convectively unstable disturbances as well as the range that will trigger a case of absolute instability. Solution: Let us consider the normal modes of the form Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -52-

u( x, t ) = Aei ( x −t ) u = −i Aei ( x −t ) = −iu t u = i Aei ( x −t ) = i u x  2u = − 2 Aei ( x −t ) = − 2u x 2

Substituting the partial differentiations in the equation,

−iu + ci u +  2u − ( −  0 ) u = 0 −i + i c +  2 − ( −  0 ) = 0

 −  c + i 2 − i ( −  0 ) = 0  = f ( ) f ( ) =  c − i 2 + i ( −  0 )

f  ( ) = c − 2i

f  (0) = c f (0) = i ( −  0 )

 −  0 = −if (0) (a) The flow is asymptotically stable when    0 because all modes decay exponentially. (b) The flow is convectively unstable if    0 and c  0 because the modes which grow exponentially also propagate. The flow is absolutely unstable if    0 and c = 0 .

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P5-43 Using cylindrical polar coordinates (r, , z) in conjunction with a velocity vector

V = ( vr , v , vz ) and pressure P, the continuity and momentum equations for inviscid, incompressible motion reduce to  vr v v v v v 2  P + vr r +  r + vz r −   = − r r  z r  r  t



1  1 v v ( rvr ) +  + z = 0 r r r  z

and

v v v v v v  v    + vr  +   + vz  + r  r r  z r  t

v v v v  v   z + vr z +  z + vz  r r  z  t

1 P  =− r  

P  =− z 

Let us now consider the case of a rotating axisymmetric flow between two concentric cylinders, where the velocity field can be prescribed by V = V0 +  v(x, t ) with 0   1 . The basic motion in this problem may be captured by a strictly swirling (azimuthal) velocity component of the form, V0 (r ) = [0,V (r ),0] . (a) Start by writing the disturbances as v(x, t ) = (uˆ, vˆ, wˆ ) = (u, v, w)ei z + cit and show that the radial disturbance must satisfy a linear, second-order, differential equation, namely, d 2u 1 du u  2 B  1 d 2 2 + − 2 −   + 2  u = 0, where B = 3 (r V ) . 2 dr r dr r  ci  r dr

(b) A journal bearing may be modeled as being composed of an inner cylinder of radius ri and an outer cylinder of radius ro  ri . Assuming that the steady viscous motion in the cylindrical annulus −1 is expressible as V (r ) = C1r + C2 r , determine the appropriate constants and then proceed to deduce B.

(c) By fixing the longitudinal wavenumber α, and assuming that B can be negative, discuss the stability characteristics of this motion. Solution: i z + ci t i z + ci t i z + ci t (a) Vr =  ue , V = V (r ) + òve , Vz =  e

V vx Vr =  ci uei z + cit ,  =  ci vei z + cit , =  ciei z + cit t t t Vr du i z + cit V V (r ) Vz , , =  iei z + cit = e = r z dr r dr

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From continuity equation:

 Vr Vr  Vz + + =0   r r  z 

du i z + ci t  uei z + ci t e + +  i ei z + ci t = 0 dr r

(Eq. 1)

From momentum balance,

Vr V2 1 P − =− t r  r

 ci uei z + cit

( v(r ) +  ve −

 ci u −

=0

r

 ci uei z + c t − i

)

i z + ci t 2

2V (r ) ei z + ci t =0 r

2V (r )v =0 r

(Eq. 2)

V V V V 1 P + Vr  + r  = − t r r r  i z + ci t

 ci e

 ci e

i z + ci t

 ci + u

)

V (r ) +  ei z + cit dV (r ) i z + cit +  ue =0 dr r

i z + ci t

dV (r )  e + dr

+  ue

+  ue

(

i z + ci t

dV (r ) uV (r ) + =0 dr r

i z + ci t

r

uV ( r )

=0

(Eq. 3)

Vz P =− t z

 ci ei z + cit = 0

 ci = 0

(Eq – 4)

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Substituting  = 0 in (Eq. 1), we get du u + =0 dr r

(Eq. 5)

Differentiating (Eq. 5) with respect to r, we get d 2u 1 du u + − =0 dr 2 r dr r 2

(Eq. 6)

From (Eq. 3), we get

v=−

 dV (r ) u  u dr + r V (r ) 

1  ci

Substituting  in (Eq. 2), we get

 ci u +

2 V (r )  dV (r ) u  u + V (r )  = 0   ci r  dr r 

Multiplying the above equation with

 ci

, we get

 dV (r ) 2V 2 (r )  2 V ( r ) + =0  dr r  

 2u +

u rci2

 2u +

u  dV (r )  2V (r )r 2 + 2V 2 (r )r  = 0  rc  dr 

 2u +

u 1 d 2 ( r V (r ) ) = 0 ci2 r 3 dr

3 2 i

 2 B  + 2 ci 

 u = 0 

where B =

(Eq. 7)

1 d 2 ( r V ( r )) . r 3 dr

Subtracting (Eq. 7) from (Eq. 6), we get Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -56-

d 2u 1 du u  2 B + − −  + 2 dr 2 r dr r 2  ci

 u = 0 

(Ans. a)

Hence proved. −1 (b) V (r ) = C1r + C2 r Boundary conditions are

At r = ri ,  = i = C1 +

C2 ri 2

At r = r0 ,  = o = C1 +

C2 ro2

C C  i − 0 =  22 − 22  r0   ri   − 0  2 2 C2 =  2i r r 2  i 0  r0 − ri 

Let  =

r o , R = i 1 ro i

 1−   C2 = i ri 2  2 1 − R  Then C1 becomes   − R2  C1 = i  2   1− R    − R2   1−   1 V (r ) = i  r + i ri 2  2  2 1 − R  r  1− R 

B=

1 d 2 ( r v (r ) ) r 3 dr

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B=

C  1 d  2 r C1r + 2    3 r dr   r 

B=

1 d C1r 3 + C2 r  3 r dr

B=

1 3r 2C1 + C2  3  r

B=

1 r3

 2   − R2   1−   + i ri 2  3r i  2  2  1 − R    1− R  

 ri 2  i   3 2 B=  − R + (1 − )  ) 2  ( 3 r 1 − R   r 

(c) B  0

  R2 Hence, the motion becomes unstable. Therefore, the stability criterion is

1 d 2 ( r V (r ) )  0 . r 3 dr

P5-44 Assuming strictly two-dimensional flow conditions with a spanwise wave number of n = 0 in the z direction and a basic motion that is prescribed by a two-component velocity vector of the form V0 ( x, y) = [U ( x, y),V ( x, y),0] , (a) show that the linearized Navier–Stokes equations with biglobal stability disturbance modes, given by Eqs. (5-82)–(5-85), can be readily reduced into

u v + =0 x y

(continuity)

  2u  2u  U U u u 1 p u+ v +U +V + =   2 + 2  ( x -mom. ) x y x y  x  x y    2v  2v  V V v v 1 p −i v + u+ v +U +V + =   2 + 2  ( y -mom. ) x y x y  y  x v  −i u +

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where pˆ ( x, y, z, t ) = p( x, y)e−i t and v( x, y, z, t ) = (uˆ, vˆ,0) = [u( x, y), v( x, y),0]e−i t . (b) Using a disturbance stream function that satisfies the linearized continuity equation identically, eliminate u and υ from the remaining momentum equations by substituting  y

u=

and

v=−

 x

Show that the resulting momentum equations can be rearranged into

  3 1 p  U  U   2  2  3  = i − + −U − V 2 +  2 + 3  ( x -mom.)  x y x y y x xy y  x y y    3 1 p  V  V   2  2  3  = −i − + +U 2 +V −  3 +  ( y -mom.)  y x x y y x x xy xy 2   x (c) The pressure can be eliminated by taking the partial derivative of the x-momentum pressure gradient with respect to y and setting it equal to the partial derivative of the y-pressure gradient with respect to x. Bearing in mind that the basic flow motion represented by (U, V) must satisfy continuity, show that the two momentum equations can be consolidated into a single fourth-order partial differential equation for the disturbance stream function viz.

  2  2    2V  2U     2U  2V   i  2 + 2  =  − + −   y   xy y 2  x  xy x 2  y  x   3   3  3   3    4  4  4  +U  3 +  + V  3 + 2  −  4 + 4 + 2 2 2  xy 2  x y   x y x y   x  y Solution: Assumptions: The flow is two dimensional ( = 0 ) .

i. ii. iii.

Spanwise wave number n = 0 . Two-component velocity vector is of the form V0 ( x, y) = [U ( x, y),V ( x, y),0] .

(a) Navier–Stokes equation with biglobal stability disturbance modes is given by u v + + inw = 0 (continuity) x y

−i u +

(5-82)

  2u  2u  U U U u u 1 p u+ v+ w +U +V + inWu + =   2 + 2 − n 2u  ( x-mom.) (5-83) x y z x y  x y  x 

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  2v  2v  V V V v v 1 p u+ v+ w +U +V + inWv + =   2 + 2 − n 2v  ( y -mom.) (5-84) x y z x y  y y  x  2 2  w  w  W W W w w p −i w + u+ v+ w +U +V + inWw + in =   2 + 2 − n 2 w  ( z -mom.) (5-85) x y z x y  y  x 

−i v +

Substituting the assumptions in the above equations, the continuity equation becomes u v + = 0 (continuity) (Eq. 1) x y

x-mom and y-mom equations become   2u  2u  U U u u 1 P −i u + u+ v +U +V + =   2 + 2  ( x-mom) (Eq. 2) x y x y  x y   x −i v +

  2v  2v  V V v v 1 P u+ v +V +V + =   2 + 2  ( y -mom) (Eq. 3) x y x y  y y   x

where pˆ ( x, y, z, t ) = p( x, y)e−i t and v( x, y, z, t ) = (uˆ, vˆ,0) = [u( x, y), v( x, y),0]e−i t .

Hence proved.

(Ans. a)

(b) Disturbance stream function is given by u=

 y

u  2  2u  3 =  2 = 2 x xy x x y u  2  2u  3 = 2  2= 3 y y y y v=−

 x

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v  2  2v  3 =− 2  2 =− 3 x x x x

v  2  2v  3 =−  2 =− y xy y xy 2 Substituting the partial derivatives of u and v in Eqs. (1) and (2), we get −i

 U   +  y x  y

 U +  y

  −  x

  2    2  + U + v    2    xy   y

 1 P   3  3  + =  +   2 3    x  x y y 

  2    2    3 1 P  U    U     3  = i −  − v  2  +   2 + 3  ( x-mom) (Eq. 4)  +   −U   x y x  y  y  x   xy   y   x y y 

Hence proved.

(Ans. b)

  2    2  1 P   3  3     V    V    −i  − =  3 + +   + + −  +U  2  +V  − xy 2   x  x  y  y  x   x   xy   x  x 1 P  V   = −i −   y y x  y

  2    2  U    + + V + V    2     y  x   xy   x

   3  3  −  +   3  ( y -mom) (Eq. 5) xy 2    x

Hence proved.

(Ans. b)

(c) To eliminate pressure term, (Eq. 4) is partially differentiated with respect to y, and we get   4 1 2 P  2 U  2 V   3  2  4  = i − + − U − V +  +  x 2y 2 y 4   xdy y 2 x y 2 y 2 x xy 2 y 3  

(Eq. 6)

(Eq. 5) is partially differentiated with respect to x, then we get   4 1 2 p  2  2V  V  2  3  2  4  = −i − + + V + V −  +  4   xdy x 2 x 2 y y x 2 x3 x 2y x 2y 2   x

(Eq. 7)

Equating (Eq. 6) and (Eq. 7), we get

  2  2    2V  2U     2U  2V   i  2 + 2  =  − + −   y   xy y 2  x  xy x 2  y  x   3   3  3   3    4  4  4  +U  3 +  + V  3 + 2  −  4 + 4 + 2 2 2  xy 2  x y   x y x y   x  y

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Hence proved.

(Ans. c)

5-45 The Taylor profile, which has been used to describe the ideal gaseous motion in slab-type rocket motors (Taylor 1956), corresponds to a two-dimensional self-similar profile in a porous channel [Griffond and Casalis (2001)]. When the channel half height is taken to be unity, the basic flow stream function may be written as  0 ( x, y) = x sin( 12  y) for 0  x  X N ,0  y  1, where

X N represents the length of the porous channel (see Fig. P5-45).

FIGURE P5-45 Assuming a wall injection speed of unity, the basic flow velocity may be expressed as V0 ( x, y) = U ( x, y)i + V ( y) j with U = 12  x cos( 12  y) and V = − sin( 12  y ) . As shown in the problem above, the linearized Navier–Stokes equations with two-dimensional biglobal stability disturbance modes (and n = 0 in the z direction) can be consolidated into a single, fourth-order, partial differential equation for the disturbance stream function  , where ˆ =  ( x, y )e V ( y) is independent of x , the stream function formulation can be further reduced into   2  2 i  2 + 2 y  x

− i t

. Since

   3  2U   2U   3  = − + + U +   3  y 2 x xy y xy 2    x

  3   4  4  3   4  + V  3 + 2  −  4 + 4 + 2 2 2  x y  y x y   y  x

The basic flow boundary conditions require uniform wall-normal injection, no slip at the headwall, and symmetry with respect to the midsection plane, y = 0 , specifically:   0  0 ( x,1) = 1; ( x,1) = 0 (wall-normal injection)  y  x   0   0 (0, y ) = (0, y ) = 0 (no slip at headwall)   x  y     2 0  0 ( x, 0) = ( x, 0) = 0 (V = U / y = 0 at y = 0) y 2   x Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -62-

(a) Recalling that the instantaneous motion is recoverable from  =  0 +ˆ , and that disturbances are not allowed to alter the basic flow conditions at the boundaries, show that the disturbances must satisfy:    ( x,1) = ( x,1) = 0 (wall-normal injection)  y  x     (0, y ) = (0, y) = 0 (no slip at headwall)  y  x    2  ( x, 0) = 2 ( x, 0) = 0 (midsection symmetry) y   x You may also impose the outflow condition suggested by Theofilis, Duck and Owen (2004) by setting a simple extrapolation relation at the domain’s exit plane, x = X N , namely,

 ( X N , y) =

X N − X N −2 X − XN  ( X N −1 , y) + N −1  ( X N −2 , y) X N −1 − X N −2 X N −1 − X N −2

(b) Using the Taylor basic flow profile and a numerical solver of your choice, find the eigenvalues for n = 0, X N = 8 , and  = 1/ 900 = 1.1110−3m2 /s . (c) Optional: Compare your findings to those of Griffond and Casalis (2001), “On the nonparallel stability of the injection induced two-dimensional Taylor flow.” Solution: (a) First, the two components of the mean flow are expressible by U = 12  x cos( 12  y) and V = − sin( 12  y ) We also know that that the instantaneous velocity and flow streamfunction are related through   and V = − U= y x Since the instantaneous motion is recoverable from  =  0 +ˆ =  0 + ( x, y)e−i t , we may differentiate  with respect to x and y to obtain   0 ˆ  0  −i t  0 ˆ  0  − i t and  = = + = + e + = + e y y y y y x x x x x For a porous channel with uniform wall injection, the fundamental boundary conditions on the instantaneous field are the same as those on the mean flow (because the “disturbances are not allowed to alter the basic flow conditions at the boundaries”). We have, for any time t , (1) A uniform injection and no-slip condition at the sidewall, U ( x,1, t ) = 0 and V ( x,1, t ) = −1 (2) A prescribed zero injection pattern at the headwall, U (0, y, t ) =V (0, y, t ) = 0

U ( x, 0, t ) = V ( x, 0, t ) = 0 y By conveying these boundary conditions to the streamfunction, we get Sidewall: (3) A vanishing speed at the centerline,

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 0   ( x,1, t ) = ( x,1) + ( x,1)e−i t x x x

and

 0   ( x,1, t ) = ( x,1) + ( x,1)e − i t y y y

or

−V ( x,1, t ) = 1 + U ( x,1, t ) = 0 +

   ( x,1)e −i t  1 = 1 + ( x,1)e −i t ; t  ( x,1) = 0 x x x

   ( x,1)e −i t  0 = 0 + ( x,1)e −i t ; t  ( x,1) = 0 y y y

Headwall:  0    0   (0, y, t ) = (0, y) + (0, y)e−i t and (0, y, t ) = (0, y ) + (0, y )e − i t y y y x x x or    −V (0, y, t ) = 1 + (0, y)e −i t  0 = 0 + (0, y)e −i t ; t  (0, y) = 0 x x x

U (0, y, t ) = 0 +

   (0, y)e −i t  0 = 0 + (0, y)e −i t ; t  (0, y) = 0 y y y

Centerline:  2 0  2  2  0   −i t ( x, 0, t ) = ( x, 0) + 2 ( x, 0)e−i t and ( x, 0, t ) = ( x, 0) + ( x, 0)e 2 2 y y y x x x or    −V ( x, 0, t ) = 1 + ( x, 0)e −i t  1 = 1 + ( x, 0)e −i t ; t  ( x, 0) = 0 x x x U  2  2  2 ( x, 0, t ) = 0 + 2 ( x, 0)e −i t  0 = 0 + 2 ( x, 0)e −i t ; t  ( x, 0) = 0 y y y y 2

(b) Eigenvalues for n = 0, X N = 8 , and  = 1/ 900 = 1.1110−3m2 /s may be obtained using a QZ solver with Chebychev discretization and N=50 with application to the incompressible Taylor mean flow profile as the basic flow solution. Results are illustrated below.

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Here, the hollow squares (□) represent eigenmodes obtained from the present incompressible stability solver. (c) Based on Griffond and Casalis (2001), “On the nonparallel stability of the injection induced two-dimensional Taylor flow,” the stability approach is based on the Local-Non-Parallel (LNP) approach. Although the LNP approach is one dimensional, it takes into account the fact that changes in the axial direction are slower than changes in the normal direction. No such assumptions are made using the biglobal approach, which is truly two dimensional. Predictions based on the biglobal approach are therefore expected to outperform in accuracy those of the LNP approach. There are several differences between the two methodologies. Griffond and Casalis (2001) solve a local stability problem, investigating how a location in space responds to small disturbances. However, the biglobal approach is a global method that accounts for the entire flow at once. Therefore, a specific point in the solution domain can be predicted to be locally unstable, yet globally stable. Additionally, since the LNP approach is essentially one dimensional, it assumes the form of the spatial growth or decay in the remaining spatial dimensions. For the present case, Griffond and Casalis (2001) assume that  ( x, y, t ) =  ( y ) xr +ii exp ( −i t ) where  represents the real temporal frequency and  =  r + i i combines the spatial growth rate

 r and the spatial eigenfrequency  i ; on the other hand, the biglobal approach employs the following ansatz:

 ( x, y, t ) =  ( x, y ) exp i ( x −  r t )  exp ( it )

where  is the spatial mode number in the axial direction and  =  r + i i stands for the complex eigenmode. In short, the LNP approach represents a one-dimensional local or spatial approach, whereas the biglobal approach represents a global temporal approach.

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P5-46 Assuming strictly axisymmetric flow conditions with an azimuthal wave number of m = 0 and a basic motion that is prescribed by a two-component velocity vector of the form V0 (r, z) = U r (r ),0,U z (r, z) , (a) show that the linearized Navier–Stokes equations with biglobal stability disturbance modes, which are given by Eqs. (5-86)–(5-89), can be readily reduced into ur ur uz + + = 0 (continuity) r r z −i ur + U r

  2u 1 ur ur  2ur  ur U r u 1 p + ur + U z r −  2r + − 2 + 2 =− (r -mom.) r r z r r r z   r  r

−i uz + U r

  2u 1 uz  2uz uz U z u U z + ur + U z z + uz −  2z + + 2 r r z z r r z  r

 1 p ( z -mom.) =−  z 

where pˆ (r , , z, t ) = p(r , z )e −i t and v(r ,  , z , t ) = ( uˆr , 0, uˆ z ) = ur (r , z ), 0, u z (r , z )  e −i t . (b) Using a disturbance stream function that satisfies the linearized continuity equation identically, eliminate ur and uz from the remaining radial and axial momentum equations by substituting ur = −

1  r z

and

uz =

1  r r

(c) The pressure can be eliminated by taking the partial derivative of the radial pressure gradient with respect to z and setting it equal to the partial derivative of the axial pressure gradient with respect to r. Show that the problem reduces to a single fourth-order partial differential equation for the disturbance stream function, namely, the one advanced by Chedevergne et al. (2006):

  2 1   2 i  2 − + r r z 2  r

  3  U r U z 2U r   2  3  2U r 1 U r  2U z 1 U z = U + + − + U + − + −  r z   r 3  r z r  r 2 z 3  r 2 r r rz r z   3  U r U r U z   2  1 U z  2U z   U z  2  3 +U r + − + + + − − + U   z   rz 2  r r z  z 2  r r r 2  z r r z r 2 z   4 2  3 3  2 3   4 2  3  4  −  4 − + − +2 2 2 − +  r r 3 r 2 r 2 r 3 r r z r r z 2 z 4   r

Solution: Assumptions: i.

Axisymmetric flow condition ( u = 0 )

ii.

Azimuthal wave number ( m = 0)

iii.

Two-component velocity vector is of the form V0 (r, z) = U r (r ),0,U z (r, z) .

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    r

(a) Navier–Stokes equations with biglobal stability disturbance modes is given by u u ur ur + + im  + z = 0 r r r z −i ur + U r

(continuity) (5-86)

U u u U r 2U u ur U r u U r 1 p + ur + im  r +  − + U z r + uz + r r r r  r z z  r

  2u 1 ur u  2u  2im =   2r + − (1 + m2 ) 2r − 2 u + 2r  r r r r z   r −i u + U r

(r -mom.) (5-87)

u U U u u U U r u urU u U m + ur + im   +  + + + U z  + uz +i p r r r r  r r z z r

  2u u  2u  1 u 2im =   2 + − (1 + m2 ) 2 + 2 ur + 2  ( -mom.) (5-88) r r r r z   r −i u z + U r

U u u U z u z U z u U z 1 p + ur + im  z +  + U z z + uz + r r r r  z z  z

  2u 1 uz m2  2u  =   2z + − 2 uz + 2z  ( z -mom.) (5-89) r r r z   r

Substituting the assumptions in the above equations, the continuity equation becomes ur ur uz + + =0 r r z

(continuity) (Eq. 1)

r-momentum and z-momentum equations become −i ur +

  2u 1 ur ur  2ur  U r ur U r u 1 p + ur + U z r −  2r + − 2 + 2 =− (r − mom) r r z r r r z   r  r

(Eq. 2) −i uz + U r

 d 2u 1 uz  2uz  uz U z u U z 1 P + ur + U z z + uz −  2z + + 2 =− ( z − mom) r r dz z r r z   z  r

(Eq. 3) Hence proved.

(Ans. a)

(b) Disturbance stream functions are as follows: Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -67-

ur = −

1  r z

and

uz =

1  r r

Let us consider the r-momentum equation: ur = −

1  r z

ur 1  1  2 = 2 − r r z r r z  2 ur 1  2 2  1 d 2 1  2 2  2 2  1  = − − + = − − r 2 r 2 r z r 3 z r r 2 z r 2 rdz r 2 rdz r 3 z r r 2 z

ur = −

1  r z

ur 1  2 =− z r z 2  2ur 1  3 =− 2 dz r z 3

Substituting the partial derivatives of ur in (Eq. 2), we get r-momentum as follows: −

 1  1  2  1  U r 1 P 1  1  2 = i + Ur  2 − − − U z   r r z r z 2  r z r rz  r z r

 2  2 2  1  3 1  1  2 1  1  3  −  2 − 3 − + − + − 2 3 2 3 3   r rz r z r r  z r z r rdz r z r z  −

 1  1  2  1  ur 1 P 1  1  2 = i + Ur  2 − − − U z   r r z r z 2  r z r rx  r z r  1  2 1  2 1   −  2 − − (r − mom) 2 3   r rz r r z r z 

(Eq. 4)

Let us consider the z-momentum equation: 1  uz = r r

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u z 1  1  2 =− 2 + r r r r r 2  2u z 1  2 2  1  3 1  2 2  2 d 2 1  3 =− 2 + + − = − + r r r 2 r 3 r r r 3 r 2 r 2 r 3 r r 2 r 2 r r 3

uz =

1  r r

u z 1  2 = z r r z

 2u z 1  3 = z 2 r r z 2

Substituting the partial derivatives of uz in (Eq. 3), we get the z-momentum equation as follows: −

 1  1  2 1 P 1  = −i + U r − 2 + 2  z r r  r r r r

 1  U z  1  2  1  uz − + U z   r z r +   r rz  r r z

 2  2  2 1  3 1  1  2 1  3  −  3 − 2 + − + + 2 r r 3 r 3 r r 2 r 2 r rdz 2   r r r r

−

1 P 1  = −i + Vr  z r r

 1  1  2  1  U z  1  2  1  uz − + − + U z   r 2 r r r 2  r z r +    r ry   r z  1  1  2 1  3 1  3  −  3 − 2 + + ( z − mom) (Eq. 5) 2 r r 3 r rdz 2   r r r r

Hence proved.

(Ans. b)

(c) To eliminate pressure term, (Eq. 4) is partially differentiated with respect to z, we get

−

1 2 P 1  2 U r  2 U r  3 1  2 ur uz  3 1  2 uz = i + − − − −  rdz r z 2 r 2 z 2 r rz 2 r z 2 r r z 3 r z 2 z  1  3 1  4 1  4  −  2 − − 2 2 2 4   r rz

r r z

r z 

r 2 P  2 U r  2  3  2 U r  3  2 U z − = i 2 + −Ur − −Uz 3 − 2  rz z r z 2 rz 2 z 2 r z z z

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 1  3  4  4  −  − − 2 r 2z 2 z 4   r rdz

(Eq. 6)

(Eq. 5) is partially differentiated with respect to r, we get −

1 2 P = −i  rz

 1  2  1   1  2 2  1  3 1  2  − + U + 3 + − r − 2  r r 2 2 r r 2  r r r r 3 r 2 r 2    r r  1  1  2  U r 1 U z  2 U z 1   1  2U z + − 2 + − + − 2  r r zr r r 2 z z r r 2  r r r r  r

 1  3 1  2  1  2 U z U z +U z  − + + 2 2 z  r r z r rz  r rz r +

 1  2 1    r r 2 − r 2 r   

 1  2 3  1  3 2  2 1  4 1  3 1  4 1   2U z 1  2  −  3 − − + + − + − 2 r r rz r 4 r r 2 r 3 r 3 r 2 r r 4 r 2 r 3 r dr 2 dz 2 r 2 r z 2   r r

  2  1   2  2 2   3  U r   2 1   r 2 P − = −i  2 − + U r − + 2 + + − 2  r z r r  r r r 3  r  r 2 r r   r  r r   3 U z  2 1 U z    2U z 1  2   2 U z − + − +Uz  2 − + r zr r r z z r 2  r z r rdz  rdz r  3  2 3  2  3  4 U z   2 1     2U z  4 1  3  + − + −  2 − 3 − + + − 2 z  r 2 r r  r rdz r r r 2 r 3 r 4 r 2z 2 r 2 r z 2   r r

(Eq. 7) Equating (Eq. 6) and (Eq. 7), we get   2 1   2 i  2 − + r r z 2  r

  3  U r U z 2U r   2  3  2U r 1 U r  2U z 1 U z   = U + + − + U + − + −   r z   r 3  r z r  r 2 z 3  r 2 r r rz r z  r   3  U r U r U z   2  1 U z  2U z   U z  2  3 +U r +− + + − 2  − +Uz 2  2 + 2 rz  r r z  z r  z r r z r z  r r   4 2  3 3  2 3   4 2  3  4  −  4 − + − + 2 − +  r r 3 r 2 r 2 r 3 r r 2z 2 r r z 2 z 4   r

Hence proved.

(Ans. c)

P5-47 The Taylor–Culick profile, which has been repeatedly used to describe the ideal gaseous motion in solid rocket motors (Culick 1966), corresponds to an axisymmetric, self-similar profile in a porous tube. When the tube radius is taken to be unity, the basic flow stream function may be Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -70-

written as  0 = z sin( 12  r 2 ) for 0  r  1 and 0  z  Z N , where Z N represents the length of the porous tube (see Fig. P5-47a).

FIGURE P5-47a Assuming a wall injection speed of unity, the basic flow velocity may be expressed as V0 (r , z ) = U r er + U z e z with U r = −r −1 sin( 12  r 2 ) and U z =  z cos( 12  r 2 ) . As shown in the problem above, the linearized Navier–Stokes equations with biglobal stability disturbance modes can be reduced into a single, fourth-order, partial differential equation for the disturbance stream − i t

function  , where ˆ =  (r , z )e . Following Chedevergne et al. (2012), the basic flow boundary conditions require uniform wall-normal injection, no slip at the headwall, and axisymmetry with respect to r = 0 , specifically:  0   0 (wall-normal injection)  r (1, z ) = 0; z (1, z ) = 1   0   0 (r , 0) = ( r , 0) = 0 (no slip at headwall)  z  r   3 0  0  0 (0, z ) = (0, z ) = 0 (U r = U z / r = 0)  3 (0, z ) = r z  r

 =  +ˆ

0 (a) Recalling that the instantaneous motion is recoverable from , and that disturbances must be prevented from altering the basic flow conditions at the boundaries, show that the disturbances must satisfy:    (wall-normal injection)  r (1, z ) = z (1, z ) = 0     (r , 0) = (r , 0) = 0 (no slip at headwall)  z  r   3   (0, z ) = (0, z ) = 0 (axisymmetry)  3 (0, z ) = r z  r You may also impose the outflow condition suggested by Theofilis, Duck and Owen (2004) by setting a simple extrapolation relation at the domain’s exit plane, z = Z N , namely,

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 (r , Z N ) =

Z N − Z N −2 Z −Z  (r, Z N −1 ) + N −1 N  (r, Z N −2 ) Z N −1 − Z N −2 Z N −1 − Z N −2

(b) Using the Taylor–Culick basic flow and a numerical solver of your choice, show that one of the eigenvalues for

 = 40.4 − 9.16i.

m = 0, Z N = 8

2 −4 2 and  = 1/1975 m /s  5 10 m /s

is approximately

(c) Obtain the “boomerang” set of eigenvalues for Z N = 8 and Z N = 10 using  = 5 10−4 m2 /s . (d) Compare your results to those of Chedevergne et al. (2012) provided in Fig. P5-47b and explain the differences.

FIGURE P5-47b Solution: (a) First, the two components of the mean flow are expressible by U r = −r −1 sin( 12  r 2 ) and U z =  z cos( 12  r 2 ) We also know that that the instantaneous velocity and flow streamfunction are related through 1  1  and U z = Ur = − r r r z Since the instantaneous motion is recoverable from  =  0 +ˆ =  0 + ( x, y)e−i t , we may differentiate  with respect to z and r ,   0 ˆ  0  −i t   0 ˆ  0  −i t and = = + + = = + + e e r z r z r z r z r z For a porous tube with uniform wall injection, the fundamental boundary conditions on the instantaneous field are the same as those on the mean flow (because the “disturbances must be prevented from altering the basic flow conditions at the boundaries”). We have, for any time t , (1) A uniform injection and no-slip condition at the sidewall, U z (1, z , t ) = 0 and U r (1, z , t ) = −1 (2) A prescribed zero injection pattern at the headwall, U z (r , 0, t ) = U r (r , 0, t ) = 0 Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -72-

 2U z (0, z, t ) = U z (0, z, t ) = U r (0, z, t ) = 0 y 2 By conveying these boundary conditions to the Stokes streamfunction, we get Sidewall:  0  0     (1, z, t ) = (1, z, t ) = (1, z ) + (1, z ) + (1, z )e−i t and (1, z )e−i t r z r z r z or    −rU r (1, z ) = 1 + (1, z )e −i t  1 = 1 + (1, z )e −i t ; t  (1, z ) = 0 z z z

(3) A vanishing speed at the axis of symmetry,

rU z (1, z ) = 0 +

   (1, z )e −i t  0 = 0 + (1, z )e −i t ; t  (1, z ) = 0 r r r

Headwall:  0  0     (r , 0, t ) = (r , 0, t ) = (r , 0) + (r , 0) + (r , 0)e−i t and (r , 0)e−i t r z r z r z or    −rU r (r , 0) = 1 + (r , 0)e −i t  0 = 0 + (r , 0)e −i t ; t  ( r , 0) = 0 z z z

rU z (r , 0) = 0 +

   (r , 0)e −i t  0 = 0 + (r , 0)e −i t ; t  ( r , 0) = 0 r r r

Centerline:  3 0  3  3  0   (0, z , t ) = (0, z ) + (0, z )e −i t (0, z, t ) = (0, z ) + (0, z )e−i t and 3 3 3 r r r z z z or    −rU r (0, z ) = 1 + (0, z )e −i t  1 = 1 + (0, z )e −i t ; t  (0, z) = 0 z z z r 3  2U z  3  3  3 − i t −i t (0, z ) = 0 + (0, z ) e  0 = 0 + (0, z ) e ;  t  (0, z ) = 0 2 r 2 r 3 r 3 r 3

(b) Using the Taylor–Culick basic flow and a numerical solver, one can show that among the eigenvalues for m = 0, Z N = 8 and  = 5 10−4m2 /s is approximately  = 40.4 − 9.16i. (c) The “boomerang” set of eigenvalues for Z N = 8 and Z N = 10 may be obtained using

 = 5 10−4m2 /s and a numerical solver Results are illustrated below. In the figure below, we compare the biglobal stability frequency spectra for a porous tube driven by the Taylor–Culick mean flow motion at a crossflow Reynolds number of Re = 1975 , m = 0 , r = 1 , and using (a) ZN = 8 and (b) 10. Here, both (□) and (■) reflect eigenmodes obtained from Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -73-

the present incompressible stability solver whereas (▲) represent eigenmodes obtained from the incompressible biglobal stability solver developed by Chedevergne et al. (2012).

(d) One may note a slight leftward shift of the eigenmodes obtained from the incompressible biglobal stability solver in comparison to the results obtained by Chedevergne et al. (2012). The leftward shift may be ascribed to the use of a slightly different formulation. The additional modes shown here are due to the use of the QZ algorithm and a larger discretization stencil, thus leading to a broader spectrum of modes than obtained previously through the Arnoldi algorithm used by Chedevergne et al. (2012); the latter has the capability of pinpointing modes that are in the neighborhood of a userspecified value.

P5-48 Identify the operators that appear in the biglobal stability disturbance equations using cylindrical polar coordinates, which are given by Eqs. (5-94)–(5-97), in order to provide the source coefficients for the eigenvalue problem represented by Aij f i =  Bij f i , where σ and fi denote the eigenvalues and corresponding eigenvectors, while Aij and Bij refer to

 Ac ,ur   Ar ,ur Aij =  A   , ur  Az ,u  r

Ac ,u

Ac ,u2

Ar ,u

Ar ,uz

A ,u

A ,u2

Az ,u

Az ,uz

Ac , p   Ar , p  A , p   Az , p 

and

 Bc ,ur   Br ,ur Bij =  B   , ur  Bz ,u  r

Bc ,u

Bc ,uz

Br ,u

Br ,uz

B ,u

B ,uz

Bz ,u

Bz ,uz

Bc , p   Br , p  B , p   Bz , p 

Show that

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U U U im  U r    A ,ur = + − 2 2 + + im  + U z   1  Ar ,ur = U r  r r r r r r z  Ac ,ur = r + r   2 2 2 U  1 U U r    1  1+ m     + + + U z + im  −  2 + − 2 + 2   A ,u = U r  A = im  r r  r z r r r r z    c ,u   r r    2 2 2  1  1+ m   1 U r 2U im    −  2 + − 2 + 2  A = − + 2  r ,u r r r z   Ac ,uz =    r r  r r2  z    U U r 1  im  Ac , p = 0   Ar ,uz = Ar , p = A ,uz = A , p =   z  r z r And

 U z 1 U z Az ,ur = Az ,u =  r r    U   U z +Uz + + im   Az ,uz = U r r z z r  2 2 2   1  m   1  − 2 + 2  Az , p = −  2 + r r r z   z   r and that the only nonvanishing values of Bij are Br ,ur = B ,u = Bz ,uz = i . Solution: Cylindrical coordinates are regrouped into operators and written as follows:

  1  im    +  ur +   u +   uz = 0 (continuity) (5-94)  r r   r   z     2 1  1 + m2  2  U  U r  U + + im + U −  − 2 + 2   ur  r  2+ z r z r r r z    r  r r

1    1 U r 2U 2 im   U  + − + 2  u +  r  u z +   p = (i )ur r r   z   r    r 

(r − mom.) (5-95)

   2 1  1 + m2  2   imU 1 U U r   U U 2 im  + − u + U + + + −  − 2 + 2  + U z  u  r  2+  2  r r r  r r  r r r r z  z   r  r  r

 U +   z  U z   r

  1 U z  ur +    r 

 im    p = (i )u  uz +    r 

( − mom.) (5-96)

   2 1  m2  2   U   U z  u + U + im + U + −  − +  2+  uz z    r r z z r r r 2 z 2     r  r

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1   +  p = (i )u z   z 

( z -mom ) (5-97)

Eigenvalue problem is represented by Aij f i =  Bij f i , where σ and fi denote the eigenvalues and corresponding eigenvectors, while Aij and Bij refer to

 Ac ,ur   Ar ,ur Aij =  A   , ur  Az ,u  r

Ac ,u

Ac ,u2

Ar ,u

Ar ,uz

A ,u

A ,u2

Az ,u

Az ,uz

Ac , p   Ar , p  A , p   Az , p 

and

 Bc ,ur   Br ,ur Bij =  B   , ur  Bz ,u  r

Bc ,u

Bc ,uz

Br ,u

Br ,uz

B ,u

B ,uz

Bz ,u

Bz ,uz

Bc , p   Br , p  B , p   Bz , p 

Equations (5-94)–(5-97) are rewritten with operator matrices elements, and we get

( A ) u + ( A ) u + ( A ) u c ,ur

r

c ,u

c ,uz

= 0 (Eq. 1)

z

( A ) u + ( A , ) u + ( A ) u + ( A ) P = ( B ) u r ,ur

r

r u

r ,u z

z

r ,P

r ,ur

r

(Eq. 2)

( A )u + ( A )u + ( A )u + ( A ) P = (B )u

(Eq. 3)

( A )u + ( A )u + ( A )u + ( A ) p = ( B )u

(Eq. 4)

 ,ur

r

z ,ur

r

 ,u

z ,u





 ,u z

z ,u z

z

z

 ,P

z, p

 ,u

z ,u z



z

Comparing the continuity equation (5-94) with (Eq. 1),

  1 Ac,ur =  +   dr r  Ac ,u =

im r

Ac ,uz =

 z

Ac , P = 0

Bc ,ur = Bc ,u = Bc ,u z = Bc , p = 0

Comparing the r-momentum equation (5-95) with (Eq. 2),

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Ar ,ur = U r

  2 1  1 + m2  2  U  U r  + + im  + U z −   2 + − 2 + 2 r r r z r z   r r r

Ar ,u =

1 U r 2U 2 im − + 2 r  r r

Ar ,u z =

U r z

Ar , p =

1   r

Br ,ur = i Br ,u = Br ,uz = Br , p = 0

Comparing the  -momentum equation (5-96) with (Eq. 3),

A ,ur =

U U 2 im + − 2 r r r

A ,u = U r

  2 1  1 + m2  imU 1 U U r    + + + − 2  2 + − 2 + 2  +Uz r r r  r r r r z  z  r

A ,uz =

U z

A ,P =

im r

B ,u = i B ,ur = B ,uz = B , p = 0

Comparing the z-momentum equation (5-97) with (Eq. 4),

Az ,ur =

U z r

Az ,u =

1 U z r 

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Az ,uz = U r Az , P =

  2 1  m2  2  U   U z + im  + U z + −  2 + − 2 + 2 r r z z z   r r r r

1   z

Bz ,uz = i

Bz ,ur = Bz ,u = Bz , p = 0

Hence proved.

(Ans)

P5-49 Using Eq. (5-106), provide the linear coordinate transformations that are needed to map an axisymmetric problem defined in the (r, z) domain from the 0 ≤ r ≤ 1 and 0 ≤ z ≤ ZN intervals to the Chebyshev domain of [−1, 1]. Solution: Coordinate transformation is necessary to redistribute the collocation points within an interval for the purpose of giving high resolution to regions of very rapid change. Using Eq. (5-106), map an axisymmetric problem for r  ( a, b ) and z  ( c, d ) . The two auxiliary variables are  and  .

r=

b a ( + 1) − ( − 1) 2 2

=

2r − ( a + b ) b−a

z=

d c ( + 1) − ( − 1) 2 2

=

2z − ( c + d ) d −c

and  2  = r b − a   2  = z d − c 

Domain that consists 0 ≤ r ≤ 1 and 0 ≤ z ≤ ZN may be reconditioned as follows: r=

1   =2 ( + 1)  = 2r − 1 2 r 

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and

z=

ZN 2z  2  = ( + 1)  = − 1 2 ZN z Z N 

(Ans)

P5-50 Using Eq. (5-106), provide the linear coordinate transformations that are needed to map a cylindrical polar problem defined in the (r, θ, z) domain from the 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, and 0 ≤ z ≤ ZN intervals to the Chebyshev domain of [−1, 1]. Solution: Using Eq. (5-106), map a cylindrical polar problem for r, θ, and z where r  ( a, b ) ,

  ( c, d ) , and z  ( e, f ) . The three auxiliary variables are  , , and  . b a r = ( + 1) − ( − 1) 2 2 d c  = ( + 1) − ( − 1) 2 2 f e z = ( + 1) − ( + 1) 2 2

2r − ( a + b) b−a 2 − (c + d ) = d −c 2 z − (e + f ) = f −e

=

 2  = r b − a   2  and =  d − c   2  = z f − e 

Domain that consists of 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π, and 0 ≤ z ≤ ZN intervals may be reconditioned as follows:

1 r = ( + 1) 2  =  ( + 1) Z z = N ( + 1) 2

 = 2r − 1   = −1  =

2z −1 ZN

  =2 r 2  1  = and     2  = z Z N 

(Ans)

P5-51 Identify the block matrices that must be used to convert the biglobal stability disturbance equations in cylindrical polar coordinates, based on Eqs. (5-94)–(5-97), to an eigenvalue problem of the form Aij f i =  Bij f i , where σ and fi denote the eigenvalues and corresponding eigenvectors. The block matrices appear as

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 Ac ,ur   Ar ,ur Aij =  A   , ur  Az ,u  r

Ac ,u

Ac ,uz

Ar ,u

Ar ,uz

A ,u

A ,uz

Az ,u

Az ,uz

Ac , p   Ar , p  A , p   Az , p 

and

 Bc ,ur   Br ,ur Bij =  B   ,ur  Bz ,u  r

Bc ,u

Bc ,uz

Br ,u

Br ,u2

B ,u

B ,uz

Bz ,u

Bz ,uz

Bc , p   Br , p  B , p   Bz , p 

Show that the only nonvanishing values of Bij are Br ,ur = B ,u = Bz ,uz = iI N and that

  im  U U    A ,ur =  r + r  − 2 r 2  ii ii  A = (U ) D r + (U ) D z +  U r + im U   r ,u r r ii N z ii N   r ii  1   r   1 U  r A ,u = (U r )ii DNr +     Ac ,ur = DN + r   r  ii  r 2 DNr 1 + m 2 ii   z 2  − ( DN ) + − 2 + ( DN )   im   r rii  +  U r + im U  + U D z ii   A = c ,u   ( z )ii N    rii r ii  r U  im  1 U r    Ar ,u =  −2  + 2 2  Ac ,u2 = DNz    r 2 D r 1 + m2 2 r ii rii  r    − ( DN ) + N − 2 + ( DNz )  rii rii  Ac , p = 0 1 r      U r  A = A = D r , u r , p N     2   z ii   A ,u =  U  A , p = im t    rii  z ii

  dU z   1 U z  Az ,ur =    Az ,u =    r ii  r  ii   U   U z + im   and  Az ,uz = (U r )ii DNr + (U z )ii DNz +  r ii  z    2 2 1 m2 1 − ( DNr ) + DNr − 2 + ( DNz )  Azp = DNz rii rii    

Solution: Partial derivatives with respect to r and z will be expressed in pseudo-spectral differentiation matrix as follows:

  = DNr , = DNz r z When applying domain mapping and differentiation matrices to the answer for Problem 5-48, we can easily extract the block matrices element as follows: For continuity equation:

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Ac ,ur = DNr + Ac ,u =

1 rii

im rii

Ac,uz = DNz Ac , P = 0 Bc ,ur = Bc ,u = Bc ,u z = Bc , p = 0

For r-momentum equation: 2  r 2 DNr 1 + m2  u imU  Ar ,ur = (U r )ii DNr + (U z )ii DNz +  r + −  − 2 + ( DNz )  ( DN ) +  r ii rii rii  r  

im  1 ur 2U  Ar ,u =  −  + 2 2 r ii rii  r  1  U  Ar ,uz =  r  , Ar , P = DNr   z ii Br ,ur = iI N

Br ,u = Br ,uz = Br , p = 0

For θ-momentum equation:

 U U  2 im A ,ur =   +   − 2 r ii rii  r 2  r 2 DNr 1 + m2  U u   1 U  z A ,u = (U r )ii DNr +  im  + r  +  + U D −  − 2 + ( DNz )  ( DN ) + z N  r r ii  r  ii rii rii   

 U  A ,uz =     y ii

A , p =

im  rii

B ,u = iI N B ,ur = B ,uz = B , p = 0

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For z-momentum equation:

 U  Az ,ur =  z   r ii  1 U z  Az ,u =    r  ii

Az ,uz

2  r 2 DNr m2  imU U z  z = ( ur )ii D +  + − 2 + ( DNz )   + (U z )ii DN −  ( DN ) + z ii rii rii  r  

Az , P =

r N

1



DNz

Bz ,uz = iI N Bz ,ur = Bz ,u = Bx , p = 0

Hence proved.

(Ans)

P5-52 Reduce the block matrices of Eqs. (5-94)–(5-97) that you just derived assuming axisymmetric disturbances with no azimuthal variations and m = 0. Solution: Equations for axisymmetric disturbance with no azimuthal variation and m = 0 were shown in the answer for problem 5-46. The block matrices appear as follows:

 Ac ,ur  Aij =  Ar ,ur   Az ,ur As we know

Ac ,uz Ar ,uz Az ,uz

 Bc ,ur Ac , p    Ar , p  and Bij =  Br ,ur   Az , p   Bz ,ur

Bc ,uz Br ,uz Bz ,uz

Bc , p   Br , p   Bz , p 

  = DNr , = DNz r z

For continuity equation:

Ac ,ur = DNr +

1 rii

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Ac,uz = DNz Ac , p = 0 Bc ,ur = Bc ,uz = Bc , p = 0

For r-momentum equation: 2 2  Dr 1  U  Ar ,ur = (U r )ii DNr +  r  −   ( DNr ) + N + ( DNz ) − 2  + (U z )ii − DNz rii rii   r ii 

 U  Ar ,uz =  r   z ii Ar , P =

1



DNr

Br ,ur = iI N

Br ,uz = Br , p = 0

For z-momentum equation:

 U  Az ,ur =  z   r ii 2 2  Dr  U  Az ,uz = (U r )ii DNr +  z  −    DNr ) + N + ( DNz )  + (U z )ii DNz rii  z ii  

Az , P =

1



DNz

Bz ,uz = iI N Bz ,ur = Bz , p = 0

Hence proved.

(Ans)

P5-53 Consider a problem with a simple axisymmetric viscous-flow configuration, namely, a cylindrical fluid domain that extends over 0  r  1 and 0  z  Z N . Assume infinite impedance

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(hardwall) boundaries everywhere along r = 0, r = 1, and z = 0 , and that an outflow condition exists at z = Z N , where a zero normal stress requirement based on Eq. (5-114) translates into

u uz ur u + = 0; = 0; 2 z − p = 0 r z z z Show that the boundary conditions on the ( ur , u , uz , p ) disturbance eigenfunctions can be written as u z  u (0, z ) = 0 (0, z ) = 0 ur (0, z ) = 0 r  u (1, z ) = 0 u z (1, z ) = 0 ur (1, z ) = 0 u (r , 0) = 0 u (r , 0) = 0 u z (r , 0) = 0  r  ur p ( r, Z N ) u u u z ( r, Z N ) = − z ( r, Z N )  ( r, Z N ) = 0 ( r, Z N ) =  r z z 2  z

Solution:

p(0, z ) =0 r −− −− −−

(centerline) (sidewall) (headwall) (exit plane)

u uz ur u + = 0; = 0; 2 z − p = 0 r z z z

It is a simple axisymmetric flow. Therefore, at the centerline the boundary conditions are

ur ( 0, x ) = u ( 0, x ) = 0, 

uz ( 0, z ) = 0 r

uz =P z

 2u z P  = zdr r



P (0, z ) = 0 r

In headwall and sidewall no slip boundary conditions are applied. ur (1,, z ) = u (1, z ) = u z (1, z ) = 0

(sidewall)

ur (r , 0) = u (r , 0) = ux (r , 0) = 0

(Headwall)

In exit plane outflow conditions are valid.

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uz u =− r r dz ur u ( r, zN ) = − z ( r, zN ) x r u ( r, zN ) = 0 z

u z p = z 

p ( r, zN ) uz ( r, zN ) = z 2 Hence proved.

(Ans)

P5-54 Using the cylindrical polar formulation of P5-51–P5-53 and a numerical solver of your 2 1 choice, investigate the stability of the Taylor–Culick profile corresponding to  0 = z sin( 2  r )

for

0  r  1, 0  z  Z N

.

Here

V0 (r , z ) = U r er + U z e z

with

U r = −r −1 sin( 12  r 2 )

and

U z =  z cos( 12  r 2 ) . Using m = 0 and  = 5 10−4m2 /s find the spectrum of eigenvalues for Z = 10 ZN = 8 and N . Compare your results to those of Chedevergne et al. (2012) provided in Fig. P5-47b. Solution: This problem follows the solution to P5-53. In the figure below, we compare the biglobal stability frequency spectra for a porous tube driven by the Taylor–Culick mean flow motion at a crossflow Reynolds number of Re = 2000 , m = 0 , r = 1 , and using aspect ratios of (a) Z N = 8 and (b) Z N = 10 . We assume infinite impedance (hardwall) boundaries everywhere along r = 0 , r = 1 and z = 0 . An outflow condition exists at z = Z N , as described in P5-53. Here, both (□) and (■) reflect eigenmodes obtained from the present incompressible stability solver whereas (▲) represent eigenmodes obtained from the incompressible biglobal stability solver developed by Chedevergne et al. (2012).

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P5-55 Using the cylindrical polar formulation of P5-51–P5-53, investigate the stability of the Cecil–Majdalani profile, which extends the Taylor–Culick motion to spinning rocket motors and porous tubes (see Fig. P5-55). As shown by Cecil and Majdalani (2017) in their AIAA awardwinning paper, the basic motion can be described using V0 (r , z ) = U r er + U er + U z e z with

U r = −r −1 sin( 12  r 2 ) , U =  r −1 1 − exp ( − 14  r 2 Re )  , and U z =  z cos( 12  r 2 ) for 0  r  1 , 0    2 , and 0  z  Z N . Here  = Vs / Vw represents the tangential speed of the spinning chamber relative to the wall injection speed, and Re = Vwa /  stands for the injection Reynolds number based on the wall injection speed Vw , chamber radius a, and kinematic viscosity  . When the injection speed and chamber radius are taken to be unity, the Reynolds number becomes equivalent to the reciprocal of the kinematic viscosity, Re = 1/  and the Hertzian frequency of rotation may be deduced from f s = Vs / (2 a).

Using m = 0 and  = 5 10−4m2 /s , find the

spectrum of eigenvalues for Z N = 8 and Z N = 10 . By comparing your results to those of Chedevergne et al. (2012) in Fig. P5-47b. what can you say about the effect of axial rotation on the stability of the basic motion? You may use the boundary conditions described in P5-53.

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FIGURE P5-55 Solution: Using an incompressible stability solver with m = 0 and  = 5 10−4m2 /s , and the boundary conditions described in P5-53, we obtain the spectrum of eigenvalues for Z N = 8 and Z N = 10 . By comparing our results to those of Chedevergne et al. (2012) in Fig. P5-47b. we see that the effect of axial rotation reduces the stability of the basic motion.

Biglobal stability frequency spectra for a porous tube driven by the Cecil–Majdalani mean flow motion at a crossflow Reynolds number of Re = 1975 , m = 0 , r = 1 , and using (a) XN = 8 and (b) 10. Here, both hollow and solid square symbols, (□) and (■), represent eigenmodes obtained from the present incompressible stability solver. The difference between them is that the solid squares, (■), illustrate the characteristic ‘boomerang’ pattern of eigenvalues associated with the Taylor–Culick mean flow motion. The difference here is that the eigenmodes appear to be slightly more hydrodynamically unstable in comparison to the Taylor–Culick mean flow motion studied by Chedevergne et al. (2012). The slight increase in the stability growth rate  i in the present case may be attributed to the additional swirling component that accompanies the Cecil–Majdalani profile.

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P5-56 Identify the operators that appear in the compressible biglobal stability disturbance equations in cylindrical polar coordinates, which are given by Eqs. (5-155)-(5-160), in order to provide the source coefficients for the eigenvalue problem represented by Aij fi =  Bij fi , where  and fi denote the eigenvalues and corresponding eigenvectors, while Aij and Bij refer to  Ac ,    Ar ,   Aij =  A ,   Az ,    Ae, 

Ac ,ur

Ac,u

Ac ,uz

Ar ,ur

Ar ,u

Ar ,uz

A ,ur

A ,u

A ,uz

Az ,ur

Az ,u

Az ,uz

Ae,ur

Ae,u

Ae,uz

Ac ,T   Ar ,T  A ,T   Az ,T   Ae,T 

and

 Bc ,    Br ,   Bij =  B ,   Bz ,    Be, 

Bc ,ur

Bc,u

Bc ,uz

Br ,ur

Br ,u

Br ,uz

B ,ur

B ,u

B ,uz

Bz ,ur

Bz ,u

Bz ,uz

Be,ur

Be,u

Be,uz

Bc ,T   Br ,T  B ,T   Bz ,T   Be,T 

Show that U U r U r 1 U U z     Ac ,  = r + r + r  + z + U r r + im r + U z z   U r  U r U U r U r U2     T0 +U r + +Uz − + T0   + R  Ar ,  =  r r  z r  r   r  t   U U U U U rU  R  T0   U +U r + +Uz + (density terms)  A ,  =   + r  imT0 +    t  r r    z r       U z U U z U z     U  T + +Uz + R  0 + T0   Az ,  =  z +U r  r r  z  z   t  z   T T U T T U     Ae,  = c p 0 + cv  U r 0 +  0 + U z 0  − RT0  U r + U z + im   t r r  z  z r     r  0    1  Ac ,ur = r + 0  r + r       4   2 1  1   2 m2  U   U r    Ar ,ur = 0  U r + im  + U z + −  − + −   2+  r z r  r r r 2  z 2 r 2   r   3  r (radial terms)   A =   U + U  −  im  7 +   0     , ur r  3r  r r   r   T P U z   1  2  −  + Ae,ur =  0 c p 0 − 0 −  r   Az ,ur =  0 r 3  r z r z  r r 

  U  1 0 im im   7   Ar ,u = 0  r − 2U  −   −   Ac,u = r  + 0 r r   3r  r r       2 1  1  2 4m 2   imU + U r  1 U   A =  U + + U + −  − + −   ,u  2+  (tangential terms) 0 r z r z r   r r r 2 z 2 3r 2   r  r   0 c p T0 1 P0  Az ,u = 0 U z −  im  Ae,u = − −   r  3r z r  r   Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -88-

 0 U U r   2  im  + 0 Ar ,uz = 0 − A ,uz = 0 −  Ac ,uz = z z z 3 r z z 3r z  2 2 2   U   U z  1  m 4     (axial terms) + im  + U z + −  2 + − 2 +  Az ,uz = 0  U r   r r z z  r r r 3 z 2    r   T P  Ae,uz = 0 c p 0 − 0 −  z z z  

and      R       Ar ,T = R  0 + 0  A ,T =  im 0 + 0  Az ,T = R  0 + 0   Ac ,T = 0 r  r   z   r  z   2 2 2  A =  c  U  + im U + U   − R  U 0 + U 0 + U 0  − k   + 1  − m +    2  0 v r z z  r r  e,T r z  z r   r r r 2 z 2   r   r 

(temperature terms) where the dissipation function may be reconstructed from 4  U U 1 U U z   1  2  U  U r U z   =  2 r − r − − +   + Ur  − −    3  r r r  z  r r  r   z    r

r

 im  U U r    1  U    U +2    +  r − U   +  z +  r   z  z     r  r  r   U 4  2  U 1  U  U r U z  im     1   1 U z U =    + Ur  − − + 2    +  r − U    −  +  +   3  r   z  r r   z    r r   r   r   r



  z   

 U z U r    1 U z U  im  U r U r 1 U   4  U = 2 z − − − + 2  + + +   3  z r r r   z z  r  r  z  r   r and that the only nonvanishing values of Bij are Bc,  = i, Br ,ur = B ,u = Bz ,uz = i 0 , Be,T = i 0cv , and z

Be,  = −iRT0 .

Solution: We are asked to determine the operators that appear in the block matrices for the stability analysis. To start, we first expand the given equation: Aij fi =  Bij fi  Ac ,    Ar ,  A   ,  Az ,    Ae, 

Ac ,ur

Ac ,u

Ac ,u z

Ar ,ur

Ar ,u

Ar ,uz

A ,ur

A ,u

A ,uz

Az ,ur

Az ,u

Az ,uz

Ae,ur

Ae,u

Ae,uz

Ac ,T    Bc ,      Ar ,T  u  Br ,   r  A ,T u  =   B ,      Bz ,  Az ,T   u z      Be,  Ae,T   T 

Bc ,ur

Bc ,u

Bc ,u z

Br ,ur

Br ,u

Br ,u z

B ,ur

B ,u

B ,uz

Bz ,ur

Bz ,u

Bz ,uz

Be,ur

Be,u

Be,uz

Bc ,T     Br ,T   u   r B ,T  u    Bz ,T   u z   Be,T   T 

After expanding out, we can collect the following 5 equations: Ac ,   + Ac ,ur ur + Ac ,u u + Ac ,uz u z + Ac ,T T =  ( Bc ,   + Bc ,ur ur + Bc ,u u + Bc ,u z u z + Bc ,T T )

(

Ar ,   + Ar ,ur ur + Ar ,u u + Ar ,u z u z + Ar ,T T =  Br ,   + Br ,ur ur + Br ,u u + Br ,u z u z + Br ,T T

)

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( T =  (B T =  (B

A ,   + A ,ur ur + A ,u u + A ,uz u z + A ,T T =  B ,   + B ,ur ur + B ,u u + B ,uz u z + B ,T T Az ,   + Az ,ur ur + Az ,u u + Az ,uz u z + Az ,T

z,

Ae,   + Ae,ur ur + Ae,u u + Ae,uz u z + Ae,T

e, 

 + Bz ,ur ur + Bz ,u u + Bz ,uz u z + Bz ,T T )

)

 + Be,ur ur + Be,u u + Be,u z uz + Be,T T )

It may be useful at this juncture to clarify the convention used to define the subscripts. While the first subscript denotes the type of equation to which a term belongs (e.g., continuity, r -,  -, z momentum, or energy), the second subscript denotes the primitive variable associated with that term. For example, the coefficient for the density in the continuity equation is labeled as Ac ,  . Bearing this in mind, we proceed by considering the continuity equation [Eq. (5-155)]:  0  im   U r U r 1 U U z   1   r + r + r  + z + U r r + r U + U z z   +  r + 0  r + r   ur      1       +  0 + im0  u +  0 + 0  u z = i (continuity) r   z    z By applying this labeling convention, we identify the following  U U r U r 1 U U z     1 Ac ,ur = 0 + 0  +  Ac ,  = + + + + Ur + im  + U z r r r  z r r z r  r r   1 0 im  Ac ,u = Ac ,uz = 0 +  0 + 0 Ac,T = 0 Bc,  = i r  z z r This specification can be easily seen by comparing the continuity equation to the expanded matrix equation. We are simply matching these two equations and identifying their coefficients. The same strategy can be extended to the momentum and energy equations. The r -momentum equation [Eq. (5-156)] is provided below:  U r U r U U r U r U2     T0 + Ur + +Uz − + T0      + R r r  z r  r    r  t    4  2 1  U   U r  1    2 m2       U + im + U + −  + − + − 2    ur   2  0 r z  2   2 r z r  r r r   z r     3  r    r   U r   2     U    0  im   7   +  0  r − 2U  − − u + − + 0  T = i 0 ur ( r -mom.)  uz + R       0 z 3 r z  r   3r  r r    r  r   

By matching the coefficients against the expanded matrix form, we are able to identify  U U r U U r U r U2     T0 Ar ,  =  r + U r + +Uz − + T0   + R r r  z r  r   r  t  4   2 1  1    2 m2   U   U r   Ar ,ur = 0 U r + im  + U z + −  − + −   2+  r z r  r r r 2   z 2 r 2    r  3  r   U   U r   2    im    7  Ar ,u = 0  r − 2U  − − Ar ,T = R  0 + 0  Br ,ur = i 0 A =  −   r ,u z 0 r   r  z 3 rz  3r  r r   r

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The same may be repeated for the  -momentum equation [(Eq. 5-157)], specifically:  U   U U  im  7   U U U U U rU  R  T0    t + U r r + r  + U z z + r  + r  imT0 +     +  0  r + r  − 3r  r + r  ur              2 1  1  2 4m2    imU + U r  1 U  +  0  U r + +Uz + −  2 + − + −   u  r z r   r r r 2 z 2 3r 2     r  r   im   R  U +  0  − u z +  im0 + 0  T = i 0 u (  -mom.)  z 3r z  r   

As before, by matching the coefficients against the expanded matrix form, we retrieve U U U U U rU  U A ,  =   + U r + +Uz + r r  z r  t U  im  7   U A ,ur = 0   +   −  +  r  3r  r r   r

 imU + U r  1 U  A ,u = 0  U r + +Uz + r z r   r A ,uz =  0

U im  − z 3r z

A ,T =

T0   R  + r  imT0 +     

  2 1  1  2 4m 2   −  − + −  2+   r r r 2 z 2 3r 2    r

  R im  0 + 0   r  

B ,u = i 0

The z -momentum equation [Eq. (5-158)] follows suit. We have:  U z U z U U z U z     T0  t + U r r + r  + U z z  + R  z + T0 z        2  U z   1     0 U z im   +  0 −  + − u   ur +  r 3  r z rz   3r z   r       2 1  m2 4  2    im  U z  +  0  U r + U + U z + −  − +  2+   uz z z  r r r 2 3 z 2     r r  r    + R  0 + 0  T = i 0 u z z   z

( z -mom.)

As such, we extract the following coefficients U z U U z U z    U  T Az ,  =  z + U r + +Uz + R  0 + T0   r r  z  z   t  z 2 0 U z im  U z   1    − Az ,ur = 0 −  +  Az ,u = r  3r z r 3  r z r z 

 im  U z  Az ,uz = 0  U r + U + U z + z z  r r

  2 1  m2 4  2   −  − +  2+   r r r 2 3 z 2    r

    Az ,T = R  0 + 0  Bz ,uz = i 0 z   z

Lastly, we may express the energy equation [(Eq. 5-159)] as follows: Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill LLC. -91-

 T0 T0    im    T0 U T0  c p t + cv U r r + r  + U z z  − RT0 U r r + U z z + r U          0 c p T0 1 P0  T P T P     +   0 c p 0 − 0 −  r  ur +  − −   u +  0 c p 0 − 0 −  z  uz r r z z      r  r  

   2 1  m2  2    U 0   im     +  0cv  U r + U + U z  − R U r 0 + U z 0 +  − k − +  2+  T z  r z r   r r r 2 z 2    r r    r = i ( 0cvT − RT0  ) (energy)

The corresponding coefficients are T0 T    im   T U T  + cv  U r 0 +  0 + U z 0  − RT0  U r + U z + U  t r r  z  z r  r   0c p T0 1 P0 T0 P0 T0 P0 Ae,ur =  0 c p − −  r Ae,u = − − z − −  Ae,uz = 0 c p r r z z r  r   U     2 1  m2  2   im     Ae,T = 0 cv  U r + U + U z  − R  U r 0 + U z 0 +  0  − k  2 + − +  z  r z r    r r r r 2 z 2   r r  Be,T = i 0cv Be,  = −iRT0 Ae,  = c p

As for the dissipation functions given by Eq. (5-160), they may be expressed as U U r U z   4  U U 1 U U z   1  2 U =  2 r − r − − +  +2 r − −   3  r r r  z  r r  r  r r z    im  U 1 U r U   U z U r    +2    + − + + r  r   r z  z   r  r

r

 U U U r U z  4im  2 U 1 U r U    1   1 U z U +2 r − − + 2   + − +  − +    3r  r  r r z  r  r   r r   r  z  r  U z U r   im  1 U z U    z 4  U z U r U r 1 U   = 2 − − − + 2  + + +   3  z r r r   z z  r r  r  z    r



=

  z   

P5-57 Identify the block matrices that must be used to convert the compressible biglobal stability disturbance equations in cylindrical polar coordinates, based on Eqs. (5-155)-(5-160), to an eigenvalue problem of the form Aij fi =  Bij fi , where  and fi denote the eigenvalues and corresponding eigenvectors. The block matrices appear as:  Ac ,    Ar ,   Aij =  A ,   Az ,    Ae, 

Ac ,ur

Ac,u

Ac ,uz

Ar ,ur

Ar ,u

Ar ,uz

A ,ur

A ,u

A ,uz

Az ,ur

Az ,u

Az ,uz

Ae,ur

Ae,u

Ae,uz

Ac ,T   Ar ,T  A ,T   Az ,T   Ae,T 

and

 Bc ,    Br ,   Bij =  B ,   Bz ,    Be, 

Bc ,ur

Bc,u

Bc ,uz

Br ,ur

Br ,u

Br ,uz

B ,ur

B ,u

B ,uz

Bz ,ur

Bz ,u

Bz ,uz

Be,ur

Be,u

Be,uz

Bc ,T   Br ,T  B ,T   Bz ,T   Be,T 

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Show that the only nonvanishing values of Bij are Bc,  = iI N , Br ,ur = B ,u = Bz ,uz = i 0 I N , Be,T = i 0cv I N , Be,  = −iRT0 I N and that   Ac ,      Ar ,     A ,    A  z,    Ae,  

U  1 U U z  U U = r + r + + + im   + (U r )ii DNr + (U z )ii DNz r r  z r ii  r  U  T0   U r U U r U r U2  =  r +U r + +Uz − + (T0 )ii DNr   + R   r r  z r ii  r ii   t U U U U U rU  R  U  T   =   +U r + +Uz + + im (T0 )ii +  0    r r  z r ii rii   t   ii   T   U z U U z U z   U z = +U r + +Uz + R  0  + (T0 )ii DNz   r r  z ii  t  z ii   T U T T   T   U   = c p  0  + cv  U r 0 +  0 + U z 0  − R (T0 )ii (U r )ii DNr + (U z )ii DNz + im     r r  z ii  t ii   r  ii  

(density)   Ac ,ur   A  r ,ur    A ,ur    Az ,u  r

 1    =  0  + (  0 )ii  DNr +  rii   r ii   U   4   U   = (  0 )ii (U r )ii DNr + im    + (U z )ii DNz +  r   −    DNr  r ii   r ii  3  

( )

2

+

1 r 1 DN − 2  + DNz rii rii 

( )

2

−

m 2   rii2 

 U  im  7 1 U = (  0 )ii   +   −   2 + DNr  r ii 3  rii rii  r  z    DN  T   P   U z  = (  0 )ii  + DNr DNz  Ae,ur = (  0 )ii c p  0  −  0  −  r  −   r ii 3  rii  r ii  r ii 

(radial terms)   Ac ,u   A   ,u       Az ,u  

im  1 7  1 U r 2U  Ar ,u = (  0 )ii  − −   DNr − 2   r ii 3  rii rii   r   im (U )ii + (U r )ii 1  U   = (  0 )ii (U r )ii DNr + + (U z )ii DNz +     rii rii   ii   2 2  Dr 1 4m 2  −   DNr + N − 2 + DNz − 2  rii rii 3rii   =

1   0  + im  0  rii   ii

( )

=

( 0 )ii  U z  rii

(tangential)

( )

im z    −  3r DN  ii ii

Ae,u = c p

( 0 )ii  T0  rii

 1 P0     −  r   −   ii  ii

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  Ac ,u z    A ,uz   A  z ,u z    

     U  =  0  + (  0 )ii DNizi Ar ,u z = (  0 )ii  r  − DNr DNz  z ii 3  z ii im z  U  = (  0 )ii    −  DN 3rii  z ii  U = (  0 )ii (U r )ii DNr + im    r  r 2  D m2 4 −   DNr + N − 2 + rii 3 rii 

( )

(axial terms)

  U z   z  + (U z )ii DN +  z    ii  ii 2 T P   DNz  Ae,u z =   0 c p 0 − 0  −  z z z ii  

( )

and   Ac ,T   A   ,T    Ae,T    

=0 =

    Ar ,T = R  0  + ( 0 )ii DNr   r ii 

  R im0 + 0   rii   ii

    Az ,T = R  0  + ( 0 )ii DNz   z ii 

  U  = ( 0 )ii cv (U r )ii DNr + im    + (U z )ii DNz   r ii    r 2 DNr m 2  U 0    − R U r 0 + U z 0 +  − k − 2 + DNz  DN +  r z r  ii rii rii  

( )

(temperature terms)



( ) 2



where the dissipation function may be reconstructed from U U r U z  1  4  U U 1 U U z  r  2 U =  2 r − r − − DN +  +2 r − −    3  r r r  z ii r r z ii rii   r    4  2  U  U 1  U 1  U r U z  im    =   + Ur  − − + 2   +  r − U    DNr −   3  r   z ii rii r   rii  ii   r   r r

   1 U z U  z  + DN  +  z ii    r    U  z 4  U z U r U r 1 U  z U r  r  1 U z U  im  = 2 − − − DN + 2  z + DN +  +    3  z r r r  ii z ii z ii rii   r   r

Solution: This problem involves converting our previous matrix formulation to a form that can be conveniently applied to a grid. To start, any primitive or mean flow variable will be evaluated at the ii location, which stands for the ( r , z ) grid location, as we assume an axisymmetric configuration for our biglobal formulation. Next, all derivatives are converted to derivative operators according to the following terminology:  = DNr r

 = DNz z

2     =   = DNr 2  r  r  r

( )

2

2    =   = DNz 2  z  z  z

( )

2

By applying these notations to the above set of matrix equations, we obtain:

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  Ac ,      Ar ,     A ,    A  z,    Ae,  

U  1 U U z  U U = r + r + + + im   + (U r )ii DNr + (U z )ii DNz r r  z r ii  r  U  T0   U r U U r U r U2  =  r +U r + +Uz − + (T0 )ii DNr   + R   r r  z r ii  r ii   t U U U U U rU  R  U  T   =   +U r + +Uz + + im (T0 )ii +  0    r r  z r ii rii   t   ii   T   U z U U z U z   U z = +U r + +Uz + R  0  + (T0 )ii DNz   r r  z ii  t  z ii   T U T T   T   U   = c p  0  + cv  U r 0 +  0 + U z 0  − R (T0 )ii (U r )ii DNr + (U z )ii DNz + im     r r  z ii  t ii   r  ii  

(density)   Ac ,ur   A  r ,ur    A ,ur    Az ,u  r

 1    =  0  + (  0 )ii  DNr +  rii   r ii   U   4   U   = (  0 )ii (U r )ii DNr + im    + (U z )ii DNz +  r   −    DNr  r ii   r ii  3  

( )

2

+

1 r 1 DN − 2  + DNz rii rii 

( )

2

−

m 2   rii2 

 U  im  7 1 U = (  0 )ii   +   −   2 + DNr  r ii 3  rii rii  r  z    DN  T   P   U z  = (  0 )ii  + DNr DNz  Ae,ur = (  0 )ii c p  0  −  0  −  r  −   r ii 3  rii  r ii  r ii 

(radial terms)   Ac ,u   A   ,u       Az ,u  

im  1 7  1 U r 2U  Ar ,u = (  0 )ii  − −   DNr − 2   r ii 3  rii rii   r   im (U )ii + (U r )ii 1  U   = (  0 )ii (U r )ii DNr + + (U z )ii DNz +     rii rii   ii   2 2  Dr 1 4m 2  −   DNr + N − 2 + DNz − 2  rii rii 3rii   =

1   0  + im  0   rii   ii

( )

=

( 0 )ii  U z  rii

(tangential)

( )

im z    −  3r DN  ii ii

Ae,u = c p

( 0 )ii  T0  rii

 1 P0     −  r   −   ii  ii

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  Ac ,u z    A ,uz   A  z ,u z    

     U  =  0  + (  0 )ii DNz Ar ,u z = ( 0 )ii  r  − DNr DNz  z ii 3  z ii im z  U  = (  0 )ii    −  DN 3rii  z ii  U = (  0 )ii (U r )ii DNr + im    r  r 2  D m2 4 −   DNr + N − 2 + rii 3 rii 

( )

  U z   z  + (U z )ii DN +  z    ii  ii 2 T P   DNz  Ae,u z =   0 c p 0 − 0  −  z z z ii  

(axial terms)

( )

and   Ac ,T   A   ,T    Ae,T    

=0 =

    Ar ,T = R  0  + ( 0 )ii DNr   r ii 

  R im0 + 0   rii   ii

    Az ,T = R  0  + ( 0 )ii DNz   z ii 

  U  = ( 0 )ii cv (U r )ii DNr + im    + (U z )ii DNz   r ii    r 2 DNr m 2  U 0    − R U r 0 + U z 0 +  − k − 2 + DNz  DN +  r z r  ii rii rii  

( )

(temperature terms)



( ) 2



As for the dissipation function, its components can be expressed as: U U r U z  1  4  U U 1 U U z  r  2 U =  2 r − r − − DN +  +2 r − −    3  r r r  z ii r r z ii rii   r    4  2  U  U  1  U 1   1 U z U  z   U r U z  im    =   + Ur  − − + 2   +  r − U    DNr −  +  + DN     3  r   z ii rii r   rii   r  z ii  ii   r    r   U  z 4  U z U r U r 1 U  z U r  r  1 U z U  im  = 2 − − − DN + 2  z + DN +  +    3  z r r r  ii z ii z ii rii   r   r r

Lastly, the right-hand side block matrix has to be converted. Recalling that the derivatives are a Kronecker product, the B terms have to be arranged to ensure that the equations are properly configured as per Eq. (5-108). The right-hand side is multiplied by the identity matrix I N to reproduce the correct distribution. This is accomplished by taking: Bc ,  = iI N Br ,ur = i 0 I N B ,u = i 0 I N Bz ,uz = i 0 I N Be,T = i 0cv I N Be,  = −iRT0 I N

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CHAPTER 6. INCOMPRESSIBLE TURBULENT MEAN FLOW

6-1 By direct substitution of fluctuating variables, Eqs. (6-6), into the x-momentum relation and time-averaging using (6-7), develop the x-component of the three-dimensional “Reynolds” equation (6-12) and reduce to a two-dimensional turbulent boundary layer. The key to this development is to write the convective acceleration in Eq. (6-11) in the “conservative” form mentioned just below. For the x direction, we obtain

(V ) u =

   ( uu ) + ( vu ) + ( wu ) x y z

Then introduce u = u + u, etc. into the full x-momentum relation, Eq. (6-11):

   ( u + u) + ρ ( uu + uu + u2 ) + ρ ( uv + uv + uv + uv ) t x y   + ρ ( uw + uw  + uw + uw  ) = − ( p + p ) + ρg x + μ 2 ( u + u  ) z x

ρ

(1)

Now take the time-average of Eq. (1), using the averaging relations, Eq. (6-7), leaving:  u  u u u    ρ +u +v +w + ( u u  ) + ( u v ) + ( uw  )  x y z x y z  t  p = − + ρg x + μ 2 u x

(Ans. 3-dimensional)

Since the first four terms on the left equal ρ ( du/dt ) , this relation is, as desired, the x-component of the mean (turbulent) momentum equation, Eq. (6-12). For a steady two-dimensional boundary layer, /t = 0, /z = 0, w = 0, and the y-derivative (across the layer) is the dominant one. The relation above reduces to  u  u  dp 2u ρ u +v + ( u v )  = − + ρg x + μ 2 y y dx y  x 

(Ans. 2-d T.B.L.)

where the pressure gradient could be replaced by dp/dx = −ρU ( dU/dx ) in the freestream. This is the desired result and is equivalent to Eq. (6-21b) in the text, page 411.

6-2 Using the Navier-Stokes equation in cylindrical coordinates (Appendix B), develop a three-dimensional Reynolds-stress equation for z-momentum and simplify it to axial steady turbulent boundary flow along me outside of a circular cylinder. This is somewhat more complicated than Prob. 6-1 above, because of the added complexity of the cylindrical geometry. The axial convective acceleration can be written in a “conservative” form, using the continuity relation, Eq. (B-3), as follows: Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -1-

( V  ) vz = vr

vz vθ vz v 1  1   + + vz z = ( rvr vz ) + ( vθ v z ) + ( v z v z ) r r θ z r r r θ z

The full Navier-Stokes z-momentum equation is given by Eq. (B-8): p  v  ρ  z + ( V  ) vz  = − + ρg z + μ2 vz , z  t 

where  2 =

1  1  2 r + +   r r  r  r 2 θ 2 z 2

Introduce fluctuating variables into this relation, using ( vr , vθ ,vz ) = ( u,v,w ) for convenience in the notation:  ρ  ρ  ( w + w) + ( ruw + ruw  + ruw + ruw  ) + ( vw + vw  + vw + vw  ) t r r r θ   + ρ ( ww + 2ww  + w w  ) = − ( p + p ) + ρg z + μ 2 ( w + w  ) z x

ρ

Now take the time-average of this relation, using the averaging rules, Eqs. (6-7), leaving:

w v w w 1  1    w ρ +u + +w + ( ruw ) + ( vw ) + ( w w  ) r r θ z r r r θ z  t  p = − + ρg z + μ 2 ( w ) (Ans. 3-dimensional) z This is the desired 3-D result. For steady axisymmetric flow along the outside wall of a circular cylinder, /t = 0, /θ = 0, vθ = “v” = 0, and the dominant derivative is  /r across the boundary layer. Further, if boundary-layer displacement is (presumed) small, there is no pressure gradient. Further neglect gravity along the cylinder. The 3-D equation above reduces to w 1  μ   w   w ρ w +u + ( ruw)  r  r r r  z  r r  r 

(Ans. 2-d T.B.L.)

where, as mentioned, u represents v r and w represents v z . We may rewrite this in terms of a cylindrical boundary-layer “turbulent shear stress”.

τ = τlam + τ tub , where τlam = μ

w and τ tub = −ρuw r

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The axisymmetric cylinder boundary-layer flow then may be written as w  1   w p w +u (r τ) (Ans. 2-dim. T.B.L.) = r  r r  z This has exactly the same form as laminar axisymmetric cylinder boundary-layer flow from Eq. (4-195b), page 299 of the text.

6-3 Consider turbulent flow past an isothermal flat plate of length L and width b, with constant ( ρ,μ,cp ,k ) . Assume Pr = 1, that is, δ u = δ T , with uniform stream velocity U and stream temperature Te at the plate leading edge. At the trailing edge, x = L, the mean velocity and temperature may be approximated by one-seventh power-law profiles:

u T − Tw  y     U Te − Tw  δ  With no further information, estimate the drag force and total heat transfer of one side of the plate, in terms of the boundary layer thickness δ and other flow parameters. 1/7

The flat-plate control-volume analysis of Section 4-l is appropriate for turbulent flow also, as long as the mean profiles are known. The drag force is given by Eq. (4-6), page 218 of the text: δ

F =  ρu ( U − u ) b dy = ρδbU 0

1

2

 η (1 − η )dη = 72 ρδbU 1/7

7

1/7

2

(Ans. a)

0

In like manner, the total heat transfer is given, for either laminar or turbulent flow, by the flatplate thermal integral analysis, Eq. (4-19), neglecting kinetic energy changes: δ

1

dQ =  ρcp u ( T − Te ) b dy = ρδc p U ( Tw − Te )  η1/7 (1 − η1/7 )dη dt 0 0 or:

dQ 7 = ρδbcp U ( Tw − Te ) dt 72

(Ans. b)

The analysis would be complete if we had an accurate estimate for the growth of the boundary layer thickness, δ(x). For example, use of Prandtl’s flat plate estimate from Eq. (6-72), δ/L = 0.37/ Re L1/5 , gives the dimensionless drag force formula CD =

2 ( 7/72 )( 0.37 ) 0.072 2F   (turbulent flat-plate flow) 2 ρU bL Re1/5 Re1/5 L L

which is popular in the literature, e.g., Kays & Crawford (1980), page 175. Similarly, the Stanton number could be approximated as

CH =

0.37 ( 7/72 ) 0.036 dQ/dt = = bLρUcp Re1/5 Re1/5 L L

This formula (which assumes Pr = 1 ) is given on p. 213 of Kays & Crawford (1980).

6-4 The first experiment of Clauser (1954) used air at 24°C and 1 atm. Velocity data at the first station, x = 6.92 ft, are given below. Taking δ = 3.5 in. and dU/dx = 1.06 sec−1 at this Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -3-

station, analyze the data to find (a) the inner law and wall shear stress; (b) the outer law and Clauser’s parameter β and Coles’ parameter P ; and (c) the logarithmic overlap. y, in

u, ft/s

y, in

u, ft/s

0.1

16.14

0.8

22.88

0.15

17.02

0.9

23.70

0.2

17.54

1.0

24.38

0.25

18.16

1.25

26.51

0.3

18.69

1.5

28.21

0.4

19.60

2.0

31.22

0.5

20.49

2.5

32.27

0.6

21.24

3.0

32.44

0.7

22.03

3.5

32.50

For air at 24C and 1 atm, take ρ = 0.00230 slug/ft 3 and v = 0.000165 ft 2 /s. A preliminary plot (not shown) of u versus log(y) reveals that the first five or six points form a straight line - the logarithmic overlap. We may thus test the log-law relation, u 1  yv*   ln   + B, with κ = 0.41 and B  5.0 v* κ  v 

for the first seven data points y and u(y) to estimate the value of friction velocity v*. The results are shown in the following table: y ( in.) = v* ( ft/s ) =

y+ =

0.1

0.15

0.2

0.25

0.3

0.4

0.5

1.091

1.080

1.068

1.078

1.072

1.080

1.094

55

82

109

136

164

218

273

All these points lie in the usual logarithmic overlap range 30  y+  300. Therefore, at this station, we conclude that the friction velocity and wall shear stress are v* = 1.08 ft/s τ w = ρv *2 = ( 0.00230 )(1.08) = 0.0027 psf 2

(Ans. a)

The first data point is not close enough to the wall to demonstrate an inner law.

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To establish Clauser’s parameter β, we need to estimate δ*. Numerical integration (by the trapezoidal rule) of the given velocity data give the estimate δ* = 0.59 in. Then β=

δ* dp δ*  dU  ( 0.59/12 )( 32.50 )(1.06 ) = −ρU = 1.45 (Ans. b) = 2 2  τ w dx ρv*  dx  (1.08)

Coles and Hirst (1968) give β  1.358 at this station for their data reduction. To establish Coles’ parameter P , we need to fit the wake law, Eq. (6-47) to the velocity-profile data. Well, the fit is good in the intermediate region, 0.5 in < y < 2 in, but the edge of the boundary layer is overpredicted with such a fit. This is seen in the following graph of the data in u+ versus log(y+) coordinates:

The wake law for P  2.4 (Ans. b) fits very well nearer the wall but overshoots the last three data points. The last point, y = δ = 3.5 in, is best fit by P  1.4. Coles and Hirst (1968) do not give an estimate of P for this station. The graph on above shows that the logarithmic overlap fits the data very well. However, Coles’ law does not fit the entire profile but only its interior portion. Coles reported similar edge discrepancies for other experimental boundary layer profiles. With P uncertain, we cannot estimate  * accurately from the wake-law formula (6-48). With a best-fit P  2.4, the formula gives  * = 0.96 inches, or about 50% too high.

6-5 Analyze the velocity data of Prob. 6-4 to achieve a power-law overlap approximation, as in Eq. (6-50). Find  and C and compare with Eq. (6.50). To save work, use Eq. (6-69) to estimate the momentum thickness. At this station the wall shear stress is approximately 0.0027 lb/ft 2 . (HINT: Use only the first 10 data points, the rest are well into the wake region.) Solution: For air at 24C, take  = 0.00230 slug/ft 3 and v = 0.000165 ft 2 /s. The data from Prob. 6-4 may be plotted as follows in log-log coordinates. The initial slope is about 0.16. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -5-

Based on the first 10 data points, the log-law is approximately u + = 7.8 ( y + )

0.16

, that is, C  7.8

and   0.16. From Eq. (6-69), approximately  / = 7/72 , or  = 7 ( 3.5 in ) /72 = 0.34 inches.

Thus Re = U /v = ( 32.5 ft/s )( 0.34/12 ft ) / ( 0.000165 ft 2 /s )  5600. Then Fig. 6-14 and Eq. (6.50) predict

C  3 + 0.62 ln ( 5600)  8.3 ( 6% high ) and  = 1.24/ln (5600 )  0.14 (10% low ) The given data τ w = 0.0027 lb/ft 2 is not needed here but could be found from the power-law.

6-6 For developed turbulent pipe flow, assuming that the log-law is va1id across the entire tube, find a formula for centerline velocity as a function of Darcy friction factor . The desired formula is hidden within Eq. (6-53), which computes average pipe-flow velocity by integrating the log-law across the entire pipe: 3   1 av* u av = v*  ln +B−  v 2κ  κ

( 6-53)

The terms in boldface are, in fact, equal to u max /v* (neglecting the slight ‘wake’ at the centerline). Thus we may rearrange the above equation as follows:

u max = u av +

u max 3 3 v* 3  v*, or: = 1+ = 1+   2κ u av 2κ u av 2κ  8 

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where the last substitution follows from the definition of Darcy friction factor, 2  = 4Cf = 8v*2 /u av . Taking κ = 0.41, we have the final desired relation: u max  1 + 1.29  u av

(Ans.)

6-7 Water at 20C flows through a smooth 3-cm-diameter pipe at 30 m3 /hr. Assuming developed flow, estimate (a) the wall shear stress; (b) the pressure drop per meter; and (c) the centerline velocity. What is the laminar-flow rate? What flow rate gives τ w = 100 Pa? For water at 20C take ρ = 998 kg/m3 and μ = 0.001 kg/m-s. The average velocity in the pipe is 3 Q ( 30/3600 ) m /s u av = = = 11.79 m/s A ( π/4 )( 0.03 m )2

whence Re D =

ρu av D ( 998 )(11.79 )( 0.03) = = 353, 000 μ ( 0.001)

The flow is clearly turbulent ( Re  2000) . Since the walls are ‘smooth’, the Darcy friction factor is given by Eq. (6-54) for developed smooth-wall pipe flow:

(

)

1 = 2.0 log10 ReD  − 0.8, or  = 0.01403 if ReD = 353, 000 

The wall shear stress is then given by 1 2 2 1 τ w =  ρu av = ( 0.01403)   ( 998)(11.79 ) = 243 Pa 8 8

( Ans. a )

The pressure drop in developed flow is given by p = τ w

4 (1.0 ) 4Δx = ( 243)  32, 400 Pa/m D 0.03

( Ans. b )

Finally, the centerline velocity is estimated from our formula derived in Prob. 6-6 above:

(

)

u max  u av 1 + 1.29  = (11.79 ) 1 + 1.29 

( 0.01403)  = 13.6 m/s ( Ans. c )

The flow rate would have to be much less for the flow to be laminar: ReD =

ρu av D 998u av ( 0.03) =  2000 if u av  0.0668 m/s, μ 0.001 or: Q transition = u av A = ( 0.0668 )( π/4 )( 0.03) = 4.72 E-5 m3 /s = 0.17 2

m3 hr

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Finally, finding the flow rate for which wall shear equals 100 Pa is vexing, because we know neither the velocity nor the friction factor and thus have to iterate. However. Eq. (6-55) gives us the estimate τ w proportional to Q7/4 . Thus a good first guess would be

Q100Pa

 100  = Q243Pa    243 

4/7

 0.6, or Q  0.6 ( 30 )  18 m3 /hr

Try this value with the numerical procedure above: Q = 18

m3 m ; u av = 7.074 ; Re D = 211800;  = 0.01546; τ w = 96.5 Pa hr s

Not bad. [The correct power on τ w = CQ n should have been n  1.81. ] Anyway, we can interpolate between these two shear stresses to find that τ w = 100 Pa if

6-8

Re D = 216000, or: Q = 0.00510

m3 m3 = 18.4 s hr

The overlap region of Clauser’s velocity profile in Prob. 6-4 may be fit by a power-law

estimate y +  7.5 ( u + ) . Use this result to estimate the wall shear stress in lbf /ft 2 and the shape 1/ 6

factor H. Solution: This is close to, but slightly different from, the power-law found in Prob. 6-5,

u + = 7.8 ( y+ ) . Find the wall shear stress by writing this in terms of v * : 0.16

1/ (1+ )



 uv *a  u  yv *  =C  , or: v* =  a  v*  v   Cy 

with C = 7.5 and  =

1 6

Apply this formula to the first 10 points in the velocity table in Prob. 6-4. The results are

v* = 1.11  0.04 ft/s

Ans.

τ w =  v *2 = 0.0028  0.0002 lb f /ft 2

Problem 6-5 stated that the wall shear stress was approximately 0.0027 lbf/ft2, or 4% lower. For the shape factor estimate, we need to compute displacement and momentum thicknesses. Write the power law in the form u /U = 1/6 , = y / . Then 1

1  * =   (1 − ) d =  ; 7 0 1/6

1

 =   1/6 (1 −1/6 ) d = 0

3  28

The ratio of these two is the desired power-law shape factor estimate:

H=

 * (1/7 )  4 = =  1.33  ( 3/28)  3

Ans.

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6-9 Analyze developed turbulent flow through a smooth-wall square duct, using the loglaw, the double symmetry, and a suitable assumption about distribution of wall shear stress. Compare your result for friction factor  with the hydraulic-radius concept. By double symmetry, we need only consider one-eighth of the duct cross-section, shown as the shaded area below. Although the flow in this region is three-dimensional (secondary circulation toward the corners), the mean velocity u in the duct-axis direction is well represented by the log-law, Eq. (6-38a), based on distance “y” from the wall, as sketched in the figure.

In general, however, the wall shear stress varies with z along the wall, being zero in the corners and rather flat near z = a/2. The total volume flow rate in the duct would be 8 times the flow rate in the shaded area:

z  1 yv*   Q = 8  dz   v* ( z )  ln + B dy  v κ   0 0 a/2

If, as a first approximation, we assume constant v* ( z ) around the cross-section, the integration is easily performed:

u av =

 1 a+ Q 3 av* = v* + B −  , where a + = av  ln 2 a 2κ  v κ 2

Now, by definition, the friction factor  for any duct is such that v*av /u av =

(  / 8 ).

Therefore the above relation may be rewritten as a friction factor formula:

(8/ ) 

1  ln Rea κ 

(  /32 )  + B −

3 , 2κ

Rea =

u av a v

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For κ = 0.41 and B = 5.0, this may be written in numerical form as

(

)

1  1.99 log10 Rea  − 1.02 

But these are exactly the same numbers which result for turbulent pipe flow - see the middle of page 426 of the text. This simple theory thus predicts that a square duct friction is equivalent to pipe friction if Rea = ReD , that is, if “a” is the effective diameter. In fact, the ‘hydraulic diameter’ of a square duct does equal its side length, a: D h ( square ) =

4A 4a 2 = =a P 4a

Therefore, a first-order use of the log-law in square-duct flow leads to the formula

(

)

1  2.0 log10 Rea  − 0.8, a = side length 

(Ans.)

If we modified the theory to have a variable v*(z) which satisfies the physical condition that τw = 0 ( or v* = 0) in the duct corners, the friction formula might be improved. However, we 1/7 can’t guess how v*(z) varies: perhaps v*/v*max = ( 2z/a ) could be tried? Meanwhile, the ‘effective diameter’ concept of page 428 is known to work well.

6-10 In the overlap layer, turbulent shear is dominant, and the effect of viscosity is small. Suppose that we neglect  and replace Eq. (6-32) by the approximate gradient relationship du  fcn ( y , w ,  ) dy

Show, by dimensional analysis, that this leads directly to the logarithmic overlap law (6-38a). Solution: There are four variables and their dimensions in the {MLT} system are: Variable

du /dy

y

w



Dimension

T −1

L

ML−1T −2

ML−3

With 4 variables and 3 primary dimensions, we expect only one Pi group, and it is: P1 =

du y  du v* v* = constant, or: = constant = C1 dy  w dy y y Integrate:

u = C1v *ln ( y ) + C2

If the constants are arranged so that the logarithm has a dimensionless argument, we obtain

u /v* = u + =

1 ln ( y + ) + B κ

6-11 Use the log-law to analyze developed turbulent flow in a smooth annulus of inner radius ri and outer radius ro. Compare the friction result with hydraulic-diameter theory. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -10-

The (turbulent) velocity profile in the annular region would look like this:

In general, the outer-wall shear stress is somewhat less than the inner-wall shear, and the radial position rm of the maximum velocity (the zero-shear line) is somewhat toward the inner wall rather than in the exact middle of the annular region. This is seen by making a force balance on the outer region and comparing to the inner region, as follows:

Inner region: p π ( rm2 − ri2 ) = τi 2πri Δx Outer region: p π ( ro2 − rm2 ) = τ o 2πro Δx Divide one by the other to get the ratio of inner to outer shear stress: 2 2 τo ri ( ro − rm ) = τi ro ( rm2 − ri2 )

(1)

Meanwhile, we can write a log-law profile for each region, using wall coordinates y i and yo, respectively, as shown in the figure on (previous) page 175 of this Manual: Inner region:

ui 1 y v*  ln i i + B, v*i = v*i κ v

( τi /ρ )

Outer region:

uo 1 y v*  ln o o + B, v*o = v*o κ v

( τo /ρ )

These two must meet at equal (maximum) velocity at the point of zero shear:

 1 ( rm − ri ) v*i   1 ( r − r ) v*o  + B = v*o  ln o m + B At r = rm: u max = v*i  ln v v κ  κ 

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(2)

Equations (1) and (2) above can be solved simultaneously to relate the shear stress ratio to the position of the maximum velocity. Obviously the algebra is quite complex, however. The volume flow rate can be computed by integrating over each region: rm

ro

ri

rm

Q =  u i 2πrdr +  u o 2πrdr

(3)

where u i and u o are given by their respective log-laws. The average velocity and hydraulic diameter are given by

Q V= 2 π ( ro − ri2 )

Dh =

4π ( ro2 − ri2 ) 2π ( ro + ri )

= 2 ( ro − ri )

from which we may form the Reynolds number Re = VDh /v. Finally, the pressure drop is directly related to inner and outer shear stress by an overall force balance: pπ ( ro2 − ri2 ) = 2π ( τ o ro + τi ri ) x

(4)

These relations may be combined to find the overall pressure drop as a function of flow rate for any particular annulus size. The dimensionless pressure drop can be taken as the (overall) Darcy friction factor,  : =

p 2D h x ρV 2

(5)

For a given ratio ( ro /ri ) ,  would vary with Re. The above relations constitute a complete “log-law theory” of developed turbulent flow in an annulus. As a first approximation, assume the maximum velocity to be in the middle of the annular region, rm = ( ri + ro ) /2. Then Eq. (2) above predicts that

v*i = v*o = v*;

τi = τ o

This cannot be exactly true because it violates the force balance, Eq. (1) above. Proceeding anyway, the volume flow for this approximation becomes  1  hv*  1 Q = 2πv*h ( ro + ri )  ln  + B− , κ κ  v  1 1 or: V/v* = ln ( h + ) + B − κ κ

where h =

1 ( ro − ri ) , 2

Since h = Dh /4, this latter relation is equivalent to a friction-factor/Reynolds-number equation:

(8/ ) = or:

1 1 1  ln  ReDh Λ/8  + B − κ 4 κ 

(

)

1/  = 1.99log10 ReDh Λ − 1.19

(6) (Ans.)

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We may compare this to the hydraulic-diameter approximation from pipe theory, page 426 of the text:

(

1  1.99 log10 Re Dh 

)

Λ − 1.02

(7)

The difference is slight. Some numerical values can be listed as follows:

Re =  ( Eq. 6 ) =

1E4 0.0353

1E5 0.0200

1E6 0.0127

1E7 0.00875

1E8 0.00635

 ( Eq. 7 ) = Difference =

0.0334

0.0192

0.0123

0.00850

0.00620

6%

4%

3%

3%

2%

Jones and Leung (1981) give a more accurate way to compute annulus friction.

6-l2 Modify the flat-plate integral analysis of Sec. 6-6.1 by combining Eq. (6-67) with Eq. (6-71) rather than (6-69). Numerical integration may be necessary. Compare this friction estimate with Eq. (6-70). We are to combine the momentum relation (6-67) with the wall-wake correlation (6-71):

Ct =

2 dθ =2 ; 2  dx

θ 3.54 22.21  − 2 ; δ  

Reδ =   κ (  − 7.2 ) 

where the last relation is the wall-wake profile evaluated at y = δ (see p. 429 of text). These relations constitute a first-order differential equation relating δ ( x ) to  ( x ) , which we may integrate numerically. One possible formulation would be dF 1 = 2, dRex 

where F = exp  κ (  − 7.2 )  ( 3.54 − 22.21/ )

With “F” known at the next station, we iterate to find the new value of  . Later, if desired, we can compute Reδ from the algebraic relation above. To solve this differential equation, we need an initial value for  at an initial Reynolds number. Strictly speaking,  = 0 at Re x = 0 , but this blows up in our equation. Therefore we start at the extremely low (non-turbulent) value  = 7.2 and assume that Re x = 0 there. The calculations then proceed very smoothly. The results may be tabulated and compared with the standard power-law approximation, Eq. (6-70):

Re x =

1E5

3E5

1E6

3E6

1E7

3E7

1E8

Cf =

0.00533

0.00437

0.00358

0.00301

0.00252

0.00216

0.00185

0.00521

0.00446

0.00375

0.00321

0.00270

0.00231

0.00194

Cf ( 6-70 ) =

The differences are only a few percent, therefore we don’t bother with this more complicated algebraic formulation.

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6-13

Devise a scheme to compute Cf ( x ) and δ ( x ) in the turbulent region of flat-plate flow

when there is an initial region of laminar flow up to a transition value Retr. One could equate total thickness δ or momentum thickness θ at the transition point, thus defining an “effective” origin x 0 for the ensuing turbulent boundary layer: 6/7 1/2 6/7 Equate δ : 5.0 Re1/2 tr = 0.16 Re XD ; Equate θ: 0.664 Re tr = 0.01555 Re XD

Both give about the same estimate of the “virtual” turbulent origin x 0 , but equating θ is more realistic. Downstream of Retr, local boundary layer parameters are computed based on the effective local Reynolds number Re x,eff = Re x − Re tr + Re XD

For example,

δ 0.16 = 1/7 x eff Re x,eff

and

Cf =

0.027 Re1/7 x,eff

Schlichting (l979, page 641) suggests modifying our turbulent drag formulas as follows: CD ( L  x tr ) =

0.031 A − , where A depends upon Retr Re1/7 ReL L

One computes “A” by equating the above formula to the laminar drag at transition, 1/2 C D = 1.328 / Re1/2 tr . For example, if Retr = 500,000 ( or 1E6 ) , then A = 1400 ( or 2980 ) . Let us compare these two approaches for an assumed value of Re1/2 tr = 500, 000. From equating θ, the effective origin is Rexo = 168000. List some tabulated values of friction by the two methods at various Reynolds numbers:

Re x =

5E5

7E5

1E6

2E6

4E6

1E7

1E8

Re x,eff =

1.68E5

3.68E5

6.68E5

1.668E6

3.668E6 9.668E6

9.967E7

Cf =

0.00484

0.00433

0.00397

0.00349

0.00312

0.00271

0.00194

0.00395

0.00375

0.00340

0.00308

0.00270

0.00194

Fully-turbulent flow:

Cf =

0.00414

The wall shear, at first higher, approaches the fully-turbulent results far downstream.

Re x = δ/x =

5E5

7E5

1E6

2E6

4E6

1E7

1E8

0.00964

0.0135

0.0157

0.0172

0.0169

0.0155

0.0115

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Fully-turbulent flow:

δ/x =

0.0245

0.0234

0.0222

0.0201

0.0182

0.0160

0.0115

At first much thinner, δ ( x ) also approaches the fully-turbulent value downstream.

6-14 Water at 20C and 1 atm flows at 6 m/s past a smooth flat plate 1 m long and 60 cm wide. Estimate (a) the trailing edge boundary layer thickness, (b) the trailing edge wall shear stress, and (c) the drag of one side of the plate, if Re tr = 1 E6. For water at 20C and 1 atm, take ρ = 998 kg/m3 and μ = 0.001 kg/m-s. Then

ReL =

ρUL ( 998)( 6.0 )(1.0 ) = = 5,988,000 μ ( 0.001)

Thus the first one-sixth of the plate flow is laminar, and the rest turbulent. If we ignore the laminar portion and assume fully turbulent flow, then the power-law Eqs. (6-70) give

δ ( L) =

0.16 L ( 0.16 )(1.0 ) δ =  0.0172 m, δ* = = 0.0022 m 1/7 1/7 ReL 8 ( 5988000 )

Cf ( L ) =

0.027 = 0.0029, or: Re1/7 L

7 CD = Cf ( L ) = 0.00339, or: 6

2 1 τ w ( L ) = ( 0.0029 )   ( 998 )( 6 ) = 52 Pa 2 1 Drag = ( 0.00339 )   ( 998 ) ( 6 2 )( 0.6 m 2 )  36.5 N 2

We don’t have a ‘log-law’ estimate for δ ( L ) , but Eq. (6-78) predicts Cf ( L )  0.00278

or τw ( L ) = 50 Pa, and Eq. (6-81) predicts CD = 0.00320, or Drag  34.4 N. If we account for the laminar leading-edge now, using the methods of Prob. 6-13 above, for Re x,tr = 1E6, we equate momentum thicknesses there to obtain Re xo = 36600, or x ( origin )  0.161 m. Then Leff = 0.839 m, and the thickness formula predicts ReL,eff =

ρULeff 0.16Leff 0.16 ( 0.839 ) = 5025000, or δ ( L ) = =  0.0148 m 1/7 μ Re1/7 ( 5025000 ) L,eff

Similarly, the trailing edge shear stress is estimated by

Cf ( L ) =

0.027 2 1  0.00298, or: τ w ( L ) = ( 0.00298)   ( 998)( 6 ) = 53.5 Pa 1/7 Re L,eff 2

Finally, using Schlichting’s idea for drag force with a laminar region (see p. 141 above),

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CD =

0.031 A 0.031 1440 −  − = 0.003095, 1/7 1/7 ReL ReL ( 5988000 ) 5988000 1 Drag = ( 0.003095)   ( 998) ( 62 )( 0.6 m 2 )  33.4 N 2

or:

These latter ‘laminar-leading-edge’ results would be the preferred engineering estimates.

6-15

Repeat Prob. 6-14 if the plate average roughness is 0.1 mm. Estimate B at x = L.

From Prob. 6-14 above, ReL = 5.99E6. With roughness height k = 0.0001 m, L/k = 10,000, and it is difficult to estimate if the flow is “fully rough”. From Eq. (6-61), the criterion for fullyrough flow would be

k+ =

kv* kU v* Rek = =  60, or: Rek  60λ = 1200 or so v v U λ

For our case, Rek = ρUk/v = ( 998)( 6)( 0.0001) / ( 0.001)  599, thus the flow is in the “intermediate roughness” regime and (the complicated) Eq. (6-82) should apply. Let us first concentrate on the wall shear stress at the trailing edge. We substitute Rex = 5.99E6 and ( x/k ) = 10,000 in Eq. (6-82) and iterate for λ. The result is

λ ( L )  22.07 = or:

U 6 , v* = = 0.272 m/s, v* ( L ) 22.07 τ w ( L ) = ρv*2 = ( 998 )( 0.272 ) = 74 Pa 2

(Ans. b)

Then k + ( L ) = kv*/v = 27.1 (intermediate roughness) and Eq. (6-62) predicts that

ΔB ( L ) 

1 1 ln (1 + 0.3k + ) = ln 1 + 03 ( 27.1)   5.4 κ 0.41 

(Ans. d)

The law-of-the-wall with roughness, Eq. (6-60) then predicts (if we neglect any ‘wake’),

 998 ( δ )( 0.272 )  U 1  δv*  1  ln  ln   + 5.0 − 5.4  + B − B = 22.07 = v* κ  v  0.41  0.001  or:

δ ( L ) = 0.37 m (Ans. a)

This is more than twice as thick as the smooth-wall boundary layer in Prob. 6-14. To compute the drag - for which there is no simple formula since the plate has intermediate roughness - we evaluate wall shear at various positions along the plate, using Eq. (6-62). Some results may be tabulated as follows:

x ( m) =

0.01

0.05

0.1

0.2

0.4

0.6

0.8

1.0 m

τ w ( Pa ) =

201

136

117

101

88

81

77

74 Pa

These are very well fit by the power-law approximation Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -16-

τ w ( Pascals )  75 x ( m )

−0.2

0.01  x  1.0 m

,

[If the entire plate had been “fully rough”, the wall shear would be correlated by Equation (6-83) in the text, but that formula is about ±10% in error.] Integration of the above formula gives an approximation for the drag on one side of the plate: L

Drag =  τ w b dx = 0

1m

−0.2  75x ( 0.6) dx = 0

75 ( 0.6 ) = 56 N 0.8

(Ans. c)

With 0.1-mm roughness, the plate has about 60% more drag compared to Prob. 6-14. [Equation (6-84b) is not bad, either, predicting Drag = 53 N for “fully rough flow”.]

6-16 Derive Stevenson’s law-of-the-wall with suction or blowing, Eq. (6-86). The derivation assumes (1) constant normal velocity v w near the wall; and (2) flat-plate flow, that is, dp/dx = 0. Then Eq. (6-85) applies:

τ = τ w + ρv w u = μ t

du du , where μ t  ρκ 2 y 2 dy dy

and τ w = ρv*2

We use the mixing-length eddy-viscosity model, Eq. (6-88), without the van Driest damping factor. Non-dimensionalize the above equation by dividing by τ w to obtain + w

+

1+ v u = κ

2

(y )

+ 2

2

 du +  u + yv* + v w + , vw =   , where u = , y = v* v v*  dy + 

Separate the variables and integrate

du +

 1 + v 

+ w

1/ 2

u + 

=

1/ 2 dy + 2 1 1 + v +w u +  = ln ( y + ) + constant , or: + +  κy vw κ

The “constant” was adjusted by Stevenson (1963) so that, in the limit as v w = 0, we recover the no-blowing law-of-the-wall, Eq. (6-38a). Thus we split the “constant”: 1/ 2 2  1 1 + v+w u + ) − 1  ln ( y+ ) + B, κ = 0.41 and B = 5.0 + (  κ vw 

(Ans.)

This is Stevenson’s law-of-the-wall with suction or blowing, Eq. (6-86).

6-17 Rewrite Stevenson’s law-of-the-wall with suction, Eq. (6-86), in the form of a ratio of local friction Cf to the no-blowing value, Cfo and relate this to a “blowing parameter”,

B = ( 2vw ) / ( UeCf ) . Plot this in the range −0.5  B  2.0 and compare to the correlation

formula Cf / Cfo = ln (1 + B )  / B, from Kays and Crawford (1980, p. 181).

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First let us note that the quantity ( v +w u + ) in Eq. (6-86), when evaluated at y = δ, u = U, is none other than the blowing parameter B: vw U vw U2 vw v U = = = 2 / Cf = B, the blowing parameter v* v* U v*2 U + w

+

Further, the quality v +w itself is related to loca1 skin friction:

v+w =

vw vw U vw 1/2 = = ( 2 / Cf ) v* U v* U

Then Eq. (6-86) - see the bottom of page 182 of this Manual - may be rewritten as

2 ( Cf /2 ) B

1/2

(1 + B )1/2 − 1 = 1 ln ( δ + ) + B = ( 2 / Cfo )1/2 if δ+  δo+   κ

where the last substitution follows since [ln(δ + ) / κ + B] exactly equals the no-blowing ratio U/v* - we assume (without proof) that this dimensionless thickness is not affected by the blowing. [A similar assumption is made by Kays and Crawford (1980, p. 181).] We apologize that italic B means ‘blowing parameter’ while non-italic B means the log-law constant, 5.0. Clean up the above equation and solve for the skin friction ratio: 2

1/2 2  Cf / Cfo =  (1 + B ) − 1   B 

to be compared to Kays & Crawford: Cf / Cfo =

ln (1 + B ) B

(1)

(2)

(Ans.)

It is interesting that the two formulas do not look alike at all, but they are very similar numerically. [Kays and Crawford (1980, pp. 180–181) make a different assumption about the constancy or eddy viscosity, rather than the constancy of boundary-layer thickness which we assumed.] The two formulas are plotted versus B for −0.5  B  2.0 below. The two agree with each other and also with available suction/blowing flat-plate friction data.

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We see that suction ( B  0) increases friction (and heat transfer), while blowing ( B  0) decreases friction (and heat transfer). This was also the case for laminar flat-plate flow (see Fig. 4-l5b) - but turbulent flow does not “blow off” at finite B. 6-18 Water at 20C flows through a smooth permeable pipe of diameter 8 cm. The volume flow rate is 0.06 m3/s. Estimate the wall shear stress, in pascals, if the wall velocity is (a) 0.01 m/s blowing; (b) 0 m/s; and (c) 0.01 m/s suction. To avoid excessive iteration, assume that the ratio of average to centerline velocity is 0.85. Solution: For water at 20C,  = 998 kg/m3 and  = 0.0010 kg/m-s. First find the average velocity and the (casually estimated) maximum velocity: V=

Q 0.06 m3 /s = = 11.94 m /s A π ( 0.04 m )2

;

Vmax 

V = 14.04 m/s 0.85

Now, for blowing and/or suction, apply Stevenson’s law, Eq. (6-86) at the centerline:

2 vw+

1/ 2     Rv *  1 + Vmax  ln  1 + vw  − 1 =  + 5, v*    0.41   

vw+ =

vw v*

For y = R = 0.04 m and Vmax = 14.04 m/s, try vw = ( a ) + 0.01 m/s; ( b )  0; and ( c ) − 0.01 m/s. Solve for v*. EES is excellent for this type of iteration. The three results are:

(a)

vw = 0.01 m/s, v* = 0.418 m/s,  w =  v*2 = ( 998 )( 0.418 )  174 Pa

Ans. (a)

(b)

vw = 0 m/s, v* = 0.483 m/s,  w =  v*2 = ( 998)( 0.483)  233 Pa

Ans. (b)

(c)

vw = −0.01 m/s, v* = 0.551 m/s,  w =  v*2 = ( 998)( 0.551)  303 Pa

Ans. (c)

2

2

2

These values are only approximate because V is not exactly equal to 85% of Vmax .

6-19 Use Stevenson’s relation (6-86) to develop a formula for wall friction in pipe flow with blowing or suction. Apply your result to Prob. 6-7, water flowing at 30 m3 /hr in a 3-cmdiameter smooth pipe, modified so that the wall blowing rate is 3 cm/s. Solution: Recall from Prob. 6-7 that uav = 11.79 m/s, D = 3 cm,  = 998 kg/m 3 , and  = 0.0010 kg/m-s. The Reynolds number is ReD = 353,000. One way is to proceed exactly as in Prob. 6-18, that is, apply Stevenson’s law at the centerline, y = R:

2 vw+

1/2     Rv *  1 + umax  ln  1 + vw  + 5,  − 1 = v*    0.41   

vw+ =

vw v*

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If the wall velocity vw is known, and if umax can be approximately related to uav , the on1y unknown in this relation is the friction velocity v*. By definition,  w =  v *2 . The writer cannot solve for v* because it appears in 4 places in the above correlation. However, one can iterate and plot the variation of  w with vw using, say, a spreadsheet. For our particular data from Prob. 6-7 with vw = 0.03 m/s, we estimate umax  1.15uav = 13.6 m/s and obtain

2v *  0.03 m/s 13.6 m/s   1 +  0.03  v* v* 

1/ 2

  ( 998 )( 0.015 ) v *  1 − 1 = ln  +5 0.001  0.41  

By iteration or EES or whatever, we obtain the result v* = 0.333 m/s, and

 w = v *2 = (998 kg/m3 ) ( 0.333 m/s ) = 111 Pa 2

Ans.

This is 54% less wall shear than the result of 243 Pa obtained for the impermeable pipe in Prob. 6-7. This is strong blowing, vw+ = +0.09, off the top of the chart in Fig. 6-22. An alternate approach is to use the wall shear ratio developed in Eq. (6-172) of the textbook:

 w ( blowing )

 w ( no blowing )

=

Cf

=

Cf0

 

e −1

, where  =

2vw umax C f 0

For a smooth impermeable pipe with ReD = 353,000, 0 = 0.014 from Eq. (6-54) and hence C f 0 = 0 /4 = 0.0035. This gives  = 2 ( 0.03 m/s ) / (13.6 m/s )( 0.0035 )  = 1.26. Thus

 w ( blowing )

 w ( no blowing )

=

 

e −1

=

1.26 1.26 = = 0.50 e − 1 2.525 1.26

This is close to our estimate by the first (Stevenson) method. The second estimate is

 w  0.50  w0 = 0.50 ( 243 Pa ) = 122 Pa

Ans.

6-20 As an alternative to Eq. (6-62), Bergstrom et al. (2002) suggest the following formula for the downshift of the log-law (6-60) due to uniform surface roughness of height k:

B 

1 ln ( k + ) − 3.5 κ

for

k +  4.2

First compare this correlation, with a sketch or graph, to Eq. (6-62). Then apply this correlation to derive a formula for pipe friction factor  similar to Eq. (6-64). Solution: The two formulas are compared in the figure below. Bergstrom’s formula cannot be used below k + = 4.2, for then it predicts negative (unrealistic) B.

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Since k + is independent of y, we simply subtract Bergstrom’s B formula from the smooth pipe analysis of Eq. (6-53): 3  1 3   1 Rv *  1 Rv * uav = v *  ln + B − B − + B − ln k + + 3.5 −  = v *  ln  v 2  v  2    where

Rv *  = Re D , v 32

uav 8 = , v* 

and

k+ =

k  Re D D 32

Use the traditional constants  = 0.41 and B = 5.0, clean up, and use base-10 logarithms. As if by magic, the Reynolds number cancels out, with the final result 1  D  2.0log10   + 1.7  k

Ans.

In other words, Bergstrom’s roughness formula, which does not level out towards zero as k + decreases, predicts a fully rough wall friction for all Reynolds numbers. Its predictions, though, are considerably smaller, by 10% to 25%, than the fully rough predictions of the Colebrook formula (6-65) as Re D → .

6-21 Use numerical quadrature to evaluate and sketch Eq. (6-96), the van Driest turbulent velocity profile, for zero pressure gradient. Compare with Eq. (6-41). Solution: Both formulas are models of the wall-law and log-law combined: y+ +

Van Driest: u =

 0

Spalding:



2dy +

1 + 1 + 4 y + 1 − exp ( − y + / A)  2

2



2 1/2

y + = u + + e− B e Z − 1 − Z − Z 2 / 2 − Z 3 / 6  ,

( 6 − 96 ) Z = u+

( 6 − 41)

The latter is purely algebraic, if implicit in u + , while the former may be integrated by, say, the Runge-Kutta Subroutine of Appendix C in the text or with a spreadsheet.

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The two formulas may be compared in the following chart:

There is little to choose between the two models as far as accuracy is concerned. However, the van Driest model is more flexible, since it may be made to fit other conditions - pressure gradient, blowing/suction, roughness, etc. - by changing the value of the clamping constant “A”.

6-22 Use the log-law, Eq. (6-38a), to analyze Couette flow between plates as shown. Since there is no pressure gradient, the shear stress is constant in the fluid between the plates. The profile is broken into an upper and a lower part, as shown below. The lower log-law begins at u = 0 and rises to u = U/2 at the center, y = h/2. The upper part begins at u = U and drops down to u = U/2 at the center.

In the both upper and lower regions, the log-law must satisfy the centerline condition

U/2 1  hv*   ln  +B v* κ  2v 

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This is a relation between shear stress, velocity, and plate separation distance. In dimensionless friction-factor/Reynolds-number from, we may rewrite this as 1/2

 Reh     4f 



1 1 ln κ  2

( Reh f )  + B,

where Reh =



τ h Uh and f = w v μU

This relation is implicit but easily computed. When Uh/v = 1E5, the result for f is

f = 51.1

(Ans.)

Other values can be listed as follows:

Uh/v =

τ w h/μU =

1E5 51.1

1E6 337

1E7 2375

1E8 17595

The velocity profile may be plotted by computing U+ = Re/ f and h + =

1E9 135270

( Re*f ).

Then plot

the log-law, Eq. (6-38a) as follows, in two parts:

Lower layer, 0  y  h/2: Upper layer, h/2  y  h:

1  + y ln  h  + B κ  h 1  h−y U + − u + = ln  h + +B κ  h 

u+ =

The results are plotted as follows for various Reynolds numbers:

6-23

Use the log-law, Eq. (6-38a), to analyze flow about a cylinder of radius R rotating at

angular rate

ω

in an infinite fluid. Sketch Cf = 2τ w /ρω R versus Re = ωR 2 /v. 2

2

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The circumferential velocity profile vθ ( r ) must equal Rω at the surface, dropping off to zero as r gets very large. However, the “log-law” does not have this asymptotic behavior and will need a little help. Near the wall, we can write the log-law as follows: Rω − vθ 1  ( r − R ) v *   ln  +B v* κ  v 

(1)

However, we do not know where to stop this formula, that is, we do not know the thickness

“ δ ” where vθ  0. A crude but effective remedy is to assume that δ equals the thickness of a flat-plate boundary layer of length L = 2πR. From Eq. (6-70), thus,

δ 0.16  2πR ( 2πR 2ω/v )1/ 7

(2)

Assume that, at this value of ( r − R ) , vθ = 0 in Eq. (1) above. That is, 1/2

Rω  2  1  δv*  =   = ln   + B, v*  Cf  κ  v 

δ from Eq. ( 2 ) above

This completes the analysis. For example, if Re = ωR /v = 1E5, Eq. (2) predicts that 2

  0.1493 R, whence the log-law predicts Rω/v* = 21.01, or Cf = 2/ ( 21.01) = 0.00453. We may tabulate a few friction factors as follows: 2

Re =

1E4

1E5

1E6

1E7

1E8

1E9

δ/R =

0.2074

0.1493

0.1074

0.00773

0.00556

0.00401

Cf =

0.00713

0.00453

0.00311

0.00225

0.00170

0.00133

The italic values are probably not really turbulent flow. A nice curve-fit to the remaining values is a one-seventh power law for Re  1E5 : Rotating cylinder turbulent flow: Cf  0.023 1/7 Re

(Ans.)

The writer knows of no other “theory” for this flow - even as crude as the one above. However, friction on rotating cylinders has been measured by Theodorsen and Regier (1944), who recommend the following correlation for the skin friction coefficient: the text for an example), they recommend the following correlation for the skin friction coefficient of a rotating cylinder in turbulent flow:

(

)

1  4.07 log10 Re Cf − 0.6 Cf

Theodorsen & Regier (1944 ) 

This experimental curve-fit differs by about ±10% from the power-law analysis found above. The two may be plotted versus Reynolds number as follows: Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -24-

We see that the agreement is quite satisfactory, considering the simplifications made.

6-24 Consider the flat-plate flow of Prob. 6-14. Use any integral method from Section 6-8 to compute τ w (x), δ*(x), and H(x) along the plate surface. Compare your results with traditional algebraic flat-plate formulas. Begin at x = 4 cm with Cf = 0.005. We choose Head’s entrainment method, Eqs. (6-123) and (6-124), d −0.6169 ( UeθH1 ) = Ue F ( H1 ) , where F = 0.0306+ ( H 1 − 3.0 ) dx

H1  3.3 + 0.8234 ( H − 1.1)

−1.287

for H 1  1.6

(6-123) (6-124)

to be combined with the Kármán integral relation, Cf = dθ/dx when U e = constant. We have U e = 6 m/s, with v = 0.001/ 998 m 2 /s. Begin at x = 0.04 m with Cf = 0.005. Take H(0)  1.3 (flat-plate turbulent flow at low Reynolds number). Estimate θ(0) from power-law theory, Eqs. 1/ 7 (6-69,70): Rex = (998)(6)(0.04) /(0.001) = 239500, whence δ/x  0.16/Re x = 0.0273 and θ/x  (7 / 72)(d/x), or  (0)  0.00011 m. With H(0) = 1.3, Eq. (6-124) above predicts H1 (0) = 9.83. We are then ready to begin the integration of the Head-method differential equations. The computed results Cf (x) and θ( x) can be compared with the power-law flatplate formulas: Power-laws: Cf 

0.027 7 Re11/ x

θ 0.0156  x Re1/x 7

(1, 2)

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A comparison may be given in the following graph:

The agreement is good enough for an integral method - I believe the Head-method skin friction is too high because we started at too low a Reynolds number (239500), where the flow is barely turbulent and the Head correlations are not well documented. The computed shape factors in Head’s method may be tabulated as follows:

x ( m) =

0.08

0.20

0.30

0.40

0.50

0.60

0.80

1.00

H



H1



1.406 7.65

1.410 7.33

1.400 7.49

1.393 7.64

1.386 7.76

1.381 7.86

1.372 8.04

1.365 8.20

We don’t have an algebraic formula for H(x) for a flat plate: power-law theory assumes a constant value of about H = 1.3. We could, if necessary, estimate H(x) using Cf (x) and Coles wall-wake profile correlation, Eqs. (6-48), taking  = 0.5. We won’t do that now. The Kármán-based theory (p. 451 of the text) does this and gives similar good agreement.

6-25 Use any integral method from Section 6-8 to analyze the Howarth freestream velocity distribution, U ( x ) = U0 (1 − x/L ) , to compute Cf ( x ) and the separation point for ReL = U0 L/v = (a) 1 E6; (b) 1 E7 ; and (c) 1 E8. (Recall that xsep /L ( laminar ) = 0.120. ] Let us again choose Head’s entrainment method. Eqs. (6-123) and (6-124), for our computation. We need a starting point, which we take to be the value of x where Cf = 0.005 as computed from flat-plate theory (the initial stages of the Howarth distribution being at low,

nearly zero, β). At this initial x0 we take H0  1.4 and compute θ 0 by flat-plate theory, say, Eqs. (6-69,70). Then we program the Kármán integral relation, Eqs. (6-28), and Head’s two correlations (6-123,124). The results for local skin friction Cf ( x ) may be plotted as follows:

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We see that turbulent flow is much more resistant to separation, and the separation point is about four times further downstream than for laminar flow and increases slightly with Reynolds number. Other methods in the text give similar results. We compare here the Head separationpoint results and the Kármán-based theory on p. 451 of the text:

Re L =

( x/L)Head  ( x/L)Karman 

1 E6

1 E7

1 E8

0.44

0.49

0.53

0.40

0.45

0.50

They are not the same. Turbulent boundary layer prediction methods give varying results. 6-26 For potential flow past a cylinder, U ( x ) = 2U0 sin ( x/a ) , use any convenient turbulent boundary method to compute the separation point for U0a / v = (a) I E6; (b) I E7; and (c) I E8. Compare with the laminar separation point of  = 105 (Fig. 4-24b). We choose the Kármán-based method, p. 456 of the text, starting at a small value of x such that

Cf0  0.004 to 0.006 depending upon Reynolds number. Take H0  1.4 and θ0 given by flat-

plate theory, Eqs. (6-69) and (6-70). Then program this method and plot the local skin friction, as follows:

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Head’s entrainment method (Sect. 6-8.1.2) was not used because, to this writer, it is difficult to get started when there are no known initial conditions - the method is sensitive to H0 ,H10 ,

and θ 0 . It works fine when initial data is given - see Prob. 6-27 below. The plot above indicates that, depending upon Reynolds number, the separation point lies between 130 and 140, or about three to four times further down the back side of the cylinder than for laminar flow. This is typical of the increased resistance to separation of a turbulent boundary layer.

6-27 Use any convenient turbulent boundary layer method from Section 6-8 to compute skin friction for the Clauser II experiment, Flow 2300 of Coles and Hirst (1968), using kinematic viscosity of 0.000165 ft2/s. The measured freestream velocities are

x ( ft ) =

7.5

9.0

11.0

12.67

16.17

19.17

23.92

26.67

U ( ft/s ) =

26.1

24.8

23.5

22.8

21.3

20.2

18.9

18.1

These velocities, as expected of “near-equilibrium” turbulent boundary layers, fit a power-law distribution with an accuracy of 1% :

U ( x )  47.3x −0.29 ,

x in feet and U in ft/s

The sparse initial data, only Cf ( 0) and U ( x ) , suggests the method of Das (1988), page 458 of the text, which keys on Cf ( x ) without thickness or shape factor information. However, for those who would like to use, say, the Head entrainment (Sect. 6-8.1.2) or Kármán type (Sect. 6-8.1.1) method, we add here the initial values of momentum thickness and shape factor

at x = 7.5 ft, as taken from Coles and Hirst (1968), p. 213: θ0 = 0.609 in, H 0 = 1.788. We have programmed all three methods and plot the theoretical and actual skin friction distributions as follows:

We see that Das’s method is in fairly good agreement, while the Head and Kármán methods jump away from the initial point to new curves more or less parallel to the data. Almost all the Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -28-

thickness-based methods used at the 1968 Stanford Conference had the same trouble - see p. 545 of Kline et al. (1968). The trouble seems to be that Clauser’s initial data are incompatible with the commonly used correlations between friction, momentum thickness, and shape factor. For example, if Cfo = 0.00127 and H0 = 1.788, the Ludwieg-Tillmann formula (6-117) predicts that Reθo = 13340, or 66% higher than Clauser’s measured value of 8032 (Coles and Hirst 1968, p. 214). Das’s method does not use any input but Cfo and the wall-wake law.

6-28

Use any integral method from Section 6-8 to compute Cf ( x ) for the experiment of

Moses, Case 5 [Flow 4000 of Coles and Hirst (1968)]. Take v = 0.000166 ft 2 /s and

Cfo = 0.00471 at x = 0 ft.

The measured freestream velocities are as follows:

x ( ft ) =

0.0 0.162 0.323 0.484 0.646 0.865 1.058 1.303 1.516 1.734 1.979 2.198

U ( ft/s ) = 82.0 76.26 69.09 61.36 55.49 51.14 49.54 48.93 48.65 48.79 48.86 48.51 We thus have a “relaxing” flow, with an initial strong adverse gradient which levels off to nearconstant freestream velocity. The Das method is available, or you may use a Kármán-type or Head entrainment method, in which case we need initial values of momentum thickness and shape factor. From Coles and Hirst (1968), page 385, Moses Flow 4000, θo = 0.0137 in and H o =1.62. We integrate these methods for an assumed 5th-order polynomial curve fit to U(x):

U ( ft/s )  82.17 − 47.409x − 2.6711x 2 + 29.209x 3 −14.105x 4 + 2.0027x5 , ( x in ft ) This fit has a correlation of 0.995 when compared to the velocity data above. The results of these three theories are shown below and compared with Moses’ skin friction data:

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We see that agreement with experiment is not that great. Head’s method drops down to separation, Cf  0, at x = 1 ft, which was a difficulty of many of the methods tested at the 1968 Stanford Conference. The Kármán-type and Das methods are better but not outstanding. This experiment was hard to predict because of the ‘relaxing’ freestream.

6-29 Use the similarity concepts of Section 6-9 to derive power-law expressions for the variation of total mass flux with x for (a) a plane jet; (b) a round jet: and (c) a plane mixing, layer. Compare with laminar flow results. In general, the total mass flow equals

 ρu dA

over the cross-section of the flow. In all three

cases ( a,b,c ) above, u = Umax fcn ( η) , where η = σy/x. Our three results are: +b

(a) Plane jet: m =

 ρu dy = Const U

max

b

b

C1 C2 x = Cx1/2 x1/2

(Ans. a)

The related result for the laminar plane jet, Eq. 4-107, is mass flow proportional to x1/3. +b

(b) Round jet: m =  ρu 2πr dr = Const Umax b2  b

C1 C2 x 2 = Cx x

(Ans. b)

The same result holds for the laminar round jet, Eq. 4-208, mass flow proportional to x. +b

(c) Plane mixing layer: m =

 ρu dy = C ( U

2

− U1 ) b = C1C2 x = Cx

(Ans. c)

−b

The result for the laminar mixing layer, Eq. 4-92, is mass flow proportional to x1/2 .

6-30 At a certain section of a developed turbulent plane water jet, the maximum velocity is 3 m/s and the mass flow is 800 kg/s per meter or width. At 2 m further downstream, estimate (a) the jet width; (b) maximum velocity; and (c) mass flow. Assuming developed flow, we use Eq. (6-151) to evaluate the mass flow: +

m=

+

 xη  2ρU x 2 − ρu dy = ρ− U maxsech ( η) d  σ  = σmax

For the plane jet, σ = 7.67, so we may evaluate the local position for the given mass flow: x=

(800 kg/s/m )( 7.67 )  1.025 m mσ = 2ρU max 2 ( 998 kg/m3 ) ( 3 m/s )

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Then at 2 m further downstream, or x = 3.025 m, we may use these formulas from Section 6-9.1.1 to evaluate the new flow conditions, as follows: (a) Width b = x tan (13) = ( 3.025)( 0.23) = 0.694 m

(Ans. a)

1/2

(b) Maximum velocity Umax (c) Mass flow =

6-31

 1.025  = U x −1.025    3.025 

 1.75

m s

(Ans. b)

2ρU max x 2 ( 998 )(1.75 )( 3.025 ) kg = = 1370 σ 7.67 s-m

(Ans. c)

Air at 20C and 1 atm issues at 0.001 kg/s from a 4-mm diameter orifice into still air.

At 1 m downstream, estimate (a) maximum velocity; (b) jet width; and (c)

μ t /μ.

For air take ρ  1.2 kg/m . Estimate the air velocity issuing from the orifice: 3

*

m 0.001 m UO = =  66.3 2 ρA (1.2 )( π/4 )( 0.004 ) s

Then an estimate of the jet momentum issuing from the orifice is J=



ρu 2dA  ρUo2 A0 = (1.2 )( 66.3)

orifice

2

π kg-m 2 ( 0.004 )  0.0663 2 4 s

We then use this value of J to estimate, at x = 1 m, jet properties, using Sect. 6-9.1.2:

7.4 ( J/ρ ) = x

1/2

At x = 1 m: U max

b1% 

( 7.4 )( 0.0663/1.2 ) = (1.0 )

1/2

= 1.74

xη1% (1.0 )( 2.993)   0.20 m 15.2 15.2

m s

(Ans. a)

(Ans. b)

μ t KρU max b ( 0.018)(1.2 )(1.74 )( 0.20 ) = = = 410 μ μ 0.000018 (Ans. c) The mass flow at this station is approximately 4πρU max x 2 / (15.2 ) = 0.114 kg/s, or about one 2

hundred times greater than the orifice outlet mass flow.

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6-32 A long 5-m-diameter vertical cylinder is placed in the ocean where the current is 60 cm/s across the cylinder. At 1 km downstream of the cylinder, estimate (a) the wake width; and (b) the wake velocity defect. From Sect. 6-9.1 of the text, we have the following far-wake estimates for a cylinder: y1/2 = 0.275 ( xθ )

u max = 1.75 ( θ/x )

1/2

1/2

where θ is the wake momentum thickness, assumed constant. The drag force on the cylinder is related to θ as follows: F = ρU 2θ = CD

1 ρU 2 D, 2

or: θ =

CD D 2

Now estimate the diameter Reynolds number of the cylinder in 20C seawater:

ReD = ρUD/μ = (1025)( 0.6)( 5.0) / ( 0.0011)  2.8 E6. From Fig. 3-38a, estimate CD = 0.5.

Then a crude estimate for cylinder wake momentum thickness is θ = ( 0.5)( 5) / 2  1.25 m. Our wake thickness estimate at x = 1000 m is thus y1/2 = 0.275 (1000 )(1.25) 

1/2

This corresponds to a total wake thickness u max = 1.75 U ( θ/x )

1/2

= 10 m

(Ans. a)

b1%  2.6y1/2  26 m. Similarly,

= 1.75 ( 0.6 ) 1.25 /1000

1/2

 0.037

m s

(Ans. b)

These are clearly very crude estimates, based on uncertain drag and wake correlations.

6-33

Evaluate the temperature law-of-the-wall, Eq. (6-164), numerically, using the van

Driest eddy viscosity (6-90) and assuming

Prt = 1.0. Compare with Eqs. (6-165,166).

We are to integrate Eq. (6-164) with Prt = 1.0: y+

μ dy + T  , with t = 1/Pr + μ t /μ μ 1+ 0 +

ζ

, ζ = 2κ 2 y+ 2 1 − e − y  (1 + 2ζ )

+

/26

 

2

This quadrature is easily performed with Subroutine RUNGE of Appendix C. As we reach, say, y+ = 1000, the values of T+ become logarithmic in y+ and have the form of Eq. (6-165). We may list some computed values for various molecular Prandtl numbers:

Pr = A ( Pr ) =

0.7 2.38

1.0 5.28

2.0 12.97

3.0 19.28

6.0 34.87

10. 52.19

30. 119.6

100. 291.6

300. 657.6

1000. 1608.

These agree reasonably well (±10% in the range 1.0  Pr  100 ) with the accepted correlations 2/3 for A ( Pr ) , Eqs. (6-166) and (6-167). They fit the formula A ( Pr )  13Pr − 8.0 to within Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -32-

±10% except for the values at Pr = 300 and 1000. These theoretical values depend upon the particular eddy-viscosity correlation chosen. 6-34 Air at 20C and 1 atm flows at 60 m/s past a smooth flat plate 1 m long and 60 cm wide. If the plate surface temperature is 50C, estimate the total heat loss in watts from one side of the plate. For air at a mean temperature of 35C, take

ρ = 1.15 kg/m3 ,μ = 1.88 E-5 kg/m-s,

Pr = 0.71,cp = 1005 J/kg-K, and k = 0.0265 W/m-K. Then the plate Reynolds number is ReL =

ρUL (1.15)( 60 )(1.0 ) = = 3.67 E6 μ 1.88E-5

( turbulent flow )

Thus Sect. 6-10.3 applies for turbulent flow. The skin friction coefficient at x = L is

Cf =

0.455 = 0.00301 ln ( 0.06 )( 3.67E6 )  2

Then the trailing edge Stanton number is given by Eq. (6-168): qw 0.001505 = 0.00168 = 2/3 1/2 ρUcp ΔT 1+13 ( 0.71) − 1 ( 0.001505)   or: q w ( x = L )  0.00168 (1.15 )( 60 )(1005 )( 50 − 20 ) = 3486 W/m 2

Ch =

The mean heat transfer over the whole plate should be about 15% higher, or q mean = 4010 W/m 2 . Then the heat loss from one side of the plate should be approximately dQ W  = q mean Aplate =  4010 2  (1.0 m )( 0.6 m )  2400 W dt m  

(Ans.)

This is a relatively modest rate (compared to water flow in the next problem) because air has modest specific heat, low density, and low thermal conductivity.

6-35

Repeat Prob. 6-34 above for water flow at U = 6 m/s.

For water at a mean temperature of 35°C, use Appendix A to find ρ = 993 kg/m3 , μ = 7.19E-4 kg/m-s, k = 0.624 W/m-K, c p = 4177 J/kg-K, and Pr = 4.8. The plate Reynolds number is

Re L =

ρUL ( 993)( 6.0 )(1.0 ) =  8.29 E6 μ 0.000719

( turbulent flow )

Thus Sect. 6-10.3 applies for turbulent flow. The skin friction coefficient at x = L is Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -33-

Cf =

0.455 = 0.000264 ln ( 0.06 )( 8.29E6 )  2

Then the trailing edge Stanton number is given by Eq. (6-168): qw 0.00132 = 0.000706 = 2/3 1/2 ρUcp ΔT 1+13 ( 4.8) − 1 ( 0.00132 )   or: q w ( x = L ) = 0.000706 ( 993)( 6.0 )( 4177 )( 50-20 ) = 527, 000 W/m2

Ch ( x = L ) =

The mean heat transfer over the whole plate should be about 15% higher, or q mean = 6.06E6 W/m2. Then the heat loss from one side of the plate should be approximately dQ W  = q mean Aplate =  606,000 2  (1.0 m )( 0.6 m )  364, 000 W dt m  

(Ans.)

This is 150 times higher than the (faster) air flow in Prob. 6-34. The reason is that water has very high specific heat and thermal conductivity, plus relatively low viscosity, making it one of the best heat-convecting fluids known. 6-36 Modify Prob. 6-34 (airflow at U = 60 m/s ) is the plate is permeable, with a uniform blowing velocity of 2 cm/s. From Prob. 6-34 we know that ReL = 3.67E6 and Pr = 0.71. The appropriate blowing correlation is Eq. (6-172):

Ch ( blowing ) vw / U ζ  ζ , where ζ = Ch ( no blowing ) e − 1 Ch ( no blowing )

(6-172)

We have already computed Ch ( no blowing ) = 0.00168 (see p. 200 of this Manual). Then the blowing parameter is approximately ζ = ( 0.02 / 60.0 ) / ( 0.00168) = 0.20. Equation (6-168) then predicts

Ch Ch ( blowing ) 0.20 =  0.20  0.904, or: Cho 0.00168 e −1 or:

Ch  0.00152 =

qw , ρUcp ΔT

q w ( x = L ) = ( 0.00152 )(1.15)( 60 )(1005)( 50 − 20 ) = 3160 W/m2

In this case, the blowing effect is slight (−10%) - corresponding to “F” = v w /U  0.000333 in Figure 6-45 of the text. The mean heat transfer on the plate should be about 15% higher, or q mean = 3620 W/m 2 . The heat loss from one side of the plate should be dQ  cm W   blowing  =  3630 2  (1.0 m )( 0.6 m )  2180 W  with 2 dt  s m   

(Ans.)

6-37 Water at 20C and 1 atm flows at 4 kg/s into a smooth tube of diameter 3 cm and length 2 m. If the wall temperature is 40C, estimate the average outlet water temperature. Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -34-

For water at a mean temperature of 30C, take ρ = 995 kg/m3 ,μ = 7.79E-5 kg/m-s, k = 0.617 W/m-K, cp = 4177 J/kg-K, and Pr = 5.4. The tube Reynolds number is

ReD =

4 ( 4.0 ) 4m = = 213,000 πμD π ( 7.97E-4 )( 0.03)

( turbulent flow )

From Eq. (6-65), the turbulent flow friction factor is

1  2.0log10  2130001/2  − 0.8, or:   0.0154 1/2  Then Eq. (6-169) is used for the (smooth) pipe Stanton number, with  /8  0.00193:

h av 0.00193 = 0.000883 = , 2/3 1/2 ρVav cp 1+13 ( 5.4 ) − 1 ( 0.00193)   W m m or: h av = ( 0.000893)( 995 )( 5.69 )( 4177 ) = 20900 2 since Vav = = 5.69 m -K ρA s

Ch =

The mean heat transfer must be compatible with an energy balance. Let T0 be the outlet water temperature. Then the total heat transferred is

( T − 40 ) − ( 20 − 40 ) dQ = mcp ( T0 − 20 ) = h av πDLΔTmean , where ΔTmean = 0 dt  T − 40  ln  0  20 − 40 

(1)

That is, the concept of “logarithmic-mean temperature difference” holds for turbulent pipe flow just as it did for laminar flow, page 123 of the text. Rather than solve Eq. (1) iteratively, we rearrange it into the following explicit form:  h πDL   ( 20900 ) π ( 0.03)( 2.0 )  T0 − 40 = exp  − av  = exp  −  = 0.790, 20 − 40 mc p  ( 4.0 )( 4177 )   

or:

Toutlet = 40 + ( 20 − 40 )( 0.790 ) = 24.2°C

(Ans.)

This corresponds to Tmean = 17.82C, or a total heat transfer dQ/dt  70100 W.

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6-38 Reconsider the liquid film flowing down an inclined plane from Prob. 3-15. This time let the flow be turbulent and of constant depth h. Using the log-law for the velocity, (a) find an expression for the bottom wall shear stress as related to  , g, , , h, and the surface velocity Us, (b) Find a formula for the flow rate Q per unit width and compare to laminar flow, Q  h3 . Solution: Let the width into the paper be b. Neglect air friction at the top surface. For steady flow, the bottom shear force balances the x-directed weight:

Fig. P6-38

 w Lb =  gLbh sin  , or:

 w =  gh sin 

A turbulent boundary layer of thickness  = h forms. From Eq. (6-69),

2 w 0.020  , or:  w  0.01 5/6 1/6U 11/6 h−1/6 1/6 2 U ( Uh /  )

Ans. (a)

6 Or we could solve for U  [200 gh7/6 sin   1/6  −1/6 ]6/11. The flow rate Q would be Q  Uhb. 7 7/11 18/11 Since U is proportional to h , Q is proportional to h Ans. (b)

6-39 Use the method of Ambrok, Eq. (6-178), to estimate the overall heat transfer on an isothermal circular cylinder at ReD = 1 E6 and Pr = 1.0. Compare your result with the experimental value Nu D = 1400. Discuss the discrepancy, if any.

For Pr = 1.0, ro =  ( 2-D flow ) , T = constant, and ρ = constant, Eq. (6-178) reduces to

Ch =

h (x) 0.0295v0.2 = 0.2 ρU ( x ) cp  x    U ( x ) dx  0 

For freestream velocity near the cylinder surface, we assume the inviscid relation, U ( x ) = 2Uosin ( x/R ) , where U o is the approach velocity and R the cylinder radius. [We know this formula does not work on the back side of the cylinder, but we have little better.]

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Ambrok’s formula above then becomes Ch ( x ) =

0.0295v0.2 x    2U o sin ( x/R ) dx  0 

0.2

=

0.0295 ( U 0 D/v ) 1 − cos ( x/R ) 

0.2

0.2

Noting that C h is non-dimensionalized by U ( x ) , not U0 , we may, finally, rewrite this as

h (x)D sin ( x/R )  0.059 Re0.8 D 0.2 k 1 − cos ( x/R ) 

(Ans.)

This result has at least two flaws: (1) near the leading edge, x/R  50, the flow should be laminar; and (2) on the back side, x/R  120, the flow is separated and boundary layer theory should not apply. Nevertheless, we may plot the local heat transfer as follows:

We have added some pseudo “data points”, based on experiments performed at somewhat lower Reynolds numbers  2E5 [see, e.g., Giedt (1949)]. The actual flow would be laminar near the front and separated near the back. The experimental Nusselt number Nu D is about 1400, while the integrated average of Eq. (6-178) in the graph above is about 2560, or 80% larger. Unfortunately, the writer knows of no better theory than this for overall heat transfer in flow past a blunt body.

6-40 Investigate modifying the finite-difference methods of Section 4-7 for a twodimensional turbulent boundary layer, using an eddy viscosity model. Comment. The turbulent continuity equation is the same as for laminar flow, so the model is the same, Eq. (4-148). If ( m, n ) denote the ( x, y ) directions, respectively, we have

vm+1,n  vm+1,n-1 −

Δy ( u m+1,n − u m,n + u m+1,n−1 − u m,n−1 ) 2Δx

(4-148)

where ( u, v ) denote time-mean velocities, of course. This formula presumes knowledge of the u’s at the new station ( m + 1) . Meanwhile, the x-momentum equation has the same type of Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -37-

acceleration and pressure-gradient terms as for laminar flow, but the shear terms now contains strongly variable nonlinear coefficients. That is,

u

u u dU   u  +v =U + ( v + v t )  x y dx y  y 

(6-21b)

The last term requires us to account for the y-variation of eddy viscosity in the model:

  u  1  ( v + v t )n+1/2 ( u m,n+1 − u m,n ) ( v + v t )n −1/2 ( u m,n − u m,n −1 )  v + v = −   ( ) t y  y  y  y y  That is, we either model the derivative of vt or we evaluate vt at ( n + 1/2) and ( n −1/2 ) , respectively, as shown above. If we use an explicit model, that is, a modification of Eq. (4-146), the step size x will be so limited that the computer-intensive turbulent-flow calculation will be too lengthy. Therefore we adopt an implicit model modified from Eq. (4-149), as follows:

−α n +1/2 u m+1,n+1 + (1 − α n +1/2 − α n −1/2 ) u m+1,n − α n −1/2 u m+1,n −1 = u m,n + β ( u m,n+1 − u m,n −1 ) + where α n +1/2 =

( v + v t )n +1/2 y2 u m,n x

and α n −1/2 =

( v + v t )n −1/2 y2 u m,n x

and β =

U 2m +1 − U m2 2u m,n

u m,n x 2u m,n y

Only the coefficients on the left change. We must evaluate the eddy viscosity at points midway between n and n + 1, or “ n + 1/2 ”, and midway between n and n −1, or “ n − 1/2 ”. The model can then be solved for the three unknown u’s with the TDMA, Section 4-7.2.1. We may step along with this model with x limited only by numerical accuracy, that is, changes in u and v should be no greater than about 5 per cent per step. The transverse step size y is limited by the need for accuracy near the wall, where gradients are very large. Since the mesh in this model is rectangular, we will need many y ’s to model the entire boundary layer, yet keep the viscous sublayer accurate - 1000 y ’s are not unusual.

6-41 Apply your ‘turbulent finite difference’ technique of Prob. 6-40 to the computation of Cf ( x ) for 0  x < L in fully turbulent flat-plate flow with ReL = 1 E6. The model used is implicit and is described in Prob. 6-40 above. For flat-plate flow, we take U = constant, or dU/dx = 0. It remains only to specify the step sizes and the eddy viscosity model. Although the van Driest ‘damping’ model of Eq. (6-90) is probably the most widely used in commercial boundary layer codes, it makes the local shear term proportional to the square of local gradient u/y and is thus highly nonlinear. We can use that model, and perhaps the students should be encouraged to do so. An alternate near-wall formula, which does not use a mixing-length model, was proposed by G. L. Mellor at a symposium in 1968 (see p. 477 of the 1st edition of the text):

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( κy ) μt  for the inner and overlap layers μ ( κy+ )3 + 328.5 + 4

Mellor (1968):

This model is easy to apply. We can then switch over in the outer or ‘wake’ layer to Clauser’s model, Eqn. (6-93): Clauser (1956):

μ t  0.016ρUδ*

in the outer layer

The cross-over point is where the two formulas are equal to each other, as in Fig. 6-23 of the text. [No ‘intermittency’ correction was made to Clauser’s formula.] The formula requires an estimate of local displacement thickness to complete the calculation. Using the continuity and momentum models of Prob. 6-40, this Mellor/Clauser model was applied to flat-plate flow with the following parameters (arbitrary units):

U = 1 unit; L=1 unit; v =1E-6 units; x = 0.01 unit; y = 0.00003 unit The choice of x was strictly for accuracy (100 steps along the plate), since the implicit (TDMA) method is unconditionally stable. The choice of y was made for two reasons: (1) to ensure that at least three mesh points would be in the viscous sublayer, and (2) to ensure that the thickest boundary layer (at x = L ) could be encompassed with about 600 total mesh points in the y direction. The computations - over 600 y points and 100 x stations - took about 30 minutes on a small personal computer, with a program written in the BASIC language. The computed velocity profiles at three stations - x/L = 0.1,0.4, and 1.0 - are plotted in law-of-the-wall coordinates as follows:

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We see that the profile shapes are excellent and closely resemble typical flat-plate flow profiles, as in Fig. 6-9 of the text. The log-layer is very well represented for κ = 0.41. The only complication in the computation was the need to know δ* in order to use Causer’s outer-layer eddy viscosity. Since x was small, we simply evaluated δ* by integrating the previous profile (at x − x ) and used it instead of iterating the local profile to match eddy viscosity with actual displacement thickness. The error is very slight. The following computed parameters can be listed at the three stations plotted above:

x/L = Re x =

0.1 1 E5

0.4 4 E5

1.0 1 E6

δ99%/ x =

0.0255

0.0186

0.0155

Cf =

0.00505

0.00372

0.00326

=

0.55

0.75

0.75

δ/x ( Eq. 6-70 ) :

0.0309

0.0253

0.0222

Cf ( Eq. 6-70) :

0.00521

0.00428

0.00375

Cf ( Eq. 6-78) :

0.00601

0.00447

0.00376

The values of Coles’ parameter  are slightly larger than the accepted value of 0.45 for turbulent flat-plate flow. [This is an artifact of the Mellor eddy-viscosity formula.] The computed boundary layer thickness is about 25% thinner than predicted by the simple power-law formula of Eq. (6-70). [The power-law extrapolates along the log-layer and overestimates the thickness.] Similarly, the computed skin friction is lower than the traditional formulas, Eqs. (6-70) and (6-78) - here the problem may simply be an inaccuracy of the Mellor/Clauser model for such low ‘turbulent’ Reynolds numbers.

6-42 Use the log-law, Eq. (6-38a), to analyze turbulent flow near a rotating disk of radius R and angular velocity ω . Let the crossflow velocity w be given by a parabolic hodograph assumption: w  u tan β (1 − y/δ ) , β  12 2

Compute local skin friction C f = 2τ w / ( ρω2 r 2 ) and moment coefficient C M = 4M/ρω2 R 5 versus

Re = ωR 2 /v and compare with Kármán’s formulas, CM = 0.146/Re1/5 and Cf = 0.053/Re1/5 (see Fig. (3-30)]. First let us estimate the skin friction by using the flat-plate formula, Eq. (6-68), which is based upon the local log-law:

Cf =

0.020 , Re1/5 δ

where Reδ =

( ωr ) δ = Re v

R

r δ RR

for this case

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Now we need to estimate local boundary layer thickness by, say, modifying Eq. (6-70):

δ 0.16  r ( ωr 2 /v )1/7 Putting these two together gives an estimate for local skin friction in terms of local r: Cf =

0.020    Re η  0.16  R  ( Re η2 )1/7 R  

1/6

   

0.0271



( Re η )

2 1/7

,

where, η =

R

r R

We will use this in computing the moment coefficient. At the point r = R, we have: Cf ( r = R ) 

0.0271 Re1/7 R

(Ans. a)

We may compare this with two accepted correlations for this rotating disk problem: Kármán (1921): Cf 

0.053 ; Theodorsen (1944 ) : Re1/5 R

(

1 = −2.05 + 4.07 log10 Re Cf Cf

)

Some computed results are given as follows:

Re : Cf : Our formula: Kármán Theodorsen

1 E5

1 E6

1 E7

1 E8

1 E9

0.00523 0.00530 0.00535

0.00377 0.00334 0.00333

0.00271 0.00211 0.00226

0.00195 0.00133 0.00162

0.00140 0.00084 0.00122

The present formula is in fairly good agreement - Theodorsen’s being the most accurate correlation, since it is fitted to actual rotating-disk data. To compute the moment on the disk, we integrate the torque caused by shear stress on each circular strip of the disk: R

M =  τ w r ( 2πr dr )

-see Eq. ( 3-190 ) of the text

0

Now introduce τ w  C f (1/2 ) ρ ( ωr )

2

from our power-law formula above and rewrite in

dimensionless form: 1

0.0271 4 0.072 η dη = 1/7 1/7 2/7 ReR η Re 0

CM = 4π 

(Ans. b)

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We may compare this result with von Kármán’s (1921) formula, CM  0.146 / Re1/5 R ;

Re : CM : Our formula: Kármán

1 E5

1 E6

0.0139 0.0146

0.0100 0.00921

1 E7

1 E8

1 E9

0.00720 0.00581

0.00518 0.00367

0.00373 0.00231

The two agree for low Reynolds numbers but then our formula falls higher as Re increases. Experimental data for rotating disks [Theodorsen and Regier (1944)] lie between these two estimates. Kármán’s formula came, not from wall friction estimates, but from the angular momentum theorem applied to the disk velocity profiles. If y is normal to the disk, δ

M=

 ( r  V ) ρVn dA = ( ru ) ρw 2πr dy CS

0

where the radial or ‘crossflow’ velocity w(y) was related to u(y) by the hodograph formula at 2 the beginning of this problem. [Actually Kármán used just (1 − y/δ ) rather than (1 − y/δ ) . ] Kármán assumed a one-seventh power-law for the profile u(y): 1/7 u  ( rω ) 1 − ( y/δ )  ,  

plus

w  u tanβ (1 − y/δ )

2

After carrying out the integration and using a one-fifth power-law estimate for boundary layer thickness δ , the result is Kármán’s formula, CM  0.146 / Re1/5 R . Goldstein (Proc. Cambridge Philo. Soc., v. 31, pt. 2, 1935, pp. 232–241) improved the analysis somewhat by assuming that u(y) followed the logarithmic law, Eq. (6-38a), rather than a one-seventh power-law.

6-43 Consider a two-dimensional flat-walled diffuser as in Fig. P6-43. Assume incompressible flow with a U ( x ) and one-dimensional freestream velocity entrance velocity U 0 at x = 0 . The entrance height is W and the constant depth into the paper is b = 4W . Assume turbulent flow at x = 0, with momentum thickness  / W = 0.02, H ( 0) = 1.3, and U0W / v = 105. Using the method of Head. Sect. 6-8.1.2, numerically estimate the angle  for which separation will occur at x = 1.5 W . Solution: By one-dimensional continuity, the average velocity U ( x ) is given by

U 0bW = Ub (W + 2 x tan  ) , or:

U=

U0 1 + ( 2 x tan  ) / W

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We solve Head’s method for a given  and the initial conditions until separation occurs. We try a range of values of  , and the results, for U 0W /v = 105 , are:



( x /W )separation

10 1.56

8 1.99

6 2.73

12

1.28

So, according to Head’s theory, separation occurs at   10.4. Ans. The plot is below:

6-44 Solve Prob. 6-43 instead by the Kármán-based method of Sect. 6-8.1.1. If time permits, investigate three different Reynolds numbers, U 0W / v = 105 ,106 , and 107 . Assume turbulent flow at x = 0, with momentum thickness  / W = 0.02, H ( 0) = 1.3, and U 0W / v = 105. Do you expect the separation angle  to increase as Reynolds number increases? Solution: Yes, I would make a guess that separation angle increases as Reynolds number increases. The Kármán-oriented method is a bit algebraically heavy, since the momentum integral (6-28) and skin friction correlation (6-120) are supplemented by a shape factor relation (6-119a), a - correlation (6-121), and the definition of  (6-122). In fact, the method has difficulty matching the given initial conditions, because the adverse gradient dU /dx is very strong for any significant diffuser angle  , so  is large and H cannot be as low as 1.3. Anyway, the Kármán-based method predicts separation near 4, as in this table for the given Reynolds number U 0W / v = 105.



( x / W )separation

3 3.4

4

1.6

5 0.8

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A plot of the (rather low) skin friction coefficients for  = 4 is as follows:

The effect of Reynolds number slight, but, yes, an increase: For ReW = 106 , sep = 4.2. For

ReW = 106 , sep = 4.4. The Kármán-based method is poor for this diffuser.

6-45 Solve Prob. 6-43 again by Head’s method if the diffuser is an expanding cone with an inlet pipe of diameter W. This time there is no “depth into the paper b”. For a given Reynolds number, do you expect a cone to have a smaller separation angle  than a flat-walled diffuser? Solution: This time use axisymmetric one-dimensional continuity to estimate U ( x ) :

U0

 4

W2 =U

 4

(W + 2 x tan  )

2

or: U 

,

U0 1 + ( 2 x tan  ) / W 

2

Since U drops twice as fast as in the flat-walled diffuser, yes, indeed, I do expect a cone to have a smaller separation angle  for a given distance, in this case, x = 1.5L. Although the initial boundary layer is rather thick, here assume that Head’s two-dimensional theory is satisfactory, so that we don’t have to put together an axisymmetric thick-layer theory. Running an array of diffuser angles gives the following separation lengths:



( x / W )separation

3 2.39

4

1.76

5 1.39

6 1.15

The desired length of L = 1.5W lies between four and five degrees. We interpolate.

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So, with Head’s theory, cone separation occurs at   4.6. Ans. The plot is below:

6-46 Sink flow, Fig. 4-16, if begun far out with a turbulent boundary layer, may relaminarize as it approaches the origin. Show that the acceleration parameter K crit relates to sink flow and also relates to the Reynolds number of the flow. According to K crit data, what Reynolds number should cause sink flow to relaminarize? Solution: For a two-dimensional sink, as shown in the figure, the stream velocity takes the form U = −B / x , where B is a constant (see Fig. 4-16 of the text). From Eq. (6-142) the dimensionless acceleration parameter is defined as

K=

v dU with K crit  3 10−6 for reverting to U 2 dx

Fig. P6-46

laminar flow. Evaluate K for sink flow: K sink =

v

( B / x)

2

d  B  vx 2 B v = = constant − = dx  x  B 2 x 2 B

Thus, in sink flow, the acceleration parameter is constant. Now evaluate the local Reynolds number and relate it to K:

Re x =

Ux B x B 1 = = = v xv v K

Ans.

Interestingly, Re x is the inverse of K. The critical Reynolds number for relaminarization is Re x ,crit =

1 1 =  330, 000 Kcrit 3E − 6

Ans.

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6-47 As an improvement to Stevenson’s inner velocity-law with suction and blowing, Eq. (6-86), Oljaca and Sucec (1997) offer the following inner/overlap/outer velocity profile formula based upon wall transpiration data compiled by Donald Coles in 1971:

u+   +

2



f+

vw+  2 4  4 2 2   +  f + f   4  2 

where  =

1



ln ( y + ) + A

The function f  3 ( y /  ) − 2 ( y /  ) is the wake shape as in Eq. (6-46). The constant A  5 2

3

for moderate blowing or suction. Show how this profile differs from the Fig. 6-22 curves for vw+ = 0.02 and also compare it with Fig. 6-44a for vw+ = 0.0038. Solution: We are to compare the above formula with Stevenson’s formula (6-86); 1/2 2  1 1 + vw+u + ) − 1 = ln ( y + ) + B + (   vw 

We take A = B = 5.0 and  = 0.41 and  = 0.5 (flat plate). Since the Oljaca/Sucec formula requires a specific boundary layer thickness  , we plot these only for the  +  1000, as in the Fig. 6-44a data. The results are shown in the following plot:

We see that the Oljaca/Sucec formula, in spite of its algebraic complexity, is simply the Stevenson formula plus a wake. The data go much higher, i.e.,  is much larger.

6-48 Consider as a starting point the shear stress expression in the viscous sublayer, i.e., u  = −  u v = const y Make the necessary assumptions and simplifications leading to the linear relation u + = y + . Solution: Considering the starting point, u  = −  u v = const y Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -46-

The linear relationship is valid only in the viscous sublayer, where turbulent fluctuations (u′, v′) are damped out. So we can assume u = Const y If we integrate the above relation, we get

 =

 dy = du   u= y 



Now we use 𝜏 = 𝜌𝑢𝜏2 , where 𝑢𝜏 = velocity at the edge of the viscous sublayer. u=

 u2 u u y =  y  u 

We can name the LHS as u + and RHS as y + , which are the nondimensional velocity and length. So, u + = y + .

(Ans)

6-49 Consider as a starting point the shear stress expression in the log layer, i.e., u −  u v = const y Make the necessary assumptions and simplifications, including the mixing-length approximation, that result in the logarithmic relation u + =  −1 ln y + + B .

 =

Solution: Continuing from the solution of problem 6-48, + − u + = f ( ) . the outer layer profile is given as uinf

Here,  =

 and  is the boundary layer thickness. 

u = uinf at y = 

This relation is obtained from the dimensional analysis. For the inner layer, u+ = g ( y+ ) Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -47-

Let  + =

 UT v

 y + =  +

Now adding the two equations above for outer and inner layer, + uinf = f ( ) + g ( + )

Differentiating this equation with respect to  + , we get + ' uinf =  g ' ( + )

where ′ means differentiation with respect to  + . Now differentiating this equation with respect to  , 0 = g ' ( + ) +  + g '' ( + )  g ' ( y + ) + y ' g '' ( y + ) = 0

This can be written as d  + dg  y =0 dy +  dy + 

Integrating with respect to

y + , we get

 + dg  = const y +   dy 

This will be named 1 . 

dy 1 = + + dy  y g=

1



ln ( y + ) + B

From the inner layer equation, we know that

g = u+ .

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1

 u+ =



ln ( y + ) + B

 u + =  −1 ln ( y + ) + B

(Ans)

6-50 The average velocity for a turbulent pipe flow can be evaluated using the expression uav =

Q 1 = A  a2



a

0

u (r )dA =

1  a2



a

0

u (r )2 r dr =

2 a2



a

0

u ( y )(a − y )dy

where the simple y = a − r coordinate transformation is used, with y being the dimensional distance from the wall. Assuming that the law of the wall is accurate all the way to the wall, i.e., neglecting the very thin viscous sublayer, substitute the log-law, the average velocity definition to prove that

u + =  −1 ln y + + B , into

uav 1  av*  3 = ln  + B− * v     2

Hint: Before substituting u =  ln y + B into the average velocity integral, express it first as u ( y ) 1  yv*  = ln  + B v*     +

−1

+

Solution: uav = =

Q 1 = A  a2

1  a2

uav =



a

0

2 a2



a

0

u (r )dA

u (r )2 dr



a

0

u ( y )(a − y )dy

(eq-1)

where y = a − r and y is the distance from the wall. We know that u + =  −1 ln ( y + ) + B . Now in the dimensional form u ( y)

*

=

 y *  ln  + B     1

 u ( y) =

 *  y *  * ln   + B    

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Substituting the value of u ( y ) in (Eq-1),

 *   *   *  ln  y  + B   ( a − y ) dy     

uavg =

2 a2

uavg =

2  *a a   *   * ln y dy      a 2   0  v   



a

0



a

0

a  *  ln  y   + * B  ( a − y )dy 0 v   C

B

A

To simplify, integrate A, B, and C separately. a   *  2  *a a   *   2  *a a ln y dy = ln y − ln ( ) 0    dy A. 2     a   0  v   a 2   0  v 

Using  ln ( x ) dx = x ln ( x ) − x

2 = 2 a

a  *a   a   *      ( y ln ( y ) − y )0 +  y  ln         y  0    

2 = 2 a

 *a   a *      a ln   − a      y   

a  *   * a  *  *  a ln  y   ydy =   y  ln ydy +  y  ln   dy  B. 0  0  v   0 v 

x 2 ln x x 2 − Using  x  ln xdx = 2 4  *   y2    *  y 2 ln y y 2   B =  −  + ln        2 4 0  v   2 0   a

a

 *   a 2 ln a a 2  a 2   *   B =  −  + ln       2 4  2  v   B=

 *  a2   *  a2   ln  a   −   2  v 4

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* C. C = v ( B )



a

0

(a − y )dy a

y2  = v ( B )  ay −  2 0 *

 a 2   *( B)a 2 = B a2 −  = 2 2 

Now substituting the values of A, B, and C in eq-2, we get

uavg

=

2 = 2 a

2 a2

 *a   a *     *  a 2   *  a 2   *( B)a 2   a ln   − a   −  ln  a   −  + 2     y      2  v  4 

 2  *  a *    * 2  1   *  1   * B 2  ln  a  − 1 − a  ln  a   +  a        2  v  4  2  

 1  v*  1  1  B  = 2v*  ln  a  − 1 −  +   2  v    u  2  

uavg

*

=

 a *  3 ln  + B−   v  2 1

(Ans)

6-51 The wall-friction factor in a pipe of radius a is defined as

Cf =

 2 w  4  uav2

where Λ is the Darcy friction factor. (a) Show that the following identities hold:

Cf uav 2 av* (2a)uav = and = Re , where ReD = D v* Cf  8  (b) Substitute these identities into our previous result, namely, uav 1  av*  3 = ln  + B− * v     2

Then using κ = 0.41, B = 5.0, and log 10 (x) = ln(x)/ln10, derive Prandtl’s 1935 Darcy friction factor formula for smooth pipes, namely, Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -51-

(

1/  = 1.99log10 Re D

)

 − 1.02;

  4C f

Note that Prandtl slightly adjusted the constants in this correlation to better fit pipe-friction measurements at low Reynolds numbers, thus leading to Eq. (6-54). Solution:

2   uav2 uav 2 = = * Cf 2 w 

a.

Because  * (shear velocity) =

so uav* =

2 Cf

v

Now, Re D

=

w ,  (Ans)

Cf 8

=

(2a)uav 1  v 2uav

(2a)uav



2  uav2  8

 

a *  = because  * = w   Hence,

b.

av*



= ReD

Cf 8

(Ans)

uav 1  av*  3 = ln  + B− * v     2



2X 4 1    3 = ln  Re D  + B −    8 4  2



 1 1  3  8= ln  Re D   + 5 − 0.41  2  0.41  32 



1 1 1 3   5 = ln  Re D   + −  0.41 8  32 8 0.82 8 

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(

)

1 = 1.99 log10 Re D  − 1.02 

(Ans)

6-52 One of the simplest ways to recapture the Reynolds stress is to time-average the conservation form of the axial momentum equation in Cartesian coordinates. (a) Start by expanding  (u 2 ) x

+

(uv) 1 dp  2u =− + 2 (axial momentum in conservation form) y  dx y

Then using the continuity equation, show that the result is identical to the traditional form:

u

u u 1 dp  2u +v = − + 2 (steady, incompressible, axial momentum form) x y  dx y

b) By substituting the Reynolds decomposed variables,

u = u + u , v = v + v , p = p + p , into

the conservation form, time-averaging, cancelling terms that average out, and assuming that streamwise variations in show that

u u  are much slower than the cross-streamwise variations in u v  ,

(

u u 1 dp  2u  u +v =− + 2 + −u v x y  dx y y

)

Hint: The time-averaged products of mean and fluctuating quantities, such as

uu , uv , vu  , and vv , all vanish identically. Similarly, the time-averaged fluctuating quantities vanish unless they are squared or multiplied by other fluctuating quantities. Finally, the timeaveraged mean quantities are the mean quantities. Solution: (a)

 (u 2 ) x

+

(uv) 1 dp  2u =− +v 2 y  dx y

(Eq-1)

u 2 (uv) u v v + = 2v + u + v x y x y y  u v  u u  u + u  + u + v y  x y  x For incompressible flow, continuity equation can be written as

u v + = 0 (for 2D flow) x y Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -53-

 ( v2 ) x

+

 u v   (uv) u v =u + +u +v y x y  x y  equal to 0

u 2  (uv) u v + =u +v x y x y

(Eq-2)

From (Eq-1) and (Eq-2),

u u −1 dp  2v u +v = +v 2 x y  dx y

(Ans)

u = u + u  (b) v = v + v p = p + p For steady incompressible 2D flow, the continuity equation is

u v + = 0 . (Eq-3) x y

u u −1 dp  2u +v = + v 2 .(Eq-4) Axial momentum equation is u x y  dx y Substitute decomposed variables in (Eq-3) and (Eq-4).

  u + u  ) + ( v + v ) = 0 ( x y

( u + u )

(Eq-5)

2   −1 d    (u + u ) + ( v + v ) ( u + u  ) = p + p + v + v u + u ) ( ) ( ) 2 ( x y  dx y

'  u u      v u (u ) + v +u v +u )+v +v + v ( u  + u ) ( x y x x y y 2 1 d    =− p + p ) + ( v + v ) (u + u ) (  dx y

(Eq-6)

 uv = ( u + u  )( v + v ) equal to uv + uv + vu + uv .

All the terms linear in fluctuations variables are zero, i.e., u' = v' = 0 , but this is not true for nonlinear terms. uv = uv + u'v

So, the continuity equation for the decomposed variables is Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -54-

u v + =0 x y From (Eq-6),

u u  1 dp  2u u +v + (uv ) = − + v 2 (because all other linear terms become zero) x y y  dx y u

u u 1 dp  2u  +v =− + 2 + −uv x y  dx y y

( )

(Ans)

The equation is very similar to the original equation except the term

 −uv which is y

( )

rooted to Reynold’s stress. It introduces the coupling between the mean and fluctuating part.

6-53 The average velocity for a turbulent channel flow can be evaluated using the expression uav =

Q 1 h = u ( y )dy A h 0

where h represents the distance from the wall to the midsection plane of the channel. Assuming that the law of the wall is accurate all the way to the wall, i.e., neglecting the very thin viscous sublayer, substitute the log-law, u =  ln y + B , into the average velocity +

−1

+

definition to prove that uav 1  hv*  1 = ln  + B− * v     

Solution: uav =

Q 1 h = u ( y )dy A h 0

Given log-law, u =  ln y + B +

−1

+

( )

* u + is the dimensionless velocity, i.e., velocity divided by the shear velocity 

u+ =

u

*

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y + is the wall coordinate i.e., distance from the wall made dimensionless by shear velocity ( * ) and kinematic viscosity ( ) . y *  y = and  * = w , where  is the density of the flowing fluid and  w is the wall   +

shear. 

u

*

=

 y *  ln  + B   v  1

Given formula for average velocity = uav = 1  u ( y ) dy , h

h



u

*

u=

=

0

 y *  ln  + B   v  1

 *  y *  * ln   + B   v 

 uavg =

 1 h   *  y *  ln  + B * dy    h 0  v  

 uavg =

* h



h

0

 y *  B * ln  dy +  h  v 



h

0

dy

Let’s assume h  y *  I =  ln  dy 0  v  h  y *  I =  1 ln  dy 0  v 

Integrating by parts, we get

  y *   d  ln     y *   h   v  I = ln    1dydy    1 dy  − 0 v dy     h

h   y *   1 * I = ln    ydy  y  − 0  y *  v   v  0    v 

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h

h   y *   I = ln   y  − 0 dy   v  0 h

  y *   h I = ln   y  −  y 0   v  0  h *   I = h ln  −h  v 

And



h

0

 uav =

dy = h * h



h

0

 y *  B * ln   dy + h  v 



h

0

dy

 h *   B * u*   uav =  h ln  (h)  − h + hk   v   h  h *   B uav 1   * =  h ln   − h  + (h)  hk   v   h  h *  1 1 B  h ln   h + (h) − hk h  v  hk



uav

=



uav

 h *  1 = ln  − + B   v  

*

*

1

uav 1  hv*  1 = ln  + B− * v     

(Ans)

6-54 Because a channel is not round, it is necessary to use the concept of a hydraulic diameter on which the Reynolds number can be based. For a sufficiently wide channel of height 2h, (a) show that the hydraulic diameter is Dh = 4 A / P . Next, (b) using ReDh = (4h)uav /  , and recalling that uav / v  (8 / ) *

1/2

, show that

hv*



=

1 4huav 8 

 1 = Re Dh 2 8 2



(c) In addition, by substituting κ = 0.41, B = 5.0, and log 10 (x) = ln (x)/ln10 into Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -57-

uav 1  hv*  1 = ln  + B− * v     

derive the Darcy friction factor formula for smooth channels, namely,

(

1/  = 2.0log10 Re Dh

)

 − 1.19

Solution:

4 A 4  w  2h = P 2 ( w + 2h ) Since the weight of the channel is very large when compared to the height, the height can be neglected w >>(2h). (a) Dh =

Dh =

4  ( w  2h ) = 4h 2w

Hence, the hydraulic diameter of the channel is 4h.

(Ans)

1/2

8 (b) uav = v*    Write the relation for Reynold’s number

Re Dh =

( 4h ) uav 

Substituting the expression for uav, 1

Re Dh

( 4h )  *  8  2 =   



Now rearrange the equation

v*



=

Re Dh 4



 8

h * Re Dh  =   8 2 h *



=

1 8 2

Re Dh 

Thus, the equation is

hv*



=

1 8 2

Re Dh  .

(Ans)

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(c) Write the expression for ratio of velocity:

uav

*

=

1



ln

h * 1 +B− v 

Substitute the values. 8 1 1  1  = ln  Re Dh   + 5 −  0.41  8 2 0.41  

2  828   1 =  2.44  ln  Re Dh  8 2 

    + 2.561 

Use the relation, log 10 (x) = ln (x)⧸ln10.  2.828   1  =  2.44  ln(10)   log    + log Re Dh   8 2   

(



(



)) )  + 2.561 



)

2.828 = −5.92 + 5.62 log Re Dh  + 2.561 

(

1 = 2.0 log10 Re Dh 

)

 − 1.19

(Ans)

6-55 Determine the beginning, ending, and maximum entrance lengths in a circular duct of diameter D = 0.03 m with air flowing between 1 m/s (beginning) and 50 m/s (ending). Repeat the analysis in a square duct of 0.03 m height. For air, use a density of 1.2 kg/m3 and a viscosity of 1.8 × 10−5 kg/(m ⋅ s) . Recall that the hydraulic diameter can be calculated using Dh = 4A/P. Calculate the velocity leading to the maximum entrance lengths in both circular and square ducts. Solution: Area of the circular duct =

 D2 4

=

  ( 0.03)

2

4

Ac = 0.0007065 m 2

Perimeter of the circular duct = Pc = 2 r  0.03  Pc = 2   = 0.094 m  2  Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw Hill LLC. -59-

Area of the square As = a 2 = ( 0.03) = 0.0009 m 2 2

Perimeter of the square = Ps = 4a = 4  0.03 = 0.12 m For square, Dh =

4 As 4  0.0009 = = 0.03m Ps 0.12

Required length =

Area 0.0009 = Dh 0.003

l = 0.03m For circular duct, Dh =

4 Ac 4  0.0007065 = Pc 0.094

Dh = 0.03m

Required length =

l=

Area Dh

0.0007065 = 0.023m 0.03

So, the beginning length in both cases is same, i.e., l = 0.03m

(Ans)

But the ending length = 0.023 m.

(Ans)

6-56 Because of the popularity of power-law profiles in turbulent boundary layer studies, Majdalani (2018) introduced a generic mean-flow pattern of the form: u  F ( ) =  1/ q Ue

where ξ ≡ y/δ represents the fractional distance within the turbulent boundary layer, and the power-law exponent q = {5, 6, 7, 8, 9} can be used to prescribe the value that best captures the motion under consideration. (a) Based on Eq. (6-28), show that the normalized displacement and momentum thicknesses, η* and θ*, as well as the shape factor H, can be expressed as direct functions of q, namely,

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 * * 1 1 =  (1 − F )d =   0  q +1   *  1 q   = 0 F (1 − F )d =  (q + 1)(q + 2)    * * q + 2 = = H   * q 

(normalized displacement thickness) (normalized momentum thickness) (momentum shape factor)

(b) Recognizing that most wall shear stress correlations, such as Eqs. (6-68) and (6-72), can be specified as functions of Reδ, Majdalani wrote a generic expression of the form 1/ m

    w = Cw  u    uav  2 av

2Cw Re1/ m

Cf =

or

where Cw and m are characteristics of the problem at hand. Assuming no pressure gradients, apply the momentum-integral relation, given by Eq. (6-67), to obtain 1/ m

d  Cw    =   dx  *  uav 

(c) After separating variables and integrating from δ(0) = 0 to δ(x) , show that the boundary layer thickness may be written as a function of the local Reynolds number, Rex = uav x/ν , namely,

 x

m

 C m + 1  m +1 ; a   w*  1  m  m +1

a

=

Rex

(d ) Recalling that the friction coefficient may be defined as C f = 2 w / (  uav2 ) , show that 1

 m *Cwm  m+1 b Cf = ; b  2   1 m +1   m +1 Re x

1

(e) Recalling that the displacement thickness may be specified  * =   (1 − F )d , show that 0

* x

m

=

1  Cw m + 1  m +1 ; c   1 q +1  * m  m +1

c

Re x

1

(f) Recalling that the momentum thickness may be deduced from  =   F (1 − F )d  , show 0

that

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 x

m

=

q  Cw m + 1  m +1 ; d   1 (q + 1)(q + 2)   * m  m +1

d

Re x

(g) Using the continuity equation v = −

 y udy , integrate the axial velocity profile to show x 0

that 1/ q

v y y m =e   ; e Ue x   (m + 1)(q + 1) (h) Recognizing that Prandtl’s 1927 power-law model uses Cw = 0.0233, q = 7, and m = 4, determine the characteristic coefficients (a, b, c, d, e) and compare your results to Eq. (6-72). (i) Recognizing that the wall–wake law of Coles uses Cw = 0.010, q = 7, and m = 6, determine the characteristic coefficients (a, b, c, d, e) and compare your results to Eq. (6-70). Solution: u  F ( ) =  1/ q , where ξ ≡ y/δ Ue

(a)

d  dU e C f + (2 + H ) = dx U e dx 2

(Given Eq 6-28)

From momentum-integral analysis, we know that Cf = 2

d dx

Also, we know that u+ 

1 2  y  ln ( x + ) + B + f  k k  

Since equilibrium factor  = 0 ,  = 0.45 . Evaluation of the profile function (Eq. 6-47) at y =  then gives  2  * = c   f Ue

1/2

  

=

1   *  2 ln  + B+ k    k

1/2   Cf    2.44ln Re    + 7.2   2  

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By analogy with the following equation,

 1 av* 3  uav = v *  ln +B−   2k  k This relation represents a friction relation between Cf and Re . It is laborious algebraically. So, C f = 0.020Re−1/6 or in terms of wall shear stress, 1/ 

     = 0.010  u    uav   2 av

Substituting all the above equations, we get

0.020 Re−1/6  2

d  7  7 d Re  = dx  72  36 d Re x

Integrating with knowledge that  = 0 at x = 0 , we get Re  0.16 Re6/7 x

 x

=

0.16 0.027 and C f = 1/7 Re x Re1/7 x

And with k = 0.41 and  = 0.45 1/2

 2   3  54 22.21 = − 2 , where  =   C      f 

2 We know Prandtl in the year 1927 used   = 0.0233 uaV  / uav 

1/4

C f  0.0466Re6−1/4 . Therefore,

 x

=

0.058 0.37 and C f  . 1/5 Re1/5 Re x x

1/q

Therefore,

u  y =  Ue   

or

and  s  y 

u = ( )1/ q . Ue

In the given Eq 6-28, we know that 



0



 * =  1 −

u  * dy =  Ue 

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1 u   y   * =  1 − d   0  Ue    

Therefore, displacement thickness becomes normalized displacement thickness as

* =

1 * u  1 =  1 − d = 0  q +1  Ue 

Also,  = 



0

(Ans)

u  u  * 1 − dy =  Ue  Ue 

u  u   y 1−  d    0U Ue   f  e 

* = 

1

Therefore, momentum thickness becomes normalized momentum thickness as

* 

1 u   u  q = 1 −  d =  0U  Ue  ( q + 1)( q + 2 ) e 

(Ans)

Also, momentum shape factor is

 * * q + 2 H= = =  * q

(Ans)

1/ m

   (b)  w = Cw  u    uav 

or C f =

2 av

2CW Re1/m 

Assuming that there are no pressure gradients and applying momentum-integral, we get C d d =* = 1/w m dx dx Re

*

C d Re C = 1/w m or Re1/ m d Re = * d Re x  d Re x Re

Using the initial condition  (0) = 0 , we can integrate and get



Re

0

Re1/ m d Re =

Cw



*



Re x

0

d Re x

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Or Re =

uav



m

m  C m + 1  m+1 m +1 =  w*  Re x   m 

Solving for  ( x ) , rearranging, and simplifying, we get m



 C m + 1  m +1 =  *  x  m 

1 Re

1 m +1 x

Differentiating, substituting, and simplifying, we get 1/ m

d  Cw    =   dx  *  uav 

(Ans)

m



 C m + 1  m +1 =  w*  x  m 

1 1

Rexm +1 1

2C C f = 1/wm = Re

*

 m *Cwm  m +1  1  m +1  Re m +1  2

x +1

m

*

1  Cw m + 1  m +1 = = And   x x q +1  * m 



1 1

Rexm +1 m

*

q  Cw m + 1  m +1 = =   x x (q + 1)(q + 2)   * m 

1 Re

1 m +1 x

These relations can be conveniently expressed as

 x

a

=

(Ans. c)

1 m +1 x

Re

m

 C m + 1  m+1 where a   w*   m  Cf =

b 1 m +1 x

(Ans. d)

Re

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 m C  where b  2    m +1  *

* x

m w

1 m +1

c

=

Re

(Ans. e)

1 m +1 x

m

1  Cw m + 1  m+1 where c    q +1  * m  And

 x

=

d Re

(Ans. f)

1 m +1 x

m

q  Cw m + 1  m+1 where d    (q + 1)(q + 2)   * m  Note that the transverse velocity can be obtained from the continuity equation, namely,

v  y u  y y  =−  dy = −    dy , and so Ue x 0 U e x 0    1/ q

1/ q

v m y y =   U e (m + 1)(q + 1) x   

 y  y 1/7    ; Prandtls 10 x    = 1/7  3  y  28 x    ; wall-wake law 

(Ans. g)

The validity of the expressions for (a, b, c, d, e) can be conformed by taking Prandtl’s Cw = 0.0233, q = 7, and m = 4 to reproduce, within the round-off error affecting Eq. (6-72), the following correlations:

 x

=

0.381 1 5 x

Re

Cf =

 0.0371 0.0593  * 0.0477 and = = 1 1 1 x x 5 5 Rex5 Re x Rex

(Ans. h)

The same may be repeated for the wall-wake law. Using Cw = 0.010, q = 7, and m = 6, we may bypass intermediate calculations and recover

 x

=

0.162 1 7 x

Re

Cf =

0.0271  * 0.0203  0.0158 = and = 1 1 1 x x 7 7 Re x Rex7 Re x

(Ans. i)

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