Solutions Manual for Viscous Fluid Flow 4th Edition By Frank White, Joseph Majdalani

Solutions Manual for Viscous Fluid Flow 4th Edition By Frank White, Joseph Majdalani

,Solutions Manual for Viscous Fluid Flow 4th Edition By Frank White, Joseph Majdalani




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
(1.205)(18)(0.014) = 16, 800
ReD = VD/ = (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/ s2-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)068 = 8.83 E-6 Pa-s
n



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




1-2 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 m2/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
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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,Solutions Manual for Viscous Fluid Flow 4th Edition By Frank White, Joseph Majdalani




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




1-4 Given, without proof, the Poiseuille-paraboloid laminar-pipe-flow formula from
Chap. 3, u = (C/)(R2 − r2), find the wall shear stress if umax = 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:

umx = u ( r = 0) = CR2/ = 30 m/s = C(0.005) / (0.3) , or: C = 3.6E5 N/ m 2 -s2
2
( )
With C thus known for this data, we may evaluate wall shear stress by differentiation:
 2RC 
 =  u = = 2RC = 2 (0.005)(3.6E5) = 3600 Pa (Ans.)
wall   
r r=0  

We should check the Reynolds number ReD but we don’t know the density. But “oil” is usually
in the range   900 kg/m3. Then ReD = umaxD/ = (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 Cis 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 .


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, Solutions Manual for Viscous Fluid Flow 4th Edition By Frank White, Joseph Majdalani




Assuming no-slip, the shear stress  = V =
(1.5 kg/m  s)(17 10−3 m/s) = 8.5 Pa . (Ans.)
h (310 −3
m/s)




1-6 Given a plane unsteady viscous flow in polar coordinates:

C  r2 
vr = 0; v =   − 4vt 
1− exp
r
  







Compute the vorticity and sketch some profiles of vorticity and velocity.
From Appendix B, the vorticity is
1 C  r2 
 =
z (rv ) = exp −
 4vt 
r r 2vt
 

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 (x0 , y0 ) at time ( t = 0). From the geometric
requirement for two-dimensional streamlines at any instant,



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