, 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.)
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 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
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
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw
Hill LLC. -1-
,From Fig. 1-9 at this Reynolds number, estimate a drag coefficient of 1.1. Then
1
Fdrag = CD V2 ( DL ) = 1.1( 0.5)(1.205)(12 ) ( 0.008)(1.0 ) = 0.76 N /m
2
(Ans.)
2
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:
u 2RC
wall = = = 2RC = 2 ( 0.005 )( 3.6E5 ) = 3600 Pa (Ans.)
r r =0
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-
, V (1.5 kg/m s ) (17 10−3 m/s )
Assuming no-slip, the shear stress = = = 8.5 Pa . (Ans.)
h ( 3 10 −3
m/s )
1-6 Given a plane unsteady viscous flow in polar coordinates:
C r 2
v r = 0; v = 1 − exp −
r 4vt
Compute the vorticity and sketch some profiles of vorticity and velocity.
From Appendix B, the vorticity is
1 C r2
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,
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw
Hill LLC. -3-
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.)
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 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
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
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw
Hill LLC. -1-
,From Fig. 1-9 at this Reynolds number, estimate a drag coefficient of 1.1. Then
1
Fdrag = CD V2 ( DL ) = 1.1( 0.5)(1.205)(12 ) ( 0.008)(1.0 ) = 0.76 N /m
2
(Ans.)
2
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:
u 2RC
wall = = = 2RC = 2 ( 0.005 )( 3.6E5 ) = 3600 Pa (Ans.)
r r =0
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-
, V (1.5 kg/m s ) (17 10−3 m/s )
Assuming no-slip, the shear stress = = = 8.5 Pa . (Ans.)
h ( 3 10 −3
m/s )
1-6 Given a plane unsteady viscous flow in polar coordinates:
C r 2
v r = 0; v = 1 − exp −
r 4vt
Compute the vorticity and sketch some profiles of vorticity and velocity.
From Appendix B, the vorticity is
1 C r2
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,
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior consent of McGraw
Hill LLC. -3-
