College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
April 29 Homework Solutions
9.9
Approximately how fast can the wind blow past a 0.25-in-diameter twig if viscous effects
are to be of importance throughout the entire flow field (i.e. Re < 1)? Explain. Repeat for a
0.004-in-diameter hair and a 6-ft-diameter smokestack.
Determine the velocity that keeps the Reynolds number below 1 using the kinematic viscosity of
standard air from Table 1.7,  = 1.57x10-4 ft2/s. For the twig with D = 0.25 in = 0.0208 ft.
1.57 x10 4 ft 2
VD

7.54 x10  3 ft
s
Re 
1  V  


D
0.0208 ft
s
For the human hair with D = 0.004 in = 3.33x10-4 ft
1.57 x10 4 ft 2
VD

0.471 ft
s
Re 
1  V  


4

D 3.33 x10 ft
s
For the smokestack with D = 6 ft
1.57 x10 4 ft 2
VD

2.62 x10  5 ft
s
Re 
1  V  


D
6 ft
s
We see that smaller distances allow larger velocities to keep viscous forces always important.
This is due to the relations implied in the Reynolds number as the ratio of inertia forces to viscous
 V = (AV)V = AV2. The
forces. The inertia forces (really momentum) are proportional to m
viscous forces are the product of the viscous shear stress, (du/dy) times the area. If D is a
characteristic dimension for viscous forces (and V is a characteristic velocity), the viscous forces
are proportional to (V/D)A. Dividing the representation of the inertia forces by the one for the
viscous forces givers the usual Reynolds number, Re =VD/. In the derivation above we see
that the viscous forces are proportional to the area divided by the characteristic distance where
the inertia forces are only proportional to the area. Thus a smaller characteristic distance means
larger viscous forces (compared to inertia forces) and we can have a larger velocity and still have
viscous forces important throughout the flow field with the smaller dimension.
Notice that even at the dimension of a twig, the inertia forces will become important at any
reasonable velocity above a dead calm.
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
April 29 homework solutions
9.12
ME 390, L. S. Caretto, Spring 2008
Page 2
Water flows past a flat plate with an upstream velocity of U = 0.02 m/s. Determine the
water velocity a distance of 10 mm from the plate at distances of x = 1.5 m and x = 15 m
from the leading edge.
The Reynolds number for these two locations can be found from the usual formula, Re = Vx/,
with the kinematic viscosity of water,  = 1.12x10-6 m2/s from Table 1.6.
0.02 m
x
Ux
s
Re 

 2.68 x10 4 , 2.68 x10 4 for x  1.5 m, 15 m 

6
2
 1.12 x10 m
s


So the flow is laminar at both locations and we can get the velocity profile from the Blasius
solution on in Table 9.1 on page 500. The similarity parameter, , is defined as y(U/x)1/2 =
(y/x)Re1/2. For y = 10 mm = 0.01 m the value of this parameter at x = 1.5 m is

y
0.01 m
Re 
2.68 x10 4  1.091
x
1.5 m
Interpolation in the table gives the value of f’() at this value of .
f ' (  1.091)  0.2647 
0.3898  0.2647
1.091  0.8  0.359
1.2  0.8
For this value of f’ we find the local velocity u from the definition of f’ = u/U.
u  Uf ' 
0.02 m
0.359  0.00718 m
s
s
For y = 10 mm = 0.01 m the value of the  parameter at x = 15 m is

y
0.01 m
Re 
2.68 x105  0.345
x
15 m
Interpolation in Table 9.1 gives the value of f’() at this value of .
f ' (  0.345)  0.0 
0.1328  0.0
0.345  0.0  0.1145
0.4  0.0
For this value of f’ we find the local velocity u from the definition of f’ = u/U.
u  Uf ' 
0.02 m
0.1145  0.00229 m
s
s
April 29 homework solutions
9.39
ME 390, L. S. Caretto, Spring 2008
Page 3
Determine the drag on a small circular disk of 0.01 ft diameter moving 0.01 ft/s through oil
with a specific gravity of 0.87 and a viscosity 10,000 times that of water. The disk is
oriented normal to the upstream velocity. By what percent is the drag reduced if the disk
is oriented parallel to the flow?
The drag force, D = FD, is given by the following equation.
1
1

FD  U 2 A  U 2 D 2
2
2
4
We can find the Reynolds number to determine the correct equation to use for the drag coefficient
CD. The density of the oil is the product of its specific gravity and the density of standard water
from Table 1.6:  = (0.87)(1.94 slug/ft3) = 1.688 slug/ft3. The viscosity is given as 10,000 times
that of water; assume that this applies to the dynamic viscosity. Again using the data for standard
water in Table 1.6 we find the oil viscosity as  = (10,000)(2.34x10-5 lbf·s/ft2) = 0.234 lbf·s/ft2 =
0.234 slug/ft·s. We can then compute the Reynolds number.
1.688 slug 0.01 ft
0.01 ft 
s
UD
ft 3
Re 

 7.21x10  4
0.234 slug

ft  s
Since Re < 1 we can use the correlation equations in Table 9.4 on page 524 of the text for flow
normal to the disk and flow parallel to the disk
C D, normal 
20.4
20.4

 2.83x10 4
Re 7.21x10  4
C D, parallel 
13.6
13.6

 1.886 x10 4
Re 7.21x10  4
Note that the definition of the Reynolds number used in each equation is the same. The drag
force in each orientation is now found, noting that the area, A, is the same for both orientations: A
= D2/4 = (0.01 ft)2/4 = 7.854x10-5 ft2,
FD, normal 
C D, normal
2


2
2
2.83x10 4 1.688 slug  0.01 ft 
 5 2 lb f  s
U A 

 7.854 x10 ft
2
1 slug  ft
 s 
ft 3
2
FD,normal = 1.88x10-4 lbf
FD, parallel 
C D, parallel
2
U 2 A 


2
2
1.886 x10 4 1.688 slug  0.01 ft 
 5 2 lb f  s
7
.
854
x
10
ft


2
1 slug  ft
 s 
ft 3
FD,parallel = 1.25x10-4 lbf
The percent reduction is (1.88 – 1.25)/1.88 = 33.3% . Note that we could have found this
by simply comparing the constants in the drag-coefficient equations, (20.4 – 13.6)/20.4 = 33.3%,
since the Reynolds numbers and all other terms in the drag force equations are the same for both
orientations.
9.42
A 38.1-mm-diameter, 0.0245-N, table-tennis ball is released from the bottom of a swimming
pool. Find the terminal velocity in its rise to the surface.
The terminal velocity occurs when the upwards buoyant force, F B =
downwards forces of drag, FD = CD
result.
U2A/2,
Vball water, is balanced by the
and weight, W = 0.0245 N. This gives the following
April 29 homework solutions
ME 390, L. S. Caretto, Spring 2008
W  FD  0.0245 N 
Page 4
CD


U 2 D 2  FB  V water  D 3  water
2
4
6
The drag coefficient for a sphere depends on the Reynolds number as shown in Figure 9.21(a) on
page 526 of the text. The Reynolds number depends on the unknown velocity, U, in the force
balance equation. Without a specific equation for CD(Re) we have an iterative solution.
Substituting the definition of Reynolds number and algebraic manipulation gives our force
balance equation as shown below.
2
0.0245 N 
C D   Re   2



D  0.0245 N  C D  Re 2  D 3 

2  D  4
8
6
We can solve this equation for the CDRe2 product.
D 3 
0.0245 N
0.06239 N 4 D 3 g
C D Re 2 
 6 

 2
 2
3 2
 2


8
8
Substituting the kinematic viscosity and density of standard water from Table 1.6 into the
equation above we have the following relationship between drag coefficient and Reynolds
number.
C D Re 2 
0.06239 N

2

3
0.06239 N  kg  m2
4D g
4
1N  s


2
2
3 
3
999 kg  1.12 x10  6 m 2 

s
m 3 

CD 
Re
30,000
31,000
32,000
33,000
34,000
35,000
36,000
37,000
38,000
39,000
40,000
CD
0.696
0.652
0.612
0.575
0.542
0.511
0.483
0.457
0.434
0.412
0.391
m
0.0381 m3 9.80655
2
s
 1.12 x10


s

6
2 2
 6.263x108
m 


6.263x108
Re 2
The correct drag coefficient will satisfy this equation and the relationship
between CD and Re in Figure 9.21(a). Calculating the drag coefficient as
a function of Reynolds number from this equation in a spreadsheet gives
the results shown in the table at the left. Plotting these results on Figure
9.21(a) shows an intersection of these data with the correlation equation
for CD(Re) at a Reynolds number of 3.7x104. This gives the unknown
velocity as follows.
1.12 x10 6 m 2
37,000 1.1 m
 Re
s
U


D
0.0381 m
s
The answer is reported to only two significant figures because this is the
limit of accuracy for reading the correlation equation from Figure 9.21(a)
April 29 homework solutions
9.56
ME 390, L. S. Caretto, Spring 2008
Page 5
The aerodynamic drag on a truck can be reduced
by the use of appropriate air deflectors. A
reduction in drag coefficient from CD = 0.96 to CD =
0.70 corresponds to a reduction of how many
horsepower needed at a highway speed of 65
mph?
From the diagram we see that the frontal area of the
truck is (10 ft)(12 ft) = 120 ft2. The drag power is the
product of the drag force times the velocity.
PD  UFD 
CD
AU 3
2
The percent reduction in drag is simply the percent
reduction in drag coefficient = (0.96 – 0./70)/0.96 =
27%. The actual power savings is found as follows,
using the fact that all factors except the drag
coefficient are the same in both cases.
3
C
 C
 AU
C D1  C D 2 2 
PD   D AU 3    D AU 3  
2
 2
2  2
1
3
1 lb f  s 2
1  0.00238 slug 
hp  s
2  65 mi 44 ft s 




120
ft
0
.
96

0
.
70



2 
slug  ft 550 ft  lb f
ft 3
 h 30 mi h 



PD = 58.4 hp
9.70
The United Nations building in New York is approximately 87.5-m wide and 154-m tall. (a)
Determine the drag on this building if the drag coefficient is 1.3 and the wind speed is a
uniform 20 m/s. (b) Repeat your calculations if the velocity profile against the building is a
typical profile for an urban area (see Problem 9.17) and the wind speed halfway up the
building is 20 m/s.
When the speed is a uniform 20 m/s, the drag force on the building, with it’s area of (77.5 m)(154
m) = 13,475 m2 is found as follows. The standard air density of 1.23 kg/m 3 from Table 1.8 is used
in this calculation.
2
FD 


2
CD
1.3 1.23 kg  20 m 
2 N s
U 2 A 
13
,
475
m
 4.31x10 6 N


3
2
2 m  s 
1 kg  m
Problem 9.17 defines wind profiles as
proportional to the elevation above
ground level, y, raised to a power that
depends on the setting. For urban
settings the power is 0.4 so the general
relationship between wind velocity and
elevation is given by the following
equation. In this equation u is the
velocity and u0 is a known velocity at
an elevation y0.
April 29 homework solutions
ME 390, L. S. Caretto, Spring 2008
 y 
u  u0  
 y0 
Page 6
0.40
 Cy 0.40
The constant C in this equation can be found from the data that the wind speed is 20 m/s at the
half-height of the building = (154 m)/2 = 77 m.
 1 
C  u0  
 y0 
0.40
20 m  1 



s  77 m 
0.40

3.519 m 0.60
s
The total drag on the building is found by integrating the drag force from ground level to the top of
the building. We assume that all factors except the wind velocity are constant over the building
area and that there are no changes along the building width, b. Thus the integral over the area
becomes an integral over the vertical dimension, y, from y = 0 to y = the building height, H = 154
m.
FD, total 
 FD ( A)dA  
Area
Area
CD 2
u dA 
2
yH

y 0
CD 2
C
u d by   D b
2
2
yH
u
2
dy
y 0
Entering the velocity profile u = Cy0.4 into this equation and integrating gives.
FD, total
C
 D b
2
yH

C
u dy  D b
2
y 0
2
yH
 Ay 
y 0
0.4 2
C
dy  D B 2b
2
yH
y
y 0
0.8
 y1.8 
C
dy  D B 2b 

2
 1.8 
H

0
CD 2
B bH 1.8
3.6
Substituting the known values of the quantities on the right side of this equation gives the total
drag force for the variable velocity profile.
FD,total 
CD 2
1.3  3.519 m 0.6
C bH 1.8 
3.6
3.6 
s
2
2
 1.23 kg
1.8 N  s = 4.17x106 N




87
.
5
m
154
m

1 kg  m
m3
