College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Solutions to Exercise Ten – Flow in Pipes Two
1
The square flat plate shown in the top figure
to the right is cut into four equal-sized
pieces and arranged as shown in the bottom
figure at the right. Determine the ratio of the
drag on the original plate to the drag on the
places in the configuration shown in the
second case. Assume laminar boundary
layer flow. Explain your answer physically.
(Problem and figure from Munson, et al.,
Fluid Mechanics.)
The drag force on a rectangular plate of length, L, and width, b, is given by the following equation,
where we use the relationship between the drag coefficient and the Reynolds number for a flat
plate in laminar flow..
1
1.328 1
1.328 1
FD  C D A U 2 
A U 2 
Lb U 2  0.664 b U 3 L
2
2
Re 2
UL

For the original case (a), L = b =
.
FD( a )  0.664b U 3 L  0.664 U 3  0.664 U 3 3
For the second case (b), L =
.
FD (b)
 , the total drag force is.
4 and b =  4 , the total drag force is.

0.664 U 3 3 FD ( a )
3
 0.664b U L  0.664
U 4 

4
2
2
3
Thus the drag force is smaller in the second case. This is due to the growth of the boundary layer
with downstream distance. More of the plate area is exposed to a region of thicker boundary
layer which gives a smaller drag in case (b).
2
A 500-N cube of specific gravity SG = 1.8 falls
through water at a constant speed U. Determine U
if the cube falls as oriented in the two figures at
the right. (Problem and Figure P9.45 from Munson,
et al., Fluid Mechanics.)
If the speed is constant, the acceleration is zero so the
sum of the vertical forces must be zero. There are
three vertical forces: (1) the weight of the body, W, (2)
the buoyant force, FB, and the drag force, FD. In both
cases the weight is 500 N and the buoyant force is the specific gravity of the water times the
volume of the cube. The drag force, FD = CDAU2/2, where CD is found from Figure 9.29. That
figure also gives the expression for the frontal area, A = D2 for both orientations of the cube.
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise nine solutions
ME 390, L. S. Caretto, Spring 2008
Page 2
We don’t know the size of the cube, which is required to determine the buoyant force, but we can
find it from the data given on the weight of the cube and its specific gravity.
Wcube   cube D 3  SG cube  water D 3
 D3
Wcube

SG cube  water 3
500 N
 0.3049 m
9800 N
1.8 3
m
The buoyant force is the volume of the cube times the specific gravity of the displaced water.
Fb   water D 3 
9800 N
0.3049 m3  277.8 N .
3
m
The drag force is evaluated as follows.
2
 46.435 N  s 2  2
1
1
2 1  999 kg  1 N  s  2


U
FD  C D A U 2  C D D 2 U 2  C D 0.3049 m 
U

C

D

2
2
2  m3  kg  m 
m2


The dimensions of the constant in this equation mean that the force will be in newtons if the
velocity is in m/s. We can now set our force balance. The weight is a downward force that is
balanced by the upward forces of buoyancy and fluid drag.
 46.435 N  s 2  2
U
 500 N  277.8 N  C D 

m2


500 N  277.8 N
4.786 m 2 1
 2.188 m  1
U



2
2
CD 
s
s
 46.435 N  s 
 CD

C D 
2
m


W  FB  FD
From Figure 9.29 on page 537, we see that the drag coefficients for cases (a) and (b) are 0.80
and 1.05, respectively. Thus for case (a) the velocity is
 2.188 m  1
 2.188 m  1
U 

 2.45 m/s


s
s

 CD 
 0.80
And, for case b,
 2.188 m  1
 2.188 m  1
U 

 2.13 m/s


s
s

 CD 
 1.05
As expected, the greater drag coefficient exerts the greater drag force, giving the lower velocity.
3
If for a given vehicle it takes 20 hp to overcome aerodynamic drag at 65 mph estimate the
horsepower required at 75 mph.
To make this estimate, assume that the drag coefficient is the same at both speeds. Then the
power to overcome drag, P = CDAFU3/2 is proportional to U3. If all other factors are equal we
can write 20 hp = A(65 mph)3 and the unknown power, P75 = A(75 mph)3. Taking the ratio of
these two proportionalities gives.
3
 75 mph 
P75
  1.536 
 
20 hp  65 mph 
P75  1.53620 hp   30.7 hp
Exercise nine solutions
4
ME 390, L. S. Caretto, Spring 2008
Page 3
A 2-in-diameter sphere weighing 0.14 lbf is
suspended by the jet of air shown in the figure at
the right. The drag coefficient for the sphere is 0.5.
Determine the reading on the pressure gage if
friction and gravity effects can be neglected for the
flow between the pressure gage and the nozzle exit.
Here we have to make two separate analyses. First we
have to determine the velocity that generates the drag
force that keeps the ball in equilibrium. Once we know
this velocity we can apply the Bernoulli equation
between the pressure gage (at point 1) and the nozzle
outlet (at point 2) to determine the pressure reading on
the gage. We are justified in applying the Bernoulli
equation by the statement that friction effects can be
neglected for the flow between the pressure gage and the nozzle exit.
Setting the drag force equal to the weight of the ball gives the following result for the given values
of CD = 0.5 and D = 2 in = 0.1667 ft, and using  = 0.00238 slug/ft3 for standard air.
1

1
W  FD  C D A V22  C D D 2 V22
2
4
2
8W
 V2 

C D D 2 
80.14 lb f
1 slug  ft
lb f  s 2
103.9 ft

0.00238 slug
s
 0.50.1667 ft 2
3
ft
We can find V1 from the continuity equation.
V1 A1  V2 A2
 V1  V2
A2 103.9 ft  0.3 ft 2  51.93 ft



2 
A1
s
s
 0.6 ft 
As planned we can now write Bernoulli’s equation between points 1 and 2.
V22
p1 V12
z2 

z  
 2g 1  2g
p2
We are given that gravity effects are negligible, which is usually the case for gas flows. To
neglect gravity forces we set z2 – z1 = 0. In addition, we can set p2 = 0 since this point is open to
the atmosphere. With these changes, we can find the pressure at the gage, p1.
 V22 V12   2
1 0.00238 slug
  V2  V12  
p1   

2
ft 3
 2g 2g  2
 103.9 ft  2  51.93 ft  2  1 lb f  s 2
 
 

s
s
 
  slug  ft

p1 = 9.63 lbf/ft2 = 9.63 psf