College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Solutions to Exercise Seven – Control Volume Analysis of Momentum
and Energy
1
A 100-ft-wide
river with a flow
rate of 2400
ft3/s flows over
a rock pile as
shown in the
figure at the
right.
Determine the direction of flow and the head loss associated with the flow across the rock
pile. (Problem 5.87 and Figure P5.87 copied from Munson et al., Fluid Mechanics text.)
We know that the head loss must be positive so we can assume a flow direction and compute the
head loss. If the head loss is negative, we have assumed the incorrect direction.
Start by assuming that the flow goes from point 1 to point 2 (from right to left) and apply the
energy equation to a control volume bounded by the top and bottom of the river and vertical lines
at section 1 and section 2. Note that our equation has to have point 1 as the assumed inlet and
point 2 as the assumed outlet.
p2


V22
p V2
 z 2  1  1  z1  hs  hL
2g
 2g
Along the top of the river, there is essentially no difference in pressure between two liquid
elevations because the air density is negligible. Thus, we have p1 = p2 = atmospheric pressure.
There is no shaft work so hs = 0. We can find the velocities from the stated flow rate and the
areas.
2400 ft 3
6 ft
Q
s
V1 


A1 4 ft 100 ft 
s
2400 ft 3
12 ft
Q
s
V2 


A2 2 ft 100 ft 
s
Rearranging the energy equation to solve for head loss with p2 – p1 = hs = 0 and substituting the
velocities just found and the elevations shown in the diagram gives the head loss shown below.
2
2
 6 ft   12 ft 

 

2
2
V1  V2
s   s 

hL 
 z1  z 2 
 4 ft  2 ft  0.32 ft
2g
 32.174 ft 
2

s2


Since this head loss is positive our original assumption that the river flows from right to left
is correct, and the head loss, hL = 0.32 ft .
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise seven solutions
2
ME 390, L. S. Caretto, Spring 2008
Page 2
If a ¾ hp motor is required to produce a 24-in
stream of air having a velocity of 40 ft/s as
shown in the figure at the right, estimate (a)
the efficiency of the fan and (b) the thrust of
the supporting member on the conduit
enclosing the fan. (Problem 5.88 and figure
taken from Munson, Fluid Mechanics.)
This problem is similar to example 5.24 starting
on page 239 of the text. In that example the
following equation is used to provide “a
reasonable estimate of the efficiency”.
Useful effect


Work input
wshaft  loss
net in
wshaft

hs  hL
hs
net in
We can find the numerator of this efficiency definition from the energy equation. 1
po


Vo2
p V2
 gz o  i  i  gz i  wshaft  loss
2

2
net in
For this fan, po = pi = 0 because both are atmospheric pressure, and zo = zi, so the pressure and
elevation terms vanish. In addition we can assume that the air entering the duct from the outer
atmosphere has such a low velocity that Vi2 << Vo2, and can be neglected. With these
assumptions, we have the following result.
2
wshaft
net in
Vo2 1  40 ft 
800 ft 2
 loss 
 
 
2
2 s 
s2
The shaft work in this equation is defined as the power input divided by the mass flow rate. If we
use the standard density of air = 0.00238 slug/ft3 from Table 1.7 in the inside front cover, we have
the following mass flow rate from the continuity equation.
0.00238 slug 40 ft  24 in 
m  VA  V

4
s
4
ft 3
D 2
2
2
 ft 
0.2991 slug

 
s
 12 in 
We can then compute the shaft work from the input power of ¾ hp and this mass flow rate.
wshaft 
net in
W shaft
net in
m
0.75 hp 

550 ft  lb f 1 slug  ft
hp  s
lb f  s 2
1379 ft 2

0.2991 slug
s2
s
We now have all the required information to compute the estimate of the efficiency.

wshaft  loss
net in
wshaft
net in
1
800 ft 2
s2

1379 ft 2
s2
 = 58.0%
Note that we could also use the equation with shaft head and head loss; then we would have to
divide the shaft work and loss term by g. The efficiency would have the same result.
Exercise seven solutions
ME 390, L. S. Caretto, Spring 2008
Page 3
To obtain the force, we use the usual momentum balance equation.
N inlets
mVk cv Noutlets
   oVo AoVk ,o    iVi AiVk ,i   Fx
t
o 1
i 1
Here the problem is steady and we have only one inlet and one outlet so there is a common mass
flow rate as found above. The only velocity components are in the direction of flow, which we call
the x-direction here. This gives our momentum equation as
 Vx,o  Vx,i    Fx
m
In the previous part we neglected the inlet velocity and we will continue to do so here, giving V x,I =
0. The pressure at both the inlet and exit are zero, so the only force that we have to consider is the
anchoring force, FAx.
FAx  m V x ,o
2
0.2991 slug 40 ft 1lb f  s

s
s slug  ft
Fax = 12.0 lbf
3. Air flows past an object in a pipe of 2-m diameter and exits as a free jet as shown in the
figure at the right. The
velocity and pressure
upstream are uniform
at 10 m/s and 50 N/m2,
respectively. At the
pipe exit, the velocity
is nonuniform as
indicated. The shear
stress along the wall is
negligible. (a)
Determine the head
loss associated with a
particle as it flows from the uniform velocity upstream of the object to a location in the
wake at the exit plane of the pipe and the total head loss for the section shown. (b)
Determine the force that the air puts on the object.
The wake has a uniform velocity of 4 m/s so we can use the simple head loss equation to look at
the effect from the uniform inlet flow of 10 m/s to the uniform wake exit flow of 4 m/s. This head
loss equation is
po


Vo2
p V2
 z o  i  i  z i  hshaft  hL
2g
 2g
net in
The elevation of the inlet and outlet are the same so zo = zi and the outlet is open to the atmosphere
so that po = 0. There is no shaft work input. With all these considerations our head loss equation
becomes.
2
 10 m   4 m 
50 N

 

2
2
2
pi Vi  Vo
s   s 

m
hL  


9.80665 m
12.0 N

2g
2
3
s2
m
2
hL = 8.45 m
In order to find the head loss for the entire flow we have to consider the nonuniform velocity profile
at the exit. To do this we use the nonuniform flow equation based on the average velocity and the
Exercise seven solutions
ME 390, L. S. Caretto, Spring 2008
Page 4
kinetic energy coefficient, , that accounts for the nonuniform velocity profile. (See equation 5.89
on page 243 of the text.)
po


 oVo2
2g
 zo 
pi


 iVi 2
2g
 z i  hshaft  hL
net in
The kinetic energy coefficient is defined by equation 5.86 on page 243 of the text. That equation
can be modified as shown below for constant density and a circular geometry where A = R2 and
dA = rdrd.
 V  V  ndA
2

Area
m V
2

  V 2 V  n dA
Area
V AV
2

 V V  ndA  V
1
3
V A Area
2 R
1
2
  V V  n rdrd
2
R
3
2
0 0
Since the outlet velocity is normal to the exit plane the Vn dot product is simply the exit velocity, V.
Since the exit velocity profile does not depend on , we can integrate over this variable giving a
factor of 2 in the numerator. This gives the following.

1
3
V R 2
2 R
2
  V V  nrdrd 
0 0
2
2
V 2Vrdr  3 2  V 3 rdr
3
2 
V R 0
V R 0
R
R
The average velocity is defined in equation 5.7 on page 195. We can perform the same
manipulations on that equation that we did on the equation for the kinetic energy coefficient. This
gives the following result.
V 
  V  n dA
Area
A


 V  n dA
Area
A

1
A
2 R
2
  V V  n rdrd 
0 0
2
2
Vrdr  2  Vrdr
2 
R 0
R 0
R
R
The velocity profile at the exit has V = 4 m/s from r = 0 to r = 0.5 m and V = 12 m/s from r = 0.5 m to
r = 1 m. This gives the following result for the average velocity.
R
0.5 m
1m
1m

0.5 m 4 m

12 m
2 4m r2
12 m r 2


rdr  
rdr  

 
2
s
s 2 0.5 m 
 0 s
 1 m   s 2 0
0.5 m

2
2
2


2
4 m 0.5 m   0 12 m 1 m   0.5 m 
10 m


2 
2
s
2
s
1 m  s

2
2
V  2  Vrdr 
R 0
1m2
Note that we could have found this result from a simple application of the continuity equation.
Because the density and area at the exit are the same as those at the inlet, the average velocity at
exit must be the same as the average velocity at inlet.
We can now apply the velocity profile into the integral for . (We know that i = 1 for the uniform
inlet velocity profile.)
3
3
1m
R
0.5 m  4 m 3

2
2
s 3  4 m  r 2
 12 m 
3
  3 2  V rdr 
  
 rdr   
 rdr  


3
V R 0
s 
s 
500 m5  s  2
 10 m 
2 


0 
0.5 m 



 1 m 
 s 
3 2 1m 
3
3

0.5 m2  0  1728 m3 1 m2  0.5 m2   1.312
 12 m  r
  s 5  64 m



3
2
s3
2
 s  2 0.5 m  500 m  s


0.5 m
0
Exercise seven solutions
ME 390, L. S. Caretto, Spring 2008
Page 5
Using this value in the non-uniform head loss equation with the previous observations that z0 = zi,
and po = hs = 0 gives the following result.
2
 10 m 
 10 m 
50 N
1
 1.312


2
2
p  V   oVo
s 
 s 
m2
hL  i  i i

 
9.80665 m
12.0 N

2g
2
3
s2
m
2
hL = 2.58 m
To determine the force that the air exerts on the object we use the momentum equation, whose
general form is
N inlets
mVk cv Noutlets
   oVo AoVk ,o    iVi AiVk ,i   Fx
t
o 1
i 1
In this case the flow is steady and we effectively have two outlets: the first (with subscript o1) has a
velocity of 4 m/s and the second (subscript o2) hash a velocity of 12 m/s. The inlet and both outlets
have a flow in the x direction only so the main velocity, V, is the same as the x-component, Vx for
each location. We thus have our momentum balance as
 o1Vo21 Ao1   o 2Vo22 Ao 2   iVi 2 Ai   Fx
The exit pressure is zero and the inlet pressure of 50 N/m 2 creates a force in the +x direction. We
are asked to find the force that the fluid exerts on the object, so this will be –Rx. There will also be a
wall shear stress force, but we are told that this force is negligible. With the additional assumption
that the density is the same at inlet and outlet 2, our equation becomes
 Vo21 Ao1  Vo22 Ao 2  Vi 2 Ai    Fx  pi Ai  Rx
We will use the standard density of air,  = 1.23 kg/m3 from Table 1.8 in the inside front cover. The
inlet area is (1 m)2 = 3.1416 m2. The inner area at the exit, with a velocity of 4 m/s is (0.5 m)2 =
0.78540 m2. The outer area at the exit, with a velocity of 12 m/s, is [(1 m)2(0.5 m)2] = 2.3562 m2.
Using these areas and the data for pressure and velocities gives the reaction force as follows.




50 N
3.1416 m 2 
2
m
2
2
2

1.23 kg  4 m 
 12 m 
 10 m 
2
2
2
 0.78540 m  
 2.3562 m  
 3.1416 m

3
m  s 
 s 
 s 
Rx  pi Ai   Vo21 Ao1  Vo22 Ao 2  Vi 2 Ai 





 1N  s2

 kg  m

Rx = 110.7 N
2
We have already made this assumption in solving for the head loss. If the exit pressure is the
standard atmospheric pressure of 101325 N/m 2 the inlet pressure will be 101375 N/m 2 which is
only a 0.05% difference in pressure (and hence a 0.05% difference in density).