College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Solutions to Exercise Six – Control Volume Analysis of Momentum and
Energy
1
A static thrust test stand is to be
designed for testing a specific
jet engine. Knowing the
conditions shown below for a
typical test, estimate a nominal
thrust to design for. (Problem
copied from Munson et al., Fluid
Mechanics text; partial figure
with dotted line as control
volume copied from solution
manual.)
Intake air velocity = 700 ft/s.
intake cross sectional area = 10 ft2;
intake static temperature = 480 R;
exhaust gas velocity = 1640 ft/s;
intake static pressure = 11.4 psia;
exhaust gas pressure = 0 psi.
We can apply the momentum conservation to the control volume indicated by the dashed line.
There is one inlet and one outlet. The forces are the thrust force on the test stand, F Ax, the
pressure forces. All forces and flows are in the x direction so we have only one momentum
equation. We can start with the general momentum equation..
N inlets
mVk cv Noutlets
   oVo AoVk ,o    iVi AiVk ,i   Fx
t
o 1
i 1
We will assume that the process is steady so the time derivative vanishes and we will write the x
component of velocity with the symbol u. With the forces described in the paragraph before the
general equation we have our x momentum equation as follows.
 2 u 2 A2 u 2  1u1 A1u1  FA, x  p1 A1  p 2 A2
We can also apply the continuity equation for this system, again assuming a steady process so
that our continuity equation becomes.
  1V1 A1   2V2 A2  1u1 A1   2u2 A2
m
Substituting the mass flow rate into our momentum equation gives.
m u 2  m u1  m u 2  u1   FA, x  p1 A1  p 2 A2
We can compute the mass flow rate from the data on the inlet.
11.4 psia 
m  1u1 A1 
144 lb f
P1
13.97 slugs
psia  ft 2  700 ft 
2
u1 A1 

 10 ft 
1714 ft  lb f
RT1
s
480 R   s 
slug  R
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348


Phone: 818.677.6448
Fax: 818.677.7062
Exercise six solutions
ME 390, L. S. Caretto, Spring 2008
Page 2
We have to determine the pressures acting on the control volume. We are given p1 in psia and p2
in psi(gage).. The atmospheric pressure acts on the cylindrical control volume shown in the red
short-and-long-dashed line. At the left, the atmospheric pressure acts over the entire inlet area.
At the right, the atmospheric pressure acts over the outlet area and part of the engine cover. The
action of atmospheric pressure to the left and to the right over the same area cancels so we must
use gage pressures in our force balance. We are given the outlet gage pressure of zero and we
can convert the inlet pressure to a gage pressure by assuming a standard atmospheric pressure
of 14.696 psia. This gives the inlet pressure, p1 = 11.4 – 14.696 = –3.296 psi(gage). We are also
given p2 = 0 psi(gage). With these pressures, the mass flow rate found above, and the given inlet
and outlet velocities, our momentum equation gives the thrust force as follows.
m u 2  u1   FA, x  p1 A1  p 2 A2
FA, x
 FA, x  m u 2  u1   p1 A1  p 2 A2
2
 3.296 lb f 144 in 2
13.97 slugs  1640 ft 700 ft  lb f  s



10 ft 2  0A2


2
2
s
s  slug  ft
in
ft
 s


FA,x = 1.79x104 lbf .
2
Water is added to the tank shown in the
figure at the right. (Problem 5.56 and Figure
P5.56 copied from Munson et al., Fluid
Mechanics text.) The water is added
through a vertical pipe to maintain a
constant water level. The tnak is placed on
a horizontal plane that has a frictionless
surface. Determine the horizontal force, F,
required to hold the tank stationary.
(0)
(1)
(2)
As usual we start with the general momentum
conservation equation. Here we apply it in the
horizontal, x, direction, using the notation that V x = u. We see that there the inlet flow has no
velocity component in the x direction so we have only two inlets and no outlets. We are not told
about any variation with time so we will assume that this is a steady problem. The general
equation is given below.
N inlets
mVk cv Noutlets
   oVo AoVk ,o    iVi AiVk ,i   Fx
t
o 1
i 1
With the assumptions stated above this equation becomes
 Fx 
Ninlets
mVx cv Noutlets
  oVo AoVx,o   iVi AiVx,i  0  1V1 A1Vx,1  2V2 A2Vx, 2  0
t
o1
i 1
We are not given the velocities, but we can find them from Bernoulli’s equation, written between
the top surface of the liquid (point 0) and each of the outlet jets (points 1 and 2). For both
applications of Bernoulli’2 equation we know that p 0 = p1 = p2 = 0 since both the top and each exit
are open to the atmosphere. Furthermore, because of the large area at the top of the liquid we
can assume that V02 is small compared to either V12 or V12. For the two outlets, with z0 – z1 = 1 m
and z0 – z2 = 2 m, we have
z0  z1  
p0  p1

2
2
V02  V12
0  0 0  V1

1m

0  V1  2 g 1 m 
2g

2g
Exercise six solutions
 z0  z 2  
p0  p2

ME 390, L. S. Caretto, Spring 2008
Page 3
2
2
V02  V22
0  0 0  V2

 2 m

0  V2  2 g 2 m   2V1
2g

2g
The values of V1 and V2 can then be found.
V1  2 g 1 m  2
9.80665 m
1 m  4.429 m
2
s
s
V2  2V1 
6.263 m
s
We can now compute the force from the momentum balance, using the density of water as
999 kg/m3, converting the areas to square meters, which gives A 1 = 0.00125 m2 and A2 =
0.000625 m2, and noting that V1,x = -V1 and Vx,2 = V2.
999kg 1 N  s 2  4.429 m
 4.429 m 
0.00125 m 2  

3

m
kg  m  s
s


6.263 m
 6.263 m 

0.000625 m 2 

s
s 


Fx  1V1 A1Vx ,1   2V2 A2Vx , 2 



Fx = 0.0 N
We can solve the problem using the algebraic results for V1 = –V1,x and V2 = V2,x while noting that
 = 1 = 2, and A2 = A1/2. This calculation, below, shows that the force is exactly zero.
Fx  1V1 A1Vx ,1   2V2 A2Vx , 2   

3


2 g 1 m A1 
2


2 g 2 m A1 2  0

2
Assuming a frictionless,
incompressible, one-dimensional
flow of water through the
horizontal tee connection
sketched in Figure P5.55 at the
right, estimate values of the x and
y components of the force exerted
by the tee on the water. Each pipe
has an inside diameter of 1 m.
(Problem 5.55 and figure taken
from Munson, Fluid Mechanics.)
We can apply the momentum conservation to the control volume drawn along the pipe walls and
ending at the circular sections shown in the diagram. Here we have two coordinate directions, x
and y, so we will need two momentum equations. The first, for the x-direction forces can be
simplified by noting that the inlet at (2) and the outlet at (3) have no x momentum component; in
addition, the x momentum component at the section (1) inlet, u1 = –V1, where V1 is the velocity
magnitude. Also, the pressure, p1 is pointing in the negative x direction. This gives.
m 3 u 3   m 2 u 2   m 1 u1   m 1u1  m 1V1  FR , x  p1 A1
In the momentum equation for the y direction forces we note that y-momentum components v2
and v3 are the negatives of the velocity magnitudes at both inlet section 2 and the outlet section 3
(v2 = –V2 and v3 = –V3) and y-momentum at section 1 is zero.
m 3 v3   m 2 v 2   m 1 v1   m 3V3  m 2V2  FR , y  p3 A3  p 2 A2
Exercise six solutions
ME 390, L. S. Caretto, Spring 2008
Page 4
We can determine the mass flow rates from data given at sections 1 and 3, using the density of
water as 999 kg/m3, and the constant pipe diameter of 1 m (giving an area of (1 m)2/4 = 0.7854
m2.
m 1  1V1 A1 


999 kg 6 m
4708 kg
0.7854 m 2 
3
s
s
m
m 3   3Q3 
999 kg 10 m 3 9990 kg

s
s
m3
From mass conservation, we can find the mass flow at section 2 as the difference between these
two mass flow rates.
m 2  m 3  m 1 
9990 kg 4708 kg 5282 kg


s
s
s
We can find the magnitude of the unknown velocities at sections 2 and 3 from the continuity
equation.
5282 kg
m 2
6.733 m
s
V2 


 2 A2 999 kg
s
0.7854 m 2
3
m


10 m 3
Q
10 m 3 12.73 m
s
V3  3 

A3 0.7854 m 2 s
s
We have the value of the pressure at section 1, p1 = 200 kPa, but we do not know the pressures
at points 2 and 3. However, we are told that the flow is incompressible and frictionless.
Therefore, we can apply the Bernoulli equation along a streamline between point 1 and point 3 to
obtain the following equation (noting that both points are at the same elevation so z 3 – z1 = 0.
V32  V12
g z 3  z1   p3  p1  
0 
2
p3  200 kPa
V32  V12
p3  p1  
2
1 kN
999 kg 1  12.73 m   6 m 

 


2
s
kPa  m
m 3 2 
  s 
2
2
 kN  s 2
137 kN


m2
 1000 kg  m
We can also apply Bernoulli’s equation to a streamline from point 2 to point 3. Repeating the
analysis above gives the following result.
2
2
V22  V32 137 kN 999 kg 1  6.733 m   12.73 m   kN  s 2
195.3 kN
p 2  p3  



 
 

2
3
2
2 
s
s
m
m
m2
 
  1000 kg  m
We now have all the data required to find the reaction forces from our momentum equations. We
first find the x-direction reaction.
FR , x  m 1u1  p1 A1 
4708 kg 6 m kN  s 2
200 kN

0.7854 m 2  185 kN .
s
s 1000 kg  m
m2
The y-direction reaction force is


Exercise six solutions
ME 390, L. S. Caretto, Spring 2008
Page 5
5282 kg 6.733 m kN  s 2
s
s
1000 kg  m
2
9990 kg 12.73 m kN  s
195.3 kN
137 kN


0.7854 m 2 
0.7854 m 2
2
s
s
1000 kg  m
m
m2
FR , y  m 2V2  m 3V3  p 2 A2  p3 A3 




FR,y = –45.8 kN .
The forces that we have solved for are the forces on the control volulme. The forces of the water
on the tee, which we were asked to find are just the negatives of the values given above.