College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Solutions to Unit 12 Quiz – Do one of the following problems
1. Air flows steadily and isentropically from standard atmospheric conditions to a receiver
pipe through a converging duct. The cross section area of the throat of the converging
duct is 0.05 ft2. Determine the mass flowrate through the duct if the receiver pressure is
(a) 10 psia; (b) 5 psia.
This problem is similar to example 11.5 in the text. We compute the mass flow rate as the
product of density times velocity times area at the throat. The density is related to the Mach
number and the velocity is the Mach number times the speed of sound.
This problem is for a duct that has a converging section only. It cannot become supersonic
because there is no increase in area beyond the minimum throat area. We first have to
determine if the flow is choked or not. To do this we compute the critical pressure ratio, p* from
the equation below, using the value of k = 1.4 for air and a stagnation pressure equal to the
standard atmospheric pressure of 14.696 psia.
 2 
p*  p0 

 k 1
k ( k 1)
1.4 (1.4 1)
 2 
 14.696 psia 

 1.4  1 
 7.763 psia
If the receiver pressure is less than p*, the flow is choked and we know that the throat Mach
number is one; if the receiver pressure is greater than p*, the flow is not choked; the throat
pressure is the receiver pressure and we have to compute the throat Mach number from this
pressure.
The initial receiver pressure of 10 psia is greater than p* = 7.763 psia so the flow is not choked;
the throat pressure is 10 psia and we can compute the throat Mach number by rearranging
equation 11.59 in the text to solve for the Mach number.
(1.41) 1.4
( k 1) k


 2  p0 
 2  14.696 psia 

Ma  
 1  
 1  0.7625
 

 k  1  p 
 1.4  1  10 psia 


Equation 11.60 gives the density at this Mach number. The density of standard air, 0 = 0.00238
slug/ft3 is taken from Table 1.7.
1 1 k 
 k 1

   0 1 
Ma 2 
2



1 11.4 
0.00238 slug  1.4  1

0.76252 
1 
3
2


ft

0.001808 slug
ft 3
In order to find the velocity at the throat as V = c(Ma) we have to find c which requires the
temperature. This is found from equation 11.56, using the standard temperature from table 1.7,
T0 = 59oF = 518.67 R.
 k 1

T  T0 1 
Ma 2 
2


Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
1
 1.4  1

 518.67 R 1 
0.76252 
2


Mail Code
8348
1
 464.6 R
Phone: 818.677.6448
Fax: 818.677.7062
Unit twelve quiz solutions
ME 390, L. S. Caretto, Spring 2008
Page 2
We can now compute the sound speed using the gas constant for air from Table 1.7: R = 1716
ft·lbf/slug·R.
 1716 ft  lb f
c  kRT  1.40
 slug  R

1 slug  ft 1057 ft
464.6 R 


s
lb f  s 2

So our speed, V = c(Ma) = (1057 ft/s)(0.7625) = 805.6 ft/s.
We now have all the terms necessary to compute the mass flow rate at the throat.
m  VA 


0.001808 slug 805.6 ft
0.05 ft 2  0.0728 slug/s
3
s
ft
In the second case, the receiver pressure of 5 psia is less than p* so the throat pressure is p* =
7.763 psia and the throat Mach number is 1. We use the same equations as above to find the
density and temperature at the throat.
1 1 k 
 k 1

   0 1 
Ma 2 
2



1 11.4 
0.00238 slug  1.4  1 2 
1 
1 
2


ft 3
 k 1

T  T0 1 
Ma 2 
2


1
 1.4  1 2 
 518.67 R 1 
1 
2



0.001509 slug
ft 3
1
 432.2 R
Since the Mach number is one, the velocity at the throat is the sound speed at the temperature
just found. Once we find the velocity we can compute the mass flow rate at the throat.
 1716 ft  lb f
V  c  1.40
 slug  R
m  VA 

1 slug  ft 1019 ft
432.2 R 


2
s
lb

s

f


0.001509 slug 1019 ft
0.05 ft 2  0.0769 slug/s
3
s
ft
2. Water flows in a rectangular channel with a flowrate
per unit width of q = 1.5 m2/s and a depth of 0.5 m at
section (1). The head loss between sections (1) and
(2) is 0.03 m. Sketch the specific energy diagram for
this flow and indicate the relative location of states
(1) and (2) on this diagram. Is it possible to have a
head loss of 0.06 m? Explain.
We can apply the energy equation to the diagram shown
above. Here we use the surface of the liquid as the
points to apply the energy equation. This means that p1 = p2 = 0. We assume that the slope is
very small so that we can use the fluid depth as the elevation. This gives z1 = y1 = 0.5 ft and z2 =
y2. With these simplifications our energy equation becomes.
V22
p1 V12
z2 

z  
h h
 2g 1  2g s L
p2

V22
V12
y2 
 y1 
 hL
2g
2g
We replace the velocity, V, by the flow per unit width: V = Q/A = (qb)/(yb) = q/y. This gives the
following result in our energy equation. Note that Q = qb is constant as the flow changes depth.
If the width, b, is constant, then q = Q/b, is also constant. Thus we do not have to subscript q in
the equation below.
Unit twelve quiz solutions
ME 390, L. S. Caretto, Spring 2008
q y2   E  y  q y1   h  E  h
V22
 y2 
2
1
L
1
L
2g
2g
2g
2
y2 
Page 3
2
 E2  E1  hL
Although the specific energy changes in the flow as a result of head loss, the relationship
between specific energy and depth does not change. At any point in the flow we must have
2
2

q y
E  y
2g
 1.5 m 2 


s 
0.1147 m 3

 y
 y
9.80665 m 2
y2
2
y
s2
We can compute the values of E for various values of y and plot the results as shown in the figure
to the left. In particular, we know the value of the depth, y = 0.5 m at section (1) in the flow.
Thus, the specific energy at this point can be found as
0.1147 m3
0.1147 m3
E1  y1 
 0.5 m 
 0.959 m
y12
0.5 m2
Problem 10.22 Solution
This confirms the value that we
can read from the chart for y =
0.5 m. According to the energy
equation, the head loss means
that the energy at station 2 is
0.03 m less than the energy at
station (1). This gives E2 =
0.929 ft. This specific energy
corresponds to the following a
depth of 0.550 m. This is found
by solving the final equation
below, using a numerical root
finder to find y. Both points are
shown on the energy curve with
the value of their depts.
indicated
3
Depth, y, m
2.5
2
1.5
1
0.550
0.5
0.5
0
0.0
0.5
1.0
1.5
2.0
2.5
Specific Energy, E, m
3.0
E2  E1  0.03 m  0.929 m  y2 
0.1147 m3
 y23  0.929 y22  0.1147  0
2
y2
We see that the lower energy
corresponds to a higher depth with the same flow.
If we had a head loss of 0.06 m from station 1 the energy at station (2) would be 0.959 m –
0.06 m = 0.859 m. This is only possible if the minimum energy is less than 0.859 m. To check
this we can compute the minimum energy by first finding the critical depth for this energy, y c. and
then applying the formula that Emin = 1.5 yc.
Unit twelve quiz solutions
13
 q2 
yc   
 g 
  1.5 m 2  2 
 
 
 s  

9.80665 m 


s2


ME 390, L. S. Caretto, Spring 2008
Page 4
13
 0.6122 m
Emin 
3
3
yc  0.6122 m   0.9183 m
2
2
Since Emin = 0.9183 m we cannot have a head loss of 0.06 m because this would make the
specific energy less than its minimum value.