College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
May 6 Compressible-Flow-Homework Solutions
11.67
The stagnation pressure ratio across a normal shock in an ideal gas flow is 0.6. Determine
the Mach number of the flow entering the shock if the gas is air.
For air k = 1.4 and we can use Figure D-4 to get the Mach number, Max, from the stagnation
pressure ratio, p0.y/p0.x. Alternatively we can solve equation 11.156 for p0.y/p0.x as a function of
Max by an iterative technique to find the value of Max that gives p0.y/p0.x = 0.6. The latter
approach gives Max = 2.26 for p0.y/p0.x = 0.6.
11.68
Just upstream of a normal shock in an ideal gas flow, Ma = 3.0, T = 600 R, and p = 30 psia.
Determine the values of Ma, T0, T, p0, p, and V downstream of the shock if the gas is (a) air;
(b) helium.
This solution will use equations for both air and helium. Note that for air we can use Figure D.4
for k = 1.4 to obtain these results for air, except for the stagnation temperature. For air, k = 1.4
and R = 1716 ft·lbf/slug·R; for helium, k = 1.66 and R = 1.242x104 ft·lbf/slug·R.
The downstream Mach number, May is found from equation 11.149.
For air with k = 1.4:
Ma y 
For helium with k = 1.66:
2
2
32 
k 1 
1.4  1  0.375
2
2
2Ma x
23
1
1
1.4  1
k 1
Ma x2 
Ma y 
 
2
2
32 
k 1 
1.66  1  0.521
2
2Ma x
2 32
1
1
1.66  1
k 1
Ma x2 
 
The value of T0 is constant across a shock. The upstream value of T 0 can be calculated from
equation 11.56 using the data given for T x and Max. This is the same as the downstream value.
This relationship for air can be found in Figure D.1, not D.4.)
 k 1

 1.4  1 2 
For air with k = 1.4: T0  T 1 
Ma 2   600 R 1 
3 
2
2




1
 1680 R
 k 1

 1.66  1 2 
For helium with k = 1.66:: T0  T 1 
Ma 2   600 R 1 
3 
2
2




1
 2382 R
The temperature ratio across a shock is found from equation 11.151; the results for air and
helium are shown below.
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
May 6 CMP homework solutions
 Ty

T
 x
 Ty

T
 x
ME 390, L. S. Caretto, Spring 2008
Page 2
 



 Air
2

  1.4  1 2  21.4 32

 k 1
2  2kMax
Ma x 
 1 1 
3 
 1
1 
 

2
2

 k  1
 1.4  1

  2.679


2
2 2
2
k  1 Ma x
1.4  1 3  1
1
21.4  1
2k  1



 He
2

  1.66  1 2  21.66 32

 k 1
2  2kMax
Ma x 
 1 1 
3 
 1
1 
 

2
2

 k  1
 1.66  1

  3.643


2
2 2
2
k  1 Ma x
1.66  1 3  1
1
21.66  1
2k  1
 
 
 
 
 
Multiplying these ratios by T x = 600 R gives Ty = 1607 R for air; Ty = 2186 R for helium
The pressure ratio across a shock is found from equation 11.151.
 
2kMax2  k  1
21.4 32  1.4  1
For air: p y  p x
 px
 30 psia 
 310 psia
px
k 1
1.4  1
py
 
2kMax2  k  1
21.66 32  1.66  1
For helium: p y  p x
 px
 30 psia 
 330 psia
px
k 1
1.66  1
py
We can find the stagnation pressure after the shock by using equation 11.59 for the ratio of
pressure to stagnation pressure. Note that this equation requires the use of the local
(downstream) Mach number. The results for air and helium are.
p0, y  p y
p0, y
p0, y  p y
p0, y
py
py
 k 1

 p y 1 
Ma 2 
2


k /( k 1)
 1.4  1

 310 psia 1 
0.375 2 
2


1.4 /(1.41)
 k 1

 p y 1 
Ma 2 
2


k /( k 1)
 1.66  1

 330 psia 1 
0.5212 
2


1.66 /(1.661)
For air, p0,y = 362 psia; for helium, p0,y = 409 psia
The speed is simply found as the product of the Mach number and the sound speed.
For air:
 1716 ft  lb f
V  cMa  Ma kRT  0.375 1.4
 slug  R

1 slug  ft
1607 R 
=934 ft/s

2
lb

s

f
 1.242 x10 4 ft  lb f
For helium: V  cMa  Ma kRT  0.521 1.66

slug  R

11.71

2186 R 1 slug  ft =3500 ft/s

lb f  s 2

An aircraft cruises at a Mach number of 2.0 at an altitude of 15 km. Inlet air is decelerated
to a Mach number of 0.4 at the engine compressor inlet. A normal shock occurs in the
inlet diffuser upstream of the compressor inlet at a section where the Mach number is 1.2.
May 6 CMP homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 3
For isentropic diffusion, except across the shock, and for standard atmosphere determine
the stagnation temperature and pressure of the air entering the engine compressor.
This is the problem of an isentropic flow from an inlet condition of Mach 2 to a shock at Mach 1.2,
followed by another isentropic flow to Mach 0.4. The entire process takes place in the diffuser of
the engine; the “inlet” for this flow is the diffuser inlet and the outlet is the diffuser outlet whidch is
the same as the compressor inlet.
The diffuser inlet conditions are standard air at an altitude of 15 km for which we can find the
following properties for the standard atmosphere in Table C.2: T = –56.50oC = 216.65 K and p =
1.211x104 N/m2 = 12.11 kPa.
Since the stagnation temperature is constant for both isentropic flows and across a normal shock,
the stagnation temperature at the entry to the compressor at (Ma = 0.4) is the same as the
stagnation temperature at the Ma = 2 inlet where T = 216.65 K. We can find this stagnation
temperature from equation 11.56 using k = 1.4 for air. (We could also use Figure D.1 for
isentropic flow to find T0.)
 k 1

 1.4  1 2 
T0  T 1 
Ma 2   216.65 K 1 
2 
2
2




1
 390 K
To find the stagnation pressure at the compressor inlet we have to analyze the three sections of
the flow: (1) isentropic to the shock, (2) across the shock, and (3) isentropic to the compressor
inlet. The isentropic flow has the same stagnation pressure so we can use equation 11.59 to
relate two different pressures in the same isentropic flow. In particular, the pressure just
upstream of the shock wave, p2, can be found from the inlet pressure, p1 = 12.11 kPa.
k /( k 1)
k /( k 1)
 k 1

 k 1

p0, upstream  p1 1 
Ma12 
 p 2 1 
Ma 22 

2
2




k /( k 1)
1.4 /(1.4 1)
1.4  1 2 
 k 1


 39.07 kPa
Ma12 
2 
1
 1
2
2


p2  p1 
 12.11 kPa
 1  k  1 Ma 2 
 1  1.4  11.2 2 



2 
2
2




We can use equation 11.151 for the pressure ratio across a shock to get p3, the start of the next
isentropic flow region.
p3  p2
py
px
 p2
 
2kMax2  k  1
21.4 1.2 2  1.4  1
 39.07 kPa
 59.13 kPa
k 1
1.4  1
We can find the Mach number, Ma3, downstream from the shock by using equation 11.149.
Ma3 
2
2
1.2 2 
k 1 
1.4  1  0.842
2
2
2Ma 2
2 1.2
1
1
1.4  1
k 1
Ma 22 
 
The stagnation pressure at this point, the start of the isentropic flow to the compressor inlet, will
be the same as the stagnation pressure at the end of the isentropic flow, the compressor inlet.
Thus we can use equation 11.59 that relates pressure to stagnation pressure to find the common
stagnation pressure for the isentropic flow, which will be the desired stagnation pressure at the
diffuser exit, which is the compressor inlet.
May 6 CMP homework solutions
 k 1

p0  p1 
Ma 2 
2


11.72
k /( k 1)
ME 390, L. S. Caretto, Spring 2008
 1.4  1

 59.13 kPa1 
0.842 2 
2


Page 4
k /( k 1)
 94.1 kPa
Determine, for the air flow through the frictionless and adiabatic converging-diverging
duct of Example 11.8, the ratio of duct exit pressure to duct inlet stagnation pressure that
will result in a standing normal shock at (a) x = +0.1 m; (b) x = +0.2 m; (c) x +0.4 m. How
large is the stagnation pressure loss in each case?
Example 11.8 computes the isentropic flow through a converging-diverging duct whose area is
given by the following equation: A = 0.1 + x2,
x (m)
A (m2) Ma
T/T0
p/p0
where x is in meters and A is in square meters.
+0.1
0.11
1.37
0.73
0.33
The entrance and exit of the duct are located at x
+0.2
0.14
1.76
0.62
0.18
= -0.5 m and x = +0.5 m, respectively. (The throat
+0.4
0.26
2.48
0.45
0.06
is at x = 0.) The table at the left shows the results
from Example 11.8 for the locations specified in this problem for a supersonic flow in the
diverging region.
For a standing shock at the various locations specified in the problem the upstream Mach number
into the shock is the Mach number at the particular location found in the isentropic calculation in
the table above. We start our calculations by using this Mach number to find the shock relations
from Figure D.4. At the location x = +0.1 m, the Example 11.8 table gives a Mach number of
1.37; using this value as Max = 1.37 in Figure D.4, we find the Mach number downstream from
the shock as May = 0.75 and the stagnation pressure ratio across the shock p0.y/p0.x = 0.96.
Downstream from the shock, the new isentropic flow has a new value of A*, the area at which a
given isentropic flow would become sonic. This new value of A* is used to characterize the
isentropic flow following the shock. From Figure D.1, we see that A/A* is 1.1 for the value of May
= 0.75. Since the area at this location is 0.11 m 2, the new value of A* = A/(A/A*) = (0.11 m 2)/1.1 =
0.1 m2. The exit area at x = 0.5 m is found from the area formula for this particular duct to be A exit
= 0.1 m2 + (0.5 m)2 = 0.35 m2. So, at the exit A/A* = (.35 m 2)/(0.1 m2) = 3.5. For this exit value of
A/A*, and subsonic flow, Figure D.1 gives a Mach number of 0.17 and a p/p0 value of 0.98 at the
exit.
To find the ratio of duct exit pressure, pexit, to inlet stagnation pressure, p0,in, we note that the
stagnation pressure in an isentropic flow is constant. Thus the inlet stagnation pressure is the
stagnation pressure just upstream of the shock and the exit stagnation pressure is the stagnation
pressure just downstream from the shock; we previously found the ratio of these two stagnation
pressures across the shock to be 0.96. So the ratio of exit pressure to inlet stagnation pressure
is found as follows:
p 0, y
p0,exit
pexit
p
p
 exit
 exit
 0.960.98  0.94
p0,in
p0,exit p0,in
p0,exit p0, x
To compute the loss of stagnation pressure across the shock we have to use the data from
Example 11.8 that the stagnation pressure upstream of the shock was 101 kPa. We can then
find the stagnation pressure loss as follows.
p0  p0, x  p0, y  p0, x  p0, x
p 0, y
p 0, x

p 0, y
 p 0, x 1 

p 0, x


  101 kPa1  0.96  4 kPa


We repeat the same calculations for x = +0.2 m, where the Example 11.8 table gives the Mach
number as 1.76. Entering Figure D.4 with Max = 1.76 gives May = 0.62 and p0,y/p0,x = 0.83.
For the isentropic flow downstream from the shock with Ma = 0.62 we find A/A* = 1.16 from
Figure D.1. The area at x = +0.2 m is 0.14 m 2, so the A* value for the isentropic flow to the exit is
May 6 CMP homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 5
(0.14 m2) / 1.16 = 0.1207 m2. At the exit area of 0.35 m 2, A/A* = (0.35 m2) / (0.1207 m2) = 2.9. At
this exit value of value of A/A*, we find (p/p0)exit = 0.97. We then find the ratio of exit pressure to
inlet stagnation pressure and the stagnation pressure loss as before.
p 0, y
p0,exit
pexit
p
p
 exit
 exit
 0.97 0.83  0.8
p0,in p0,exit p0,in
p0,exit p0, x
p0  p0, x  p0, y  p0, x  p0, x
p 0, y
p 0, x

p 0, y
 p 0 , x 1 

p 0, x


  101 kPa1  0.83  17 kPa


We repeat the same calculations for x = +0.4 m, where the Example 11.8 table gives the Mach
number as 2.48. Entering Figure D.4 with Max = 2.48 gives May = 0.515 and p0,y/p0,x = 0.51.
For the isentropic flow downstream from the shock with Ma = 0.515 we find A/A* = 1.3 from
Figure D.1. The area at x = +0.4 m is 0.26 m2, so the A* value for the isentropic flow to the exit is
(0.26 m2) / 1.3 = 0.2 m2. At the exit area of 0.35 m 2, A/A* = (0.35 m2) / (0.2 m2) = 1.75. At this
exit value of value of A/A*, we find (p/p0)exit = 0.92. We then find the ratio of exit pressure to inlet
stagnation pressure and the stagnation pressure loss as before.
p 0, y
p0,exit
pexit
p
p
 exit
 exit
 0.920.51  0.47
p0,in
p0,exit p0,in
p0,exit p0, x
p0  p0, x  p0, y  p0, x  p0, x
p 0, y
p 0, x

p 0, y
 p 0 , x 1 

p 0, x


  101 kPa1  0.51  50 kPa


We see that lower exit pressures create the standing shock wave closer to the exit of the duct;
this results in a stronger shock wave with a larger loss of stagnation pressure.
11.73
A normal shock is positioned in the diverging portion of a frictionless, adiabatic
converging-diverging air flow duct where the cross section area is 0.1 ft2 and the local
Mach number is 2.0. Upstream of the shock, p0 = 200 psia and T0 = 1200 R. If the duct exit
area is 0.15 ft2, determine the exit area temperature and pressure and the duct mass
flowrate.
We have to determine ratios across the shock and for the frictionless, adiabatic (isentropic) flow
following the shock. The downstream Mach number from the shock can be found from Figure
D.1 or from the following equation.
Ma y 
2
k 1 
2
Ma 2y 
2 Ma y
k 1
1
22 
2
1.4  1  0.577
2
 
22
1
1.4  1
The stagnation pressure ratio across the shock is found from Figure D.1 as p0,y/p0,x = 0.72. Since
we are given data on area, we have to determine the new A* for the isentropic flow following the
shock. At the value of May = 0.577 we can find the value of A/A* = 1.22 from Figure D.1. Since
the area at the shock location is 0.10 ft2, the new value of A* = (0.1 ft2)/1.22 = 0.0822 ft2. At the
exit area of 0.15 ft2, the value of A/A* = (0.15 ft2) / (0.0822 ft2) = 1.825, we can find the Mach
number from Figure D.1 to be Maexit = 0.34.
We know the stagnation temperature T0, which is constant for both isentropic flow and across a
shock; thus the exit stagnation temperature is the value of 1200 R upstream from the shock. We
can find the exit temperature from Figure D.1 or the equation below.
May 6 CMP homework solutions
Texit
ME 390, L. S. Caretto, Spring 2008
 k 1
2 
 T0 1 
Maexit

2


1
 1.4  1
0.342 
 1200 R 1 
2


Page 6
1
 1173 R
The stagnation pressure ratio across the shock can be found from Figure D.4 to be 0.72; thus the
stagnation pressure downstream from the shock is (200 psia)(0.72) = 144 psia. This stagnation
pressure is constant for the isentropic flow from the shock to the exit. We can use Figure D.1 to
find the ratio of pressure to stagnation pressure at the exit Mach number of 0.34: p/p0 = 0.92.
Thus the exit pressure is (0.92)(144 psia). Pexit = 132 psia .
  VA , where V = cMa =
The exit mass flow rate is found from the usual equation: m
Ma(kRT)1/2, and  = P/RT.
 1716 ft  lb f
V  cMa  Ma kRT  0.34 1.4
 slug  R

1 slug  ft 570 ft
1173 R 


s
lb f  s 2

132 lb f 144 in 2

0.00952 slug
P
in 2
ft 2


RT 1716 ft  lb f
ft 3
1173 R 
slug  R
m  VA 


0.00952 slug 570 ft
0.15 ft 2  0.81 slug/s
3
s
ft
Using equations instead of Figure D.1 and D.4 to solve the problem, including an iterative solution
of the equation to find the Mach number for a given value of A/A*, results in a final mass flow rate
of 0.814 slug/s.