Chapter 12
Compressible Flow
Problem 12.1
Methane at 25oC ( = 518 J/kg/K, = 1 31) is owing in a pipe at 400 m/s.
Is the ow subsonic, sonic, supersonic, or hypersonic?
Solution
The speed of sound in methane is


= = 1 31 × 518 × 298
= 450 m/s
Because the velocity is less than the speed of sound, the ow is subsonic.
Problem 12.2
Air ( = 1716 ft-lbf/slug/oR, = 1 40) with a velocity of 1500 ft/s, a pressure
of 10 psia, and a temperature of 40oF passes through a normal shock wave. Find
the velocity, pressure, and temperature downstream of the shock wave.
Solution
First nd the upstream Mach number and then use relationships for normal shock
waves. The speed of sound is
p
= = 1 4 × 1716 × (460 + 40)
= 1096 ft/s
109
110 CHAPTER 12. COMPRESSIBLE FLOW
The upstream Mach number is
1500
1 = = = 1 37
1096
The Mach number behind the shock wave is
2
( 1) 1 + 2
2
2 =
2
2 1 ( 1)
0 4 × 1 372 + 2
=
2 × 1 4 × 1 372 0 4
= 0 567
2 = 0 753
The temperature ratio across the wave is
1 + 1
2 1 2
2
=
1 1 2
1 + 2
2
1 + 0 2 × 1 372
=
1 + 0 2 × 0 7532
= 1 24
Thus the temperature is
2 = 1 24 1
= 1 24 × 500
= 620oR = 160o
The pressure ratio is
2
2 1 + 1
=
2
1 1 + 2
1 + 1 4 × 1 372
=
1 + 1 4 × 0 7532
= 2 022
Thus the pressure is
2 = 2 022 1 = 2 022 × 10
= 20 22 psia
The speed of sound behind the shock wave is

2 = 1 4 × 1716 × 620 = 1220 ft/s
The velocity behind the shock wave is
2 = 2 2
= 0 753 × 1220
= 919 ft/s
111
Problem 12.3
Air ( = 287 J/kg/K, = 1 4) at 800 kPa and 20oC exhausts through a trun-
cated nozzle with an area of 0.6 cm2 to a back pressure of 100 kPa. Calculate the
ow rate.
Solution
First nd out if the exit condition is sonic or subsonic. The exit pressure for a
sonic nozzle would be

µ Å› 1
+ 1
= = 1 23 5 = 1 89
2
so
800
= = 423 kPa
1 89
Since the exit pressure is larger than the back pressure, the ow at the exit will be
sonic.
The ow rate is
=

where the conditions are evaluated at the exit (sonic condition). The exit temper-
ature is found from
+ 1
= = 1 2
2
Thus the exit temperature is
273 + 20
= = 244 K
1 2
The sonic velocity at this temperature is

= 1 4 × 287 × 244 = 313 m/s
The exit density is
423 × 103
= = = 6 04 kg/m3
287 × 244
The ow rate is
= 6 04 × 0 6 × 10 4 × 313

= 0 113 kg/s
112 CHAPTER 12. COMPRESSIBLE FLOW
Problem 12.4
A rocket nozzle is designed to expand exhaust gases ( = 300 J/kg/K, = 1 3)
from a chamber pressure of 600 kPa and total temperature of 3000 K to a Mach
number of 2.5 at the exit. The throat area is 0.1 m2 Find the area at the exit,
the exit pressure, the exit velocity, and the mass ow rate.
Solution
The relationship for the ratio of the nozzle area to the throat area is
+1
à !2( 1)
1
1 + 2
1
2
=
+1

2
For a Mach number of 2.5
µ Å›3 83
1 1 + 0 15 × 2 52
=
2 5 1 15
= 2 95
Thus the exit area is
= 2 95 × 0 1 = 0 295 m2
The exit pressure is obtained from

µ Å› 1
1
= 1 + 2
2
= (1 + 0 15 × 2 52)4 33
= 17 5
The exit pressure is
600
= = 34 3 kPa
17 5
The exit temperature is obtained from
1
= 1 + 2
2
= 1 + 0 15 × 2 52
= 1 94
The exit temperature is
3000
= = 1546 K
1 94
The speed of sound at the exit is
p

= = 1 3 × 300 × 1546
= 776 m/s
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so the exit velocity is
= = 2 5 × 776 = 1940 m/s
The density at the exit is
34 3 × 103
= = = 0 074 kg/m3
300 × 1546
The mass ow is
= = 0 074 × 1940 × 0 295

= 42 3 kg/s
Problem 12.5
An airplane is ying through air ( = 1716 ft-lbf/slug/oR, = 1 4) at 600 ft/s.
The pressure and temperature of the air are 14 psia and 50oF. What are the pres-
sure and temperature at the stagnation point? (Assume the stagnation process is
isentropic.)
Solution
The stagnation pressure corresponds to the total conditions if the process is isen-
tropic. The speed of sound in air at this condition is
p
= = 1 4 × 1716 × (460 + 50)
= 1107 ft/s
The Mach number is
600
= = = 0 542
1107
The total temperature is
1
= (1 + 2)
2
= 510 × (1 + 0 2 × 0 5422)
= 540oR =80oF
The total pressure is

µ Å› 1
1
= 1 + 2
2
= 14 × (1 + 0 2 × 0 5422)3 5
= 17 1 psia
114 CHAPTER 12. COMPRESSIBLE FLOW
Problem 12.6
Carbon dioxide ( = 189 J/kg/K, = 1 3) ows through a Laval nozzle. A
normal shock wave occurs in the expansion section where the Mach number is 2
and the cross-sectional area is 1 cm2 The exit area is 1.5 cm2 The total pressure
before the shock wave is 400 kPa. Find the area at the throat and the exit pressure.
Solution
The ratio of the nozzle area to the throat area is
+1
à !2( 1)
1
1 + 2
1
2
=
+1

2
µ Å›3 83
1 1 + 0 15 × 22
=
2 1 15
= 1 77
The throat area is
1
= = 0 56 cm2
1 77
After the normal shock wave, the ow will be subsonic. The Mach number just
downstream of the normal shock is
2
( 1) 1 + 2
2
2 =
2
2 1 ( 1)
0 3 × 22 + 2
=
2 × 1 3 × 22 0 3
= 0 317
2 = 0 563
The change in pressure across the shock wave is
2
2 1 + 1
=
2
1 1 + 2
1 + 1 3 × 22
=
1 + 1 3 × 0 5632
= 4 39
The change in total pressure across the normal shock wave is

à ! 1
1 + 2
2 2 1 2
2
=
1 1 1 2
1 + 1
2
µ Å›4 33
1 + 0 15 × 0 5632
= 4 39 ×
1 + 0 15 × 22
= 0 701
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The total pressure for the subsonic ow behind the shock wave is
2 = 400 × 0 701 = 280 kPa
The subsonic ow in the expansion section will now decrease in Mach number as
the area increases. The Mach number at the exit must be obtained from the
relationship between area-ratio and Mach number that was used previously. First,
we have to nd the area where the ow would become sonic for the conditions
downstream of the normal shock wave.
+1
à !2( 1)
1
1 + 2
1
2
=
+1

2
µ Å›3 83
1 1 + 0 15 × 0 5632
=
0 563 1 15
= 1 24
so the area where sonic ow would occur is
1
= = 0 806 cm2
1 24
and the area ratio of the exit is then
1 5
= = 1 86
0 806
We now have to nd the subsonic Mach number that corresponds to this area ratio.
µ Å›3 83
1 1 + 0 15 2
1 86 =
1 15
This equation must be solved iteratively. The result is 0.336. The exit pressure is
then
2
= (1 + 0 15 × 0 3362)4 33

= 1 075
So the exit pressure is
280
= = 260 kPa
1 075
116 CHAPTER 12. COMPRESSIBLE FLOW