12-Practice Problems.pdf

Chapter 12

Compressible Flow

Problem 12.1

Methane at 25 o C ( ! = 518 J/kg/K, " = 1#31 ) is ß owing in a pipe at 400 m/s.

Is the ß 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 ß ow is subsonic .

Problem 12.2

Air ( ! = 1716 ft-lbf/slug/ o R, " = 1#40) with a velocity of 1500 ft/s, a pressure

of 10 psia, and a temperature of 40 o F passes through a normal shock wave. Find

the velocity, pressure, and temperature downstream of the shock wave.

Solution

First Þ nd the upstream Mach number and then use relationships for normal shock

waves. The speed of sound is

$ =

"!% =

1#4×1716×(460+40)

= 1096 ft/s

109

110

CHAPTER 12. COMPRESSIBLE FLOW

1096 = 1#37

The Mach number behind the shock wave is

& 2 = (" " 1)& 1 +2

$ = 1500

"(" "1)

= 0#4×1#37 2 +2

2×1#4×1#37 2 "0#4

= 0#567

& 2 = 0#753

2"& 1

The temperature ratio across the wave is

% 2

% 1

= 1+ ! ! 2 & 1

1+ !! 2 & 2

= 1+0#2×1#37 2

1+0#2×0#753 2

= 1#24

Thus the temperature is

% 2 = 1#24% 1

= 1#24×500

= 620 o R = 160 o (

The pressure ratio is

) 2

) 1

= 1+"& 1

1+"& 2

= 1+1#4×1#37 2

1+1#4×0#753 2

= 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

The upstream Mach number is

& 1 = '

111

Problem 12.3

Air ( ! = 287 J/kg/K, " = 1#4) at 800 kPa and 20 o C exhausts through a trun-

cated nozzle with an area of 0.6 cm 2 to a back pressure of 100 kPa. Calculate the

ß ow rate.

Solution

First Þ nd out if the exit condition is sonic or subsonic. The exit pressure for a

sonic nozzle would be

) "

" +1

!!1

= 1#2 3#5 = 1#89

) " = 800

1#89 = 423 kPa

Since the exit pressure is larger than the back pressure, the ß ow at the exit will be

sonic.

The ß ow rate is

ú* = + " ,$ "

where the conditions are evaluated at the exit (sonic condition). The exit temper-

ature is found from

% "

= "+1

= 1#2

Thus the exit temperature is

% " = 273+20

1#2

= 244 K

The sonic velocity at this temperature is

$ " =

1#4×287×244 = 313 m/s

The exit density is

+ " = ) "

!% "

= 423×10 3

287×244 = 6#04 kg/m 3

The ß ow rate is

ú* = 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 m 2 # Find the area at the exit,

the exit pressure, the exit velocity, and the mass ß ow rate.

Solution

The relationship for the ratio of the nozzle area to the throat area is

, "

= 1

1+ ! ! 1

2 & 2

!+1

2(!!1)

For a Mach number of 2.5

, "

= 1

2#5

1+0#15×2#5 2

1#15

3#83

= 2#95

Thus the exit area is

, $ = 2#95×0#1 = 0#295 m 2

The exit pressure is obtained from

) "

) =

1+ " " 1

!!1

& 2

= (1+0#15×2#5 2 ) 4#33

= 17#5

The exit pressure is

) $ = 600

17#5 = 34#3 kPa

The exit temperature is obtained from

% = 1+ " " 1

& 2

= 1+0#15×2#5 2

= 1#94

The exit temperature is

% $ = 3000

1#94 = 1546 K

The speed of sound at the exit is

$ $ =

"!% $ =

1#3×300×1546

= 776 m/s

% "

113

so the exit velocity is

' $ = & $ $ $ = 2#5×776 = 1940 m/s

The density at the exit is

+ $ = ) $

!% $ = 34#3×10 3

300×1546 = 0#074 kg/m 3

The mass ß ow is

ú* = + $ ' $ , $ = 0#074×1940×0#295

= 42#3 kg/s

Problem 12.5

An airplane is ß ying through air ( ! = 1716 ft-lbf/slug/ o R, " = 1#4 ) at 600 ft/s.

The pressure and temperature of the air are 14 psia and 50 o F. 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

$ =

"!% =

1#4×1716×(460+50)

= 1107 ft/s

The Mach number is

& = '

$ = 600

1107 = 0#542

The total temperature is

% " = %(1+ " " 1

& 2 )

= 510×(1+0#2×0#542 2 )

= 540 o R = 80 o F

The total pressure is

1+ " " 1

!!1

) " = )

& 2

= 14×(1+0#2×0#542 2 ) 3#5

= 17#1 psia

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