Chapter 12.3 : Compressible Flow - Notes, Chemical, Engineering, Semester

# Chapter 12.3 : Compressible Flow - Notes, Chemical, Engineering, Semester Notes - Chemical Engineering

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Chapter 12 Compressible Flow

Review Problems

12-131 A leak develops in an automobile tire as a result of an accident. The initial mass flow rate of air
through the leak is to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow of air through the hole is isentropic.
Properties  The gas constant of air is R = 0.287 kPa ·m
3
/kg ·K. The specific heat ratio of air at room
temperature is k = 1.4.
Analysis The absolute pressure in the tire is
kPa 314 94 220
atm gage
= + = + = P P P
The critical pressure is, from Table 12-2,
kPa 94 > kPa 166 = kPa) 314 )( 5283 . 0 ( 5283 . 0 *
0
= = P P
Therefore, the flow is choked, and the velocity at the exit of the hole is the sonic speed. Then the flow
properties at the exit becomes

3
) 1 4 . 1 /( 1
3
) 1 /( 1
0
*
3
3
0
0
0
kg/m 327 . 2
1 4 . 1
2
) kg/m 671 . 3 (
1
2
kg/m 671 . 3
K) 298 )( K kg / m kPa 287 . 0 (
kPa 314
=
?
?
?
?
?
?
+
=
?
?
?
?
?
?
+
=
=
· ·
= =
- - k
k
RT
P
? ?
?

K 3 . 248 K) (298
1 4 . 1
2
1
2
0
*
=
+
=
+
= T
k
T
m/s 9 . 315 K) 3 . 248 (
kJ/kg 1
s / m 1000
K) kJ/kg 287 . 0 )( 4 . 1 (
2 2
*
=
?
?
?
?
?
?
?
?
· = = = kRT c V
Then the initial mass flow rate through the hole becomes
kg/min  0.554 = kg/s 0.00924 = m/s) /4](315.9 m) (0.004 )[ kg/m 327 . 2 (
2 3
p ? = = AV m &
Discussion The mass flow rate will decrease with time as the pressure inside the tire drops.

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution
permitted only to teachers and educators for course preparation.  If you are a student using this Manual,
you are using it without permission.
12-83
Page 2

Chapter 12 Compressible Flow

Review Problems

12-131 A leak develops in an automobile tire as a result of an accident. The initial mass flow rate of air
through the leak is to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow of air through the hole is isentropic.
Properties  The gas constant of air is R = 0.287 kPa ·m
3
/kg ·K. The specific heat ratio of air at room
temperature is k = 1.4.
Analysis The absolute pressure in the tire is
kPa 314 94 220
atm gage
= + = + = P P P
The critical pressure is, from Table 12-2,
kPa 94 > kPa 166 = kPa) 314 )( 5283 . 0 ( 5283 . 0 *
0
= = P P
Therefore, the flow is choked, and the velocity at the exit of the hole is the sonic speed. Then the flow
properties at the exit becomes

3
) 1 4 . 1 /( 1
3
) 1 /( 1
0
*
3
3
0
0
0
kg/m 327 . 2
1 4 . 1
2
) kg/m 671 . 3 (
1
2
kg/m 671 . 3
K) 298 )( K kg / m kPa 287 . 0 (
kPa 314
=
?
?
?
?
?
?
+
=
?
?
?
?
?
?
+
=
=
· ·
= =
- - k
k
RT
P
? ?
?

K 3 . 248 K) (298
1 4 . 1
2
1
2
0
*
=
+
=
+
= T
k
T
m/s 9 . 315 K) 3 . 248 (
kJ/kg 1
s / m 1000
K) kJ/kg 287 . 0 )( 4 . 1 (
2 2
*
=
?
?
?
?
?
?
?
?
· = = = kRT c V
Then the initial mass flow rate through the hole becomes
kg/min  0.554 = kg/s 0.00924 = m/s) /4](315.9 m) (0.004 )[ kg/m 327 . 2 (
2 3
p ? = = AV m &
Discussion The mass flow rate will decrease with time as the pressure inside the tire drops.

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution
permitted only to teachers and educators for course preparation.  If you are a student using this Manual,
you are using it without permission.
12-83
Chapter 12 Compressible Flow
12-132 The thrust developed by the engine of a Boeing 777 is about 380 kN. The mass flow rate of gases
through the nozzle is to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow of combustion gases through the
nozzle is isentropic. 3 Choked flow conditions exist at the nozzle exit. 4 The velocity of gases at the nozzle
inlet is negligible.
Properties  The gas constant of air is R = 0.287 kPa ·m
3
/kg ·K, and it can also be used for combustion gases.
The specific heat ratio of combustion gases is k = 1.33.
Analysis The velocity at the nozzle exit is the sonic speed, which is determined to be
m/s 6 . 335 K) 295 (
kJ/kg 1
s / m 1000
K) kJ/kg 287 . 0 )( 33 . 1 (
2 2
=
?
?
?
?
?
?
?
?
· = = = kRT c V
Noting that thrust F is related to velocity by V m F & = , the mass flow rate of combustion gases is determined
to be
kg/s  1132 =
N 1
kg.m/s 1
m/s 335.6
N 000 , 380
2
?
?
?
?
?
?
?
?
= =
V
F
m &
Discussion The combustion gases are mostly nitrogen (due to the 78% of N
2
in air), and thus they can be
treated as air with a good degree of approximation.

12-133 A stationary temperature probe is inserted into an air duct reads 85 °C. The actual temperature of air
is to be determined.
Assumptions 1 Air is an ideal gas with constant specific heats at room temperature. 2 The stagnation
process is isentropic.
Properties The specific heat of air at room temperature is c
p
= 1.005 kJ/kg ·K.
Analysis The air that strikes the probe will be brought to a complete stop, and thus it will undergo a
stagnation process. The thermometer will sense the temperature of this stagnated air, which is the
stagnation temperature. The actual air temperature is determined from
T
250 m/s
C 53.9 ° = ?
?
?
?
?
?
· ×
- ° = - =
2 2
2 2
0
s / m 1000
kJ/kg 1
K kJ/kg 005 . 1 2
m/s) (250
C 85
2
p
c
V
T T
Discussion Temperature rise due to stagnation is very significant in high-speed flows, and should always
be considered when compressibility effects are not negligible.

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc.  Limited distribution
permitted only to teachers and educators for course preparation.  If you are a student using this Manual,
you are using it without permission.
12-84
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