Chapter 12.2 : Compressible Flow - Notes,Chemical, Engineering, Semester Chemical Engineering Notes | EduRev

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Chemical Engineering : Chapter 12.2 : Compressible Flow - Notes,Chemical, Engineering, Semester Chemical Engineering Notes | EduRev

 Page 1


Chapter 12 Compressible Flow 
 
Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) 
 
12-96C The characteristic aspect of Rayleigh flow is its involvement of heat transfer. The main 
assumptions associated with Rayleigh flow are: the flow is steady, one-dimensional, and frictionless 
through a constant-area duct, and the fluid is an ideal gas with constant specific heats. 
 
112-97C The points on the Rayleigh line represent the states that satisfy the conservation of mass, 
momentum, and energy equations as well as the property relations for a given state. Therefore, for a given 
inlet state, the fluid cannot exist at any downstream state outside the Rayleigh line on a T-s diagram. 
 
12-98C In Rayleigh flow, the effect of heat gain is to increase the entropy of the fluid, and the effect of 
heat loss is to decrease it. 
 
12-99C In Rayleigh flow, the stagnation temperature T
0
 always increases with heat transfer to the fluid, but 
the temperature T decreases with heat transfer in the Mach number range of 0.845 <  Ma < 1 for air. 
Therefore, the temperature in this case will decrease. 
 
12-100C Heating the fluid increases the flow velocity in subsonic flow, but decreases the flow velocity in 
supersonic flow. 
 
12-101C The flow is choked, and thus the flow at the duct exit will remain sonic. 
 
 
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.   
12-56
Page 2


Chapter 12 Compressible Flow 
 
Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) 
 
12-96C The characteristic aspect of Rayleigh flow is its involvement of heat transfer. The main 
assumptions associated with Rayleigh flow are: the flow is steady, one-dimensional, and frictionless 
through a constant-area duct, and the fluid is an ideal gas with constant specific heats. 
 
112-97C The points on the Rayleigh line represent the states that satisfy the conservation of mass, 
momentum, and energy equations as well as the property relations for a given state. Therefore, for a given 
inlet state, the fluid cannot exist at any downstream state outside the Rayleigh line on a T-s diagram. 
 
12-98C In Rayleigh flow, the effect of heat gain is to increase the entropy of the fluid, and the effect of 
heat loss is to decrease it. 
 
12-99C In Rayleigh flow, the stagnation temperature T
0
 always increases with heat transfer to the fluid, but 
the temperature T decreases with heat transfer in the Mach number range of 0.845 <  Ma < 1 for air. 
Therefore, the temperature in this case will decrease. 
 
12-100C Heating the fluid increases the flow velocity in subsonic flow, but decreases the flow velocity in 
supersonic flow. 
 
12-101C The flow is choked, and thus the flow at the duct exit will remain sonic. 
 
 
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.   
12-56
Chapter 12 Compressible Flow 
12-102 Fuel is burned in a tubular combustion chamber with compressed air. For a specified exit Mach 
number, the exit temperature and the rate of fuel consumption are to be determined.  
Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an 
ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional 
effects) are valid. 2 Combustion is complete, and it is treated as a heat addition process, with no change in 
the chemical composition of flow. 3 The increase in mass flow rate due to fuel injection is disregarded.   
Properties We take the properties of air to be k = 1.4, c
p
 = 1.005 kJ/kg ·K, and R = 0.287 kJ/kg ·K.    
Analysis The inlet density and mass flow rate of air are  
3
1
1
1
 kg/m 787 . 2
K) 00 kJ/kgK)(5 (0.287
 kPa 400
= = =
RT
P
? 
   kg/s 207 . 2 m/s) 70 ]( 4 / m) (0.12 )[ kg/m 787 . 2 (
2 3
1 1 1
= = = p ? V A m
c air
&
The stagnation temperature and Mach number at the inlet are  
K 4 . 502
/s m 1000
 kJ/kg 1
K kJ/kg 005 . 1 2
m/s) 70 (
K 500
2
2 2
2 2
1
1 01
= ?
?
?
?
?
?
· ×
+ = + =
p
c
V
T T 
Q
&
COMBUSTOR 
TUBE 
P
1
 = 400 kPa 
T
1
 = 500 K 
 
V
1
 = 70 m/s 
T
2
, V
2
 
 
 m/s 2 . 448
 kJ/kg 1
s / m 1000
K) K)(500 kJ/kg 287 . 0 )( 4 . 1 (
2 2
1 1
=
?
?
?
?
?
?
?
?
· = = kRT c 
 1562 . 0
m/s 2 . 448
m/s 70
Ma
1
1
1
= = =
c
V
 
The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-15): 
Ma
1
 = 0.1562:      T
1
/T
*
  = 0.1314,       T
01
/T
*
 = 0.1100,    V
1
/V
*
 = 0.0566    
Ma
2
 = 0.8:       T
2
/T
*
  = 1.0255,       T
02
/T
*
  = 0.9639, V
2
/V
*
 = 0.8101    
The exit temperature, stagnation temperature, and velocity are determined to be 
804 . 7
1314 . 0
0255 . 1
 
/
/
 
*
1
*
2
1
2
= = =
T T
T T
T
T
 ?  K 3903 = = = ) K 500 ( 804 . 7 804 . 7 
1 2
T T 
763 . 8
1100 . 0
9639 . 0
 
/
/
 
*
01
*
02
1 0
2 0
= = =
T T
T T
T
T
 ?  K 4403 ) K 4 . 502 ( 763 . 8 763 . 8 
01 2 0
= = = T T 
31 . 14
0566 . 0
8101 . 0
 
* /
* /
1
2
1
2
= = =
V V
V V
V
V
 ?    m/s 1002 ) m/s 70 ( 31 . 14 31 . 14
1 2
= = = V V 
Then the mass flow rate of the fuel is determined to be 
 kJ/kg 3920 K ) 4 . 502 4403 )( K kJ/kg 1.005 ( ) (
01 02
= - · = - = T T c q
p
 
 kW 8650 ) kJ/kg 3920 )(  kg/s 2.207 (
air
= = = q m Q &
&
 
kg/s 0.222 = = =
 kJ/kg 39,000
 kJ/s 8650
fuel
HV
Q
m
&
& 
Discussion Note that both the temperature and velocity increase during this subsonic Rayleigh flow with 
heating, as expected. This problem can also be solved using appropriate relations instead of tabulated 
values, which can likewise be coded for convenient computer solutions.  
 
 
 
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.   
12-57
Page 3


Chapter 12 Compressible Flow 
 
Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) 
 
12-96C The characteristic aspect of Rayleigh flow is its involvement of heat transfer. The main 
assumptions associated with Rayleigh flow are: the flow is steady, one-dimensional, and frictionless 
through a constant-area duct, and the fluid is an ideal gas with constant specific heats. 
 
112-97C The points on the Rayleigh line represent the states that satisfy the conservation of mass, 
momentum, and energy equations as well as the property relations for a given state. Therefore, for a given 
inlet state, the fluid cannot exist at any downstream state outside the Rayleigh line on a T-s diagram. 
 
12-98C In Rayleigh flow, the effect of heat gain is to increase the entropy of the fluid, and the effect of 
heat loss is to decrease it. 
 
12-99C In Rayleigh flow, the stagnation temperature T
0
 always increases with heat transfer to the fluid, but 
the temperature T decreases with heat transfer in the Mach number range of 0.845 <  Ma < 1 for air. 
Therefore, the temperature in this case will decrease. 
 
12-100C Heating the fluid increases the flow velocity in subsonic flow, but decreases the flow velocity in 
supersonic flow. 
 
12-101C The flow is choked, and thus the flow at the duct exit will remain sonic. 
 
 
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.   
12-56
Chapter 12 Compressible Flow 
12-102 Fuel is burned in a tubular combustion chamber with compressed air. For a specified exit Mach 
number, the exit temperature and the rate of fuel consumption are to be determined.  
Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an 
ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional 
effects) are valid. 2 Combustion is complete, and it is treated as a heat addition process, with no change in 
the chemical composition of flow. 3 The increase in mass flow rate due to fuel injection is disregarded.   
Properties We take the properties of air to be k = 1.4, c
p
 = 1.005 kJ/kg ·K, and R = 0.287 kJ/kg ·K.    
Analysis The inlet density and mass flow rate of air are  
3
1
1
1
 kg/m 787 . 2
K) 00 kJ/kgK)(5 (0.287
 kPa 400
= = =
RT
P
? 
   kg/s 207 . 2 m/s) 70 ]( 4 / m) (0.12 )[ kg/m 787 . 2 (
2 3
1 1 1
= = = p ? V A m
c air
&
The stagnation temperature and Mach number at the inlet are  
K 4 . 502
/s m 1000
 kJ/kg 1
K kJ/kg 005 . 1 2
m/s) 70 (
K 500
2
2 2
2 2
1
1 01
= ?
?
?
?
?
?
· ×
+ = + =
p
c
V
T T 
Q
&
COMBUSTOR 
TUBE 
P
1
 = 400 kPa 
T
1
 = 500 K 
 
V
1
 = 70 m/s 
T
2
, V
2
 
 
 m/s 2 . 448
 kJ/kg 1
s / m 1000
K) K)(500 kJ/kg 287 . 0 )( 4 . 1 (
2 2
1 1
=
?
?
?
?
?
?
?
?
· = = kRT c 
 1562 . 0
m/s 2 . 448
m/s 70
Ma
1
1
1
= = =
c
V
 
The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-15): 
Ma
1
 = 0.1562:      T
1
/T
*
  = 0.1314,       T
01
/T
*
 = 0.1100,    V
1
/V
*
 = 0.0566    
Ma
2
 = 0.8:       T
2
/T
*
  = 1.0255,       T
02
/T
*
  = 0.9639, V
2
/V
*
 = 0.8101    
The exit temperature, stagnation temperature, and velocity are determined to be 
804 . 7
1314 . 0
0255 . 1
 
/
/
 
*
1
*
2
1
2
= = =
T T
T T
T
T
 ?  K 3903 = = = ) K 500 ( 804 . 7 804 . 7 
1 2
T T 
763 . 8
1100 . 0
9639 . 0
 
/
/
 
*
01
*
02
1 0
2 0
= = =
T T
T T
T
T
 ?  K 4403 ) K 4 . 502 ( 763 . 8 763 . 8 
01 2 0
= = = T T 
31 . 14
0566 . 0
8101 . 0
 
* /
* /
1
2
1
2
= = =
V V
V V
V
V
 ?    m/s 1002 ) m/s 70 ( 31 . 14 31 . 14
1 2
= = = V V 
Then the mass flow rate of the fuel is determined to be 
 kJ/kg 3920 K ) 4 . 502 4403 )( K kJ/kg 1.005 ( ) (
01 02
= - · = - = T T c q
p
 
 kW 8650 ) kJ/kg 3920 )(  kg/s 2.207 (
air
= = = q m Q &
&
 
kg/s 0.222 = = =
 kJ/kg 39,000
 kJ/s 8650
fuel
HV
Q
m
&
& 
Discussion Note that both the temperature and velocity increase during this subsonic Rayleigh flow with 
heating, as expected. This problem can also be solved using appropriate relations instead of tabulated 
values, which can likewise be coded for convenient computer solutions.  
 
 
 
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.   
12-57
Chapter 12 Compressible Flow 
12-103 Fuel is burned in a rectangular duct with compressed air. For specified heat transfer, the exit 
temperature and Mach number are to be determined.  
Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal 
gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) 
are valid.   
Properties We take the properties of air to be k = 1.4, c
p
 = 1.005 kJ/kg ·K, and R = 0.287 kJ/kg ·K.    
Analysis The stagnation temperature and Mach number at the inlet are 
 m/s 2 . 347
 kJ/kg 1
s / m 1000
K) K)(300 kJ/kg 287 . 0 )( 4 . 1 (
2 2
1 1
=
?
?
?
?
?
?
?
?
· = = kRT c 
  m/s 4 . 694 m/s) 2 . 347 ( 2 Ma
1 1 1
= = = c V
K 9 . 539
/s m 1000
 kJ/kg 1
K kJ/kg 005 . 1 2
m/s) 4 . 694 (
K 300
2
2 2
2 2
1
1 01
= ?
?
?
?
?
?
· ×
+ = + =
p
c
V
T T 
The exit stagnation temperature is, from the energy equation , ) (
01 02
T T c q
p
- =
T
2
, Ma
2 
 
P
1
 = 420 kPa 
T
1
 = 300 K 
 
Ma
1
 = 2
 
q = 55 kJ/kg
 
K 6 . 594
K kJ/kg 1.005
 kJ/kg 55
K 539.9 
01 02
=
·
+ = + =
p
c
q
T T 
The maximum value of stagnation temperature T
0
*
 occurs at Ma = 1, and its value can be determined from 
Table A-15 or from the appropriate relation. At Ma
1
 = 2 we read T
01
/T
0
*
 = 0.7934. Therefore, 
K 5 . 680
7934 . 0
K 9 . 539
7934 . 0
 
01 *
0
= = =
T
T 
The stagnation temperature ratio at the exit and the Mach number corresponding to it are, from Table A-15,     
8738 . 0
K 5 . 680
K 6 . 594
 
*
0
02
= =
T
T
      ?      Ma
2
 = 1.642  
Also,  
Ma
1
 = 2  ?  T
1
/T
*
  = 0.5289    
 Ma
2
 = 1.642 ?  T
2
/T
*
  = 0.6812    
Then the exit temperature becomes 
288 . 1
5289 . 0
6812 . 0
 
/
/
 
*
1
*
2
1
2
= = =
T T
T T
T
T
   ?  K 386 = = = ) K 300 ( 288 . 1 288 . 1 
1 2
T T 
Discussion Note that the temperature increases during this supersonic Rayleigh flow with heating. This 
problem can also be solved using appropriate relations instead of tabulated values, which can likewise be 
coded for convenient computer solutions.  
 
 
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.   
12-58
Page 4


Chapter 12 Compressible Flow 
 
Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) 
 
12-96C The characteristic aspect of Rayleigh flow is its involvement of heat transfer. The main 
assumptions associated with Rayleigh flow are: the flow is steady, one-dimensional, and frictionless 
through a constant-area duct, and the fluid is an ideal gas with constant specific heats. 
 
112-97C The points on the Rayleigh line represent the states that satisfy the conservation of mass, 
momentum, and energy equations as well as the property relations for a given state. Therefore, for a given 
inlet state, the fluid cannot exist at any downstream state outside the Rayleigh line on a T-s diagram. 
 
12-98C In Rayleigh flow, the effect of heat gain is to increase the entropy of the fluid, and the effect of 
heat loss is to decrease it. 
 
12-99C In Rayleigh flow, the stagnation temperature T
0
 always increases with heat transfer to the fluid, but 
the temperature T decreases with heat transfer in the Mach number range of 0.845 <  Ma < 1 for air. 
Therefore, the temperature in this case will decrease. 
 
12-100C Heating the fluid increases the flow velocity in subsonic flow, but decreases the flow velocity in 
supersonic flow. 
 
12-101C The flow is choked, and thus the flow at the duct exit will remain sonic. 
 
 
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.   
12-56
Chapter 12 Compressible Flow 
12-102 Fuel is burned in a tubular combustion chamber with compressed air. For a specified exit Mach 
number, the exit temperature and the rate of fuel consumption are to be determined.  
Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an 
ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional 
effects) are valid. 2 Combustion is complete, and it is treated as a heat addition process, with no change in 
the chemical composition of flow. 3 The increase in mass flow rate due to fuel injection is disregarded.   
Properties We take the properties of air to be k = 1.4, c
p
 = 1.005 kJ/kg ·K, and R = 0.287 kJ/kg ·K.    
Analysis The inlet density and mass flow rate of air are  
3
1
1
1
 kg/m 787 . 2
K) 00 kJ/kgK)(5 (0.287
 kPa 400
= = =
RT
P
? 
   kg/s 207 . 2 m/s) 70 ]( 4 / m) (0.12 )[ kg/m 787 . 2 (
2 3
1 1 1
= = = p ? V A m
c air
&
The stagnation temperature and Mach number at the inlet are  
K 4 . 502
/s m 1000
 kJ/kg 1
K kJ/kg 005 . 1 2
m/s) 70 (
K 500
2
2 2
2 2
1
1 01
= ?
?
?
?
?
?
· ×
+ = + =
p
c
V
T T 
Q
&
COMBUSTOR 
TUBE 
P
1
 = 400 kPa 
T
1
 = 500 K 
 
V
1
 = 70 m/s 
T
2
, V
2
 
 
 m/s 2 . 448
 kJ/kg 1
s / m 1000
K) K)(500 kJ/kg 287 . 0 )( 4 . 1 (
2 2
1 1
=
?
?
?
?
?
?
?
?
· = = kRT c 
 1562 . 0
m/s 2 . 448
m/s 70
Ma
1
1
1
= = =
c
V
 
The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-15): 
Ma
1
 = 0.1562:      T
1
/T
*
  = 0.1314,       T
01
/T
*
 = 0.1100,    V
1
/V
*
 = 0.0566    
Ma
2
 = 0.8:       T
2
/T
*
  = 1.0255,       T
02
/T
*
  = 0.9639, V
2
/V
*
 = 0.8101    
The exit temperature, stagnation temperature, and velocity are determined to be 
804 . 7
1314 . 0
0255 . 1
 
/
/
 
*
1
*
2
1
2
= = =
T T
T T
T
T
 ?  K 3903 = = = ) K 500 ( 804 . 7 804 . 7 
1 2
T T 
763 . 8
1100 . 0
9639 . 0
 
/
/
 
*
01
*
02
1 0
2 0
= = =
T T
T T
T
T
 ?  K 4403 ) K 4 . 502 ( 763 . 8 763 . 8 
01 2 0
= = = T T 
31 . 14
0566 . 0
8101 . 0
 
* /
* /
1
2
1
2
= = =
V V
V V
V
V
 ?    m/s 1002 ) m/s 70 ( 31 . 14 31 . 14
1 2
= = = V V 
Then the mass flow rate of the fuel is determined to be 
 kJ/kg 3920 K ) 4 . 502 4403 )( K kJ/kg 1.005 ( ) (
01 02
= - · = - = T T c q
p
 
 kW 8650 ) kJ/kg 3920 )(  kg/s 2.207 (
air
= = = q m Q &
&
 
kg/s 0.222 = = =
 kJ/kg 39,000
 kJ/s 8650
fuel
HV
Q
m
&
& 
Discussion Note that both the temperature and velocity increase during this subsonic Rayleigh flow with 
heating, as expected. This problem can also be solved using appropriate relations instead of tabulated 
values, which can likewise be coded for convenient computer solutions.  
 
 
 
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.   
12-57
Chapter 12 Compressible Flow 
12-103 Fuel is burned in a rectangular duct with compressed air. For specified heat transfer, the exit 
temperature and Mach number are to be determined.  
Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal 
gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) 
are valid.   
Properties We take the properties of air to be k = 1.4, c
p
 = 1.005 kJ/kg ·K, and R = 0.287 kJ/kg ·K.    
Analysis The stagnation temperature and Mach number at the inlet are 
 m/s 2 . 347
 kJ/kg 1
s / m 1000
K) K)(300 kJ/kg 287 . 0 )( 4 . 1 (
2 2
1 1
=
?
?
?
?
?
?
?
?
· = = kRT c 
  m/s 4 . 694 m/s) 2 . 347 ( 2 Ma
1 1 1
= = = c V
K 9 . 539
/s m 1000
 kJ/kg 1
K kJ/kg 005 . 1 2
m/s) 4 . 694 (
K 300
2
2 2
2 2
1
1 01
= ?
?
?
?
?
?
· ×
+ = + =
p
c
V
T T 
The exit stagnation temperature is, from the energy equation , ) (
01 02
T T c q
p
- =
T
2
, Ma
2 
 
P
1
 = 420 kPa 
T
1
 = 300 K 
 
Ma
1
 = 2
 
q = 55 kJ/kg
 
K 6 . 594
K kJ/kg 1.005
 kJ/kg 55
K 539.9 
01 02
=
·
+ = + =
p
c
q
T T 
The maximum value of stagnation temperature T
0
*
 occurs at Ma = 1, and its value can be determined from 
Table A-15 or from the appropriate relation. At Ma
1
 = 2 we read T
01
/T
0
*
 = 0.7934. Therefore, 
K 5 . 680
7934 . 0
K 9 . 539
7934 . 0
 
01 *
0
= = =
T
T 
The stagnation temperature ratio at the exit and the Mach number corresponding to it are, from Table A-15,     
8738 . 0
K 5 . 680
K 6 . 594
 
*
0
02
= =
T
T
      ?      Ma
2
 = 1.642  
Also,  
Ma
1
 = 2  ?  T
1
/T
*
  = 0.5289    
 Ma
2
 = 1.642 ?  T
2
/T
*
  = 0.6812    
Then the exit temperature becomes 
288 . 1
5289 . 0
6812 . 0
 
/
/
 
*
1
*
2
1
2
= = =
T T
T T
T
T
   ?  K 386 = = = ) K 300 ( 288 . 1 288 . 1 
1 2
T T 
Discussion Note that the temperature increases during this supersonic Rayleigh flow with heating. This 
problem can also be solved using appropriate relations instead of tabulated values, which can likewise be 
coded for convenient computer solutions.  
 
 
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.   
12-58
Chapter 12 Compressible Flow 
12-104 Compressed air is cooled as it flows in a rectangular duct. For specified heat rejection, the exit 
temperature and Mach number are to be determined.  
Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal 
gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) 
are valid. 
Properties We take the properties of air to be k = 1.4, C
p
 = 1.005 kJ/kg ·K, and R = 0.287 kJ/kg ·K.    
Analysis The stagnation temperature and Mach number at the inlet are 
 m/s 2 . 347
 kJ/kg 1
s / m 1000
K) K)(300 kJ/kg 287 . 0 )( 4 . 1 (
2 2
1 1
=
?
?
?
?
?
?
?
?
· = = kRT c 
  m/s 4 . 694 m/s) 2 . 347 ( 2 Ma
1 1 1
= = = c V
K 9 . 539
/s m 1000
 kJ/kg 1
K kJ/kg 005 . 1 2
m/s) 4 . 694 (
K 300
2
2 2
2 2
1
1 01
= ?
?
?
?
?
?
· ×
+ = + =
p
c
V
T T 
The exit stagnation temperature is, from the energy equation , ) (
01 02
T T c q
p
- =
 
P
1
 = 420 kPa 
T
1
 = 300 K 
 
Ma
1
 = 2
 
q = -55 kJ/kg
 
T
2
, Ma
2 
K 2 . 485
K kJ/kg 1.005
 kJ/kg -55
K 539.9 
01 02
=
·
+ = + =
p
c
q
T T 
The maximum value of stagnation temperature T
0
*
 occurs at Ma = 1, and its value can be determined from 
Table A-15 or from the appropriate relation. At Ma
1
 = 2 we read T
01
/T
0
*
 = 0.7934. Therefore, 
K 5 . 680
7934 . 0
K 9 . 539
7934 . 0
 
01 *
0
= = =
T
T 
The stagnation temperature ratio at the exit and the Mach number corresponding to it are, from Table A-15,     
7130 . 0
K 5 . 680
K 2 . 485
 
*
0
02
= =
T
T
      ?      Ma
2
 = 2.479  
Also,  
Ma
1
 = 2  ?  T
1
/T
*
  = 0.5289    
 Ma
2
 = 2.479 ?  T
2
/T
*
  = 0.3838    
Then the exit temperature becomes 
7257 . 0
5289 . 0
3838 . 0
 
/
/
 
*
1
*
2
1
2
= = =
T T
T T
T
T
   ?  K 218 = = = ) K 300 ( 7257 . 0 7257 . 0 
1 2
T T 
Discussion Note that the temperature decreases and Mach number increases during this supersonic 
Rayleigh flow with cooling. This problem can also be solved using appropriate relations instead of 
tabulated values, which can likewise be coded for convenient computer solutions.  
 
 
 
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.   
12-59
Page 5


Chapter 12 Compressible Flow 
 
Duct Flow with Heat Transfer and Negligible Friction (Rayleigh Flow) 
 
12-96C The characteristic aspect of Rayleigh flow is its involvement of heat transfer. The main 
assumptions associated with Rayleigh flow are: the flow is steady, one-dimensional, and frictionless 
through a constant-area duct, and the fluid is an ideal gas with constant specific heats. 
 
112-97C The points on the Rayleigh line represent the states that satisfy the conservation of mass, 
momentum, and energy equations as well as the property relations for a given state. Therefore, for a given 
inlet state, the fluid cannot exist at any downstream state outside the Rayleigh line on a T-s diagram. 
 
12-98C In Rayleigh flow, the effect of heat gain is to increase the entropy of the fluid, and the effect of 
heat loss is to decrease it. 
 
12-99C In Rayleigh flow, the stagnation temperature T
0
 always increases with heat transfer to the fluid, but 
the temperature T decreases with heat transfer in the Mach number range of 0.845 <  Ma < 1 for air. 
Therefore, the temperature in this case will decrease. 
 
12-100C Heating the fluid increases the flow velocity in subsonic flow, but decreases the flow velocity in 
supersonic flow. 
 
12-101C The flow is choked, and thus the flow at the duct exit will remain sonic. 
 
 
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.   
12-56
Chapter 12 Compressible Flow 
12-102 Fuel is burned in a tubular combustion chamber with compressed air. For a specified exit Mach 
number, the exit temperature and the rate of fuel consumption are to be determined.  
Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an 
ideal gas with constant properties through a constant cross-sectional area duct with negligible frictional 
effects) are valid. 2 Combustion is complete, and it is treated as a heat addition process, with no change in 
the chemical composition of flow. 3 The increase in mass flow rate due to fuel injection is disregarded.   
Properties We take the properties of air to be k = 1.4, c
p
 = 1.005 kJ/kg ·K, and R = 0.287 kJ/kg ·K.    
Analysis The inlet density and mass flow rate of air are  
3
1
1
1
 kg/m 787 . 2
K) 00 kJ/kgK)(5 (0.287
 kPa 400
= = =
RT
P
? 
   kg/s 207 . 2 m/s) 70 ]( 4 / m) (0.12 )[ kg/m 787 . 2 (
2 3
1 1 1
= = = p ? V A m
c air
&
The stagnation temperature and Mach number at the inlet are  
K 4 . 502
/s m 1000
 kJ/kg 1
K kJ/kg 005 . 1 2
m/s) 70 (
K 500
2
2 2
2 2
1
1 01
= ?
?
?
?
?
?
· ×
+ = + =
p
c
V
T T 
Q
&
COMBUSTOR 
TUBE 
P
1
 = 400 kPa 
T
1
 = 500 K 
 
V
1
 = 70 m/s 
T
2
, V
2
 
 
 m/s 2 . 448
 kJ/kg 1
s / m 1000
K) K)(500 kJ/kg 287 . 0 )( 4 . 1 (
2 2
1 1
=
?
?
?
?
?
?
?
?
· = = kRT c 
 1562 . 0
m/s 2 . 448
m/s 70
Ma
1
1
1
= = =
c
V
 
The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-15): 
Ma
1
 = 0.1562:      T
1
/T
*
  = 0.1314,       T
01
/T
*
 = 0.1100,    V
1
/V
*
 = 0.0566    
Ma
2
 = 0.8:       T
2
/T
*
  = 1.0255,       T
02
/T
*
  = 0.9639, V
2
/V
*
 = 0.8101    
The exit temperature, stagnation temperature, and velocity are determined to be 
804 . 7
1314 . 0
0255 . 1
 
/
/
 
*
1
*
2
1
2
= = =
T T
T T
T
T
 ?  K 3903 = = = ) K 500 ( 804 . 7 804 . 7 
1 2
T T 
763 . 8
1100 . 0
9639 . 0
 
/
/
 
*
01
*
02
1 0
2 0
= = =
T T
T T
T
T
 ?  K 4403 ) K 4 . 502 ( 763 . 8 763 . 8 
01 2 0
= = = T T 
31 . 14
0566 . 0
8101 . 0
 
* /
* /
1
2
1
2
= = =
V V
V V
V
V
 ?    m/s 1002 ) m/s 70 ( 31 . 14 31 . 14
1 2
= = = V V 
Then the mass flow rate of the fuel is determined to be 
 kJ/kg 3920 K ) 4 . 502 4403 )( K kJ/kg 1.005 ( ) (
01 02
= - · = - = T T c q
p
 
 kW 8650 ) kJ/kg 3920 )(  kg/s 2.207 (
air
= = = q m Q &
&
 
kg/s 0.222 = = =
 kJ/kg 39,000
 kJ/s 8650
fuel
HV
Q
m
&
& 
Discussion Note that both the temperature and velocity increase during this subsonic Rayleigh flow with 
heating, as expected. This problem can also be solved using appropriate relations instead of tabulated 
values, which can likewise be coded for convenient computer solutions.  
 
 
 
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.   
12-57
Chapter 12 Compressible Flow 
12-103 Fuel is burned in a rectangular duct with compressed air. For specified heat transfer, the exit 
temperature and Mach number are to be determined.  
Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal 
gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) 
are valid.   
Properties We take the properties of air to be k = 1.4, c
p
 = 1.005 kJ/kg ·K, and R = 0.287 kJ/kg ·K.    
Analysis The stagnation temperature and Mach number at the inlet are 
 m/s 2 . 347
 kJ/kg 1
s / m 1000
K) K)(300 kJ/kg 287 . 0 )( 4 . 1 (
2 2
1 1
=
?
?
?
?
?
?
?
?
· = = kRT c 
  m/s 4 . 694 m/s) 2 . 347 ( 2 Ma
1 1 1
= = = c V
K 9 . 539
/s m 1000
 kJ/kg 1
K kJ/kg 005 . 1 2
m/s) 4 . 694 (
K 300
2
2 2
2 2
1
1 01
= ?
?
?
?
?
?
· ×
+ = + =
p
c
V
T T 
The exit stagnation temperature is, from the energy equation , ) (
01 02
T T c q
p
- =
T
2
, Ma
2 
 
P
1
 = 420 kPa 
T
1
 = 300 K 
 
Ma
1
 = 2
 
q = 55 kJ/kg
 
K 6 . 594
K kJ/kg 1.005
 kJ/kg 55
K 539.9 
01 02
=
·
+ = + =
p
c
q
T T 
The maximum value of stagnation temperature T
0
*
 occurs at Ma = 1, and its value can be determined from 
Table A-15 or from the appropriate relation. At Ma
1
 = 2 we read T
01
/T
0
*
 = 0.7934. Therefore, 
K 5 . 680
7934 . 0
K 9 . 539
7934 . 0
 
01 *
0
= = =
T
T 
The stagnation temperature ratio at the exit and the Mach number corresponding to it are, from Table A-15,     
8738 . 0
K 5 . 680
K 6 . 594
 
*
0
02
= =
T
T
      ?      Ma
2
 = 1.642  
Also,  
Ma
1
 = 2  ?  T
1
/T
*
  = 0.5289    
 Ma
2
 = 1.642 ?  T
2
/T
*
  = 0.6812    
Then the exit temperature becomes 
288 . 1
5289 . 0
6812 . 0
 
/
/
 
*
1
*
2
1
2
= = =
T T
T T
T
T
   ?  K 386 = = = ) K 300 ( 288 . 1 288 . 1 
1 2
T T 
Discussion Note that the temperature increases during this supersonic Rayleigh flow with heating. This 
problem can also be solved using appropriate relations instead of tabulated values, which can likewise be 
coded for convenient computer solutions.  
 
 
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.   
12-58
Chapter 12 Compressible Flow 
12-104 Compressed air is cooled as it flows in a rectangular duct. For specified heat rejection, the exit 
temperature and Mach number are to be determined.  
Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal 
gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) 
are valid. 
Properties We take the properties of air to be k = 1.4, C
p
 = 1.005 kJ/kg ·K, and R = 0.287 kJ/kg ·K.    
Analysis The stagnation temperature and Mach number at the inlet are 
 m/s 2 . 347
 kJ/kg 1
s / m 1000
K) K)(300 kJ/kg 287 . 0 )( 4 . 1 (
2 2
1 1
=
?
?
?
?
?
?
?
?
· = = kRT c 
  m/s 4 . 694 m/s) 2 . 347 ( 2 Ma
1 1 1
= = = c V
K 9 . 539
/s m 1000
 kJ/kg 1
K kJ/kg 005 . 1 2
m/s) 4 . 694 (
K 300
2
2 2
2 2
1
1 01
= ?
?
?
?
?
?
· ×
+ = + =
p
c
V
T T 
The exit stagnation temperature is, from the energy equation , ) (
01 02
T T c q
p
- =
 
P
1
 = 420 kPa 
T
1
 = 300 K 
 
Ma
1
 = 2
 
q = -55 kJ/kg
 
T
2
, Ma
2 
K 2 . 485
K kJ/kg 1.005
 kJ/kg -55
K 539.9 
01 02
=
·
+ = + =
p
c
q
T T 
The maximum value of stagnation temperature T
0
*
 occurs at Ma = 1, and its value can be determined from 
Table A-15 or from the appropriate relation. At Ma
1
 = 2 we read T
01
/T
0
*
 = 0.7934. Therefore, 
K 5 . 680
7934 . 0
K 9 . 539
7934 . 0
 
01 *
0
= = =
T
T 
The stagnation temperature ratio at the exit and the Mach number corresponding to it are, from Table A-15,     
7130 . 0
K 5 . 680
K 2 . 485
 
*
0
02
= =
T
T
      ?      Ma
2
 = 2.479  
Also,  
Ma
1
 = 2  ?  T
1
/T
*
  = 0.5289    
 Ma
2
 = 2.479 ?  T
2
/T
*
  = 0.3838    
Then the exit temperature becomes 
7257 . 0
5289 . 0
3838 . 0
 
/
/
 
*
1
*
2
1
2
= = =
T T
T T
T
T
   ?  K 218 = = = ) K 300 ( 7257 . 0 7257 . 0 
1 2
T T 
Discussion Note that the temperature decreases and Mach number increases during this supersonic 
Rayleigh flow with cooling. This problem can also be solved using appropriate relations instead of 
tabulated values, which can likewise be coded for convenient computer solutions.  
 
 
 
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.   
12-59
Chapter 12 Compressible Flow 
12-105 Air is heated in a duct during subsonic flow until it is choked. For specified pressure and velocity 
at the exit, the temperature, pressure, and velocity at the inlet are to be determined.  
Assumptions The assumptions associated with Rayleigh flow (i.e., steady one-dimensional flow of an ideal 
gas with constant properties through a constant cross-sectional area duct with negligible frictional effects) 
are valid.   
Properties We take the properties of air to be k = 1.4, c
p
 = 1.005 kJ/kg ·K, and R = 0.287 kJ/kg ·K.    
Analysis Noting that sonic conditions exist at the exit, the exit temperature is 
 m/s 620 m/s)/1 620 ( /Ma
2 2 2
= = = V c 
2 2
kRT c = ?  m/s 620
 kJ/kg 1
s / m 1000
K) kJ/kg 287 . 0 )( 4 . 1 (
2 2
2
=
?
?
?
?
?
?
?
?
· T 
It gives T
2
 = 956.7 K. Then the exit stagnation temperature becomes 
K 1148
/s m 1000
 kJ/kg 1
K kJ/kg 005 . 1 2
m/s) 620 (
K 7 . 956
2
2 2
2 2
2
2 02
= ?
?
?
?
?
?
· ×
+ = + =
p
c
V
T T 
 
P
2
= 270 kPa 
V
2
= 620 m/s 
 
Ma
2
= 1
 
q = 60 kJ/kg
 
P
1
  
T
1
  
 
Ma
1
 
 
The inlet stagnation temperature is, from the energy equation , ) (
01 02
T T c q
p
- =
K 1088
K kJ/kg 1.005
 kJ/kg 60
- K 1148 
02 01
=
·
= - =
p
c
q
T T 
The maximum value of stagnation temperature T
0
*
 occurs at Ma = 1, and its value in this case is 
T
02
 since the flow is choked. Therefore, T
0
*
 = T
02
 = 1148 K. Then the stagnation temperature ratio at the 
inlet, and the Mach number corresponding to it are, from Table A-15,     
9478 . 0
K 1148
K 1088
 
*
0
01
= =
T
T
      ?      Ma
1
 = 0.7649  
The Rayleigh flow functions corresponding to the inlet and exit Mach numbers are (Table A-15): 
Ma
1
 = 0.7649:      T
1
/T
*
  = 1.017, P
1
/P
*
 = 1.319,  V
1
/V
*
 = 0.7719    
Ma
2
 = 1:       T
2
/T
*
  = 1, P
2
/P
*
  = 1,  V
2
/V
*
 = 1    
Then the inlet temperature, pressure, and velocity are determined to be 
017 . 1
1
 
/
/
 
*
1
*
2
1
2
= =
T T
T T
T
T
  ?  K 974 = = = ) K 7 . 956 ( 017 . 1 017 . 1 
2 1
T T 
319 . 1
1
 
/
/
 
*
1
*
2
1
2
= =
P P
P P
P
P
  ?  kPa 356 = = = ) kPa 270 ( 319 . 1 319 . 1
2 1
P P 
7719 . 0
1
 
* /
* /
1
2
1
2
= =
V V
V V
V
V
  ?    m/s 479 = = = ) m/s 620 ( 7719 . 0 7719 . 0
2 1
V V 
Discussion Note that the temperature and pressure decreases with heating during this subsonic Rayleigh 
flow while velocity increases. This problem can also be solved using appropriate relations instead of 
tabulated values, which can likewise be coded for convenient computer solutions.  
 
 
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.   
12-60
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