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. 12-60Read More

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