Page 1 Chapter 12 Compressible Flow Normal Shocks in Nozzle Flow 12-68C No, because the flow must be supersonic before a shock wave can occur. The flow in the converging section of a nozzle is always subsonic. 12-69C The Fanno line represents the states which satisfy the conservation of mass and energy equations. The Rayleigh line represents the states which satisfy the conservation of mass and momentum equations. The intersections points of these lines represents the states which satisfy the conservation of mass, energy, and momentum equations. 12-70C No, the second law of thermodynamics requires the flow after the shock to be subsonic.. 12-71C (a) decreases, (b) increases, (c) remains the same, (d) increases, and (e) decreases. 12-72C Oblique shocks occur when a gas flowing at supersonic speeds strikes a flat or inclined surface. Normal shock waves are perpendicular to flow whereas inclined shock waves, as the name implies, are typically inclined relative to the flow direction. Also, normal shocks form a straight line whereas oblique shocks can be straight or curved, depending on the surface geometry. 12-73C Yes, the upstream flow have to be supersonic for an oblique shock to occur. No, the flow downstream of an oblique shock can be subsonic, sonic, and even supersonic. 12-74C Yes. Conversely, normal shocks can be thought of as special oblique shocks in which the shock angle is ß = p/2, or 90 o . 12-75C When the wedge half-angle d is greater than the maximum deflection angle ? max , the shock becomes curved and detaches from the nose of the wedge, forming what is called a detached oblique shock or a bow wave. The numerical value of the shock angle at the nose is be ß = 90 o . 12-76C When supersonic flow impinges on a blunt body like the rounded nose of an aircraft, the wedge half-angle d at the nose is 90 o , and an attached oblique shock cannot exist, regardless of Mach number. Therefore, a detached oblique shock must occur in front of all such blunt-nosed bodies, whether two- dimensional, axisymmetric, or fully three-dimensional. 12-77C Isentropic relations of ideal gases are not applicable for flows across (a) normal shock waves and (b) oblique shock waves, but they are applicable for flows across (c) Prandtl-Meyer expansion waves. 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. 12-35 Page 2 Chapter 12 Compressible Flow Normal Shocks in Nozzle Flow 12-68C No, because the flow must be supersonic before a shock wave can occur. The flow in the converging section of a nozzle is always subsonic. 12-69C The Fanno line represents the states which satisfy the conservation of mass and energy equations. The Rayleigh line represents the states which satisfy the conservation of mass and momentum equations. The intersections points of these lines represents the states which satisfy the conservation of mass, energy, and momentum equations. 12-70C No, the second law of thermodynamics requires the flow after the shock to be subsonic.. 12-71C (a) decreases, (b) increases, (c) remains the same, (d) increases, and (e) decreases. 12-72C Oblique shocks occur when a gas flowing at supersonic speeds strikes a flat or inclined surface. Normal shock waves are perpendicular to flow whereas inclined shock waves, as the name implies, are typically inclined relative to the flow direction. Also, normal shocks form a straight line whereas oblique shocks can be straight or curved, depending on the surface geometry. 12-73C Yes, the upstream flow have to be supersonic for an oblique shock to occur. No, the flow downstream of an oblique shock can be subsonic, sonic, and even supersonic. 12-74C Yes. Conversely, normal shocks can be thought of as special oblique shocks in which the shock angle is ß = p/2, or 90 o . 12-75C When the wedge half-angle d is greater than the maximum deflection angle ? max , the shock becomes curved and detaches from the nose of the wedge, forming what is called a detached oblique shock or a bow wave. The numerical value of the shock angle at the nose is be ß = 90 o . 12-76C When supersonic flow impinges on a blunt body like the rounded nose of an aircraft, the wedge half-angle d at the nose is 90 o , and an attached oblique shock cannot exist, regardless of Mach number. Therefore, a detached oblique shock must occur in front of all such blunt-nosed bodies, whether two- dimensional, axisymmetric, or fully three-dimensional. 12-77C Isentropic relations of ideal gases are not applicable for flows across (a) normal shock waves and (b) oblique shock waves, but they are applicable for flows across (c) Prandtl-Meyer expansion waves. 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. 12-35 Chapter 12 Compressible Flow 12-78 For an ideal gas flowing through a normal shock, a relation for V 2 /V 1 in terms of k, Ma 1 , and Ma 2 is to be developed. Analysis The conservation of mass relation across the shock is 2 2 1 1 V V ? ? = and it can be expressed as ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = = = 1 2 2 1 2 2 1 1 2 1 1 2 / / T T P P RT P RT P V V ? ? From Eqs. 12-35 and 12-38, ? ? ? ? ? ? ? ? - + - + ? ? ? ? ? ? ? ? + + = 2 / ) 1 ( Ma 1 2 / ) 1 ( Ma 1 Ma 1 Ma 1 2 2 2 1 2 1 2 2 1 2 k k k k V V Discussion This is an important relation as it enables us to determine the velocity ratio across a normal shock when the Mach numbers before and after the shock are known. 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. 12-36 Page 3 Chapter 12 Compressible Flow Normal Shocks in Nozzle Flow 12-68C No, because the flow must be supersonic before a shock wave can occur. The flow in the converging section of a nozzle is always subsonic. 12-69C The Fanno line represents the states which satisfy the conservation of mass and energy equations. The Rayleigh line represents the states which satisfy the conservation of mass and momentum equations. The intersections points of these lines represents the states which satisfy the conservation of mass, energy, and momentum equations. 12-70C No, the second law of thermodynamics requires the flow after the shock to be subsonic.. 12-71C (a) decreases, (b) increases, (c) remains the same, (d) increases, and (e) decreases. 12-72C Oblique shocks occur when a gas flowing at supersonic speeds strikes a flat or inclined surface. Normal shock waves are perpendicular to flow whereas inclined shock waves, as the name implies, are typically inclined relative to the flow direction. Also, normal shocks form a straight line whereas oblique shocks can be straight or curved, depending on the surface geometry. 12-73C Yes, the upstream flow have to be supersonic for an oblique shock to occur. No, the flow downstream of an oblique shock can be subsonic, sonic, and even supersonic. 12-74C Yes. Conversely, normal shocks can be thought of as special oblique shocks in which the shock angle is ß = p/2, or 90 o . 12-75C When the wedge half-angle d is greater than the maximum deflection angle ? max , the shock becomes curved and detaches from the nose of the wedge, forming what is called a detached oblique shock or a bow wave. The numerical value of the shock angle at the nose is be ß = 90 o . 12-76C When supersonic flow impinges on a blunt body like the rounded nose of an aircraft, the wedge half-angle d at the nose is 90 o , and an attached oblique shock cannot exist, regardless of Mach number. Therefore, a detached oblique shock must occur in front of all such blunt-nosed bodies, whether two- dimensional, axisymmetric, or fully three-dimensional. 12-77C Isentropic relations of ideal gases are not applicable for flows across (a) normal shock waves and (b) oblique shock waves, but they are applicable for flows across (c) Prandtl-Meyer expansion waves. 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. 12-35 Chapter 12 Compressible Flow 12-78 For an ideal gas flowing through a normal shock, a relation for V 2 /V 1 in terms of k, Ma 1 , and Ma 2 is to be developed. Analysis The conservation of mass relation across the shock is 2 2 1 1 V V ? ? = and it can be expressed as ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = = = 1 2 2 1 2 2 1 1 2 1 1 2 / / T T P P RT P RT P V V ? ? From Eqs. 12-35 and 12-38, ? ? ? ? ? ? ? ? - + - + ? ? ? ? ? ? ? ? + + = 2 / ) 1 ( Ma 1 2 / ) 1 ( Ma 1 Ma 1 Ma 1 2 2 2 1 2 1 2 2 1 2 k k k k V V Discussion This is an important relation as it enables us to determine the velocity ratio across a normal shock when the Mach numbers before and after the shock are known. 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. 12-36 Chapter 12 Compressible Flow 12-79 Air flowing through a converging-diverging nozzle experiences a normal shock at the exit. The effect of the shock wave on various properties is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one- dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Properties The properties of air are k = 1.4 and R = 0.287 kJ/kg·K. Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Then, P 01 = P i = 1 MPa T 01 = T i = 300 K Then, K 7 . 166 1)2 - (1.4 + 2 2 K) 300 ( Ma ) 1 ( 2 2 2 2 1 01 1 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? - + = k T T 2 i 1 AIR Shock wave V i ˜ 0 and MPa 1278 . 0 300 166.7 MPa) 1 ( 4 . 0 / 4 . 1 ) 1 /( 0 1 01 1 = ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = - k k T T P P The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A-14. For Ma 1 = 2.0 we read 6875 . 1 and , 5000 . 4 , 7209 . 0 , Ma 1 2 1 2 02 02 2 = = = = T T P P P P 0.5774 Then the stagnation pressure P 02 , static pressure P 2 , and static temperature T 2 , are determined to be P 02 = 0.7209P 01 = (0.7209)(1.0 MPa) = 0.721 MPa P 2 = 4.5000P 1 = (4.5000)(0.1278 MPa) = 0.575 MPa T 2 = 1.6875T 1 = (1.6875)(166.7 K) = 281 K The air velocity after the shock can be determined from V 2 = Ma 2 c 2 , where c 2 is the speed of sound at the exit conditions after the shock, V 2 = Ma 2 c 2 = m/s 194 = ? ? ? ? ? ? ? ? · = kJ/kg 1 s / m 1000 K) K)(281 kJ/kg 287 . 0 )( 4 . 1 ( ) 5774 . 0 ( Ma 2 2 2 2 kRT Discussion We can also solve this problem using the relations for normal shock functions. The results would be identical. 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. 12-37 Page 4 Chapter 12 Compressible Flow Normal Shocks in Nozzle Flow 12-68C No, because the flow must be supersonic before a shock wave can occur. The flow in the converging section of a nozzle is always subsonic. 12-69C The Fanno line represents the states which satisfy the conservation of mass and energy equations. The Rayleigh line represents the states which satisfy the conservation of mass and momentum equations. The intersections points of these lines represents the states which satisfy the conservation of mass, energy, and momentum equations. 12-70C No, the second law of thermodynamics requires the flow after the shock to be subsonic.. 12-71C (a) decreases, (b) increases, (c) remains the same, (d) increases, and (e) decreases. 12-72C Oblique shocks occur when a gas flowing at supersonic speeds strikes a flat or inclined surface. Normal shock waves are perpendicular to flow whereas inclined shock waves, as the name implies, are typically inclined relative to the flow direction. Also, normal shocks form a straight line whereas oblique shocks can be straight or curved, depending on the surface geometry. 12-73C Yes, the upstream flow have to be supersonic for an oblique shock to occur. No, the flow downstream of an oblique shock can be subsonic, sonic, and even supersonic. 12-74C Yes. Conversely, normal shocks can be thought of as special oblique shocks in which the shock angle is ß = p/2, or 90 o . 12-75C When the wedge half-angle d is greater than the maximum deflection angle ? max , the shock becomes curved and detaches from the nose of the wedge, forming what is called a detached oblique shock or a bow wave. The numerical value of the shock angle at the nose is be ß = 90 o . 12-76C When supersonic flow impinges on a blunt body like the rounded nose of an aircraft, the wedge half-angle d at the nose is 90 o , and an attached oblique shock cannot exist, regardless of Mach number. Therefore, a detached oblique shock must occur in front of all such blunt-nosed bodies, whether two- dimensional, axisymmetric, or fully three-dimensional. 12-77C Isentropic relations of ideal gases are not applicable for flows across (a) normal shock waves and (b) oblique shock waves, but they are applicable for flows across (c) Prandtl-Meyer expansion waves. 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. 12-35 Chapter 12 Compressible Flow 12-78 For an ideal gas flowing through a normal shock, a relation for V 2 /V 1 in terms of k, Ma 1 , and Ma 2 is to be developed. Analysis The conservation of mass relation across the shock is 2 2 1 1 V V ? ? = and it can be expressed as ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = = = 1 2 2 1 2 2 1 1 2 1 1 2 / / T T P P RT P RT P V V ? ? From Eqs. 12-35 and 12-38, ? ? ? ? ? ? ? ? - + - + ? ? ? ? ? ? ? ? + + = 2 / ) 1 ( Ma 1 2 / ) 1 ( Ma 1 Ma 1 Ma 1 2 2 2 1 2 1 2 2 1 2 k k k k V V Discussion This is an important relation as it enables us to determine the velocity ratio across a normal shock when the Mach numbers before and after the shock are known. 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. 12-36 Chapter 12 Compressible Flow 12-79 Air flowing through a converging-diverging nozzle experiences a normal shock at the exit. The effect of the shock wave on various properties is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one- dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Properties The properties of air are k = 1.4 and R = 0.287 kJ/kg·K. Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Then, P 01 = P i = 1 MPa T 01 = T i = 300 K Then, K 7 . 166 1)2 - (1.4 + 2 2 K) 300 ( Ma ) 1 ( 2 2 2 2 1 01 1 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? - + = k T T 2 i 1 AIR Shock wave V i ˜ 0 and MPa 1278 . 0 300 166.7 MPa) 1 ( 4 . 0 / 4 . 1 ) 1 /( 0 1 01 1 = ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = - k k T T P P The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A-14. For Ma 1 = 2.0 we read 6875 . 1 and , 5000 . 4 , 7209 . 0 , Ma 1 2 1 2 02 02 2 = = = = T T P P P P 0.5774 Then the stagnation pressure P 02 , static pressure P 2 , and static temperature T 2 , are determined to be P 02 = 0.7209P 01 = (0.7209)(1.0 MPa) = 0.721 MPa P 2 = 4.5000P 1 = (4.5000)(0.1278 MPa) = 0.575 MPa T 2 = 1.6875T 1 = (1.6875)(166.7 K) = 281 K The air velocity after the shock can be determined from V 2 = Ma 2 c 2 , where c 2 is the speed of sound at the exit conditions after the shock, V 2 = Ma 2 c 2 = m/s 194 = ? ? ? ? ? ? ? ? · = kJ/kg 1 s / m 1000 K) K)(281 kJ/kg 287 . 0 )( 4 . 1 ( ) 5774 . 0 ( Ma 2 2 2 2 kRT Discussion We can also solve this problem using the relations for normal shock functions. The results would be identical. 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. 12-37 Chapter 12 Compressible Flow V i ˜ 0 shock wave AIR 1 i 2 P b i 2 P b ressible flow and normal shock shock wave V i ˜ 0 AIR 1 12-80 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic, P 01 = P i = 2 MPa It is specified that A/A* =3.5. From Table A-13, Mach number and the pressure ratio which corresponds to this area ratio are the Ma 1 =2.80 and P 1 /P 01 = 0.0368. The pressure ratio across the shock for this Ma 1 value is, from Table A-14, P 2 /P 1 = 8.98. Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be P 2 =8.98P 1 = 8.98 ×0.0368P 01 = 8.98 ×0.0368 ×(2 MPa) = 0.661 MPa Discussion We can also solve this problem using the relations for compressible flow and normal shock functions. The results would be identical. 12-81 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic, P 0x = P i = 2 MPa It is specified that A/A* = 2. From Table A-13, the Mach number and the pressure ratio which corresponds to this area ratio are the Ma 1 =2.20 and P 1 /P 01 = 0.0935. The pressure ratio across the shock for this M 1 value is, from Table A-14, P 2 /P 1 = 5.48. Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be P 2 =5.48P 1 = 5.48 ×0.0935P 01 = 5.48 ×0.0935 ×(2 MPa) = 1.02 MPa Discussion We can also solve this problem using the relations for comp functions. The results would be identical. 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. 12-38 Page 5 Chapter 12 Compressible Flow Normal Shocks in Nozzle Flow 12-68C No, because the flow must be supersonic before a shock wave can occur. The flow in the converging section of a nozzle is always subsonic. 12-69C The Fanno line represents the states which satisfy the conservation of mass and energy equations. The Rayleigh line represents the states which satisfy the conservation of mass and momentum equations. The intersections points of these lines represents the states which satisfy the conservation of mass, energy, and momentum equations. 12-70C No, the second law of thermodynamics requires the flow after the shock to be subsonic.. 12-71C (a) decreases, (b) increases, (c) remains the same, (d) increases, and (e) decreases. 12-72C Oblique shocks occur when a gas flowing at supersonic speeds strikes a flat or inclined surface. Normal shock waves are perpendicular to flow whereas inclined shock waves, as the name implies, are typically inclined relative to the flow direction. Also, normal shocks form a straight line whereas oblique shocks can be straight or curved, depending on the surface geometry. 12-73C Yes, the upstream flow have to be supersonic for an oblique shock to occur. No, the flow downstream of an oblique shock can be subsonic, sonic, and even supersonic. 12-74C Yes. Conversely, normal shocks can be thought of as special oblique shocks in which the shock angle is ß = p/2, or 90 o . 12-75C When the wedge half-angle d is greater than the maximum deflection angle ? max , the shock becomes curved and detaches from the nose of the wedge, forming what is called a detached oblique shock or a bow wave. The numerical value of the shock angle at the nose is be ß = 90 o . 12-76C When supersonic flow impinges on a blunt body like the rounded nose of an aircraft, the wedge half-angle d at the nose is 90 o , and an attached oblique shock cannot exist, regardless of Mach number. Therefore, a detached oblique shock must occur in front of all such blunt-nosed bodies, whether two- dimensional, axisymmetric, or fully three-dimensional. 12-77C Isentropic relations of ideal gases are not applicable for flows across (a) normal shock waves and (b) oblique shock waves, but they are applicable for flows across (c) Prandtl-Meyer expansion waves. 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. 12-35 Chapter 12 Compressible Flow 12-78 For an ideal gas flowing through a normal shock, a relation for V 2 /V 1 in terms of k, Ma 1 , and Ma 2 is to be developed. Analysis The conservation of mass relation across the shock is 2 2 1 1 V V ? ? = and it can be expressed as ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = = = 1 2 2 1 2 2 1 1 2 1 1 2 / / T T P P RT P RT P V V ? ? From Eqs. 12-35 and 12-38, ? ? ? ? ? ? ? ? - + - + ? ? ? ? ? ? ? ? + + = 2 / ) 1 ( Ma 1 2 / ) 1 ( Ma 1 Ma 1 Ma 1 2 2 2 1 2 1 2 2 1 2 k k k k V V Discussion This is an important relation as it enables us to determine the velocity ratio across a normal shock when the Mach numbers before and after the shock are known. 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. 12-36 Chapter 12 Compressible Flow 12-79 Air flowing through a converging-diverging nozzle experiences a normal shock at the exit. The effect of the shock wave on various properties is to be determined. Assumptions 1 Air is an ideal gas with constant specific heats. 2 Flow through the nozzle is steady, one- dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Properties The properties of air are k = 1.4 and R = 0.287 kJ/kg·K. Analysis The inlet stagnation properties in this case are identical to the inlet properties since the inlet velocity is negligible. Then, P 01 = P i = 1 MPa T 01 = T i = 300 K Then, K 7 . 166 1)2 - (1.4 + 2 2 K) 300 ( Ma ) 1 ( 2 2 2 2 1 01 1 = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? - + = k T T 2 i 1 AIR Shock wave V i ˜ 0 and MPa 1278 . 0 300 166.7 MPa) 1 ( 4 . 0 / 4 . 1 ) 1 /( 0 1 01 1 = ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = - k k T T P P The fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions listed in Table A-14. For Ma 1 = 2.0 we read 6875 . 1 and , 5000 . 4 , 7209 . 0 , Ma 1 2 1 2 02 02 2 = = = = T T P P P P 0.5774 Then the stagnation pressure P 02 , static pressure P 2 , and static temperature T 2 , are determined to be P 02 = 0.7209P 01 = (0.7209)(1.0 MPa) = 0.721 MPa P 2 = 4.5000P 1 = (4.5000)(0.1278 MPa) = 0.575 MPa T 2 = 1.6875T 1 = (1.6875)(166.7 K) = 281 K The air velocity after the shock can be determined from V 2 = Ma 2 c 2 , where c 2 is the speed of sound at the exit conditions after the shock, V 2 = Ma 2 c 2 = m/s 194 = ? ? ? ? ? ? ? ? · = kJ/kg 1 s / m 1000 K) K)(281 kJ/kg 287 . 0 )( 4 . 1 ( ) 5774 . 0 ( Ma 2 2 2 2 kRT Discussion We can also solve this problem using the relations for normal shock functions. The results would be identical. 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. 12-37 Chapter 12 Compressible Flow V i ˜ 0 shock wave AIR 1 i 2 P b i 2 P b ressible flow and normal shock shock wave V i ˜ 0 AIR 1 12-80 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. 3 The shock wave occurs at the exit plane. Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic, P 01 = P i = 2 MPa It is specified that A/A* =3.5. From Table A-13, Mach number and the pressure ratio which corresponds to this area ratio are the Ma 1 =2.80 and P 1 /P 01 = 0.0368. The pressure ratio across the shock for this Ma 1 value is, from Table A-14, P 2 /P 1 = 8.98. Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be P 2 =8.98P 1 = 8.98 ×0.0368P 01 = 8.98 ×0.0368 ×(2 MPa) = 0.661 MPa Discussion We can also solve this problem using the relations for compressible flow and normal shock functions. The results would be identical. 12-81 Air enters a converging-diverging nozzle at a specified state. The required back pressure that produces a normal shock at the exit plane is to be determined for the specified nozzle geometry. Assumptions 1 Air is an ideal gas. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Analysis The inlet stagnation pressure in this case is identical to the inlet pressure since the inlet velocity is negligible. Since the flow before the shock to be isentropic, P 0x = P i = 2 MPa It is specified that A/A* = 2. From Table A-13, the Mach number and the pressure ratio which corresponds to this area ratio are the Ma 1 =2.20 and P 1 /P 01 = 0.0935. The pressure ratio across the shock for this M 1 value is, from Table A-14, P 2 /P 1 = 5.48. Thus the back pressure, which is equal to the static pressure at the nozzle exit, must be P 2 =5.48P 1 = 5.48 ×0.0935P 01 = 5.48 ×0.0935 ×(2 MPa) = 1.02 MPa Discussion We can also solve this problem using the relations for comp functions. The results would be identical. 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. 12-38 Chapter 12 Compressible Flow 12-82 Air flowing through a nozzle experiences a normal shock. The effect of the shock wave on various properties is to be determined. Analysis is to be repeated for helium under the same conditions. Assumptions 1 Air and helium are ideal gases with constant specific heats. 2 Flow through the nozzle is steady, one-dimensional, and isentropic before the shock occurs. Properties The properties of air are k = 1.4 and R = 0.287 kJ/kg·K, and the properties of helium are k = 1.667 and R = 2.0769 kJ/kg·K. Analysis The air properties upstream the shock are shock wave Ma 1 = 2.5, P 1 = 61.64 kPa, and T 1 = 262.15 K Fluid properties after the shock (denoted by subscript 2) are related to those before the shock through the functions in Table A-14. For Ma 1 = 2.5, AIR i 2 1 1375 . 2 and , 125 . 7 , 5262 . 8 , Ma 1 2 1 2 1 02 2 = = = = T T P P P P 0.513 Ma 1 = 2.5 Then the stagnation pressure P 02 , static pressure P 2 , and static temperature T 2 , are determined to be P 02 = 8.5261P 1 = (8.5261)(61.64 kPa) = 526 kPa P 2 = 7.125P 1 = (7.125)(61.64 kPa) = 439 kPa T 2 = 2.1375T 1 = (2.1375)(262.15 K) = 560 K The air velocity after the shock can be determined from V 2 = Ma 2 c 2 , where c 2 is the speed of sound at the exit conditions after the shock, m/s 243 = ? ? ? ? ? ? ? ? · = kJ/kg 1 s / m 1000 K) K)(560.3 kJ/kg 287 . 0 )( 4 . 1 ( ) 513 . 0 ( Ma = Ma = 2 2 2 2 2 2 2 kRT c V We now repeat the analysis for helium. This time we cannot use the tabulated values in Table A-14 since k is not 1.4. Therefore, we have to calculate the desired quantities using the analytical relations, 0.553 = ? ? ? ? ? ? ? ? - - × × - + = ? ? ? ? ? ? ? ? - - - + = 2 / 1 2 2 2 / 1 2 1 2 1 2 1 ) 1 667 . 1 /( 667 . 1 5 . 2 2 ) 1 667 . 1 /( 2 5 . 2 1 ) 1 /( Ma 2 ) 1 /( 2 Ma Ma k k k 5632 . 7 553 . 0 667 . 1 1 5 . 2 667 . 1 1 Ma 1 Ma 1 2 2 2 2 2 1 1 2 = × + × + = + + = k k P P 7989 . 2 2 / ) 1 667 . 1 ( 553 . 0 1 2 / ) 1 667 . 1 ( 5 . 2 1 2 / ) 1 ( Ma 1 2 / ) 1 ( Ma 1 2 2 2 2 2 1 1 2 = - + - + = - + - + = k k T T () ) 1 /( 2 2 2 2 2 1 1 02 2 / Ma ) 1 ( 1 Ma 1 Ma 1 - - + ? ? ? ? ? ? ? ? + + = k k k k k P P () 641 . 9 2 / 553 . 0 ) 1 667 . 1 ( 1 553 . 0 667 . 1 1 5 . 2 667 . 1 1 667 . 0 / 667 . 1 2 2 2 = × - + ? ? ? ? ? ? ? ? × + × + = Thus, P 02 = 11.546P 1 = (11.546)(61.64 kPa) = 712 kPa P 2 = 7.5632P 1 = (7.5632)(61.64 kPa) = 466 kPa T 2 = 2.7989T 1 = (2.7989)(262.15 K) = 734 K m/s 881 = ? ? ? ? ? ? ? ? · = = = kJ/kg 1 s / m 1000 K) K)(733.7 kJ/kg 0769 . 2 )( 667 . 1 ( ) 553 . 0 ( Ma Ma 2 2 2 2 2 2 y kRT c V 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. 12-39Read More

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