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Page 1: slutions 5 - web.engr.uky.edumenguc/me220/solutions/slutions 5.pdf3-141 (a) On the P-vdiagram the constant temperamre process through the state P = 280 kPa, v= 0.06 m3/kg as pressure

Page 2: slutions 5 - web.engr.uky.edumenguc/me220/solutions/slutions 5.pdf3-141 (a) On the P-vdiagram the constant temperamre process through the state P = 280 kPa, v= 0.06 m3/kg as pressure

Page 3: slutions 5 - web.engr.uky.edumenguc/me220/solutions/slutions 5.pdf3-141 (a) On the P-vdiagram the constant temperamre process through the state P = 280 kPa, v= 0.06 m3/kg as pressure

Page 4: slutions 5 - web.engr.uky.edumenguc/me220/solutions/slutions 5.pdf3-141 (a) On the P-vdiagram the constant temperamre process through the state P = 280 kPa, v= 0.06 m3/kg as pressure

Page 5: slutions 5 - web.engr.uky.edumenguc/me220/solutions/slutions 5.pdf3-141 (a) On the P-vdiagram the constant temperamre process through the state P = 280 kPa, v= 0.06 m3/kg as pressure

Page 6: slutions 5 - web.engr.uky.edumenguc/me220/solutions/slutions 5.pdf3-141 (a) On the P-vdiagram the constant temperamre process through the state P = 280 kPa, v= 0.06 m3/kg as pressure

Page 7: slutions 5 - web.engr.uky.edumenguc/me220/solutions/slutions 5.pdf3-141 (a) On the P-vdiagram the constant temperamre process through the state P = 280 kPa, v= 0.06 m3/kg as pressure

Page 8: slutions 5 - web.engr.uky.edumenguc/me220/solutions/slutions 5.pdf3-141 (a) On the P-vdiagram the constant temperamre process through the state P = 280 kPa, v= 0.06 m3/kg as pressure

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