non-flow device that will compress air isothermally at a temperature of 500 °C from a pressure of 5.0 MPa to a final pressure of 15.0 MPa. If the device uses 200 kJ/kg of work input, is it
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Answer:
m = P1V1/RT1
= (100 × 0.3)/(0.287 × 293.2)
= 0.3565 kg
T2/T1
= (P2/P1)^[(n-1)/n]
= 293.2(800/100)^(0.167)
= 414.9 K
W
= (P2 v2 – P1 v1)/(1-n)
= R(T2-T1)/(1-n)
= 0.287(414.9-293.2)/(1-1.20)
= -174.6 kJ/kg
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