Solution Continue (a)Properties at each state, v=const Table A-17 for ideal gas air T1=300K,p1=1bar → u1=214.07k/kgVr1=621.2 Relative volume:V2 Process1-2:③ U2= Ul621.2 1= =77.65 dT 8 A-c=0 dT Table A-17 for air T Vr2 →T2=673K,u2=491.2k/kg 5= =300*(8)14=689.2K EoS → L 673K P2 PIT V2 =(1am） 8=17.95bars 300K Process2-3:⑦ EoS→p3=P2 T3-(17.95 bars) 2000K 53.3 bars 673K/ Table A-17 for air T3>u3=1678.7K,v3=2.776 上游充通大 May15,2018 19 HANGHAI HAO TONG LINIVERSITYMay 15, 2018 19 Solution Continue (a) Properties at each state, T1=300K, p 1=1bar  u 1=214.07 kJ/kg, vr1=621.2 Process 1‐2: S Table A‐17 for ideal gas air vr2  T2=673 K, u 2=491.2 kJ/kg  Table A‐17 for air EoS Process 2‐3: V EoS  T3  u 3=1678.7K, v Table A‐17 for air r3=2.776 Relative volume: 1 1 1.4 1 2 1 2 * 300*(8) 689.2 K v T T v           2 1 2 g 1 d ln 0 T v T T v sc R T v     0 r g 1 d ln T v T T v c R T   