Energy balance analysis example_cont 国 air alone as the system,energy balance 0 0 (△KE+△PE+△U)ar=Q-W Q=W+△Uair→ Q=W+mair(△lair) 个 =4.72kJ+11.34kJ=16.06kJ Const.p =i nair(△uair) pdv=p(V2-Vi) given W=p(V2-Vi) 一Piston Force balance 10N/m2 1kJ Patm I bar =(1.049bar)(.045m2) =4.72kJ 1bar 103Nm System mpiston =45 kg boundary PApiston-mpiston PamApiston Apiston=.09 m2 for part(a) 9=9.81m/s2. Air p= Apiston 十Patm slowly (45 kg)(9.81 m/s2)1 bar mair =.27 kg Good D= 10N/m2 I bar 1.049 bar V2-1=.045m3 .09m2 insulator △4ir=42kJ/kg. (a)air alone 上游充通大 March 7,2019 9 SHANGHAI JIAO TONG UNIVERSITYMarch 7, 2019 9 Energy balance analysis example_cont. air alone as the system, energy balance Const. p Force balance given