Energy balance analysis example_cont. air piston as the system,energy balance 0 (△KE+△PE+△U)air+(KE+△PE+△H)piston=Q-W different p ! Q=W+(△PE)piston+(△U)air =4.5kJ+.22kJ+11.34kJ =16.06kJ W=pdV=Pam(V2-Vi) = mair(△uair) =(1bar)(.045m 105N/m2 1kJ (APE)piston=mpiston 1bar =4.5kJ 103Nm Piston =(45kg)(9.81m/s2)(0.5m)=.22kJ System Patm =I bar boundary for part (b) mpiston =45 kg -Apiston =.09 m2 4:=上-y=045m =.5m Air Apiston .09m2 Q:identical！ g=9.81m/s2. W:different!！ mair =.27 kg V2-V1=.045m3 △Mai=42kJ/kg. (b)air piston March 7,2019 10March 7, 2019 10 Energy balance analysis example_cont. air + piston as the system, energy balance Q: identical !! W: different !! different p !!