Example #1 on 1st law (Adiabatic) W.in=10 kJ (Boundary work) △E=8kJ Wsh.in=8 kJ △E=10kJ Cout=3 kJ (Adiabatic) 1 Cout =500 kJ △E=(15-3)+6 =18kJ U =800 kJ Wsh.in=6 kJ U2=? Wsh,in=100 kJ Fluid Cin =15 kJ March 7,2 SHANGHAI JIAO TONG UNIVERSITY
March 7, 2019 6 Example #1 on 1st law (Boundary work)
Example #2 on 1st Law What if you rolled a window AC into a well-insulated room that has no other air conditioner,put it on a table and plugged it in?Would you expect the room temperature to increase,decrease,or remain the same? AC System 上游充通大学 March 7,2019 7 SHANGHAI JLAO TONG UNIVERSITY
March 7, 2019 7 Example #2 on 1st Law What if you rolled a window AC into a well-insulated room that has no other air conditioner, put it on a table and plugged it in? Would you expect the room temperature to increase, decrease, or remain the same? System
Example #3 on 1st law Known:Data are provided for air contained in a vertical piston- cylinder fitted with an electrical resistor.Neglect friction. Find:heat transfer from the resistor to the air for (a)air alone;(b)air piston. Schematic and Given Data: -Piston Piston Patm=1 bar System System mpiston =45 kg boundary boundary Apiston=.09 m2 for part (b) for part (a) g=9.81m/s2. Air Air slowly Good mair=.27 kg V2-V1=.045m3 insulator △lair=42kJ/kg. (a)air alone (b)air piston 上游充通大 March 7,2019 8 SHANGHAI JIAO TONG UNIVERSITY
March 7, 2019 8 Example #3 on 1st law Known: Data are provided for air contained in a vertical piston– cylinder fitted with an electrical resistor. Neglect friction. Find: heat transfer from the resistor to the air for (a) air alone; (b) air + piston. Schematic and Given Data: slowly Good insulator g =9.81 m/s2 . air alone air + piston
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 UNIVERSITY
March 7, 2019 9 Energy balance analysis example_cont. air alone as the system, energy balance Const. p Force balance given