Thermodynamics

Chapter 4 | 185Analysis We take the contents of the tank as the system (Fig. 4–30). This isa closed system since no mass crosses the system boundary during theprocess. We observe that there is shaft work done on the system.(a) The amount of paddle-wheel work done on the system isW sh W # 2545 Btu>hsh¢t 10.02 hp210.5 h2a b 25.45 Btu1 hpUnder the stated assumptions and observations, the energy balance on thesystem can be expressed asE in E out ¢E system⎫ ⎪⎬⎪⎭⎫⎪⎬⎪⎭Net energy transferby heat, work, and massChange in internal, kinetic,potential, etc., energiesW sh,in ¢U m 1u 2 u 1 2 mc v,avg 1T 2 T 1 2As we pointed out earlier, the ideal-gas specific heats of monatomic gases(helium being one of them) are constant. The c v value of helium is determinedfrom Table A–2Ea to be c v 0.753 Btu/lbm · °F. Substituting thisand other known quantities into the above equation, we obtain25.45 Btu 11.5 lbm2 10.753 Btu>lbm # °F21T2 80°F2T 2 102.5°F(b) The final pressure is determined from the ideal-gas relationP 1 V 1T 1 P 2V 2T 2where V 1 and V 2 are identical and cancel out. Then the final pressurebecomes50 psia180 4602 R P 21102.5 4602RP 2 52.1 psiaDiscussion Note that the pressure in the ideal-gas relation is always theabsolute pressure.P, psiaHem = 1.5 lbmP 2 2T 1 = 80°FP 1 = 50 psiaW sh501V 2 = V 1VFIGURE 4–30Schematic and P-V diagram forExample 4–8.