Thermodynamics

464 | ThermodynamicsAssumptions 1 Air is an ideal gas. 2 The kinetic and potential energies arenegligible. 3 The properties of air at the inlet remain constant during theentire charging process.Analysis We take the rigid tank combined with the compressor as the system.This is a control volume since mass crosses the system boundary duringthe process. We note that this is an unsteady-flow process since the masscontent of the system changes as the tank is charged. Also, there is only oneinlet and no exit.The minimum work required for a process is the reversible work, whichcan be determined from the exergy balance applied on the extended system(system immediate surroundings) whose boundary is at the environmenttemperature of T 0 (so that there is no exergy transfer accompanying heattransfer to or from the environment) and by setting the exergy destructionterm equal to zero,X in X out X destroyedS 0 (reversible) ¢X system⎫ ⎪⎬⎪⎭⎫⎪⎬⎪⎭⎫⎪⎪⎪⎬⎪⎪⎪⎭Net exergy transfer Exergy Changeby heat, work, and mass destruction in exergyNote that f 1 c 1 0 since the initial air in the tank and the air enteringare at the state of the environment, and the exergy of a substance at thestate of the environment is zero. The final mass of air and the exergy of thepressurized air in the tank at the end of the process arem 2 P 2VRT 2 P 0 1v 2 v 0 2 T 0 1s 2 s 0 2We note thatX in X out X 2 X 1W rev,in m 1 c 1Q0 m2 f 2 m 1 f 1Q 0W rev,in m 2 f 211000 kPa2 1200 m 3 210.287 kPa # m 3 >kg # K21300 K2 2323 kgf 2 1u 2 u 0 2 Q0 1since T 2T 0 2 P0 1v 2 v 0 2 T 0 1s 2 s 0 2 V 2 0 Q2 gz 0 Q22P 0 1v 2 v 0 2 P 0 a RT 2P 2 RT 0P 0b RT 0 a P 0P 2 1 b1since T 2 T 0 2T 2Q 0 P 2P 2T 0 1s 2 s 0 2 T 0 a c p ln R ln b RTT 0 P 0 ln 1since T0 P 2 T 0 20Therefore,f 2 RT 0 a P 0 1 b RT 0 ln RT 0 a ln P 0 1 bP 2 P 0 P 0 P 2 10.287 kJ>kg # K21300 K2 aln1000 kPa100 kPa 120.76 kJ>kgP 2P 2100 kPa1000 kPa 1 b