Thermodynamics

Chapter 8 | 439The exergy content of the compressed air can be determined fromX 1 mf 1We note thatTherefore,and m c1u 1 u 0 2 Q0 P 0 1v 1 v 0 2 T 0 1s 1 s 0 2 V 2 0 Q1 gz 0 Q1 d2 m3P 0 1v 1 v 0 2 T 0 1s 1 s 0 24P 0 1v 1 v 0 2 P 0 a RT 1P 1 RT 0P 0b RT 0 a P 0P 1 1 b1since T 1 T 0 2T 0 1s 1 s 0 2 T 0 a c p ln T 1T 0 R ln P 1P 0b RT 0 ln P 1P 01since T 1 T 0 2f 1 RT 0 a P 0P 1 1 b RT 0 ln P 1P 0 RT 0 a ln P 1P 0 P 0P 1 1 b 10.287 kJ>kg # K21300 K2 aln1000 kPa100 kPa 120.76 kJ>kg100 kPa1000 kPa 1 bX 1 m 1 f 1 12323 kg2 1120.76 kJ>kg2 280,525 kJ 281 MJDiscussion The work potential of the system is 281 MJ, and thus a maximumof 281 MJ of useful work can be obtained from the compressed airstored in the tank in the specified environment.EXAMPLE 8–8Exergy Change during a Compression ProcessRefrigerant-134a is to be compressed from 0.14 MPa and 10°C to 0.8MPa and 50°C steadily by a compressor. Taking the environment conditionsto be 20°C and 95 kPa, determine the exergy change of the refrigerant duringthis process and the minimum work input that needs to be supplied tothe compressor per unit mass of the refrigerant.Solution Refrigerant-134a is being compressed from a specified inlet stateto a specified exit state. The exergy change of the refrigerant and the minimumcompression work per unit mass are to be determined.Assumptions 1 Steady operating conditions exist. 2 The kinetic and potentialenergies are negligible.Analysis We take the compressor as the system (Fig. 8–25). This is a controlvolume since mass crosses the system boundary during the process.Here the question is the exergy change of a fluid stream, which is thechange in the flow exergy c.T 0 = 20°CCOMPRESSORP 1 = 0.14 MPaT 1 = –10°CFIGURE 8–25T 2 = 50°CP 2 = 0.8 MPaSchematic for Example 8–8.