14 kjkg 11584 table a21 the enthalpy of the air at the

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Unformatted text preview: and thus determine the exit temperature. The property that can be determined with minimal effort in this case is h2a since the isentropic efficiency of the compressor is given. At the compressor inlet, T1 285 K → h1 (Pr1 285.14 kJ/kg 1.1584) (Table A–21) The enthalpy of the air at the end of the isentropic compression process is determined by using one of the isentropic relations of ideal gases, Pr2 Pr1 P2 P1 1.1584 800 kPa 100 kPa 9.2672 and Pr2 9.2672 → h2s 517.05 kJ/kg (Table A–21) Substituting the known quantities into the isentropic efficiency relation, we have hC h2s h2a h1 → 0.80 h1 (517.05 285.14) kJ/kg (h2a 285.14) kJ/kg Thus, h2a 575.03 kJ/kg → T2a 569.5 K (Table A–21) (b) The required power input to the compressor is determined from the energy balance for steady-flow devices, 80 0 kP a T,K P2 = 800 kPa 2a T2a AIR COMPRESSOR · m = 0.2 kg/s T2s 2s Actual process Isentropic process Pa 100 k FIGURE 7–53 Schematic and T-s diagram for Example 7–15. 285 P1 = 100 kPa T1 = 285 K 1 s2s = s1 s cen54261_ch07.qxd 11/18/03 9:57 AM Page 317 317 CHAPTER 7 · E in · Wa, in · Wa, in · m h1 · E out · m h2a · m (h2...
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