Thermodynamics

374 | ThermodynamicsEXAMPLE 7–15Effect of Efficiency on Compressor Power InputAir is compressed by an adiabatic compressor from 100 kPa and 12°C to apressure of 800 kPa at a steady rate of 0.2 kg/s. If the isentropic efficiencyof the compressor is 80 percent, determine (a) the exit temperature of air and(b) the required power input to the compressor.Solution Air is compressed to a specified pressure at a specified rate. For agiven isentropic efficiency, the exit temperature and the power input are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas. 3 Thechanges in kinetic and potential energies are negligible.Analysis A sketch of the system and the T-s diagram of the process are givenin Fig. 7–53.(a) We know only one property (pressure) at the exit state, and we need toknow one more to fix the state and thus determine the exit temperature. Theproperty that can be determined with minimal effort in this case is h 2a sincethe isentropic efficiency of the compressor is given. At the compressor inlet,The enthalpy of the air at the end of the isentropic compression process isdetermined by using one of the isentropic relations of ideal gases,andSubstituting the known quantities into the isentropic efficiency relation, we haveThus,T 1 285 KSh 1 285.14 kJ>kg1Table A–1721P r1 1.15842P r2 P r1 a P 2800 kPab 1.1584 aP 1 100 kPa b 9.2672P r2 9.2672Sh 2s 517.05 kJ>kgh C h 2s h 1h 2a h 1S0.80 1517.05 285.142 kJ>kg1h 2a 285.142 kJ>kgh 2a 575.03 kJ>kgS T 2a 569.5 KP 2 = 800 kPaT, KT 2a2a800 kPaAIRCOMPRESSORm · = 0.2 kg/sT 2s2sActual processIsentropic process100 kPaFIGURE 7–53Schematic and T-s diagram forExample 7–15.P 1 = 100 kPaT 1 = 285 K2851s 2s = s 1s