Thermodynamics

Chapter 7 | 369P, kPaP 2 = 900 kPa900Isentropic (k = 1.4)Polytropic (n = 1.3)AIRCOMPRESSORP 1 = 100 kPaT 1 = 300 KTwo-stageIsothermalw comp 1001vFIGURE 7–47Schematic and P-v diagram forExample 7–13.Assumptions 1 Steady operating conditions exist. 2 At specified conditions,air can be treated as an ideal gas. 3 Kinetic and potential energy changes arenegligible.Analysis We take the compressor to be the system. This is a control volumesince mass crosses the boundary. A sketch of the system and the T-s diagramfor the process are given in Fig. 7–47.The steady-flow compression work for all these four cases is determined byusing the relations developed earlier in this section:(a) Isentropic compression with k 1.4:w comp,in kRT 1k12>k1k 1 caP 2b 1 dP 1 11.4210.287 kJ>kg # K21300 K21.4 1 263.2 kJ/kg(b) Polytropic compression with n 1.3:w comp,in nRT 1n12>n1n 1 caP 2b 1 dP 1 11.3210.287 kJ>kg # K21300 K21.3 1 246.4 kJ/kg(c) Isothermal compression:P 2900 kPaw comp,in RT ln 10.287 kJ>kg # K21300 K2 lnP 1 100 kPa 189.2 kJ/kg11.412>1.4900 kPac a100 kPa b 1 d11.312>1.3900 kPac a100 kPa b 1 d(d) Ideal two-stage compression with intercooling (n 1.3): In this case, thepressure ratio across each stage is the same, and its value isP x 1P 1 P 2 2 1>2 31100 kPa21900 kPa24 1>2 300 kPa