Thermodynamics

364 | ThermodynamicsIn gas power plants, the working fluid (typically air) is compressed in thegas phase, and a considerable portion of the work output of the turbine isconsumed by the compressor. As a result, a gas power plant delivers less network per unit mass of the working fluid.EXAMPLE 7–12Compressing a Substance in the Liquid versusGas PhasesDetermine the compressor work input required to compress steam isentropicallyfrom 100 kPa to 1 MPa, assuming that the steam exists as (a) saturatedliquid and (b) saturated vapor at the inlet state.Solution Steam is to be compressed from a given pressure to a specifiedpressure isentropically. The work input is to be determined for the cases ofsteam being a saturated liquid and saturated vapor at the inlet.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potentialenergy changes are negligible. 3 The process is given to be isentropic.Analysis We take first the turbine and then the pump as the system. Bothare control volumes since mass crosses the boundary. Sketches of the pumpand the turbine together with the T-s diagram are given in Fig. 7–43.(a) In this case, steam is a saturated liquid initially, and its specific volume isv 1 v f @ 100 kPa 0.001043 m 3 >kg1Table A–52which remains essentially constant during the process. Thus,2w rev,in vdP v 1 1P 2 P 1 21 10.001043 m 3 1 kJ>kg2311000 1002 kPa4a1 kPa # b m3 0.94 kJ>kg(b) This time, steam is a saturated vapor initially and remains a vapor duringthe entire compression process. Since the specific volume of a gas changesconsiderably during a compression process, we need to know how v varieswith P to perform the integration in Eq. 7–53. This relation, in general, is notreadily available. But for an isentropic process, it is easily obtained from theTP 2 = 1 MPaP 2 = 1 MPa21 MPaPUMPCOMPRESSOR(b)FIGURE 7–43Schematic and T-s diagram forExample 7–12.P 1 = 100 kPa(a) Compressinga liquidP 1 = 100 kPa(b) Compressinga vapor(a)21100 kPa1s