Thermodynamics

372 | ThermodynamicsT,°CP 1 = 3 MPaT 1 = 400°CSTEAMTURBINE2 MW14003 MPa100 50 kPa2s2Actual processIsentropic processFIGURE 7–50Schematic and T-s diagram forExample 7–14.P 2 = 50 kPaT 2 = 100°Cs 2s = s 1sAnalysis A sketch of the system and the T-s diagram of the process are givenin Fig. 7–50.(a) The enthalpies at various states areState 1:State 2a:The exit enthalpy of the steam for the isentropic process h 2s is determined fromthe requirement that the entropy of the steam remain constant (s 2s s 1 ):State 2s:Obviously, at the end of the isentropic process steam exists as a saturatedmixture since s f s 2s s g . Thus we need to find the quality at state 2s first:andP 1 3 MPAT 1 400°C fh 1 3231.7 kJ>kgs 1 6.9235 kJ>kg # K1Table A–62P 2a 50 kPaT 2a 100°C fh 2a 2682.4 kJ>kg1Table A–62P 2s 50 kPaS s f 1.0912 kJ>kg # K1s 2s s 1 2s g 7.5931 kJ>kg #1Table A–52Kx 2s s 2s s f 6.9235 1.0912 0.897s fg 6.5019h 2s h f x 2s h fg 340.54 0.897 12304.72 2407.9 kJ>kgBy substituting these enthalpy values into Eq. 7–61, the isentropic efficiencyof this turbine is determined to beh T h 1 h 2a 3231.7 2682.4 0.667, or 66.7%h 1 h 2s 3231.7 2407.9(b) The mass flow rate of steam through this turbine is determined from theenergy balance for steady-flow systems:E # in E # outm # h 1 W # a,out m # h 2aW # a,out m # 1h 1 h 2a 21000 kJ>s2 MW a1 MW b m# 13231.7 2682.42 kJ>kgm # 3.64 kg/s