4版講稿例題a466398英

1、A466398（英）A steam power plant operates on a cycle with pressures andtemperature as designated in following Fig. The efficiency of the turbine is 86% and the efficiency of the pump is 80%. Determine the thermal efficiency of this cycle. Assume that Each process is steady state with no changes in kine

2、tic or potential energy.12AnalysisThis cycle is an actual one deviates from an ideal cycle. The most important of the losses are due to the turbine, the pump, the pipes, and the condenser. For the turbine we haveControl volume: TurbineInlet state:p5,T5known; state fixedExit state: p6known.From the f

3、irst and the second laws we havewt= h5 - h6 ,= s5s6sThe efficiency of the turbine ish5 - h6wth=th- hh- h56 s56 sFrom the steam tables we geth5 = 3169.1kJ/kg,s5 = 6.7235kJ /(kg K)= s5 = 6.7235kJ /(kg K)s6s= 0.6490kJ /(kg K) + x4 (8.1481- 0.6490)kJ /(kg K)= 0.8098x6sh6s = h6s + x6s (h6s- h6s )= 191.8k

4、J/kg + 0.8098(2583.72 -191.76)kJ/kg = 2129.5kJ/kgwt= hT (h5 - h6s )= 0.86(3169.1- 2129.5)kJ/kg = 894.1kJ/kg Fog the pump we haveControl volume: PumpInlet state:p1,T1 known; state fixedExit state: p2known.3The first law givesh2 - h1wP = h2 - h12vdp = v( p2- p1 )=1The Second law gives s2s= s1The pump

5、efficiency is- h1- h1= h2 s= h2 shPh- hwP21- h1 = v( p2- p1 )= h2sThereforewPhhPP= 0.001009m /kg(5000 -10)kPa =36.3kJ/kg0.8= wt - wP = 894.1kJ/kg - 6.3kJ/kg = 887.8kJ/kgwnet4 Finely,fog the boiler :Control volume: BoilerInlet state:p3,T3known; state fixed.Exit state:p4,T4known; state fixed.qH = h4 - h3The first law isSubstituting givesqH= h4 -