Solution
- At the inlet of turbine, P3=6MPa, 100% saturated vapor x3=1, from saturated table A-5, h3=hg=2784.3(kJ/kg), s3=sg=5.89(kJ/kg K)
- From 3-4, isentropic expansion: s3=s4=5.89 (kJ/kg K)
- From 4-1, isothermal process, T4=T1=45.8°C (why?)
From table A-5, when T=45.8°C, sf4=0.6491, sfg4=7.5019, hf4=191.8, hfg4=2392.8
x4 = (s4-sf4)/sfg4 = (5.89-0.6491)/7.5019 = 0.699
h4 = hf4+x4* hfg4 = 191.8+0.699(2392.8) = 1864.4 (kJ/kg)
- At the inlet of the pump: saturated liquid h1=hf1=191.8
qout = h4-h1=1672.6(kJ/kg)
- At the outlet of the pump: compressed liquid v2=v1=vf1=0.00101(m3/kg)
work input to pump Win = h2-h1 = v1 (P2-P1) = 0.00101(6000-10) = 6.05
h2 = h1 + v1 (P2-P1) =191.8 + 6.05 = 197.85 (kJ/kg)
- In the boiler, qin=h3-h2=2784.3-197.85=2586.5(kJ/kg)