In a Rankine cycle with a single stage of reheat, steam leaves the boiler at 3.5

Question

In a Rankine cycle with a single stage of reheat, steam leaves the boiler at 3.5 MPa and 400degC and expands to .5 MPa in the high pressure turbine, after which is is reheated to 400degC and expanded to 7.5 kPa in the low-pressure turbine. Water is discharged from the condensor to the pump at 30degC. If the turbines and pump are adiabatic, with isentropic efficiencies of 80% each, and the combined power output of the two turbines is 10MW, find: A) the mass flow rate of the system B) the power output required by the pump C) the thermal efficiency of the cycle

Explanation / Answer

P1 = 3.5 MPa T1 = 400 degC h1 = 3223.14 kJ/kg s1 = 6.8426 kJ/kgK P2 = 0.5 MPa s2 isentropic = s1 = 6.8426 kJ/kgK h2 isentropic = 2757.5 kJ/kg h2 actual = 3223.14 - 0.8*(3223.14 - 2757.5) = 2850.628 kJ/kg P3 = 0.5 MPa T3 = 400 degC h3 = 3271.84 kJ/kg s3 = 7.7944 kJ/kg P4 = 7.5 kPa s4 isentropic = s3 = 7.7944 kJ/kgK h4 isentropic = 2431.21 kJ/kg h4 actual =3271.81 - 0.8*(3271.81 - 2431.21) = 2599.33 kJ/kg T5 = 30 degC x5 = 0 (Assuming liquid water only is handled by pump, not liq vapour mixture) h5 = 125.66 kJ/kg s5 = 0.436509 kJ/kgK P5 = 4.2417 kPa v5= 0.00100431 m3/kg P6 = 3.5 MPa s6 isentropic = s5 = 0.436509 kJ/kgK v = 0.00100431 m3/kg W pump isentropic = vdP (work consuming) = 0.00100431*(3500 - 4.2417) = 3.5108 kJ/kg W pump actaul = 3.5108/0.8 = 4.3885 kJ/kg Therefore h6 actual = 125.66 + 4.3885 = 130.0485 kJ/kg 10*1000 kW = m(3223.14-2850.628 + 3271.81 -2599.33) = 9.5694 kg/s W pump = 4.3885*9.5694 = 41.995 kW Eff = (1044.992-4.3885)/((3223.14-130.0485)+(3271.84-2850.628)) = 29.610% Refer http://www.waterproperties.eu/ for thermodynamic tables

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