A utility plant runs an ideal Rankine cycle with water. The boiler pressure is 3

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Solution: 1: 45C , x = 0 => h1 = 188.42 , v1 = 0.00101 , Psat = 9.6 kPa 3: 3.0MPa , 450C => h3 = 3337.2 , s3 = 7.0051 C.V. Pump Rev adiabatic -wP = h2 - h1 ; s2 = s1 since incompressible it is easier to find work as -wp = ? v dP = v1 (P2 - P1) = 0.00101 (3000 - 9.6) = 3.020 => h2 = h1 - wp = 188.42 + 3.020 = 191.44 C.V. Boiler : qh = h3 - h2 = 3337.2 - 191.44 = 3144.79 C.V. Turbine : wt = h3 - h4 ; s4 = s3 s4 = s3 = 7.0051 = 0.6386 + x4 (7.5261) => x4 = 0.8459 => h4 = 188.42 + 0.8459 (2394.77) = 2214.2 wt = 3337.2 - 2214.2 = 1123 kJ/kg C.V. Condenser : qL = h4 - h1 = 2214.2 - 188.42 = 2025.78 kJ/kg ?cycle = wnet / qH = (wt + wp) / qH = (1123 - 3.0) / 3145.3 = 0.356

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