Question

Steam generated in a power plant at a pressure of 8,600 kPa and temperature of
600°C is fed to a turbine. Exhaust from the turbine enters a condenser at 10 kPa, where it is
condensed to saturated liquid, which is then pumped to the boiler. The turbine and the pump in the Rankine cycle operate at 80% and 75% efficiency, respectively.

a. Determine the rates of heat transfer in the boiler

b.Determine the rate of heat transfer in the condenser

QH 2 Boler W, (turbine) 1 Turbine W, (pump) 3 Condenser_ 4 2c

Answer 1

The enthalpy of saturated liquid at 10 kpa, H_3 = 191.8 KJ/Kg therefore Q_out = H_3 - H_2 = 191.8 - 2496.1 = -2304.5 kJ/kg properties of saturated liquid at 10 kpa, 45 degrees c, V = 0.0010 m^3/kg, Cp = 4.178 kJ/kg.k therefore W_p (Insentropic) = delta PV = 0.0010 times (8600 - 10) = 8.676 kJ/kg pump efficiency eta_p is given as 0.75 eta_p = (delta H)_s/delta H therefore (delta H) = 8.676/0.75 = 11.57 kJ/Kg delta H = H_1 - H_4 = 11.57 H_1 = H_4 + 11.57 = 191.8 + 11.57 = 203.37 kJ/kg \r\n from quality, H^1_2 = H^l_2 + x^1_2 (H^v_2 - H^l_2) {from steam tables} H^1_2 = 191.8 + 0.844 times (2584.8 - 191.8) = 2211.492 kJ/kg therefore H^1_2 = 2211.492 kJ/kg We know, eta_turbine = delta H/(delta H)_s (delta H)_s = H^1_2 - H_1 = 2211.492 - 3634.58 = -1423 kJ/kg eta_Turbine = 0.8 = delta H/-1423 therefore delta H = H_2 - H_1 = 0.8 times (-14723) = -1138.4 therefore