Question

An Actual Steam Power Cycle A steam power plant operates on the cycle. If the isentropic efficiency of the turbine is 87 percent and the isentropic efficiency of the pump is 85 percent, determine (a) the thermal efficiency of the cycle and (b) the net power output of the plant for a mass flow rate of 15 kg/s. 15.9 MPa 35°C 15.2 MPa 625°C Boller 15 MPa 5 t 600 C 16 MPa 4 Wturb,out Turbine M7, 0.87 pump.in Pump lp 0.85 6 10 kPa 9 kPa 38°C 2s Condenser

Answer 1

At (1) 9 kPa h_f = 183.21 kJ/kg (ep_2) = 4.18 kJ/kg middot k, v = 0.001009 m^3/kg T_sat = 43.79 degree C h_1 = h_f - (C_p)_r (T_sat - 38 degree) = 159 kJ/kg h_23 = h_1 + v middot Delta P = 159 + 0.001009 (16 times 1063 - 9) = 175.13 kJ/kg But, eta_p = 85% = h_23 - h_1/h_2 - h_1 so, h_2 = 177.98 kJ/kg At - 3, 15.9 MPa, 35 degree C hf_3 = 1645 middot 6 kJ/kg middot T_sat = 347.4 degree C h_3 = h_f3 - (Cp)_2 (347.4 degree - 35 degree) = 348.768 kJ/kg \r\n at - 4, 15.2 MPa, 625 degree C (Superheated state) h_4 = 3646.61 kJ/kg At - 5 15 MPa, 600 degree C (Superheated state) h_5 = 3579.77 kJ/kg S_5 = 6.6756 kJ/kg-k At - 6, 10 kPa h_f_6 = 191.75 kJ/kg s_f_6 = 0.6489 kJ/kg-k h_fg_6 = 2393.25 kJ/kg s_fg_c = 7.5021 kJ/kg.k Considering iseatropic expansion, s_5 = s_6 6.6756 = s_f_6 + x s_fg_6 6.6756 = 0.6489 + x(7.5021) x = 0.8 h_65 = h_f6 + x h_fg6 = 2114.33 kJ/kg