Answers
T_1 = 500 degree c & P_1 = 3 MPa T_2 = 50 degree c from steam tables at P_1 = 3 MPa T_sat = 233.9 degree c therefore T_1 > T_sat => superheated vapour at T_1 = 500 degree c & P_1 = 3 MPa h_1 = 3456.5 kJ/kg: s_1 = 7.2338 kJ/kg at T_3 = 50 degree c, P_sat = 0.1235 bar therefore pressure is remains same at state '2' also therefore S_f = 0.7033 kJ/kgk: S_g = 8.0763 kJ/kg k therefore turbine is isentropic, s_1 = s_2 s_f < s_2 < s_g => saturated light vapour therefore s_2 = s_f + x_2 (s_g - s_f) => x_2 7.2338 - 0.7038/8.0763 - 0.7038 x_2 = 0.885 leftarrow quality at condenser inlet at T_3 = 50 degree c, v_f = 1.0121 times 10^-3 m^3/kg s_3 = s_f = 0.7038 kJ/kg k therefore w_p = v_f (p_4 - p_3) = 1.0121 times 10^-3 (3 times 10^6 - 0.1235
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