Question

3. Figure P8.78 shows a vapor power cycle that provides process heat and produces power. The steam generator produces vapor at 500 lbf/in2, 800°F, at a rate of 8x104 Ib/h. Eighty-eight percent of the steam expands through the turbine to 10 lbf/in.2 and the remainder is directed to the heat exchanger. Saturated liquid exits the heat exchanger at 500 lbf/in.* and passes through a trap before entering the condenser at 10 lbf/in.2 Saturated liquid exits the condenser at 10 lbf/in.2 and is pumped to 500 lbf/in2 before entering the steam generator. The turbine and pump have isentropic efficiencies of 85% and 89%, respectively. Determine (a) the process heat production rate, in Btu/h. (b) the thermal efficiency of the cycle. Pi = 500 lbf/in? Ti = 800°F i m, = 8 x 104 lb/h = 85% Steam generatorI Turbineに.. w, P2 = 10 lbf/in? Heat exchanger P4 500 lbf/in.2 saturated liquid out Condenser Pump Ps = 10 lbf/in.2 saturated liquid Tp = 89% Fig. P8.78

Answer 1

Properties at solution and 800 F h_1 = 1413.5 Btu/lb S_1 = 1.6586 Btu/lbR. m_s = 8 times 10^4 lb/hr mass of steam compounded in turbine(mT) m_T = 88 % m_s = 0.88 times 10^4 m_T = 70400 lb/hr may of stream for properly heating(m_p) m_p = 8times 10^4 - 70400 = 9600 lb/nr properties of 10 lb/v^2 h_3 = h_f_2 = 161.36 Btu/lb s_f_2 = 0.2838.Btu/lb R h_f_g_2 = 982.44 Btu/lb s_f_2 = 1.50489 Btu/lb v_f_3 = 0.016589 ft^3/lb properties at 500 lbt/ui^2 and satuvated liquid h_4 = 449.84 Btu/lb mu_r = 85% = 0.85 mu_p = 89% = 0.89 process heat (Q_p) = m_p (h_1 - h_4) = 9600 times (1412.5 -449.84) Q_p = 9251136 Btu/nr s_1 = s_2s = 1.6586 = s_f_2 + x_2 s_f_2 1.6586 = 0.2838 + x_2 times 1.50489 x_2 = 0.9135 h_2s = h_f_2 + x_2 h_f_g2 = 161.36 + 0.9135 times 982.44 h_2s = 1058.9