Question

In Geothermal-steam power cycle, the source of heat is the hot water from well. The heat exchanger receive hot water at Tg1= 105 oC and the water leaves at 40 oC as shown in Figure 3. Steam exits the boiler at 40 bar, 300 oC, and it exits the turbine at 1 bar. Saturated liquid water exits the condenser at 1 bar. The plant operate to generate 20 MW electric power with generator efficiency of 0.95. The isentropic efficiency of the turbine is 0.85, and pump isentropic efficiency is 0.8.

The system is insulated so you can disregard heat transferred to surroundings.

a) Calculate the mass flow rate of the steam b) Calculate the mass flow rate of the hot water from the geothermal well c) The rejected heat from the condenser d) Calculate the thermal efficiency of the cycle consider only the turbine output power as a useful power

p1=40 bar Tl-300 °C Heat exchanger Turbine Generator 20 MW Tgl-105 °C Tg2-40 °C 1 bar Condenser Figure 3 Table 3 Data Numeric Unit value Hot water to the heat exchanger temperature,Tgl105 Hot water to the well temperature, Tg2 Steam pressure p4 p Steam temperature ti Pressure after turbine and condenserp2-ps Wet vapour quality after condenser Turbine isentropic efficiency,nsT Pump isentropic efficiency, ? Generator efficiency, ? G oC 40 40 300 oC ar oC ar (saturated liquid). x30 0.85 0.8 0.95

Answer 1

given T_g1 = 105 degree c T_g2 = 40 degree P_1 = p_4 = 40 bar = 4 mpa T_1 = 300 degree c P_2 = P_3 = 1 bar = 0.1 mpa x_3 = 0 eta T = 0.85 eta_p = 0.8 eta_a = 0.95 as generator efficiency = 95 % = 0.95 power generated by turbine w_T = w _alpha = electric power eta_alpha = n = 20 /0.95mw = 21.053 mw first from steam tubic for P_1 = 4 mpa T_1 = 300 degree c } h_1 = 2961.7 KJ/kg s_1 = 6.3639 KJ/kg k now , p_2 = 0.1 mpa s_2s = s_1 = 6.3639 KJ/kg k and s_2s = s_f @ 0.1 mpa + x_2s s_fg @ 0.1 mpa therefore 6.3639 = 1.3028 + x_2s (6.0562) therefore x_2s = 0.8357 hence , h_2s = hf @ 0.1 mpa + x_2s hfg @ 0.1 mpa = 417 .51 + x_2 s (2257.5) = 2304.1 KJ/kg from eta_T = 0.85 eta_T = h_1 - h_24 h_1 - h_2s therefore 0.85 = 2961.7 - h_29 /2961.7 - 2304 . 1 therefore h_2a = 2402.74 KJ/kg h_3 = h_f @ 0.1 mpa = 417.51 KJ/kg v_3 = v_f @ 0.1 mpa = 0.001043 m^3 /kg