Chapter 6, Problem 6.61P

6.107 Figure P6.107 provides the schematic of a heat pump using Refrigerant 134a as the working fluid, together with steady-state data at key points. The mass flow rate of the refrigerant is 7 kg/min, and the power input to the compressor is 5.17 kW. (a) Determine the co- efficient of performance for the heat pump. (b) If the valve were re- placed by a turbine, power could be produced, thereby reducing the power requirement of the heat pump system. Would you recommend this power-saving measure? Explain. She P2 = P3 = 9 bar Tz = 60°C Saturated liquid Condenser Expansion W = 5.17 kW Compressor valve Evaporator m= 7 kg/min P1 =P4 = 2.4 bar FIGURE P6.107

6.111 Steam enters a two-stage turbine with reheat operating at steady state as shown in Fig. P6.111. The steam enters turbine 1 with a mass flow rate of 120,000 lb/h at 1000 lbf/in.², 800°F and expands to a pressure of 60 lbf/in. From there, the steam enters the reheater where it is heated at constant pressure to 350°C before entering tur- bine 2 and expanding to a final pressure of 1 lbf/in.? The turbines operate adiabatically with isentropic efficiencies of 88% and 85%, respectively. Kinetic and potential energy effects can be neglected. Determine the net power developed by the two turbines and the rate of heat transfer in the reheater, each in Btu/h. Qin P3 = 60 lbf/in.2 T = 350°C P2 = 60 lbf/in.2 Reheater W net Turbine 1 Turbine 2 Nu = 88% Ni2 = 85% P4 =1 lbf/in.2 P1 = 1000 Ibf/in.2 T = 800°F m = 120,000 lb/h FIGURE P6.111

6.110 Figure P6.110 shows a simple vapor power plant operating at steady state with water as the working fluid. Data at key locations are given on the figure. The mass flow rate of the water circulating through the components is 109 kg/s. Stray heat transfer and kinetic and potential energy effects can be ignored. Determine a. the net power developed, in MW. b. the thermal efficiency. c. the isentropic turbine efficiency. t2 d. the isentropic pump efficiency. e. the mass flow rate of the cooling water, in kg/s. f. the rates of entropy production, each in kW/K, for the turbine, condenser, and pump. P = 100 bar T = 520°C %3D Power out Turbine P2 = 0.08 bar 2 = 90% %3D Steam Cooling water in at 20°C generator Condenser Pa= 100 bar T= 43°C Cooling water out at 35°C 4. Pump 3 P3 0.08 bar Saturated liquid Power in FIGURE P6.110 2. www