8-142 8-189 An insulated rigid tank is connected to a piston-cylinder device with zero clearance that is maintained at constant pressure. A valve is opened, and some steam in the tank is allowed to flow into the cylinder. The final temperatures in the tank and the cylinder are to be determined. Assumptions 1 Both the tank and cylinder are well-insulated and thus heat transfer is negligible. 2 The water that remains in the tank underwent a reversible adiabatic process. 3 The thermal energy stored in the tank and cylinder themselves is negligible. 4 The system is stationary and thus kinetic and potential energy changes are negligible. Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From the steam tables (Tables A-4 through A-6), v 1 v g @500 kPa 0.37483 m 3 /kg 500 kPa ½ ¾ u1 u g @500 kPa 2560.7 kJ/kg sat.vapor ¿s s g @500 kPa 6.8207 kJ/kg K 1 P1 150 kPa ½ ° s 2 s1 ¾ sat.mixture °¿ v 2, A P2 T2, A Tsat @150 kPa 111.35 qC s 2, A s f 6.8207 1.4337 x 2, A v f x 2, Av fg s fg 0.9305 0.001053 (0.9305)(1.1594 0.001053) 1.0789 m 3 /kg u f x 2, A u fg u 2, A 5.7894 466.97 (0.9305)(2052.3 kJ/kg ) 2376.6 kJ/kg The initial and the final masses in tank A are m1, A VA v1, A m2, B m1, A m2, A Thus, 0.4 m3 and m2, A 1.067 kg 0.37483 m3/kg VA v 2, A 0.4 m3 1.0789 m3/kg 0.371 kg 1.067 0.371 0.696 kg (b) The boundary work done during this process is Wb,out ³ 2 1 P dV PB V 2, B 0 PB m 2, Bv 2, B Taking the contents of both the tank and the cylinder to be the system, the energy balance for this closed system can be expressed as E Eout in Net energy transfer by heat, work, and mass Wb, out 'Esystem 150 kPa Change in internal, kinetic, potential, etc. energies 'U 'U A 'U B Wb,out 'U A 'U B or, Sat. vapor 500 kPa 0.4 m3 PB m2, Bv 2, B m2u2 m1u1 A m2u2 B m2, B h2, B m2u2 m1u1 A 0 0 0 Thus, h 2, B m1u1 m 2 u 2 A 1.0672560.7 0.3712376.6 m 2, B 0.696 2658.8 kJ/kg At 150 kPa, hf = 467.13 and hg = 2693.1 kJ/kg. Thus at the final state, the cylinder will contain a saturated liquid-vapor mixture since hf < h2 < hg. Therefore, T2, B Tsat @150 kPa 111.35qC PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.