Thermodynamics

342 | ThermodynamicsEXAMPLE 7–4Entropy Change during a Constant-PressureProcessA piston–cylinder device initially contains 3 lbm of liquid water at 20 psiaand 70°F. The water is now heated at constant pressure by the addition of3450 Btu of heat. Determine the entropy change of the water during thisprocess.Solution Liquid water in a piston–cylinder device is heated at constantpressure. The entropy change of water is to be determined.Assumptions 1 The tank is stationary and thus the kinetic and potentialenergy changes are zero, KE PE 0. 2 The process is quasi-equilibrium.3 The pressure remains constant during the process and thus P 2 P 1 .Analysis We take the water in the cylinder as the system (Fig. 7–13). This isa closed system since no mass crosses the system boundary during theprocess. We note that a piston–cylinder device typically involves a movingboundary and thus boundary work W b . Also, heat is transferred to the system.Water exists as a compressed liquid at the initial state since its pressure isgreater than the saturation pressure of 0.3632 psia at 70°F. By approximatingthe compressed liquid as a saturated liquid at the given temperature, theproperties at the initial state areState 1:P 1 20 psiaf s 1 s f @ 70°F 0.07459 Btu>lbm # RT 1 70°F h 1 h f @ 70°F 38.08 Btu>lbmAt the final state, the pressure is still 20 psia, but we need one more propertyto fix the state. This property is determined from the energy balance,E in E out ¢E systemNet energy transferby heat, work, and massChange in internal, kinetic,potential, etc., energiessince U W b H for a constant-pressure quasi-equilibrium process. Then,State 2:⎫ ⎪⎬⎪⎭⎫⎪⎬⎪⎭Q in W b ¢UQ in ¢H m 1h 2 h 1 23450 Btu 13 lbm2 1h 2 38.08 Btu>lbm2h 2 1188.1 Btu>lbmP 2 20 psiah 2 1188.1 Btu>lbm fs 2 1.7761 Btu>lbm # R1Table A-6E, interpolation2TP = const.2FIGURE 7–13Schematic and T-s diagram forExample 7–4.Q inH 2 OP 1 = 20 psiaT 1 = 70°F1s 1s 2s