Thermodynamics

Chapter 7 | 341EXAMPLE 7–3Entropy Change of a Substance in a TankA rigid tank contains 5 kg of refrigerant-134a initially at 20°C and 140 kPa.The refrigerant is now cooled while being stirred until its pressure drops to100 kPa. Determine the entropy change of the refrigerant during this process.Solution The refrigerant in a rigid tank is cooled while being stirred. Theentropy change of the refrigerant is to be determined.Assumptions The volume of the tank is constant and thus v 2 v 1 .Analysis We take the refrigerant in the tank as the system (Fig. 7–12). Thisis a closed system since no mass crosses the system boundary during theprocess. We note that the change in entropy of a substance during a processis simply the difference between the entropy values at the final and initialstates. The initial state of the refrigerant is completely specified.Recognizing that the specific volume remains constant during thisprocess, the properties of the refrigerant at both states areState 1:State 2:The refrigerant is a saturated liquid–vapor mixture at the final state sincev f v 2 v g at 100 kPa pressure. Therefore, we need to determine thequality first:Thus,P 1 140 kPaf s 1 1.0624 kJ>kg # KT 1 20°C v 1 0.16544 m 3 >kgP 2 100 kPaf v f 0.0007259 m 3 >kg1v 2 v 1 2 v g 0.19254 m 3 >kgx 2 v 2 v f 0.16544 0.0007259v fg 0.19254 0.0007259 0.859s 2 s f x 2 s fg 0.07188 10.859210.879952 0.8278 kJ>kg # KThen the entropy change of the refrigerant during this process is¢S m 1s 2 s 1 2 15 kg2 10.8278 1.06242 kJ>kg # K 1.173 kJ/KDiscussion The negative sign indicates that the entropy of the system isdecreasing during this process. This is not a violation of the second law,however, since it is the entropy generation S gen that cannot be negative.m = 5 kgRefrigerant-134aT 1 = 20°CP 1 = 140 kPa∆S = ?HeatT12s 2s 1v = const.sFIGURE 7–12Schematic and T-s diagram forExample 7–3.