Unformatted text preview: tial state of the refrigerant is completely specified.
Assumptions The volume of the tank is constant and thus v2 v1.
Analysis Recognizing that the specific volume remains constant during this
process, the properties of the refrigerant at both states are State 1: P1
T1 140 kPa
υ1 1.0532 kJ/kg · K
0.1652 m3/kg State 2: P2 100 kPa
(υ2 υ1) υf
υg 0.0007258 m3/kg
0.1917 m3/kg The refrigerant is a saturated liquid–vapor mixture at the final state since
vf v2 vg at 100 kPa pressure. Therefore, we need to determine the quality
first: x2 υ2 υf
υ fg 0.1652
0.0007258 0.861 Thus, s2 sf x2sfg 0.0678 (0.861)(0.9395 0.0678) 0.8183 kJ/kg · K Then the entropy change of the refrigerant during this process is S m(s2 s1) (5 kg)(0.8183
1.175 kJ/K 1.0532) kJ/kg · K Discussion The negative sign indicates that the entropy of the system is decreasing during this process. This is not a violation of the second law, however,
since it is the entropy generation Sgen that cannot be negative. υ= con s...
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