(1) Determine the entropy change of refrigerant
R-134a
Take refrigerant R-134a as a system (system A). It
enters the evaporator as a saturated liquid-vapor mixture, which is state
1.
P1 = 200 kPa
T1 = Tsat. @ 200 kPa =
-10.09oC
Refrigerant R-134a leaves the evaporator as saturated vapor, which
is state 2.
P2 = 200 kPa
T2 = Tsat. @ 200 kPa =
-10.09oC
During this process, refrigerant R-134 absorbs 150 kJ heat.
Q = 150 kJ
Hence, the entropy change for refrigerant R-134a from process 1-2 is
Temperature T is a constant in this case, and the above equation can
be integrated as
Substituting Q and T into the above equation yields,
ΔSA = 150(1000)/(-10.09
+ 273) = 570.56 J/K
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