(1) Determine the entropy change of refrigerant
R134a
Take refrigerant R134a as a system (system A). It
enters the evaporator as a saturated liquidvapor mixture, which is state
1.
P_{1} = 200 kPa
T_{1} = T_{sat. @ 200 kPa} =
10.09^{o}C
Refrigerant R134a leaves the evaporator as saturated vapor, which
is state 2.
P_{2} = 200 kPa
T_{2} = T_{sat. @ 200 kPa} =
10.09^{o}C
During this process, refrigerant R134 absorbs 150 kJ heat.
Q = 150 kJ
Hence, the entropy change for refrigerant R134a from process 12 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,
ΔS_{A} = 150(1000)/(10.09
+ 273) = 570.56 J/K
