Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid / saturated vapor mixture at a pressure of 160 kPa. The refrigerant absorbs 180 kJ of heat from the cooled space, which is maintained at -5°C, and leaves as saturated vapor at the same pressure. For this process, determine ...
a.) the entropy change of the refrigerant
b.) the entropy change of the cooled space
c.) the total entropy change
3 comments:
graham 4:48 PM
a.)
Because the process is isothermal, the refrigerant behaves like a thermal reservoir and you can evaluate deltaS of the refrigerant using the definition of entropy.
b.)
The T of the refrigerated space is also constant, so you treat it as a thermal reservoir as well.
You only need the R-134a tables and the P in order to determine the T of the refrigerant.
For part (c), my instinct is that deltaS(univ) is zero because deltaS(ref) + deltaS(space) = 0. Obviously this is not correct. Could you give me some further hints?
Carina 1:47 AM
Both the refrigerant and the thermal reservoir operate at constant (but different) temperatures. Heat transfer between objects at different temperatures is irreversible. Therefore, we expect: Sgen = deltaSuniv > 0.
Also, because T=constant for both the refrigerant and the reservoir, you can evaluate deltaS for each from the definition of entropy. The Q is the same in each case but the T is different. You should/will find that the entropy of the refrigerant increases by a larger amount than the entropy of the reservoir decreases. As a result, deltaSuniv > 0.
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