Question

Refrigerant-134a enters the coils of the evaporator of a refrigeration system as a saturated liquid–vapor mixture at a pressure of 140 kPa. The refrigerant absorbs 180 kJ of heat from the cooled space, which is maintained at −10°C, and leaves as saturated vapor at the same pressure. Determine (a) the entropy change of the refrigerant, (b) the entropy change of the cooled space, and (c) the total entropy change for this process.

Answer 1

Answer:

A) ΔS_refrigerant = 0.70754 Kj/K

B) ΔS_space = -0.68441 Kj/K

C) ΔS_total = 0.02313 Kj/K

Explanation:

A) From he table attached, at Pressure of 140 KPa, and by interpolation, we get, Temperature of T = -18.77°C

Converting to degree kelvin yields;

T = -18.77 + 273 = 255.23 K

Formula for entropy change of refrigerant is given as;

ΔS_refrigerant = Q_in/T_refrigerant

We are given Q = 180 KJ

Thus, ΔS_refrigerant = 180/255.23 = 0.70754 Kj/K

B) Formula for entropy change of cooled space is given as;

ΔS_space = Q_out/T_s pace

T_space = -10°C = 273 - 10 = 263K

Thus, ΔS_space = -180/263 = -0.68441 Kj/K

C) the total entropy change would be;

ΔS_total = ΔS_refrigerant + ΔS_space

Thus,

ΔS_total = 0.70754 - 0.68441 = 0.02313 Kj/K