In a vapor-compression refrigeration cycle, R-134a exits the evaporator as satur

Question

In a vapor-compression refrigeration cycle, R-134a exits the evaporator as saturated vapor at 20lbf/in2. The refrigerant enters the condenser at 160 lbf/in2 and 140?F, and then exits as a saturated liquid at the same pressure. There is no significant heat transfer between the compressor and its surroundings, and the refrigerant passes through the evaporator with a negligible change in pressure. If the refrigerant capacity is five tons, determine:
a) The mass flow rate of refrigerant, in lbm/min (Ans: 18.61 lbm/min)
b) The power input to the compressor in Btu/min (Ans: 428.4 Btu/min)
c) The coefficient of performance (Ans: 2.33)
d) The isentropic compressor efficiency (Ans: 80.6%)

Explanation / Answer

5 tons = 5*200 Btu/min = 1000 Btu/min

For R-134a, at 20 lbf/in2 we get h1 = hg = 101.39 Btu/lb, s1 = sg = 0.2227 Btu/lb-R

For 160 lbf/in2 and 140 deg F, we get h2 = 124.41 Btu/lb

For 160 lbf/in2 and s = 0.2227 Btu/lb-R, we get h2_is = 119.95 Btu/lb

At 160 lb/in2, h3 = hf = 47.65 Btu/lb

During throttling process, h4 = h3 = 47.65 Btu/lb.

At 20 lbf/in2, we get hf = 10.89 Btu/lb

x4 = (h4 - hf)/(hg hf) = (47.65 - 10.89)/(101.39 - 10.89) = 0.406

a) Qin = 1000 = m(h1 - h4)

m = 1000/(101.39 - 47.65) = 18.61 lbm/min

b) Power input to compressor = m(h2 - h1) = 18.61*(124.41 - 101.39) = 428.4 Btu/min

c) Coefficient of Performance = 1000/428.4 = 2.33

d) Isentropic efficiency = (h2_is - h1)/(h2 - h1) = (119.95 - 101.39)/(124.41 - 101.39) = 0.806 = 80.6 %

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