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Question

An air refrigeration system operating in open air cycle is required to produce 25 ton refrigeration with a cooler pressure of 12 bar and the refrigerator pressure of 1 bar. The temperature of air entering the compressor is at 0C and entering the expander is 25C. Assume the isentropic efficiency for compressor to be 90% and polytropic expansion for expander to be n = 1.35. The COP of refrigeration system is (Take γ for air = 1.4 and cp for air = 1.005 kJ/kgK) .
(Correct upto 3 decimal places)
  1. 0.739

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Solution

The correct option is A 0.739


P2P1=12

T3=25C=298 K

T1=0C=273 K

T2T1=(P2P1)γ1γ=(12)0.2857

T2=555.245 K

T4T3=(P4P3)n1n=1(12)1.3510.35

T4=156.47 K

ηc=T2T1T2T1

a=555.245273T2273

T2=586.60K

So, W=cp(T2T1)

Wc=1.005(586.60273)=315.168 kJ/kg

WT=n(n1)R(T3T4)=1.350.35×0.287×(298156.47)=156.673 kJ/kg

Wnet=WcWT=158.494 kJ/kg

COP=Q2Wnet=cp(T1T4)158.494=1.005×(273156.47)158.494=0.739


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