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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 0∘C and entering the expander is 25∘C. 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)0.739

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Solution

## The correct option is A 0.739 P2P1=12 T3=25∘C=298 K T1=0∘C=273 K T2T1=(P2P1)γ−1γ=(12)0.2857 T2=555.245 K T4T3=(P4P3)n−1n=1(12)1.35−10.35 T4=156.47 K ηc=T2−T1T2−T1 a=555.245−273T′2−273 T′2=586.60K So, W=cp(T′2−T1) Wc=1.005(586.60−273)=315.168 kJ/kg WT=n(n−1)R(T3−T4)=1.350.35×0.287×(298−156.47)=156.673 kJ/kg Wnet=Wc−WT=158.494 kJ/kg COP=Q2Wnet=cp(T1−T4)158.494=1.005×(273−156.47)158.494=0.739

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