SSC JE Mechanical Engineering 11 Oct 2020 Shift 1 Papers

Question : 173 of 200

Marks: +1, -0

Ideal refrigeration system is used to cool the system at 5°C. Heat rejection happens at 100°C. If ambient temperature is 30°C, COP of the system is:

Solution:

T_1 = 5°C = 278 K, T_2 = 100°C = 373 K

COP=Q_1/\WR=T_1/{T_2-T_1}

COP = 278\/95 = 2.93

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