SSC JE Mechanical Engineering 25 Sep 2019 Shift 1 Paper

Question : 168 of 200

Marks: +1, -0

The specific enthalpies of a refrigerant in a vapour compression refrigeration cycle are:
Entrance of evaporator = 100 kJ/kg, Exit of evaporator = 1200 kJ/kg, Entrance of condenser = 1300 kJ/kg. Its coefficient of performance (COP) is:

12
9.24
11
10.56

Validate

Solution:

Given:

h_1

(Exit of evaporator) = 1200 kJ/kg

h_2

(Entrance of condenser) = 1300 kJ/kg

h_4

(Entrance of evaporator) = 100 kJ/kg

COP = (h_1 -h_4) \/ (h_2 -h_1)

COP = (1200-100) / (1300-1200) = 11

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