Efficiency of the Carnot engine is given as 80%. If the cycle direction be reversed, what will be the value of coefficient of performance of reversed Carnot cycle refrigerator?

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Question

This question was previously asked in

SSC JE ME Previous Paper 16 ( Held on: 1 March 2017 Evening)

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Answer 1

Concept:

Carnot Cycle:

SSC JE me 1

Reversed Carnot Cycle:

SSC JE me 2

\({η _{carnot}} = \frac{{{T_H} - {T_L}}}{{{T_H}}}\)

By reversing the Carnot cycle we can have either heat pump or refrigerator.

COP of Reversed Carnot cycle (H.P):

\(CO{P_{HP}} = \frac{{{T_H}}}{{{T_H} - {T_L}}} = \frac{1}{η }\)

COP of Refrigerator:

\(CO{P_{Ref}} = \frac{{{T_L}}}{{{T_H} - {T_L}}} = CO{P_{HP}} - 1\)

Calculation:

Given: η = 0.8

Using the relation \(CO{P_{HP}} = 1 + CO{P_R} = \frac{1}{{{η _{HE}}}}\)

\( \Rightarrow CO{P_R} = \frac{1}{{0.8}} - 1 = 0.25\)