It is proposed to build refrigeration plant for a cold storage to be maintained at –3ºC. The ambient temperature is 27ºC. If 5 × 10⁶ kJ/h of energy is to be continuously removed from the cold storage, the minimum power required to run the refrigerator will be

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BPSC AE Paper V Mechanical 2024 Official Paper

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Answer 1

Concept:

Minimum power required is calculated using the COP of a Carnot refrigerator:

\( \text{COP}_{\text{Carnot}} = \frac{T_L}{T_H - T_L} \)

Then, power input is calculated by: \( W = \frac{Q_L}{\text{COP}} \)

Calculation:

Given:

Cold storage temperature, \( T_L = -3^\circ C = 270~K \)

Ambient temperature, \( T_H = 27^\circ C = 300~K \)

Energy to be removed, \( Q_L = 5 \times 10^6~\text{kJ/h} = \frac{5 \times 10^6}{3600} = 1388.89~\text{kW} \)

Now,

\( \text{COP} = \frac{270}{300 - 270} = \frac{270}{30} = 9 \)

\( W = \frac{Q_L}{\text{COP}} = \frac{1388.89}{9} = 154.3~\text{kW}\)