• question_answer The operating temperature of a cold storage is $-2{}^\circ C,$ Heat leakage form the surrounding is 30 kW for the ambient temperature of $40{}^\circ C.$ the actual COP of the refrigeration plant used is one fourth that of an ideal plant working between the same temperatures. The power required to drive the plant is: A) 1.86 kW           B) 3.72 kWC) 7.44 kW           D) 18.60 kW

${{(COP)}_{th}}=\frac{{{T}_{2}}}{{{T}_{1}}-{{T}_{2}}}=\frac{271}{313-271}=6.4524$ ${{(COP)}_{act}}=\frac{6.4524}{4}=1.613$ Worked required $\frac{Q}{{{(COP)}_{act}}}=\frac{30}{1.613}=18.60\,\,kW$