Subtopic: Refrigeration and cold storage basics and applications
Question Statement:
A refrigerator with a COP of 3.2 uses 2.4 kg min⁻¹ refrigerant extracting 150 kJ kg⁻¹ heat in the evaporator. Assuming compressor efficiency of 85% the minimum size of the motor is
Options:
(A) 0.5 hp
(B) 1.5 hp
(C) 2.0 hp
(D) 3.0 hp
Official Answer Key:
D
Step-by-Step Solution & Explanation
Refrigeration load, Q_evap = ṁ×q = (2.4/60)×150 = 6.0 kW. Ideal compressor work, W_ideal = Q_evap/COP = 6.0/3.2 = 1.875 kW. Actual (shaft) work accounting for compressor efficiency, W = 1.875/0.85 = 2.206 kW = 2.96 hp → minimum standard motor size ≈3.0 hp.
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