GATE AG 2013 • Q25 MCQ • 1 Mark

Section 7: Dairy and Food Engineering

Preservation of Food

Subtopic: Refrigeration and cold storage basics and applications

Question Statement:

The horse power of the motor running the compressor of a refrigerator having COP of 4.5 and extracting 200 kJ kg⁻¹ of evaporating heat with 1.5 kg min⁻¹ refrigerant flow rate is

Options:

  • (A) 0.5
  • (B) 1.0
  • (C) 1.5
  • (D) 2.0

Official Answer Key:

C

Step-by-Step Solution & Explanation

Refrigeration effect (evaporator heat load), Q_evap = 200 kJ/kg × (1.5/60) kg/s = 200×0.025 = 5.0 kW. Compressor work, W = Q_evap/COP = 5.0/4.5 = 1.111 kW. Converting to horsepower: 1.111/0.7457 = 1.49 ≈ 1.5 HP. Answer: (C).

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