Subtopic: Preservation of food by cooling and freezing
Question Statement:
Preservation of Food: Kinetics of microbial death – pasteurization and sterilization of milk and other liquid foods; preservation of food by cooling and freezing; refrigeration and cold storage basics and applications.
A chiller working on mechanical vapour compression refrigeration (COP = 4.5) is used for cooling 12500 kg of fresh cow milk (cp = 3.8 kJ/kg·K) from 30°C to 4°C in 3 hours. Assuming ideal compression, the power consumed by the electric motor (kW) and the tonnage of refrigeration (TR), respectively, are
Options:
(A) 25.4 and 32.5
(B) 25.4 and 114.3
(C) 32.5 and 25.4
(D) 114.3 and 25.4
Official Answer Key:
A
Step-by-Step Solution & Explanation
Total heat to be removed = m·cp·ΔT = 12500 × 3.8 × (30−4) = 12500×3.8×26 = 1,235,000 kJ.
Refrigeration load (rate), Q̇ = 1,235,000 kJ / (3×3600 s) = 1,235,000/10,800 = 114.35 kW.
Power consumed by the motor = Q̇/COP = 114.35/4.5 = 25.41 kW.
Tonnage of refrigeration: 1 TR = 3.5167 kW, so TR = 114.35/3.5167 = 32.52 TR. Answer: (A) 25.4 and 32.5.
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