Subtopic: Refrigeration and cold storage basics and applications
Question Statement:
A cold storage takes 5 hours to bring down the temperature of 100 metric tons of potato from 35°C to 8°C. Specific heat of potato = 3.1 kJ/kg·°C. COP = 3.66; latent heat of vaporization of R-22 (at −10°C evaporation) = 230 kJ/kg. Neglecting respiration heat load and power loss, find the refrigerant flow rate and compressor power input.
Options:
(A) 121.3 kg/min and 127.1 kW
(B) 124.7 kg/min and 121.3 kW
(C) 127.1 kg/min and 121.3 kW
(D) 124.7 kg/min and 127.1 kW
Official Answer Key:
A
Step-by-Step Solution & Explanation
Total heat to be removed = m·cp·ΔT = 100,000×3.1×27 = 8,370,000 kJ, over 5 h (18,000 s) ⇒ cooling load = 465.0 kW. Refrigerant flow rate = load/latent heat = 465.0/230 = 2.0217 kg/s = 121.3 kg/min. Compressor power = load/COP = 465.0/3.66 ≈ 127.1 kW.
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