Subtopic: Preservation of food by cooling and freezing
Question Statement:
In a cold store of 30 m × 15 m × 15 m size, 4000 tonnes of potato having the specific heat of 3.62 kJ kg⁻¹ K⁻¹ and heat of respiration of 20 W m⁻³ is kept at 30 °C. Potato is required to be cooled to 2 °C in 30 days. Neglecting other sources of heat, the capacity of the refrigeration plant required is
Options:
(A) 6 TR
(B) 38 TR
(C) 44 TR
(D) 83 TR
Official Answer Key:
D
Step-by-Step Solution & Explanation
Cooling load (product sensible heat, spread over 30 days): Q_cool = m·c_p·ΔT = (4,000,000 kg)(3.62 kJ/kgK)(30−2 K) = 4.0544×10⁸ kJ, over t=30×86400 s = 2.592×10⁶ s ⟹ rate = 156.4 kW. Respiration heat (steady, volumetric): V_store = 30×15×15 = 6750 m³; Q_resp = 20 W/m³ × 6750 = 135,000 W = 135.0 kW. Total load = 156.4+135.0 = 291.4 kW. In tons of refrigeration (1 TR = 3.517 kW): 291.4/3.517 = ≈82.9 ≈ 83 TR.
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