Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses
Question Statement:
In a cascade refrigeration system, the COPs of the cooling cycle and cascade are 3.7 and 4.2, respectively. If tonnage (1 TR = 3.52 kW) of the cooling cycle is 15 and the cascade removes 70% of the total heat rejected in the liquid receiver of the cooling cycle, then the powers required by the cooling and cascade compressors in hp are _____ and _____, respectively.
Options:
(A) 67.07, 46.95
(B) 52.08, 39.35
(C) 19.13, 14.99
(D) 14.27, 11.18
Official Answer Key:
C
Step-by-Step Solution & Explanation
Cooling-cycle refrigeration load: Q_cool = 15 TR × 3.52 kW/TR = 52.8 kW.
Cooling-cycle compressor work: W_cool = Q_cool/COP_cool = 52.8/3.7 = 14.27 kW.
Heat rejected by the cooling cycle at the liquid receiver (cascade condenser) = Q_cool + W_cool = 52.8+14.27 = 67.07 kW.
Heat absorbed by the cascade evaporator = 70% of this rejected heat = 0.7×67.07 = 46.95 kW.
Cascade compressor work: W_cascade = 46.95/COP_cascade = 46.95/4.2 = 11.18 kW.
Converting both compressor powers to hp (1 hp = 0.746 kW):
W_cool = 14.27/0.746 = 19.13 hp; W_cascade = 11.18/0.746 = 14.99 hp. Matches option (C) exactly. (Note the intermediate kW values 67.07 and 46.95 numerically coincide with the hp figures in distractor option (A) — a trap for candidates who forget the final kW→hp conversion.)
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