--- title: "Refrigerator COP 4: power input and cost" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 5 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:refrigerator-cop4-freezer-power-cost" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/refrigerator-cop4-freezer-power-cost site: BCT Engineering --- # Refrigerator COP 4: power input and cost Fundamental of Thermodynamics & Heat Transfer · Chapter 5: Second Law of Thermodynamics - Asked in: 2074 Bhadra, 8 marks - Video: [Thermodynamics - 6-4 Refrigerators and Heat Pumps - examples](https://www.youtube.com/watch?v=tf_ey50En3c) — Engineering Deciphered - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/refrigerator-cop4-freezer-power-cost - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5 ## Question A refrigerator having a COP of 4 maintains the freezer compartment at -3°C by removing heat at a rate of 10800 kJ/kg and rejects heat to the surroundings at 27°C. Determine the power input to the refrigerator and compare it with minimum theoretical power input. If the electricity cost 10/kWh, determine the actual and minimum theoretical cost per day for effective operation of 12h/day. ## Short answer Cooling load Q_L = 10800 kJ/h = 3 kW. Actual power = Q_L/COP = 0.75 kW. Carnot COP = T_L/(T_H−T_L) = 270.15/30 = 9.005, minimum power = 0.333 kW. Over 12 h/day at Rs 10/kWh, cost is Rs 90 actual and Rs 40 minimum. ## Answer **Reading of the data.** The heat removal rate is given as "10800 kJ/kg"; a rate must have units of energy per time, so it is taken as **10800 kJ/h** (= 3 kW). The cost is taken as Rs 10 per kWh. **Given:** COP = 4, T_L = −3 + 273.15 = 270.15 K, T_H = 27 + 273.15 = 300.15 K, Q_L = 10800 kJ/h = 3 kW, operation 12 h/day. #### Actual power input COP = Q_L/W, so W_actual = Q_L/COP = 3/4 = **0.75 kW** #### Minimum (theoretical) power A reversible (Carnot) refrigerator has COP_max = T_L/(T_H − T_L) = 270.15/30 = 9.005 W_min = Q_L/COP_max = 3/9.005 = **0.333 kW** #### Comparison W_actual/W_min = 0.75/0.333 = 2.25. The actual refrigerator needs about 2.25 times the minimum power. Its COP (4) is 44 % of the Carnot COP (9.005). #### Cost per day (12 h) - Actual energy = 0.75 × 12 = 9.0 kWh. Cost = 9.0 × 10 = **Rs 90 per day** - Minimum energy = 0.333 × 12 = 4.0 kWh. Cost = 4.0 × 10 = **Rs 40 per day** **Answer:** actual power 0.75 kW, minimum 0.333 kW; daily cost Rs 90 (actual) against Rs 40 (theoretical minimum). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is 10800 kJ/kg read as 10800 kJ/h? A. A rate needs energy per time, and 10800 kJ/h = 3 kW B. Because kg is a typing error for s C. Because 10800 kJ/kg is already kW D. Because cost is per hour Answer: A. A rate must have units of energy per time; 10800/3600 = 3 kW. (Key point: Units of heat rate) 2. Actual power input is: A. 12 kW B. 0.75 kW C. 3 kW D. 1.33 kW Answer: B. W = Q_L/COP = 3/4 = 0.75 kW. (Key point: Actual power) 3. Minimum theoretical power uses COP_max = T_L/(T_H − T_L) = 9.005. W_min is: A. 3 × 9.005 kW B. 0.75 kW C. 0.75/9.005 kW D. 0.333 kW Answer: D. W_min = 3/9.005 = 0.333 kW. (Key point: Minimum power) 4. With Rs 10/kWh and 12 h/day, daily cost for actual and minimum are: A. Rs 90 and Rs 40 B. Rs 9 and Rs 4 C. Rs 75 and Rs 33 D. Rs 40 and Rs 90 Answer: A. Actual 0.75 × 12 × 10 = 90; minimum 0.333 × 12 × 10 = 40. 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