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Refrigerator COP 4: power input and cost

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.

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.

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).

✓ Answer checked by Claude Opus 5.5 (high effort)

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