Cold storage at −2 °C (271 K) with ambient 40 °C (313 K): heat leakage is 30 kW. The actual COP of the plant is one-fourth of the ideal (Carnot) COP between these temperatures. Calculate the required input power to drive the plant.

Introduction / Context:
Refrigeration problems often benchmark real systems against the reversed Carnot cycle. Here we compare an actual plant to the ideal COP to determine the shaft power needed for a specified cooling load.

Given Data / Assumptions:

• Tc = −2 °C = 271 K (cold space).
• Th = 40 °C = 313 K (ambient sink).
• Cooling load Q_L = 30 kW (heat leakage to be removed).
• Plant COP_actual = (1/4) * COP_Carnot, same temperature limits.

Concept / Approach:
For a Carnot refrigerator, COP_Carnot = Tc / (Th − Tc). The actual COP is a fraction of this. Once COP_actual is known, power input follows from W = Q_L / COP_actual.

Step-by-Step Solution:
Compute temperature difference: Th − Tc = 313 − 271 = 42 K.COP_Carnot = Tc / (Th − Tc) = 271 / 42 ≈ 6.452.COP_actual = (1/4) * 6.452 ≈ 1.613.Required power W = Q_L / COP_actual = 30 / 1.613 ≈ 18.6 kW.

Verification / Alternative check:
If the plant had ideal performance (COP ≈ 6.45), the power would be ≈ 4.65 kW. One-fourth of this COP multiplies power by four, giving ≈ 18.6 kW—consistent with the calculation.

Why Other Options Are Wrong:

• 1.86, 3.72, 7.44 kW: These correspond to dividing 30 kW by larger COPs than the stated degraded COP; they underestimate the needed power.
• 24.0 kW: Overestimates the power; would imply an even poorer COP than specified.

Common Pitfalls:
Using Celsius in the COP_Carnot formula; forgetting that a lower actual COP raises required power for the same cooling load.

Final Answer:
18.6 kW