- 2072 Magh · 8 marks
AC with COP 4 for hall: minimum theoretical power
Short answer
Total cooling load 720 kJ/min = 12 kW. Actual power = Q_L/COP = 3 kW. Carnot COP = 293.15/15 = 19.54, so minimum theoretical power = 12/19.54 = 0.614 kW.
An air conditioning unit having COP of 4 maintains a hall at 20°C in a day when the outside temperature is 35°C. The thermal load consists of heat energy entering through the wall at a rate of 600 KJ/min and from the occupants, computers and lighting at rate of 120 KJ/min. Determine the power required to drive the unit and compare it with the minimum theoretical power required.
Given: COP = 4, T_L = 20 °C = 293.15 K (hall), T_H = 35 °C = 308.15 K (outside).
Cooling load
Heat from walls = 600 kJ/min Heat from occupants, computers, lighting = 120 kJ/min Total Q_L = 720 kJ/min = 720/60 = 12 kW
The unit must remove this much heat to hold the hall at 20 °C.
Actual power required
COP = Q_L/W, so W_actual = 12/4 = 3.0 kW
Minimum theoretical power
For a reversible (Carnot) refrigerator:
COP_max = T_L/(T_H − T_L) = 293.15/(308.15 − 293.15) = 293.15/15 = 19.54
W_min = Q_L/COP_max = 12/19.54 = 0.614 kW
Comparison
W_actual/W_min = 3.0/0.614 = 4.9
The actual unit needs about 4.9 times the minimum power (about 2.39 kW more). Its COP of 4 is only 20 % of the Carnot COP of 19.54, because of irreversibilities.
Answer: actual power = 3 kW; minimum theoretical power = 0.614 kW.
✓ Answer checked by Claude Opus 5.5 (high effort)
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