---
title: "AC with COP 4 for hall: minimum theoretical power"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 5
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:ac-cop4-hall-minimum-power"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/ac-cop4-hall-minimum-power
site: BCT Engineering
---

# AC with COP 4 for hall: minimum theoretical power

Fundamental of Thermodynamics & Heat Transfer · Chapter 5: Second Law of Thermodynamics

- Asked in: 2072 Magh, 8 marks
- Video: [Problems on Heat Pump and Refrigerator](https://www.youtube.com/watch?v=pG3Qltn_k6I) — Keshav Jadhav
- 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/ac-cop4-hall-minimum-power
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-5

## Question

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.

## 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.

## Answer

**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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The cooling load is:
   A. 600 kJ/min only
   B. 720 kJ/min = 12 kW
   C. 480 kJ/min
   D. 12 kJ/min
   Answer: B. Wall gain plus internal gains: 600 + 120 = 720 kJ/min = 12 kW. (Key point: Cooling load)

2. Actual power to drive the unit with COP 4 is:
   A. 48 kW
   B. 0.614 kW
   C. 3 kW
   D. 12 kW
   Answer: C. W = Q_L/COP = 12/4 = 3 kW. (Key point: Actual power)

3. Which temperatures give COP_max for the unit?
   A. T_L = 308.15 K, T_H = 293.15 K
   B. T_L = 293.15 K (hall), T_H = 308.15 K (outside): COP_max = 19.54
   C. T_L = 20 and T_H = 35 in °C
   D. COP_max = T_H/(T_H − T_L)
   Answer: B. Cooling COP_max = T_L/(T_H − T_L), with absolute temperatures. (Key point: Carnot COP)

4. The minimum power is 0.614 kW. Compared to this, the actual unit needs about:
   A. 4.9 times more
   B. the same power
   C. 20 times more
   D. 2 times more
   Answer: A. 3.0/0.614 = 4.9; the actual COP is only about 20% of Carnot. (Key point: Comparison)

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