---
title: "Refrigeration, TR, COP and applications"
subject: "Applied Thermodynamics"
chapter: 3
key: "semester_3/applied_thermodynamics:refrigeration-definition-applications"
source: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/refrigeration-definition-applications
site: BCT Engineering
---

# Refrigeration, TR, COP and applications

Applied Thermodynamics · Chapter 3: Refrigeration

- Asked in: year not given, 4 marks
- Video: [Unit of refrigeration(TR)|COP|Refrigeration effect|Refrigeration](https://www.youtube.com/watch?v=xmnjapH4of4) — PK Classes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_thermodynamics/refrigeration-definition-applications
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_thermodynamics/chapter-3

## Question

Define refrigeration. Explain the terms ton of refrigeration and coefficient of performance (COP), and list five applications of refrigeration.

## Short answer

Refrigeration removes heat from a space to maintain temperature below surroundings. One ton of refrigeration equals 12000 Btu/h (3.517 kW). COP is the ratio of refrigerating effect to work input. Applications include domestic refrigerators, cold storage, air-conditioning, ice manufacture, and medical/vaccine storage.

## Answer

**Refrigeration** is the process of removing heat from a body or space so that its temperature is lowered below that of the surroundings and maintained there. It needs work (or heat) input and rejects heat to a higher-temperature sink.

#### Ton of refrigeration (TR)
One TR is the rate of heat removal needed to freeze 1 ton (2000 lb, the US short ton) of water at 0 °C into ice at 0 °C in 24 hours.

$$
1\ \text{TR} = \frac{2000 \times 144\ \text{Btu}}{24\ \text{h}} = 12000\ \text{Btu/h} = 211\ \text{kJ/min} = 3.517\ \text{kW}
$$

(With a metric tonne of ice per day the value is about 3.87 kW.)

#### Coefficient of performance
$$
\text{COP} = \frac{\text{refrigerating effect}}{\text{work input}} = \frac{Q_L}{W}
$$
For a refrigerator the COP can exceed 1. The reversed Carnot (maximum) value is $T_L/(T_H - T_L)$.

#### Applications
1. Domestic refrigerators and freezers.
2. Cold storage of food, fruits, vegetables, meat and milk; dairy plants.
3. Air-conditioning of buildings and vehicles.
4. Manufacture of ice, and chilling in breweries and chemical plants.
5. Medical use: vaccine and blood storage, and industrial process cooling.

## Check yourself

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

1. Refrigeration is the process of
   A. adding heat to a space to keep it warm
   B. converting heat fully into work
   C. compressing air only
   D. removing heat from a body or space to keep it below the temperature of surroundings
   Answer: D. It lowers and maintains temperature below the surroundings and requires work input. (Key point: Refrigeration definition)

2. One ton of refrigeration (TR) is equal to
   A. 3.517 kW (211 kJ/min)
   B. 1 kW
   C. 35.17 kW
   D. 0.3517 kW
   Answer: A. 1 TR = 12000 Btu/h = 211 kJ/min = 3.517 kW. (Key point: Ton of refrigeration)

3. The COP of a refrigerator is
   A. work input divided by refrigerating effect
   B. refrigerating effect divided by work input and can exceed 1
   C. always less than 1
   D. heat rejected divided by heat absorbed
   Answer: B. COP = Q_L/W. (Key point: COP)

4. The maximum (reversed Carnot) COP is
   A. (TH - TL)/TL
   B. TH/(TH - TL)
   C. TL/(TH - TL)
   D. TL/TH
   Answer: C. It depends only on the temperature limits. (Key point: Carnot COP)

## Related in Applied Thermodynamics

- [Ideal vapour compression cycle for 5 TR](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_thermodynamics%3Avcr-ideal-cycle-numerical) — Applied Thermodynamics, chapter 3

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