Library / Thermodynamics / THE-08

Thermodynamics

Refrigeration coefficient of performance

Cooling delivered per unit of work put in.

Ref: THE-08
Rev: v1
Field: Thermodynamics
Result: COP

{COP}_{R} = \frac{Q_{C}}{W_{in}}

Use positive cycle energy magnitudes. Q_C is heat removed from the cold space and W_in the work (or power) input. Energy balance gives Q_H = Q_C + W_in.

A refrigerator moves heat from cold to hot; the COP is how much heat it removes for each unit of energy it uses.

Calculator

Solve for

COPCoefficient of performanceComputed

Q_CCooling effect

0.11,000

W_inPower input

0.1500

Heat rejected Q_H8 kW
Power per kW of cooling0.3333 kW/kW

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Coefficient of performance () current value 3

Q_CCooling effectKey driver

0.05500

W_inPower input

0.01260

A COP of 3 means 3 kW of cooling for 1 kW of electrical power, possible because the machine moves heat rather than making it.
The COP can exceed 1; efficiency cannot.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing ()	Local step	Change in result
Q_C Cooling effect	0.1 to 1,000 kW	0.05	500	499.9	+10%	+10 %
W_in Power input	0.1 to 500 kW	0.012	60	59.99	+10%	−9.09 %

Sweep of Power input

Power input (kW)	Coefficient of performance
0.1	60
0.29	20.69
0.8409	7.135
2.438	2.461
7.071	0.8485
20.5	0.2926
59.46	0.1009
172.4	0.0348
500	0.012

Sweep one variable

Vary

Coefficient of performance ()Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

COP

Coefficient of performance

Result ·

What the equation gives you. Everything else feeds it.

Q_C

Cooling effect

kW · typical 0.1 to 1,000

Heat removal rate from the cold space. Directly proportional.

W_in

Power input

kW · typical 0.1 to 500

Compressor power input. Inversely proportional.

Worked example with the values above

Cooling effect Q_C = 6 kW.
Work input W_in = 2 kW.
COP = 6 / 2 = 3.

Assumptions and limits

Steady operation.
Positive energy magnitudes.

Common mistakes

Treating COP as an efficiency below 100%.
Using heat rejected instead of cooling effect.

Statistics and uncertainty

Measured COP values from field tests typically carry ±5–10% because of load and ambient variation.

Where it is used

Cooling plant and air-conditioning sizing.
Equipment-cooling system checks.

Maths you need

Energy balance: QH = QC + W
Heat engine: reverse cycle

Related formulas

Practise this formula