Find Cop Calculations

Device Type	Typical COP Range (Heating)	Typical COP Range (Cooling – EER/SEER related)
Air Source Heat Pumps (ASHP)	2.0 – 4.5 (depending on outdoor temp)	2.5 – 5.0 (EER/SEER)
Ground Source Heat Pumps (GSHP)	3.0 – 5.0	3.5 – 5.5 (EER/SEER)
Refrigerators	–	2.0 – 4.0
Air Conditioners (Room)	–	2.5 – 3.5 (EER)
Air Conditioners (Central)	–	3.0 – 5.0 (SEER/EER)
Absorption Chillers	0.5 – 1.2 (uses heat input)	0.5 – 1.2

What is Coefficient of Performance (COP)?

The Coefficient of Performance (COP) is a measure of the efficiency of a heat pump, refrigerator, or air conditioning system. It is defined as the ratio of the desired heat output (for heating) or heat removed (for cooling) to the work input required to achieve that output. A higher Coefficient of Performance (COP) indicates a more efficient system, meaning it provides more heating or cooling for the same amount of energy input (work).

The Coefficient of Performance (COP) is dimensionless, as it’s a ratio of two energy values (or power values) in the same units. It’s crucial for comparing the efficiency of different heating and cooling technologies.

Who should use it?

Engineers, HVAC technicians, homeowners considering new heating/cooling systems, and anyone interested in energy efficiency should understand and use the Coefficient of Performance (COP). It helps in selecting more energy-efficient appliances, which can lead to significant cost savings and reduced environmental impact.

Common misconceptions about COP

A common misconception is that the Coefficient of Performance (COP) can never be greater than 1 (or 100% efficiency). While this is true for converting work directly to heat (like in an electric resistance heater, where COP is ideally 1), heat pumps and refrigeration cycles move heat rather than converting work directly to the desired heat effect. They use the work input to transfer a larger amount of heat from a source to a sink, so their Coefficient of Performance (COP) values are often greater than 1, typically between 2 and 5 for common devices.

Coefficient of Performance (COP) Formula and Mathematical Explanation

The formula for the Coefficient of Performance (COP) depends on whether the system is used for heating or cooling.

For Heating (e.g., Heat Pump):

The desired output is the heat delivered to the hot reservoir (Qh), and the input is the work (W).

COP_heating = Qh / W

Where Qh is the heat delivered to the hot space, and W is the work input.

For Cooling (e.g., Refrigerator, Air Conditioner):

The desired output is the heat removed from the cold reservoir (Qc), and the input is the work (W).

COP_cooling = Qc / W

Where Qc is the heat removed from the cold space, and W is the work input.

In both cases, W is the energy (e.g., electricity) consumed by the compressor or other work-input components.

Variables Table

Variable	Meaning	Unit	Typical Range
COP_heating	Coefficient of Performance for heating	Dimensionless	1 – 5+
COP_cooling	Coefficient of Performance for cooling	Dimensionless	1 – 5+
Qh	Heat delivered to the hot reservoir	Joules (J), Watts (W), BTU/hr	Varies greatly
Qc	Heat removed from the cold reservoir	Joules (J), Watts (W), BTU/hr	Varies greatly
W	Work input to the system	Joules (J), Watts (W), BTU/hr	Varies greatly

It’s important that Qh, Qc, and W are measured in the same units (or consistent power units) when calculating the Coefficient of Performance (COP).

Practical Examples (Real-World Use Cases)

Example 1: Heat Pump in Winter

A heat pump is used to heat a house. It consumes 2000 Watts (2 kW) of electrical power (work input, W) and delivers 7000 Watts of heat (Qh) to the inside of the house.

Inputs:

Qh = 7000 W
W = 2000 W

Coefficient of Performance (COP)_heating = Qh / W = 7000 W / 2000 W = 3.5

Interpretation: This heat pump delivers 3.5 units of heat energy for every 1 unit of electrical energy consumed. This is much more efficient than an electric resistance heater (COP ≈ 1).

Example 2: Refrigerator

A refrigerator’s compressor consumes 150 Watts (W) to remove 450 Watts of heat (Qc) from the inside of the refrigerator.

Inputs:

Qc = 450 W
W = 150 W

Coefficient of Performance (COP)_cooling = Qc / W = 450 W / 150 W = 3.0

Interpretation: The refrigerator removes 3 units of heat energy from its interior for every 1 unit of electrical energy the compressor uses. Learn more about refrigerator energy costs.

How to Use This Coefficient of Performance (COP) Calculator

Select System Type: Choose whether you are calculating the Coefficient of Performance (COP) for a “Heating” system (like a heat pump in winter) or a “Cooling” system (like a refrigerator or air conditioner). This selection determines the formula used and the label for the first input.
Enter Useful Heat Output/Cooling Effect: Input the amount of heat the system delivers (for heating) or removes (for cooling). Make sure to note the units (e.g., Joules or Watts).
Enter Work Input: Input the amount of work (usually electrical energy) the system consumes to achieve the heating or cooling effect. Use the same units as the heat output/cooling effect.
View Results: The calculator will instantly display the Coefficient of Performance (COP), the system type, the input values, and the formula used. The chart will also update to visualize the energies.
Reset (Optional): Click “Reset” to return to default values.
Copy Results (Optional): Click “Copy Results” to copy the main result and intermediate values to your clipboard.

When reading the results, a higher Coefficient of Performance (COP) means better efficiency. For example, a heat pump with a COP of 4 is more efficient than one with a COP of 3.

Key Factors That Affect Coefficient of Performance (COP) Results

Several factors influence the Coefficient of Performance (COP) of heat pumps and refrigeration systems:

Temperature Difference (Lift): The larger the temperature difference between the heat source and the heat sink (the “lift” the system has to work against), the lower the Coefficient of Performance (COP). For a heat pump, this is the difference between the outdoor and indoor temperatures. For an AC, it’s between the indoor and outdoor temperatures.
Operating Temperatures: The absolute temperatures of the source and sink also matter. Heat pumps are less efficient at very low outdoor temperatures.
Compressor Efficiency: The efficiency of the compressor, which does the work, directly impacts the overall Coefficient of Performance (COP).
Heat Exchanger Design: The effectiveness of the evaporator and condenser (heat exchangers) in transferring heat affects efficiency. Larger or more efficient heat exchangers can improve the Coefficient of Performance (COP).
Refrigerant Type: The thermodynamic properties of the refrigerant used in the cycle influence the system’s performance.
System Maintenance: Dirty filters, low refrigerant charge, or poorly maintained components can significantly reduce the actual Coefficient of Performance (COP) of a system. Consider our HVAC sizing calculator for proper system selection.
Defrost Cycles (for Heat Pumps): In cold, humid conditions, heat pumps need to run defrost cycles, which temporarily reduce the heating output and overall seasonal Coefficient of Performance (COP) (reflected in HSPF).

Frequently Asked Questions (FAQ) about Coefficient of Performance (COP)

1. Can the Coefficient of Performance (COP) be less than 1?

For standard vapor-compression heat pumps and refrigerators, the COP is almost always greater than 1 when operating under normal conditions. However, in extremely adverse conditions or for certain types of systems like absorption chillers, the COP can be less than 1. An electric resistance heater has a COP of 1.

2. Is a higher COP always better?

Yes, a higher Coefficient of Performance (COP) means the system is more efficient at converting work input into the desired heating or cooling effect, leading to lower energy consumption for the same output.

3. What is the difference between COP and EER/SEER?

COP is a general ratio of output to input. EER (Energy Efficiency Ratio) and SEER (Seasonal Energy Efficiency Ratio) are specific types of COP used for cooling systems (like air conditioners) under standardized test conditions. EER is usually measured at a specific outdoor temperature, while SEER is an average over a range of seasonal temperatures. What is SEER?

4. How does outdoor temperature affect the COP of a heat pump?

As the outdoor temperature drops, a heat pump has to work harder to extract heat, and its Coefficient of Performance (COP) decreases. At very low temperatures, it might switch to auxiliary heating.

5. What is HSPF?

HSPF (Heating Seasonal Performance Factor) is similar to SEER but for the heating mode of heat pumps. It represents the total heating output over a season divided by the total electrical energy consumed, including defrost cycles and auxiliary heat. It’s related to the average Coefficient of Performance (COP) over a heating season.

6. How can I improve my system’s COP?

Regular maintenance, cleaning filters, ensuring proper refrigerant charge, and sealing air leaks in your home can help maintain or slightly improve your system’s operational Coefficient of Performance (COP).

7. Does the unit of energy (Joules vs. Watts) matter for COP?

As long as you use the same units for both the heat output/removal and the work input, the Coefficient of Performance (COP) will be the same dimensionless value. Watts are Joules per second, so if you use Watts for both, it’s a ratio of powers, which is equivalent.

8. What is a good COP for a heat pump?

A good modern air-source heat pump might have a Coefficient of Performance (COP) between 2.5 and 4.5 depending on conditions, while ground-source heat pumps can achieve COPs between 3.5 and 5 or even higher. Consider a heat pump savings analysis.