The coefficient of performance (or COP) is a measure of energy efficiency. It is the ratio of the energy output to work (energy) input.

## Examples

The COP of a heat pump expresses the proportion of electricity (work) in a heat pump that is being converted into desired heat. For example, if a heat pump has a coefficient of.8, then the amount of heat coming from the heat pump is 80% of the energy going into the system; 20% is lost.

In a heat pump, the COP can be calculated by taking the heat transferred by the heat pump (Qh) over the power input (W). For example, if the heat pump requires a power input of 500 Watts, and the heater has an output (Qh) of 350 Watts, then the heat pump has a COP of 350 Watts/500 Watts. The watts cancel and we get a COP of.7.

In a refrigerator system, heat is moved from one location to another. Work is input into the compressor of the system, heat is removed from a source, like a cold reservoir, at the evaporator, and heat is released at the condenser. The COP of a refrigerator can be calculated by taking the cooling effect (Ql), which is the heat removed from the cold source, over the work input.

The COP of a Refrigerator:

$COP=Q/W$ Where $Q$ is the heat transferred and $W$ is the work done by the heat pump

The COP of a heat pump and COP of a refrigerator are related by: $COP_{HP}=COP_{R}+1$ ## Heat pump at work

The COP of a heat pump decreases as the outside temperature drops. As the temperature decreases, there is less heat pumped and therefore the COP is lowered. This is because the heat is being created by extreme pressurization of the inside channels of the pump, which will heat the air inside the pump immensely. This heat will then transfer from the system into the surrounding environment as heat always travels to colder spaces. The extreme pressurization of heat inside this part of the system is counterbalanced by extreme depressurization of another part of the system, the channels of the pump that are outside of the heated room. The transfer of heat into these channels requires more heat in the outside environment than inside these outside channels, as heat transfer only moves from hot to cold. So if the outside environment is already very cold, there is not much of a potential for heat transfer, meaning that this heat can't be used to heat the inside environment either. To sum it up, a heat pump can only be effective if the pump is out-competing the temperature of the outside environment to force heat transfer into the system.

## COP benefit

This means that $COP=Q_{C}/W$ , where W is the work and Qc is the heat sucked out of the cold reservoir (for a refrigerator)

$COP=Q_{h}/W$ Where Qh is the heat pushed into the hot reservoir (for a heat engine) cost

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