THC Science

The coefficient of performance, or COP (sometimes CP), of a heat pump is the ratio of the output heat to the supplied work or

$COP={\frac {|Q|}{W}}$
where Q is the useful heat supplied by the condenser and W is the work consumed by the compressor. (Note: COP has no units, therefore in this equation, heat and work must be expressed in the same units.)

According to the first law of thermodynamics, in a reversible system we can show that $Q_{hot}=Q_{cold}+W$ and $W=Q_{hot}-Q_{cold}$ , where $Q_{hot}$ is the heat taken in by the cold heat reservoir and $Q_{cold}$ is the heat given off by the hot heat reservoir.
Therefore, by substituting for W,
$COP_{heating}={\frac {Q_{hot}}{Q_{hot}-Q_{cold}}}$
For a heat pump operating at maximum theoretical efficiency (i.e. Carnot efficiency), it can be shown that ${\frac {Q_{hot}}{T_{hot}}}={\frac {Q_{cold}}{T_{cold}}}$ and $Q_{cold}={\frac {Q_{hot}T_{cold}}{T_{hot}}}$ , where $T_{hot}$ and $T_{cold}$ are the temperatures of the hot and cold heat reservoirs respectively.

Hence,
$COP_{heating}={\frac {T_{hot}}{T_{hot}-T_{cold}}}$
Similarly,
$COP_{cooling}={\frac {Q_{cold}}{Q_{hot}-Q_{cold}}}={\frac {T_{cold}}{T_{hot}-T_{cold}}}$

It can also be shown that $COP_{cooling}=COP_{heating}-1$ . Note that these equations must use the absolute temperature, such as the Kelvin scale.

$COP_{heating}$ applies to heat pumps and $COP_{cooling}$ applies to air conditioners or refrigerators. For heat engines, see Efficiency.

Example

A geothermal heat pump operating at $COP_{heating}$ 3.5 provides 3.5 units of heat for each unit of energy consumed (e.g. 1 kW consumed would provide 3.5 kW of outputted heat). The outputted heat comes from both the heat source and 1kW of input energy, so the heat-source is cooled by 2.5kW, not 3.5 kW.

A heat pump of $COP_{heating}$ 3.5, such as in the example above, could be less expensive to use than even the most efficient gas furnace.

A heat pump cooler operating at $COP_{cooling}$ 2.0 removes 2 units of heat for each unit of energy consumed (e.g. such an air conditioner consuming 1 kW would remove heat from a building's air at a rate of 2 kW).

The COP of heat pumps compares favorably with high-efficiency gas-burning furnaces (90-99% efficient), and electric heating (100%), but the full costs of the energy consumed must be considered, and energy from gas is typically much less expensive than that from electricity.

THC Science

Bringing Science to the Cannabis Conversation!

Cannabis Ruderalis

Example

See also

Leave a Reply