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The coefficient of performance, or COP (sometimes CP), of a heat pump is the ratio of the change in heat at the "output" (the heat reservoir of interest) to the supplied work:## Derivation

## 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 output heat). The output heat comes from both the heat source and 1 kW of input energy, so the heat-source is cooled by 2.5 kW, not 3.5 kW. ## See also

- $COP\; =\; frac$
{Delta W}>

- $|Delta\; Q|$ is the change in heat at the heat reservoir of interest, and
- $Delta\; W$ is the work consumed by the heat pump.

(Note: COP has no units, therefore in this equation, heat and work must be expressed in the same units.)

The COP for heating and cooling are thus different, because the heat reservoir of interest is different. When one is interested in how well a machine cools. For example, the COP is the ratio of the heat removed from the cold reservoir to input work. However, for heating, the COP is the ratio of the heat removed from the cold reservoir plus the heat added to the hot reservoir by the input work to input work:

- $COP\_\{heating\}=frac$
- $COP\_\{cooling\}=frac$

>{Delta W}

- $Delta\; Q\_\{cold\}$ is the heat moved from the cold reservoir (to the hot reservoir).

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 given off by the hot heat reservoir and $Q\_\{cold\}$ is the heat taken in by the cold 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, at maximum theoretical efficiency,

- $COP\_\{heating\}=frac\{T\_\{hot\}\}\{T\_\{hot\}-T\_\{cold\}\}$

- $COP\_\{cooling\}=frac\{Q\_\{cold\}\}\{Q\_\{hot\}-Q\_\{cold\}\}\; =frac\{T\_\{cold\}\}\{T\_\{hot\}-T\_\{cold\}\}$

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. Values for actual systems will always be less than these theoretical maximums.

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.

- Seasonal energy efficiency ratio (DSEER)
- Thermal efficiency
- Vapor-compression refrigeration
- Air conditioner
- HVAC
- Heat Pump

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Last updated on Tuesday October 07, 2008 at 09:07:27 PDT (GMT -0700)

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This article is licensed under the GNU Free Documentation License.

Last updated on Tuesday October 07, 2008 at 09:07:27 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

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