Editing Heat pump (section)

== Principle of operation ==

[[File:Refrigerator-cycle.svg|thumb|A: indoor compartment, B: outdoor compartment, I: insulation, 1: condenser, 2: expansion valve, 3: evaporator, 4: compressor]]{{Main|Heat pump and refrigeration cycle|Vapor-compression refrigeration}}

Heat flows spontaneously from a region of higher temperature to a region of lower temperature. Heat does not flow spontaneously from lower temperature to higher, but it can be made to flow in this direction if [[Work (physics)|work]] is performed. The work required to transfer a given amount of heat is usually much less than the amount of heat; this is the motivation for using heat pumps in applications such as the heating of water and the interior of buildings.<ref name="R&M">G. F. C. Rogers and Y. R. Mayhew (1957), ''Engineering Thermodynamics, Work and Heat Transfer'', Section 13.1, Longmans, Green & Company Limited.</ref>

The amount of work required to drive an amount of heat Q from a lower-temperature reservoir such as ambient air to a higher-temperature reservoir such as the interior of a building is:
<math display="block">W = \frac{ Q}{\mathrm{COP}}</math>
where
* <math>W </math> is the [[Mechanical work|work]] performed on the [[Working fluid selection|working fluid]] by the heat pump's compressor.
* <math> Q </math> is the [[heat]] transferred from the lower-temperature reservoir to the higher-temperature reservoir.
* <math>\mathrm{COP}</math> is the instantaneous [[coefficient of performance]] for the heat pump at the temperatures prevailing in the reservoirs at one instant.

The coefficient of performance of a heat pump is greater than one so the work required is less than the heat transferred, making a heat pump a more efficient form of heating than electrical resistance heating. As the temperature of the higher-temperature reservoir increases in response to the heat flowing into it, the coefficient of performance decreases, causing an increasing amount of work to be required for each unit of heat being transferred.<ref name=R&M/>

The [[Heat pump and refrigeration cycle#Coefficient of performance|coefficient of performance, and the work required]] by a heat pump can be calculated easily by considering an ideal heat pump operating on the [[Carnot cycle#Reversed Carnot cycle|reversed Carnot cycle]]:
*If the low-temperature reservoir is at a temperature of {{cvt|270|K|C}} and the interior of the building is at {{cvt|280|K|C}} the maximum theoretical coefficient of performance is 28. This means 1 joule of work delivers 28 joules of heat to the interior. The one joule of work ultimately ends up as [[thermal energy]] in the interior of the building and 27 joules of heat are moved from the low-temperature reservoir.{{refn|group=note|1=As explained in [[Coefficient of performance]] TheoreticalMaxCOP = (desiredIndoorTempC + 273) ÷ (desiredIndoorTempC - outsideTempC)  = (7+273) ÷  (7 - (-3)) = 280÷10 = 28 <ref name="physics.stackexchange.com">{{Cite web |title=Is there some theoretical maximum coefficient of performance (COP) for heat pumps and chillers? |url=https://physics.stackexchange.com/questions/350074/is-there-some-theoretical-maximum-coefficient-of-performance-cop-for-heat-pump |access-date=2024-02-22 |website=Physics Stack Exchange |language=en}}</ref>}}
*As the temperature of the interior of the building rises progressively to {{cvt|300|K|C}} the coefficient of performance falls progressively to 10. This means each joule of work is responsible for transferring 9 joules of heat out of the low-temperature reservoir and into the building. Again, the 1 joule of work ultimately ends up as thermal energy in the interior of the building so 10 joules of heat are added to the building interior.{{refn|group=note|1= As explained in [[Coefficient of performance]]  TheoreticalMaxCOP = (desiredIndoorTempC + 273) ÷ (desiredIndoorTempC - outsideTempC)  = (27+273) ÷  (27 - (-3)) = 300÷30 = 10<ref name="physics.stackexchange.com"/>}}

This is the theoretical amount of heat pumped but in practice it will be less for various reasons, for example if the outside unit has been installed where there is not enough airflow. More data sharing with owners and academics—perhaps from [[heat meter]]s—could improve efficiency in the long run.<ref>{{Cite web |last=Williamson |first=Chris |date=2022-10-13 |title=Heat pumps are great. Let's make them even better |url=https://medium.com/all-you-can-heat/heat-pumps-are-great-lets-make-them-even-better-d508b8e3a751 |access-date=2024-02-22 |website=All you can heat |language=en}}</ref>