
Coefficient of efficiency
Equation[edit]
The equation is:
where
- is the heat supplied to or removed from the reservoir.
- is the work consumed by the heat pump.
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, 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 input work) to the input work:
- is the heat removed from the cold reservoir.
- is the heat supplied to the hot reservoir.
Derivation[edit]
According to the first law of thermodynamics, in a reversible system we can show that and , where is the heat transferred to the hot reservoir and is the heat collected from the cold heat reservoir.
Therefore, by substituting for W,
For a heat pump operating at maximum theoretical efficiency (i.e. Carnot efficiency), it can be shown that
andwhere and are the temperatures of the hot and cold heat reservoirs respectively. Note that these equations must use an absolute temperature scale, for example, Kelvin or Rankine.
At maximum theoretical efficiency,
which is equal to the reciprocal of the ideal efficiency for a heat engine, because a heat pump is a heat engine operating in reverse. Similarly,
Note that the COP of a heat pump depends on its duty the heat rejected to the hot sink is greater than the heat absorbed from the cold source, so the heating COP is 1 greater than the cooling COP.
applies to heat pumps and applies to air conditioners or refrigerators. For heat engines, see Efficiency. Values for actual systems will always be less than these theoretical maximums. In Europe, the standard tests for ground source heat pump units use 35 °C (95 °F) for and 0 °C (32 °F) for . According to the above formula, the maximum achievable COP would be 8.8. Test results of the best systems are around 4.5. When measuring installed units over a whole season and accounting for the energy needed to pump water through the piping systems, seasonal COP's are around 3.5 or less. This indicates room for improvement.
Improving COP[edit]
As the formula shows, the COP of a heat pump system can be improved by reducing the temperature gap minus at which the system works. For a heating system this would mean two things: 1) reducing the output temperature to around 30 °C (86 °F) which requires piped floor, wall or ceiling heating, or oversized water to air heaters and 2) increasing the input temperature (e.g. by using an oversized ground source or by access to a solar-assisted thermal bank ). For an air cooler, COP could be improved by using ground water as an input instead of air, and by reducing temperature drop on output side through increasing air flow. For both systems, also increasing the size of pipes and air canals would help to reduce noise and the energy consumption of pumps (and ventilators). The heat pump itself can be improved by increasing the size of the internal heat exchangers relative to the power of the compressor, and to reduce the system's internal temperature gap over the compressor. This latter measure, however, makes such heat pumps unsuitable to produce output above roughly 40 °C (104 °F) which means that a separate machine is needed for producing hot tap water.
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