# Coefficient of performance

The coefficient of performance or COP (sometimes CP or CoP) of a heat pump, refrigerator or air conditioning system is a ratio of useful heating or cooling provided to work required. Higher COPs equate to lower operating costs. The COP usually exceeds 1, especially in heat pumps, because, instead of just converting work to heat (which, if 100% efficient, would be a COP of 1), it pumps additional heat from a heat source to where the heat is required. For complete systems, COP calculations should include energy consumption of all power consuming auxiliaries. COP is highly dependent on operating conditions, especially absolute temperature and relative temperature between sink and system, and is often graphed or averaged against expected conditions.

## Equation

The equation is:

${\rm {COP}}={\frac {Q}{W}}$

where

• $Q\$  is the useful heat supplied or removed by the considered system.
• $W\$  is the work required by the considered system.

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 which the heat removed from the cold reservoir plus the input work bears to the input work:

${\rm {COP}}_{\rm {heating}}={\frac {|Q_{H}|}{W}}={\frac {|Q_{C}|+W}{W}}$
${\rm {COP}}_{\rm {cooling}}={\frac {|Q_{C}|}{W}}$

where

• $Q_{C}\$  is the heat removed from the cold reservoir.
• $Q_{H}\$  is the heat supplied to the hot reservoir.

## Derivation

According to the first law of thermodynamics, in a reversible system we can show that $Q_{H}=Q_{C}+W$  and $W=Q_{H}-Q_{C}$ , where $Q_{H}$  is the heat transferred to the hot reservoir and $Q_{C}$  is the heat collected from the cold reservoir.
Therefore, by substituting for W,

${\rm {COP}}_{\rm {heating}}={\frac {Q_{H}}{Q_{H}-Q_{C}}}$

For a heat pump operating at maximum theoretical efficiency (i.e. Carnot efficiency), it can be shown that

${\frac {Q_{H}}{T_{H}}}={\frac {Q_{C}}{T_{C}}}$  and $Q_{C}={\frac {Q_{H}T_{C}}{T_{H}}}$

where $T_{H}$  and $T_{C}$  are the absolute temperatures of the hot and cold heat reservoirs respectively.

At maximum theoretical efficiency,

${\rm {COP}}_{\rm {heating}}={\frac {T_{H}}{T_{H}-T_{C}}}$

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,

${\rm {COP}}_{\rm {cooling}}={\frac {Q_{C}}{Q_{H}-Q_{C}}}={\frac {T_{C}}{T_{H}-T_{C}}}$

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.

${\rm {COP}}_{\rm {heating}}$  applies to heat pumps and ${\rm {COP}}_{\rm {cooling}}$  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 ${T_{H}}$  and 0 °C (32 °F) for ${T_{C}}$ . 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.  Air source air conditioners efficiency is derived from "dry bulb" air temperature used 20 °C (68 °F) for ${T_{H}}$  and 7 °C (44.6 °F) for ${T_{C}}$ .

## Improving COP

As the formula shows, the COP of a heat pump system can be improved by reducing the temperature gap $T_{hot}$  minus $T_{cold}$  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  ). Accurately determining thermal conductivity will allow for much more precise ground loop  or borehole sizing, resulting in higher return temperatures and a more efficient system. 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) by decreasing the speed of fluid which in turn lower the Re number and hence the turbulence (and noise) and the head loss (see hydraulic head). The heat pump itself can be improved by increasing the size of the internal heat exchangers which in turn increase the efficiency (and the cost) relative to the power of the compressor, and also by reducing the system's internal temperature gap over the compressor. Obviously, this latter measure makes such heat pumps unsuitable to produce high temperatures which means that a separate machine is needed for producing hot tap water.

## Example

A geothermal heat pump operating at a ${\rm {COP}}_{\rm {heating}}$  of 3.5 provides 3.5 units of heat for each unit of energy consumed (i.e. 1 kWh consumed would provide 3.5 kWh of output heat). The output heat comes from both the heat source and 1 kWh of input energy, so the heat-source is cooled by 2.5 kWh, not 3.5 kWh.

A heat pump with ${\rm {COP}}_{\rm {heating}}$  of 3.5, such as in the example above, could be less expensive to use than even the most efficient gas furnace except in areas where the electricity cost per unit is higher than 3.5 times the cost of natural gas (e.g. Connecticut or New York City).

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

Given the same energy source and operating conditions, a higher COP heat pump will consume less purchased energy than one with a lower COP. The overall environmental impact of a heating or air conditioning installation depends on the source of energy used as well as the COP of the equipment. The operating cost to the consumer depends on the cost of energy as well as the COP or efficiency of the unit. Some areas provide two or more sources of energy, for example, natural gas and electricity. A high COP of a heat pump may not entirely overcome a relatively high cost for electricity compared with the same heating value from natural gas.

For example, the 2009 US average price per therm (100,000 British thermal units (29 kWh)) of electricity was $3.38 while the average price per therm of natural gas was$1.16. Using these prices, a heat pump with a COP of 3.5 in moderate climate would cost $0.97 to provide one therm of heat, while a high efficiency gas furnace with 95% efficiency would cost$1.22 to provide one therm of heat. With these average prices, the heat pump costs 20% less to provide the same amount of heat. At 0 °F (-18 °C) COP is much lower. Then, the same system costs as much to operate as an efficient gas heater. The yearly savings will depend on the actual cost of electricity and natural gas, which can both vary widely.

The above example applies only for an air-source heat pump. The above example assumes that the heat pump is an air-source heat pump moving heat from outside to inside, or a water-source heat pump that is simply moving heat from one zone to the other. For a water-source heat pump, this would only occur if the instantaneous heating load on the condenser water system exactly matches the instantaneous cooling load on the condenser water system. This could happen during the shoulder season (spring or fall), but is unlikely in the middle of the heating season. If more heat is being withdrawn by the heat pumps that are in heating mode than is being added by the heat pumps that are in cooling mode, then the boiler (or other heat source) will add heat to the condenser water system. The energy consumption and cost associated with the boiler would need to be factored in to the above comparison. For a water-source system, there is also energy associated with the condenser water pumps that is not factored in to the heat pump energy consumption in the example above.

## Seasonal efficiency

A realistic indication of energy efficiency over an entire year can be achieved by using Seasonal COP or Seasonal Coefficient of Performance (SCOP) for heat. Seasonal energy efficiency ratio (SEER) is mostly used for air conditioning. SCOP is a new methodology that gives a better indication of expected real-life performance, using COP can be considered using the "old" scale. Seasonal efficiency gives an indication on how efficient a heat pump operates over an entire cooling or heating season.