Coefficient of performance
The coefficient of performance 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, 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. Performance of Absorption refrigerator chillers is typically much lower, as they are not heat pumps relying on compression, but instead rely on chemical reactions driven by heat.
Equation
The equation is:where
- is the useful heat supplied or removed by the considered system.
- is the work required by the considered system.
where
- is the heat removed from the cold reservoir.
- is the heat supplied to the hot reservoir.
Derivation
Therefore, by substituting for W,
For a heat pump operating at maximum theoretical efficiency, it can be shown that
where and are the thermodynamic temperatures of the hot and cold heat reservoirs respectively.
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.
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.
Similarly, the COP of a refrigerator or air conditioner operating at maximum theoretical efficiency,
applies to heat pumps and applies to air conditioners and refrigerators. 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 for and 0 °C 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.
The COP of an air source air conditioner is derived using dry-bulb temperature of 20 °C for and 7 °C for.
Improving COP
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 which requires piped floor, wall or ceiling heating, or oversized water to air heaters and 2) increasing the input temperature. 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 by decreasing the speed of fluid which in turn lower the Re number and hence the turbulence and the head loss. The heat pump itself can be improved by increasing the size of the internal heat exchangers which in turn increase the efficiency 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.COP of Absorption chillers can be improved by adding a second or third stage. Double and triple effect chillers are significantly more efficient than single effect, and can surpass a COP of 1. They require higher pressure and higher temperature steam, but this is still a relatively small 10 pounds of steam per hour per ton of cooling.
Example
A geothermal heat pump operating at a of 3.5 provides 3.5 units of heat for each unit of energy consumed. 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 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.
A heat pump cooler operating at a of 2.0 removes 2 units of heat for each unit of energy consumed.
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 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.
The COP of a heat pump or refrigerator operating at the Carnot efficiency has in its denominator the expression TH - TC. As the surroundings cool the denominator increases and COP reduces. Therefore the colder the surroundings, the lower the COP of any heat pump or refrigerator. If the surroundings cool, say to 0 °F, COP falls in value below 3.5. 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, 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 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.