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Energy & Renewable Energy Engineering DISCUSSION
A heat pump brochure claims a COP of 4, meaning 4 kWh of heat for every 1 kWh of electricity. That sounds like 400 percent efficiency, which should be impossible, since an electric resistance heater is already 100 percent efficient.
Where does the extra energy come from, is there a theoretical upper limit to COP, and why does the same unit have a much lower COP in cold weather or when it has to supply hotter water?
Energy is conserved; the heat pump does not create heat, it moves it. The electricity drives a compressor that carries heat from a cold source (outside air, ground or water) to a warmer place. The balance is Q_delivered = Q_from_source + W. With 1 kWh of electrical work and 3 kWh collected from the outside air, 4 kWh arrives indoors.
COP = Q_delivered / W = 4 is therefore not an efficiency in the sense of output over total input, because the 3 kWh from the environment is not counted as an input you pay for. A resistance heater converts work directly to heat, so its COP is 1, the worst case for using electricity to heat.
The upper limit comes from the second law. For an ideal reversible cycle, COP_Carnot = T_hot / (T_hot - T_cold), with temperatures in kelvin. Delivering heat at 35 °C (308 K) from a source at 5 °C (278 K): 308 / 30 = 10.3.
Real machines reach roughly 40 to 50 percent of the Carnot value, because the refrigerant has to be colder than the source and hotter than the sink for heat to flow, and the compressor, fans and pumps have their own losses. Half of 10.3 is about 5, which is in line with good units under those mild conditions.
The same formula explains the fall in cold weather and with hot water. What matters is the temperature lift. Delivering 55 °C (328 K) from -5 °C (268 K): Carnot COP = 328 / 60 = 5.5, and 45 percent of that is about 2.5. The lift has doubled from 30 to 60 K and the COP has halved.
Design consequences follow directly: underfloor heating or large radiators running at 35 to 40 °C give a much better COP than small radiators needing 60 °C or more, and a ground source at a steady few degrees above freezing beats air at -10 °C. Air-source units also spend energy defrosting the outdoor coil in cold, humid weather, which the instantaneous COP figure does not show.
When comparing products, check the conditions behind the quoted COP. A single figure is measured at one test point, stated as source and delivery temperatures, for example outdoor air at 7 °C and water at 35 °C. The seasonal figure (SCOP or a seasonal performance factor) averages over a heating season for a given climate and is the better guide to running cost. In a cold climate it is noticeably lower than the mild-weather COP.
Running cost per kWh of heat is the electricity price divided by the seasonal COP. At a seasonal COP of 3, heat costs one third of the electricity price per kWh, which you can compare with the fuel price divided by the boiler efficiency for gas or oil.
It is also worth following the energy back to the power station. If the electricity comes from a thermal plant at about 40 percent efficiency, then 1 kWh of fuel gives 0.4 kWh of electricity, and a heat pump with a COP of 3.5 turns that into 1.4 kWh of heat. Burning the same fuel in a boiler at 90 percent efficiency gives 0.9 kWh. The heat pump is still ahead, and the advantage grows as more of the grid supply comes from wind, solar or hydro.
With resistance heating the same chain yields only 0.4 kWh of heat per kWh of fuel, which is why direct electric heating is the poorest use of thermally generated electricity. On that basis a COP above about 2.3 is needed just to match a good boiler.
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