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Mechanical Engineering DISCUSSION
A heat pump data sheet claims a COP of 4: 4 kW of heating for 1 kW of electrical input. An electric heater gives exactly 1 kW of heat per kW of electricity, and I was taught that nothing is more than 100 percent efficient.
Where do the other 3 kW come from when it is 0 °C outside, is there a theoretical limit to the COP, and why does the same unit have a lower COP on a colder day or with hotter radiators?
A heat pump does not convert electricity into heat, it uses work to move heat that already exists from a cold place to a warm one. The energy balance is Q_hot = Q_cold + W: 4 kW delivered indoors = 3 kW taken from the outside air + 1 kW of compressor work. Energy is conserved.
COP is not an efficiency in the sense of output over input of the same energy stream, so it is not limited to 1. It is heat delivered divided by work supplied. Outside air at 0 °C is at 273 K and contains plenty of thermal energy; the machine only needs something colder than the air to collect it.
The second law of thermodynamics sets the limit. For an ideal (Carnot) heat pump, COP = T_hot / (T_hot - T_cold) with temperatures in kelvin. Delivering heat at 35 °C from a source at 0 °C: 308 / 35 = 8.8. Delivering at 55 °C from -10 °C: 328 / 65 = 5.0.
Real machines reach very roughly half of the ideal figure, which is where data sheet values of 3 to 5 come from. The formula also answers your last question: the COP depends on the temperature lift, not on the temperatures themselves. A colder day or hotter radiators both increase the lift, and the COP falls.
The mechanism is the vapour-compression cycle, the same one as in a refrigerator. In the outdoor coil the refrigerant is at low pressure and boils at, say, -8 °C; being colder than the 0 °C air, it absorbs heat from it. The compressor raises the vapour's pressure, and with it the temperature at which it condenses, to about 40 °C. In the indoor heat exchanger it condenses and gives up both the collected heat and the compressor work. An expansion valve drops the pressure again and the cycle repeats.
Heat always flows from hot to cold at each exchanger. The work input is what creates the two pressure levels that make that possible.
Those example temperatures show why real performance is well below Carnot for the air and water temperatures. Each heat exchanger needs a temperature difference to transfer heat, so the refrigerant works between about -8 °C and 40 °C while the useful lift is only 0 °C to 35 °C. The ideal COP for the refrigerant's own temperatures is 313 / 48 = 6.5, not 8.8. Compressor losses, fan and pump power and, in damp cold weather, defrosting the outdoor coil take it down to around 4.
The practical consequence is that large heat exchangers and low flow temperatures matter more than anything else. Underfloor heating at 35 °C gives a markedly better seasonal COP than radiators that need 55 °C.
A fair comparison with burning fuel has to include how the electricity was made. If it comes from a thermal power station at about 40 percent efficiency, a COP of 4 gives 0.4 × 4 = 1.6 units of heat per unit of fuel energy, against roughly 0.9 for a good boiler burning the fuel directly. With a COP of 2.5 the two are about level.
When reading data sheets, check the rating point. A COP is quoted at a specific outdoor and water temperature, and the seasonal figure (SCOP) over a whole heating season is the one that predicts energy use.
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