About this time of year in my region people often look at the air conditioning compressor pumping heat to the outside air on one side of their house while their swimming pool heater tries to pump heat into the water on the other side, and wonder why they're paying twice to move heat. Inevitably someone thinks of submerging their compressor coils, but then they run the numbers and realize that it won't be too many days before their swimming pool is uncomfortably hot.

But this leads to an interesting engineering question: What is the practical breakeven point between air-source and water-source heat pumps (for cooling indoor air)? I.e.: Is it possible, due to its higher thermal conductivity, for a water sink at a higher temperature to be more efficient for cooling than an air sink at a lower temperature?

More specifically: Suppose we have an outdoor compressor with surface area x pumping y BTUs of heat. I am assuming that because water has something like 20 times the thermal conductivity of air that it would be a more efficient heat sink even when it's somewhat hotter than the air sink. (Is that correct?) But how much more efficient? I.e., at what temperature differential is air as efficient as water?

(Or is this not a practical question? For example, we run fans to force more air through air-cooled compressors. That takes energy, but maybe it raises the effective thermal conductivity to the point that it is breakeven with a convective water sink ... even including the energy to drive the fan?)

  • $\begingroup$ Water has a lot of thermal mass. 75.375 ±0.05 J/mol·K Which means that for 18 liters of water it takes a 750W heating element around 1 minute and 40 seconds to heat it from 20 °C to 30 °C. $\endgroup$ May 27, 2016 at 9:29
  • $\begingroup$ What is the question here? You appear to be mixing thermal storage capacity with thermal transfer rates. $\endgroup$ May 27, 2016 at 13:07
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    $\begingroup$ The question is about thermal transfer efficiency. I just edited in an attempt to clarify. If you ask a physicist he'll say, "It doesn't matter whether it's water or air, it takes more energy to pump heat over a wider temperature range. So if the outside water is hotter than the air it's less efficient." But I'm wondering if that ignores the costs associated with transfer to a medium with vastly different thermal conductivity. $\endgroup$
    – feetwet
    May 27, 2016 at 16:02

1 Answer 1


The hot-side air coil of an AC installation (the part that heats the outdoor air) is a compromise between effectivity of heat transfer from the refrigerant to the air, the size, and the cost. Heat transfer to air typically comes with a significant thermal resistance, for two reasons. Firstly, air is a bad heat conductor. Secondly, air has a low heat capacity, so the air will heat up as it passes through the air coil.

A quick Google didn't give me typical values of the effective thermal resistance of the air coil. If you have an IR thermometer, you could measure the temperature of the coil and compare it to the air temperature at the outdoor inlet. It wouldn't surprise me if the coil is 20 K higher in temperature than the outdoor air, assuming that the AC is running at full power.

If you put a coil designed for air into water (assuming that corrosion etc. is not a problem), with some means of circulating the water, I think you can safely assume that the coil temperature will be very close to the water temperature. So, the break-even point would be when the water temperature is, say, 19 K higher than the air temperature.

  • $\begingroup$ Very good idea: I just went outside to measure my relatively new and efficient (meaning, among other things, large) residential AC compressor. The return line was over 50C, ambient air was just under 30C, and I couldn't find any spot on the exterior coil that was more than 4C higher than ambient! So, when exchanger size is not a significant constraint, it appears that the state-of-the art in air-exchange heat pumps is pretty efficient! (And, therefore, breakeven temperature for a water sink using the same coil could not be much higher.) $\endgroup$
    – feetwet
    May 29, 2016 at 21:30
  • $\begingroup$ I'm not sure what the higher temperature of the return line means. Losing another 16 K in something else than the air coil sounds like an unnecessary efficiency loss. If you used an IR thermometer, did you make sure to point it at painted surfaces, not bare metal? $\endgroup$ May 30, 2016 at 5:59
  • $\begingroup$ Sorry, "return line" is unclear: The 50C segment is leaving the compressor and going into the condenser coil, and is certainly within the design parameters of the system. (Now that I think of it I should have just measured the temperature on the high-pressure line leaving the condenser.) $\endgroup$
    – feetwet
    May 30, 2016 at 12:42

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