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The "efficiency" for heat pumps and chillers is called coefficient of performance (COP), because there is no conversion of electrical energy to heat energy but only a transportation. This COP can be way higher than 1, a heat pump can supply more heat energy than you put electrical energy into it. I once read that typical systems achieve COPs up to 3 or 4.

But how is the COP limited? I know, it depends on the temperature difference, the cooling fluid and some more factors, but is there a theoretical maximum? If I had a super duper cooling fluid with all the properties I need, could I build a chiller (or heat pump, I guess it works both ways pretty similar, because it only depends on which site of the system you use) with a COP of 10? Or 50? Or a few hundred? Or is there some fundamental law of physicis (probably some thermodynamic stuff) that limits the COP to a maximum value? If so, how high would that approximatly be?

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  • $\begingroup$ I think this is an interesting study. This is practically available that almost all heat pumps push 2 to 4.5 or in some cases even 6 times the heat energy the pump consume to do this. In lay man language a heat pump can push nE kw heat from a low temperature to high temperature side just by requiring E ke of energy as input. Now if some one reverse this and use a heat engine to produce energy then this nE ke heat available on hot side will do the work through a heat engine then the output given by the heat engine will be efficiency(say f) x nE ke. It means, if principle of conversation of ener $\endgroup$ Commented Sep 5, 2021 at 4:24

2 Answers 2

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In a heat pump, $Q_1$ amount of heat is removed from the system maintained at temperature $T_1$, and $Q_2>Q_1$ amount of heat is dumped into the ambient at temperature $T_2>T_1$. If work input is $W$, then energy conservation requires $Q_2-Q_1=W$.

The entropy change of the universe is given by \begin{align} \Delta S=-\frac{Q_1}{T_1}+\frac{Q_2}{T_2} \end{align} Second law requires $\Delta S\geq 0$ which implies \begin{align} -\frac{Q_1}{T_1}+\frac{Q_2}{T_2} & \geq 0 \\ \frac{Q_2}{Q_1} & \geq\frac{T_2}{T_1} \\ \frac{Q_2}{Q_1}-1 & \geq \frac{T_2}{T_1}-1 \\ \frac{W}{Q_1} & \geq \frac{\Delta T}{T_1}\quad (\Delta T\equiv T_2-T_1)\\ \therefore\quad \textrm{C.O.P.}& \leq\frac{T_1}{\Delta T}\quad (\textrm{C.O.P.}\equiv Q_1/W) \end{align}

Therefore for specified temperature of system and ambient above relation gives theoretical bounds on C.O.P. See that nothing forbids C.O.P. from becoming less than one.

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  • $\begingroup$ Ah, thanks. That means, to bring an example, if I would want to heat my home to a temperature of 21°C at an ambient temperature of 10°C, my COP would be limited to COP < (21K + 273K) / (21 - 10) K = 26.7. Thats not so bad. Thanks again for that interesting insight =) $\endgroup$
    – jusaca
    Commented Aug 9, 2017 at 7:46
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    $\begingroup$ @jusaca In your example, COP would be defined as $Q_2/W$, while the definition I have used in my answer applies to refrigerator viz. $Q_1/W$. In your case the bounds would be COP $\leq T_2/\Delta T$, as you have done correctly in your example. $\endgroup$
    – Deep
    Commented Aug 9, 2017 at 12:43
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    $\begingroup$ Maybe you should rewrite the formulas using T_l and T_h for T-low and T-high $\endgroup$
    – JanKanis
    Commented Sep 2, 2019 at 8:59
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    $\begingroup$ @jusaca The short version is ideal COP = T-high / delta-T. But for your high temperature you need to take the temperature of the heating element or water temperature, depending on what kind of heating system you use. This will be higher than the 21°C you want to keep your home at. $\endgroup$
    – JanKanis
    Commented Sep 2, 2019 at 9:59
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    $\begingroup$ Or for the layman, to put it simply: TheoreticalMaxCOP = (desiredIndoorTempC + 273) ÷ (desiredIndoorTempC - outsideTempC) $\endgroup$
    – Dan W
    Commented Mar 24, 2020 at 21:35
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There is no theoretical upper limit. The equation is the reciprocal of the carnot equation for heat pumps/the refrigeration cycle: "Q_refrig = T_max / (T_max - T_min); \\ T in Kelvin"

Take from it what you will, but actually utilising a high COP from a heat pump is impractical because it requires extremely small temperature gradients. If the difference between T_max and T_min is 0.1 deg K and your T_max is 275 deg K, then your COP will be 2750 in a perfect world, but it's useless since, who needs to maintain a 0.1 deg k temp difference that efficiently?

Plus you have to remember friction. So once friction is factored in, at very low temperature gradients, the actual "sum(total)" energy being moved is actually very small. This affects the measure and calculation of COP.

You might have a COP of 2750, but if you're only moving 1 watt of energy and overcoming bearing resistance requires 2 watts of energy....

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