Is there some theoretical maximum coefficient of performance (COP) for heat pumps and chillers?

Question

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 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 (for a heat pump, I guess it works both ways pretty similarly because it only depends on which side of the system you use) with a COP of $10$, or $50$, or a few hundred?

Or is there some fundamental law of physics (probably some thermodynamic stuff) that limits the COP to a maximum value? If so, how high would that approximately be?

Answer 1

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.

Answer 2

There is no theoretical upper limit. The equation is the reciprocal of the Carnot equation for heat pumps/the refrigeration cycle: $$Q_{\text{refrig}} = \frac{T_{\text{max}}}{(T_{\text{max}} - T_{\text{min}})},$$ with $T$ in Kelvin ($\text{K}$).

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 $T_{\text{max}}$ and $T_{\text{min}}$ is $0.1 \ \text{K}$, and your $T_{\text{max}}$ is $275 \ \text{K}$, then your COP will be $2750$ in a perfect world, but it's useless since who needs to maintain a $0.1 \ \text {K}$ temperature 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 you're only moving $1 \ \text{Watt}$ of energy and overcoming bearing resistance requires $2 \ \text{Watts}$ of energy...