I have read that heat pumps are more efficient than generating heat directly for heating homes, until the temperature gets down to a critical value.

I just asked how to calculate temperature of gasses: Behavior of gasses, ideal and otherwise

and have a little octave code

function T2 = compressedgas(cp, cv, P1, V1, T1, P2)
  gamma = cp/cv; % approx 7/5 for dry air
  R = 8.3;
  V2 = V1 * ( P1/P2)^(1/gamma)
  n = P1 * V1 / (R * T1);  
  T2 = P2 * V2 / (n * R);

For air at STP, T = 298, P = 102kPa, compressed to P2 = 714kPa, I get a temperature of 1005 K given gamma = 1.4

Assuming a pressurized tank allowed to radiate heat until it cooled down, this seems like a very efficient heat source indeed.

When I change the assumption to very cold air T = 248 the temperature is still 839K.

This seems hot enough that I don't understand why heat pumps should ever be less efficient than directly heating the house, so can anyone identify the engineering details that make this difficult?

  • $\begingroup$ I don't understand what the octave code does? Isn't it sufficient to calculate the Carnot efficiency? An ideal heat pump will always be better than a machine that generates heat directly. However, real heat pumps are not ideal and that is one reason generating heat directly might be better. $\endgroup$ Oct 22, 2013 at 1:47
  • $\begingroup$ I assume the reason not to compress cold air is that there is insufficient heat content for the effort of compressing it. How would I calculate Carnot efficiency? Work should be $W= \int P dV$ ? $\endgroup$
    – Dov
    Oct 22, 2013 at 11:24
  • 1
    $\begingroup$ Efficiency is probably not the only issue. As the outdoor temperature gets colder ice starts forming on the equipment, which requires defrost cycles to melt. At some point it's not worth the effort to keep it clear, so it makes more sense to just heat the air directly. $\endgroup$
    – Mark
    Oct 23, 2013 at 22:07
  • 1
    $\begingroup$ Here is a nice pedagogical paper on this subject, which might help make things clearer: bayes.wustl.edu/etj/articles/AJP00180.pdf $\endgroup$
    – N. Virgo
    Jan 22, 2014 at 6:18

1 Answer 1


Heat pump should transfer heat from outside into the house. It should not generate heat (ideally) it should only force the heat to move.

One joule of work executed by the heat pump can transfer several joules of heat - for example 4 joules (it is the reason why the heat pump is efficient "source" of thermal energy). This ratio is called Coefficient of Performance or COP.

The bigger is the difference between input and output temperature -> the more work must the heat pump do to transfer the heat. Therefore the COP is decreasing.

If the COP did not decrease with increasing temperature difference, it would be possible to construct a perpetual motion machine of the second kind.

If work required to compress the gas was included into the computation the COP would be visible. And in order to achieve good COP the much lower output temperature would be needed.

EDIT: Computation of $COP$ with example.

We will start from ideal heat engine modeled by Carnot cycle. Carnot cycle has efficiency $$\eta = \frac{T_H - T_C}{T_H}$$

Let's assume $T_H=310K$, $T_C=270K$ and assume $100J$ of heat $Q$ will be delivered from hot reservoir to the heat engine.

As a result $$\eta \times 100J = \frac{310 - 270}{310} \times 100J = 12.9J$$ of work $W$ will be done by heat engine and $87.1J$ of heat will end up in cold reservoir.

What happens when we reverse the process? (Carnot cycle is reversible)

In the reversed process $87.1J$ of heat will be taken from cold reservoir, $12.9J$ of work will be delivered to the engine (now the heat pump) and $100J$ of heat will end up in the hot reservoir.

The COP is $$COP = \frac{Q}{W} = \frac{100J}{12.9J} = 7.75$$

And in general $$COP_{ideal} = \frac{Q}{W} = \frac{Q}{\eta \times Q} = \frac{1}{\eta} = \frac{T_H}{T_H - T_C}$$


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