B Does a good heat engine make a bad refrigerator?

Hi all,

the efficiency ##\eta## of a generic heat engine working between two temperatures is bound from above by the efficiency ##\eta_{\rm C}## of a Carnot machine working between the same temperatures.
That is, if the temperatures are the same, a (ideal) Carnot machine is better than any (real) machine.

I assumed that a "Carnot refrigerator" would be also better than a regular refrigerator operating between the same temperatures. Without thinking too much I thought that the coefficient of performance ##\epsilon## of an air conditioner (AC) or heat pump (HP) would obey the same kind of inequality as the efficiency.

Upon further thinking this does not seem the case, because I find
[tex]\epsilon_{\rm AC}=\frac{1-\eta}{\eta}[/tex]
[tex]\epsilon_{\rm HP}=\frac{1}{\eta}[/tex]
Looking at these relations it seems that the better the heat engine, the worse the device obtained by reversing it. If the Carnot engine is the best possible machine, is the Carnot fridge the worst possible fridge?

Is this really the case, or am I getting it wrong?

Maybe it does no make sense to connect ##\eta## and ##\epsilon##, because engines and heat pumps (or air conditioners) work in different temperature ranges?
I did this because in texbooks I see many examples/exercises discussing what happens by reversing the cycle of a heat machine.
So the Carnot engine is the most efficient cycle between two given reservoirs. If you are pumping heat from the cold reservoir to the hot reservoir then the Carnot engine requires the least work, and if you are dumping heat from the hot reservoir to the cold reservoir then it gives the most work output.

I suspect that your equations are looking at different reservoirs. Better to use formulas that explicitly list the heat from/to each reservoir and the work.
Hi Dale, thanks for your help. Let me give you some more context.

I derived these formulas while thinking to problems along the lines "A scientist claims that he designed a heat engine working between the temperatures ##T_{\rm H}## and ##T_{\rm L}## which produces a work ##L## by extracting an amount of heat ##Q_{\rm H}## from a hot reservoir [...] Is the scientist's claim reasonable?"

In order to assess the scientist's claim one derives the efficiency of his machine and compares it against the efficiency of a Carnot machine working between the same two temperatures. And if ##\eta>\eta_{\rm C}## one calls the fraud police, right?

So, I was wondering how would I formulate and solve a similar problem where the scientist's claim is about a refrigerator rather than an engine.
I initially thought that the line of reasoning would be the same, under the assumption that ##\epsilon_C## would be larger than ##\epsilon##.

Now I'm not sure that is really the case. I am confused by my calculations above. I'm not sure they're really relevant, though.

What should I compare in this case? Would it be correct to assume ##\epsilon_C>\epsilon##?


What should I compare in this case? Would it be correct to assume ##\epsilon_C>\epsilon##?
Yes, carnot efficiency is always the best possible.

But a heat pump and heat engine have different goals, so the optimal temperature is different for each.
Hi Russ, thanks for your insight. I see...

I guess what confuses me is the details of reversing the same device while it operates between the same temperatures.

Suppose I have a real engine that operates by absorbing ##|Q_{\rm H}|## at ##T_{\rm H}## and produces a work ##|L|##. Its efficiency is
[tex]\eta=\frac{|L|}{|Q_{\rm H}|} < \eta_{\rm C} = 1-\frac{T_{\rm C}}{T_{\rm H}} [/tex]
Probably the point where I'm mistaken is that when I invert the cycle, and use the machine as a heat pump between the same temperatures, I cannot expect that the same amount of work ##|L|## has the effect of pumping the same heat ##|Q_{\rm H}|## as before into the hot reservoir.
This seems reasonable. If this was the case I could operate a heat pump exactly with the work I get from the thermal engine.
While this might be possible for an ideal machine, for a "real" machine there must be some kind of loss...

Also, if the above was possible I would get
[tex]\epsilon=\frac{|Q_{\rm H}|}{|L|} > \epsilon_{\rm C} = \left(1-\frac{T_{\rm C}}{T_{\rm H}}\right)^{-1} [/tex]
which is exactly the opposite of what you say (and I believe to be true).

The only sensible conclusion I can think of is that the heat pumped by the real machine must obey
[tex]|Q'_{\rm H}|< \epsilon_{\rm C} |L| [/tex]
and hence ## |Q'_{\rm H}|<|Q_{\rm H}|##. That is: if I invert the cycle and inject work into the machine, the heat pumped into the hot reservoir is less than that absorbed by the engine when producing the same amount of work.

Am I making sense?
That is: if I invert the cycle and inject work into the machine, the heat pumped into the hot reservoir is less than that absorbed by the engine when producing the same amount of work.
Yes, that is correct. The efficiency of a real heat engine is ##\frac{L}{Q_H}<\frac{T_H-T_C}{T_H}## and the COP of a real heat pump is ##\frac{Q_H}{L}<\frac{T_H}{T_H-T_C}##. So if you use a real engine to generate ##L_e## work from ## Q_H## heat and then use ##L_p## work in a heat pump to move the same ##Q_H## heat back then we have


So it always takes more work to put the heat back than you got out initially.
ok... so my original starting point, that inverting a heat machine would completely reverse the heat and work fluxes , was a misconception.
It actually makes a lot of sense, and I strongly suspected that my initial conclusions could not be right.
Thanks everybody for helping me getting rid of this misconception.

I do not know if it's just me, but I think that the (high-school) books I have do not really stress this point, which seems kind of important to me.
I mean, the concept is there... Carnot's theorem is stated, and there are inequalities involving coefficients of performance, but they are brief and dropped very casually.
On the other hand, the exercises in the same textbooks often require to reverse a cycle, and I do not recall any of them pointing out what I or Dale derived.
There is even one exercise requiring to show one of the equations in my OP, which kind of started my misconception.

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