# Carnot efficiency, coefficient of performance for a refrigerator

I was reading that a new type of refrigerator might reach a coefficient of performance (COP) of 10. This seems quite the achievement and the authors state that their approach might achieve a Carnot efficiency of 75%. (I could not get hold of the original source as it is a part of an expensive book series "Zimm C, Jastrab A, Sternberg A, Pecharsky V K, Gschneidner K Jr, Osborne M and Anderson I 1998 Adv. Cryog. Eng. 43 1759".)

Now I am a bit confused. Wikipedia states that the COP should be the inverse of the Carnot efficiency but than these numbers do not fit together or am I missing something?

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do you know the temperatures at which those figures where achieved? Does Carnot efficiency of 75% mean effiency = 75%, or 75% of the Carnot efficiency at the relevant temepratures? high COP implies low $\delta T$, the other way round for a heat engine. So I agree, the numbers don't fit. –  mart Aug 9 '12 at 14:02
The exact temperatures are not stated, they achieved a $\Delta T$ of 10 Kelvin in the room temperature range. It just says "a Carnot efficiency approaching 75%" in a review article by the same authors ("Recent developments in magnetocaloric materials" by Gschneidner et al.) –  Alexander Aug 9 '12 at 14:09
well, 10K and a cOP of 10 would mean the upper temperature is 100K ... there must be a mistake somwhere –  mart Aug 9 '12 at 14:44

Coefficient of performance (COP) talks about the amount of heat transferred for the amount of power used. When two points are close together in temperature, you can move a lot of heat for not a lot of energy expended. This is the principle behind using heat pumps for heating a house, for example. You take the heat from outside (which is "free"), and pump it into the house (with a $\Delta T$ of maybe 20 degrees). This means that in principle you could get almost 15 J of heating power for 1 J of pumping power, since the COP would be

$$COP = \frac{T_{hot}}{T_{hot}-T_{cold}} = \frac{293}{20}$$

Now the Carnot efficiency (which is the inverse of the COP as you pointed out) tells us how much of the heat difference can be converted to work. But Carnot cycle requires "infinitely slow" operation, and in practice you cannot achieve that. You might achieve just 75% of the efficiency of an ideal Carnot engine.

But if that is the interpretation, then if you run at 75% of the Carnot cycle efficiency, then a COP of 10 with 75% efficiency corresponds to a Carnot efficiency of $(0.75*1/10)=7.5\%$. "In the room temperature range" that is indeed too high for just 10 K difference.

The point I am trying to make: there is a difference between "theoretical COP" and "COP achieved" - while the former is the inverse of the Carnot Efficiency, you normally don't get close.

However, there is another way to look at this. Although I don't have access to the source you read, I wonder whether your quote

the authors state that their approach might achieve a Carnot efficiency of 75%

actually means "if you use this approach to a heat engine, you might be able to create a heat engine that runs at 75% efficiency" (which would be quite a bit higher than most heat engines used today). That is a completely different, but valid, reading of that sentence. In that case the numbers don't have to agree - we just have to say "gosh, that's quite a good approach to heat engines", and then believe this would scale / be optimized at some point in the future to "approach" 75%.

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@EnergyNumbers - I am sure I did not mean that. Did some major editing - see if you like it better now? –  Floris Aug 28 '14 at 18:27