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It seems that the only condition used in proving that the Carnot engine is the most efficient is that it is reversible. More specifically, the Carnot engine can be run in reverse as a refrigerator. Furthermore, it is asserted that all reversible engines will have the same efficiency.

However, isn't any closed loop on a PV diagram reversible? The arrows can simply be drawn in the reverse way to create a refrigerator. If any closed loop is reversible then why does the specific Carnot engine (a specific loop) have the highest efficiency?

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However, isn't any closed loop on a PV diagram reversible? The arrows can simply be drawn in the reverse way to create a refrigerator. If any closed loop is reversible then why does the specific Carnot engine (a specific loop) have the highest efficiency?

This was exactly the question I asked myself ten years ago :-) The problem is that often students do not appreciate the whole statement: Carnot's engine is operating between two temperatures (heat sources). Any circle on the PV-plane is reversible if you have many heat sources. In the case of many heat sources, you may also know that you do not talk about the efficiency of the engine, but you talk about the Clausius' equality: $$\sum_{i} \frac{Q_i}{T_i}= 0.$$ Note that $T_i$ is the temperature of the $i$th heat source (this is a very important point often missed!), which equals the temperatures of the system when they are in reversible contact. This is not true if the process is irreversible: you have heat flow from hot sources to the (colder) engine. Then one has the Clausius' inequality: $$\sum_{i}\frac{Q_i}{T_i}<0.$$

So, in short: Carnot's engine is the only reversible engine operating between two temperatures.

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  • $\begingroup$ I was just about to post the answer ;) congratz of fast explanation. $\endgroup$ – Cheshire Cat Nov 27 '14 at 19:20
  • $\begingroup$ @Cheshire: Nice to know you ;-) $\endgroup$ – ttuethao Nov 27 '14 at 19:25
  • $\begingroup$ sorry this might be trivial but can you explain why loops on the PV plane need many heat sources? $\endgroup$ – curiousgeorge Dec 27 '14 at 18:48
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    $\begingroup$ Take any segment of such a cycle where there is a heat exchange. For the segment to be reversible, the heat can only be exchanged when the temperatures of the system and of the heat source are the same (otherwise the segment can not be reverted: heat always flows from hot to cold). Therefore, apart from adiabatic segments, the cycle requires as many sources as its temperatures over the cycle, which is theoretically infinite. In practice, there are only finite number of sources, the best way to make a reversible process is to approximate it by a series of isothermal and adiabatic segments. $\endgroup$ – ttuethao Dec 29 '14 at 11:14
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All the answers here are wrong to an extent - or at least very misleading. The Carnot cycle is not the highest efficiency cycle of all possible cycles, it is only one of an infinity of cycles all of which exhibit the highest possible efficiency. There is nothing particularly special about the Carnot cycle, except that it is a simple cycle, that it is relatively easy to conceptualize and thus it makes a good teaching example, and that it is the cycle Carnot chose to use to explain the concept so it has historical precedence. Any cycle made up entirely of reversible processes will be a reversible cycle and all of these will have the same highest possible efficiency - the Carnot efficiency. As an example of one - look at the Stirling Cycle with an ideal regenerator.

Also, the Carnot cycle has relatively little enclosed area on the P-V diagram so it does relatively little work per cycle making it a relatively poor cycle to implement in real machinery. Thus, there are not a lot of intentionally designed Carnot Cycle engines lying about.

We do everyone a disservice propagating the misconception that the Carnot Cycle is the one best cycle of all possible cycles efficiency wise. It is only one of many.

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  • $\begingroup$ I suggest changing the CAPITALIZATION in this answer. Nobody likes people "screaming" at them (that's what capitalization reads like to many, including myself). Otherwise, good answer. +1 $\endgroup$ – Danu Jan 6 '16 at 21:01
  • $\begingroup$ I thought the Carnot cycle was the most efficient gas cycle. Does any other cycle have an efficiency greater or equal to $1-\frac{T_L}{T_H}$? $\endgroup$ – kilojoules Oct 15 '17 at 1:36
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The Carnot engine is incapable of producing any power utilizing the Carnot cycle.

The Carnot cycle only defines the limit of efficiency for a heat engine operating between a high temperature heat source and a lower temperature heat sink.

It defines the limit of efficiency, since it would use up all of the potential heat energy available to transfer from the high temperature heat source to the lower temperature heat sink.

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    $\begingroup$ Well, except that it can... $\endgroup$ – Jon Custer Nov 17 '16 at 23:23

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