Here is a diagram of the system in question (adapted from the diagram on this page):

enter image description here

All of the work produced by the Carnot engine is used to drive the Carnot refrigerator, which is simply a Carnot engine operating in reverse. Accordingly, no work is performed outside the system. Despite the fact that no work is performed, such a system would create a perpetual flow of heat from the hot reservoir, through the Carnot engine, through the cold reservoir, and back up through the Carnot refrigerator. Also, because the Carnot engine is driving another Carnot engine in reverse, the net entropy increase is zero, ie all the heat that flows from the hot reservoir to the cold one to drive the Carnot engine is pumped back to the hot reservoir by the Carnot refrigerator.

I realize that such a perpetual flow of heat is only theoretical. Any physical realization would lose heat due to friction. I'm also aware that my question seems similar to a naive attempt at perpetual motion, but that is not at all the focus of my question. I am trying carefully distinguish between a flow of heat that does no work and any motion that produces work.

The focus of my question is simply whether such a theoretical system would indeed create a perpetual flow of energy with no increase in entropy. If so, this would mean that while perpetual motion (aka perpetual work) is theoretically impossible, perpetual flow (that does no work) is possible.

I'm also curious as to why such a system has not been discussed in the thermodynamic literature, even if merely to dismiss it as a possibility. I searched pretty extensively (using both Google and Google Scholar) and found no discussion of such a system, which is why I am now asking here. The only related discussion concerned a similar system that tried to bleed off some of the work produced by the Carnot engine to drive an external system. Such a configuration is clearly an attempt at a perpetual motion machine.

  • $\begingroup$ Systems like this are discussed quite often in textbooks. For instance, they are the mechanism by which you prove the equivalence of various pairs of the classical statements of the second law. $\endgroup$ Jul 25 '18 at 17:21
  • $\begingroup$ Yes, but see my response to Philip below. $\endgroup$
    – Nick Gall
    Jul 25 '18 at 20:55

"I searched pretty extensively (using both Google and Google Scholar) and found no discussion of such a system." I expect you know that the set-up in your diagram is the one used in the traditional development of thermodynamics to show that the Second Law (Clausius version) implies that all reversible engines operating between the same two temperatures have the same efficiency, and that it is greater than for an irreversible engine. See Zemansky's Heat and Thermodynamics, Pippard's Classical Thermodynamics, Fermi's Thermodynamics...

I think that you would have a perpetual flow of energy in such a theoretical situation, but that it is theoretical in just as damning a way as the perpetual motion of a simple pendulum could be called 'theoretical'. It won't happen! This is for a number of reasons (which, arguably, have as much right to be called 'theoretical' in that we have a pretty good understanding of them!)

  • $\begingroup$ I agree that the theoretical "perpetual heat flow" in the system I described is exactly analogous to the theoretical "perpetual conversion between kinetic and potential energy" of an (idealized) simple gravity pendulum: wikiwand.com/en/Pendulum_(mathematics)#/Simple_gravity_pendulum I believe both are useful "intuition pumps". Your mileage may very. $\endgroup$
    – Nick Gall
    Jul 25 '18 at 20:44
  • $\begingroup$ Also, I'm aware that my configuration is used to explain a different point, ie the equal efficiency of reversible heat engines. I'm just (mildly) surprised that while there is plenty of discussion of the perpetual conversion of energy with an ideal pendulum, there is little or no discussion of the perpetual flow of heat between a perfectly balanced engine and refrigerator. $\endgroup$
    – Nick Gall
    Jul 25 '18 at 20:54

By combining the two cycles you may theoretically be able to continually move the same amount of Q from the high temperature reservoir to the low and back again, but there is a cost. The overall change in entropy (the systems plus the two reservoirs) will always be > 0. That means each time you carry out the cycles it will result in some energy not available to do work in the current environment.

The Carnot Cycle can only approach a zero total entropy change but can never actually equal zero. And the reason for that is no matter how small you keep the temperature differentials between the working fluid and the reservoirs during the isothermal processes and how small you keep the pressure differentials during the adiabatic processes and how slowly you carry out all the processes you cannot make them zero, because then the processes would stop. And as long as temperature and pressure disequilibrium exist, the will be some degree of irreversibility and therefore entropy increase. The cycle is an idealization intended to establish an upper limit to the efficiencies of real heat engines and refrigerators.

We can illustrate this by looking at the isothermal expansion for the heat engine cycle alone.

Let $TH$ equal the temperature of the working fluid (the system) during reversible isothermal expansion.

Let $TH + dT$ equal the temperature of the high temperature reservoir (the surroundings) in order to enable heat transfer Q. The differential temperature $dT$ is made as small as possible to approach a reversible process but can never actually equal zero (because then there there would be no heat transfer).

Now let’s look at the entropy changes:

$\Delta S (system)=+Q/TH$

$\Delta S (surroundings)=-Q/(TH + dT)$

$\Delta S (total)=+Q/TH - Q/(TH + dT) >0$ for all $dT>0$

So the only way total entropy change can be zero is if $dT=0$ which would prohibit heat transfer.

If you go through the same with the isothermal compression you will again get a positive total entropy change. The same for the adiabatic processes. For the adiabatic processes it is the pressure differentials that cause the processes to be irreversible. Some finite differential is needed for the processes to occur.

  • $\begingroup$ "The Carnot Cycle can only approach a zero total entropy change but can never actually equal zero." This claim contradicts everything I have ever read about the Carnot Cycle. Every description of the Carnot Cycle states that it is BOTH that one full cycle does not increase entropy AND that isotermal expansion is indeed theoretically possible, ie the volume of the working fluid can increase without any increase in temperature. See eg hyperphysics.phy-astr.gsu.edu/hbase/thermo/carnot.html#c2 $\endgroup$
    – Nick Gall
    Jul 25 '18 at 20:30
  • $\begingroup$ Nick, quoting from same Hyperphysics website: "in order to approach the Carnot efficiency, the processes involved in the heat engine cycle must be reversible and involve no change in entropy. This means that the Carnot cycle is an idealization, since no real engine processes are reversible and all real physical processes involve some increase in entropy". $\endgroup$
    – Bob D
    Jul 25 '18 at 22:27
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    $\begingroup$ Sorry, Bob, I thought I made it clear that I am only interested in ideal aka theoretical Carnot cycles, not real engine processes. I said in the description of my question, "I realize that such a perpetual flow of heat is only theoretical." I agree that all real physical processes involve some increase in entropy. $\endgroup$
    – Nick Gall
    Jul 25 '18 at 22:40
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    $\begingroup$ @NickGall I thought it was pretty obvious that you knew the Carnot cycle was an idealization just by reading your question. You did make it pretty clear. $\endgroup$ Jul 26 '18 at 12:09
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    $\begingroup$ OK we all agree the Carnot Cycle is an idealization just like the perpetual motion of a simple pendulum as brought up by Phillip Wood (not sure I saw his answer when I submitted mine). I guess what I should have said is a real cycle composed of two isothermal and two adiabatic processes can not result in perpetual heat flow. That't the only point I was trying to make. Anyway, its been an interesting discussion. $\endgroup$
    – Bob D
    Jul 26 '18 at 13:39

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