How many reversible thermodynamic cycles are there between two heat reservoirs?

  • #1
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Summary:
I find some versions of Carnot's theorem generically referring to "a reversible thermodynamic cycle between two reservoirs". Does that necessarily mean "a Carnot cycle"?
Hi,
I was revisiting my (high school level) understanding of thermodynamic cycles and I think I still have some doubts. Last year and more recently I posted a few questions which surely helped me, but I think I need more clarifications.

In a nutshell, what I'd like to know is the following: suppose you have a reversible heat engine that operates (i.e. exchanges heat) between two reservoirs.

Is that machine necessarily a Carnot engine? That is a machine operating according to Carnot's cycle (adiabatic, isothermal, adiabatic, isothermal)?

Would this mean that any non-Carnot engine transferring heat between two reservoirs is not reversible, not even in the ideal situation where mechanical friction causing dissipation has been entirely removed?

For instance: suppose you have a heat engine operating between two reservoirs via a isobaric-adiabatic-isobaric-adiabatic cycle. Suppose all the mechanical parts are perfectly ideal, with no friction and hence no dissipation.
That would not be a reversible engine anyways?
Is this because of the non reversibility of spontaneous heat flow?

I'm asking all this because my textbook gives a version of Carnot's theorem generically referring to "a reversible heat engine operating between two temperatures", whereas other versions of the theorem I found explicitly refer to "a Carnot engine".

Now, a Carnot engine surely fits the description. But the wording in my book kind of leaves other possibilities open, like there were reversible cycles other than Carnot's.

Thanks a lot for your help.
Franz
 

Answers and Replies

  • #2
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"I find some versions of Carnot's theorem generically referring to 'a reversible thermodynamic cycle between two reservoirs'. Does that necessarily mean 'a Carnot cycle'? "


No, there are other cycles that fit the bill. Among them: Ideal Stirling, Ideal Ericsson
 
  • #3
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Hi mfig,
thanks for your reply. That would explain the phrasing in my textbook, which however won't give any clue about reversible thermodynamic cycles other than Carnot's. Also, different textbooks only mention the Carnot cycle in the Carnot theorem.

I have a few more questions. Please bear with me, I'm keeping all of this at the high-school level.

Assuming that all the ideal cycles you mention work between the same temperatures, your reply should mean that their efficiency is the same as in a Carnot cycle, right?

I'm asking because I tried to work out the efficiency of an isobaric-adiabatic-isobaric-adiabatic cycle. It seems to me that the result I found is even smaller than the efficiency in the Carnot cycle.

Since this cannot be right, I thought that perhaps I was wrong in assuming that such cycle is reversible.
I thought that this might have to do with the fact that the isobaric processes won't happen at a fixed temperature, and hence the heat flow would be irreversible.

This way my result won't go against Carnot's theorem.

Of course another possible explanation could be an error in my calculations.

Then again my reasoning could be too naive, and the situation with this cycle is more complex than I'm imagining.

In your reply to one of my recent posts you mentioned that in a Stirling cycle only the isothermal transformations exchange heat with the reservoirs. Perhaps I should take something like that into account also in the cycle I've examined (isobaric-adiabatic-isobaric-adiabatic)?
I really cannot see how, though. Here there are only two heat exchanges ....
 
  • #4
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Hi mfig,
Assuming that all the ideal cycles you mention work between the same temperatures, your reply should mean that their efficiency is the same as in a Carnot cycle, right?

I'm asking because I tried to work out the efficiency of an isobaric-adiabatic-isobaric-adiabatic cycle. It seems to me that the result I found is even smaller than the efficiency in the Carnot cycle.

In your reply to one of my recent posts you mentioned that in a Stirling cycle only the isothermal transformations exchange heat with the reservoirs. Perhaps I should take something like that into account also in the cycle I've examined (isobaric-adiabatic-isobaric-adiabatic)?
I really cannot see how, though. Here there are only two heat exchanges ....

Yes, you need to include the fact that regenerater heat interactions do not take place between the operating fluid and the thermal reservoirs, but are internal, and redo the calculation of the efficiency. You should arrive at the Carnot efficiency for the ideal Stirling engine.
 
  • #5
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Ok, perhaps I'm starting to see the light. Apologies for being so slow, and thanks a lot for your help!

If you have a reason for disregarding the isochoric processes (the regenerator), the heat-exchanging processes are the same as in the Carnot cycle.

I guess that the reason would be similar for the Ericsson cycle, which consists in isobaric-isothermal-isobaric-isothermal transformations. If you keep the isobaric processes "internal", so that only the isothermal transformations exchanges heat with the reservoirs, once again you have the efficiency of a Carnot cycle.

But if one was to calculate the efficiency in a more naive way, just by analysing the heat exchanges involved in the cycle, the efficiency would be different. It should be worse than in a Carnot cycle, right?

But what about the following cycles? For the sake of simplicity let us assume that the mechanical friction can be entirely removed:
  • isobaric-adiabatic-isobaric-adiabatic
  • isochoric-adiabatic-isochoric-adiabatic
  • isobaric-adiabatic-isochoric-adiabatic
In those cycles the heat-exchanging processes are only two, and they are not isothermal.
Should I assume that in their practical applications I could (ideally) make it so that the isobaric/isochoric exchanges are internal, by using a regenerator or a similar device?

Would that be the same with other cycles, like the ones with more than 2 heat-exchanging processes, none of which is isothermal? For instance isobaric-isochoric-isobaric-isochoric?

Or else, for building an ideally reversible cycle you always need two isothermal processes plus two other processes which "compensate" each other through some device, like a regenerator?
 
  • #6
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Ok, perhaps I'm starting to see the light. Apologies for being so slow, and thanks a lot for your help!

If you have a reason for disregarding the isochoric processes (the regenerator), the heat-exchanging processes are the same as in the Carnot cycle.

I guess that the reason would be similar for the Ericsson cycle, which consists in isobaric-isothermal-isobaric-isothermal transformations. If you keep the isobaric processes "internal", so that only the isothermal transformations exchanges heat with the reservoirs, once again you have the efficiency of a Carnot cycle.

But if one was to calculate the efficiency in a more naive way, just by analysing the heat exchanges involved in the cycle, the efficiency would be different. It should be worse than in a Carnot cycle, right?

That wouldn't be naive, that would be wrong. Thermal efficiency for a power cycle is defined as the ratio of input heat to output work. Internal interactions are neither of these, and so can not be counted in the calculation. It's just a matter of definition.

But what about the following cycles? For the sake of simplicity let us assume that the mechanical friction can be entirely removed:
  • isobaric-adiabatic-isobaric-adiabatic
  • isochoric-adiabatic-isochoric-adiabatic
  • isobaric-adiabatic-isochoric-adiabatic
In those cycles the heat-exchanging processes are only two, and they are not isothermal.
Should I assume that in their practical applications I could (ideally) make it so that the isobaric/isochoric exchanges are internal, by using a regenerator or a similar device?

The first cycle you describe is the Brayton cycle. The thermal efficiency of this cycle can be improved by a regenerator that takes the turbine output and uses it to heat the compressor output.

The second cycle you describe is the Otto cycle, and as far as I know this cannot be benefited from the same approach.

The third cycle you describe is the Diesel cycle, and the same goes for this one.
 
  • #7
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That wouldn't be naive, that would be wrong. Thermal efficiency for a power cycle is defined as the ratio of input heat to output work. Internal interactions are neither of these, and so can not be counted in the calculation. It's just a matter of definition.

Right, I (sort of) understand that. But, as I say, I'm taking the viewpoint of a high-school student reading a textbook that just mentions
[tex]
\eta = 1- \frac{|Q_{\rm C}|}{|Q_{H}|}
[/tex]
Most textbook problems involve some of the cycles listed above, without even mentioning the fact that some processes are internal, or a regenerator.
If the student wanted to calculate their efficiency, he/she would use the formula naively.

By the way "Fundamentals of Physics", by Halliday, Resnick & Walker, seems to have this naive approach in mind when it states that "the efficiency of an ideal Stirling engine is lower than that of a Carnot engine operating between the same two temperatures" (see here).
This makes my head spin. Wouldn't this statement mean that a Stirling cycle is not reversible?

This is what my question is about.


The first cycle you describe is the Brayton cycle. The thermal efficiency of this cycle can be improved by a regenerator that takes the turbine output and uses it to heat the compressor output.

Ok, but does this mean that the Brayton cycle can be made (ideally) reversible with the aid of a regenerator, and hence its efficiency would be the same as a Carnot cycle between the same temperatures?
Even if, unlike the Stirling and Ericsson cycles, none of its processes is isothermal?

The second cycle you describe is the Otto cycle, and as far as I know this cannot be benefited from the same approach.
The third cycle you describe is the Diesel cycle, and the same goes for this one.

Then must one conclude that the efficiency of these cycles is necessarily smaller than that of a Carnot cycle between the same temperatures?
Does this mean that these last two cycles are not reversible, even in the ideal case where no friction is present?

And what about a isobaric-isochoric-isobaric-isochoric cycle?
 
  • #8
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Right, I (sort of) understand that. But, as I say, I'm taking the viewpoint of a high-school student reading a textbook that just mentions
[tex]
\eta = 1- \frac{|Q_{\rm C}|}{|Q_{H}|}
[/tex]
Most textbook problems involve some of the cycles listed above, without even mentioning the fact that some processes are internal, or a regenerator.
If the student wanted to calculate their efficiency, he/she would use the formula naively.

We really can't help it if someone uses a formula naively, can we? I am not sure what you want from me on this... perhaps some method to prevent people from misusing an equation? There is no such method. The best we can do is explain what the symbols mean and what they don't mean. There are two external reservoirs, one that provides a heat sink ##Q_c## at temperature ##T_c## and the other which provides a heat source ##Q_h## at temperature ##T_h##. That is what the formula is talking about. Any other use is not right - whether someone is in high school or not.


By the way "Fundamentals of Physics", by Halliday, Resnick & Walker, seems to have this naive approach in mind when it states that "the efficiency of an ideal Stirling engine is lower than that of a Carnot engine operating between the same two temperatures" (see here).
This makes my head spin. Wouldn't this statement mean that a Stirling cycle is not reversible?

They appear to be discussing a case where the regenerator is not treated as 100% efficient, a requirement I mentioned explicitly in your last post on a similar topic. That's fine, and yes if the regenerator is not treated as 100% efficient then it is a source of irreversibilities - as in a real Stirling engine. In either case, they are not simply using the formula you show naively. They are considering internal irreversibilities and explaining how that will lower the efficiency.


Ok, but does this mean that the Brayton cycle can be made (ideally) reversible with the aid of a regenerator, and hence its efficiency would be the same as a Carnot cycle between the same temperatures?
Even if, unlike the Stirling and Ericsson cycles, none of its processes is isothermal?

No. The regenerator in a Brayton cycle doesn't work the way the (ideal) regenerator works in a Stirling engine.

Then must one conclude that the efficiency of these cycles is necessarily smaller than that of a Carnot cycle between the same temperatures?
Does this mean that these last two cycles are not reversible, even in the ideal case where no friction is present?

Yes, that's right.


And what about a isobaric-isochoric-isobaric-isochoric cycle?

Where did you see such a cycle?
 
  • #9
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I am not sure what you want from me on this...
I was just trying to explain what I meant by naive application. Of course I understand it can be wrong.

They appear to be discussing a case where the regenerator is not 100% efficient,

I really fail to see this. Maybe this is obvious for someone that is familiar with the Stirling engine. But in that paragraph there is no mention of the word regenerator. They seem to suggest that, since we cannot ignore the heat exchanges occurring in the isochoric processes, we cannot derive the formula for the efficiency of the Carnot cycle from the general one.

That's fine, and yes if the regenerator is not 100% efficient then it is a source of irreversibilities - as in a real Stirling engine.

That is not what they are saying, though.
Not that I'm holding you responsible for what they say, but it seems to me they are speaking of an ideal reversible case: "Thus reversible heat transfers (and corresponding entropy changes) occur in all four of the processes .... does not apply to an ideal Stirling engine ...".
They do refer to the real case later on, when they say "Real Stirling engines have even lower efficiency".

No. The regenerator in a Brayton cycle doesn't work the way the (ideal) regenerator works in a Stirling engine.

Ok... So, does this mean that even with a regenerator an ideal Brayton cycle wouldn't attain the maximum possible (Carnot) efficiency? Hence can we say that the ideal Brayton cycle is not reversible?

Yes, that's right.
Ok, so since there is no friction, irreversibility is in the heat exchanges, right?

Where did you see such a cycle?

In basic problems about thermodynamic cycles given in high school textbooks (I see one my version of Halliday, Resnick and Walker, and one in Cutnell, Johnson, Young, Stadler).
That is actually one of the simplest cycles one can draw in the [tex](V,p)[/tex] space.
Actually in the book by Halliday et al there is a problem asking the efficiency of such cycle, which turns out to be less than that of the Carnot cycle.

My original question stems from such textbook problems, where
  • there is no mention of regenerators,
  • [itex]Q_h[/itex] is the sum of the heats absorbed in all the processes of the cycle,
  • [itex]Q_c[/itex] is the sum of the heats released in all the processes of the cycle
It seems to me that in this context no cycle can attain the efficiency of the Carnot cycle. According to Carnot's theorem, this would imply that all cycles except Carnot's have some source of irreversibility (irreversible heat exchanges).

Then, digging a little deeper, one can say that in the Stirling and Ericsson cycles the non isothermal exchanges can be made "internal" and hence their efficiency equals the one in Carnot's cycle.

Also, surely the above cycles have the names you mentioned and appear in "modern" engines.
But I think they also can be thought to describe more primitive machines.
For instance, the isothermal-isochoric-isothermal-isochoric cycle is involved in the Stirling engine. But one could think of a very rudimentary machine which works according to that cycle without regenerators.
That would be impractical, probably in the same way original heat machines were (I'm thinking the very first ones, used perhaps for lifting stuff).
Now, one can assume that is possible to get rid of all mechanical dissipation, thus making the machine "ideal".
Still, its efficiency would be less than Carnot, because of the irreversible heat exchanges.

Sorry, I'm not trying to be difficult just for the sake of it.
I really want to understand what is really meant by "reversible" and what the Carnot theorem is really saying.
 
  • #10
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But in that paragraph there is no mention of the word regenerator. They seem to suggest that, since we cannot ignore the heat exchanges occurring in the isochoric processes, we cannot derive the formula for the efficiency of the Carnot cycle from the general one.

They don't have to mention a regenerator for there to be one... this is a Stirling engine we are discussing, right?? That's what a Stirling engine is... I suppose that one could make the same design with 4 thermal reservoirs instead of a regenerator, but then that really is a different kind of Stirling engine than is usually discussed! Even Stirling's patent for his engine discusses this. (see the Wikipedia article).

That is not what they are saying, though.
Not that I'm holding you responsible for what they say, but it seems to me they are speaking of an ideal reversible case: "Thus reversible heat transfers (and corresponding entropy changes) occur in all four of the processes .... does not apply to an ideal Stirling engine ...".

Yes, it is. Their ideal is not the same as one addressed in most thermodynamics courses or texts (See Moran and Shapiro for instance). In these, the regenerator is treated as 100% efficient as an ideal. Halliday et al are not treating that way. Different authors are free to make such stipulations as long as they are explicit.

Ok, so since there is no friction, irreversibility is in the heat exchanges, right?

This sounds confused. There are irreversibilities, but not because there is no friction. The irreversibilities in the Brayton cycle come from irreversible heat transfers from the reservoirs.

I really want to understand what is really meant by "reversible" and what the Carnot theorem is really saying.

Reversible literally means that the process can be reversed, and when the process is reversed there are no changes in either the system or the surroundings. Entropy is produced in an irreversible process. The last term in the following expression is the entropy production term.

##\Delta S = \int\frac{Q}{T} + \dot \sigma##

If there are no irreversibilities, then the change in entropy is equal to the entropy transferred only. If the process is adiabatic, then there is no change in entropy. An ideal Carnot engine has no entropy production term. Neither does the ideal Stirling cycle where the regenerator is stipulated to be 100% efficient. This is arbitrary, and of course not realistic, but neither is an ideal Carnot engine! What Carnot's theorem is saying, essentially, is that there is no heat engine that can produce more work between the same two reservoirs than an engine which operates without producing entropy. That's it. All real engines have entropy producing processes.
 
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  • #11
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They don't have to mention a regenerator for there to be one... this is a Stirling engine we are discussing, right?? That's what a Stirling engine is..

It seems to me that you are suggesting that the book I linked assumes that everyone knows the details of a Stirling engine, regenerator and all.
That makes no sense.
We use that book in high school, and it explains in detail much simpler stuff. Why should the book give the internal workings of a Stirling engine for granted?

Also, I'm actually discussing a isothermal-isochoric-isothermal-isochoric thermodynamic cycle. I agree that this is how a Stirling engine works.
But are you saying that any isothermal-isochoric-isothermal-isochoric cycle represents a Stirling engine? I am not sure of that. What if there is no regenerator? I can still think of a (arguably very crappy) machine working along that cycle.

Yes, it is. Their ideal is not the same as one addressed in most thermodynamics courses or texts (See Moran and Shapiro for instance). In these, the regenerator is treated as 100% efficient as an ideal. Halliday et al are not treating that way. Different authors are free to make such stipulations as long as they are explicit.
Not exactly sure what you mean.
And, as I said, I'm trying to keep this at the high school level. You are quoting a book for a thermodynamics course which is surely more specialized and advanced than that.

This sounds confused. There are irreversibilities, but not because there is no friction. The irreversibilities in the Brayton cycle come from irreversible heat transfers from the reservoirs.
That is exactly what I meant. Sorry for the confusion. I'm not a native English speaker and it's past 1 am here.

I think there is a misunderstanding here.
I'm trying to keep this at a level accessible to a high school student, and I'm thinking about closed thermodynamic cycles consisting of four processes.

You are making this a bit too technical and specialistic (good luck explaining integrals to 17 year old students) and thinking cycles applied to specific engines.

I know that what it matters to an engineer is building working and efficient engines, but here I'm trying to lay it out the basics, so that the average high school student is able to grasp them.
 
  • #12
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I've just seen this discussion so I'm a bit late. There was a similar discussion on PF and on ResearchGate before.

Anyway I would say the following: Forget regenerators to start with.
A process that runs between exactly two temperatures has to have two isotherms at those temperatures.
If you connect those isotherms by anything but adiabats you involve more than two temperatures!
This is true for both reversible and irreversible processes.
I would say the Carnot cycle is the only possible cycle between exactly two temperatures.

It's true that you can connect the isotherms with isochors for example and then you can use a regenerator to collect the heat you get from the isochoric compression and reuse it in the isochoric expansion.
In my opinion that is just another Carnot engine.
All you need to do is define the regenerator as part of the system and the isochors become adiabats.

You can look at the following paper for more discussion:
A Simple Approach to Heat Engine Efficiency
Carl Salter (2000)
Journal of Chemical Education 77(8), pp. 1027 -1030
 
  • #13
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… then you can use a regenerator to collect the heat you get from the isochoric compression and reuse it in the isochoric expansion.
should read: "… then you can use a regenerator to collect the heat you get from the isochoric pressure decrease and reuse it in the isochoric pressure increase."
 

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