# Reversible cycle approximated by Carnot cycles

1. Aug 14, 2015

### DavideGenoa

My textbook gives the definition of a reversible transformation as a transformation that can be inverted by effectuating only infinitesimal changes in the surroundings. I admit that I have no idea of what infinitesimal means in a rigourous mathematical language, therefore the definition is quite confusing to me.

Then the book prove that the entropy of a Carnot cycle is zero and, since a reversible cycle can be approximated by sum of many Carnot cycles, the entropy of a reversible cycle is zero too.

How can we see that a reversible cycle can be arbitrarily approximated by Carnot cycles?

Thank you so much for any help in understanding!!!

2. Aug 14, 2015

### techmologist

It sounds like you are talking about the proof of the Clausius inequality. Namely, that if during a cycle C a system is successively in contact with heat sources at temperatures T1, T2,...., Tn, and if the system absorbs heat Qj (which can be positive or negative) from source Tj, then for the cycle we have

$$\sum_{j=1}^n \frac {Q_j}{T_j} \leq 0$$

This can be shown by combining the cycle C with n reversible Carnot cycles C1, C2,...., Cn. The Carnot cycle Cj operates between some reference temperature T0 and Tj and is sized so that it delivers heat Qj to source Tj and absorbs heat Q0,j from source T0. The net result of cycle C combined with the Carnot cycles is to absorb heat

$$Q_0 = \sum_{j=1}^n Q_{0,j}$$

from source T0 and convert it into work. Kelvin's postulate (of the second law) says that this quantity can't be positive--because then you'd have a process that did nothing other than convert heat from one temperature into work.

For each Carnot cycle, we have that

$$\frac{Q_{0,j}}{Q_j} = \frac {T_0}{T_j}$$

Thus
$$Q_0 = \sum_{j=1}^n Q_{0,j} = T_0 \sum_{j=1}^n \frac {Q_j}{T_j} \leq 0$$

If the cycle C is reversible, then you can reverse the whole combined process and prove that Q_0 >= 0 also. Combining it with the above inequality, you get that

$$\sum_{j=1}^n \frac {Q_j}{T_j} = 0$$

Last edited: Aug 14, 2015
3. Aug 15, 2015

### DavideGenoa

Thank you so much, techmologist! Well, I know that for a Carnot cycle operating between temperatures $T_C$ (cold) and $T_H$ (hot) we have $\frac{Q_H}{T_H}+\frac{Q_C}{T_C}$, and the same applies for the combination of $n\ge 1$ Carnot cycles, but I do not understand why any reversible cycle can be approximated by Carnot cycles $C_{1,n},...,C_{n,n}$ (say that the $n$-th set of Carnot cycles used to approximate is composed by $n$ cycles) so that

$\oint\frac{\delta Q}{T}=\lim_{n\to\infty} \sum_{k=1}^n \frac{Q_{C,k,n}}{T_{C,k,n}}+\frac{Q_{H,k,n}}{T_{C,k,n}}$

I understand that $\frac{Q_{C,k,n}}{T_{C,k,n}}+\frac{Q_{H,k,n}}{T_{C,k,n}}=0$, but I do not understand why $\oint\frac{\delta Q}{T}=\lim_{n\to\infty} \sum_{k=1}^n \frac{Q_{C,k,n}}{T_{C,k,n}}+\frac{Q_{H,k,n}}{T_{C,k,n}}$, why the cycle is approximated by $C_{1,n},...,C_{n,n}$...

I heartily thank you, techmologist, and anybody wishing to add any answer!!!​

4. Aug 15, 2015

### techmologist

You are welcome!

At least in the proof I gave, I don't think it the cycle C was really approximated by the Carnot cycles C1, C2,...., Cn. It is simply combined with them to make one complex cycle C + C1 + C2 + .... + Cn so that we can prove Clausius' inequality

$$\sum_{j=1}^n \frac {Q_j}{T_j} \leq 0$$

When you pass to the limit of a continuum of different heat sources, from each of which the system absorbs dQ(T) of heat, this inequality becomes

$$\int_C \frac{dQ}{T} \leq 0$$

When the cycle C is reversible, the inequality becomes an equality:

$$\int_C \frac{dQ}{T} = 0$$

Since the cycle C here is reversible, we have that dQ/T = dS. The above equation says that the total change in entropy of a reversible cycle is zero, which is what you wanted to prove. Is it possible that your book was speaking informally when it said you can conclude this by approximating the cycle C by a sum of Carnot cycles? Does it actually offer a proof?

5. Aug 16, 2015

### DavideGenoa

I fully understand your proof. My problem is that I want to prove that $\int\frac{\delta Q}{T}=0$ for any reversible cycle, not only Carnot cycles.
My book says that a reversible cycle $C$ can be approximated by the sum of Carnot cycles, but it does not say what that means rigourously and does not give a proof, which is the reason why I am wondering how we can prove that...
Thank you so much again!

6. Aug 16, 2015

### Chandra Prayaga

The best way to understand this is with the help of a picture, given in many textbooks, such as the classic book on heat & thermodynamics by Sears and Zemansky. I also found the same picture and explanation in one of the MIT lectures online:
http://curricula2.mit.edu/pivot/book/ph2103.html?acode=0x0200

7. Aug 16, 2015

### techmologist

Ahh. That is a different picture than what I had in mind. Different reasoning. I bet that is exactly what DavideGenoa is talking about. The thing is, I don't quite follow it. It is obvious to me that the area of the reversible cycle can be arbitrarily approximated by the Carnot cycles in the way shown, so that the total work is equivalent. But it isn't clear to me that the heats absorbed are the same. No matter how thin the Carnot cycles are made, the adiabatic parts never actually follow the path. What you end up with is a very jagged path approximating a smooth path. So you end up needing another argument that any small reversible path is equivalent to a small isothermal path followed by a small adiabatic path. Which it is, but I thought that was pretty much what we are trying to prove.

8. Aug 16, 2015

### techmologist

Well, that is actually what I proved, although I admit it is hard to follow that proof without a picture to look at. Maybe I can dig one up. It is a tricky proof, one that I had to go over several times before I got the point of it. I know that Feynman Lectures Vol 1 has a picture of what I'm talking about, if you can get hold of that.

In fact what I showed was even more general. That for any cycle at all, reversible or irreversible, we have that

$\int_C \frac{dQ}{T} \leq 0$

In the special case of reversible cycles (which don't have to be Carnot cycles), you can also conclude that

$\int_C \frac{dQ}{T} \geq 0$

simply by reversing the cycle, which has the effect of replacing dQ with -dQ. That is, if in the forward cycle the system absorbs heat dQ at temperature T, then in the reverse cycle the system rejects heat dQ (or absorbs -dQ) at temperature T. The only way that both inequalities can hold is if

$\int_C \frac{dQ}{T} = 0$

There is a subtle point in the interpretation of the Clausius inequality. The temperature T that appears in the integral is the temperature of the heat source that the system is in contact with at the moment. It is not necessarily the temperature of the system. In fact, for an irreversible process, the system generally does not have a well-defined temperature at most points along the path. If the process is reversible, the system is always arbitrarily close to an equilibrium state and has a well defined temperature. By definition of equilibrium, that temperature must equal T, the temperature of the heat source it's in contact with.

9. Aug 16, 2015

### techmologist

The main idea of the proof is a direct application of Kelvin's statement of the second law, that no cycle has as its only effect the conversion of heat from a single temperature into work. In order to convert heat into work, you have to have access to at least one other (lower) temperature where you can dump out some waste heat. What the proof does is show that if

$$\int_C \frac{dQ}{T} > 0$$

for a cycle, then it is possible to combine that cycle with a bunch of Carnot cycles in such a way that you violate Kelvin's postulate. That is assumed not to be possible, so Clausius' theorem results:

$$\int_C \frac{dQ}{T} \leq 0$$

EDIT: I got the proof from Enrico Fermi's little book on thermodynamics, if that helps. I don't think it includes a picture with the proof, though.

Last edited: Aug 16, 2015
10. Aug 16, 2015

### Chandra Prayaga

That is where the idea of "infinitesimal" come in. As the "width" of each small Carnot cycle goes to zero, the path is no longer jagged, and becomes identical to the jagged path.

11. Aug 16, 2015

### techmologist

I'll have to think about it for a while. Since the tiny corners of the carnot cycles are not generally chords of the curve of reversible path, it is not clear to me in what sense the limit of the Carnot cycles converges to the reversible path. I recognize that the thermodynamic effect is the same, but I still think that that is what we are attempting to demonstrate here. I'm not sure that it can be assumed.

12. Aug 16, 2015

### techmologist

Here's one I found with a google image search. Tried to attach it to my last post but it just showed a red X.

For every amount of heat that the system S absorbs from a particular heat source, there is a Carnot cycle sized so that it will deliver the same amount of heat to that source. In effect, the Carnot cycles are undoing what the system does to the external heat sources, with the help of an additional heat source T0. After all is said and done, there may be some net exchange of heat with the source T0, but no net exchange of heat with the other sources. For a reversible cycle, the total exchange of heat at temperature T0 is also zero. So in that sense the Carnot cycles exactly cancel everything the system S has done, and can be said to approximate the cycle it performs.

13. Aug 17, 2015

### DavideGenoa

As to your proof of $\oint \frac{\delta Q}{T}\le 0$, techmologist, I completely misunderstood it because I thought I could interpretate it as a proof of the validity of the above said approximation, but I think it much clearer now to me, thanks to the image.
My book also says that $\oint \frac{\delta Q}{T}< 0$ when the cycle is irreversible. Have you got a proof for that too?

Exactly, that is plot very similar to my book's. My problem is that I cannot graps the mathematical reason why we can consider a reversible cycle as the "limit of a series of sets of Carnot cycles" approximating it (what I mean by this wording is what I wrote in my second post) or, as techmologist puts it:
I heartily you both, as well as anyone else wishing to join the thread!

14. Aug 18, 2015

### techmologist

Right. The argument in that MIT lecture is very different from the one I presented. That's why I was reluctant to say that the cycle is being approximated by a combination of Carnot cycles in the argument I gave. Rather, the cycle C is being supplemented by several Carnot cycles, C1, ... , Cn. But in the MIT lecture, they are definitely approximating the cycle by thin Carnot cycles that operate between temperatures T and T+dT. It is more of a geometrical argument, combined with some thermodynamic reasoning. I get the basic idea of it. And I think that is the usefulness of that particular argument. It shows you "at a glance", without a long chain of reasoning, how Carnot cycles can be used to build up more general cycles.

Carnot cycles form the logical basis of thermodynamics. They are used to define the thermodynamic temperature. And when the Carnot cycle has an ideal gas as its working fluid, it can be used to show that the thermodynamic temperature is the same as the ideal gas temperature that appears in the equation of state pV = nRT.

Next they are used as in the above argument to show that

$$\int_C \frac{dQ}{T} = 0$$

for reversible cycles. Which shows that dQ/T (reversible) is the differential of some function of the state only (which does not depend on the path taken). That function is the entropy S. This is the definition of entropy.

Well, provided that you agree that we have already shown that

$$\int_C \frac{dQ}{T} \leq 0$$

for any cycle, and that

$$\int_C \frac{dQ}{T} = 0$$

for reversible cycles in particular, then for irreversible cycles the integral must be strictly negative. I see what you are getting at though. What we showed is that the integral is zero for reversible paths. You are asking, are all paths for which the integral is zero reversible? I am pretty sure the answer is yes. I think if you had an irreversible cycle for which $\int_C \frac{dQ}{T} = 0,$ then it should be possible to take a small part of that cycle and close it with a reversible path (making a different cycle) so that for the resulting cycle C', $\int_{C'} \frac{dQ}{T} > 0$. And that can't be done.

Maybe someone else could weigh in on that, as I am sort of guessing.

Last edited: Aug 18, 2015
15. Aug 18, 2015

### techmologist

Crap. I messed that up. That was supposed to say...

Obviously the integral is typically not zero for non-cyclic paths, even when they are reversible. In that case it is just the change in entropy.

While I'm at it, my claim that Carnot cycles are the logical basis of thermodynamics doesn't sound right either. Obviously the logical basis is the first law and the various statements of the second law. Carnot cycles are the primary theoretical tool that allows you to draw interesting conclusions from those laws.

16. Aug 19, 2015

### DavideGenoa

Thank you very, very much!