# How to tell if a thermodynamic cycle is reversible without calculating entropy change?

Consider the Carnot cycle, consisting of two reversible, isothermal processes and two isentropic processes. It is reversible, pretty much by definition.

Now consider the Lenoir cycle, consisting of an isochoric compression (heat addition), followed by an isentropic expansion, followed by an isobaric compression (heat loss). I calculated the entropy created by this cycle and found it to be strictly positive.

However it's not clear intuitively why this cycle should be irreversible. Is heat change at constant volume or at constant pressure necessarily irreversible?

• What do you mean by "I calculated the entropy of this cycle and found it to be strictly positive"? You mean the entropy change of the system? But that is strictly $0$ by definition of cycle, since entropy is a state function. Or you mean the entropy of the surroundings? But then it is not clear to me how you calculated that. – valerio Jun 4 '17 at 19:28
• I mean the entropy created which is the change in entropy of the two heat sources plus the change in entropy of the system which is zero because as you said it's a cycle. I calculated the change in entropy of the sources by the standard formula Q/T where T is the constant temperature of the source and Q is the heat transfered. – Joshua Benabou Jun 4 '17 at 22:28
• Show us the details of what you did to calculate the entropy change of the sources. Did you take into account the temperature changes of the system during the first and third steps and the need for the surroundings to match these changes? – Chet Miller Jun 5 '17 at 0:57
• Ok I will post my calculations. I did take into account that the temperature of the system changes however I don't see why this would imply that the temperature of the heat sources would need to change. As far as I understood a heat source is constant temperature. If we wish to transfer heat to the system we place the heat source in thermal contact with the system until the temperature of the system becomes that of the source. No? – Joshua Benabou Jun 5 '17 at 1:05
• If you want to do it reversibly, you have to use a continuous sequence of reservoirs, running from an initial temperature to a final temperature. Otherwise, entropy will be generated within the gas, and transferred to the constant temperature reservoir, such that, in the end, the combined changes in the entropies of the reservoirs will be positive. This is what you showed. – Chet Miller Jun 5 '17 at 11:59

If the isochoric and isobaric transformation are performed reversibly, i.e. quasistatically and without heat dissipation caused by friction or other effects, then your cycle will be reversible.

This is true for every thermodynamic cycle you can draw in the $$PV$$ plane: if every step is performed reversibly, then the cycle is reversible.

The peculiarity of the Carnot cycle is that it is the only reversible engine that operates between two heat sources only. You can easily see how many different heat sources you are using if you draw the cycle into the $$TS$$ diagram (picture from Wikipedia): In this case, it is easy to verify that the change in entropy of the surroundings is

$$\Delta S_{surr} = -\frac{Q_H}{T_H}+\frac{Q_C}{T_C} =0$$

So that the engine is indeed reversible. But now let's take your Lenoir cycle in the $$TS$$ diagram (picture from Wikipedia): As you can see, during $$1 \rightarrow 2$$ and $$3 \rightarrow 1$$ you are cutting infinitely many isotherms. The formula you have to use is in this case

$$\Delta S_{surr} = -\int_1^2 \frac{\delta Q}{T} + \int_3^1 \frac{\delta Q}{T}$$

But this time you cannot take out $$T$$ from the integral like you would do with a Carnot cycle, because it is not a constant.

What you can do is to assume that step $$1 \rightarrow 2$$ and $$3 \rightarrow 1$$ are performed reversibly: in this case, $$\Delta S_{surr}=0$$ by definition.

• my understanding is that a heat exchange between a constant temperature heat resevoir and a system is reversible if and only if it occurs isothermally (as any heat exchange across a temperature difference which is not infinitesimal creates entropy). My first question is can the lenoir cycle be realized using two heat sources? I don't see why not? The conclusion is that, if it is realized using two heat sources, the cycle will thus transfer heat across isothermally, and thus the cycle will be irreversible. – Joshua Benabou Jun 5 '17 at 11:06
• This makes sense, because when you calculate the efficiency of the Lenoir cycle operating between two heat reservoirs, you find it to be strictly less than that of the carnot cycle operating between the same two heat reservoirs. Furthermore, this whole thing about ensuring reversibility by using an infinity of heat resevoirs makes sense theoretically, but I'm guessing it's impossible to realize in practice. – Joshua Benabou Jun 5 '17 at 11:11
• The Lenoir cycle does not operate between two heat reservoirs: it operates between an infinity of them, because you cut infinitely many isotherms when performing it. A cycle operating between two and only two heat sources must consist of isotherms and isentropics only, and the only reversible cycle respecting this condition is the Carnot cycle. – valerio Jun 5 '17 at 11:32
• Think about it in the $TS$ diagram: what is the only cycle you can draw consisting of two horizontal lines (the isotherms representing the two heat sources) and as many vertical lines as you want? The answer is a rectangle, i.e. the Carnot engine. I repeat: this is the only reversible cycle that operates between exactly two heat sources. – valerio Jun 5 '17 at 11:34
• I still don't understand why the Lenoir cycle can't operate between two heat reservoirs? Perhaps you mean to stay the reversible Lenoir cycle requires an infinity of heat reservoirs, in which case I agree. Or are you claiming that it is impossible to construct a heat engine using two heat reservoirs whose pv-graph is that of the lenoir cycle? – Joshua Benabou Jun 5 '17 at 11:44