2
$\begingroup$

I have this homework question:

"Show that any reversible engine operating between T1 and T2 is a carnot engine."

I think I have a solution, but it feels very hand-wavy. We know that any process that can be represented as a loop in the PV plane is reversible as the net entropy change will be zero. We must operate between two specifice temperatures, so the loop must comprise of two isotherms a T1 and T2. So the question is what curves join the isotherms. As a heat engine comprises of energy input at constant temperature, there will be no energy change between the isotherms. So the curves connecting the isotherms must be adiabatic curves. So we have a carnot cycle.

Is this sufficient? I don't know why, but I doubt it.

$\endgroup$
1
  • 1
    $\begingroup$ The question is phrased rather loosely. I think the only meaning, in this context, that can be given to the phrase "operating between T1 and T2" is "exchanging heat only at T1 and at T2". In this case the possibilities are only (1) Carnot cycle, or (2) a cycle involving an entropy-changing process as the system moves between T1 and T2 without exchanging heat. But such a process cannot help but be irreversible. Hence Carnot is the only cycle that fits the description. $\endgroup$ Commented Nov 5, 2018 at 21:30

5 Answers 5

5
$\begingroup$

The Carnot cycle is the unique reversible cycle working between two reservoirs of temperature $T_1$ and $T_2$, such that when the engine is in contact with reservoir 1, evolution is isothermal at temperature $T_1$, isothermal at $T_2$ when the engine is in contact with reservoir 2 and adiabatic (isentropic) elsewhere. There are no other possibilities with these "boundary conditions" i.e. interface to the outside World.

However, there are many other possibilities for the interface to the outside World. The engine could make contact with many more than two different temperature reservoirs during a cycle, or heat could be added at constant volume (e.g. from detonation of a chemical reaction inside a rigid vessel, as approximately happens during a Diesel cycle).

$\endgroup$
1
  • $\begingroup$ Should I infer from the last para that there exists such arbitrary reversible non-Carnot engine which operates between more than two temperatures? If so, how would then efficiency be defined? Efficiency is defined for engines working between two temperatures in all the literatures I'm accustomed with. How would we define efficiency for such an engine operating between more than two temps. and would it still be less than Carnot's which operates between two temps? $\endgroup$
    – user36790
    Commented Jun 13, 2016 at 3:38
2
$\begingroup$

You reasoning is right: to make it even more simple, just draw the process in the $TS$ plane.

If you want to work with only two heat sources at two different temperature, you are allowed to draw only:

  • Exactly two isotherms (horizontal lines)
  • An infinite number of isentropics (vertical lines), because they don't exchange heat.

What is the only closed loop you can draw with two horizontal lines a number from $0$ to $\infty$ of vertical lines?

Answer: a rectangle, also known as the Carnot cycle.

enter image description here

$\endgroup$
1
  • $\begingroup$ By invoking $T$ throughout the whole cycle your reasoning is restricted to quasistatic processes. Since the question concerned "show that Carnot is the only one" you need to add some further reasoning to deal with the possibility of moving between $T_H$ and $T_C$ by a path involving non-equilibrium states. $\endgroup$ Commented Nov 5, 2018 at 21:24
1
$\begingroup$

Another way to show this is through the non-existence of perpetual motion, the Feynman way. As he says, we first assume that perpetual motion and hence the creation of energy is not possible. Next, let engine $A$ be a reversible engine working between temperatures $T_1$ and $T_2$ where $T_1>T_2$ and let it absorb heat $Q_1$ frome the reservoir at $T_1$ and give out heat $Q_2$ at reservoir at temperature $T_2$ doing work $W=Q_1-Q_2$ in the process.

Consider another reversible engine $B$ working between the same temperature but having different efficiency. On taking heat $Q_1$ from the hotter reservoir, let us suppose it does work $W'>W$ and therefore gives out heat $Q'$ at the colder reservoir which is less than $Q_2$. After operating engine $B$ for one cycle, we could use $A$ in reverse to siphon out $Q_2$ from colder reservoir, submit heat $Q_1$ at the hotter reservoir with a work input of $W$. But since $W'>W$, therefore the net result of oprating $B$ followed by reverse $A$ will be to have drawn heat $Q_2-Q'$ from the colder reservoir and completely converted it into work $W'-W$, without any other change or entropy since the engines are reversible. But this contradicts the well know statement of second law of thermodynamics:-

Clausius staement-"Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time."

This argument shows that working between any two temperatures, irrespective of the nature of the engine, a reversible engine shows the maximum efficiency and all reversible engines must show the same efficiency (Carnot principle) to not violate second law of thermodynamics. This idea, originally Carnot's proves that all reversible engines have the same efficiency as a Carnot engine. Your argument uses certain assumption about the working of a reversible engine which might not be universally true.

$\endgroup$
2
  • $\begingroup$ Thanks for the detailed response. What exactly am I assuming that "might not be universally true"? $\endgroup$ Commented Sep 20, 2013 at 3:37
  • 2
    $\begingroup$ @KieranCooney You are assuming that there are only four quasistaic processes in a reversible engine, but that is not true. You can design a reversible engine which performs several transitions, which may not be adiabatic and isothermal only. You may use a complex setup with several compartments and mechanical constraints which would ultimately complicate the PV loop of the system by including several processes unlike only 4 used in Carnot engine. But if it is reversible and operates between 2 reservoirs at different temperatures, it's efficiency must be = to Carnot efficiency. $\endgroup$ Commented Sep 20, 2013 at 7:01
1
$\begingroup$

Clausius' statement about heat not being able to flow spontaneously from a cold body to a warm body is sufficient to prove that no engine can have an efficiency greater than that of a perfectly reversible engine. But it's not enough to prove that the Carnot engine is the only reversible engine. For example, there could be a perfectly reversible engine where the gas at initial volume $V_1$ and temperature $T_1$ draws heat from a hot reservoir, expands doing work against constant pressure till the gas reaches a higher temp $T_2$, then makes contact with a cold reservoir to contract isochorically to a lower temperature $T_3$ and lower pressure $P_2$, then makes contact with an even colder reservoir to bring down the temperature isobarically until the volume contracts to the original volume $V_1$, and then makes contact with a hot reservoir to heat up isochorically back to temperature $T_1$ (the original state). This seems to be as reversible an engine as the Carnot engine. The only difference is that here more than two heat reservoirs are involved. In fact, a Carnot engine is the only possible engine which can operate between just two temperatures; all other engines require one or more heat source or heat sink at some intermediate temperatures. This has a bearing on the issue and make the two not comparable. It can be graphically demonstrated that, between the two outermost temps (i.e. between the maximum and minimum temperatures), the Carnot engine is the most efficient - see this excellent video.

$\endgroup$
0
$\begingroup$

I suspect the expression "operating between T1 and T2" actually means "operating between heat reservoirs with temperatures T1 and T2". But even then I am not sure "any reversible engine operating between heat reservoirs with temperatures T1 and T2 is a Carnot engine." As far as I know, a Carnot engine is an engine "that operates on the reversible Carnot cycle" (http://en.wikipedia.org/wiki/Carnot_heat_engine ), which cycle consists of two isothermal processes and two adiabatic processes. However, it seems that more complex reversible processes can exist that use the same isothermal processes (but maybe different parts of them) and more than two adiabatic processes (e.g., T1S1-T1S2-T2S2-T2S3-T1S3-T1S4-T2S4-T2S1-T1S1). so maybe the condition of the problem lacks some additional requirement.

$\endgroup$
4
  • 1
    $\begingroup$ "it seems that more complex reversible processes can exist that use the same isothermal processes (but maybe different parts of them) and more than two adiabatic processes " Would you draw for me a sketch of this in a $TS$ diagram? Because I believe that the process you are describing is absolutely impossible. $\endgroup$
    – valerio
    Commented Jun 5, 2017 at 8:37
  • $\begingroup$ @valerio92: You can draw the sketch yourself using the points specified in my example (T1S1-T1S2-T2S2-T2S3-T1S3-T1S4-T2S4-T2S1-T1S1). I don't see why this process is "absolutely impossible"- care to explain? $\endgroup$
    – akhmeteli
    Commented Jun 5, 2017 at 14:18
  • 2
    $\begingroup$ But the process you describe is basically two Carnot cycles connected by an isotherm that gives a trivial contribution. So maybe I should rephrase: it is absolutely impossible to think of a reversible cyclic process working between only two temperatures which is different from a sequence of Carnot cycles connected by isotherms. $\endgroup$
    – valerio
    Commented Jun 5, 2017 at 14:30
  • $\begingroup$ @valerio92: Trivial or not trivial, this is not a Carnot cycle, so I stand by my answer. And maybe you should rephrase your answer. $\endgroup$
    – akhmeteli
    Commented Jun 6, 2017 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.