0
$\begingroup$

enter image description here

Does the following question conflict that fact that efficiency of a reversible engine is independent of processes involved?

$\endgroup$
  • $\begingroup$ Doesn't that "fact" only apply to a reversible Carnot engine? $\endgroup$ – Chet Miller Feb 6 '18 at 18:30
2
$\begingroup$

The Carnot theorem states the maximum efficiency of an heat engine is that of the Carnot heat engine and this depends only on the temperatures of the two heat reservoirs. In addition to that, there is a corollary (found it also here) that states the following: "all reversible engines that operate between the same two heat reservoirs have the same efficiency". This means that a reversible engine can have the same efficiency of a Carnot engine but provided it to exchange heat only with two heat reservoir. As an example, the Stirling cycle made up by two same-volume and two isothermal transformations, achieves the same efficiency of the Carnot cycle (read below the edit).

Then the answer: the two reversible cycle you shown in the figure do not fit the assumptions since there must be more than two heat reservoirs in order to allow the system to follow the cycle (in the transformations $A-B$ and $B-C$ respectively for the left and right engine). So they have efficiency that is less than that of a Carnot engine.

Edit: As noted in the comments, in a Stirling cycle there is exchange of heat also in the isochoric transformations, so that actually the engine works with more than two heat reservoir. However, the amounts of heat absorbed during the jump from the lower to the upper temperature and that released during the cooling are equal and sum to zero, so that the effective reservoirs are only those at the highest and lower temperature. It seems that the important point is that in a Stirling cycle you actually need only two heat reservoirs in order to follow the transformations involved (in particular, you can perform the isochoric transformations without using other reservoirs). In this sense I think that must be read the statement of the corollary above.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How is Stirling cycle involving only two reservoirs and being reversible at the same time, when the two isochoric processes involve heat transfer at a finite temperature difference? $\endgroup$ – Akhil Feb 7 '18 at 14:01
  • $\begingroup$ You're right. In the Stirling cycle there are heat transfers during the isochoric transformations and I forgot it, sorry. However, the example of the Stirling cycle still works because the heat lost during the cooling isochore is equal to that absorbed on the other isochore. I will edit answer. $\endgroup$ – ndrearu Feb 7 '18 at 15:02
0
$\begingroup$

Only the Carnot cycle has the Carnot efficiency. The Carnot cycle is a rectangle in a TS diagram. These two processes are not Carnot cycles.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Doesn't every reversible engine have the same efficiency? $\endgroup$ – Akhil Feb 6 '18 at 18:44
  • $\begingroup$ A Carnot engine works between isotherms connected by adiabatics. These two cycles have only one isotherm. The other path involves smaller, intermediate temperatures. $\endgroup$ – Pieter Feb 6 '18 at 18:59
  • $\begingroup$ So these cycles must be irreversible because heat transfer occurs at a finite temperature difference. Is that right? $\endgroup$ – Akhil Feb 6 '18 at 19:12
  • $\begingroup$ No, these cycles are not irreversible. One could run them either way. $\endgroup$ – Pieter Feb 6 '18 at 19:33
  • $\begingroup$ If heat transfer takes place at a finite temperature difference then it must not be a quasistatic process, right? $\endgroup$ – Akhil Feb 6 '18 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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