0
$\begingroup$

I was reading the following article: https://www.researchgate.net/publication/305025128_Can_the_Efficiency_of_an_Arbitrary_Reversible_Cycle_be_Equal_to_the_Efficiency_of_the_Enclosing_Carnot_Cycle_Part_B

Right at the end in the conclusion, the following statement is made:

"Since we already showed above that the efficiency of cycle in Fig. 4 is equal to the efficiency of the enclosing Carnot cycle, it follows that if the efficiency of a reversible cycle is less than the efficiency of the enclosing Carnot cycle, then it is also possible for the efficiency of a reversible cycle to be more than the efficiency of the enclosing Carnot cycle."

How does that make sense? Why must it be possible for the reverse of a statement be true if the statement is true?

I was trying to find an answer to the following question:

Why the most efficient heat engine (carnot cycle), consists of 4 stages, 2 adiabatic ones and 2 isothermal ones. I can't seem to find a proof on why this is the most efficient thermodynamic cycle for a engine, only that if we assume a greater efficiency, we get trouble.

$\endgroup$
3
  • 2
    $\begingroup$ It was my understanding that all reversible cycle operating between two fixed temperature reservoirs have the same efficiency. That is, all the heat is added at a single temperature and all heat rejected at a single temperature. This paper looks at heat additions/rejections occurring isothermally but at different temperatures. Hopefully Chet Miller will see this and give a critique on the paper you cite. $\endgroup$
    – Bob D
    Dec 18 '19 at 20:46
  • $\begingroup$ the enclosing Carnot cycle has always more efficient than any other reversible cycle, see reference and "sort of proof" in physics.stackexchange.com/questions/300347/…. $\endgroup$
    – hyportnex
    Dec 18 '19 at 22:28
  • $\begingroup$ @BobD I agree with hyportnex. $\endgroup$ Dec 19 '19 at 0:04
0
$\begingroup$

A very optimist paper actually :).

What they've said is: in Fig. 4 and Fig. 6 Heat absorbed is the same as both the cycle's top-edges are the same but work by 6 is lesser. And that heat in Fig.7 is the same too but work is more than 4.

And then the Efficiency of 4 is the same as "Enclosing" Carnot Cycle, I'm not sure if it's talking about fig2 or the Rectangle enclosing Fig.4 so ill take both since it doesn't matter.

let S1 be entropy at B,

$$\eta_{fig2}=1-\frac {T_LS_1}{T_HS_1}=1-\frac {T_H}{T_L}$$

$$\eta_{Enclosing \ fig4}=1-\frac {T_L(S_1+S_x)}{T_H(S_1+S_x)}=\eta_{fig2}$$

let Sx be entropy at D' $$\eta_{fig4} = 1-\frac{T_xS_x+T_LS_1}{T_HS_1+T_xS_x}$$

They Claimed: $$\eta_{fig4} = \eta_{fig2}$$

$$T_H(T_HS_1 + T_xS_x)=T_L(T_xS_x+T_LS_1)$$

$S_1(T_L+T_H)=-S_xT_x$ or $T_H = T_L$

Both the results are Super Absurd. This claim is flawed. They must've backed this claim with some rigour (which would have made them see why it's absurd).

Carnot Cycle's the most efficient because any other hypothetically more efficient cycle will result in the violation of the Second Law. Intuitively Carnot Cycle is a rectangle between two given temperatures in the T-S diagram and the rectangle covers the most area in a given Q.

See this to get an intuitive idea of why it's the most efficient. Rigourous proof are all over the internet, you can easily find one. KhanAcademy

[Courtesy - GrizzleBear10 ] This Paper was published into vixra.org and is not necessarility a serious paper. This probably never appeared in any journal or is taken to be a "Research Paper" as we use the phrase in an everyday sense. So don't worry. It's false.

$\endgroup$
1
  • $\begingroup$ This paper was published by a government-run Research Institute. What this makes me think is - Either my answer is horrendously wrong or the paper has a huge flaw. Since this is a paper published by such a research body, ill confirm once again with my Physics Sir and then 1. Delete answer if its wrong or 2. See what can be done to report this. I think disproving something like Carnot Engine might obviously be flawed but I'm not really sure. $\endgroup$ Dec 19 '19 at 9:11

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.