Can the Efficiency of an Arbitrary Reversible Cycle be Equal to the Efficiency of the Enclosing Carnot Cycle? 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.
 A: 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.
