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I have to prove that the Carnot cycle is the most efficient by comparing it with a arbitrary cycle working within the temperature range of $T_L$ to $T_H$ in a ST-diagram. How do I do this?

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How do I prove that the Carnot cycle is the most efficient graphically?

See the diagrams below.

Fig 1 shows an arbitrary cycle on a T-S diagram.

Fig 2 shows a Carnot cycle on a T-S diagram operating between the same high and low temperature extremes and having the same range of entropy (the Carnot graph can be superimposed on top of the arbitrary cycle to show this).

We will assume that both the arbitrary and Carnot cycles are reversible.

For any such arbitrary cycle we can see that area enclosed by the T-S diagram of the Carnot cycle is greater than that for any arbitrary cycle operating between the same maximum and minimum temperatures and range of entropies. This means the net work of the Carnot cycle is greater than the arbitrary cycle operating between the two temperature extremes an range in entropy difference.

In order to show that, in addition to their being more work done, the Carnot cycle is the most efficient we need to take into consideration entropy.

First, we know that for both heat engine cycles shown in the figure we have the following:

$$W_{Net}=Q_{IN}-Q_{OUT}$$

Where $W_{Net}$ is the net work done in the cycle, $Q_{IN}$ is the total heat added during the cycle, and $Q_{OUT}$ is the total heat rejected during the cycle.

The efficiency for both heat engine cycles is given by:

$ζ=$ Net work done ($W_{NET})$ / Gross heat added ($Q_{IN})$

Combining the last two equations we get, for both cycles

$$ζ=1-\frac {Q_{OUT}}{Q_{IN}}$$

So now we need to show that $\frac {Q_{OUT}}{Q_{IN}}$ for the arbitrary cycle is greater than that for the Carnot cycle in order to show that $ ζ_{Carnot}> ζ_{Arbitrary}$

To do this we start with the definition of a differential change in entropy, $dS$, which is defined as, for a reversible transfer of heat:

$$dS=\frac {dQ}{T}$$

Or,

$$dQ=TdS$$

Where $dS$ is a differential change in entropy, $dQ$ is a differential reversible transfer of heat and $T$ is the temperature at which the heat transfer occurs.

We now return to the diagram for the arbitrary cycle, Fig 1.

Heat in, $Q_{IN}$:

Note that with the exception of heat transfer in at the maximum temperature $T_H$, for a given $dS$ each differential transfer of heat occurs at a temperature $T$ less than $T_H$. That means each $dQ_{IN}$ for the arbitrary cycle is less than $dQ_{IN}$ for the Carnot cycle except where it occurs at the maximum temperature of the arbitrary cycle. Adding them all up, $Q_{IN}$ for the arbitrary cycle is less than $Q_{IN}$ for the Carnot cycle.

Heat out, $Q_{OUT}$:

Note that with the exception of heat transfer out at the minimum temperature $T_L$, for a given $dS$ each differential transfer of heat occurs at a temperature $T$ greater than $T_L$. That means each $dQ_{OUT}$ for the arbitrary cycle is greater than $dQ_{OUT}$ for the Carnot cycle except where it occurs at the minimum temperature of the arbitrary cycle. Adding them all up, $Q_{OUT}$ for the arbitrary cycle is greater than $Q_{OUT}$ for the Carnot cycle.

Summing it up, since $Q_{IN}$ for the arbitrary cycle is less than $Q_{IN}$ for the Carnot cycle, and since $Q_{OUT}$ for the arbitrary cycle is greater than $Q_{OUT}$ for the Carnot cycle, we conclude that

$\frac {Q_{OUT}}{Q_{IN}}$ for the arbitrary cycle is greater than $\frac {Q_{OUT}}{Q_{IN}}$ for the Carnot cycle, and therefore $ζ_{Carnot}>ζ_{Arbitrary}$.

Hope this helps

enter image description here

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  • $\begingroup$ Thank you for the clear answer! But why do both cycles have the same range of entropy? Is it possible to have different ranges of entropy? $\endgroup$ – Henri Södergård Dec 6 '18 at 7:17
  • $\begingroup$ @HenriSödergård Good question. They don’t have to have the same range but if one wants to compare the net work done in the Carnot cycle to the arbitrary cycle, one needs to use the same operating parameters, to compare apples to apples. If one is only interested in the efficiency, then it is not necessary. $\endgroup$ – Bob D Dec 6 '18 at 13:27
  • $\begingroup$ For example, if I chose to only look at the middle third of the range of the Carnot cycle, the net work would be less than the arbitrary cycle, but the ratio $\frac {Q_{OUT}}{Q_{IN}}$ for the Carnot cycle would still be less than the arbitrary cycle. Less work done, but still done more efficiently. Had I done it that way I’m sure someone would question why the Carnot work was less. Hope this clarifies things for you. $\endgroup$ – Bob D Dec 6 '18 at 13:27
  • $\begingroup$ Bob D, Yours is essentially the proof drawn up here physics.stackexchange.com/questions/300347/… but do you have any views on my question? $\endgroup$ – hyportnex Feb 18 at 15:34
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The area enclosed by your cycle in your T-S-diagram is the net work you gain/lose in this cycle.

In the T-S-diagram, the Carnot cycle will first move straight to the right (Isothermal Expansion), then straight down (Isentropic Expansion), straight to the left (Isothermal Compression) before finally closing the circle in a straight up line (Isentropic Compression).

The efficiency of a cycle is defined as: $$\mu=\frac {W}{Q_H}=\frac {Q_H-Q_C} {Q_H}=1- \frac {Q_C}{Q_H}$$

$Q_C$ is the area beneath the rectangle, whereas $W$ is the area within.

For maximum efficiency, we see that $W$ must be at a maximum while keeping $Q_C$ at a minimum, which results in a rectangle drawn between $T_c$ and $T_h$.

Without altering $T_c$ or $T_h$, we are not able to draw a cycle enclosing a larger area for $W$ while keeping the area for $Q_C$ the same.

Including a picture from wikipedia for better understanding: enter image description here

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  • $\begingroup$ what is the reason you cannot have an arbitrary cycle in wich the two vertical lines are curved in a way that increases the area? $\endgroup$ – Wolphram jonny Dec 5 '18 at 13:36
  • $\begingroup$ @Wolfram_jonny see the sketch of the proof here physics.stackexchange.com/questions/300347/… $\endgroup$ – hyportnex Dec 5 '18 at 14:02
  • $\begingroup$ My answer lacked a bit of detail, I've edited it to hopefully provide a better explanation as to why it must be a rectangle. $\endgroup$ – Hubblenaut Dec 5 '18 at 14:07
  • $\begingroup$ Your answer might be better with a diagram $\endgroup$ – Aaron Stevens Dec 5 '18 at 14:11
  • $\begingroup$ @Hubblenaut Don't you mean for maximum efficiency $Q_C$ must be at a minimum while $Q_H$ at a maximum? $\endgroup$ – Bob D Dec 5 '18 at 14:17

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