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I understand Carnot's theorem, specifically the result that "the efficiency of all reversible engines operating between the same two temperatures is the same." The proof is by contradiction and you can find it all over the web.

However, the theoretical Otto or Brayton cycles are reversible, and have efficiency less than the Carnot efficiency. I've gone through the efficiency calculation myself for Otto, and got the same result that is all over the web. Indeed, one can construct all sorts of theoretical reversible cycles which are less efficient than Carnot.

I do not understand the existence of reversible cycles with efficiency less than Carnot. If they exist, even in theory, we could run the output from a Carnot engine into one of them (say, a Brayton or Otto heat pump), and move heat continually from a cold reservoir into a hot one with no external input.

Indeed, this is exactly the contradiction at the center of Carnot's theorem leading to its main result, that all reversible processes have the same efficiency. Even worse, the more inefficient the non-Carnot process (which we turned into a heat pump), the more we violate the second law of thermodynamics.

To put numbers on it, say $T_\text{hot} = 400 \ \text{K}$ and $T_\text{cold} = 300 \ \text{K}$, so the Carnot efficiency is 0.25. Our Carnot engine is turning 1000 W of input heat into 250 W of work and discharging 750 W into the cold reservoir. If the 250 Watts of work is running a heat pump, a reversed cycle which would have had engine efficiency 0.1, then the heat pump is drawing 2250 W of heat out of the cold reservoir and discharging 2500 W into the hot one. The result is 1500 Watts being transferred from the cold reservoir to the hot one with no external input. Obviously impossible.

The issue seems to be that reversing the inefficient engine creates a really efficient heat pump. So I know that's where the error lies, but not exactly what the error is. Can anyone clarify?

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  • $\begingroup$ Please tell more about the reversed cycle. $\endgroup$ Feb 9, 2021 at 21:23
  • $\begingroup$ Hello Dennis. I've updated my answer. Hope it helps. $\endgroup$
    – Bob D
    Feb 9, 2021 at 23:11

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The reversible Otto and Brayton cycles don’t operate between two fixed temperature reservoirs as does the Carnot cycle. That makes them less efficient.

Regarding your statement:

"The efficiency of all reversible engines operating between the same two temperatures is the same."

The key point is the reversible cycle has to operate between two fixed temperature reservoirs. The Otto and Brayton cycles do not operate between two fixed temperature reservoirs. They operate between a series of reservoirs between the maximum and minimum temperatures.

The Carnot efficiency is

$$\eta =1-\frac{T_L}{T_H}$$

Where $T_H$ is a single high temperature reservoir where all heat is taken from and $T_L$ is a single low temperature reservoir where all heat is rejected.

For the Otto and Brayton cycles $T_H$ is replaced by $T_{Hm}$ where $T_{Hm}$ is the mean temperature at which heat is taken in. That value is less than $T_H$.

Then $T_L$ is replaced by $T_{Lm}$ where $T_{Lm}$ is the mean temperature at which heat is rejected. That value is always greater than $T_L$ of the Carnot cycle.

So if you plug in the mean high and low temperatures of the Otto and Brayton cycles into the Carnot efficiency equation you will find that the efficiency is always less than the Carnot efficiency.

I have provided TS diagrams of the idealized reversible Otto and Brayton cycles gas cycles with that of the Carnot Cycle, for both heat engines and reverse heat engines, in case they may be of help to you.

The issue seems to be that reversing the inefficient engine creates a really efficient heat pump. So I know that's where the error lies, but not exactly what the error is. Can anyone clarify?

In fact, a less efficient heat engine does make a more "efficient" heat pump, and vice versa. This may see counter intuitive until you look at it closely. Consider the following:

First, the “efficiency” of the heat pump is actually called its Coefficient of Performance (COP) and equals the heat delivered to the high temperature environment divided by the work required. In your example, it would be 2500/250 = 10. If your reverse your heat pump cycle to make it a reversible heat engine cycle the efficiency of the heat engine would be 0.1. This means the relationship between the COP of the heat pump (HP) and the efficiency of the heat engine (HE) is

$$COP_{HP}=\frac{1}{\eta _{HE}}$$

Which gives the interesting, but important, result that the lower the efficiency of the heat engine the greater the COP of the heat engine when operating in reverse as a heat pump, and vice versa. As I said, this may seem counter intuitive until you look at it closely.

You know the efficiency of a reversible heat engine cycle is greater the larger the difference in temperature between the high and low temperature reservoirs. The greater that difference in temperature, the more output work you get for the same heat input.

On the other hand, for the heat pump the greater the difference in temperature the more work is required to move the same amount of heat from the low temperature environment to the high temperature environment. It’s kind of the equivalent of pulling a weight up a hill. The higher the hill, the more work required to move the same weight up the hill. Or, to put it another way, moving heat from a low temperature to a high temperature is not a natural thing, just like things don’t naturally roll up hill. It takes work. And the greater the temperature difference the harder it is (the more work required) to move the same amount of heat.

This is one of the reasons that heat pumps are used primarily in temperate climates where the difference between the indoor and outdoor temperature is not too great.

Hope this helps.

enter image description here

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  • $\begingroup$ Hmm... I understand this, although the intermediate temperatures along the non-isothermal processes are completely determined (in theory) by the thermodynamic equations of state (and may be expressed only in terms of T_hot and T_cold). I also understand that the existence of non-isothermal, non-adiabatic processes in the cycle would make it hard to construct in practice with only two heat reservoirs. That said, strictly in theory, we could still consider the reverse of a less-than-Carnot efficient engine, that would create a super heat pump and violate the theorem. Or where does that go wrong? $\endgroup$
    – Dennis
    Feb 9, 2021 at 22:28
  • $\begingroup$ @Dennis Not sure I follow you. The point is the two cycles in question take in heat over a range of temperatures between a max and min and reject heat over a range of temperatures between a max and min, whereas the Carnot cycle takes in all its heat at a single high temperature and rejects all its heat at a single low temperature. That's what makes it the most efficient reversible cycle. $\endgroup$
    – Bob D
    Feb 9, 2021 at 22:32
  • $\begingroup$ Bob, my problem is not that there exist cycles which are less efficient than Carnot. It is that, if I manage to reverse one of these cycles, it will create a heat pump that is more efficient than a Carnot heat pump. This super heat pump, in combination with a Carnot engine, would violate the Carnot theorem and 2nd law of thermo. So it must be impossible. Yet, online it is repeatedly stated that idealized Otto or Brayton cycles (among many others) are reversible. What gives? $\endgroup$
    – Dennis
    Feb 9, 2021 at 22:40
  • $\begingroup$ @Dennis Do you realize that efficiency of a heat pumps is its Coefficient of Performance (COP).? $\endgroup$
    – Bob D
    Feb 10, 2021 at 13:45
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    $\begingroup$ @Dennis "Well, Bob, since we're still discussing this... an Otto engine accomplishes its isochoric heating (2 to 3 in your diagram) by combusting gas inside the chamber. This is not reversible". You have got to be kidding. I have thought all along we were talking about ideal reversible cycles, not real combustion engines. Any cycle can be "reversed" but that doesn't make it a reversible cycle. A reversible cycle generates no entropy. I have had enough. You are free to withdraw your ACCEPT. Goodbye. $\endgroup$
    – Bob D
    Feb 14, 2021 at 18:50
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Because you don't consider entropy. In reality, things are not the same as in the theoretical formulas. Principle I puts an equivalence between different sizes - state and process. Principle II comes with a supplement that takes into account reality. The Carnot cycle is an ideal one. Otto is real.

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  • $\begingroup$ "In reality, things are not the same as theoretical formulas". I'm sorry, but ... are you kidding me? The Carnot theorem is a theoretical statement about idealized processes. We can just as easily discuss a theoretical Otto cycle as a theoretical Carnot cycle, as is done in countless websites and class materials available online. I understand very well that entropy increases in any real-world process. The question was, since there seem to exist reversible cycles whose efficiency is less than Carnot, why don't they violate the theorem by implying the existing of extremely efficient heat pumps? $\endgroup$
    – Dennis
    Feb 9, 2021 at 22:24
  • $\begingroup$ I'm sorry but I can't make sense of your answer. Perhaps you can elaborate to help me out. $\endgroup$
    – Bob D
    Feb 9, 2021 at 23:13

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