Apparently, in order to prove that an engine must be as efficient as a Carnot engine if reversible is because apparently for its efficiency in either direction:

$\eta_E \le (1-T_2/T_1)$

$\eta_R \ge (1-T_2/T_1)$

I assume (as this is my copying down the solution to this exercise) that the subscripts $E$ and $R$ stand for engine and refrigerator respectively, as they should be the same process but in reverse directions.

However, while I agree this is true for any engine $E$:

$\eta_E \le (1-T_2/T_1)$

I don't see how this is the case:

$\eta_R \ge (1-T_2/T_1)$

If $\eta_E$ can't exceed $(1-T_2/T_1)$ which is the efficiency of a Carnot engine, then why can its efficiency in the reverse direction?

From there, my lecturer says that for this to be true for the same engine, we require $\eta_E = \eta_R$ otherwise it would gain efficiency by changing the direction of the process which is nonsensical.

This implies that apparently if a heat engine is reversible it must have the efficiency of a Carnot engine.

Perhaps this is because $\eta_E = \eta_R$ and since $\eta_E$ cannot exceed $(1-T_2/T_1)$ then it must be the case that the only possible solution is $\eta_E = \eta_R = (1-T_2/T_1)$ which demands that the efficiency in either direction be that of the Carnot efficiency if reversible? Either way I don't see how if for a reversible engine if $\eta_E \le (1-T_2/T_1)$ then $\eta_R \ge (1-T_2/T_1)$.

  • $\begingroup$ If I write the efficiencies of $\eta_{E}$ and $\eta_{R}$ in terms of $T_{c}$ and $T_{h}$, then $\eta_{E}=\frac {T_{h}-T_{c}}{T_{h}}$, and $\eta_{R}=\frac {T_{c}}{T_{h}-T_{c}}$ since the refrigeration cycle is a heat engine in running in reverse. $\endgroup$ – Cinaed Simson May 17 '19 at 2:24
  • $\begingroup$ I see. Why does this imply $\eta_R \ge (1-T_2/T_1)$? $\endgroup$ – sangstar May 17 '19 at 12:45

Why a reversible engine's efficiency must be equal to that of a Carnot engine

Just want to point out that this statement applies to a reversible engine operating between two fixed thermal reservoirs, not just any reversible engine..

don't see how this is the case: $$πœ‚π‘…β‰₯(1βˆ’π‘‡2/𝑇1)$$

It isn't the case.

Your equation for the Carnot refrigerator efficiency is incorrect. This efficiency is called the coefficient of performance (COP) and is given by


It equals the heat transferred out of the low temperature reservoir divided by the work required to cause the transfer. It is a number greater than 1.

From there, my lecturer says that for this to be true for the same engine, we require πœ‚πΈ=πœ‚π‘… otherwise it would gain efficiency by changing the direction of the process which is nonsensical.

I agree it would be nonsensical if in fact that is what the lecturer said. Clearly the heat engine efficiency is less than 1 and the refrigerator COP is greater than 1. In any case the comparison of the heat engine and refrigerator to prove the Carnot efficiency is the maximum possible normally goes something like the following: (See the diagram below).

The output of a Carnot heat engine is used as the input of the Carnot refrigerator. In the diagram $Q_H$ and $Q_L$ are the same for both the heat engine and the refrigerator. The work out of the heat engine equals the work in to the refrigerator so that the overall net work done for the combination of the two is zero.

The proof that the maximum heat engine efficiency is the Carnot heat engine is a proof by contradiction.

Let's say we are told that the heat engine will produce the same amount of work out to run the refrigerator with less heat in $Q_H$, which would make it more efficient than the Carnot engine. But that would also mean the heat engine extracts less heat from the hot reservoir than the refrigerator exhausts, and since $Q_L$ is the same for both, the net effect will be heat transferred from cold to hot without any outside assistance. Heat transfer from cold to hot without net work done would be a violation of the second law.

Hope this helps.

enter image description here

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The performance of refrigerators can be quantified using the coefficient of performance:

$$ COP_\text{R} \equiv \frac{\text{Useful out}}{\text{Required in}} = \frac{Q_\text{lo,in}}{W_\text{net,in}} = \frac{Q_\text{lo,in}}{Q_\text{hi,out}-Q_\text{lo,in}}. $$

Like the efficiency of a heat engine, the COP of a refrigerator has an upper bound:

$$ COP_\text{R} \leq \frac{T_\text{lo}}{T_\text{hi}-T_\text{lo}} = \frac{1}{\frac{T_\text{hi}}{T_\text{lo}}-1}. $$

For whatever reason, your professor chose to instead define an efficiency for the refrigerator in the same way as one would for a heat engine:

$$ \eta_\text{R} = \frac{W_\text{net,in}}{Q_\text{hi,out}} = \frac{Q_\text{hi,out}-Q_\text{lo,in}}{Q_\text{hi,out}}. $$

With a little math, we can convert the limit on $COP_\text{R}$ into a limit on $\eta_\text{R}$: \begin{align} \overbrace{\frac{Q_\text{lo,in}}{Q_\text{hi,out}-Q_\text{lo,in}}}^{COP_\text{R}} &\leq \frac{1}{\frac{T_\text{hi}}{T_\text{lo}}-1} \\ \frac{Q_\text{lo,in}}{Q_\text{hi,out}-Q_\text{lo,in}}+1 &\leq \frac{1}{\frac{T_\text{hi}}{T_\text{lo}}-1}+1 \\ \frac{Q_\text{lo,in}}{Q_\text{hi,out}-Q_\text{lo,in}}+\frac{Q_\text{hi,out}-Q_\text{lo,in}}{Q_\text{hi,out}-Q_\text{lo,in}} &\leq \frac{1}{\frac{T_\text{hi}}{T_\text{lo}}-1}+\frac{\frac{T_\text{hi}}{T_\text{lo}}-1}{\frac{T_\text{hi}}{T_\text{lo}}-1} \\ \frac{Q_\text{hi,out}}{Q_\text{hi,out}-Q_\text{lo,in}} &\leq \frac{\frac{T_\text{hi}}{T_\text{lo}}}{\frac{T_\text{hi}}{T_\text{lo}}-1} \\ \frac{Q_\text{hi,out}}{Q_\text{hi,out}-Q_\text{lo,in}} &\leq \frac{1}{1-\frac{T_\text{lo}}{T_\text{hi}}} \\ \underbrace{\frac{Q_\text{hi,out}}{Q_\text{hi,out}-Q_\text{lo,in}}}_{\eta_\text{R}} &\geq 1-\frac{T_\text{lo}}{T_\text{hi}}. \end{align}

This math illustrates that the definition of $\eta_\text{R}$ is correct in the sense that it is mathematically consistent. Despite this, I'm of the opinion that $\eta_\text{R}$ is wrong in that it is an unusual and unhelpful way of thinking about refrigerator performance. Most textbooks instead use the COP - perhaps refer to yours if you don't like your lecturer's way of presenting this material.

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