A Efficiency of Two Carnot Engines Operating in Series

[SteveMaryland]

This has been discussed in a previous thread https://www.physicsforums.com/threads/efficiency-of-two-carnot-engines-in-series.173879/

Th conventional answer is that the series efficiency is the same as one engine running between T max and T min but I am skeptical.

I have attached my calculations as a PDF. My calculation says that the series efficiency is the sum of the two individual efficiencies minus the "one engine" efficiency. Ed Jaynes seems to have another answer also, but it differs from mine (see attached screenshot of jayne's paper; I tried uploading Jaynes paper but too large...).

 

[Andrew Mason]

The overall efficiency of a series of Carnot engines operating between each of n constant temperature reservoirs is just:
$\eta=\frac{W}{Q_{in}}=\frac{(Q_{1}-Q_2)+(Q_2-Q_3).... (Q_{n-1}-Q_n)}{Q_1}$ which reduces to:
(1) $\eta=\frac{(Q_1-Q_n)}{Q_1}=1-\frac{Q_n}{Q_1}$

Since it is a Carnot cycle, $\Delta S=\Sigma_1^n \Delta S_i=0$ over one complete cycle where $\Delta S_i$ is the change in entropy of reservoir no. i.

Since the reservoirs are at constant temperature for any given reservoir no. i:
$\Delta S=\frac{Q_i}{T_i}=\frac{Q_{i-in}}{T_i}-\frac{Q_{i-out}}{T_i}$.

But we know that for all the intermediate reservoirs, heat flow in=heat flow out so:
$Q_i=0$.

So $\Sigma_2^{n-1}\Delta S_i=0$. Therefore, there is a change in entropy of only the first and last reservoirs.

We conclude therefore, that: $\Delta S=\Delta S_1+\Delta S_n=0$. And since $\Delta S_1=\frac{-Q_1}{T_1}$ and $\Delta S_n=\frac{Q_n}{T_n}$, then $\frac{Q_n}{T_n}=\frac{Q_1}{T_1}$

So:
(2) $\frac{Q_n}{Q_1}=\frac{T_n}{T_1}$.

Substituting (2) into (1) gives:

(3) $\eta=1-\frac{T_n}{T_1}$ which is the same as a single Carnot engine operating between those temperatures.

AM

 

[SteveMaryland]

Thank you. That is what I hear, but can you comment on what Jaynes talking about?

 

[Andrew Mason]

I was addressing your skepticism about the conventional answer as to the series efficiency. I am not sure why you would want to express that in terms of the efficiencies of each engine but just do this:

(1) $\eta_{12}=\frac{T_1-T_2}{T_1}$ so $T_2=T_1-T_1\eta_{12}$
(2) $\eta_{23}=\frac{T_2-T_3}{T_2}$ so $T_3=T_2-T_2\eta_{23}$
(3) $\eta_{13}=\frac{T_1-T_3}{T_1}$

Then substitute in (3) the value for $T_3$ in (2) resulting in:
$\eta_{13}=\frac{T_1-(T_2-T_2\eta_{23})}{T_1}$. Then substituting the value for $T_2$ in (1) gives:

$\eta_{13}=\frac{T_1-([T_1-T_1\eta_{12}]-[T_1-T_1\eta_{12}]\eta_{23})}{T_1}$.

This results in:
$\eta_{13}=1-(1-\eta_{12})-\eta_{23}(1-\eta_{12})=1-[(1-\eta_{12})(1-\eta_{23})]$
$\eta_{13}=\eta_{12}+\eta_{23}-\eta_{23}\eta_{12}$

AM

 

[SteveMaryland]

Thank you again. That is what Jaynes said. My assumption was that the (combined) efficiency of two engines in series would be = the product of the (separate) efficiencies of each engine, which would always be less than the 1-3 efficiency. But apparently (because of zero net entropy?) that is not the case.

 

[Andrew Mason]

My first post explains it. It is just a matter of calculating efficiency. If you think about it, the combined efficiency of two equally efficient engines would not be less than the efficiency of each engine operating between the same two temperatures. And when two equally efficient engines are in series, operating between reservoirs whose temperature difference is proportionately the same (800K-400K and 400K-200K for example) they should be more efficient because more work is being done but no new heat is flowing.

AM

 

[SteveMaryland]

Agreed. I think my confusion derives from conflating "thermal series" with "mechanical series".

In the above we are connecting Carnot engines in both thermal series and mechanical series. This is a different case from where two Carnot engines were connected in mechanical series but operate in thermal parallel.

Nevertheless, expressing the 1-3 efficiency in terms of 1-2 and 2-3 efficiencies, as Jaynes did, looks very counter-intuitive. Who knew that the net series efficiency could be expressed as the sum minus the product?

 

[hutchphd]

but no new heat is flowing.

I think there is a subtlety here. Suppose the engines are not identical. Tacitly you are assuming that the intermediate reservoir maintains a steady temperature (without external input of heat). This is where one must invoke Carnot explicity.

 

[SteveMaryland]

Further consideration - we may indeed have two Carnot engines in "thermal series" but they may or may not be in "mechanical series". If the "work output shafts" of the two Carnot engines are not mechanically connected, then the total output work is indeed Wa + Wb, but if the shafts are mechanically connected, the work output becomes a single "combined" work, and if one engine is "weak", it may act as a drag on the other and reduce the "combined" work to < Wa +Wb.

Jaynes was talking about a case where the two engine shafts were indeed "hooked together" mechanically. If so, and if one of the engines is "weak" (and Eb is certainly weaker than Ea because Eb is operating between a lesser dT than Ea) then Eb may "be a drag" on the total work output if the engines are constrained to be mechanically linked.

 

[hutchphd]

The way they are "hooked together" need not be specified. The condition that they each act as a Carnot engine is the stated requirement. The exact mechanical linkage may in fact be difficult, but these are theoretical constructs.