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I'm a little confused about something. All reversible engines have the same efficiency, or one could drive the other to move more heat in the reverse direction. Also, no engine has an efficiency greater than a Carnot engine. Finally, a Carnot engine is reversible. If all these statements are true, then isn't it true that all reversible engines operate at Carnot efficiency? Where am I misinterpreting the logic?

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If all these statements are true, then isn't it true that all reversible engines operate at Carnot efficiency?

It is true provided the reversible engines operate between the same pair of temperatures. The statements that lead to your conclusion are true for engine that operates between two temperatures. So is the conclusion based on them; all reversible engines operating between the same pair of temperatures have the same efficiency $$ 1-\frac{T_{cold}}{T_{hot}} $$ where $T_{hot}$ is the temperature of the heater and $T_{cold}$ is the temperature of the cooler.

Where am I misinterpreting the logic?

I do not think you do.

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  • $\begingroup$ But that is the Carnot efficiency. So if the efficiency of any engine operating between $T_{cold}$ and $T_{hot}$ is $\eta = 1 - T_{cold}/T_{hot}$, then it must be true that every reversible engine has the same efficiency as the Carnot engine. I don't see where the logic is breaking down. $\endgroup$ – abalter Aug 1 '14 at 20:07
  • $\begingroup$ I've edited my answer. No logic is breaking down as far as I can see. $\endgroup$ – Ján Lalinský Aug 1 '14 at 22:52
  • $\begingroup$ I still don't get it. All reversible engines have the same efficiency. Then why don't all reversible engines have the Carnot efficiency? $\endgroup$ – abalter Aug 2 '14 at 16:40
  • $\begingroup$ If they work between two temperatures, they do have the Carnot efficiency. $\endgroup$ – Ján Lalinský Aug 3 '14 at 0:05
  • $\begingroup$ Thanks! I was thinking about it, and I think I figured out why I was surprised about that. In physics textbooks, you often derive the $\eta = 1 - T_{cold}/T_{hot}$ specifically for the Carnot cycle. But it's because you can use the idea gas laws etc. for that specific cycle. But, it really is more general to any reversible cycle. $\endgroup$ – abalter Aug 5 '14 at 19:22
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Simple model isobaric/isovolemic heat engine

I believe that the above sketch represents an example of a simple heat engine that can be made to be reversible, but has varying ideal efficiencies depending upon the particular choice of V1, V2, P1, P2. This engine does not operate between two constant temperature reservoirs but operates adjacent to a reservoir whose temperature is continually changing. Going clockwise from the lower-left hand corner the segment from (V1, P1) to (V1, P2) is isochoric (constant-volume) but experiences increasing pressures as heat is transferred from a reservoir with a temperature slightly above that in the engine chamber (for purposes of reversibility). Simultaneously the pressure in the ambient environment is gradually and steadily increased to match the pressure inside the chamber to maintain an isochoric state (this is not quite the same as having a sealed chamber which would result in an irreversible process).

The second segment from (V1, P2) to (V2, P2) occurs as heat continues to be transferred reversibly from a reservoir with the temperature slightly above that in the engine chamber - this segment's process is isobaric and produces both work on the environment as well as an increase in the internal energy of the gas in the chamber. The third segment from (V2, P2) to (V2, P1) occurs as heat is slowly and reversibly transferred to the reservoir (whose temperature now is maintained slightly below that in the chamber) - it is an isochoric process due to a gradually decreasing ambient pressure. The final segment from (V2, P1) to (V1, P1) is isobaric and results in work being done ON the system as well as a simultaneous decrease in the internal energy of the system.

It is possible to determine the efficiency of this system by determining the ratio of the work performed divided by the total heat into the system. It is easy to see that the total net work performed is (P2 - P1)*(V2 - V1). It is possible to determine the heat in (as well as the heat out) by using:

$\Delta$U=Q - W, or equivalently Q = $\Delta$U + W. Also, $\Delta$U = $\frac{3}{2}$nR$\Delta$T (we will assume n=1 for simplicity from now on). Some equivalent formulas are $\Delta$U = $\frac{3}{2}$$\Delta$PV (assuming constant volume) and $\Delta$U = $\frac{3}{2}$P$\Delta$V (assuming constant pressure). We have all we need to proceed.

Analyzing segment by segment:

Segment 1: $Q_{in}$ = $\Delta$U + W. Because this is isochoric, W = zero, so $Q_{in}$ = $\Delta$U = $\frac{3}{2}$$\Delta$PV = $\frac{3}{2}$(P2 - P1)*V1.

Segment 2: $Q_{in}$ = $\Delta$U + W. There is an increase in internal energy ($\Delta$U) = $\frac{3}{2}$P$\Delta$V = $\frac{3}{2}$P2*(V2-V1). There is also work performed: W = P2*(V2 - V1), so the total heat in for this segment is $Q_{in}$ = $\frac{5}{2}$P2*(V2-V1).

The above means that for this engine the Total $Q_{in}$ = $\frac{3}{2}$(P2 - P1)V1 + $\frac{5}{2}$P2(V2-V1). The efficiency can therefore be calculated as

Efficiency = $\frac{Total Net Work}{Total Heat Input}$=$\frac{(P2 - P1)*(V2 - V1)}{\frac{3}{2}(P2 - P1)*V1 + \frac{5}{2}P2*(V2-V1)}$ = $\frac{(P2 - P1)*(V2 - V1)}{(P2 - P1)*(V2 - V1)+(\frac{3}{2}P2V2 + P1V2 - \frac{5}{2}P1V1)}$

The denominator was rewritten in an equivalent form in the last term on the right to demonstrate that, as expected, the efficiency is always <100% because the term $(\frac{3}{2}P2V2 + P1V2 - \frac{5}{2}P1V1)$ is always positive for P2 > P1 and V2 > V1.

The ideal efficiency of this heat engine varies according to the actual values of P1, P2, V1, and V2. Here is a snippet from a spreadsheet that shows this:

Spreadsheet of Efficiencies by Varying P1, P2, V1, V2

Although it is not necessary to go any further to demonstrate the efficiency piece, it is instructive to notice a few other facts. Actually we already know the Total $Q_{out}$ because by the conservation of energy for the entire cycle Total $Q_{in}$ - Total $Q_{out}$ = Net Work, so doing the math we find that Total $Q_{out}$ = Total $Q_{in}$ - Net Work = $\frac{3}{2}(P2 - P1)*V2 + \frac{5}{2}P1*(V2 - V1)$. Although we have already calculated these figures, it is instructive to complete the heat flow analysis to confirm.

(I am electing to keep Total $Q_{out}$ quantities positive even though they are technically negative but remembering that they will be subtracted out at the end so that's OK):

Segment 3: $Q_{out}$ = $\Delta$U + W = $\frac{3}{2}(P2 - P1)*V2$ because W = zero.

Segment 4: $Q_{out}$ = $\Delta$U + W = $\frac{3}{2}P1*(V2 - V1) + P1*(V2 - V1) = \frac{5}{2}P1*(V2 - V1)$

Adding these together we have confirmed the Total $Q_{out}$ portion of the formula as expected.

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