# Efficiency of Stirling engine and Carnot's theorem

I want to calculate the efficiency of this Stirling cycle for an ideal gas $pV = nRT$

The mechanical work is

$$\Delta W_{12} = - \int_{V_1}^{V_2} p(V) \mathrm{d}V = -nRT_2 \ln \frac{V_2}{V_1}\\ \Delta W_{23} = \Delta W_{41} = 0\\ \Delta W_{34} = -nRT_1 \ln \frac{V_1}{V_2}$$ On the isothermal curves the change in inner energy $\Delta U = \Delta W + \Delta Q$ is zero. $$\Delta Q_{12} = - \Delta W_{12} > 0\\ \Delta Q_{34} = - \Delta W_{34} < 0$$ On the isochoric (isovolumetric) curves the heat quantities are $$\Delta Q_{23} = C_V (T_1 - T_2) < 0\\ \Delta Q_{41} = C_V (T_2 - T_1) > 0$$ The efficiency is then $$\eta = \frac{-\Delta W}{\Delta Q}$$ $\Delta Q$ is the input heat, i.e. sum of all the heat quantities $> 0$: $$\Delta Q = Q_{12}+Q_{41} = n R T_2 \ln \frac{V_2}{V_1} + C_V (T_2 + T_1)$$ $\Delta W$ is the total mechanical work: $$\Delta W = W_{12}+\Delta W_{34} = - nR(T_2 - T_1) \ln \frac{V_2}{V_1}$$

So finally the efficiency is $$\eta = \frac{T_2 - T_1}{T_2 + \frac{C_V (T_2 - T_1)}{nR \ln V_2 / V_1}} < \eta_\text{C}.$$ It is smaller than the efficiency of the Carnot cycle. But it should be equal to it if all processes are done reversibly.

The calculations are taken from a textbook (Nolting: Grundkurs Theoretische Physik 4) which actually points out this problem as a question to the reader. My only explanation is that this process is not reversible but I don't know how to tell without actually seeing how the isothermal and isochoric processes are realized.

So my questions are:

• Is this a contradiction to Carnot's theorem that the efficiency $\eta_\text{C} = 1 - T_1/T_2$ is the same for all reversible heat engines between two heat baths?
• Is this cycle reversible?
• Is there a way to say whether a process is reversible or irreversible only with a figure like the one above?
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You're miscalculating the efficiency. For an arbitrary engine cycle, you have $Q_{H} = Q_{L} + W$, and $e = \frac{W}{Q_{H}}$. You're not dividing by your input heat. –  Jerry Schirmer Sep 28 '13 at 19:41
I edited the question for clarity of what I'm calculating. The denominator in the efficiency should be the heat that is put into the engine. The way I'm calculating the heat it is the one that is $> 0$. Do you agree? –  frankundfrei Sep 28 '13 at 19:56
Related to the question of reversibility of curves in thermodynamic state space: physics.stackexchange.com/questions/78405/… –  joshphysics Sep 29 '13 at 18:25

There is no contradiction. The Stirling cycle you drew above is reversible but does not operate between two reservoirs at fixed temperatures $T_1$ and $T_2$. The isovolumetric parts of the cycle operate at continuously changing temperatures (think ideal gas law).

Addendum. Note that in thermodynamics, a heat engine is said to operate (or work) between (two reservoirs at) temperatures $T_1$ and $T_2$ provided all of the heat it absorbs or gives up is done so at one of those two temperatures.

To give credence to this definition (which is essentially implicit in most discussions of heat engines I have seen), here is a quote from Fermi's thermodynamics text:

In the preceding section we described a reversible cyclic engine, the Carnot engine, which performs an amount of work $L$ during each of its cycles by absorbing a quantity of heat $Q_2$ from a source at temperature $t_2$ and surrendering a quantity of heat $Q_1$ to a source at the lower temperature $t_1$. We shall say that such an engine works between the temperatures $t_1$ and $t_2$.

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Then I think I didn't understand what operate between two two reservoirs at fixed temperature actually means. The Carnot cycle consists of adiabatic and isothermics. temperature changes but no heat is transfered. So "operating" basically means heat transfer? You also said the cycle is reversible. Are all curves that can be drawn within the $p$-$V$-plane reversible? –  frankundfrei Sep 28 '13 at 20:22
@frankundfrei Yes; "operating between two temperatures" in this context means that all heat transfers that occur during the cycle occur at one or the other temperature. As for the question of reversibility, I'm not sure it's appropriate to call a curve in thermodynamic state space itself reversible or irreversible. I think we need to answer a more nuanced question like: when we perform some idealized physical process, and it's successive states can be well-approximated by a continuous curve in the state space, then can that process be performed in reverse? –  joshphysics Sep 28 '13 at 21:08
Could you add this definition of "operation" in your answer? I haven't come across a precise definition neither in the lecture I attended nore the textbooks I use, so it also might be helpful to other people. Thanks for your help! –  frankundfrei Sep 29 '13 at 11:29
@frankundfrei Sure thing. I made an edit. –  joshphysics Sep 29 '13 at 17:58

The Stirling cycle as you describe it is not reversible. The transfer of heat from thermal reservoirs along paths 4->1 and 2->3 is not a reversible process, because heat is being transferred between two objects at different temperatures. To reverse the process, you would need to spontaneously transfer heat from a colder to hotter reservoir, which violates the 2nd law of thermodynamics.

Stirling engines are often described as reversible, but this requires a special kind of process. Notice that the heat transferred into the engine along 4->1 is the same as the heat transferred out of the engine along 2->3 and that 4->1 and 2->3 operate between the same two temperatures. Therefore, a Carnot efficient Stirling engine can be constructed if heat is transferred isothermally within the engine along these paths. This is accomplished with a "regenerator," a thermal mass that stores the energy liberated in 2->3 and returns it to the gas along path 4->1. You can see that the regenerator has to vary continuously in temperature between T2 and T1 and exchange heat isothermally with the gas as it passes through.

Note that all reversible engines must operate with the same efficiency. This follows from the definitions of efficiency and entropy. A reversible engine operates with 0 entropy change. $\Delta S = -\frac{Q_h}{T_h} + \frac{Q_c}{T_c}$, so $\Delta S = 0$ implies $\frac{Q_h}{T_h} = \frac{Q_c}{T_c}$ or efficiency = $\frac{Q_h - Q_c}{Q_h} = \frac{T_h - T_c}{T_h}$

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