I'm trying to understand Stirling Cycle that consists of two isochoric processes and two isothermal processes. The heat transfer are given as

$$Q_1 = \nu RT_1 \mathrm{ln} \frac{V_1}{V_2}$$ $$Q_2 = C_V (T_2 - T_1)$$ $$Q_3 = \nu RT_2 \mathrm{ln} \frac{V_2}{V_1}$$ $$Q_4 = C_V (T_1 - T_2)$$

Where $T_1$ is the temperature of the hot heat reservoir, $T_2$ is the temperature of the cold heat reservoir, and $V_1$ is the higher volume, $V_2$ is the lower volume. $Q_1$ is the heat absorbed in the isothermal inflation process, $Q_2$ is the heat released in the isochoric cooling process, $Q_3$ is the heat released in the isothermal compression process, $Q_4$ is the heat absorbed in the isochoric heating process.

And the efficiency of the cycle is

$$\eta = \frac{W}{Q} = \frac{Q_1 - Q_3}{Q_1} = 1 - \frac{T_2}{T_1} $$

I am confused why $Q_2$ and $Q_4$ don't count towards the calculation of $\eta$ as they're absorbed and released by the working ingredient (gas).

  • $\begingroup$ Can you label which of these processes is which and how they relate to the various temperatures and volumes? It is currently very hard to piece together what exactly is going on. $\endgroup$ Commented Jan 13, 2018 at 17:01
  • $\begingroup$ @BySymmetry I've added it. $\endgroup$
    – iBug
    Commented Jan 13, 2018 at 17:07
  • $\begingroup$ The way you have written the equations, the efficiency should be $$\eta=\frac{Q_1+Q_2+Q_3+Q_4}{Q_1+Q_4}$$where the numerator is the amount of work done, and the denominator is the heat flowing into the working fluid. $\endgroup$ Commented Jan 13, 2018 at 18:51

3 Answers 3


The Stirling engine by common definition includes the part of the system which gives and absorbs $Q_2$ and $Q_4$ . That part is commonly called regenerator and is included as part of the system. Equivalently you can imagine that the heat flow is compared between to and from the reservoir, and everything else is part of the engine

  • $\begingroup$ But the engine won't work without $Q_2$ and $Q_4$. $\endgroup$
    – iBug
    Commented Jan 13, 2018 at 17:19
  • $\begingroup$ Yes but the reversible sterling engine includes the regenerator which absorbs and provides $Q_2$ and $Q_4$ as your working material goes through the cycle. In the formula for efficiency the heat absorbed and released by the entire engine is counted not transactions between parts of it. Imagine if i break the carnot engine material into 2 , and counted the heat absorbed and emitted by one part to another. its a similar problem here $\endgroup$
    – ssj3892414
    Commented Jan 13, 2018 at 17:32

I've always justified the non-inclusion of $Q_2$ and $Q_4$ by arguing that each morsel of heat excreted at a particular temperature in the first isochoric process is matched by an equal morsel of heat taken in at the same temperature in the second isochoric process. So the isochorics contribute nothing to the net transfer of heat between two temperatures (any more than they contribute to the work output).

[In fact, I'd say that the Stirling Cycle is a poor man's Carnot Cycle; for an ideal gas the maths is easier, so it can be used at quite an elementary level to give a taste of what the Second Law of Thermod is all about, though of course the Physics isn't as clean as for a Carnot cycle.]

Note: I'm discussing a Stirling cycle and not any form of Stirling engine.

  • $\begingroup$ I see your point. But, this is also a matter of perspective. The efficiency is more conventionally defined as the work done divided by the heat in. $\endgroup$ Commented Jan 13, 2018 at 20:08
  • $\begingroup$ @ChesterMiller The point that may not be clear is that Sterling engines are specified to provide $Q_2$ and $Q_4$ internally, such that they're not external heat flows. $\endgroup$
    – Nat
    Commented Jan 13, 2018 at 20:17
  • $\begingroup$ I see. Then Q2 and Q4 are not really exchanges of heat between the surroundings and the system (in this case the system is comprised of the two separate volumes of working fluid). $\endgroup$ Commented Jan 13, 2018 at 20:46
  • $\begingroup$ Still, if you consider the gases in the two cylinders as separate systems, Q2 and Q4 are external heat flows for each, and, by the conventional definition of engine efficiency, either Q2 or Q4 would have to be included in the equation for their separate efficiencies. However, for the overall combination, they would not. Interesting. $\endgroup$ Commented Jan 13, 2018 at 23:04
  • $\begingroup$ The argument for not including $Q_2$ and $Q_4$ because they cancel can only be used for an ideal gas indicates that there must be a problem with it. In fact, the correct efficiency is what @ChetMiller wrote above. The Carnot efficiency can only be achieved with a Carnot cycle (isothermal-adiabatic-isothermal-adiabatic) as it can be shown reasonably easily, and the Stirling's theoretical efficiency is always less than that exactly because it must absorb thermal energy reversibly in the isochoric compression stage by an infinity of reservoirs each delivering infinitesimal amount of entropy. $\endgroup$
    – hyportnex
    Commented Mar 24, 2023 at 2:49

Images of the Stirling engine'll probably help.

Here's the "alpha" version:

enter image description here


The heat flow $Q_1$ is happening in the jacket around the upper chamber while the heat flow $Q_3$ is happening in the jacket around the lower chamber.

As for $Q_2$ and $Q_4$? They both happen in the tube connecting the chambers, called the "regenerator". Going from hot-to-cold is $Q_2$ while going from cold-to-hot is $Q_4$. So, ideally, they'd cancel each other out without external action because the tube's meant to store the heat between cycles. Alternatively, could consider multiple Stirling engines working in counter-step operation, such that one supplies its $Q_2$ to match the other's $Q_4$ and vice-versa.

Since $Q_2$ and $Q_4$ are meant to cancel each other out, they're not included in characterizations of the process's efficiency. 'course, this in ideal characterizations; real-life implementations'll tend to be less well-behaved.


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