Why doesn't the heat in the two isochoric processes count towards efficiency calculation in Stirling cycle?

Question

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).

Answer 1

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

Answer 2

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.

Answer 3

Images of the Stirling engine'll probably help.

Here's the "alpha" version:

-Wikipedia

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.