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Consider the p-V diagrams of an ideal and real stirling cycle, as found here The article suggests that dead volumes reduce the efficiency of a stirling engine. On the other hand, a simple thought experiment that comparas an ideal stilring engine A with a bigger real stirling engine B, where B has dead volumes so the amount of actually working fluid in both is the same, suggests that the thermodynamic efficiency should be the same. Only the power-per-cycle should suffer from dead volume.

To clarify, dead volume means volume (in the hot or cold chamber or regenerator) not being swept by the pistons.

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4 Answers 4

Let me try one more time. Now after you confirmed that dead volume means volume not being swept by the pistons, I could explain your question the following ways:

  1. Computational way: Stirling engine is regarding thermodynamic cycles analogous to Otto cycle - two isochoric and two adiabatic processes. The difference is that in Otto machine you have internal combustion and in case of Stirling machine you have external combustion. Because of these specifics the effective design of both machines is utterly different. Still, because the thermodynamic cycles are the same, I can write down the well-known efficiency for the ideal Otto motor: $$\eta = 1 - \frac{1}{r^{\gamma-1}} $$ where $$r = \frac{V_1}{V_2}$$ is compressibility. See that efficiency is larger for larger compressibilities.
    The problem is that in case of Stirling machine you have some volume that is not included in the process, so you have add some volume $V'$ to both volumes, so compressibility is reduced to $$r = \frac{V_1+V'}{V_2+V'}.$$ Smaller compressibility means smaller efficiency.

  2. Intuitive way: Since part of the gas does not participate in the exchange of the heat, its efficiency is zero. When calculating efficiency you are taking into consideration the whole gas, so the average efficiency is smaller than efficiency of the active part of the gas.

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@mart thanks for the bounty. If you still have reservations about my answer, please post them as a comment here. I am very interested in this question as well, so I want to make it absolutelly clear for myself too, despite already obtaining bounty. –  Pygmalion Apr 23 '12 at 10:42

(Why can't I comment? I'll post this comment to dmytry as an answer. No, I'm not sure this is correct, this is an attempt to invite a more rigorous critique)

The thermodynamic efficiency of a stirling engine with dead volumes is the same as an ideal stirling engine with the same lower/higher temperature. Consider an ideal stirling engine operating between the high and low Tempreatures $T_h$ and $T_l$ Now, consider a real stirling engine with the lower effective $T_h$' and higher $T_l$' - I would assume the efficience will be the same as for an ideal stirling engeine working with those temperatures.

Dead Volumes only make the engineering challenges (getting heat through the hot chamber walls, storing the heat in the regenerator, getting heat through the walls of the cold chamber) more, um, challenging.

Considering an ideal stirling engine with a dead volume attached. The dead volume acts as a temporary storage for energy, but provides no pathway for heat to leave the system. No work is done in the dead volume.

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The issue is that you have mixing of colder and hotter air as the air enters the dead volume or leaves the dead volume. That is loss of thermodynamic efficiency; picture that you ran a heat engine off that temperature difference. –  Dmytry Apr 17 '12 at 16:47
    
so by this mixing process you get new effective higher and lower temperatures with less difference. I would argue that the new efficiency is the one of an ideal stirling engine with those effective tempreatures (see edit) if you disagree please elaborate –  mart Apr 18 '12 at 15:21
    
I don't quite get your point. Of course you can always find the lower temperature difference for which the ideal engine would make same energy output as non-ideal engine. –  Dmytry Apr 18 '12 at 16:57
    
My point was: there's no ideal stirling engine with the same effiviency as a real for which I calculate the corresponding $T_h$' and $T_l$', $T_h$' etc are real temperatures resulting from the mixing you mentioned. Your edit about the entropy clarified this. Maybe you can elobarate this in your answer. –  mart Apr 20 '12 at 7:40
    
There is no real-world uniform temperatures like this in a real-world Stirling engine. Temperatures vary in the gas. Ultimately, the one thing that can be said in general is that for same amount of heat going from same temperature heat source to same temperature heat sink, the engine with dead volumes would have lower efficiency. Any more requires to specify the type of Stirling engine in question, the placement of dead volumes, and how much mixing does happen. –  Dmytry Apr 20 '12 at 9:36

I think the issue is that the dead volume acts as a form of heat reservoir. Consider an ideal Stirling engine connected to a dead volume via pipe. During the higher pressure part of the cycle, hot air from the engine mixes with colder air in the dead volume, which will clearly lead to loss in efficiency. During the lower pressure part of the cycle, warm air mixes into the engine.

Same would apply to a practical dead volume, such as those within the displacer chamber.

edit: To clarify. Mixing colder and warmer air together increases entropy without extracting any work out of it. That process is irreversible. That mixing happens inside of the dead volume.

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It would be nice if author explained what exactly is the dead volume. If you had ever seen real Strirling machine, you'd see that it has two cylinders and two pistons. One piston is moving air from hot to cold reservoar and vice-verse, while the power piston is where gas expands and where it makes useful work. Part of this work is transfered into moving the first piston. Obviously, part of the gas in the power piston, as well as in the tube between two cylinders does not participate in heat exchange, or participate in heat exchange only partially.

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