Why Stirling engines have low power and efficiency

Discussion on Stirling or "hot air" engines (all types)
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matt brown
Posts: 488
Joined: Thu Feb 10, 2022 11:25 pm

Why Stirling engines have low power and efficiency

Post by matt brown »

Yep, title looks like clickbait, but read on. Since last summer, my deep dive on gammas has been amusing with all sorts of nuggets gleaned. Several times, Tom has called out gamma PP running on cold gas vs other engines running on hot gas. I always thought this odd, but until recently I've been mostly an alpha guy when scheming SE. However, I recently found a simple way to diagram a gamma with hot PP vs cold PP. Previous attempts were too busy and suggested a bogus slight of hand (heck, some here might still suggest this).

hot PP_1.png
hot PP_1.png (23.11 KiB) Viewed 632 times

Note PVTm values with 300-600k cycle, DP volume 2x PP volume, double dwell motion for ideal phasing, regen, values ignore all dead volumes (clearance, regen and conduit volumes). Also note the 6 bar charge pressure (when cold) and 6 bar buffer pressure. Reducing both of these pressures by 6 will yield values approaching LTD with sub ambient pressure during low end of cycle.
matt brown
Posts: 488
Joined: Thu Feb 10, 2022 11:25 pm

Re: Why Stirling engines have low power and efficiency

Post by matt brown »

After I did a response for Vincent on his LTD thread, I returned to working on this post, whereupon I had another altered state...

hot PP_2.png
hot PP_2.png (22.17 KiB) Viewed 626 times

This is similar previous hot PP gamma A except that this hot PP gamma B has PP volume equal DP volume. Now, forget A and focus on B vs S gamma. Hmmm, figs. B1, B3 and B4 are very similar S1, S2, and S4, but expansion B1-2 is drastically different expansion S1-2. Note how the same input produces 2x output for B hot PP than S cold PP. No, this doesn't mean that for an ideal 300-600k cycle, when (think if) S has Carnot = .50 then B has Carnot = 1.0 but rather when B has Carnot = .50 then S has Carnot = .25 (ouch). This simple diagram proves we've been lied to for years and that the typical Stirling pitch is bogus for gamma and beta schemes due to their cold PP. As always, the devil is in the details.
VincentG
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Joined: Tue Feb 21, 2023 3:05 pm

Re: Why Stirling engines have low power and efficiency

Post by VincentG »

The plot thickens. Good stuff Matt.
matt brown
Posts: 488
Joined: Thu Feb 10, 2022 11:25 pm

Re: Why Stirling engines have low power and efficiency

Post by matt brown »

Yep, the plot thickens. Tom has chosen to ignore conv'l thermo, but not me. Here's an update on that last graphic for those who wonder what I'm babbling about. Only a rudimentary thermo knowledge is req'd, especially PV=nRT where I'll use PV=mT since R doesn't matter and I prefer m vs n, even tho I'll be using m to denote gas mass, not gas moles. OK, here's a detail of above expansion stage...

hot PP detail.png
hot PP detail.png (26.85 KiB) Viewed 561 times

This is same as previous graphic except that I've used cc for volume (vs v) and shrunk down those lofty pressure values. So, what's happening here ??? Let's continue with previous ideal assumptions where all dead volumes are zero (clearance, regen & conduits) and where DP and PP have separate dwells. During any isothermal expansion or compression, internal energy is constant, and internal energy is linear T. So, during B1-2 hot PP expansion, gas expands 1:2 (100cc>>>200cc) and isothermal input equals work output (ignoring charge vs buffer pressure issue). Meanwhile, during S1-2 cold PP expansion, gas expands 1:1.5 (100cc>>>150cc) thru regenerator. Note that during both expansions, the same gas flow occurs, but that the results are different. If we consider both PP identical except for B has 2x the area, we can easily conclude that B hot PP produced 2x the output of S cold PP.

The question is how much heat was used for B vs S, especially since there's no simple method to measure heating while expanding thru a regenerator like Cv and Cp heats. So far, all we know is that hot 100cc B PP produced 2x cold 50cc S PP, and that both 100cc DP had the same internal energy at B1 and S1 when both were 100cc, 600k and 6m (prior expansion). Some might suggest that B simply used more heat during expansion, but if B used 2x the heat during expansion than S, then S regenerator would have no heat to remove to allow cooling for cold S PP. So, even if we 'simply' calculated heat flows for B, how would we know heat flows for S ??? Hmmm, saved by thermo basics, again !!! Recall that internal energy is constant for any isothermal expansion or compression AND that internal energy is linear T, and we can easily deduce that each 3m gas flow req'd the same source input, but that S lost 1/2 this 'isothermal' input passing thru S regenerator.

This ratio will vary depending upon thermal cycle and volume ratios, but losing 50% during just expansion phase likely explains why SE never live up to expectations. As Fool, Nobody and others have noted, energy is path dependent...

BTW, both cycles have the same total gas flow thru regenerator, just a different sequence. Interestingly, S regen heat in this example equals 300-600k Cv heat (like alpha regen) despite massive PV variation.
matt brown
Posts: 488
Joined: Thu Feb 10, 2022 11:25 pm

Re: Why Stirling engines have low power and efficiency

Post by matt brown »

Xlnt find Vincent !!! When I first hit on this hot PP "proof", I did a short search and found nothing. However, tonight, I found various studies on hot PP, but all focused on greater output and torque while ignoring efficiency. Returning to my previous graphic, it all became clear...consider each 100cc DP is comprised of 2x 50 cc volumes. The hot PP model will have two 600k 50cc volumes become two 600k 100cc volumes, but the cold PP model will have one 600k 50cc volume become one 600k 100cc volume while the other 50cc 600k volume becomes a 50cc 300k volume via the regenerator. In this manner, the hot PP model does indeed use twice the heat input of the cold PP model to produce twice the work output.

So, the hot PP model has greater output (and torque) than the cold PP model, but both have substantially the same efficiency (varies slightly depending upon charge vs buffer pressure).
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