[Translating resource] 100W class LTD Stirling engine( Lesson 5-3 )

[Bumpkin]

Hey Matt, I hadn’t noticed that 25-40 ratio — had to go back and find it to see if you weren’t hallucinating. It might be a misinterpretation or misprint, otherwise yes I’m surprised the engine runs at all at that temperature ratio. It would be working against itself as a heat-pump for much of each stroke.

Bumpkin

[matt brown]

Hey Matt, I hadn’t noticed that 25-40 ratio — had to go back and find it to see if you weren’t hallucinating. It might be a misinterpretation or misprint, otherwise yes I’m surprised the engine runs at all at that temperature ratio. It would be working against itself as a heat-pump for much of each stroke.

Bumpkin

Kiochi LTD specs.png (269.6 KiB) Viewed 5613 times

Indeed, hard to believe it runs at all, however there's all sorts of stuff that can be gleaned from this experiment. The most obvious is that at this paltry dT this massive bugger is only producing 1 watt per cycle which begs the question whether combined friction is greater than this per cycle. Years ago, I would have jumped on all these specs to glean details thruout cycle, but I'm getting lazy in old age (starting with ideal energy calcs before guessing where it went). There's still various glaring details that should jump out at everyone, like a displacer bore 10x its stroke but only 2x piston bore, yet piston stroke is 2.5x displacer stroke. Heck, the only normal thing is a piston stroke 2x its bore.

With 90 deg phasing and Scotch Yoke, I think the skimpy displacer stroke is solely to dwell displacer at TDC and BDC.

[Bumpkin]

I’d like to have some confirmation of the test temperature ratio.The ratios are very interesting to me. I’ve been thinking, (assuming?) that the compression ratio should be about half of the temperature ratio. But I’m still not too old to either learn or get more confused.

Bumpkin

[matt brown]

I’d like to have some confirmation of the test temperature ratio.The ratios are very interesting to me. I’ve been thinking, (assuming?) that the compression ratio should be about half of the temperature ratio. But I’m still not too old to either learn or get more confused.

Bumpkin

viewtopic.php?f=1&t=5412#p17284

Above post I made a year ago (regen issues, how much heat) appears to have slide past most here. Scroll down to the heat triangle part and note your comment afterwards. Per some previous comms, I spent a few years trying to reduce regen load into something simple, and my eureka moment was my heat triangle (eyes roll).

sq cycle.jpg (58.59 KiB) Viewed 5596 times

Here's a well proportioned PV of a 'sq cycle' where the thermal ratio (henceforth Tr) equals the volume ratio (aka compression ratio, henceforth Vr). Note that with a sq cycle, the pressure after compression (pt 2) equals the pressure after expansion (pt 4). With any such cycle, (ideal) regen heat (2-3) is 1.8x (ideal) input heat (3-4) when diatomic (air), but (ideal) regen heat is...only...1.08x (ideal) input heat when monatomic (helium). If you want to verify this, refer to my post and head on over to omnicalculator.com per post. Once you get this, you'll notice that a 300-600k cycle (Tr=2) with Vr=2 has the same 'relative' regen heat as a 300-1200k cycle (Tr=4) with Vr=4, yet the Tr=4 has inherently more eff thanks to Carnot. Unfortunately, Vr=4 SE doesn't exist due to mechanical constraints (so far).

Now I wouldn't drag you thru this mumbo-jumbo only to leave you wondering WTF, would I ??? Interestingly, isothermal work varies by the 2nd power of expansion, whereby for ANY given Tr, Vr=4 has 2x the output of Vr=2 (aka 2^1) and Vr=8 has 3x the output of Vr=2 (aka 2^3). Despite the mechanical constraints, this diminishing return of output is why most SE hover around Vr=2 with no interest is Vr>2. However, a 'low end' Tr=2 cycle (300-600k) with Vr=2 would (in theory) cut any regen ineff in half if mech could manage Vr=4. So, the basic idea for any isothermal cycle is to have Vr>Tr, and PV of such is the banner-like PV of typical LTD (vs the flag-like PV of comm'l SE).

Lacking expansive mech, comm'l SE push Tr>Vr, but this only results in very 'top heavy' cycles where regen dwarfs input. I'll stop here, but I could go on all night...

[matt brown]

Urieli D90 regen_1.png (99.51 KiB) Viewed 5594 times

I cracked regen load 20 yrs ago when in the yahoo energy groups, but it went past as no biggie except for one guy that picked up on it...Zig Herzog (Penn St Prof) who asked if he could use some of my stuff in his classroom. He later gave me the nod when he updated his SE pages which Urieli (finally) linked to ~1/2 dz yrs ago. Finally, an easy way to grasp regen load which was previously buried in calculus (kinda like scheming thermo with F or C vs K deg). Once you get a grip on how top heavy regen cycles are, and how regen inefficiency plays out between mono vs dia gasses, then you'll arrive at stuff similar this Urieli graph (and hope the dog doesn't stop barking)...

[matt brown]

Hey Bumpkin, are you learning more or just more confused ? I've been studying thermo for 50 yrs, so sometimes struggle to simplify stuff. As for the basics of the Stirling cycle, all you mainly need to know is that

(1) internal energy is linear absolute temp for isochoric process
(2) internal energy is constant for isothermal process where work varies to 2nd power volumetrically
(3) regen load is 1.8 for any sq cycle (when air)
(4) Carnot is king for all 'simple' compression cycles

You don't need any calculus or computer sims to estimate efficiency as a dimensionless ratio. Most know the Carnot buzz (tho some still argue it) and many know that Cv (constant volume heat capacity) is linear K (absolute temp) for an ideal gas (this avoids entropy issue/s). Some know the 2nd power of isothermal expansion, but few know regen load. Here's the simple way to game Stirling basics...

Consider 2 cycles, A & B, both cycles air (diatomic)
A is 300-600k (Tr=2) with Vr=2
B is 300-1200k (Tr=4) with Vr=2

A will have Carnot=.50 and regen 1.8x input
B will have Carnot=.75 and regen 5.4x input

In this dimensionless comparison we only know a few things, but we know Cv is linear K, so whatever finite value of heat was req'd for 300-600k 'regen' will require 3x this heat for 300-1200k regen. And since A has premise of 1.8 heat units during regen, then B will have 5.4 heat units during regen (ya know, 1.8 units 300-600k, 1.8 units 600-900k, 1.8 units 900-1200k). See any problem/s here ???...B has gained 50% via Carnot (.50>>>.75) but only via 300% more regen load. This is no problem in (ideal) Fantasyland, but up jumps the devil in reality....as regen efficiency falls from 100%, regen losses will quickly outstrip any Carnot gain via higher Tr. This explains the often touted regen load of 5-6x input, and hidden efficiency issue that plagues SE where endless parametric studies and computer sims attempt to game the ends against the middle.

[matt brown]

regen eff Tr3 Vr2.png (75.11 KiB) Viewed 5572 times

Here's another graph that shows effect of regen eff on Carnot eff. Note Vr=2 and Tr=3 with hydrogen gas (diatomic, so same as air). This is from an xlnt 2009 Canterbury thesis, but graph needs a background grid. Nevertheless, feast your Vulcan squinties on that curve...starting with Carnot = .67 @ 1.0 regen, Carnot = .60 @ .95 regen, then Carnot = .53 @ .90 regen, then Carnot = .45 @ .80 regen. So, for this cycle, with everything else ideal, when regen drops to 80% (down 1/5) then Carnot drops to 45% (down 1/3). This graph was computer generated and this otherwise xlnt thesis misses regen load completely...there's simply no way to pick it out from this graph.

This is the type of basic thermo anyone scheming SE should know.

Hey, I just noticed...a math whiz could pick out regen load from Carnot @ .00 regen, but I doubt even Elon could with a gun at his head...

[Bumpkin]

Golly Matt, you know lots of words and stuff.

Respect and thanks to gitPharm01 for this thread and your time and trouble.

Bumpkin

[VincentG]

Where's the guy sitting back and eating popcorn emoji?

[Bumpkin]

Yeah I could have said that better. I need to quit using cutesy mental short-hand in mixed crowds. Anyway Vincent, since you’re here and we’re talking ratios; you said in another way-back thread:

“I have done further testing with the power of the piston to lift weight over simmering water and ice with the second(failed) housing. Even with a significant air leak, the piston was easily able to lift a nearly one pound weight after a full compression cycle. I have calculated the compression ratio of the engine(with the exaggerated stroke of 1.2") to be just over 1.1:1. I think this is a much more useful number that the swept volume ratio, which is 1:9 in this case. The stock engine being over 1:60.”

I never figured out what you meant there as per compression ratio vs swept volume ratio. I sorta think of them (with dead-space thrown in,) as being the same thing. In the interest of common terminology, could you explain how you are figuring/arriving at each?

And Matt, apologies, but just down-thread from that you said:

“I applaud scheming Otto, but you're running into similar stuff that I did years ago. The heater area under displacer will have greater temperature AND pressure than gas on top, thus requiring some type of 'backwork' to keep displacer down. However, it still might be possible to 'balance' the displacer by vary the higher pressure bottom area vs lower pressure top area (similar sketch).”

I originally thought that the pressure difference thing was just mis-stated, or addressing the aero drag as the displacer moves, but I’m not sure now what you meant. If for instance the displacer is static, why would there be a pressure difference across it?

Thanks guys. These things are important to me regarding my own big-bore moderate temp project.

Bumpkin

[VincentG]

Bumpkin, I was rushing my calcs for sure and should go back over them. But what I was referring to was the compression ration of the power piston displacement v. the total system volume. And by swept volume ratio I was referring to the displaced volume of the power piston v. the displacer piston.

It seemed from my research that this swept volume ratio was important but in my testing this did not seem very critical.

The pressure differential under the displacer was a result of the cam/spring displacer drive system allowing the displacer to make full contact with the hot plate. The trapped air under the displacer would cause the issues Matt spoke of. But in my low buck models, the seal is not tight enough to cause an issue here. With proper machining and materials, I do think this "issue" can be used to a net benefit.

Hope this clears things up. The only thing I know from my testing is that I know almost nothing about how these engines really work.

[matt brown]

The pressure differential under the displacer was a result of the cam/spring displacer drive system allowing the displacer to make full contact with the hot plate. The trapped air under the displacer would cause the issues Matt spoke of. But in my low buck models, the seal is not tight enough to cause an issue here. With proper machining and materials, I do think this "issue" can be used to a net benefit.

Vincent, you're a stud...you got this right off. There's so many details that can slip past for years that can obscure both success and failure. This reminds me of Donald Rumfeld addressing 'knowns & unknowns' where he builds to a climax of unknown unknows.

Bumpkin, did you get it after Vincent's post ? I'm bogged down optimizing a Rider alpha scheme which lead to another amusing alpha scheme. Meanwhile, I think I've figured out an Otto solution. I sat around for 15 yrs trying to resolve the classic pressure drop issue in my old SE design (alternating HP & LP blows) and various Otto input schemes (heat valve in head), but got nowhere until last year when a bunch of new ideas emerged. So many details can slip past for years...

[Bumpkin]

OK guys. Vince, I thought you’d given one ratio: (approx 1 to 1) for compression, and then other ratios, (about 1 to 6 or 1 to 9) for the entire volume, which I can’t see the difference. And Matt, sorry I didn’t see you were talking about the spring tension external of the thermal system. I’ll go away mumbling now…

Bumpkin

[VincentG]

Bumpkin, right so the compression ratio was about 1.1 to 1 while at the same time the swept ratio was 1 to 9 or so. In other words, the displacer moves 9 times the volume of air as the power piston, but of course the displacer does not compress this air in the LTD engine. In the stock engine the displacer moves 60x the volume of air as the power piston.

I think what needs to be determined is the ideal compression ratio and swept volume ratio given a known delta T.

[VincentG]

Thanks Matt I'll have to go find that from mr. Rumsfeld.