The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Discussion on Stirling or "hot air" engines (all types)
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

There are additional factors not indicated or apparent looking at a typical PV diagram.

Real engines have actual physical mass, which has inertia and momentum.

Reciprocating engines have pistons which have weight, velocity and momentum.
pv_velocity_momentum.jpg
pv_velocity_momentum.jpg (226.8 KiB) Viewed 280 times
From TDC (far left) to midway towards BDC the piston is accelerating in velocity and storing potential energy as momentum.

Likewise on the return stroke. Atmospheric (or buffer) pressure accelerates the piston back towards TDC.

On the return stroke there is only a relatively brief upturn in pressure above atmospheric just before the piston reaches TDC.

The velocity of the piston with its stored force of momentum is enough to carry the engine through this moment of high compression at TDC, often assisted by the fact that the rotation of the crank has a substantial leverage or "mechanical advantage" at this point to effect high compression. After TDC the piston is free to move into the expansion stroke.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Fri Apr 05, 2024 6:53 am ....the big problem with those points is that Tom doesn't believe in 'constant T volume change', neither expansion nor compression. He believes in adiabatic. And he thinks adding an electric generator will change that process. Granted, real processes are somewhere between adiabatic and constant T, unfortunately adding heat during a Stirling Expansion is the only way it will work effectively. The pressure of expansion must be higher than the pressure of compression, and the only way to accomplish that is by adding heat and removing heat. It needs to be for a full cycle. He seems to always narrow the discussion to single processes ignoring the rest of them. Puts buffer pressure on for the return stroke, leaving it off for the forward stroke, etc...

I may start a tread on heat transfer and why there is no such thing as Adiabatic.
Your (miss-) characterization of my views, opinions or observations is specious.

I tend to emphasize the adiabatic expansion phase because you fellows tend to discount it entirely.

There is most certainly heat addition during expansion. To imply that I don't "believe" heat needs to be added to effect expansion of the working fluid is disingenuous straw manning.

Ideally, though, heat input should be discontinued as early as possible allowing the later part of the expansion to be carried out adiabatically.


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heat_input-expansion.jpg
heat_input-expansion.jpg (208.73 KiB) Viewed 274 times

Adiabatic expansion without continued heat input allows the heat added to accelerate the engine at the beginning of the expansion to be utilized (converted to "work" output). The "internal energy" lost to "work" reduces the pressure.
"Adding an electric generator" or other external load increases the work output accelerating the cooling of the working fluid which brings the pressure down below atmospheric pressure more quickly, to make the return stroke more forceful.

The Carnot/caloric theory did not recognize the conversion of heat into work at all. Carnot and apparently Kelvin as well, believed that ALL the heat that entered into the engine would have to be removed

Later theoreticians conceded that "some" heat is converted, concocting the unsubstantiated so-called "Carnot Limit" which has no empirical basis and no sound or logical mathematical basis.

In reality ALL the heat added to expand the working fluid is converted to work output during the course of the expansion stroke leaving the piston at BDC and the working fluid at a reduced pressure with the result that atmospheric pressure accelerates the piston back towards TDC.

The heat addition at the early part of the expansion is like the push given to a child on a swing which carries the engine through the remainder of the cycle.

The piston returns completing the cycle in much the same way a swing returns to receive the next push.

Heat addition at, and to a limited extent after TDC is the requisite "push" given to the engine to keep it moving.
Stroller
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Stroller »

All the while after TDC, while the piston is accelerated to midway, and then decelerates to BDC, the volume is increasing, and the pressure will therefore start to drop due to that, plus the displacer having already moved 90 degrees ahead to shift the working fluid away from the heat source and shield it from further heating.

But heating of the working fluid needs to recommence as soon as the displacer starts to shift it back from the cold end, which happens before the piston reaches BDC. So I'm not sure how long a 'dwell' of no heat input Tom thinks would improve the dynamics.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Stroller wrote: Sat Apr 06, 2024 4:23 am ...
But heating of the working fluid needs to recommence as soon as the displacer starts to shift it back from the cold end, which happens before the piston reaches BDC. So I'm not sure how long a 'dwell' of no heat input Tom thinks would improve the dynamics.
You are,... sort of right, if by "BDC" you mean TDC.

Generally the traditional or standard displacer motion would "to shift it (the working fluid) back from the cold end" beginning the introduction of some heat before the completion of the compression stroke and about 90° before reaching TDC (not BDC) though in most LTD engines tbe piston is inverted so FUNCTIONAL TDC is actually when the piston is all the way down.

To illustrate that I would have had to wrap the words "Heat Addition" more sharply around TDC (left side or minimum volume on the PV diagram) I did not do so simply because the simple photo editor on my phone is not that capable.

Exactly what the "ideal" dwell might be I have not precisely worked out in practice.

The main idea is to reduce "mixing" of hot and cold air as much as possible. But recent indications that there may be influence from radiant heat more than convection has me rethinking things.

I think VincentG's experiments with extended "dwell" have shown considerable improvement in his engine performance

You may want to ask VincentG, I would consider the dwell time an area of active research, but for me mostly an area for future research.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Stroller wrote: Sat Apr 06, 2024 4:23 am ..... So I'm not sure how long a 'dwell' of no heat input Tom thinks would improve the dynamics.
Imagine someone being pushed on a swing.

The "dwell" would be the entire time the swing is not being pushed.

Ideally the heat input should be variable so that it could be extended when additional power is required.

Ideally the heat input should perfectly balance the load (work output).

Also "dwell" where heat input is cut off does not, or should not mean that the working fluid is subject to some cold or outer "sink" for removing heat, rather the "cold side" need not exist at all (made of non-heat conducting material).

This has probably been best demonstrated by the little magnetic LTD with an acrylic body covered by an aerogel blanket.

https://youtu.be/lx1tet8aHJU?si=Mx8_BqdWE2ZSsQpL


https://youtu.be/F8P_g8Vwoz0?si=XLrb62GKAgYW2eoB
Stroller
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Stroller »

Thanks Tom, I'll back off and have a think about this. Apologies for crashing in on this interesting thread!
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Stroller wrote: Sat Apr 06, 2024 3:54 pm Thanks Tom, I'll back off and have a think about this. Apologies for crashing in on this interesting thread!
No need to "back off", the forum is for discussion and debate.

BTW I don't necessarily think a cold side is necessarily "bad". It can often be helpful and increase power. I t just doesn't seem to be necessary as some kind of "law of the universe.". I've had many engines that at least appear to continue running just fine, if not actually better without it.
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

That PV diagram is for an engine that is producing 0.0034 Watts, or about 0.0000046 HP.

A quote from Jake Bobowski;
The work done per cycle by this stirling engine is 0.46 mJ. The power of the stirling engine is given by the work divided by the time for each engine cycle. Equivalently, the power is the product of work and the frequency. P = W × f = 3.4 mW, which corresponds to 4.6 micro-horse-power. The power of a typical car engine is 120 horse power, 26 million times more power than this Stirling engine!
https://cmps-people.ok.ubc.ca/jbobowsk/ ... sults.html

https://cmps-people.ok.ubc.ca/jbobowsk/

Question, How long will an engine take to heat up, a 4" diameter 1/8" thick aluminum disk, from heat rejected, one degree C, that is producing only 4.6 micro horsepower, from a100 K temperature difference, how about 50K? Would it even heat up enough to measure? Would the aluminum disk cool faster than it heats up, effectively reading zero temperature change, even rejecting 90% of input heat? 0.0306 Watts?

LTD engines run on such little energy it requires very sensitive, accurate, and precise diminutive measuring tools. Are tools even available for such measurements? Micro Kelvin?

Engines that produce little power produce little rejected heat. Conclusion: Inconclusive. AKA, try again.
VincentG
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by VincentG »

Just thought it was worth pointing out that the LTD in the link above has to be about the worst performing LTD they could have picked for that test. Its power piston volume is roughly .6cc and the total pressure swing is only .1psi?! For comparison the LTD in my tests is 1.6cc and total pressure swing is over 1psi on hot water.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Sun Apr 07, 2024 6:01 am ....
Engines that produce little power produce little rejected heat. Conclusion: Inconclusive. AKA, try again.
I thought according to thermodynamics engines that produce more power (convert more heat to work) reject less heat.

Corollary: engines with low power (convert less heat to work) "reject" more heat.

Anyway feel free to set up your own experiment. I wish someone would.
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

That is totally awesome.

W=PV

Averaging that for Jakes engine:
0.0034 = 0.6•0.1•Factor

Rearranging:
Factor = 0.0034/(0.6•0.1) = 0.05666666666

Factor multipled to your engines parameters:
0.0566•1.6•1 = 0.0906666666 Watts

Or about 26.66666 times more than for Jake's.

0.090666666 if 90% of the heat is rejected:
9•0.0906666666 = 0.816 Watts rejected.

How long will it take to heat up your cold plate one degree C? How many Watts of heat would be rejected to the atmosphere from your cold plate for that one degree, this cooling it?

I point this out to stress the importance of either a PV/Indicator diagram, and or, a dynamometer measurement, also graphed, preferably both. These little engines pump through so little heat, because, they absorb so little, measured it becomes obvious. Unmeasured it becomes a bafflement trying to figure out these things.

The following is a thread by gitPharm01, that is a translation for a YouTube video for the design and construction of a LTD Stirling that produces 100 Watts. It is given as an example of how large LTD Engine's need to be to generate more than just milliwatts.

viewtopic.php?t=5513

That engine would probably be rejecting 300+ Watts, but the cold plate would be much much larger and massive. A very impressive build and design process. It probably needs a supply of 1 kW. 1 m^2 solar panel
VincentG
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by VincentG »

To be clear I was not arguing with your results. Just pointing out that the engine used in the test was not a good choice with a pv graph so small.

Off topic I do wonder though how the idealized Stirling engine pv diagram would look overlayed on the egg shaped LTD diagram. Unless one parameter is severely skewed or shrunk, it looks like the actual pv diagram has much more area than the idealized.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

One thing usually not considered is that the "temperature" of the working fluid, or any gas, is supposed to be the average kinetic energy of the gas particles.

So, in a running engine with a piston rapidly retreating down the cylinder which particles are impacting the piston?

Presumably the hottest particles are impacting the piston most, these lose energy and fall away as additional recently heated particles take over to drive the piston

It is likely a relatively few hot particles that have contacted the hot plate are responsible for creating the "pressure" that drives the engine. Likely much of the working fluid is never heated at all, and at least some of the particles impacting the piston return to the "pool" with less than the average kinetic energy.

So, it could be imagined you are in a way, shooting bullets at the piston (the hot particles) through a cloud of particles at average (ambient) temperature.

So, perhaps you could have an expanding gas that has a high partial pressure from the relatively few extremely hot particles that drive up the "average" temperature but as these few very hot particles drive the piston they quickly loose energy so that by BDC the "average" temperature and/or pressure has fallen below ambient.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

VincentG wrote: Sun Apr 07, 2024 10:25 am To be clear I was not arguing with your results. ...
We cross posted, so my previous responses were intended for "fool", though I would certainly appreciate any data gathering from additional experiments/experimenters.

Oddly, and rather unique in the annals of science, it seems any research that questions this centuries old "Carnot Limit" theory is considered taboo.
matt brown
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Fool wrote: Sun Apr 07, 2024 6:01 am
Question, How long will an engine take to heat up, a 4" diameter 1/8" thick aluminum disk, from heat rejected, one degree C, that is producing only 4.6 micro horsepower, from a100 K temperature difference, how about 50K? Would it even heat up enough to measure? Would the aluminum disk cool faster than it heats up, effectively reading zero temperature change, even rejecting 90% of input heat? 0.0306 Watts?

LTD engines run on such little energy it requires very sensitive, accurate, and precise diminutive measuring tools. Are tools even available for such measurements? Micro Kelvin?

Engines that produce little power produce little rejected heat. Conclusion: Inconclusive. AKA, try again.
Hey Fool, you've been playing with numbers the hard way, see my recent response on another thread:

http://www.stirlingengineforum.com/view ... 925#p21925

For any LTD with ambient compression, the DP/PP volume ratio must equal the cycle thermal ratio, so even 50k spread is a no-go. Instead, the heat passing thru common LTD is like baby's breath...

Modern accounting would reduce all LTD values to zero, written off as rounding errors where heat in = 0, heat out = 0, work out = 0, case closed.
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