100% efficiency (+) it it possible?

[Tom Booth]

Anyway, what I came to the forum for today is,

I was driving to a nearby city today to pick up a stove, and being a long drive, I had time to ponder and think..

Anyway it dawned on me the reason why, or rather a theory.

Why it is necessary that a Stirling engine have two dissimilar heat sources.

In other words, instead of looking at it as a "Hot" heat source and a "Cold" sink, the reality is, what we have is two sources of kinetic energy, two sources of heat, one, on an absolute temperature scale, only slightly hotter than the other

More heat is needed for the power stroke because the "extra" heat is needed for adiabatic expansion, or in other words, the extra power is used for cooling, adiabatic expansion being a kind of refrigeration.

This extra energy, used for cooling, is what results in the lesser heat source, in the form of atmospheric pressure, being able to complete the cycle

[Tom Booth]

Put another way,

There is a volume of air in the engine.

Enough heat has to be added suddenly enough that the piston is driven out with enough force that the gas is expanded enough to cool sufficiently that the pressure drops enough below atmospheric pressure that the piston is driven back.

There is not a flow of heat from hot to cold.

There are two heat sources playing against each other much like tennis players batting a ball back and forth.

I've been looking at this for a long time but had not been able to conceptualize it in just this way.

I knew the heat did not "flow" to the "sink" so why should a temperature difference be necessary?

In effect, the hotter heat source needs to have enough energy to cool the working fluid enough for the colder heat source to be temporarily hotter.

Of course, "hot", and "pressure" are just terms applied to the same thing, kinetic energy.

[Tom Booth]

If this is true, then it may be thought that all the heat of expansion is "used up" in cooling the engine enough to make atmospheric pressure effective in powering the engine on the return stroke. So this greatly reduces the efficiency of the engine,

But, while that statement is true, the initial heat used to expand the gas is, in a sense, "used up" so as to cool the gas, so that atmospheric pressure can be effective in powering the engine, the manner, or means by which the gas is cooled is by expansion of the gas against a load.

That is, the refrigeration of the gas is the consequence of the conversion of heat into work output.

[matt brown]

sq cycle.jpg (58.59 KiB) Viewed 4818 times

There is not a flow of heat from hot to cold.

There are two heat sources playing against each other much like tennis players batting a ball back and forth.

I've been looking at this for a long time but had not been able to conceptualize it in just this way.

I knew the heat did not "flow" to the "sink" so why should a temperature difference be necessary?

In effect, the hotter heat source needs to have enough energy to cool the working fluid enough for the colder heat source to be temporarily hotter.

Of course, "hot", and "pressure" are just terms applied to the same thing, kinetic energy.

Tom - you almost got it ! An ideal engine normally has no 'flow' of heat from hot to cold, since that would comprise a thermal short (a waste of heat that produces no work). This whole flow buzz is like a lot of stuff in life, an analogy that's crude and bogus. However, the main thermo buzz that causes problems is using the word heat which is a form of energy, but often conveys temperature. Engineer slang is loaded with analogies, but a physicist would never blunder like this. I checked Reflections, and Carnot's heat flow reeks of caloric theory amongst other (now) major errors. Anyways, returning to energy, heat engines operate between a high energy state (or process) and a low engine state (or process). Consider the Stirling PV above which we know has 2 isotherms, and here's where basic thermo comes in, and Alphax & Nobody dropped the ball. An isotherm has no change in internal energy, or in thermo notation an isotherm has dU=0. So, 'riding along' (yep, slang) each isotherm...no matter how 'long'...has no change in internal energy (for an ideal gas). So, if the Stirling cycle in PV above was expanded (think larger vol ratio) while the temperature ratio remained constant, this modified engine would produce more work (longer compression & expansion) from the SAME temperature ratio, but dU=0 of each isotherm would remain constant. despite more energy in, and more work out.

Now, if pt 2=300k, and pt 3=600k, orig PV (above) is a sq cycle (I love these for their simplicity). Now, let's further consider this is a closed cycle, and we can almost tell everything about this cycle. MOST IMPORTANTLY, since isotherms mean dU=0, we know that ALL the 'heat' that's added during expansion = Wpos (because internal energy did not change) AND we know that ALL the 'heat' that's removed during compression = Wneg. And the brainiacs are right that this cycle has ideal eff =.50 since the thermal ratio = 2 (Tlow=300k, Thigh=600k) but they ignored pointing out why !!! It's so simply it probably slipped their minds...dT=dU such that the temperature ratio equals the internal energy ratio.

Now I feel safe to add - despite obvious - that previous no heat flow from hot to cold in ideal engine...normally...does not apply to Stirling due to regen process which requires such.

Once your onboard internal energy vs 'heat', the next thing to grasp is how heat capacity at constant volume (Cv) and heat capacity at constant pressure (Cp) effect combining processes into cycles. Meanwhile, the PV above indicates that when a 'T2' cycle (here 300k & 600k) we (should) now know that 1 unit of Wnet AND 1 unit of Qout (waste 'heat') results from 2 units of Qin, but to discover how much 'energy' is ideally recycled during regen will require a trip over to my recent post under "regen issues...how much heat" which exposes just how 'top heavy' (more slang) regen cycles are.

Tom - it's really sad that no one in any forum could clarify eff from simple example, especially since Stirling Cycle is ripe for proof and you're well familiar with it. You're right that the academic mind set would rather flaunt their superiority than help out. 20 yrs ago, I had some interesting chats with some living legends, and when I was first trying to reduce regen energy to cycle ratio, even Urieli told me "sorry, but there's no back of the envelope answer". For the academics, it's all about grant moola propped up with goofy calculus & studies, and excluding outsiders from looking behind the curtain.

Note that their calculus often takes mathematical expressions and reduces them to simple conclusions that are correct, but this 'trip' often hides the journey and begs questions. An excellent example is how/why the thermal eff of (simple) Otto & Brayton Cycles correspond (are proportional) to their adiabatic compression ratio.

[Tom Booth]

....AND we know that ALL the 'heat' that's removed during compression = Wneg. And the brainiacs are right that this cycle has ideal eff =.50 since the thermal ratio = 2 (Tlow=300k, Thigh=600k) but they ignored pointing out why !!!

Hope you don't mind, there is, or may be elements in your post I can generally agree with, the above quoted portion though, in particular...

... not so much.

Heat is not "removed" during adiabatic compression.

Because of the timing in a Stirling engine, heat introduction begins at the same time as effective compression.

The result is a sudden concentration of heat in the working fluid. The heat of compression combined with heat addition from the heat source.

No heat is "removed". There is, for one thing, nowhere for the heat to be removed to, as it (the working gas, along with it's heat) is concentrated/compressed by adiabatic (no heat exchange) compression into the hot cylinder, or hot end of the displacer chamber.

The compression is anything but isothermal (without any temperature change) , though some heat continues being converted to mechanical action or "work", and may, in a sense be "removed" in that way: as positive work output to a load.

The idea that the compression stroke is isothermal is derived from, (or a component of) the Carnot water wheel mythos.

The work during compression is not "negative".

The work of compression, carried out by atmospheric pressure, contributes energy to the build up of the heat of compression in the working fluid, which build up of heat then results in the subsequent rapid expansion during the power stroke..That heat of compression is not lost at all.

In addition, there is a positive "work" contribution made to the rotation of the crank/flywheel if present, or to the linear alternator with it's electrical output if that is used for output.

That the flywheel momentum or some other "stored" energy is used to return the piston making this work "negative" simply does not correspond with observable facts

As a result of adiabatic expansion and cooling of the working fluid during the power stroke) the working fluid looses energy and "contracts" so that the piston is "pulled" inward or rather "pushed" inward by outside atmospheric pressure.

Contrary to conventional thinking, and what is currently being taught, there is no isothermal "heat rejection" to the atmosphere or "sink" during compression. Quite the opposite really.

I still stand by the general accuracy of the diagram I posted to another thread quite a while ago:

Resize_20220224_062218_8222.jpg (145.46 KiB) Viewed 4809 times

As depicted at #3 in the diagram, heat is introduced due to the movement of the displacer at the same time that there is heat build up due to the heat of compression. Both sources contribute to the subsequent power stroke.

No heat is lost. No work is wasted.

[Nobody]

Your diagrams have insufficient detail and the timing appears erroneous. Please supply a graph showing the precise timing of the depicted apparatus.

[Tom Booth]

It's an observation based on a quite ordinary Stirling timed to standard specifications: displacer about 90° ahead of the piston.

You can get one of these cheap engines and make your own observations.


https://youtu.be/Cu0IdJbyUfY


In reality, however, I have found that the timing can be varied or adjusted quite widely, and the engine will still run. In slightly more expensive models that allow timing adjustments.

The timing advance should, actually, vary, if possible, according to RPM, heat input, load etc.just as an IC engine.

My diagram is certainly well within the general ballpark and not much, if any different than the standard, generic 90° timing advance.

[Nobody]

From the Lenor thread. By Tom Booth

Resize_20220309_050739_9244.jpg (119.68 KiB) Viewed 4699 times

That graphic adds confusion. Are you guessing at where your point 5 is, or is that a stroke depiction for diagram 5? Stroking from 5 to 5?

[Nobody]

Here is an example that may help your engine graphics timing depiction.

By Nobody.

IMG_20220314_220836814~2.jpg (134.88 KiB) Viewed 4698 times

Sorry about the roughness of the sketch.

[airpower]

Resize_20220309_050739_9244.jpg

That graphic adds confusion. Are you guessing at where your point 5 is, or is that a stroke depiction for diagram 5? Stroking from 5 to 5?

.
That graphic suggest that 50% of the engine is a atmospheric pressure engine.

"Compression work by atmospheric pressure"

Atmosphere does not do any work, it will get back to equilibrium asap after the creation of imbalance.
That graphic also means when at rest part of the toy LTD engine is below atmospheric pressure,

Live feed of the cycle (Atm is 1 (101325 Pa) what is roughly 1 Bar (100,000 Pa, 111 m above sea-level))
viewtopic.php?f=1&t=5312#p16785

Negative pressure can be found in the galaxy next door.

[Nobody]

Yes, along with a Dark Emitting Diode.



Otherwise know as a DED.

[Tom Booth]

Resize_20220309_050739_9244.jpg

That graphic adds confusion. Are you guessing at where your point 5 is, or is that a stroke depiction for diagram 5? Stroking from 5 to 5?

#5 in my diagram depicts the piston fully extended at the end of the power/expansion stroke.

There is a "dwell" period, where the piston lingers virtually motionless (constant volume) as the flywheel continues to rotate and the general cycle continues.

So there is a start and a finish to this "dwell" when the piston is motionless and about to reverse direction (from expansion to compression).

Moving clockwise around the cycle in the PV plot, the first 5 represents the begining and the second the end of this "dwell".

As the piston is basically motionless at that point, the #5 "snapshot" in my diagram is meant to represent the entire "dwell" period 5 to 5 in the PV plot.

[Tom Booth]

.
That graphic suggest that 50% of the engine is a atmospheric pressure engine.

"Compression work by atmospheric pressure"

Correct.

Atmosphere does not do any work, it will get back to equilibrium asap after the creation of imbalance.

Your statement is self contradictory.

To "get back to equilibrium", atmospheric pressure will need to do work. (push the piston back in, driving the engine).

That graphic also means when at rest part of the toy LTD engine is below atmospheric pressure,

When the piston is at "dwell", at the extremity of the expansion stroke, yes, the internal pressure of the engine is below external atmospheric (or buffer) pressure.

The "engine" as a whole however is in continuous operation.

Negative pressure can be found in the galaxy next door.

As has been made plane, repeatedly, "negative pressure" is simply a way of saying "partial vacuum" or an internal pressure below atmospheric pressure.

The internal pressure is "negative" when the external pressure is "positive".

Likewise, as with "negative" and "positive" terminals on a battery or electrical supply "negative" does not literally mean an absence of electricity.

"Negative" is a common English word, the general meaning of which is "less than": ".Of or relating to a quantity, number, angle, velocity, or direction in a sense opposite to another of the same magnitude indicated or understood to be positive." https://www.thefreedictionary.com/negative

This is accepted usage in the English language, especially in scientific/engineering terms

If the connecting rod can be removed from the flywheel of an engine, and the piston is still able to return, it becomes obvious that the reason is not due to "stored energy" in the flywheel, but rather because of negative internal pressure.

Examples:

https://youtu.be/HUWt3YrxoB4


https://youtu.be/ErlvMZI0tlA


https://youtu.be/iOs3BADFeKI


Actually, in my own experiments, the piston returns much more easily and quickly when not required to drag the entire engine apparatus; connecting rod and flywheel; along with it.

If you have some alternative explaination for why the piston can return without stored energy in a flywheel, other than that the engines internal pressure has momentarily, at the end of the expansion stroke, become "negative" (or less than) in relation to external (positive) atmospheric pressure, I'm sure everyone would love to hear your theory.

[Tom Booth]

The demonstration displacer setup in the following video is not producing any "work" output, so the explaination for why a flywheel is not needed in a Stirling engine is incomplete, nevertheless, the video is interesting and instructive.

Heat being converted to work during expansion would be another reason why a flywheel is not needed in addition. (Thermal lag, lamina flow, thermal acoustic, and some "free piston" type heat engines often have no displacer and sometimes no regenerator)

https://youtu.be/LjjvIl1BfbQ

[airpower]

............
Likewise, as with "negative" and "positive" terminals on a battery or electrical supply "negative" does not literally mean an absence of electricity.

"Negative" is a common English word, the general meaning of which is "less than": ".Of or relating to a quantity, number, angle, velocity, or direction in a sense opposite to another of the same magnitude indicated or understood to be positive." https://www.thefreedictionary.com/negative

This is accepted usage in the English language, especially in scientific/engineering terms

........

When you have more Electrons than Protons you have a negative number.


Less than zero


Modern Science is doing space walks at pressures of very very roughly about
0.00000000000000000000088725345303 Pa
(still not negative pressure)