Stirlingengineforum.com

An integral is the area between the function's curve and the X-axis, at Y = zero.

The two figures show that the area on a P-V, indicator, diagram is the work produced by going from 1 to 2. The work produced by an adiabatic expansion is less than for an isothermal expansion at T1. Note: this is in reference to the X-axis aka zero, or absolute zero Kelvin, and zero pressure.

The problem with that is, real engines can't cool the working fluid enough to get to absolute zero. They must recycle their mechanical parts at a higher than zero temperature to get a series of work pulses, not just one. Let us say T2. Are you in agreement?


https://en.m.wikipedia.org/wiki/File:Is ... rocess.svg

https://en.m.wikipedia.org/wiki/File:Adiabatic.svg

Nobody wrote: ↑Sat Oct 23, 2021 4:36 am .... Are you in agreement?

I'm not sure where you may be going with this, but from the general drift highly unlikely

When it comes to analyzing Stirling engines, especially running at any speed, any isothermal process would appear to be ruled out, being, by definition, impossibly slow in nature.

I'm also not sure in what way you determine that an engine failing to run at absolute zero constitutes any sort of "problem".

Nobody wrote: ↑Sat Oct 23, 2021 4:36 am. .... The work produced by an adiabatic expansion is less than for an isothermal expansion

True, in theory, you could get more work out of isothermal expansion, in the short run, (if it were actually possible) to the extent that heat is added as the gas expands to compensate for cooling due to expansion, the added heat increases expansion force. however, the result would be, the piston would also have to be FORCED to return or ALL the heat added would have to be removed (and wasted) so that the piston could return.

A rapid adiabatic expansion on the other hand, results in rapid cooling of the gas so that there is a corresponding drop in pressure below atmosphere. The result being that atmospheric pressure contributes to driving the piston back to complete the cycle.

Real engines differ from their ideal theories. It is possible to study theoretically ideal theories using simple discussions. Real engines need real indicator diagrams, numerical integrations and high precision. Those three things exclude anything I can say here about them.

Ideal engines will always describe more work and better efficiency than is possible with a real engine.

Ideal, but unobtainable, thermodynamic processes:

Isothermal expansion can be approached by a very slow steady expansion. How slow that needs to be depends on increased surface area and decreased fluid thickness. That becomes a compromise with increased fluid drag.

Adiabatic expansion can be approached by higher speed. How fast that needs to be depends on reduced surface area and increased fluid thickness. That becomes a compromise between reduced heat influx to the fluid and decreased engine power out.

A properly designed regenerator stores the hot side fluid heat, and returns it after compression is completed. No need to dump that heat to the cold side. The cold side only absorbs the heat of compression, making compression easier. The regenerator is what makes it a Stirling Engine. First patent for the Stirling's.

Adiabatic expansion reversed by atmospheric pressure requires extra work to expand past atmospheric pressure. That work is stored in the MV2 of the fluid and any piston and crank it may have. No crank necessary. No piston necessary.

In regard to your engine diagram, there is a mistake in frame 5. It takes work input to cool the fluid below the cold temperature body, during that stroke. You can't get work out during that expansion.

A regenerator can absorb enough heat to return the fluid to close to the cold body temperature, but never quite to it. And close to the lowest pressure, which will already be below ambient atmospheric pressure. To get temperature below it requires more expansion and more work input against the atmosphere.

That work is supplied by the energy stored in the moving: fluid, piston, diaphragm, and flywheel. It doesn't need a flywheel or crank, or piston, or diaphragm.

Also the timing is off, in that description, when compared to working engines. Unfortunately the drawings are too simplified to explain it well. Or, IOW, you haven't explained it well. Also, need an engine running on your theory for valid peer review. Thanks.

Nobody wrote: ↑Sun Oct 24, 2021 3:02 pm In regard to your engine diagram, there is a mistake in frame 5. It takes work input to cool the fluid below the cold temperature body, during that stroke. You can't get work out during that expansion.

Well, it becomes a little bit complicated. It depends what you mean by "work input". Work input from where and to where? What is doing work on what?

Let's assume a "free piston" engine, though I know the illustration depicts a crankshaft and flywheel, let's dispense with that for the moment. Eliminate some variables to simplify things.

At the beginning of the stroke the gas starts out hot and compressed, then it expands and drives the piston out. At that point the hot expanding gas is doing work on the piston pushing it out. This expansion work continues until the internal pressure equalizes with the outside atmospheric pressure, however, the piston has weight and momentum and continues traveling out very briefly.

How does the gas inside feel about things at this point? Have the gas molecules suddenly stopped moving around and impacting the piston? Do the internal gas molecules "know" that the pressure on the other side of the piston is increasing? (Relatively speaking, the internal pressure is what is actually decreasing)

I tend to think, as long as the piston continues moving out by momentum, providing room for the gas to expand, the gas will continue to expand, continue exerting force on the piston. The work the gas was doing, as far as the gas is concerned, has not been interrupted at all.The gas is still expanding, pushing on the piston.

As an illustration, if ten people are pushing a car up a slight but gradually increasing incline, as some get tired and fall away, the few strong individuals that are left have even more work on their hands. The "pressure" they have to work against is increasing as they all get more and more fatigued.

Likewise as the gas molecules push the piston along, some molecules lose energy and get cold and fall away, as the piston travels, fewer and fewer hot air molecules are left, but some very energetic fast "hot" molecules continue "pushing" all the way to the very end until the piston actually reverses direction. Even then there is still some internal pressure, but the gas has become so exhausted of nearly all energy that now it is unable to overcome the outside pressure and the atmospheric pressure now does work on the piston and on the gas and the internal temperature of the gas soon begins to increase as the piston travels back inward.

First, at the start of the cycle, many hot molecules impacted the piston the piston is driven out and some momentum is stored in the piston. This work causes the gas to loose energy and cool. It is the work done by the gas that causes cooling of the gas. Due to the momentum of the piston that work is able to continue, briefly, beyond the point when the inside and outside pressure have reached equilibrium. At that point the work the gas has to do does not stop or even decrease, it increases. As long as the piston continues traveling outward by momentum, the internal gas is still transferring energy to the piston, loosing energy and cooling untill so completely exhausted of energy it can offer no more resistance to outside pressure By that time, the internal pressure has dropped so far below external pressure that the external pressure drives the piston inward with a force as great as the force that previously drove the piston outward.It is as if the last person left with any strength pushing the car up hill finally gave up and fell away, so the car rolled back down the incline.

In other words, I don't believe I am wrong at all. The depiction is accurate.

...

That work is supplied by the energy stored in the moving: fluid, piston, diaphragm, and flywheel. // It doesn't need a flywheel or crank, or piston, or diaphragm.

I agree with the first part, what you mean by "It doesn't need..." I don't follow. IT? It what?

Gas molecules striking the receding (outward traveling) piston continue to transfer energy to the piston. (Or diaphragm, crank and flywheel)

Well, you have said the same thing, but "expansion" of the gas "against the atmosphere" I would call work OUTPUT, not input

The stored energy in the piston is doing work on the OUTSIDE atmosphere. The "moving fluid" (expanding gas) is still doing work on the piston which does work on the atmosphere.

You say "To get temperature below it requires more expansion and more work input against the atmosphere."

If you had said: "To get temperature below it requires more expansion and more work input against the atmosphere." I would say we are in 100% agreement.

Expansion work against the atmosphere is work output by the gas, not work input, right? unless by "work input" you mean work performed or work done. The gas is pushing OUT against the atmosphere. For a brief moment when the gas has used up all it's energy to push the piston, the weakened cooled gas has just a few hot molecules left and these work very strenuously against the increasing resistance of the piston (atmospheric pressure) until the piston too, loses all of it's stored momentum. Then the atmosphere drives the piston back.

At that last moment, before the piston comes to a stop and reverses course, I believe the internal gas temperature snaps suddenly to a very low temperature, the energy (added heat input energy in the gas) having been completely exhausted.

Also the timing is off, in that description, when compared to working engines. Unfortunately the drawings are too simplified to explain it well. Or, IOW, you haven't explained it well. Also, need an engine running on your theory for valid peer review. Thanks.

I hope all the above explains it a little better.

Existing Stirling engines are, IMO, already running as described and as illustrated.

To say: "To get temperature below it requires more expansion and more work input against the atmosphere." makes no sense to me.. the gas doing expansion work against the piston/outside atmosphere cannot be work "input". That seems like a contradiction.

In a sense, it could be said that the momentum of the piston is doing additional work that allows the gas to expand, much like "pulling a vacuum" to expand and cool a gas.

I assume what you mean is that there needs to be some additional "outside" force operating to drive things to get the piston past the equalibrium point to cause the gas to cool, because "refrigeration requires work input". As "Everybody knows".

That "outside work" is provided by the momentum of the piston imparted to it by the expanding gas. Likewise if there is a flywheel with stored momentum.

Some of the momentum imparted to the piston and flywheel by the expanding gas acts to allow the gas to continue to expand and cool past the point of equalibrium, by forcing the piston to continue traveling past that point so that the gas can continue to expand.

All this is an attempt to explain an event that takes place in a fraction of a second.

That's my theory anyway.

So far, you haven't convinced me that it is wrong, inaccurate, or needs revision.

Of course, not every heat engine that might be called a Stirling engine operates in exactly this way. I'm mostly interested in how to explain a thermal Lag or free piston type Stirling operating with no crank or flywheel. But adding back a crank and flywheel doesn't necessarily alter the bare bones working principle.

As far as "You can't get work out during that expansion.", all I can say is - you can and do.

"To get temperature below it requires more expansion and more work input (output) against the atmosphere."

How could there be expansion of the gas without that expansion doing work on the (outside) atmosphere it is expanding against? How is that not work output?

If there is no crank or flywheel there is nothing outside the piston to provide work input, other than atmosphere. For that to do work, the internal pressure has to SOMEHOW drop below atmospheric pressure.

Think of it as; work/energy into the piston and work out of the piston. Work into the piston speeds it up. Work out of the piston slows it down.

W=1/2mV^2. kinetic work

W=Fd potential work

Work into the working fluid from the piston slows the piston.

The working fluid is in opposition to the atmosphere so the two need to be subtracted to get net force and direction.

Work out of the system, generator, slows the piston.

Work into the system, finger pushing, displaces or speeds up the piston, or both.

If no work is removed from the system the engine will put energy into the piston until friction equals heat in minus heat out, idling or revved up, efficiency zero.

Nobody wrote: ↑Sat Nov 13, 2021 7:56 am Think of it as; work/energy into the piston and work out of the piston. Work into the piston speeds it up. Work out of the piston slows it down.

W=1/2mV^2. kinetic work

W=Fd potential work

Work into the working fluid from the piston slows the piston.

The working fluid is in opposition to the atmosphere so the two need to be subtracted to get net force and direction.

Work out of the system, generator, slows the piston.

Work into the system, finger pushing, displaces or speeds up the piston, or both.

If no work is removed from the system the engine will put energy into the piston until friction equals heat in minus heat out, idling or revved up, efficiency zero.

Sorry, but I can make no sense out of this.

First of all "The piston" itself, isn't really doing any work at any time. It might just be considered a kind of dummy that gets pushed and pulled, but it, by and of itself does nothing. The heat/energy enters the "working fluid" (air or gas inside the engine). It's called "working fluid" for a reason, I think.

The piston does not slow down because of work or energy being taken out of it, a little momentum at times, but in general, the piston is shoved around by pressure variations. It might even be a diaphragm, in which case, any possible "momentum" is pretty much non-existent.

The highlighted part sounds reasonable.

I don't know what you mean by "work into the system". The input is heat.

You are equating heat input with "work" or kinetic energy?

OK, but I'm not sure what point you are trying to make, if any.

If you are still trying to dismantle or debunk my earlier diagram, well earlier you wrote:

In regard to your engine diagram, there is a mistake in frame 5. It takes work input to cool the fluid below the cold temperature body, during that stroke. You can't get work out during that expansion.

Sorry, but you are wrong. It is called adiabatic expansion.

When a gas expands and does work driving a piston the "working fluid" looses energy and there is a drop in temperature. This is the basis for many processes and is well documented and experimentally verifiable.

It is not an intuitive process, and not easy to understand, which was why I started the thermodynamic discussion ten years ago, it is nevertheless true.

Your statement "you can't get work out during that expansion" is wrong. Work out during expansion is what causes the loss in energy that results in the drop in temperature.

This is a quite interesting experiment by "Blade Attila" a YouTuber who is always tinkering around with Stirling engines.

He wanted to see what would happen with a thermal Lag engine if driven to operate like a heat pump.


https://youtu.be/2CnNOY1OVhc


This result, the apparent direction of heat flow when driven, is seemingly not in the same direction as it would be with other Stirling engines. (Or is it?)

Most Stirling engines have a directional aspect. That is, the engine, when running as an engine on a heat source, turns in a certain direction and heat "flows" (apparently?) In a certain direction. Or so it is generally believed or assumed.

Many of the reciprocating "free piston" type Stirling engines capable of running without a flywheel, though, can run just as well in either direction (if equiped with a flywheel). Without a flywheel, there is no actual "direction of rotation", So...

In a thermal Lag type engine, such as depicted in the above experiment, is the effect the same if the piston is driven inward by atmospheric pressure or by a linear motor to compress the gas?

What about expansion?

Is the effect the same if the piston is pulled out by a linear generator, or by the force of it's own momentum?

Why doesn't the heat input side get cold when the thermal Lag is driven as a heat pump? And why does the "sink" get cold?

Thanks to Blade Attila for having enough curiosity and ingenuity to run this unusual experiment!

The above video from Blade Attila demonstrates a thermal Lag engine working as a heat pump.

I've theorized that this heat pump behavior (cooling of the "driven end" as it is called in the video, the open end of the test tube) takes place during the normal operation of the engine. This cooling is so slight, it would never become apparent, though, without insulating the cold side of the engine, due to being exposed to the surounding ambient heat.

I made this animated gif back in 2012


http://calypso53.com/stirling/amb_eng_anim.gif


I forgot about having posted this to a "free energy" forum back then, but just came across it a few minutes ago.

https://overunity.com/13159/teslas-ambi ... #msg348512


The fluidine trap could just as easily be a diaphragm and might not even be necessary. Possibly the cold box could be pressurized?

I can't really say it would definitely work as I haven't tried it, but the Blade Attila heat pump experiment gives me a little more incentive.

The idea, of course (or maybe it might not be immediately clear) is to chill the interior of this box, possibly by pouring in a bit of liquid nitrogen or something, using ambient heat as the heat input at the end of the engine sticking out of the cold box.

If the engine really does act as a heat pump, it should keep the interior of the box cold without the need for any supplemental refrigeration.

At any rate, I don't see any reason why it shouldn't at least start up and run for a while, but for how long?

Another project/experiment on my to-do list.

Tom said, "Sorry, but you are wrong. It is called adiabatic expansion."

That "adiabatic expansion" requires work into the ambient atmosphere. The ambient pressure is higher than the pressure on the opposing side of the piston/inside. That Work comes from loss of piston momentum, or it comes in from the outside. It is the shaded area on the Senft diagram.

Expanding gas always does work, it is just less than the work needed to push the piston back into the ambient pressure. The ambient pressure is receiving work/energy from the piston. Hence, the piston loses stored work/energy, momentum, velocity, all those things. The piston slows down to the reversing point.

That is kinetic energy that could be extracted if it weren't lost cooling the working fluid. Of course, it is put back in during the return stroke, with the heat too, plus some additional frictional heat and some heat absorbed from the ambient. As shown by the temperature verses crank diagram.

Nobody wrote: ↑Fri Nov 26, 2021 8:51 am ...

That is kinetic energy that could be extracted if it weren't lost cooling the working fluid

The principle involved is: The cooling of the working fluid is a result of kinetic energy being extracted from the working fluid during expansion.

You have my sympathies, because this is a difficult concept to grasp and I struggled with it myself for quite a long time and I found it hard to accept. Trying to understand it is what led directly to my various experiments. Information on this subject is also rather difficult to find, but not impossible, you just have to look for it.

The first part of this video explains it rather well.


https://youtu.be/PMKPZuCj9a0


I sited several references near the begining of my thermodynamics thread, such as:

Another way to cool a gas is to have it do work adiabatically against a piston in an engine, and this has no temperature boundary like the Joule-Kelvin effect.

http://mysite.du.edu/~jcalvert/phys/helium.htm

The temperature drop is equivalent to the work extracted, by the first law of thermodynamics.

So to say: "That is kinetic energy that could be extracted if it weren't lost cooling the working fluid" doesn't really make sense.

The kinetic energy is not "lost cooling the working fluid".

The working fluid cools as a result of its kinetic energy having been transfered to the piston as "work extracted". The heat in the gas is converted into work to drive the piston, which loss of kinetic energy by the gas (in the form of work performed by the gas in pushing the piston) results in cooling of the gas.

The? Principal involved?

There are two principles, or processes. One is expanding the the working fluid. One is compressing the ambient. Energy on both sides. Ambient stays the same pressure for the entire stroke. Working fluid decreases through the entire stroke. At one point it becomes lower than ambient. The rest of that stroke the piston slows until it reverses.

Nobody wrote: ↑Fri Nov 26, 2021 5:44 pm The? Principal involved?
...

The principle involved in frame 5 of my engine diagram.


Previously you wrote:

In regard to your engine diagram, there is a mistake in frame 5. It takes work input to cool the fluid below the cold temperature body, during that stroke. You can't get work out during that expansion.

The diagram is an accurate representation of my belief, my observation, regarding how a Stirling engine ACTUALLY operates in many circumstances, or could operate assuming it is not supplied with excess heat and is properly load balanced, so as to obviate the necessity for dumping excess heat to a sink.

You may not agree or accept my conclusions, but it is not a "mistake" in that it does accurately portrays what I believe are accurate observations and conclusions.

After 100+ years of assuming heat engines work by having heat pass through, it is difficult to see or comprehend that heat goes into a heat engine but does not always come out the other side but can be transformed into mechanical motion.

In other words a heat engine produces work by extracting energy, which results in a kind of refrigeration.

You do get work out. You also get cooling. The "work out" is the same energy that went in as heat.

I don't expect everyone to accept that, but hopefully you can see my point.

Frame 5 of the diagram is not a "mistake", it is a representation of the principle of expansion cooling as also described and illustrated in the above video.

Whatever work the piston itself might do to expand and cool the gas, due to momentum, is only second hand. The energy from the expanding gas is what put the energy into the piston in the first place.

The only source of energy is the heat supplied. When any other form of energy appears, pressure, momentum, velocity... heat disappears, and cooling results.