if stirling engine is driven as reversed, does it work as cooler?

[Tom Booth]

You constantly throw in this red herring of liquefaction. As if gas must be cold enough and under high enough pressure to liquify before it can "contract" at all.

Personally, I like building and experimenting and trying to improve the performance of these model Stirling engines, and reporting to others that may have an interest.

I would not know about or care about all this "Carnot Limit" total BS, but the 2nd Law fanatics like yourself, keep bashing me over the head with it, day after day, year after year.

I have, IMO a little LTD Stirling that looks like it's getting 99% efficiency, or at least, less than 1% of the heat applied is measurable at the "cold reservoir" where the Carnot efficiency formula predicts there should be at least 80% of the heat flowing through.

I can tell you there is nowhere near that much heat "flowing through" ANY Stirling engine that I have ever examined.

The entire Carnot theory of what makes a Striking engine operate, by "transporting" heat is wrong. It only looked that way, and is basically true for steam engines because steam passing through a steam engine carries "latent heat of vaporization", and deposits "heat of condensation" in the condenser.

There is no phase change in a hot air engine, and so no "latent heat".

A Stirling engine does not operate by "transporting heat" from a "higher reservoir" to a "lower reservoir'.

It looks more like a Stirling engine works by pitting two relatively Hot heat sources against each other, or in cooperation with each other, like any other oscillation. The piston in a Stirling engine is like a ball bouncing back and forth between two springs of nearly equal strength. If one spring we're at "absolute zero" with no "spring" power to it whatsoever it would not work.

A Stirling engine does not require a "cold reservoir" for the "waste heat" to flow into. The whole Carnot engine concept along with it's. "limit" is a fallacy.

If you think it's the absolute truth, then go build yourself a "perfect" Carnot engine that can "remove all thermal energy," to 0K sink and leave me alone.

[Fool]

You are mixing percentage of the heat that comes in with quantity of heat coming in. Some measure of the heat coming in can be gathered by measuring the power out. Using that number one can assume an efficiency. I like that number to be between zero and 100%. Once the assumption is made a quantity of heat coming in can be calculated. If zero heat is being rejected, the engine is receiving no more heat than that being output as work. If zero power is output, one must assume that all heat coming in is being rejected because it is all being reconverted to friction. Where it is rejected and how much, become the dominant questions.

The piston in a Stirling engine is like a ball bouncing back and forth between two springs of nearly equal strength.

If the springs are equal, wouldn't the balls decay until it stopped in the middle with both springs being equal length? Wouldn't trying to extract power from that system make the decay faster? Like an oar in syrup?

To get useful power out one could heat the springs in a cycle. First one spring, then the other, driving the ball back and forth. That would lead to the springs getting hotter, so one could heat one spring and cool the other. Then reverse it to drive the ball back the other way. The power out from that would reflect the amount of heat applied. Right?

I'm a little incensed that you would call my comment a "red hearing". It seems you can bring up points that are incorrect, requiring information to disprove, then you labeled the information derogatorily, as if a derogatory comment proves something is wrong when it clearly doesn't.

I only bring up liquidation and phase change because you erroneously claim that cooling a gas pulls harder on a piston than a vacuum. And your claim of attraction only applies to gas liquid phase change. And you are constantly bring up both in engine, cyro-cooling, and other threads. Yes please let us stick to Stirling Engines area of operation. Let us stick to the area where gases act close enough to model them as ideal. Please. Change the above springs and balls to gas cylinders and a piston between them.

I've seen your calculations for your LTD engine. I disagree with them. Pons and Fleischmann found out the hard way. They later spent millions of dollars trying to prove their point, and have to this day, failed. It is hard to perform good science. Please be more receptive and professional.

[Tom Booth]

(...)
If zero heat is being rejected, the engine is receiving no more heat than that being output as work. If zero power is output, one must assume that all heat coming in is being rejected because it is all being reconverted to friction.
(...)

one must assume. Really?

Just like that.

Here is one big fallacy I've seen repeated over and over by the Carnot limit advocates, which IMO shows both a lack of comprehension of the basic principles of cause and effect as well as non-comprehension of general mechanics. The assertion of a fallacy to shore up the original fallacy.

Here is virtually the same assertion from the college physics textbook cited previously as an example of the Carnot theory being full of logical contradictions and absurdities.

Friction is mentioned several times:

What is crucial to the Carnot cycle—and, in fact, defines it—is that only reversible processes are used. Irreversible processes involve dissipative factors, such as friction and turbulence. This increases heat transfer to the environment and reduces the efficiency of the engine. Obviously, then, reversible processes are superior.

Figure_16_04_05.jpg (168.82 KiB) Viewed 8017 times

Figure 5. Real heat engines are less efficient than Carnot engines. (a) Real engines use irreversible processes, reducing the heat transfer to work. Solid lines represent the actual process; the dashed lines are what a Carnot engine would do between the same two reservoirs. (b) Friction and other dissipative processes in the output mechanisms of a heat engine convert some of its work output into heat transfer to the environment.

While Carnot engines are ideal engines, in reality, no engine achieves Carnot’s theoretical maximum efficiency, since dissipative processes, such as friction, play a role

https://pressbooks.bccampus.ca/collegep ... -restated/

If mechanical FRICTION is to be included as part of the heat "rejected to the environment" or to the "cold reservoir" I think any sane, rational individual can recognize that friction is a variable. You can have more or less friction. NASA Stirling engines have literally eliminated mechanical friction in their no-contact free piston engines. We have air bearings and magnetic bearings and so forth that reduce mechanical friction to near zero.

Diagram 5b clearly depicts that "work done against friction goes to cold reservoir"

"Work"...

Where does this "work" to produce the friction come from?

Well, to get work to produce friction you first have to convert the heat into work to produce a mechanical effect before there can be any friction.

What of an LTD running on Ice?

There the "cold reservoir" is below the engine while all the mechanical "work" takes place above the engine. Any heat from friction dissipated to the environment goes out to the ambient surroundings NOT to the ice or "cold reservoir".

Supposedly the Carnot "limit" is supposed to represent the theoretical maximum for a reversible engine that has zero friction but we have statements like:

"While Carnot engines are ideal engines, in reality, no engine achieves Carnot’s theoretical maximum efficiency, since dissipative processes, such as friction, play a role"

Does a "Carnot engine" have friction or not? If it is such a defining characteristics that it does not, then why is it not 100% efficient? Where is all the 80% of the input heat (in actual measured Joules) that supposedly MUST in one way or another, pass through to the sink supposed to be disappearing to, if not to irreversible dissipative processes like friction and turbulence?

[Tom Booth]

The "carnot theorem" has enjoyed a monopoly as the only theory of heat and heat engines for 200 years. I don't consider that something to recommend it. Real science progresses.

If zero power is output, one must assume that all heat coming in is being rejected because it is all being reconverted to friction. Where it is rejected and how much, become the dominant questions

.

"One MUST assume' again.

No, one does not need to assume any such thing.

Carnot limit theory has had a stranglehold on heat engine theory since the early 1800's I think the time has come for it to be re-evaluated, as Kelvin suggested would eventually need to be done someday when he decided to invest in the Caloric theory.

[Tom Booth]

(...)
I'm a little incensed that you would call my comment a "red hearing". It seems you can bring up points that are incorrect, requiring information to disprove, then you labeled the information derogatorily, as if a derogatory comment proves something is wrong when it clearly doesn't.

I only bring up liquidation and phase change because you erroneously claim that cooling a gas pulls harder on a piston than a vacuum. And your claim of attraction only applies to gas liquid phase change. And you are constantly bring up both in engine, cyro-cooling, and other threads. Yes please let us stick to Stirling Engines area of operation. Let us stick to the area where gases act close enough to model them as ideal.
(...)

First of all, you are completely misrepresenting my position on the topic.

I never said: "cooling a gas pulls harder on a piston than a vacuum".

Honestly, I can't even figure out what that is supposed to mean.

Cooling the gas (working fluid) in a Stirling heat engine may lower the internal pressure below atmospheric pressure so the piston can return.

Work output during expansion reduces the internal energy of the gas lowering the temperature of the working fluid so that the internal pressure falls below atmospheric pressure. Then atmospheric pressure can push the piston back, and the piston can return.

This is all very well illustrated in the Andrew Hall video.

In the steam in a barrel demonstration, it appears that condensation of the steam occurs in a "partial vacuum".

Logically the steam condenses leaving a vacuum BEFORE atmospheric pressure can crush the can. So it appears the vapor will condense on its own inside the can without the aid of atmospheric pressure, since the partial vacuum is formed while the can is still intact.

The order of events, then, logically is:

1) The water vapor condenses.
2) The condensation results in a partial vacuum.
3) Once a partial vacuum is formed atmospheric pressure crushed the can.

I don't think I ever said that anything "pulls harder on the piston". Only that the above suggests that a gas may cool and "contract" reducing its pressure which allows atmospheric pressure to push the piston back in.

My helium balloon experiment, I think proves that such an event is possible for a gas that is not anywhere near the liquefaction point as my refrigerator does not go down to 4°K yet the helium in the balloon measurably reduced pressure when cooled resulting in a size reduction.

Another observation worth noting is how slowly a gas cools by conduction. It took hours for the helium to "shrink" in the freezer. Cooling by "work" output is literally instantaneous.

You claim: "...attraction only applies to gas liquid phase change."

Real gases are always influenced by forces of molecular attraction and repulsion.

Let us stick to the area where gases act close enough to model them as ideal.

Sorry, but no.

I'm sorry if that causes you some anxiety, but I'm interested in what real gas or real air is doing inside real engines under actual circumstances in the real world.

[Fool]

"First of all, you are completely misrepresenting my position on the topic."

I only do so to point out the severity of the misleading term "contraction". Contraction is a pulling condition. Attraction is a pulling condition.

"Real gases are always influenced by forces of molecular attraction and repulsion."

No. Not above the liquid/vapor temperature and below the critical pressure. They act like gases, always with positive pressures. And temperatures for that matter.

"Sorry, but no."

Sorry, but that is exactly where these engines operate.

"I'm sorry if that causes you some anxiety,"

I think you are enjoying the sake of arguing just to argue. You have made no gains on trying to learn. You take simple concepts and obfuscate them, such as the simple difference between heat, internal energy, work, and temperature, money paycheck and bank balance.

If we were to stick to real gases, you would have to stop bringing in the analogies you are so fond of.

[Fool]

One potentially interesting or useful observation to come out of this:
....

I still need to see if cooling the cylinder with a 15 lb weight on the piston, would have the effect of causing the gas inside to "contract", lifting the 15 pound weight hanging upside down from the piston.

In theory (my theory that is), the 15 lb weight should balance or cancel the 15 pounds of atmospheric pressure.

If there is no "contraction" upon cooling of the gas in the cylinder there should be no possibility of the piston lifting the 15 lb weight due to lowering of internal pressure by cooling the cylinder.

If the gas actually in fact contracts or "clumps together" more and more due to molecular attraction as the temperature is lowered further and further then the piston should be drawn inward more and more in spite of the 15 pound weight counterbalancing atmospheric pressure.

Or will the gas only contract some when each constituent gas reaches liquefaction temperature, or not ever contract at all?

Watching a video of a 55 gallon barrel violently "imploding", it's hard to imagine such a force could be counteracted by attaching a 1" cylinder to the barrel with a piston and hanging a 15 pound weight onto the piston. With the 15 lb weight the piston should not be drawn in when the barrel is cooled. Right?

Looking at that twisted metal, I can't help but think something more than 15 psi of atmospheric pressure is involved.
.....

If the gas "never contracts", then the plastic bag should stay inflated and not shrink in size at all until the temperature of liquefaction. Right?

What you are suggesting here is that "contraction" is pulling the barrel and 15 lb weight in more forcefully than a vacuum. Hence my response.

The level of my response shows how the idea of contraction is so baseless. Go ahead run that experiment, please. Please also report your findings.

[Tom Booth]

...

"Real gases are always influenced by forces of molecular attraction and repulsion."

No. Not above the liquid/vapor temperature and below the critical pressure. They act like gases, always with positive pressures. And temperatures for that matter.

...

Perhaps you could cite some credible sources to that effect, (in reference to REAL rather than "Ideal" gases), because I believe I can easily site several seemingly credible sources that say exactly what I have stated.

The attractive forces of atoms and molecules come into play whenever those atoms or molecules are in close enough proximity, and I would venture a guess that in a closed sealed chamber such as a Stirling engine, the gas molecules are always in close enough proximity to influence each other, especially when compressed at all above 1atm. Above 1 ATM gases generally no longer behave "ideally".

But the point is relatively inconsequential for a little LTD operating at nearly 1 ATM the entire cycle.

My point is simply that with work output the working fluid cools and the pressure drops below external atmospheric pressure enough that the piston is able to return without heat rejection to a sink..

That could be by momentum, atmospheric pressure, "molecular attraction", some combination of forces, or whatever you might like to attribute it to, but if there is no heat sink, then you can't "reject" heat to a heat sink that isn't there.

So maybe it is worth considering some alternative hypothesis. I think so anyway. If you don't, why do you care?

[Tom Booth]

...

What you are suggesting here is that "contraction" is pulling the barrel and 15 lb weight in more forcefully than a vacuum. Hence my response.

The level of my response shows how the idea of contraction is so baseless. Go ahead run that experiment, please. Please also report your findings.

Your statement doesn't really make sense:

""contraction" is pulling the barrel and 15 lb weight in more forcefully than a vacuum"

The attraction of the molecules and their "clumping together" into a liquid is what creates the vacuum. The liquid then condenses inside a vacuum, or at least a partial vacuum, before the barrel collapses.

The liquid or "attraction" is not reaching up and pulling in the inside walls of the barrel, I don't imagine, but then again, why not? A molecule is a molecule, and "nature abhors a vacuum" so they say.

What I'm mainly suggesting is that it looks like and most of the people doing these demonstrations, in fact say, by cooling tbe water vapor in the can, bottle or barrel, tbe gas "contracts" into a liquid.

Even in the example of the flask with a cork in it turned upside down it appears that the water condensed into a liquid inside the flask creating a vacuum BEFORE the cork was removed and the water allowed to rush in.

So it looks as though the contraction of the gas into a liquid takes place IN A VACUUM, or in spite of having to leave a vacuum behind in the process. The vacuum thus formed does not prevent the "attraction" of the molecules from forming a liquid

The cooling is also not that extreme. Just cold water or ice water or spray from a hose, or just sitting around at ambient air temperature long enough. This is all well below the liquefaction temperature of water, but regardless, the water vapor apparently condenses creating a vacuum inside the container when cold enough.

Does that not suggest that the forces of molecular attraction are greater than the vacuum, or how could a partial vacuum ever form at all?

Suppose someone does the experiment using a VERY STRONG container with a pressure gauge.

Boil a bit of water in the container and put on the lid, then dunk it in some ice water and watch the pressure gauge.

Do you imagine the pressure guage will never go below 1 ATM because the molecules will never be attracted to each other?

I would assume the pressure would have to fall pretty far below 1atm to initiate the collapse of a steel drum.

[Tom Booth]

Actually, watching a lot of those collapsing a Barrell using atmospheric pressure videos...

It often comes as a total shock when the barrel suddenly implodes violently. Everyone in the video(s) gasps. There often seems to be little if any warning. The barrel doesn't just get crushed gradually it implodes seemingly instantaneously.

I think it would be even more remarkable if the pressure stayed at 1atm the whole time until a sudden violent implosion occurs but actually, it does kind of look that way.

I haven't ever seen the experiment done with a pressure gauge attached to the drum. It might actually be interesting to try it.

[Tom Booth]

(...)

The piston in a Stirling engine is like a ball bouncing back and forth between two springs of nearly equal strength.

If the springs are equal, wouldn't the balls decay until it stopped in the middle with both springs being equal length? Wouldn't trying to extract power from that system make the decay faster? Like an oar in syrup?

To get useful power out one could heat the springs in a cycle. First one spring, then the other, driving the ball back and forth. That would lead to the springs getting hotter, so one could heat one spring and cool the other. Then reverse it to drive the ball back the other way. The power out from that would reflect the amount of heat applied. Right?
(...)

First of all I said two springs of "nearly" equal strength. Not equal length,.

I'm basically saying that because a Stirling engine (like an LTD) can run on a very minute ∆T so the temperature of the hot and "cold" side are almost or "nearly" equal.

It might actually be possible two run a Stirling engine on two EQUAL temperature heat sources with some arrangement involving two displacers.

Peter Lindemann's invention for example:

Resize_20230823_141339_9266.jpg (146.58 KiB) Viewed 7922 times

If one displacer doesn't need a sink, probably two don't need any either, but something like that has yet to be demonstrated, so for now "almost equal" is within known, demonstrated parameters.

Your statements about cooling or alternately cooling the springs seems a bit overboard.

The same principle as I was talking about, could be demonstrated using a rubber ball. You can play handball against a wall because the ball is "elastic".

Well gases are also elastic, so it states in the physics books.

You don't need to cool a rubber ball to play hand ball against a wall. You just need to keep giving the ball an intermittent dose of energy by hitting it away towards the wall whenever it comes close enough.

In handball, the player is the heat/energy source, and you could think of tbe wall as atmospheric pressure.

You don't need a "cold reservoir" to play hand ball. You don't even need a secondary heat source. You just need a bouncy ball, a wall and an intermittent source of energy.

Why should a heat engine be any different?

[Tom Booth]

You are mixing percentage of the heat that comes in with quantity of heat coming in.

I'm not.

The "Carnot efficiency limit" itself is a ratio or percentage. It only has a bearing on quantity when applied to some particular known heat input or where there is some other known value, such as waste heat, but even then it is still just a ratio or percentage.

Some measure of the heat coming in can be gathered by measuring the power out. Using that number one can assume an efficiency.

Not enough information. You can't just "assume an efficiency" from power out alone, or maybe you could explain how that's done.

... I like that number to be between zero and 100%. Once the assumption is made a quantity of heat coming in can be calculated.

Calculations based on an "assumption" is meaningless.

If zero heat is being rejected, the engine is receiving no more heat than that being output as work. If zero power is output, one must assume that all heat coming in is being rejected because it is all being reconverted to friction...

The first statement is true enough. If the engine is load balanced.

The second statements is self contradictory. Heat can't be "reconverted" back into friction/heat unless first converted to work.

[Fool]

The PV diagram shows the entire work performed by the gas volume. The work is sum of all P•∆V. What comes out on the shaft, work output, is that depicted work minus the internal work necessary for the machine to operate. The internal work is often called losses or friction or windage.

There are also thermodynamic losses. Those losses are depicted on a PV diagram. An indicator diagram of a real engine is compared to the best known model engine cycles. The difference is the thermodynamic losses.

If a barrel is sufficiently strong enough to not crush with a full vacuum inside, we model it as a constant volume.

If filled with steam at atmospheric pressure at high enough temperature and capped. As the barrel and steam cool, a slow process, the pressure and temperature will drop. As the temperature drops low enough some of the steam will turn into liquid, condense. Liquid takes up way less volume than gas. But it doesn't all condense at once. At little condenses at first. More cooling, now at constant temperature and volume, condenses more steam, lowering the pressure still more. This leaves less and less steam to fill out the entire volume. But it does and it has a small but ever present positive pressure. It is true even at freezing temperatures where liquid turns to solid. The steam, well below freezing temperatures, sublimes and fills the volume with a minute but positive pressure.

Why does it appear to happen suddenly? A round spherical or cylindrical volume is much stronger than a distorted one. The crush pressure differential is higher when shaped more true. When it gets to the crush pressure the barrel distorts, providing a weaker shape, which distorts more and is weaker yet. Failure is magnified and the process is rapid.

Side note: To call this an implosion is somewhat misleading. It is more a sudden crushing. An implosion is, as the name suggests, an explosion. A release of energy in a confined volume. Implosions start out by an inward failure, then explode outwardly. The Titan submarine experienced an implosion from about 6000 psi. It was blown to bits. A 55 gallon barrel experiences a simple crushing from 15 psi and is still complete. Something to think about, when should it be call crush vs implosion?

[Tom Booth]

The PV diagram shows the entire work performed by the gas volume.
(...)

It's fine for you to say so. Can you provide any references?

The reason I'm skeptical is a Stirling engine is not a simple piston in a cylinder as modeled by a PV diagram.

Setting aside the more complicated issue of forces of molecular attraction and repulsion;

You have the power piston and power cylinder and the displacer and displacer cylinder.

Should we treat this as one volume?

In reality, you can have one thing going on in the power cylinder, such as expansion or contraction, while there are other things going on associated with the displacer such as heat addition, accumulation of potential energy etc.

You can have heat going into the displacer chamber theoretically increasing the temperature, pressure and volume while work is going out via the power piston theoretically lowering the "internal energy" at the same time.

Some engines, (thermoacoustic) have a throttle midway in the chamber, which makes the engine similar to a turbine, or an "open system" within a "closed system".

That is, you could, in theory, have high pressure on one side of the throttle and low pressure on the other, or you can even have both a "jet stream" or laminar flow of high pressure gas AND a low pressure zone forming within the power cylinder simultaneously.

You can have the regenerator taking in heat as "potential energy" while energy is being lost to work output via the power cylinder.

Because of the 90° phase angle, you actually have gas being heated and expanded in the displacer chamber while volume is decreasing in the power cylinder, and many other very rapid complex, dynamic interactions.

A PV diagram represents what? An "infinitely slow" "quasi-static" process that is easy to model and really, nothing like the complex reality of many simultaneous processes going on inside a Stirling engine.

A PV diagram does nothing to explain how the power piston in a Laminar Flow or thermoacoustic engine can return at all while the engine has heat being continuously supplied to the regenerator.

There must, IMO, be something, or possibly many things, which are not being accounted for that allows the piston to return and for the cycle to complete (in Stirling engines generally) without necessitating "heat rejection" to a sink during "compression". Not because I wish it were so, but because that appears to me to be an observable fact that the standard "ideal" thermodynamic models do not adequately explain.

[Tom Booth]

(...)

If a barrel is sufficiently strong enough to not crush with a full vacuum inside, we model it as a constant volume.

If filled with steam at atmospheric pressure at high enough temperature and capped. As the barrel and steam cool, a slow process, the pressure and temperature will drop. As the temperature drops low enough some of the steam will turn into liquid, condense. Liquid takes up way less volume than gas. But it doesn't all condense at once. At little condenses at first. More cooling, now at constant temperature and volume, condenses more steam, lowering the pressure still more. This leaves less and less steam to fill out the entire volume. But it does and it has a small but ever present positive pressure. It is true even at freezing temperatures where liquid turns to solid. The steam, well below freezing temperatures, sublimes and fills the volume with a minute but positive pressure.

Why does it appear to happen suddenly? A round spherical or cylindrical volume is much stronger than a distorted one. The crush pressure differential is higher when shaped more true. When it gets to the crush pressure the barrel distorts, providing a weaker shape, which distorts more and is weaker yet. Failure is magnified and the process is rapid.

Side note: To call this an implosion is somewhat misleading. It is more a sudden crushing. An implosion is, as the name suggests, an explosion. A release of energy in a confined volume. Implosions start out by an inward failure, then explode outwardly. The Titan submarine experienced an implosion from about 6000 psi. It was blown to bits. A 55 gallon barrel experiences a simple crushing from 15 psi and is still complete. Something to think about, when should it be call crush vs implosion?

So,... you seem to want to view this "minute but positive pressure", in the course of our conversation relating to Stirling engines, something that would prevent the piston from returning? Or prevent the working fluid from "contracting" or from being "compressed" by atmospheric pressure???

That seems to have been your argument.

My "theory" is that work output during expansion results in a loss of "internal energy" which lowers the pressure below atmospheric pressure so the piston can return.

You are like No No No... there is always "positive pressure" not a "vacuum".

What are we talking about? "minute but positive pressure"?

"Positive" in relation to what?

1 or 2 psi is "positive pressure" next to atmospheric pressure?

Whatever, 1, 2 or 10 psi is still below atmospheric pressure which low pressure would generally be recognized as a "partial vacuum".

A pressure of 14.9 psia is still a "vacuum" by common parlance.

To insist, as it seems you have been doing all along, that atmospheric pressure cannot push the piston back in because there is no such thing as an "absolute vacuum", is just a silly diversion and a waste of time.