Thermodynamic work vs. real work

Discussion on Stirling or "hot air" engines (all types)
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Fool
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Re: Thermodynamic work vs. real work

Post by Fool »

VincentG wrote: Wed Sep 11, 2024 8:56 am Ideally there is no gas in the vacuum engine once the valve closes. Or at least you are gaming the curve where any significant compression occurs only very near TDC.

The Stirling cycle as defined by Fool and Senft above has a fixed mass of gas that is fighting the piston all the way from BDC to TDC..
Ideally. Yes. Of course it would need to be powered by two large reservoirs. One at atmosphere pressure, and one at zero pressure. Ideally. For a "vacuum" engine, that is.

'Ideally' a fame licker Engine, (I kinda like the term flame sucker, but there, also, is no such thing as "suck" with gasses.), "sucks" in the high temperatures of a flame. Ideally at BDC the cylinder would be full of high temperature exhaust gas, Th. The valve would close. The temperature and pressure would drop to Tc. The piston would now be pushed towards TDC by the outside air pressure and low gas pressure. Internal gas pressure would build to atmospheric. When Atmospheric pressure is reached, the check valve would open allowing the momentum to purge the gas fairly unrestricted. This is, again, 'ideal'.

I'm guessing that in a real engine the speed of the piston near TDC would be slow, as it stops for just an instance at TDC. The pressure would then be very close to atmospheric. As the piston recedes 'sucking' in the flame the inside gas is lower than atmospheric, otherwise the flame would not be pushed in. Again the piston is slow near BDC. The gas pressure again would be close to atmospheric. So the indicator diagram would show the pressure on the expansion stroke to dip below the "yellow line" slightly. Maximum pressure drop would be near the middle of the stroke, maybe. If the valve is closed early, to take advantage of that pressure drop, the pressure would drop even more by BDC.

On the compression stroke, the pressure at some point in the cooling, would be much lower. But it would build until the check valve opened at slightly above atmospheric. This, depending on timing of the valve and maximum temperature of the gas would be before top dead center. It would be reflected in the percentage of maximum vacuum at BDC. If 25% of atmospheric, it would be at the 25% stroke. Wouldn't it.

The PV diagram would have a big dip below the yellow line at the beginning of the return stroke then return to the yellow line and slightly above when the check valve opens.

A check valve, or exhaust/purge valve is needed because, venting the gas out the flame sucker hole would tend to blow the flame out.

The above is just analyzing the process. It is why I'm interested in seeing a measured PV indicator diagram. It would be a gathering of data to confirm or refute the analysis.

It would be similar to a real Stirling PV showing how there is no adiabatic process, nor process on ether of the Th or Tc isotherms. And how it pulls away from those isotherms when running faster. And expands outwards towards those isotherms when loaded and running slower. And the effects of harmonic motion from both the PP and DP. More realistically.

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Fool
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Re: Thermodynamic work vs. real work

Post by Fool »

VincentG wrote:Those big wigs at my oil company should have fed me a better line of BS that was harder to unravel. I'll have a word with them at our next meeting. In the meantime I'll continue to try and break the atmospheric Watt/cc power record for hot air engines all while keeping it open source and hiding it from my big oil overlords who pay me to troll this forum.
LOL. Good one. Cheers.

Almost forgot to comment, that the maximum efficiency of a flame licker will be quite a bit lower than an ideal Carnot. It doesn't have an isothermal expansion. But who cares?

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Tom Booth
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Re: Thermodynamic work vs. real work

Post by Tom Booth »

VincentG wrote: Thu Sep 12, 2024 5:02 am
O boy you're right, just ran the numbers on a 100cc vacuum engine with a full gulp of 3000 degree F gas at BDC and already hits 1 bar at 20cc and 80 degrees F. I figured an ideal cycle like that would be much more impressive. Seems to me the vacuum engine is uninspiring after all.
...
I don't know how you "ran the numbers" but my point is, once the valve closes at (or a little before) BDC the hot gas is trapped in a big cold metal chamber and cools rapidly.

If entering at 3000°F into a 100cc chamber, atmospheric pressure, then cooled to 80°F the pressure would drop to about probably 2 or 3 psi (while piston is still near BDC) giving atmosphere a driving pressure of about 10 psi

A Stirling engine has the advantage of additional potential vacuum or "negative pressure" due to work output on expansion causing additional cooling and low pressure at BDC, but essentially the same situation: gas trapped in a cylinder at very low pressure and temperature, so atmospheric pressure becomes a real driving force.

As the piston accelerates towards TDC the (external) pressure is converted to velocity.

Personally I think it is a mistake to incorporate a relief valve as the added pressure and heat of compression at TDC would be beneficial as velocity is converted back into heat at TDC providing "adiabatic bounce" at TDC.

IMO the intake valve (in a flame licker) should open just AFTER TDC as the pressure drops back down to 1 atmosphere, to take advantage of the adiabatic bounce, then draws in fresh hot gas as the pressure drops lower, probably.

Just to reiterate: at BDC you have essentially the same situation in both engines: gas trapped in a cylinder at very low pressure and temperature, so atmospheric pressure becomes the driving force to return the piston while doing "REAL" positive + work in the process.

Not "fighting the piston all the way from BDC to TDC".
VincentG
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Re: Thermodynamic work vs. real work

Post by VincentG »

Tom Booth wrote: Sun Sep 08, 2024 8:24 pm To make this clear, view the engine as a "fire piston" with a sliding regenerator in the bottom of the chamber.

Have the piston driven by a heavy flywheel.

Now imagine whenever the piston passes the halfway point a solenoid is tripped that switches from the hot to the cold regenerator on the way up and back again on the way down.

Something like this:


Compress_20240908_231001_1637.jpg

You can have heat going in from the candle on the right to compensate for friction or crank/drive power output. But in the way up, once all the heat is exhausted the piston is carried up further cooling the gas while the cold regenerator moves into position.

The cylinder walls, of course, should be non-heat conducting and inert, not absorbing or releasing any heat.

As hopefully can be seen or imagined, there is no transfer of heat between the hot and cold regenerator/heat exchangers. At least not through the working fluid. In this arrangement heat could conduct directly through the divider, but let's just say the divider is a perfect insulator.

I have trouble wrapping my head around making one regenerator work in harmony with a running engine. Two is just bonkers, what is the major benefit that you see?
Fool
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Re: Thermodynamic work vs. real work

Post by Fool »

Tom Booth wrote:Not "fighting the piston all the way from BDC to TDC".


I beg to I differ. The piston motion/acceleration is a result of the pressure inside minus the pressure outside. The two pressures battle each other for the entire cycle. Whether the two produce positive work out or not depends on the direction of piston travel and which pressure is higher.

Acceleration of the piston from that pressure difference equates to an change in flywheel speed and energy. If the torque is in the direction that the flywheel is turning, it is positive work. If it is in opposition it is negative work.

Positive effective work is positive work delivered to the flywheel.

The outside pressure fights the position over the entire expansion stroke. It helps the piston over the entire compression stroke.

The gas helps the piston during the entire expansion stroke. It fights it for the entire compression stroke. It would be best to have it go to zero for the entire return stroke. But! As Tom has pointed out, that is impossible from simple temperature changes.

The key is to have the gas be at a higher pressure for the expansion stroke, and a lower pressure for the return stroke. It is commonly done by expansion being at a higher temperature than compression. It is done by heat conduction in, and out, to get away from zero work adiabatic bounce, zero heat, cycles.

'heat in only' just makes a bounce at a higher temperature.

Compression always fights the internal gas. It can only be
helped by the outside buffer pressure/atmosphere, and heat removal. Not eliminated.

If the buffer pressure is higher than the gas, the compression stoke adds positive energy/work to the flywheel. If the buffer pressure is lower than the gas the compression stroke takes energy from the flywheel, called negative work.

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Fool
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Re: Thermodynamic work vs. real work

Post by Fool »

VincentG wrote:I have trouble wrapping my head around making one regenerator work in harmony with a running engine. Two is just bonkers, what is the major benefit that you see?
I agree. I've tried to visualize extending expansion and compression adiabatically to extend the cycle beyond the Carnot, or Stirling, to get higher and lower Th and Tc by saving cold and hot, and reusing it later. It gets uselessly mind boggling very quickly. It then drops back to the simplistic beauty of Stirlings original regenerator which saves both hot and cold in one device. It merely depends on which way the gas flows.

As Senft says it is like breathing through a scarf wrapped around your face on a cold day. When you breath out it warms the scarf. When you breath in, the warm scarf, warms the air coming in, and the cold air cools the scarf for the next cycle. It saves heat and cold.

It is technically called a counter flow mixed fluid heat exchanger. Counter flow separated heat exchangers are awesome enough on their own.

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Tom Booth
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Re: Thermodynamic work vs. real work

Post by Tom Booth »

Fool wrote: Mon Sep 16, 2024 7:24 am
Tom Booth wrote:Not "fighting the piston all the way from BDC to TDC".


I beg to I differ. The piston motion/acceleration is a result of the pressure inside minus the pressure outside. The two pressures battle each other for the entire cycle. Whether the two produce positive work out or not depends on the direction of piston travel and which pressure is higher.

Acceleration of the piston from that pressure difference equates to an change in flywheel speed and energy. If the torque is in the direction that the flywheel is turning, it is positive work. If it is in opposition it is negative work.

Positive effective work is positive work delivered to the flywheel.

The outside pressure fights the position over the entire expansion stroke. It helps the piston over the entire compression stroke.

The gas helps the piston during the entire expansion stroke. It fights it for the entire compression stroke. It would be best to have it go to zero for the entire return stroke. But! As Tom has pointed out, that is impossible from simple temperature changes.

The key is to have the gas be at a higher pressure for the expansion stroke, and a lower pressure for the return stroke. It is commonly done by expansion being at a higher temperature than compression. It is done by heat conduction in, and out, to get away from zero work adiabatic bounce, zero heat, cycles.

'heat in only' just makes a bounce at a higher temperature.

Compression always fights the internal gas. It can only be
helped by the outside buffer pressure/atmosphere, and heat removal. Not eliminated.

If the buffer pressure is higher than the gas, the compression stoke adds positive energy/work to the flywheel. If the buffer pressure is lower than the gas the compression stroke takes energy from the flywheel, called negative work.

.
The point is, RELATIVE pressure.

So if a higher relative internal pressure can produce "REAL" work on expansion, then logically a higher relative external pressure can produce "REAL" work on "compression".

One is not so-called "positive" work and the other "negative".

Both are a result of a relative pressure differential. Both produce positive "useful" work output
Tom Booth
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Re: Thermodynamic work vs. real work

Post by Tom Booth »

VincentG wrote: Fri Sep 13, 2024 5:59 pm
Tom Booth wrote: Sun Sep 08, 2024 8:24 pm To make this clear, view the engine as a "fire piston" with a sliding regenerator in the bottom of the chamber.

Have the piston driven by a heavy flywheel.

Now imagine whenever the piston passes the halfway point a solenoid is tripped that switches from the hot to the cold regenerator on the way up and back again on the way down.

Something like this:


Compress_20240908_231001_1637.jpg

You can have heat going in from the candle on the right to compensate for friction or crank/drive power output. But in the way up, once all the heat is exhausted the piston is carried up further cooling the gas while the cold regenerator moves into position.

The cylinder walls, of course, should be non-heat conducting and inert, not absorbing or releasing any heat.

As hopefully can be seen or imagined, there is no transfer of heat between the hot and cold regenerator/heat exchangers. At least not through the working fluid. In this arrangement heat could conduct directly through the divider, but let's just say the divider is a perfect insulator.

I have trouble wrapping my head around making one regenerator work in harmony with a running engine. Two is just bonkers, what is the major benefit that you see?
The point was for illustrative purposes, not for any practical benefit necessarily. Just trying to illustrate how a Stirling engine actually works. I thought it might be easier to conceptualize.

Ordinarily the displacer would alternately cover the hot and cold plate rather than the plates themselves sliding in and out

The only point I was trying to make is that heat FLOW between the hot and cold plate is rendered impossible, because when there is "heat of compression" the cold plate is isolated. During cooling by expansion the hot plate is isolated.

I was just illustrating a different method of isolation that is easier to see or make obvious. With a displacer SEEMINGLY pushing air from side to side it APPEARS that the hot air is being transfered over to the cold side and vice-versa, so the idea that heat is "flowing through" appears to be obvious or a given, but that isn't really the case. Instead heat is being pushed or pumped from the cold side over to the hot side.

From there the heat is converted to work.
VincentG
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Re: Thermodynamic work vs. real work

Post by VincentG »

... With a displacer SEEMINGLY pushing air from side to side it APPEARS that the hot air is being transfered over to the cold side and vice-versa, so the idea that heat is "flowing through" appears to be obvious or a given.. Instead heat is being pushed or pumped from the cold side over to the hot side.
How then would you explain the operation of a displacer chamber with no work output?

All in all it's a driving engine, it shouldn't be pumping more heat than it's utilizing, otherwise it would be a driven heat pump
VincentG
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Re: Thermodynamic work vs. real work

Post by VincentG »

bump
Fool
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Re: Thermodynamic work vs. real work

Post by Fool »

Tom Booth wrote:Not "fighting the piston all the way from BDC to TDC".
Tom Booth wrote:The point is, RELATIVE pressure.
Yes. Relative pressures. Like two teams in a tug-o-war fighting each other the whole way. The pressure inside fights the pressure outside in a push-o-war the whole way for both strokes. The higher pressure wins.

The outside pressure is constant. The inside pressure changes by adding and subtracting heat. PV=nRT adiabatic Temperature and Pressure change from Volume change balances out because Volume change during expansion is equal and opposite to Volume change during compression.

You seem also to have problems with negative number representations of real items. Putting apples in a basket is adding real and positive apples. Taking apples out of a basket it subtracting real apples and the same as adding negative real apples. Putting dollars into your account is adding positive dollars. Taking dollars out of your account is adding negative dollars. Positive and negative numbers are just a direction of the flow of items.

Subtracting negative dollars is adding positive dollars.

Engineers must define which direction is positive. The opposite direction then becomes negative. It is the same kind of item, work, heat, energy, just moving in the opposite direction.

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Tom Booth
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Re: Thermodynamic work vs. real work

Post by Tom Booth »

Fool wrote: Thu Sep 26, 2024 6:42 am
Tom Booth wrote:Not "fighting the piston all the way from BDC to TDC".
Tom Booth wrote:The point is, RELATIVE pressure.
Yes. Relative pressures. Like two teams in a tug-o-war fighting each other the whole way. The pressure inside fights the pressure outside in a push-o-war the whole way for both strokes. The higher pressure wins.

The outside pressure is constant. The inside pressure changes by adding and subtracting heat. PV=nRT adiabatic Temperature and Pressure change from Volume change balances out because Volume change during expansion is equal and opposite to Volume change during compression.

You seem also to have problems with negative number representations of real items. Putting apples in a basket is adding real and positive apples. Taking apples out of a basket it subtracting real apples and the same as adding negative real apples. Putting dollars into your account is adding positive dollars. Taking dollars out of your account is adding negative dollars. Positive and negative numbers are just a direction of the flow of items.

Subtracting negative dollars is adding positive dollars.

Engineers must define which direction is positive. The opposite direction then becomes negative. It is the same kind of item, work, heat, energy, just moving in the opposite direction.

.
Baloney.

"Positive" work output is what results in actual, physical mechanical work output.

Take the Cox clock ratchet.

Most ratcheting devices only work in one direction, so it is fair to say that the ratchet only does positive work on the "forward" stroke.

The Cox clock however is rather ingenious in that it works on both the "forward" and "backward" stroke. So the clock is wound regardless if the barometric pressure is rising or falling, that is, if the relative pressure is on one side or the other of the mechanism.

440px-Cox_timepiece_winding_switch.png
440px-Cox_timepiece_winding_switch.png (30.04 KiB) Viewed 1002 times
https://en.m.wikipedia.org/wiki/Cox%27s_timepiece

Unlike an internal combustion engine that actually does only do real positive work during an explosive expansion phase a Stirling engine, similar to the Cox clock is intelligently engineered to do positive mechanical work throughout the entire cycle in both the forward expansion stroke as well as the return stroke.

If you cannot see that, IMO you are simply stupid and blind.
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Re: Thermodynamic work vs. real work

Post by VincentG »

Tom the IC engine has the same atmospheric assist on the return stroke as the Stirling. The only difference is that it becomes insignificant compared to the output of the power stroke.

If you had a hot air engine with a buffer pressure of zero bar, all the internal pressure on the power stroke would go towards work output, and internal pressure could be allowed to nearly reach zero bar. Positive work would result all the while.

The return stroke would now be all work negative from stored flywheel energy, with no assist from buffer pressure.

Overall work output should still be the same as an engine with a buffer pressure.

That is not to say there can be no benefit from buffer pressure, but for the most part it cancels out.
Fool
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Re: Thermodynamic work vs. real work

Post by Fool »

Tim Booth wrote:"Positive" work output is what results in actual, physical mechanical work output.
What do you call 'work in' when 'work out' is positive? Negative work out. Duh.

Your resistance to understand negative numbers is a very old and rejected argument, used centuries ago by ignorant archaic people.
Tom Booth wrote:so it is fair to say that the ratchet only does positive work on the "forward" stroke.
It is fair to say that ratchets 'do' zero work. They only store it. They store it in only one direction. Two ratchets can be combined to store bidirectional motion into one direction. It still only stores it in one direction. The work comes from outside of the ratchet mechanism.

A crankshaft and flywheel combination stores kinetic energy, from two directions of momentum, into one rotational momentum direction.

VincentG, you are absolutely correct. I've been pointing that out to Tom for years. Inside gas always pushes outwards. Helps when being expanded. Hurts when being compressed.

Similar for the atmosphere but oposite. Helps for compression stroke. Hurts for expansion stroke.

Expansion work = gas pressure work - atmosphere pressure work.

Compression work = - gas pressure work + atmosphere pressure work.

Or rearranging the last equation for positive gas pressure work:

Compression work = atmosphere pressure work - gas pressure work

Same thing. Fights the internal gas pressure for the entire compression stroke

Atmospheric work is equal and opposite.

Gas pressure is opposite, but not equal. Hence, heat in doesn't equal heat out.

.
Tom Booth
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Re: Thermodynamic work vs. real work

Post by Tom Booth »

VincentG wrote: Fri Sep 27, 2024 6:22 am Tom the IC engine has the same atmospheric assist on the return stroke as the Stirling. The only difference is that it becomes insignificant compared to the output of the power stroke.
What type of IC engine?

Power stroke...

Followed by exhaust stroke, valve open pushing out against atmosphere.

Intake stroke, valve open, piston must work against crankcase pressure.

Compression stroke?

Both valves closed.

Two cycle?

I don't think so.
If you had a hot air engine with a buffer pressure of zero bar, all the internal pressure on the power stroke would go towards work output, and internal pressure could be allowed to nearly reach zero bar. Positive work would result all the while.

The return stroke would now be all work negative from stored flywheel energy, with no assist from buffer pressure.
Hypothetical fantasy world.

A Stirling engine does not require a flywheel or "stored flywheel energy".

We aren't running our engines in outer space.

A rocket in outer space, which is basically what you described, has no "return stroke".
Overall work output should still be the same as an engine with a buffer pressure.

That is not to say there can be no benefit from buffer pressure, but for the most part it cancels out.
Sure, I agree, I think, though maybe you mean something else. I don't see any purpose in these hypothetical "vacuum" thought experiments.

Atmospheric pressure is the same both inside and outside the engine, so cancels out and can be ignored, before the engine has heat applied. Like a pendulum at equilibrium before it is started swinging by being moved away from equilibrium.

A Stirling engine does not operate like an IC engine from forced internal pressure or an "explosion".

A Stirling engine operates like a pendulum or any other oscillator around a point of equilibrium.

An oscillator can do "real work" in both directions.

Either way, you have a piston between two volumes of pressurized air, one higher pressure than the other. The piston moves towards the point of equilibrium, between TDC and BDC.

The momentum of the piston causes it to overshoot.

Like a pendulum is drawn down by gravity, but reaching equilibrium in the center it does not simply stop, but overshoots the equilibrium point and so continues its oscillations.

A perfect vacuum is impossible.so even your hypothetical engine that will "nearly reach zero bar" would also oscillate around a point of equilibrium... Eventually.
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