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

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

Post by Tom Booth »

VincentG wrote: Fri May 24, 2024 8:39 am Work is a function of pressure x volume.

Pressure is a function of heat applied to a gas.

But that does not mean that heat converts to work. Work is just a consequence of heat. Heat creates pressure and pressure converts to work as the heat does what heat does, attempt to reach equilibrium.
Your free, of course, to think what you want.

But the gas expanding (pressure) is just gas molecules striking the piston, transferring kinetic energy to the piston, loosing kinetic energy in the process.

U = U(T) only for an ideal gas

For most temperatures and pressures, air is basically an ideal gas.

When the gas transfers energy to the piston the gas temperature goes down Joule for Joule.

The piston is not a gas but a composite of molecules joined, so when it receives kinetic energy from the gas it moves as a composite structure. It's temperature does not increase.

So the "heat" or thermal internal energy measured by temperature of the gas is converted into mechanical motion of the composite piston as a whole structure.

The result is the piston is moved and the temperature of the gas goes down.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Just to be clear, cooling due to transfer of kinetic energy is over and above cooling that may be due to "expansion", simply spreading out the heat over a larger volume.

IMO there is no such thing as heat attempting to reach equilibrium.

The idea that heat wants to combine with cold, seeks out or finds itself compelled to flow to cold is just Caloric/Carnot theory mythology.
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

Tom Booth wrote:But the gas expanding (pressure) is just gas molecules striking the piston, transferring kinetic energy to the piston, loosing kinetic energy in the process.
This is true. You seem to also, at the same time, conveniently ignore the rest of the cycle, specifically the return stroke. During the return stroke the opposite happens. Gas molecules hit a returning piston, gaining momentum from the piston and increasing their gas molecular speed, AKA their temperature.

This is the reintroduction of the thermal energy of compression that needs to be shed as heat to the cold plate. It won't warm the plate significantly. Shedding the energy allows the return stroke to use less work than the expansion stroke work. The gas will be cooler and available again to receive Qh by the original temperature difference.

The cold plate is kept cool by thermal mass and conduction. If that conduction is cut off, the system will tend to put out zero work. It might run at a higher RPM. To challenge that process, work output must be measured. Along with other variables.

My guess is the opening of the indicator diagram, will close down to a slit. Only indicating a residual tiny heat loss. And heat intake. Essentially a zero work path. This assumes the outflowing heat is indeed cut off and hasn't found an alternate path or conductor.

The alternate path is the one I would suspect. Increased surface area by the insulation. It might just be thermal mass but doesn't appear that way from my distance away. Too far away to conclude anything about the experiment experiment other than you have a temperature anomaly. But I can see your theory and comment on that.

You have asked for proof of the Carnot Theorem in this thread. I have provided several, as have others. You dismiss the proof. Fine. You are giving points as to why you dismiss it. And I'm pointing out why I disagree with your points. It is still part of this proof.
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

"The idea that heat wants to combine with cold, seeks out or finds itself compelled to flow to cold is just Caloric/Carnot theory mythology."

Energy has nowhere to go but away from itself. Away from itself is the cold, or cooler. Energy will only stay separated if there are forces keeping it separated. Gravity and buoyancy is such a force. Energy, heat and pressure tend to expand until even.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Apparently much more heat is lost from the gas due to work output during the power stroke so the energy input from atmospheric pressure on the return stroke is not enough to require any heat rejection on the compression stroke.

This is an experimentally observable and verifiable reality regardless of your calculations or imagination to the contrary
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

A tendency is not an effort.

The general tendency for heat to disperse in all directions has been greatly overstated so as to imply that heat makes a bee line for cold just like a river flows downhill. Such a concept is utter nonsense.
VincentG
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by VincentG »

All three rockets will heat the room equally.
Mysteriously then, if the rockets are each placed within their respective piston and cylinder, and the same load differences are applied, the piston engine performing the work of lifting the weight 5ft off the ground will heat the room less?

I won't keep beating the same drum, I'm just presenting a thought experiment to gain a deeper understanding of the nature of heat.
VincentG
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by VincentG »

And Tom, I don't see any evidence that heat will not flow as fast as it can to reach equilibrium, depending on the resistance of the medium it has to work with.
matt brown
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

VincentG wrote: Fri May 24, 2024 11:04 am
All three rockets will heat the room equally.
Mysteriously then, if the rockets are each placed within their respective piston and cylinder, and the same load differences are applied, the piston engine performing the work of lifting the weight 5ft off the ground will heat the room less?

I won't keep beating the same drum, I'm just presenting a thought experiment to gain a deeper understanding of the nature of heat.
Vincent - the rockets are doing work via a direct action/reaction principle, not 'indirectly' via the gas. Note that rockets work in space without any thermo considerations.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

VincentG wrote: Fri May 24, 2024 11:10 am And Tom, I don't see any evidence that heat will not flow as fast as it can to reach equilibrium, depending on the resistance of the medium it has to work with.
If that were the case, I think the universe would have reached equilibrium eons ago.

I think this general assumption, that heat is going to automatically race towards cold as fast as it can is why Stirling engines are generally poor performers.

Designers THINK all they need for power and efficiency is to supply a temperature.difference, or MORE temperature difference, and neglect a host of other more important considerations.

The last thing the heat wants to do is enter the highly thermally resistant and insulating working fluid/air.

Well you did mention resistance.

Air, the usual working fluid, is very very resistant. Almost any other material the engine could be made of is much less resistant. The engine body, bolts, connecting rods, etc. etc. or just the surrounding air everywhere. There is nothing compelling the heat to prefer a path through the working fluid from your hot to your cold side in preference to any other path or direction available.

Then there is, in addition, dozens of other considerations.

IMO the idea that the temperature difference alone is the only thing that matters has been a severe detriment to progress. It's not only false, but encourages a lazy, negligent, thoughtless approach to engine design.

Just add some ice cubes to make it go faster.
matt brown
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Fool wrote: Fri May 24, 2024 9:56 am "The idea that heat wants to combine with cold, seeks out or finds itself compelled to flow to cold is just Caloric/Carnot theory mythology."

Energy has nowhere to go but away from itself. Away from itself is the cold, or cooler. Energy will only stay separated if there are forces keeping it separated. Gravity and buoyancy is such a force. Energy, heat and pressure tend to expand until even.
Kudos, Fool, very well worded in mere 2 lines.

I'm still waiting for Tom to explain his spin on how hot air rises despite a sealed can of air weighs the same regardless of temperature, so each molecule weighs the same regardless of temperature, right ??? (Fool, no help please)
matt brown
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

VincentG wrote: Fri May 24, 2024 8:39 am Work is a function of pressure x volume.
always
VincentG wrote: Fri May 24, 2024 8:39 am Pressure is a function of heat applied to a gas.
sometimes
VincentG wrote: Fri May 24, 2024 8:39 am But that does not mean that heat converts to work.
no direct conversion
VincentG wrote: Fri May 24, 2024 8:39 am Work is just a consequence of heat.
sometimes, but 'consequence' denotes there are constraints and limits
VincentG wrote: Fri May 24, 2024 8:39 am Heat creates pressure and pressure converts to work as the heat does what heat does, attempt to reach equilibrium.
more like...as pressure does what pressure does, attempt to reach equilibrium

When it comes to work, the driving force is always pressure and heat is merely "along for the ride" (energy balance). In our common engine context, WORK ALWAYS INVOLVES A VOLUME DIFFERENTIAL, but any temperature or pressure differential will depend upon the thermo process/es involved.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

matt brown wrote: Fri May 24, 2024 12:01 pm ...so each molecule weighs the same regardless of temperature,...
Temperature is "average kinetic energy" of the gas molecules is it not?
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

matt brown wrote: Fri May 24, 2024 12:26 pm
VincentG wrote: Fri May 24, 2024 8:39 am But that does not mean that heat converts to work.
no direct conversion
As far as "heat" is just a manifestation of the kinetic energy transfered from the hot plate to the gas molecules, and "pressure" is also just a manifestation of the same kinetic energy of the gas molecules, heat and pressure are just two different words for the same thing.

So, yes there is direct conversion of heat, kinetic energy of the gas molecules to work, kinetic energy transfered to the piston.
Last edited by Tom Booth on Fri May 24, 2024 12:40 pm, edited 1 time in total.
matt brown
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Tom Booth wrote: Fri May 24, 2024 6:55 am
Adding heat increases internal energy, work decreases internal energy.

In reality for transfers to or from a gas, work or heat are just different words to describe a transfer of kinetic energy in or out of the gas. Actually identical on a molecular or atomic level.

Fool likes to imagine some significant distinction between heat and internal energy. For a gas, there really isn't any.

The "vibrations" of the hot plate transfer energy to the gas which increases the ",vibration" of the gas. The gas transfers the same to the piston.

A simple transfer of kinetic energy

Hot plate molecule to gas molecule, gas to piston.
Xlnt description except for the caveat that adding heat TENDS to increase internal energy, and work TENDS to decrease internal energy.
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