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

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

Post by Fool »

This is an interesting question I've asked before many times on various science and physics forums. What does "rejection" actually refer to?
Plain and simply it means that the temperature of the working gas is hotter than the walls of the heat exchanger and Th, or Tc. Usually Tc, unless it is a heat pump.
How does this cause the "supplied heat" supposedly still present in the working fluid to be "rejected" if that thermal energy has already been transformed (stored as kinetic energy in the motion of the flywheel)?
The energy analysis of a full cycle doesn't care how energy leaves, enters, or is stored, only wether it is heat or work. Work can leave through the flywheel or atmospheric pressure, and return, and the internal gas won't react any differently.

During the expansion stroke, work leaves the engine equal to the heat entering, for an isothermal process.

During the compression stroke work enters, and an equal amount of heat leaves that gas. If cooler it is less than for the expansion stroke.

The two strokes are required for a complete cycle.

There are two equal and opposite processes that are needed as well. Regenerations/Displacement for the Stirling, adiabatic expansion and compression for Carnot. Ultimately they cancel so have no input on the energy analysis for a full cycle.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Wed Apr 10, 2024 12:20 pm
The more heat converted to work on expansion, the LESS heat remaining in the working fluid to necessitate "back work" to compress the gas, which would only be hot and expanded due to supplied heat (heat supplied to and entering the working fluid) that had NOT been converted to work.
Heat doesn't remain in the working fluid. What you are talking is internal energy. Internal energy remains constant during a isothermal expansion. And zero total change for a complete cycle.
Right, because:
The change in internal energy during an isothermal expansion is zero because the temperature remains constant. Internal energy is a state function that depends on temperature. In an isothermal process, the heat absorbed by the gas is equal to the work done by the gas on its surroundings
Internal energy remains the same because the heat input is immediately converted to work output, 100% so there is no added "heat"/energy left over to increase the "internal energy".

If it is ALL converted to work on the surroundings then there is no added "heat"/energy left in the working fluid that needs to be "rejected". It has already gone out as "work" Joule for Joule.

There is, however "potential energy" due to the change in position

The internal energy is the same, but the piston is at BDC instead of TDC. Because of this potential energy, similar to extending a spring, the piston returns.

Just like rolling a ball up a hill against gravity, the piston has been moved, displacing atmosphere, against atmospheric pressure. The piston returns due to the potential energy, the same way the ball will roll back down the hill due to potential energy.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Wed Apr 10, 2024 12:20 pm ....
Or the 80% energy stored in the atmospheric "buffer" outside the engine compressing the working fluid on the return stroke is the same energy being expelled from the working fluid from inside the engine.
Please leave 'buffer' and atmospheric pressure out of this discussion. I've already explained why outside effects completely cancel in the energy analysis of a complete cycle.
...
Well that's convenient. But that is exactly where you are wrong, so no wonder you want to leave it out of the discussion.

In the real world, an analysis of a real engine includes all of the real forces and energies involved.

Atmospheric pressure is real. The work load on the engine is real and does not "cancel out".

Real engines are not "reversible".

Energy goes out and does not "return to the system". Work output is work output.

You seem to want to make fine distinctions between heat input, "internal energy" and work output, as if these three things can be isolated and treated as separate objects, which is nonsense.

Heat input is transformed into internal energy which is then transformed into work output.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Wed Apr 10, 2024 12:20 pm ...
I'd really rather use math logic and experimentation.
...
Why don't you then.

Try the ideal gas law for starters.

If you expand a gas within a sealed Stirling engine isothermally, the temperature does not change, so the internal energy remains the same. The number of moles of gas is the same, so what happens to the pressure?

If the pressure starts at 1 bar and the volume doubles the pressure will drop to 0.5 bar. So atmospheric pressure will push the piston back until the pressure equalizes.

At what volume will the pressure equalize?

Right back where it started.

Unlike a steam engine where the expansion in volume is accomplished by an actual increase in mass, by the injection of steam into the cylinder.

Heat as a form of energy has no mass.
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

Internal energy remains the same because the heat input is immediately converted to work output, 100% so there is no added "heat"/energy left over to increase the "internal energy".
That is only true in reference to zero Kelvin and Zero pressure. Work would be less if considering a buffer pressure. Heat added during the expansion will be the same.

Well that's convenient. But that is exactly where you are wrong, so no wonder you want to leave it out of the discussion.
My motives are to simplify the analysis, not to be wrong, or deceptive as you seem to imply. It is you that is wrong and your next statement hopefully will point it out.

If the pressure starts at 1 bar and the volume doubles the pressure will drop to 0.5 bar. So atmospheric pressure will push the piston back until the pressure equalizes.

At what volume will the pressure equalize?

Right back where it started.
Somehow the buffer pressure has now been added, could it be that the beneficial aspect of it is providing incentive to do so when convenient. Math isn't about convenience, it is about rigor. You can't add something cause you want it's benefit, and conveniently leave it out when it hurts.

Buffer pressure hurts and helps, efficiency wise they cancel.



A double ended cylinder with a free piston in the center will move away from the side that is heated, and won't move back until that side is cooled, or the other side heated. Expansion won't cool it enough to completely reverse it. Sure it won't expand as much because of work out. Yes, it will expand further if the second side is cooled.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Thu Apr 11, 2024 4:35 am
Internal energy remains the same because the heat input is immediately converted to work output, 100% so there is no added "heat"/energy left over to increase the "internal energy".
That is only true in reference to zero Kelvin and Zero pressure.
We are talking, so far about expansion. A single process. I've already cited references confirming 100% conversion of heat to work is possible when expanding a gas in a cylinder doing work pushing a piston against atmospheric pressure or the "outside environment" in a single process. It is only in a cycle where this has been deemed to be "impossible" and this has nothing to do with zero Kelvin. That zero Kelvin idea is a consequence of the (unproven) Carnot formula. Appealing to your own theory to prove your theory is circular reasoning.
Work would be less if considering a buffer pressure. Heat added during the expansion will be the same.
No idea what your talking about. Atmospheric pressure is the "buffer" in this case.
Well that's convenient. But that is exactly where you are wrong, so no wonder you want to leave it out of the discussion.
My motives are to simplify the analysis, not to be wrong, or deceptive as you seem to imply. It is you that is wrong and your next statement hopefully will point it out.

If the pressure starts at 1 bar and the volume doubles the pressure will drop to 0.5 bar. So atmospheric pressure will push the piston back until the pressure equalizes.

At what volume will the pressure equalize?

Right back where it started.
Somehow the buffer pressure has now been added, could it be that the beneficial aspect of it is providing incentive to do so when convenient. Math isn't about convenience, it is about rigor. You can't add something cause you want it's benefit, and conveniently leave it out when it hurts.

Buffer pressure hurts and helps, efficiency wise they cancel.
Nobody is "leaving out" or "adding" buffer pressure. I stated the engine is sealed, which means it's volume of working fluid is limited to a certain number of moles of gas. as in all Stirling engines. "Buffer pressure" is the outside environment and has no influence on the moles of working fluid inside the engine.

Your ridiculous analysis puts your magical "Qcz" or the outside atmosphere "base energy" inside the engine by illegitimate, impossible mathematical gymnastics.

Yes, when the engine was constructed atmospheric air is trapped inside, but is then sealed within the engine. Moving the piston, or the gas moving the piston expands, contracts, heats up or cools or several operations simultaneously which alters the properties of the working fluid. You seem to think your Qcz inside the engine is somehow immune to these manipulations, it isn't.

Do the math. Apply the "ideal gas law" to the working fluid. The "buffer" (outside atmospheric pressure) does not change, but it stays OUTSIDE.

The Carnot limit formula grew out of a meleu of Caloric theory and steam engines. Maybe it makes sense in that context where you have mass (steam) flowing through the engine carrying latent heat. Applied to a Stirling engine it's just nonsense.
A double ended cylinder with a free piston in the center will move away from the side that is heated, and won't move back until that side is cooled, or the other side heated. Expansion won't cool it enough to completely reverse it. Sure it won't expand as much because of work out. Yes, it will expand further if the second side is cooled.
So what?

Obviously when it comes to a Stirling engine the situation is different and more complex. The piston returns though the engine is continuously heated and never cooled.

You are talking about a very slow process of heating and cooling by conduction.

In an engine completing several cycles per second there is quite obviously other factors involved. Heat converted to work output results in instantaneous cooling of the working gas.
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

Heat converted to work output results in instantaneous cooling of the working gas.
I hope you realize, that statement s just as contradictory as, 'Adiabatic Cooling'.

'Cooling', the process of removing heat, and the fraudulent statement 'conversion of heat to work', both would require 'heat' flow, no flow happens. There is, No heat in adianatics's. No heat is converted to work. Through the process of work Internal energy gets converted to kinetic energy. No heat.

Instead :
Both are a process of conversion between kinetic energies. Temperature, to or from, mgh or 1/2mv^2. Heat doesn't flow. In fact there is no such thing as heat or work. Heat doesn't exist. Work doesn't exist. Caloric theory and heat, were completely discarded about 60 years ago, after starting to be discarded 200 years ago. Please let the theory go bye bye. There are no particles of, heat or work.

The correct terms for the energies involved are, internal energy described by temperature MCvT, kinetic described by Velocity 1/2MV^2, potential described by height Mgh, electrical described by voltage and current EI, and others...

The concept of work, F•DeltaX, is a change in potential or kinetic energy. An addition or subtraction of kinetic energy. Energy exists, work doesn't. Force is not energy, and neither is distance X. It takes a change in distance to enact energy gain to a system. Work is a measure of change of something else.

The concept of heat, M•Cv•DeltaT, is a change in Internal Energy. Temperature is not Energy. It's a measure, the equivalence of distance. Mass M times Cv are equivalent to force. It seems strange but that is how the equations appear to work. Heat is a measure of something else's energy change.

When describing these systems, it is the equation that matters. Experiments just shape the equations. Without the equations there is zero understanding.

In other words, heat is never something that can be converted, it is just a measure of how much of something else that was converted. Internal energy, measured by temperature and mass is converted to kinetic energy, not even work. The term heat converted to work is a confusing erroneous colloquialism. Like calling kilowatt hours, "Watts" [sic]. Wrong in every sense. Unfortunately commonly used to the detriment of communication and science.

Furthermore, if a system performs work by expanding and the temperature drops, performing work on the system by compressing it will raise the temperature back for zero work out.
If the pressure starts at 1 bar and the volume doubles the pressure will drop to 0.5 bar. So atmospheric pressure will push the piston back until the pressure equalizes.

At what volume will the pressure equalize?

Right back where it started.
Yes, but the work output by such a scenario will be zero. Zero area inside the cyclic path equals zero energy output. Zero input too. A real world example would just be a noise making room heater requiring energy input, caused by friction.

.
VincentG
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by VincentG »

There is so much misinformation online, it makes it very hard to use common language unless you are classically trained. Even then there is likely confusion.

Fool here is a question for you.

Suppose an engine is in a perfectly insulated room. The engine is exactly 30 percent "efficient". We send 100 units of heat energy to a perfectly insulated engine running under full load to lift a mass within the room. How many units of heat will go on to heat the room?
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Fri Apr 12, 2024 5:43 am ...
If the pressure starts at 1 bar and the volume doubles the pressure will drop to 0.5 bar. So atmospheric pressure will push the piston back until the pressure equalizes.

At what volume will the pressure equalize?

Right back where it started.
Yes, but ...
But nothing. "Work" has already gone out, permanently and irretrievably.

If an engine is running, then the working fluid is doing "work" to keep the engine running in addition to whatever shaft work to power an external load.

The rest of your post is just unnecessary semantic hair splitting. We've already settled "heat" is the transfer of energy. Nobody said heat is a "particle".
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Fri Apr 12, 2024 5:43 am ....

When describing these systems, it is the equation that matters. Experiments just shape the equations. Without the equations there is zero understanding.
....
.
That may be YOUR point of view, and your entitled to it, but Personally I think that's crazy

Mathematics needs to be grounded in physical reality and tested by empirical experiment.

The Carnot Limit equation fails the test when subject to any attempt at experimental validation.
Conversion of heat into work is accomplished by means of a heat engine
A transfer of energy to or from a system by any means other than heat is called “work”.
This is widely accepted current, modern day, thermodynamics terminology. Your long rant complaining about common language like; 'Adiabatic Cooling', 'conversion of heat to work', and your statement:
In fact there is no such thing as heat or work. Heat doesn't exist. Work doesn't exist.
Is getting just a bit silly IMO.

Your hair splitting over the meaning of common thermodynamic terms is just a distraction from the main issue.

According to the ideal gas law, at the end of an expansion, since it describes a state, it doesn't really matter how it got there, if the temperature ends up the same, the number of moles are the same but the volume has increased, the pressure decreases.

To restore the pressure balance the piston returns.

This is the obvious result that is demonstrated by simple observation of a Stirling engine in operation.

A device that converts beat into work is the definition of a heat engine.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Fri Apr 12, 2024 5:43 am.
....
Furthermore, if a system performs work by expanding and the temperature drops, performing work on the system by compressing it will raise the temperature back for zero work out.
If the pressure starts at 1 bar and the volume doubles the pressure will drop to 0.5 bar. So atmospheric pressure will push the piston back until the pressure equalizes.

At what volume will the pressure equalize?

Right back where it started.
Yes, but the work output by such a scenario will be zero. Zero area inside the cyclic path equals zero energy output. Zero input too. ...

.
This is one reason, as I've said before, the conventional "Ideal" PV diagram does not show actual work output and is misleading.

How can you say: " if a system performs work by expanding and the temperature drops, performing work on the system by compressing it will raise the temperature back for zero work out."

That is a contradiction.

Zero NET external shaft work maybe, but it doesn't appear that external shaft work is even indicated by such a diagram, only pressure/volume work of the gas.

The actual conversion of supplied "thermal energy" into mechanical energy is not shown.

Logically, if as you say: " a system performs work by expanding and the temperature drops," (do you mean pressure?) "performing work on the system by compressing it will raise the temperature back".

There is indeed "heat" being converted to mechanical motion. The engine is moving. "heat" is being transformed into work. Enough so as to cause a drop in temperature and completion of a cycle.(Or if "isothermal" a drop in pressure).

Without a load, maybe you could say zero net external shaft work out from the engine, but when talking about the Carnot Limit, I believe the issue is not practical "useful work" output but a limit on the conversion of thermal energy into mechanical motion; the transfer of energy from the working fluid to the piston.

If you have an isothermal expansion that is without an external load, and an isothermal expansion WITH an external "shaft work" load on the engine, will the idealized PV diagram be any different? NO! The beginning state and ending state are the same. So you just connect the dots, a straight line.

If the piston returns with isothermal compression/contraction and the cycle continues, will the diagram now show work output?

Apparently not. And yet the "working fluid" is expanding and contracting. The engine is running. If the engine is doing external shaft work, there is still zero volume indicated by the PV diagram.

PV diagrams were not developed to show the actual conversion of thermal "heat" energy into "work" or mechanical energy. Like Carnot, Kelvin believed that ALL the heat flowed right through the engine to a "lower" level lower temperature.. But if caloric theory is not true, why should this be true?

So, an isothermal expansion and contraction could do enough actual work to lift a mountain and the PV diagram would show nothing, because there is no change in temperature.

Heat as ENERGY that can be converted to another form of energy is a reality not recognized by a PV diagram.

According to Caloric theory, without a fall in temperature there is zero work potential, so what is isothermal expansion and contraction where the heat input is matched by work output and the temperature (internal energy) remains constant?

You could have tons of heat going in and tons of work going out, and enough energy stored as potential energy to complete the cycle, all while doing additional external shaft work and the PV diagram still shows zero volume.

PV diagrams are crap science based on an obsolete theory of heat. They don't indicate the work performed. They indicate "the fall of Caloric".
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Wed Apr 10, 2024 12:20 pm ...
Tom, I maybe wrong about how you think in regard to the following, but please hear me out. I think you are battling the following beliefs :

1: 200 years of thermodynamics and engine building are somehow wrong.

2: 2000 year old mathematics, such as Calculus and graphs, are wrong.

3: Current science classroom education is wrong.

4: Your experiment can't possibly be wrong or even misunderstood.

5: You want to have greater efficiency, maybe even over unity, so everyone scoffing at such is wrong.

6: Being hands on educated is more important than classroom learning, and that no one with a classroom degree, not the easiest thing to get, has any hands on experience.

I will try to explain, in hopes that it starts people thinking.

1: A scientific theory is only as good as it is useful. If another comes along that is more useful, the previous will be scrapped.
...
Interesting, because Carnot himself, in his last days, already recognized Caloric theory was inadequate. In his own words:

"When a hypothesis no longer suffices to explain phenomena, it should be abandoned, This is the case with the hypothesis which regards caloric as matter, as a subtle fluid."

He went on to systematically cover all the recent experiments of his day that he was aware of that demolish the Caloric theory recognizing heat instead, as a form of kinetic energy. ("motion").

Years after Carnot wrote those words, Kelvin and others, effectively brought caloric theory back, making it the foundation of "ALL" of thermodynamics! (Quote from Kelvin cited earlier in the tread).

This unresolved squabble about the nature of heat has been going on for 200 years!

I thought Tesla, in his "human energy" article made a valid point, that if heat was a form of energy then there was no reason it had to "all" pass through a heat engine, it could be converted to other forms of energy so as to never reach the sink, at least in part, if not entirely.

At that time Kelvin still favored Caloric theory and believed ALL the heat passes through to the "cold reservoir" 100%.

Modern thermodynamics with the so-called 'Carnot Limit" just looks like a cheap compromise, number fudging, which has never been experimentally substantiated.

My only interest was to build a functional engine. To do so I need to know how these engines actually work

Encountering this unresolved issue I've taken, as a mostly disinterested third party, what I think is an entirely objective view.

Examining the history of the debate, I saw, that debate seemed to be all it ever amounted to. A war of words. Nobody ever conducted any experiment. Mostly because the "Carnot Limit" is practically "unfalsifiable". Nobody could build a Carnot engine so how could you ever "prove" it isn't "the most efficient engine possible".

Well, after some thought, my reasoning was, If heat is, or isn't supposed to go through the engine and is or is not "converted" and all this is too difficult to trace or something, and you can't build a Carnot engine for comparison, we do at least have real engines and you should at least be able to catch the "rejected" heat leaving the engine on the "cold" side of a REAL engine and work backwards from there. How much heat does a Stirling engine actually "exhaust"?

Seems like very straightforward, easy to perform experiments to me. Why everybody on the Science and Physics forums (and you) cry how difficult it is is beyond me

I don't care one way or the other, but I think my experiments have been carried out objectively and in good faith and I've simply reported on the results.

Anyone who cares to could easily repeat any experiments I've done, on a budget of less than $500 for a few model engines and thermal measuring devices and other easily obtainable materials.

But will anyone? No! Instead they throw me off the forums for simply sharing the results of simple experiments they all claim are "too hard". Then pat each other on the backs for adhering to "established science".

If you think my methodology is flawed, please just do some experiments yourself. You should have no problem finding all the missing heat that is supposed to be leaving the engines that for some reason I'm too incompetent to locate. Please do an experiment and prove my results are in error.

All your doing here is mud slinging, trying to sully my character and motives or ability to carry out simple experiments.

If you can do better, please do. On the Science forums, I even offered to foot the bill for the engines, but they refuse, kicking me off the forum instead for supposedly advocating "perpetual motion".egads!! another one of THEM! A heretic! Get him out!
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

VincentG wrote: Fri Apr 12, 2024 7:14 am There is so much misinformation online, it makes it very hard to use common language unless you are classically trained. Even then there is likely confusion.
That's for sure. Most physicist are very pedantic about terms to minimize confusion, but Feynman did a very good job bridging this gap.
VincentG wrote: Fri Apr 12, 2024 7:14 am Fool here is a question for you.

Suppose an engine is in a perfectly insulated room. The engine is exactly 30 percent "efficient". We send 100 units of heat energy to a perfectly insulated engine running under full load to lift a mass within the room. How many units of heat will go on to heat the room?
Hey, this looks like a trick question...

70 units of heat go to room since 30% eff was already given BUT this would assume that engine requires multiple cycles (rpm) to achieve said work.

I think I see where you're going, but if mass was lifted within 1/2 cycle (just expansion process) then engine would appear 100% eff and no heat would go to room..
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Tom Booth wrote: Sat Apr 13, 2024 1:44 am
This is one reason, as I've said before, the conventional "Ideal" PV diagram does not show actual work output and is misleading.

...it doesn't appear that external shaft work is even indicated by such a diagram, only pressure/volume work of the gas.

The actual conversion of supplied "thermal energy" into mechanical energy is not shown.

PV diagrams are crap science based on an obsolete theory of heat. They don't indicate the work performed. They indicate "the fall of Caloric".
Indeed, PV plots are limited and don't show finite energy in and out (that requires actual calcs). I assume your current PV beef is that PV plots disregard load vs no load conditions.

Tom, you've gone from trashing Carnot to trashing the kinetic theory to trashing PV plots. While hacking thermo, did you ever wonder if a thermometer actually measures temperature in linear units vs some type of log units (and that we've all been lied to for centuries).

PV plot back story - Watt's invention for comparing output between engines. Clever, yet simple, Watt guarded his secret sauce for decades.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

matt brown wrote: Sun Apr 14, 2024 8:16 pm ...
...I assume your current PV beef is that PV plots disregard load vs no load conditions.
...
My "beef" is, basically;

Carnot based his assumption that heat goes THROUGH a heat engine in tact, to power the engine just like water through a turbine, (that is, there is no material change in the "substance" of either the water or the "heat", both being some kind of fluidic substance), on Caloric theory.

Joule later proved experimentally that heat is a form of energy. "Heat" being one form of energy, "Work" another form of energy. One could be converted completely into the other. Rather than passing through a "heat engine" the heat is "consumed", or "disappears" entirely having been transformed completely into a different form of energy.

Carnot's Caloric based theories were completely incompatible with Joule's experimental results. Even Carnot himself (in his private journals written shortly before his death) concluded that Caloric theory had to be abandoned and that "heat" and "work" were interconvertible, one into the other.

In the Carnot/Caloric theory heat RUNS THROUGH a heat engine like water, in one side and out the other.

In Joules findings, heat enters the engine and completely "disappears" or is "consumed", or "goes out of existence" and is entirely replaced by "work" and vice versa. In other words, a "heat engine" is not unlike a refrigerator in that it PRODUCES a lowering in temperature. Heat goes in resulting in work along with a reduction in heat. A reduction in heat = cold. Cold is the absence of heat. The transformation of heat into work results in an absence or reduction in heat, a lowering in temperature.

This is a natural consequence of conservation of energy. "Heat" is one form of energy, "work" another. When one increases the other decreases and vice versa. The transformation of heat into work results in LESS heat. A lowering in temperature.

More than 20 years after Canots passing we find Kelvin commenting on Carnot's treatise on heat:

As quoted near the start of the thread:

In referring to heat engines generally Kelvin wrote:

if it has absorbed any heat during one part of the operations, it must have given out again exactly the same amount during the remainder of the cycle. The truth of this principle is considered as axiomatic by Carnot, who admits it as the foundation of his theory; and expresses himself in the following terms regarding it, in a note on one of the passages of his treatise:
"In our demonstrations we tacitly assume ... that the quantities of heat lost by the body under one set of operations are precisely compensated by those which are absorbed in the others. This fact has never been doubted; it has at first been admitted without reflection, and afterwards verified, in many cases, by calorimetrical experiments. To deny it would be to overturn the whole theory of heat, in which it is the fundamental principle."
(...) I shall refer to Carnot's fundamental principle, in all that follows, as if its truth were thoroughly established
The entire field of thermodynamics developed by Kelvin then, is a return to Caloric theory.

Yes, PV indicator diagrams were Watts invention. It was a simple mechanical tracing taken directly from the engine. Pressure inside the engine pushed a plunger or gauge attached to a pen, which was also attached in some way to the piston. Probably this was nothing more than a method for monitoring the engine, to regulate the heat input and avoid explosions.

With Kelvin it became abstract nonsense imbued with his reserected version of the Carnot/caloric theory.

Yes, the current PV diagrams not only don't differentiate between load and no-load conditions they are entirely abstracted from any real engine or actual physical mechanics, leading to such nonsense statements as:
...Zero area inside the cyclic path equals zero energy output. Zero input too....
If the gas is expanding and contracting and driving the piston, the engine is running tracing out a PV diagram with the pen attached to the piston, at a minimum the "working fluid" or gas, expanding and contracting is doing "work" to overcome the inertia and friction of the engine. Heat is going into the engine being converted to mechanical motion.

But, because it appears there is "Zero area inside the cyclic path" supposedly nothing is happening. No energy input???

How then do we have a continuous change in volume? How do we have a cycle at all? How do we have a running engine?

If the same engine is doing shaft work turning an electric generator that is putting out 10,000 kilowatts does the abstracted idealized PV so-called diagram represent the work being done by the expanding gas to overcome this mechanical resistance on the shaft?

Nope. No energy in, no energy out.

Joule's results were experimentally demonstrated.

Thermodynamics based on Kelvin and others resurrection of Caloric theory is shot through and through with hypothetical abstract pie in the sky nonsense that has no correspondence with the real world or any real engine.

These abstract PV diagrams of "reversible" engine cycles do not represent anything real.

There are no "reversible" engines, there is no "Carnot engine". Thermodynamics is garbage, in particular the so-called "Second Law" and specifically the so-called "Carnot Limit" equation as generally interpreted and applied which can be easily demonstrated experimentally.

Just try to find the "caloric" flowing through a Stirling engine and out to the "cold reservoir".

As Joules experiments demonstrated, it isn't there.

The "heat" DISAPPEARS!

The "heat" (molecular kinetic energy) is converted to, or transformed into the mechanical motion of the engine.

In the process of heat conversion there is a corresponding "fall" in temperature. No heat passes through the engine. Not through the "working fluid" anyway.

In a steam engine, if steam passes through the engine and comes out as steam, you have latent heat carried by the steam to the condenser, there is mass carrying latent heat passing through the engine. In an IC engine you have air and fuel with combustion gasses.

There is no mass to carry any latent heat through a Stirling engine. Heat goes in and "disappears" resulting in "work" output or mechanical motion.
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