Could Both Carnot and Tom be Correct?

[Fool]

I have demonstrated several correct ways to have stated the percentage. The important things to be sure of in mathematics are the little details and standards and precise definitions. It's the only way to learn why mathematics is so reliable. Gloss over the details and it loses its meaning.

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[Tom Booth]

I have demonstrated several correct ways to have stated the percentage. The important things to be sure of in mathematics are the little details and standards and precise definitions. It's the only way to learn why mathematics is so reliable. Gloss over the details and it loses its meaning.

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What you did is to take a statement out of context then twist and misinterpret that perfectly correct statement and insist it was wrong when it wasn't. Then try to insert your own "correct" calculation which was not correct at all for that context.

You apparently just enjoy wasting time and spreading confusion or like arguing for the sake of argument even long after you've been proven wrong.

Typical useless "Troll".

[Fool]

Last I checked it is the trolls that resort to name calling, when corrected. Grow up.

[Tom Booth]

Last I checked it is the trolls that resort to name calling, when corrected. Grow up.

Your so-called "corrections" are nothing more than a diversion and a waste of time.

Your a moronic troll at best or some kind of malicious saboteur out to derail conversation and research or both.

Either way a waste of time and internet/forum resources.

[Fool]

Blah blah blah...

[sarimshaikh]

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[matt brown]

Could Both Carnot and Tom be Correct?

I've always suspected the following, but just figured it out. Heat is being rejected in a nonmagical but different way than through the cold plate.

The question is why are both Carnot and Tom correct? The answer is buffer pressure. The derivations I've provided in the past, are valid for both an engine running in a vacuum or with a buffer pressure. Tom is working exclusively with a buffer pressure. Although the efficiency equations are not affected, the heat gets rejected in different ways depending on having a buffer pressure or not.

The fact that an isothermal power stroke is 100% efficient, heat in equals work output, only applies to zero buffer pressure, and a single stroke. Many derivations, using calculus, assumes integration to zero pressure. And or zero Kelvin. Running an atmosphere or buffer pressure, the zero in calculus now becomes the buffer pressure. Easily recalculated, should someone care to do so. Adds a simple subtraction of buffer pressure.

Buffer pressure 15 psi. Plus of minus 5 psi. 15/5 = 0.3 or 30%.

A nonzero buffer means the power stroke is no longer 100%. Example, say 30%. 70% of the heat coming in goes out as work through the piston into the cold sink known as the atmospheric pressure. This energy is lost directly to the outside air. It is gone. It doesn't get rejected as heat through the cold plate. In fact the cold plate gets colder during this stroke if insulated. The gas, by definition, isothermal, stays hot.

Both right and wrong. I usually scheme with external pressure considered a vacuum or Pmin of expansion stroke equal 1 bar. In this manner, both have no 'direct' Wneg effecting expansion process (stroke), but any external pressure will have an 'indirect' Wneg effect during expansion.

This 30/70 example misses that external "load" is constant and internal gas dynamics identical for both external vacuum and "buffer" pressures. The difference is that when external vacuum, 100% of expansion force can go to mechanical load vs when external buffer, only 30% can go to mechanical load since 70% will go to buffer load. PVT values remain constant regardless of external pressure, but simple engine designs with both ambient charge and buffer will require a flywheel to complete expansion otherwise the piston will stall mid stroke when inner and out pressures equalize.

Dismissing the magic of flywheels is bogus woke energy nonsense...

​In other words, for a 80% Carnot rule engine, the buffer pressure eats 70% of the rejected heat as atmospheric work. Leaving 10% or less to come out of the returning piston and cold plate.

Yikes, Fool needs more sleep or less grog - lol

Assuming isothermal expansion, the buffer pressure is eating 70% of Wpos during expansion with no effect on T or Qin

[Tom Booth]

This thread would be better titled "Could both Matt Brown and Fool be talking gibberish?"

The answer would be: absolutely.

[Fool]

NOT.

Matt, I was playing devil's advocate as a good scientist should. The area inside the path is the same regardless of buffer pressure or not. Tom claims he has insulated the cold plate extremely well. The engine is free running, so only the internal loads require any heat input at overcoming work. I.e., minimal amount, and very difficult to measure. But if his claim is true, and zero heat is coming out from the cold plate, for Carnot to be correct, it must come out somewhere else. I am playing with some ideas. Thanks for your input.

So if the indicator diagram doesn't change between, open and insulated cold plate, the heat must be coming out in another way. Of course, he hasn't provided any indicator diagram to shed light on this. So all we can do is guess. I am exploring how heat/Work can come out during the forward stroke by battling the buffer pressure. Perhaps it's more proof that very little heat is going in to free run the engine.

Yes I know it doesn't make any sense.

Of course, a better explanation might be that part of the cold plate gets colder, and part gets hotter. The hotter ultimately winning out. But that information hasn't been measured either.

It is also possible that the outside of the piston is pumping ambient room air in and out from under the insulation. I asked for the engine to be inverted and the flat side insulated/cooled, and the piston side heated. That too was met withe resistance and ignorance.

Science dies, when zero data is measured.

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