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

[Fool]

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You can add back 300K from room temperature. Which, as Carnot predicts, becomes 100% efficiency, and 400J out per cycle. However, you forget that going to zero Kelvin is impossible except in the ideal mathematical analysis of the process. Remember, the goal is to calculate the efficiency with the constraint that the temperature only drops to 300K and to no less than atmosphere pressure. The point I was making going to zero was that it doesn't effect calculations if you make proper accounting of the energies necessary. Adding more heat wasn't part of the definition.

You have no clue how to properly account for the energies involved. Hence you add up things willie nilly and come up with 800% efficiency. It makes about as much sense as the following.

Put 100J in, expand it to -500000000K. That is minus Kelvin.
Free move back too 300K ambient heat goes back to 300J.

n=(500000000+300+100)/(100)=5000004 or 500000400%
Wrong wrong wrong...

Stupid stupid stupid.

Comparing what is theoretically possible to what your constraints will be is a very specific process. You can't add magic energy from the atmosphere to get to 300K if you haven't expanded it below that temperature. It takes energy away from the 100J expansion momentum for it to expand beyond the 1 bar pressure. Your momentum would run out long before you got to zero Kelvin. But good luck trying.

Infact at an expansion to 1 bar, (the constraint) you won't even be down to 300K, or get out 100J. But that is a different story and only a related calculation and proof. It is beyond your mathematical abilities.

Sure, atmospheric pressure will give it back, at the expense of heat of compression, for again a zero zero gain from the attempt. So good luck trying for a partial expansion to zero Kelvin. And good luck getting the gas to only condense and freeze to the head, during a full expansion to zero Kevin.

You seem to be forgetting that you can't get any energy from the environment if you set up that temperature constraint in the definition of the theorem. As I say, please, go ahead and try. I know you will come up with the lame excuse that it is beyond your desire to build a simple Stirling Engine. Or instrument it. Bollocks. Grow up.

Measure, the work, or an indicator diagram, before you say you know anything other than, 'insulating the cold side of an engine doesn't stop it, hmmm. Why? '. The question of why, can't be answered without that data. Most likely it is an unobserved miniscule heat leak in your experiment.

[Tom Booth]

"However, you forget that going to zero Kelvin is impossible"


LOL...
It was your nonsensical "ideal" scenario, I just pointed out one of your glaring omissions which puts your whole "point" in the trash bin.

Obviously though, you could run a Stirling engine between a dewar of liquid nitrogen and "free" ambient heat, or whatever heat source you choose.

The volume of the air cooled to -320°F will still be near zero due to the gas "contracting"/condensing.

Compress_20241030_032918_8974.jpg (38.4 KiB) Viewed 794 times

https://youtu.be/KFRlFj9p9V0

So, with no back pressure from liquid efficiency should still be near 100% on both the ambient heat expansion and atmospheric pressure return.

But with "free" ambient heat as a power source, who's that concerned about efficiency?

With 100% plus efficiency, however, no heat would be transfered into the dewar from the engine. (According to the Carnot theory).

Anyway, you're now basically admitting your own argument is nonsense. Another "retraction". LOL...

[Fool]

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What are you blabbering about? I omitted nothing. You tried to add a useless side diversion from your lack of any reasonable logic. I easily destroyed it. I left it as a trap. You laughed at your own ignorance, and gleefully fell right in. Snap! Thanks for the entertainment. Only I don't think it's funny. It's a piddy.

If you run a heat engine on liquid nitrogen, it will not exceed the Carnot efficiency associated with the temperature constraints. Run a dyno test. Prove Carnot wrong. How many thousands of Watts did it take to make the liquid nitrogen? How many are you getting back? Right, milliWatts output.

n=(300-78)/300=0.74

If you get 30% or more I would congratulate you. 40% would be awesome. 50% or higher would be phenomenal.

It would be interesting even if you get 30% of the energy needed to boil off the nitrogen. But my guess would the overall efficiency is likely 10% to 1% or even less.

Maximum COP for making liquid nitrogen.

COP=(320/(320-60))-1=0.23 or 23%.

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[Tom Booth]

If your engine isn't transferring heat into the liquid nitrogen, then you only have to make it once, or infrequently.

Reportedly a quality dewar can maintain liquid nitrogen for more than six months.

[Fool]

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Yes "IF". The magic "if". Good luck with that.

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[Tom Booth]

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Yes "IF". The magic "if". Good luck with that.

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It has already been experimentally confirmed, many times.

You prefer to believe the so-called "Carnot" efficiency limit.

That's your prerogative, but Carnot was wrong.

[Fool]

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No it hasn't. You are wrong, and refuse to verify it.

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[Tom Booth]

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No it hasn't. You are wrong, and refuse to verify it.

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Verify what? lunatic.

Wrong about what? You're a crazy person obsessed with the ramblings of a natural philosopher in 1823 who immediately retracted everything he wrote as obsolete.

Unfortunately he died before his retractions could be published.

The kinetic theory and ideal gas laws were superceded by the Lennard-Jones potentials and Van der Waals forces. (and probably something else by now) You need to get a little more up to speed from 1823.

[Fool]

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I only respond to science that continues to pass the rigors of every new scientific viewpoint. You are just using guesswork. Good luck with that.

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[Tom Booth]

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I only respond to science that continues to pass the rigors of every new scientific viewpoint. You are just using guesswork. Good luck with that.

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More like you only consider old, outdated, obsolete disproven nonsense from a bygone era.

My theorizing or "guesswork" is based on observations and experiments because your so-called "science" is such a muddle of obsolete nonsense and contradictory, baseless theorizing I really have no choice.

I studied all your crap ten years or more ago and found it severely lacking and incompetent to explain observable facts.

You prefer to simply ignore facts and reality for your "ideal" fantasy land.

[Fool]

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You have failed to study the things that would help you. A pity.

Ideal theories are only one part of a formal education. Your lack of knowing that is telling. Learning how to read a phase diagram.

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[Tom Booth]

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You have failed to study the things that would help you. A pity.

Ideal theories are only one part of a formal education. Your lack of knowing that is telling. Learning how to read a phase diagram.

I've studied phase diagrams quite a bit.

Generally speaking they contain a number of "gray areas" and where they do show apparent sharp divisions can be misleading if taken literally.

They need to be supplemented by more in depth understanding, the details of molecular interaction, repulsion and attraction that you deny even exists at all with your broken record insistence that "gases only push, never pull" nonsense.

You are quite obviously a moron. Your opinions are crap. You need to get a life and stop wishing you were me. When is the last time you had an original thought or idea?

Your a brainwashed automaton, or perhaps a chat bot, as far as I can tell.

[Fool]

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I've studied phase diagrams quite a bit.

Studying one, and learning it, are two different things. Please use a diagram to explain why gasses at really low pressures and temperatures still 'push'. In fact tell me at what temperature water turns to steam, boils, at 0.001 psi? Show it on a diagram.

Your a brainwashed automaton, or perhaps a chat bot, as far as I can tell.

Grow up. If that is your defense. I win. You loose. Science comes from all directions. Your job appears to be ignorance.

Besides chatbots are becoming smarter than you. Eventually they'll even be smarter than Matt, VincentG, and myself. Then your Caloric reasons for all that is evil, will come crashing down on you from pure logic and data.

Your erroneous logic makes me think you're an uneducated human.


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[Tom Booth]

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I've studied phase diagrams quite a bit.

Studying one, and learning it, are two different things. Please use a diagram to explain why gasses at really low pressures and temperatures still 'push'. In fact tell me at what temperature water turns to steam, boils, at 0.001 psi? Show it on a diagram.

Your a brainwashed automaton, or perhaps a chat bot, as far as I can tell.

Grow up. If that is your defense. I win. You loose. Science comes from all directions. Your job appears to be ignorance.

Besides chatbots are becoming smarter than you. Eventually they'll even be smarter than Matt, VincentG, and myself. Then your Caloric reasons for all that is evil, will come crashing down on you from pure logic and data.

Your erroneous logic makes me think you're an uneducated human.


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I don't need to prove anything to you dipshit.

[Fool]

Well, right off the bat I see the same conceptual difficulty with:

Let Qhz be the internal energy from heat added to the cycle all the way from zero K Kelvin to Th.

We add 100 joules at a temperature of 400°K

As stated previously "internal energy from heat ADDED to the cycle all the way from zero K Kelvin to Th" counts the baseline "internal energy" of 300 joules as if it is carried along or included in the 100 joules actually supplied as "heat".

Then later this same baseline internal energy is supposedly "rejected" though in actuality it was never added in the first place.

You already conceded previously:

Qcz is not rejected during the running. Nor is it added. It is the base energy that DQh is added to. Earlier definition discarded.

viewtopic.php?p=21686#p21686

This is the fundamental fallacy of the whole Carnot Limit so-called "LAW". It's an accounting error.

The baseline "internal energy" between 0°K and 300°K cannot be included with "heat added".

The only way that statement begins to make sense is in the context of Caloric theory where heat is considered to be a fluid that passes through the engine and temperature is the measure of a quantity of a fluid.

Qcz=MCvTc. It is the internal energy at the start of the cycle top dead center acquired by the room temperature and pressure. 300K and 15 psi. It is before Qh 100J is added. It is 300J.

It is like having 300 dollars in the bank.

There is talk that ambient air is full of energy. This is that energy and contained by the engine's working space, TDC.

Our goal is to determine the efficiency of the engine for a full cycle in reference to thermodynamic variables. First look at the efficiency if no heat is added.

300J, 300K, 1 to 1 temp energy correlation, determined by gas mass and type Cv. If a stroke expands that gas, the gas does work. That work will be equivalent to the temperature reduction. 100K = 100J. If the expansion goes until T becomes zero Kelvin, the engine will put out a maximum of 300J. Ideally.

A zero Kelvin all real gasses would have frozen into a solid chunk. Luck of the draw, the chunk is all located on the head. The piston can now be returned with zero work used. Back at top dead center with no compression or back work needed. 100% of the initial internal energy is converted to work. But wait...

The gas is a frozen chunk, end of cycle not reached. To get there, heat must be put back into the frozen gas until it is 300K again. That will take 300J, or the same amount as the work gained. Leaving a zero energy in zero energy out. So including the base energy has no effect on power in or out of a cycle.

Now let us add 100J every cycle. Start at 300J add 100J, expand to zero Kelvin return free stroke. Return 300J from the 400J work obtained. Grand total per cycle 100J in and out, or 100% efficiency. As predicted by the Carnot formula if you go all the way to zero Kelvin.

So now we want to see how efficient a cycle is if we don't go all the way to zero. Start at 300J add 100J for a total of 400J. Expand it getting 100J of work out.

Shazam! Pushing the piston back in now requires work. It comes from the momentum obtained by the expansion. Why? Because the gas is at a minimum pressure. It is at or close to atmospheric pressure 15 psi, but not zero, infact high, near 75% of the highest value.

I've already shown that going from 400K to 300K and back to TDC staying at 300K, is going from 400J to 300J and getting 100J of work, and now minus back work.

n=work/total Joules available.
n=(400J-300J)/400J = 0.25 or 25%.
Or normalized with temperature:
n=(Th-Tc)/Th=(400K-300K)/400K=0.25 or 25%.

So how much work does it take to compress it. Well it is the work that the efficiency takes away.

100J•(1-0.25) or 75J. That work must be rejected as 'the heat of compression' or the temperature will rise above Tc=300K, and require more back work for an even lower efficiency.

This is true for any complete cycle operating between two temperatures. No specific cycle was specified, therefore it applies to all cycles, all engines.