Why a temperature differential?

[Tom Booth]

That would make VincentG's response basically correct:

VincentG wrote: ↑Jack if you heat 1 liter of 300k air to 600k and expand it adiabatically, it will reach 1 bar well before it lowers to 300k.

For such a situation. No work, no heat loss, free expansion into a vacuum.

Heating the gas to 600 without allowing it to expand initially would double the pressure.

Free expansion into a vacuum would bring the pressure back down but the temperature would stay at, or near the temperature to which it was heated.

I was reading something not too long ago about the upper atmosphere.

Apparently the air is very hot, maybe thousands of degrees, but because the density is so low you would still freeze to death, because there are so few air molecules to transfer heat.

Temperature is independent of density.

[Tom Booth]

A random quote:

Even though the temperature at the ionosphere is 1000degC you would still freeze to death if exposed.

The individual molecules have a very high energy but the pressure is so low and they are so widely spaced that you would radiate far more heat than you would receive from the occasional interaction with one of these high temperature (read fast) molecules.

https://www.eng-tips.com/viewthread.cfm?qid=121084

Not a particularly authoritative source, but it's an obscure topic and I'm done with searching.

Bottom line I think is, without reducing the internal energy of the gas through work or heat output you can let it expand indefinitely, it isn't going to reduce the temperature

[Tom Booth]

In other words; temperature, as a measure of "internal energy" is a measure of the AVERAGE kinetic energy of the individual gas molecules.

It does not matter how close together or how far apart those individual molecules are, their average kinetic energy is not going to change no matter how much room they are given to move around in.

Without loosing some energy to work or through heat transfer, the average energy level of the particles is the same regardless of how much empty space there is between them.

On another thread, Stroller wrote:

The internal energy of the entire expanding volume isn't reduced, but the internal energy per unit volume is, because the total internal energy of the gas is spread out more, into a bigger space.

By the same token, the temperature of individual molecules of air doesn't change, but the bulk temperature of the gas does reduce, because the thermometer, or finger end, isn't getting hit so often because the molecules are more spread out or rarified in the expanded volume; the pressure falls.

This is half right.

It sounds logical that expansion alone would reduce the temperature:

"the bulk temperature of the gas does reduce, because the thermometer, or finger end, isn't getting hit so often because the molecules are more spread out or rarified in the expanded volume", but experimentally this has been found to not be the case.

When the gas molecules hit the thermometer the temperature reading will not change due to expansion alone.

[Tom Booth]

Tommy, I have answered his question twice.

Well his recent response to that was: "...there was a lot of noise and many words that didn't relate to my question"

I'm guilty there as well , of misunderstanding the question and making assumptions.

Hopefully the foregoing was more helpful.

You appear to not be getting it. And that again doesn't surprise me. You go on and on proving that you don't get it.

For the record it was a simple generic, heat a volume of gas expand it until the temperature is back to the ambient starting pressure T1 Tc. There is zero logic, science, or mathematics, in the whining you are doing now, but there sure is soapboxing, following Jack around, vituperations. Back off.

As far as I know, you are not the forum owner or a moderator here.

Jack apparently did not find your answers helpful. So maybe you should mind your own beeswax.

Jack had a good question, which actually has a good answer, but my previous responses were inapplicable, partly I think because the question was a little unclear or ambiguous, and we all, or at least, speaking for myself, I made assumptions, that the question was about gas in an engine, with atmospheric pressure outside.

Expansion with zero work and zero heat loss is a different situation.

[Jack]

That whole wall of text was actually helpful. Thanks!
I've learnt something new and see that I am actually looking for something else than I thought.

But first I need to read it a few more times and let it sink in.

[Fool]

So why does a heat engine require a temperature difference? Could it be that it needs a pressure difference and heat going in and out provides those differences?

[Tom Booth]

So why does a heat engine require a temperature difference? Could it be that it needs a pressure difference and heat going in and out provides those differences?

Heat going in provides a pressure difference, sure.

But if that pressure difference results in work being done, then the work in joules is deducted from the heat input (turned "internal energy") in joules.

Conservation of energy.

You cannot input 100 joules of heat and take out 100 joules of heat and expect to produce any work, you can only expand and contract a gas, taking out all the heat, if it is doing nothing. No work.

You can input 100 joules of heat and get 50 joules of work and throw away the other 50 joules as heat if you want to.

That you have to throw away 99.5 joules of heat to get 0.5 joules of work from an input of 100 joules, in an "ultra LTD" for example (Carnot limit) is what I object to, and cannot demonstrate experimentally, even if I try.

If, as I've said before, "rejection" is interpreted as not letting the heat in due to low heat capacity that makes a little more sense but is ultimately meaningless.

You could burn down a building and run an LTD on some of the heat radiating off that 200 yards away or something.

You utilize 0.00000000001% of the heat from the burning building, but that heat (the other 99.9999999999 whatever%) is not getting anywhere near the engine.

So interpreting "rejection" as never entering the working fluid doesn't make any sense.

If it is low heat capacity, having 99.5% of the heat going through the working fluid doesn't make sense.

The whole Carnot limit proposition is silly nonsense.

[Fool]

The whole Tom Booth theory proposition is silly nonsense.

That kind of talk proves nothing. Can you feel the pain it causes.

Heat not getting into the hot plate, nor the gas, is important in understanding experimental data, however it is not the heat removal/rejection an engine needs to output work.

Yes, what comes 'out' as work is subtracted from the absorbed heat and won't need to be rejected. The work put back in for compression adds energy that needs to be rejected as heat. Subtracted from the work out makes total work output less, so the heat rejected is more. Conservation of energy don't you know?

It is total work out for the whole cycle.

When you show me work output from a LTD engine at a magnitude of half a Joule or more I'll start listening again.

When you show an input gas temperature increase, ∆T gas, equivalent to 10 Joules or more, I'll start listening again. It must be measured in the internal gas.

Until then I would advise exploring Matt's processes calculator site. Numbers help.

[Tom Booth]

The whole Tom Booth theory proposition is silly nonsense.

That kind of talk proves nothing. Can you feel the pain it causes.

LOL,.. no.

I don't think Carnot can either. He recommended the theory be abandoned

Heat not getting into the hot plate, nor the gas, is important in understanding experimental data, however it is not the heat removal/rejection an engine needs to output work.

Yes, what comes 'out' as work is subtracted from the absorbed heat and won't need to be rejected. The work put back in for compression adds energy that needs to be rejected as heat. Subtracted from the work out makes total work output less, so the heat rejected is more. Conservation of energy don't you know?

It is total work out for the whole cycle.

When you show me work output from a LTD engine at a magnitude of half a Joule or more I'll start listening again.

When you show an input gas temperature increase, ∆T gas, equivalent to 10 Joules or more, I'll start listening again. It must be measured in the internal gas.

Until then I would advise exploring Matt's processes calculator site. Numbers help.

I wish you would stop listening. Please do.

I have better things to do than field your criticisms, mischaracterizations and misrepresentations all day and night, and more I don't have words for in polite company.

You waste my time on these petty pointless debates over mostly non issues of little consequence.

I've at least gathered real experimental data. I've seen nothing of the kind in the way of empirical evidence to support the Carnot efficiency limit, from you or anyone else in the past hundred years, or however long it's been around.

Just a lot of vehement claims about its infallibility that boarders on religious fervor, fanaticism and hero worship like devotion to the great Carnot. All empty gesticulation as far as I can see, nothing substantial in support of the theory (so-called "LAW") at all.

[Tom Booth]

... The work put back in for compression adds energy that needs to be rejected as heat. ...

I disagree,

First of all, the work atmosphere does pushing the piston in is DRIVING THE ENGINE, at least until the internal pressure builds up to a pressure above the atmospheric pressure, which does not happen until very near TDC.

Further, the heat generated after that point contributes to the engines power stroke, by converting velocity to potential energy in the "air spring" working fluid and directly back into heat at just the right moment, if the timing of the engine is set properly, with a 90° advance.

That is how a driven oscillation works. The driving force is introduced with a 90° offset. That contributes to the power of the oscillations, and to that extent "work" can be withdrawn.

You can consider that just Tom Booth's unfounded theory or whatever you like, but it fits the data and experimental observations 1000 times better than anything else I've seen, especially the Carnot limit, based on a completely obsolete theory of heat.

I really don't care what some "fool" on a message board thinks.

[Fool]

Too bad. You are wrong. Your theory doesn't have any backing. Your empirical evidence lacks any power measurement, or indicator diagram, or several temperature measurements necessary for any conclusions.

It is you, that must provide the missing data. Asking us to repeat your inconclusive experiments is getting you nowhere.

[Tom Booth]

Too bad. You are wrong. Your theory doesn't have any backing. Your empirical evidence lacks any power measurement, or indicator diagram, or several temperature measurements necessary for any conclusions.

It is you, that must provide the missing data. Asking us to repeat your inconclusive experiments is getting you nowhere.

Like I said, I could care less about your opinions. I don't need an endorsement from an obvious self-proclaimed "fool" who does an excellent job of living up to his own name.

[Fool]

I'm "excellent". Cooooool!