Struggling With Internal Energy..

[Tom Booth]

A confusion between two concepts does not involve mathematics, it involves common sense.

If they are mathematical concepts, they require mathematical common sense, aka. equations.

You erroneously insinuate I have some issues with the equation: ∆U=Q-W, but I do not.

The internal energy U can change ∆ by the addition or subtraction of either heat or work.

If either heat or work are added or subtracted from the internal energy the internal energy will increase or decrease.

Good. So do you accept the following:

∆U=Q-W

∆U=U2-U1=Q-W

U2=U1+Q+W

U1=MCvTc

Up to this point.

U2 however = (as above) U1+Q-W

U1 and U2 are state functions Q and W are path functions so there can be virtually an infinite number of +Q -W interactions taking heat in and pushing work out before arriving at the state U2

All these additions and subtractions, potential or actual are "invisible" on a P:V diagram which is a "connect the dots" of state functions but hides what is or might be in between.

In other words

U1 + Q(=5001 Joules) -W(=5000 Joules) = U2

The ∆U=1Joule
Actual work output?

5000 Joules

5000 is of course an arbitrary number and could be anything, 10,000; 100,000 whatever.

U2=MCvTh

∆U=MCv(Th -Tc)=MCvTh-MCvTc

???

They are the equations that go with the first law, conservation of energy.

Fundamentally work and heat are equivalent. Different labels for one and the same unit of energy.

Work and heat are energy in transit. Energy can be quantified with the units of Joules, calories, watt hours, horsepower seconds, etc... it is best to use the same units throughout. But that doesn't make work the same as heat. An equivalence exists, yes. Equivalence, and "The same", are not the same. Mathematically speaking.

In this case they are actually the same, mathematically and physically. Heat is the exchange of kinetic energy between particles of matter. Work is the exchange of kinetic energy between particles of matter.

Work just has more simultaneity. The particles act and respond more in unison.

Heat is random particle collisions. Work is organized particles collisions.

A heat engine turns disorganized into organized particles collisions by setting up oscillations in the working fluid particle field so as to deliver organized blows to the piston like soldiers with a battering ram.

Fundamentally however, an exchange of kinetic energy between particles is an exchange of kinetic energy between particles.

You seem to think or be arguing that an increase in U that results from heat addition can only be reduced by heat removal, which misses the entire point regarding the equivalence of heat and work.

I'm arguing that adiabatic processes do not involve heat transfer. Work exchanges with internal energy. Because heat, work, and internal energy are not the same thing,

Heat = energy
Work = energy
Internal energy = a ham sandwich?

certain processes can affect two variables, effecting a proper energy balance without affecting the third. Adiabatic is where Q is Zero. The equation becomes:

∆U=W

Only if the engine cycle is "purely" 100% adiabatic.

Which, of course, is not an engine cycle.

All engine cycles begin with heat addition which generally does not fit into any neat "ideal" thermodynamic category.

What thermodynamic term describes heat converted to potential energy by compression or rapid expansion of a gas with a simultaneous temperature increase?

To put it in simple thermo terms which is unfortunately, inherently inaccurate you could say Q in occurs "before" W out, one way or another. Then, in addition, once Q in is converted to motion/velocity additional work becomes possible drawing on U.

Moving on:
Heat and work exchange when ∆U= zero.

0=Q-W
Or
W=Q

Two cases where that is true are, isothermal process, and a full cycle. For other individual single strokes, ∆U is not zero. If it's not adiabatic, Q will not be zero. And single strokes with a change in volume will have nonzero work output, could be negative.

It's the full cycle that is of Carnot concern. When adding up a full cycle, it is easy to make mistakes, leave something out.

True. Your simplistic "ideal" whatever leaves out a lot.

You don't need more math, you need a better understanding of basic thermodynamic concepts.

You don't need more math, you need a better understanding of basic thermodynamic mathematical concepts.

Nicola Tesla was right about a lot of things, however quantum mechanics has proven Tesla and Einstein wrong, it is extremely useful for atomic theory. The atomic bomb was first proven two years after Tesla died. That was about the time the last doubts of Entropy were eliminated, and cemented in the Carnot Theorem. Tesla may not have known.

The important thing in this thread is to understand the difference between Q, ∆U, W, and U and how Q added to U1 produces U2. And how taking W out of ∆U requires Qin and Qout, unless adiabatic, for a full cycle.

Your theory goes against the standard model. Two choices, one a new model needs to form. Or your tests are somehow not working.

If you want someone to agree with your theory, you need to model it. I see no way to model it, and can't help. Further more I see why current theory prohibits it, and it's more than just because it does. Valid logical, mathematical, and empirical, objections arise.

Mostly irrelevant and/or incoherent rambling and opinionations there.

It's worth pointing out however that this is wrong;

The important thing in this thread is to understand the difference between Q, ∆U, W, and U and how Q added to U1 produces U2.

That is a complete and final sentence.

However you left out W.

You said: "Q added to U1 produces U2" which is wrong but reveals your misunderstanding of the subject as well as the math.

The equation is: ∆U=Q-W

That is

Q added to U1 minus W out produces U2

Which you still fail to understand miserably. Unfortunately for all who have to endure your persistent ramblings and insistence that everyone accept your error.

You've turned ∆U=Q-W into ∆U=Q1-Q2 or something of that nature.

Still don't comprehend the equivalence of heat and work, in fact, you deny it outright, which obviously, IMO goes against very well established science.

[Tom Booth]

This, perhaps deserves separate treatment:

Fool wrote:

And how taking W out of ∆U requires Qin and Qout, unless adiabatic, for a full cycle

.

Again the equation is ∆U=Q-W

Not ∆U = Qin - Qout

That is, of course, possible, but not an engine.

Technically it is ∆U=Qnet-Wnet

The first law does not forbid ∆U=Q-W where Qin is equivalent to Wout.

True, the second law does forbid that and it is generally recognized that 100% conversion of heat into work is impossible, even in a single process in an engine.

I at least recognize expansion of a gas in a cylinder to drive a piston will involve loses to friction, vibration, noise, conduction, radiation, etc. even for a single one shot process.

So I don't contest the letter of the second law either as generally stated. 100% conversion of heat into work is impossible.

True enough.

The Carnot Limit formula as currently interpreted and applied in academia however goes far far beyond saying you can approach but never reach 100% efficiency.

The Carnot efficiency limit puts a hard cap on efficiency based on a completely arbitrary ratio based on no science or sound mathematics whatsoever.

Where there is a small ∆T that cap might be as low as something like 0.2%

Take the P-90 Ultra LTD for example running on a ∆T of 0.5°C

The thing is running on virtually no heat at all. The efficiency of Stirling engines generally is recognized to be very high.

True it is not putting out "useful work".

"Useful work" however is not what the W in ∆U=Q-W represents.

Thermodynamic work includes the motion of the running engine itself which in an isothermal expansion could be 100% theoretically or ideally.

So, the second law, generally speaking, or as broadly interpreted says that you may be able to achieve 99.99% efficiency, but not 100%

The Carnot Limit however may put the cap at 0.2% for actual known engines that have been demonstrated to run on virtually no heat at all, a barely detectable 0.5°C

What that implies is that in such a case running an "ultra LTD" 99.5% of all the heat entering U must be found flowing straight through and out the LTD to the other side to be "rejected".

The argument that the working fluid has a low heat capacity is illogical.

How does a working fluid take in transfer and dump the 99.5% waste heat to the "cold reservoir" given it only has the capacity to take in and convert 0.5% of the supplied heat.

The Carnot efficiency limit is so obviously bogus nonsense it's hard to imagine why anyone would promote it as an "unbreakable LAW".

I can't fathom how anyone ever took it seriously at all, in the first place ever.

[Tom Booth]

I also think it is clear that if the engine is running, heat is indeed entering the working fluid or U.

From the thermal image readings it also seems undeniable that heat is getting into the LTD engine in considerable abundance.

Compress_20240630_075644_4754.jpg (14.99 KiB) Viewed 2319 times

https://youtu.be/NtrYSpYD43w

For "some reason" however, the heat entering into the engine does not reach the cold plate. It enters the working fluid, expands the gas and drives the engine, but that's as far as it goes.

The heat is converted to work. Random particle motion is converted to coherent organized mechanical motion, with little if any residual heat left over to be "rejected'.

To me this is quite obvious and undeniable.

It seems far fetch to me, but perhaps the thermal image readings are misleading and internally the engine is still near room temperature. The heat is just going up around the sides outside the engine?

As I say, IMO a rather far fetched theory, the bottom plate is highly conductive metal in direct contact with hot steam. but perhaps someone should put thermocouples actually inside the engine just to make sure.

I may do that, just to verify that the working fluid is indeed taking in heat.

[Fool]

There is more that needs adjustments. Unfortunately I lack the time and energy to accomplish them. Too many

Only if the engine cycle is "purely" 100% adiabatic.

Which, of course, is not an engine cycle.

All engine cycles begin with heat addition which generally does not fit into any neat "ideal" thermodynamic category.

Q=0 for any adiabatic process. If a full cycle is composed of all adiabatic processes, Q will be zero as well.

Yes, that is an adiabatic spring, not an engine. If heat is added to the spring it's PV.pot just revolved at a higher pressure and Temperature, barring any further heat input, it will return to Th not Tc.

If heat is added at appropriate points, and removed at other appropriate places, it becomes an engine.

[Tom Booth]

There is more that needs adjustments. Unfortunately I lack the time and energy to accomplish them. Too many

Only if the engine cycle is "purely" 100% adiabatic.

Which, of course, is not an engine cycle.

All engine cycles begin with heat addition which generally does not fit into any neat "ideal" thermodynamic category.

Q=0 for any adiabatic process. If a full cycle is composed of all adiabatic processes, Q will be zero as well.

I think you are basically describing an oscillator. But you do need to set the oscillator in motion somehow, with heat or work input.

Yes, that is an adiabatic spring, not an engine.

Agreed, I think.

If heat is added to the spring it's PV.pot just revolved at a higher pressure and Temperature, barring any further heat input, it will return to Th not Tc.

The oscillations will continue at a different energy level. Sure.

If heat is added at appropriate points, and removed at other appropriate places, it becomes an engine.

Well No.

If we are talking about a heat ENGINE, then a heat engine, by definition, is a machine that converts heat into work.

Your statement:

If heat is added at appropriate points, and removed at other appropriate places, it becomes an engine.

Leaves out work, so is incomplete and does not describe an engine.

To be an "engine" at least some energy needs to go out as work.

Energy goes out as either heat or work, but if ONLY heat, then not an engine. You have heat transfer with no work output.

That is not an engine.

You can quibble about heat not being directly converted to work.

Sure, first, or concurrently, the "heat" entering into the engine becomes "internal".

Internal thermal energy or just internal energy if you like.

So the "internal energy" is converted to work.

The heat, I guess you can say is indirectly converted to work, but that does not change the equation. You can't just drop the W and still have an engine.

[Fool]

Leaves out work, so is incomplete and does not describe an engine.

To be an "engine" at least some energy needs to go out as work.

You just contradicted yourself. If it becomes an engine it is putting out work. It only does so if it absorbs heat, and rejects heat, both. Otherwise, with just one, it is just, as you say, an "oscillator". I prefer an air spring, in oscillation.

Yes an oscillator needs an energy input, constant and phased correctly for a real one, and, an initial only input for an ideal oscillators.

[Tom Booth]

Leaves out work, so is incomplete and does not describe an engine.

To be an "engine" at least some energy needs to go out as work.

You just contradicted yourself. If it becomes an engine it is putting out work. It only does so if it absorbs heat, and rejects heat, both. Otherwise, with just one, it is just, as you say, an "oscillator". I prefer an air spring, in oscillation.

Yes an oscillator needs an energy input, constant and phased correctly for a real one, and, an initial only input for an ideal oscillators.

I don't really understand your issue here.

I did not say there could not be both heat input and output, just that without work output, by definition, it is not an engine.

Your statement said nothing about work output.

If heat is added at appropriate points, and removed at other appropriate places, it becomes an engine.

No, just removing only heat is not an engine.

Whatever heat OR work is removed needs to be replaced by additional heat OR work if you want to maintain the oscillation.

If, as you wrote: "If heat is added at appropriate points, and removed at other appropriate places, it becomes an engine.

If you just add and remove heat you just have an oscillator that increases and decreased in its energy level. Vibrates faster or slower.

To be an engine some of that energy has to be converted to work output is all I said.

How is that a contradiction?

What percentage of the energy can be converted to work is a different issue. By definition, to be a "heat ENGINE" there needs to be some work output of some kind.

A theoretical "perfect" oscillator with no friction and weightless piston is not doing any "work".

[Fool]

"No, just removing only heat is not an engine."

Correct, it must be removed at appropriate points in the cycle for work to come out making it an engine.

It is redundant to say an engine with work coming out.

[Tom Booth]

"No, just removing only heat is not an engine."

Correct, it must be removed at appropriate points in the cycle for work to come out...

LOL,

You always word this in such a way as to imply that the work output is a mere side effect.

The work output is at the expense of the internal energy of the gas/air/"working fluid".

...making it an engine.

It is redundant to say an engine with work coming out.

Maybe so, in ordinary conversation, but you have promoted your "Calorics" theory in here to such a degree I think it needs to be pointed out your not so subtle use of words to imply heat is added and removed in toto, while work is simply a side effect, like free energy to do work springing out of nowhere.

100% heat in 100% heat out and some work output as a mere side effect.

The old Carnot Caloric theory of water going over a water wheel and ALL the water continuing on its way.

That was a reasonable hypothesis if heat was a fluid that actually flowed "down" in temperature, rather than just being another form of energy.

That has long ago been proven to be a fallacy. An obsolete theory of heat.

Why some people like you cling tenaciously to this old obsolete Calorics theory is beyond my comprehension.

[Fool]

You are the only one that brings up Calorics.

[Fool]

I'm not talking 100% heat out. Duh. I'm taking appropriate heat out at a specific point in the cycle.

Use Matt's calculator to see what kind of work, forward and backward you get out of each kind of cycle. As I've already shown it equates to the Carnot maximum. Try it yourself if you don't believe me. If you don't know how to do it, I can explain it better. I figured it out without any help. It took some time. I figured you can too.

Your guessing will fall apart when you put the numbers into your description.

[Tom Booth]

You are the only one that brings up Calorics.

Well somebody using the "fool" handle has:

Taking just one page for example:

viewtopic.php?p=22026#p22026

Carnot, Kelvin, Caloric and Kinetic Theories. All of which are useful if used properly, and even sometimes improperly. Education is the key to knowing how to use them. Carnot and the rest were brilliant for using what they had to work with.

...

Caloric Theory had been abandoned in favor of the Kinetic Theory. Perhaps the fact that the old mathematics was reverified was missed by you?

...

Caloric theory works just fine...

....

If a waterwheel can put out work with all the mass going through, losing some of its energy by altitude drop, a heat engine can output work if all the caloric goes straight through losing some of its energy by temperature drop.

...

Calorics are just carriers. They carry different quantities of heat as specified by the temperature.

Just one page.

Aside from that, modern "Carnot efficiency" historically was 100% based on Caloric theory and is just a watered down version of the same, and no, modern mathematics has not verified anything.

Your fake "derivation" failed miserably. Just circular reasoning.

[Tom Booth]

I'm not talking 100% heat out. Duh. I'm taking appropriate heat out at a specific point in the cycle.

Use Matt's calculator to see what kind of work, forward and backward you get out of each kind of cycle. As I've already shown it equates to the Carnot maximum. Try it yourself if you don't believe me. If you don't know how to do it, I can explain it better. I figured it out without any help. It took some time. I figured you can too.

Your guessing will fall apart when you put the numbers into your description.

I'll take a look, but calculators work how they're programed to work.

I'm a computer programmer myself, they are far from infallible and often full of glitches. Not saying that one is, but as I already mentioned, if it doesn't allow the user to specify moles it is already highly suspect or possibly not applicable to Stirling engines.

[Fool]

It's blatantly obvious you haven't even looked at that calculator site.

Page 1 of the "truth" thread is where you are the first to bring up Caloric Theory, in disgust.

You are often the first to bring it up in many threads.

I've done my share of programming as well.

"Your fake "derivation" failed miserably. Just circular reasoning."

I already covered that derogatory logical fallacy.

If you have a problem with my derivation, calling it fake doesn't disprove it. You must show the logical or mathematical flaw. You didn't, so my proof stands up from lack of disproof.

[Tom Booth]

It's blatantly obvious you haven't even looked at that calculator site.

Page 1 of the "truth" thread is where you are the first to bring up Caloric Theory, in disgust.

You are often the first to bring it up in many threads.

You advocate it and support it.

A completely disproven and obsolete theory.

You still "believe" in it is the point.

I've done my share of programming as well.

"Your fake "derivation" failed miserably. Just circular reasoning."

I already covered that derogatory logical fallacy.

If you have a problem with my derivation, calling it fake doesn't disprove it. You must show the logical or mathematical flaw. You didn't, so my proof stands up from lack of disproof.

LOL, what a joke.

Baloney. I showed exactly where you inserted your altered Carnot limit equation at the very start of your so-called derivation, then pretended to "derive" the equation from the equation itself.

Not only "fake" but IMO appears to have been intentionally misleading and deceptive.