Why a temperature differential?

[Tom Booth]

Well, according to the above instructional videos:

Here I am doing a simple experiment with a model LTD engine, measuring the heat being "expelled" and my infrared imaging camera is indicating a temperature of apparently slightly below ambient, which is around 300°K (not absolute zero).

https://youtu.be/NtrYSpYD43w

Strange to say, however, although my engine is not expelling ice chunks and has gotten nowhere near absolute zero, this could be interpreted as 100% efficiency if all the supplied heat is converted to work.

Well, which is it?

Is 100% efficiency when the engine no longer rejects any heat above ambient or when it reaches absolute zero?

This is an obvious glaring contradiction in theory.

Is 100% efficiency just utilizing all of the heat supplied above ambient or "all of the heat" period, all the way down to absolute zero?

Well, it's quite obviously just two different concepts of efficiency. Relative efficiency based on ambient temperature as the starting point and "absolute efficiency" based on?....

Well, I don't know. Some silly old idea from the obsolete Caloric theory that in this day and age can be completely disregarded.

Yet relative and absolute efficiency are routinely compounded as in the above videos with only a kind of vague surprise or sense of wonder over the obviously preposterous consequences.

So you can be "absolutely sure" my simple experiment is not real. Obviously I'm a liar and a fraud and the experiment must be fake. Certainly that is the one thing we can all be 100% certain about.

[Tom Booth]

Put simply in practical terms; the Carnot Limit postulate asserts that for 100% efficiency, to fully utilize that infinitesimal increase in temperature provided by my hand, (a veritable drop in the ocean on an absolute temperature scale), Rather than the engine utilizing the "drop", ALL the heat, all the "internal energy" of the working fluid down to absolute zero must be completely utilized.

No, that's not correct.

Carnot starts with the fuel.

To be 100% efficient as an engine, your HAND must drop to absolute zero.

You don't know what you're talking about apparently.

For an LTD running on the heat of your hand, your hand is the "hot reservoir" not the "cold reservoir". The heat source not the heat sink.

The ambient, relatively cold side is the sink or "exhaust".

My hand does not go through the engine, only the heat.

Or , well, I suppose you could say the engine needs to suck all the heat out of my hand, and by extension the entire universe to be absolutely 100% "efficient"

[Fool]

I still believe there might be a way around it, but I'm now thoroughly convinced that that can't be done with a piston engine.

Jack I think you are getting it. Excellent. I was hoping this thread would help someone. Thanks.

Your skepticism is very appropriate. I keep an open mind too. I read, and watch a lot of 'free energy' schemes. I see through most very quickly. It strengthens the mind and is fun. Be skeptical about their claims, most are reported by non scientists, a clue into why we should be skeptical.

Put simply in practical terms; the Carnot Limit postulate asserts that for 100% efficiency, to fully utilize that infinitesimal increase in temperature provided by my hand, (a veritable drop in the ocean on an absolute temperature scale), Rather than the engine utilizing the "drop", ALL the heat, all the "internal energy" of the working fluid down to absolute zero must be completely utilized.

No, that's not correct.

Carnot starts with the fuel.

To be 100% efficient as an engine, your HAND must drop to absolute zero.

Thanks for the help, in addition. The way the Carnot theorem works is if you want to get 100% of the heat of your hand converted to work the gas inside the frictionless perfect ideal cycle engine must start at zero Kelvin, and end a zero Kelvin. How many degrees a hand warms it, is not important.

n=(T.hand-zero)/T.hand

T.hand cancels with itself to equal one or 100%.

A hand could also heat the gas to a way lower temperature and still work.

n=(1K-0.0)/1K=1 or 100%

Don't need to freeze your hand, it just needs to transfer a little heat from warm hand to increase the temperature of the cold gas a little. The starting and ending point of the gas is the important temperature, zero Kelvin. I think that is what you were trying to say good one, thanks.

Utter poppycock.

Completely illogical, unreasonable, irrational nonsense.

Logical fallacy of attack on authorities, bashing, and general vituperation, noted and dismissed as unprofessional arrogant error.

Proven on what basis?

As has been demonstrated to you several times, proof is for mathematics. It has been proven. You are being foolish and ignorant to deny this.

Experimental data, measurement of Qh and Work output, can refute it only if those two values, collected meticulously and rigorously with excellent scientific laboratory practices, are outside the prediction. Scientifically 'proven' by being falsifiable and lack of false findings, not from lack of trying.

Your demands are gregarious and unreasonable, and have been satisfied for many. That is a classic symptom of science denial and cognitive dissonance. You hurt yourself and this website.

[Tom Booth]

Well, Mr anonymous "fool" your always entitled to your opinions.

Sorry, but I disagree. Other people are entitled to their opinions as well. Most sane people recognize that and are willing to live and let live. Agree to disagree.

IMO it is your relentless mission to "educate" and "correct" others use of ordinary language (word policing) and opinions (thought policing) that is disrupting and largely ruining the relative peace, harmony and cooperation that existed before you arrived.

Most people try to get along.

Your approach represented here in your own words is bullying. You will not rest until everyone else conforms to your point of view.

viewtopic.php?p=23642#p23642

To "agree to disagree" is a logical fallacy and it fails any logical relevance. It is an erroneous attempt to claim that both of the opposing points are valid, and trivializes the entire discussion. It is very demeaning. It makes the claim that both sides are nothing more than blowhard opinion. It attempts this claim, but fails, using nothing more than fallacy for evidence. Good luck with that.

[Tom Booth]

The practical everyday use of the Carnot formula as currently being taught such as in those previously posted two videos starts with "Q" entering the engine.

Q is not Th.

Th is the temperature of the heat source. My hand, for example

Q is the actual, well defined, finite quantity of heat entering the engine measured in Joules to be incorporated into the "internal energy" of the working fluid.

To use the formula Th does not need to be at "absolute zero". Obviously it can be at any temperature above Tc.

But, IMO, there is no actual clear consensus on the subject.

Nevertheless, once inside Q is treated as the entirety of the "internal energy" which for 100% efficiency would ALL need to be removed.

The single drop added has become "the ocean".

IMO, completely silly, irrational, illogical and of no practical importance to the engineer. Just (at best) an academic exercise in number juggling.

At worst, a handy weapon in the hands of the energy monopolies wishing to suppress competing technologies.

[MikeB]

You don't know what you're talking about apparently.

Tom, you are correct with that, if a little rude! I don't understand everything that goes on in a Hot air engine, and never will unless (a) I ask stupid questions (b) get helpful answers

If your definition of 'perfect' engine efficiency doesn't require the removal/transfer of ALL energy from the fuel source, what is it instead?

[Tom Booth]

You don't know what you're talking about apparently.

Tom, you are correct with that, if a little rude! I don't understand everything that goes on in a Hot air engine, and never will unless (a) I ask stupid questions (b) get helpful answers

If your definition of 'perfect' engine efficiency doesn't require the removal/transfer of ALL energy from the fuel source, what is it instead?

Let's say your heat source is a cup of warm water that was heated with 50 joules of heat.

If the little LTD runs until it utilized all 50 joules that were used to warm the cup of water I'd call that 100% efficiency.

To bring the cup of warm water down to absolute zero would require the removal of something like maybe 250,000 joules?

Just a low-ball rough estimate.

So to satisfy the Carnot limit, my LTD would need to take in and reject 249,,950 joules of heat bringing the water down to near absolute zero, Then it might happily convert the remaining 50 joules into work.

I guess that's reasonable, right?

I just think 50 joules of heat in and 50 joules of work out equals 100% efficiency maybe makes a little more sense.

[Jack]

To check my understanding of this. If a fluid is heated it will keep expanding until it reaches ambient temperatures again?
So assume you have two separate liters of fluid in a closed fully insulated space and you heat one up, you expand it until it reaches starting temperature again. Now you have one with a bigger volume than the other, the same temperature and pressure but lower density?

[Fool]

You have posted many logical fallacies, such as attacks on style and people, and vituperations, that, I mention here only in dismissal. And am concentrating on the science and errors in science.

The practical everyday use of the Carnot formula as currently being taught such as in those previously posted two videos starts with "Q" entering the engine.

Q is not Th.

Th is the temperature of the heat source. My hand, for example

Q is the actual, well defined, finite quantity of heat entering the engine measured in Joules to be incorporated into the "internal energy" of the working fluid.

To use the formula Th does not need to be at "absolute zero". Obviously it can be at any temperature above Tc.

That's a good beginning. Jules are what Q would be measured in, if Q could be measured, they are 'quantifying units'.

Let Th be non zero and higher than Tc. Yes.

But, IMO, there is no actual clear consensus on the subject.

A clear symptom of everyone struggling with these concepts.

Nevertheless, once inside Q is treated as the entirety of the "internal energy" which for 100% efficiency would ALL need to be removed.

The single drop added has become "the ocean".

That is one way to look at it. Another way is to think of it as a bank ballance. Your ballance is 300, you put in a Q of 100, the balance becomes 400. This is not a spital in an ocean. It is a very significant change in internal energy of a small mass of gas. 25% change in fact.

To calculate efficiency of your energy bank the spent amount will be the total 400 minus the final 300 or 100 spent on producing work. We already agree it will not be 100%. Producing work will produce some energy loss to ambient. The ambient is Tc. We call all of this loss, 'heat rejection to Tc'.

We are now trying to determine how much 'n' is. We can measure work by using a tachometer, and torque using a brake, length of arm, and a spring scale. Q can be measured by having it entered through electrical resistance and using volt meter, or an ammeter, or both meters, to get the information.

We can also analyze it purely mathematically. To do it mathematically some Conversions and models must be developed. These can be PV=MRT, U=MCvT, Q=MCv(Th-Tc), W=Qh-Qc, n=W/(Qh-Qc), Uh and Uc.

So how are Qc and Uc calculated? The answer revolves around the concept of the engine getting the same efficiency during a cycle,

1, From UC + Qh, to Patm, then back to Uc.
OR
2, From UC to Uh, to zero, then back to UC.


The Work going from Uh to zero can be calculated, it is just Uh. The work going from zero to Uc can be calculated, it is just Uc. So the total work for the second cycle is Uh-Uc. The heat is just Uh because we are using all the heat in the second cycle converting 400 Jules 100% in to Work 400 Joules. So for the second cycle n=(Uh-Uc)/Uh which is the same engine operating between Th and Tc as the first cycle so it should have the same efficiency, and it is n=(Qh-Qc)/Qh.

It is the same lowest common.denominator and proportion with different numbers. Like 8/10=16/20. Qh an Uh not the same but the fraction 'n' is the same ratio.

That is how the proportion can be the same for total energy and delta energy. Q is not the same as U, but the equation for efficiency is proportionally the same. It makes sense since the same engine should get the same efficiency operating on different sized paths with the same shape. Like two different sized triangles having the exact same angles but different length sides. Proportionally the same but different triangles and sizes.

Qh doesn't become an ocean, the small 3/4 cup of water gets full when four tablespoons are added. Cup moves down. Take four spoons out. Generate energy from them. Use equivalent energy of three spoons to move the cup back. More energy is needed to move heavier cup back up, I.E., the same for Tc when it is higher.

[Fool]

If the little LTD runs until it utilized all 50 joules that were used to warm the cup of water I'd call that 100% efficiency.

I just think 50 joules of heat in and 50 joules of work out equals 100% efficiency maybe makes a little more sense.

I don't know how you are measuring the 50 J of work output. None of your videos include any kind of work measurements. A flywheel on excellent bearings will keep spinning for a long time without rejecting much heat, or absorbing it. Could it be that the little engine might just be absorbing almost zero heat. And putting out zero work.

If a fluid is heated it will keep expanding until it reaches ambient temperatures again?

I calculated that theory in the "Let's beat up Carnot" thread. The expansion stops when P inside equals P outside. The temperature ends up hotter than Tc. The extra temperature holds the larger volume. It doesn't go down to Tc until pressure is well below atmospheric. Again that is a manifestation of internal energy. Work out was way less than heat in.

Gas only expands if the chamber gets larger. The enlargement of the chamber can be from internal or external forces, or a battle between the two.

[Tom Booth]

The single drop added has become "the ocean".

That is one way to look at it. Another way is to think of it as a bank ballance. Your ballance is 300, you put in a Q of 100, the balance becomes 400. This is not a spital in an ocean. It is a very significant change in internal energy of a small mass of gas. 25% change in fact.
...

The problem there, regardless of the exact numbers, is that the existing "internal energy" your claiming here to be 300 is not Q the heat added.

You basically just made my case for me. Thanks.

Starting ambient is 300, (an unrealistically low number in my opinion)

Air temperature, engine, cup of water, the entire planet perhaps, is 300°K to start with.

Say we add 100 joules to the "system", the water, the working fluid, no matter, as long as it is made to eventually go into the engine.

The 100 joules added are 100 joules added.

The heat you supply to raise the temperature is not the "balance".

The "heat" added and the ambient energy are two different things. This is plain and obvious. The heat you supply, the 100 joules is not the ambient energy that was already there, in equilibrium with all the surroundings.

You can't just add up 100 and 300 any more than you can just add together my 50 and 250,000

Making the difference smaller does not make it all alright.

I think Goofy said it well:

viewtopic.php?p=21653#p21653

efficiency is the relation between what you put in, versus what you get out.

Not what you put in versus what was already there.

If we used all that energy, the 100 added PLUS the 300 already there, that would be 400% efficiency. Like a COP it only took 100 joules to convert 400, just to be clear.

[Fool]

Depends on how you look at it. It could be, what you get out verses what you put back in.

[Tom Booth]

Depends on how you look at it. It could be, what you get out verses what you put back in.

Either way, you didn't "put back in" 400 joules.

Rephrasing the sentence doesn't change anything.

BTW, I have no problem looking at "Carnot efficiency" that way, as long as it's understood your actually talking about "all the heat" down to absolute zero.

But this is constantly abused.

"You're proposed engine would only be 2% Carnot efficiency". Don't bother! A waste of time.

It's completely disingenuous, misleading and basically an outright lie.

That is a ridiculous, unrealistic, completely misleading metric.

[Fool]

Abusing a science doesn't make the science wrong.

A lot of people think they can take the 70% wasted energy from a ICE, run it through a Stirling Engine and significantly improve the overall ICE Efficiency. The thing they don't realize is that at 300 to 500 F the best they can do, with a lot of complexity, is a few percent. It may not be cost effective. Run the numbers to check.

Now if there is free hot and cold, and no electricity, efficiency is second or lower, to power and energy out per dollar.

Carnot should be used as encouragement. It has its place.

[Tom Booth]

....

Carnot should be used as encouragement. It has its place.

Potentially, I suppose that's possible.

"Look, your engine did use all the heat supplied to it, that you provided by building a fire under it, but look, that is only a small fraction of ALL the heat in the surrounding atmosphere, maybe only 2% you are neglecting the 98% available in the ambient surroundings."

That would be using it for encouragement, but in my experience on many many messages boards over the years is that it is never used that way. It is used instead to crush out ideas. Discourage plans, warn away potential investors, label inventors as frauds and charlatan's. It is only used to mislead and discourage, abuse, censor discussion, ban members, lock threads and label thinking people as "perpetual motion" "free energy", "over unity" nuts, cranks, dreamers, delusional, etc.etc.

Just as you have been doing to me and others on this forum since your arrival.