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

[Tom Booth]

The "first law" thermal efficiency equation is:

Qh=Ql+W

Which I don't have a problem with. It does not imply that work can exceed the heat input, but neither does it suggest that the work output in Joules cannot match the heat input in Joules.

The first law does not forbid the existence of an engine more efficient than the Carnot limit, not at all.

Qh = Qc + W

Or rearranged:

W = Qh - Qc

Second Law:

Efficiency ratio:

n = W/Qh

Or combining first and second

n = (Qh-Qc)/Qh = 1- Qc/Qh

Qh and Qc have a linear relationship to Temperature. Meaning add 100 Joules get 100 K temperature increase. Add 200 Joules get 200 degrees increase.

So they are of the following form:

Qh = Th x Cv + K
Qc = Tc x Cv + K

Y intercept, constant: K = zero because, at absolute zero Kelvin, there will be no heat. Qh and Qc will be zero at zero Kelvin Temperature.

So those two simplify to:

Qh= CvTh
Qc=CvTc

Putting those into the previous equation for n, efficiency:

n = 1 - Qc/Qh

Or

n = 1 - (CvTc)/(CvTh)

The Cv's cancel, becoming one, which leaves:

n = 1- Tc/Th

That is Carnot's Theorem derived from the first and second laws. The first law, conservation of energy, manifests itself into a great many following areas. Maximum efficiency is just one of those.

Here is a more rigorous proof:

Relationship to ideal gas law:

https://en.m.wikipedia.org/wiki/Thermod ... emperature

Scroll down to the appropriate section. It uses a slightly higher level of mathematics.

There are other more complicated proofs should you care for them. It has been proven many different ways.

The Carnot Theorem is just for thermodynamic efficiency. Kinematic efficiency reduces the efficiency even further. And not following the ideal engine cycles, also reduces the efficiency.

Efficiency is a valent goal to pursue, but mostly a waste of effort for a starting concern. Power to weight ratio is a much better pursuit. Think American Drag Racer. 1/4 mile per 12 gallons. 48 gpm. Excellent.

[Tom Booth]

Thanks, but I've been through all these "proofs" and equations a hundred times already.

The way it looks to me is you start with the first law.

Then you have this completely arbitrary "Carnot" (more like Clausius) "limit" adopted as a premise, and like trying to fit a square peg in a round hole just insert an "=" sign between the two. Out pops "entropy" and all kinds of other nonsense.

It's like you adopt a premise that all men are 7 feet tall, then use that for all your calculations.

It might seem to work most of the time on the basketball court, but it's a false premise with limited application and fundamentally simply not true at all outside of some very limited circumstances.

As they say, even a stopped clock is right twice a day. You just need a very limited field of vision and ignore all the anomalies.

[Tom Booth]

Wow! Thanks for the compliment. I guess from your response that everything mathematical is correct in my supplied proof. Good. Great.

I'm sorry you were unable to follow it. Your erroneous comments on it is telling. I would be happy to explain further if you ask specific questions as to where you begin to disagree. If no questions, I'll assume you are uninterested in the truth.

[Tom Booth]

How do you define "heat"?

[Tom Booth]

Are you OK? I'm looking forward to your response.

For now; I'll go with the definition in Britannica:

Heat:

energy that is transferred from one body to another as the result of a difference in temperature.

That is of course the modern scientific definition, in contrast with the Caloric theory.

I pointed out not too long ago a passage from Kelvin:

In his "complete commentary on Carnot's theories"

He wrote regarding Carnot, heat engines and Caloric theory:

if it has absorbed any heat during one part of the operations, it must have given out again exactly the same amount during the remainder of the cycle. The truth of this principle is considered as axiomatic by Carnot, who admits it as the foundation of his theory; and expresses himself in the following terms regarding it, in a note on one of the passages of his treatise:
"In our demonstrations we tacitly assume ... that the quantities of heat lost by the body under one set of operations are precisely compensated by those which are absorbed in the others. This fact has never been doubted; it has at first been admitted without reflection, and afterwards verified, in many cases, by calorimetrical experiments. To deny it would be to overturn the whole theory of heat, in which it is the fundamental principle."
(...) I shall refer to Carnot's fundamental principle, in all that follows, as if its truth were thoroughly established

Personally the first time I came across this I found it jaw dropping. "In ALL that follows". Do you know how much that encompasses? Basically all of thermodynamics.

So the question is important. Basically all of thermodynamics, especially as it relates to heat engines is firmly rooted in Caloric theory, and for the most part this has never changed!

This "fundamental principle" is that no heat is converted into work when it powers a heat engine. "if it has absorbed any heat during one part of the operations, it must have given out again exactly the same amount during the remainder of the cycle.".

"All that follows" includes your proofs and formulas. Kelvin states it plain and clear. It's all based on the assumption that the Caloric theory was "thoroughly established".

He admits that isn't actually true. He admits Caloric theory is wrong. He admits the whole foundation of thermodynamics will need to be revised and put on a more secure footing someday. He admits heat is actually energy that is transformed into work in a heat engine, but,... he's going to continue on the foundation already accepted and established with no looking back. Leave that to somebody else someday far far in the future.

So we've been left with this mishmash of modern physics and Caloric theory just wedged together in a big baloney sandwich for our continued diligent consumption.

So, can we agree on the Encyclopedia Britain a definition of heat as "ENERGY that is transferred from one body to another as the result of a difference in temperature or not?

Because it makes a world of difference when examining those mathematical formulas, because those formulas as you've presented are impossible if the modern scientific definition of heat AS a form of ENERGY is accepted. They are only possible based on Caloric theory.

[Fool]

I'm okay with heat being a form of energy passing from hotter to colder. I am not comfortable with internal energy being called heat. Something hotter has a higher temperature and higher internal energy than that same thing at a colder temperature, everything else the same.

Sorry it took so long to reply, I get busy, tired, lazy, distracted, procrastinated, plus, I like to think about your questions, and think a lot, sometimes for days. I research and think about my answers, also often for days, sometimes longer. I want to check and recheck that I'm not missing anything.

I think the question, "what is heat", is a proverbial 'good one'. I'm glad you put out an answer before I had a chance to do so. Although I think heat needs more than that description, it isn't your definition of heat that is our problem. It appears to be what follows.

Heat can be 'described' just as you have stated. Mathematics 'defines' the variable Q as heat, and it is solely to be used in equations for models. Subscripts are used to separate different processes or values of heat.

Moving on, I have a problem with your last statements. Conservation of energy is not derived from conservation of caloric, the two tend to contradict each other. Similar to how in an inelastic collisions kinetic energy and momentum appear to not both be conserved. They are, however, caloric doesn't exist. Let us forget about caloric.

First law:

Qh = Ql + W

Has zero to do with caloric theory. It reflects the observation that energy is conserved and caloric isn't. If caloric were conserved, the equation would be of the form:

Qh = Ql

Caloric/heat in would equal caloric/heat out. Work energy would then have to spontaneously form and be pushed out magically. Since this has never been observed, we conclude energy passes through and is conserved, caloric isn't (Which agrees with the lack of observed caloric.). Done talking about caloric. Leading to many energy balance equations such as the following:

Qh = Ql +W

Which means that for work out, the ejected Ql must be smaller than Qh. The process of 'converting heat to work', isn't direct. Heat must first be converted to internal energy. This is accomplished by a rise in temperature and or pressure, or both. 'Internal energy'/'pressure', pushes out the work, and at the same time pressure decreases, decreasing internal energy. That equation doesn't, by itself, disallow Ql from being zero. It does have a serious, divide by zero, problem with Qh being zero.

A side example. That is not the only equation. Example, the following:

Qh = U + Ql + W

U = Energy saved, stored inside the engine. Also called internal energy.

The left side is energy in, the right side is energy out plus energy saved. The equals symbol means the two sides must balance, similar to putting weight on either side of a balance.

The energy balance equation can be expanded even further:

Win + Qh + Uout = Uin + Ql + Wout

U can be in the form of potential, kinetic/momentum, light, or heat, chemical and perhaps others. Nuclear decay strikes me as one I left out.

Ending with:
Since the equation, Qh=Ql+W, doesn't come from caloric theory, and you said you were comfortable with it, can we move on with accepting it as the first law of kinematic thermodynamics? Please. If more clarification is desired, please explain.

I'm hoping to get to efficiency, n =W/Qh next.

[Tom Booth]

I think I'm fine with everything so far.

Thank you for the very thorough and lucid explanations.

Also sorry about the hit-n- run comment. I think this is a very important and critical issue. Probably THE most fundamental or pivotal issue in all of thermodynamics

And thank you for taking this with the seriousness it deserves and giving it some actual careful consideration.

Feel free to carry on.

[Tom Booth]

We might look at this equation at some point:


Q/t = kA((T1-T2)/l)

Q/t = heat transfer rate
k = thermal conductivity of the material
A = cross-sectional area,
T1-T2 = temperature difference
l = thickness

And/or

Q=m⋅c⋅ΔT

Q – heat transferred
m – mass
c – specific heat
ΔT – temperature difference

And maybe

ΔT=Th−Tc

∆T= temperature difference
Th= Hot temperature
Tc= Cold temperature

Specifically, what is Q or Qt when T1=T2 ?


I mean we know, I suppose, in a heat bath of uniform temperature particle collisions continue and so on a molecule to molecule level thermal energy (or translational kinetic energy) continues to transfer but overall, is it safe to say that if T1=T2 net heat transfer is zero ?

[Fool]

Okay. Thanks.

If you are comfortable with:

Qh = Ql +W

Are you comfortable with rearranging the terms:

W = Qh - Ql

That is the exact same equation. Just solved for W, Work.
Or would you like me to go through the detailed steps?

If comfortable, I'd like to combine that equation with efficiency. (Put W=Qh-Ql into the efficiency equation for W.):

W = Qh - Ql
into:
n = W/Qh

and get:
n = (Qh-Ql)/Qh

I can give more detail on that too.

Please. Any one may ask for more detail. It would please me to expand. I would rather not lose anyone at this point, or any point. Algebra is about skipping steps if you can. They can always be put back in if needed for clarity. I wasn't really comfortable with this stuff until having two years of college. I had to use small steps. I confess.

[Tom Booth]

Just a reminder of what has been established so far:

"Heat can be 'described' just as you have stated."

Which was what exactly?

We are accepting the Britannica definition: "energy that is transferred from one body to another as the result of a difference in temperature."

"Mathematics 'defines' the variable Q as heat, Subscripts are used to separate different processes or values of heat."

So can we then also assume that the same definition for heat also applies to the various subscripts?

Qh Qc Ql Qt etc. ?

All sub-categories of heat or processes involving heat are subject to the same rules or the same definition.

No ∆T no "heat"
No transfer of energy that results from a ∆T no "hear'.

Are there any exceptions to this rule or definition?

Is there an exception for Qh for example?

Q in other words constitutes QUANTITIES of ENRGY so, we can also substitute Joules,in for Q if the quantity of heat transfered is known.

No problem with that?

No exceptions for any of the various Q subscripts ?

[Tom Booth]

Okay. Thanks.

If you are comfortable with:

Qh = Ql +W

Are you comfortable with rearranging the terms:

W = Qh - Ql

That is the exact same equation. Just solved for W, Work.
Or would you like me to go through the detailed steps?

If comfortable, I'd like to combine that equation with efficiency. (Put W=Qh-Ql into the efficiency equation for W.):

W = Qh - Ql
into:
n = W/Qh

and get:
n = (Qh-Ql)/Qh

I can give more detail on that too.

Please. Any one may ask for more detail. It would please me to expand. I would rather not lose anyone at this point, or any point. Algebra is about skipping steps if you can. They can always be put back in if needed for clarity. I wasn't really comfortable with this stuff until having two years of college. I had to use small steps. I confess.

I have no problem with the mathematics per se.

There is, however, now an issue looming on the horizon.

It is important that we have some clear definition of terms when applying these formula to the specific subject matter.

The subject matter under investigation is external combustion heat engines. A separate matter from internal combustion engines, turbines etc.

So, what does Qh for example actually represent?

The heat of combustion itself?

Of course, the heat of combustion in an IC engine is internal to the engine, so we can basically answer YES.

Combustion in an external combustion engine however takes place outside the engine, so can we still use the same definition for Qh or Q1 and so forth for an internal combustion engine as for our external combustion engine?

I would say no. These formulas cannot be broadly and indiscriminately applied but Q along with the various subscripts have to be clearly defined on a case by case basis.

Does Qh represent the total heat of combustion.? Or the heat that is transfered to the hot heat exchanger, some of which could be lost due to conduction or radiation, or are we talking only about heat that is transfered from the hot heat exchanger to the working fluid inside the engine?

There may, of course be similar issues in regard to the other subscripts.

I think it can be seen where internal combustion engines constitute a separate category of study or investigation when using these mathematical tool sets.

Qh does not necessarily represent the same quantity when applied to either an external or to an internal combustion engine.

[Tom Booth]

Similarly when talking about internal combustion engines and efficiency, we are talking mainly in terms of physical fuel consumed or burned.

Stirling or external combustion heat engines do not require the direct burning of fuel at or inside the engine.

They can be used in solar and waste heat applications for example where the connection between the fuel burned and the heat produced as a consequence is more tenuous.

Take solar for example. Ultimately the energy source is the sun which burns or fuses hydrogen.

Light reaching the earth from the sun further differentiates into various wavelengths, some of which can be absorbed and transform into heat, and some may be reflected.

I believe photovoltaic panel efficiency is calculated on the basis of total sunlight of all wavelengths not just the narrow band that can be absorbed and converted to electricity.

Of course, depending on what we actually want to know, the quantity Qh could be different, or alternatively might be further subdivided.

[Bumpkin]

When I saw the above question, “How do you define heat?” I knew it wasn’t asked of me, but I thought: (As a noun,) Molecular motion duh. That “official” Britannica definition is too clever by half. If instead of “energy that is transferred,” they were to say “energy that has the potential to be transferred,” it would be better, but why send people off into the weeds like that? I’d still say; molecular motion duh. That way I can stay out of the weeds.

Bumpkin

[Tom Booth]

When I saw the above question, “How do you define heat?” I knew it wasn’t asked of me, but I thought: (As a noun,) Molecular motion duh. That “official” Britannica definition is too clever by half. If instead of “energy that is transferred,” they were to say “energy that has the potential to be transferred,” it would be better, but why send people off into the weeds like that? I’d still say; molecular motion duh. That way I can stay out of the weeds.

Bumpkin

I don't care what the definition is, as long as it is used consistently. Words are just symbols or sounds to represent a concept. Having our nomenclature well defined can help avoid a lot of conflicts and misunderstanding.

[Tom Booth]

In this context I'm more concerned with the mathematical "Q" designation which is supposed to, I believe, have a very well defined precise scientific meaning.

BTW, to "fool", you said:

Please. Any one may ask for more detail. It would please me to expand. I would rather not lose anyone at this point, or any point.

I'm OK to move on,, but your thorough, detailed, step by step treatment of the subject is something that is quite valuable and badly needed, so for the sake of general readership, including anyone who might come along in the future, I would favor as much thoroughness and detail as possible.

Assume your listeners know nothing about the subject. Each element of any algebraic formula, for example should be defined and explained so that the equations make sense to everyone, including those who know absolutely nothing about the subject

I don't personally have any problem with your presentation so far and I think, heartily agree with it in every detail.