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To be fair, I'll say, I "checked" the validity of your derivation briefly, or rather, after some hours doing the various substitutions and whatnot on paper looking for purely "mathematical" errors, while assuming the Carnot efficiency to be correct and then plugging that in.

I haven't cross checked it using numbers assuming efficiencies other than those known to be predicted by the Carnot assumption.

Such a more in depth less "easy" analysis will take more time while also being alert to logical errors or hidden false or potentially false assumptions.

I don't know, or have not found exactly where, or what Matt means by your "switching it up".

Anyway, I"ll run through it again, or a few more times using numbers other than "Carnot efficiency" numbers.

I recall, while working out some of the checks using the "known"Carnot values that the equation I was analyzing would have accepted "ANY" value I chose.

In other words, it looked like the formula was "reverse engineering" η based on the Carnot assumption rather than the other way around.

In other words, if I ",assumed" 100% efficiency the resultant would be 100% but I ignored those apparent hiccups, as I was only looking mainly for basic math errors "at the front door" so to speak, rather than anything sneaking in through the back door, by way of "assuming" some value or equivalence without actually working it out from the "available data".

Anyway I"ll refrain from assuming your derivation is "wrong" without being able to put my finger on exactly where or how.

I mean, taking the fuel tank analogy.

Suppose, instead of your gas tank being only 1/4 full of gas it is 1/4 full of gas AND 3/4 full of an inert fluid "mixed" with the gasoline?

Then every new gallon or "Joule" "supplied" or added to the "tank" becomes diluted/contaminated by mixing with this "given" but inert/useless Qcz.

Thanks.

I like to think of the car analogy as having a full tank of fuel and only burning the top 1/4 part of the tank, then refilling it. 100 J for the added 1/4 tank. 3/4 of it 75 J are spent lugging around the extra weight of fuel never used. 1/4 or 25 J spent getting you around.

400 J of energy converted to work on the forward stroke. 300 J getting back to the start. That sets the efficiency. I'll have look at how to post the letter Etta. Qcz is kind of a penalty cost of back work or compression.

Spending only 100 J for work, means 75 J will be needed to return to the start. Proving that is similar to proving 1+1=2. It seems trivial. It even looks circular at times, as 2-1=1, and 2-1-1=0 and 0+1+1=2 all tend to come up. I never was much for mathematical proofs. All are different ways of getting to two, but two is still two.

LOL.

Superposition it is now. LOL.

Like Schrodinger's Cat?

So what now? Is the proper term now "Cat Wash". LOL indeed. Love it. Meow!

Superposition, not "Quantum Superposition"! You lost me! Tee hee hee. (Galaxy Quest "minors/miners")

https://en.m.wikipedia.org/wiki/Quantum_superposition

https://en.m.wikipedia.org/wiki/Superposition_principle

Superposition is a mathematical and engineering tool that allows effects to be added up. It is sort of like seeing further cause you are standing on the shoulders of giants.

The speed of a ball Vb, please make it wiffle ball for safety, thrown from a car as seen by a person on the ground is the speed of the car Vc plus the speed of the throw Vt. Vb=Vc+Vt superposition.

Well at least until the air friction hits it and slows it down almost immediately.

Like a six foot person being able to paint a 36 foot wall by using a 30 foot ladder.

Qhz = Qcz + DQh
400=300+100

Hee hee hee 'Cat Wash'. Too silly. Thanks.

Analogies or thought experiments are a dime a dozen. Which analogy comes closest to describing reality is the question. Ultimately resolved by observation/experimental outcome.

I like to think of the car analogy as having a full tank of fuel and only burning the top 1/4 part of the tank, then refilling it. 100 J for the added 1/4 tank. 3/4 of it 75 J are spent lugging around the extra weight of fuel never used. 1/4 or 25 J spent getting you around

One possibility.

But as said regarding heat pumps, it's easier to move than create fuel/heat.

Hopefully if I happen to always keep my tank topped off, my car engine doesn't use up 3/4's of all the gas I put in just to carry around the "extra" gas that never gets used.

Mat Brown wrote:Fool, Tom is right #1, you're "switching it up" as the kids say.

I think Tom is a little confused as to why I would be defining the number 300 several different ways. It's not my intention to identify how the starting point is at 300 J, just that it is and is tied to Tc.

Depending on how one arrives there, the end point is still 300. Building an engine in Earth's atmosphere encloses 300 J as the starting and ending point for the engines cycle. Switching it up was an attempt to explain how that number could be calculated in different ways. Like trying to get to 300 by adding up ones. Or 3 100's or 6 50's. They all end at 300. A cohesive calculation explaining the validity of the number Qcz=300 J at 300 Kelvin.

Matt Brown wrote: I only got this after my recent zero-zero comment, a buzz I haven't used in years.

I think you have the justification nailed. Integration goes to zero. 100 Joules in 100 Joules of work out for an isothermal process only applies if the area under the curve goes to zero. In my proof example, using superposition and integration and absolute temperature, 400 J in for 400 J of work out, and 300 J of back work in. Put in the atmospheric buffer pressure and forward work dwindles to 100 J and back work dwindles to 75 J, to be consistent with n = Etta.

Just to reword it for everyone, the efficiency isn't going to change if 1/4 or 100 J are used. Back work will still be 75%.

You still pay the penalty of lugging around the extra weight of fuel. Cars are more efficient on the bottom 1/4 of a fuel tank, because the extra weight is missing.

In a car fuel weight is not a big percentage of car weight. You won't see a big savings. It may not be worth the risk of running out, or moisture condensation.

In a low temperature heat engine Qcz is a big load to carry around.

Fool wrote: ↑Fri Apr 19, 2024 9:56 am ...

400 J of energy converted to work on the forward stroke. 300 J getting back to the start. That sets the efficiency. I'll have look at how to post the letter Etta. Qcz is kind of a penalty cost of back work or compression.

...

Another way of looking at it:

400 joules converted to work on the forward stroke. Emphasis on ALL 400 joules have already been converted to work during expansion. An already acknowledged possibility.

So, where does all that work go? How is it used/divided up?

#1) the heat supplied is also already gone. Used up/converted.

#2) as a result of #1 the working fluid internal energy has not changed. (But pressure and possibly temperature have dropped).

#3) the work, however much it amounts to for the return stroke has already been "paid" in full. Work/energy is returned to the gas. At least some of that returned energy can be used in restoring the original pressure and/or temperature of the working fluid.

#4) much of the return energy is in the form of velocity The piston returns due to a pressure imbalance. The piston picks up velocity on its way back to TDC.

Personally, in spite of Matt's protests to the contrary, I don't buy that expanded cooled gas molecules do not ever "contract".

Kinetic theory is an idealization based on "ideal gases" with no intermolecular or intramolecular attractive forces.

Real gases do have attractive forces.

Water vapor "contracts" or condensed onto a glass of ice tea on on a picnic table on a summers day. Quite heavily, so the water droplets run down the side.

How do the water droplets get there? The cool water vapor molecules slow down and begin to attract each other more than they repel each other due to kinetic motion. When kinetic energy is less, attraction is more on a gradient, always, at all times, REAL gas molecules are in balance between attracting and repelling each other. Otherwise water would never condense and gases could never be liquified and earth would have no atmosphere and there would be no oceans.

So on "expansion" the working fluid was expanded, "stretched" like a rubber band, storing potential energy. "Compression" then, in a Stirling engine, is mostly a returning of the "stretched out" gas back to its original position and state.

The energy to do the "stretching" came from the initial heat input. In the process of returning from a "stretched" position, the gas COOLS further. In returning to its original state the gas absorbs heat returning some of the internal energy lost during expansion.

#5} the velocity of the piston on the return stroke can be converted back into heat at TDC giving the piston some additional "adiabatic bounce", strengthening the oscillation rather than demi fishing it, if this conversion of velocity back into heat is carried out at just the right instant.

In an engine, timing, of course, is always crucial.

The return stroke, then, rather than reducing conversion of heat into work, simply makes best use of the work already generated from the supplied heat fully converted during expansion.

The return stroke is like a rock that has been lifted. The potential energy stored in the rock, when the rock is dropped, converts to velocity. It doesn't convert into heat until the rock hits the ground.

The velocity of the piston converts into heat when the piston "hits" TDC. Right when the heat is needed!

Look at a REAL PV diagram.

Pressure and temperature continue to drop after BDC on the return stroke.

On the return stroke pressure doesn't rise above outside atmospheric pressure until very close to TDC (full compression).

The piston returns the same way, even with no flywheel and no displacer and no "sink" and can even lift weight in opposition to gravity in the process:

https://youtu.be/e-7DFp_B0y4

That's one pretty heavy metal magnetic clip that working fluid is lifting on the return stroke some 10x a second or so.

Now, as far as overall "efficiency" in terms of external work. Granted, that could be zero or near zero, but that is not the issue. The issue is how much of the heat taken in by the working fluid is converted to mechanical effect or mechanical motion and so no longer manifests as the form of energy we know as "heat'.

"Back work' is still WORK not "heat".

Fool wrote: ↑Fri Apr 19, 2024 10:53 am You still pay the penalty of lugging around the extra weight of fuel. Cars are more efficient on the bottom 1/4 of a fuel tank, because the extra weight is missing.

In a car fuel weight is not a big percentage of car weight. You won't see a big savings. It may not be worth the risk of running out, or moisture condensation.

In a low temperature heat engine Qcz is a big load to carry around.

That's mere opinion/speculation.

I could say it is negligible. Also opinion/speculation.

So, I make my determinations based on observation/experiment and try to leave my and others opinions out of it.

I think it is a numbers game, not pure opinion. Qcz = 300, DQh = 100, Qcz is three times larger.

If Th is 1000 K and Tc is 100 K then it becomes :

Qcz = 100, DQh = 900 and Qcz is way less significant.

I guess that is the main reason LTD Stirlings have little power and low efficiencies.


I'm on Matt's side. Gas never contracts. There is always a positive pressure inside a gas bottle even when the gas condenses. Gas doesn't ever pull. It always has a positive pressure. Vacuums don't pull, there is nothing to do the pulling. Evacuating a container reduces the pressure until it is almost zero. It never quite gets to zero. Search for "vapor pressure of a liquid vs temperature". It will never become zero.

Your contraction theory fails to consider sublimation of a solid, liquid propane bursting into gas when the pressure is released, and the fact that condensing water on a glass fails to "pull" anything else with it. It's more like a slow moving water molecule sticks to a colder surface rather than pulling anything with it. Kinetic energy just rattles it loose.

Riddle me this, Fool.

Describe the conditions of a perfect world in which 50% or more of the energy can be extracted from a 300k to 400k cycle. What real world conditions are directly related to the low efficiency that a 100k delta gives on planet earth?

VincentG wrote: ↑Sat Apr 20, 2024 2:54 pm Riddle me this, Fool.

Describe the conditions of a perfect world in which 50% or more of the energy can be extracted from a 300k to 400k cycle. What real world conditions are directly related to the low efficiency that a 100k delta gives on planet earth?

Fool, hold my beer...

Current science says there isn't any, since energy appears to work the same thruout the universe.

Per this running 300-400k cycle example, no one (including Tom) has any issue with 300J passing thru engine from 400k to 300k. The problem occurs when trying to explain why the 100J SOURCE input fails to yield 100J work output across this cycle (ie the Carnot theorem) whereby maximum work output = 25J vs 100J. The problem is explaining how the potential work output of this 100J source input is (at least) diminished by the same ratio as the thermal ratio, whereby maximum work output = 25J despite 100J source input.

It really boils down to the internal energy of a gas. When we create a heat engine (regardless of type) we're doing so from our relative position. A more absolute position would be if we consider 400k a red army and 300k a blue army.

At a glance, it appears that when the red army = 400 and the blue army = 300 that the red army will win a battle via 400 - 300 = 100. Simple math that a politician might entertain, but a military commandeer would try to avoid. The caveat with this folksy analogy is the quantitative approach where the red and blue armies are different sizes, but all soldiers have equal "quality". However, if we use a qualitative approach where both armies have the same soldier count and consider each blue soldier only 3/4 the quality of each red soldier, then a summed battle could still have the red army win the day intact, but end at 25% fighting strength where 3/4 of the red army's fighting strength (think energy) was lost fighting the blue army.

It's all about how you approach and sum this, and what Fool is struggling with while trying to keep it simple and logical.

Fool wrote: ↑Fri Apr 19, 2024 10:45 am
Integration goes to zero. 100 Joules in 100 Joules of work out for an isothermal process only applies if the area under the curve goes to zero. In my proof example, using superposition and integration and absolute temperature, 400 J in for 400 J of work out, and 300 J of back work in. Put in the atmospheric buffer pressure and forward work dwindles to 100 J and back work dwindles to 75 J, to be consistent with n = Etta.

Just to reword it for everyone, the efficiency isn't going to change if 1/4 or 100 J are used. Back work will still be 75%.

Xlnt Fool, short and sweet !!! This is how it should be taught vs Khan video. I think the Khan style lecture evolved from academics trying to accelerate thermo edu that aligns with various student (laymen) concepts despite shortcomings.

Thanks for the kudos. Here's your Beer back. :)

The Kahn Academy video is intriguing. He talks too much. Not that I don't. Everyone learns differently. The professor I liked, weren't liked by other's, and the professors others liked, I didn't. I later learned to respect some of the professors I didn't like. It was purely because they didn't dumb down the material.

Vincent, I don't know what you want or are striving for, but to get 50% efficiency out of any heat engine would be a very good feather in someone's cap.

Carnot efficiency of 50% with DQh of 100 could be done between 100 K and 200 K. However, a real engine at those temperatures probably wouldn't get much better than about 25%, and it would have to be run on a much colder planet.

It could also run at 300 K and 600 K, for the same efficiencies, but a larger DQh.

If run at 300 K and 1200 K the Carnot Efficiency would be 75% and a real engine might be 37.5%.

Fool wrote: ↑Sat Apr 20, 2024 1:41 pm ...
I'm on Matt's side. Gas never contracts.

.... Gas doesn't ever pull. ...

I don't think "pulling" is often included in a definition of "contract".

Some definitions from some online dictionaries:

to cause to draw more closely together

to reduce to smaller size by or as if by squeezing or forcing together.

Gases have attractive forces and when cooled or brought closer together by being compressed, the attractive forces get stronger and this can moderate or ebate the pressure.

Think of it like pushing or "compressing" magnets closer and closer together. You will need to put pressure on the magnets with your fingers to push them closer and closer, but as they begin to attract each other the pressure on your fingers will grow less and less and it will be easier, or take less force to push them together.

Likewise when gases are compressed in a cylinder eventually the gas molecules start attracting and the gas will become easier to compress, and the pressure will begin to reduce as the gas molecules get closer and closer and the attractive forces begin to dominate.

Eventually, the attractive forces will cause the gas to liquify, but even before liquefaction, the pressure will start to reduce and the gas will become easier to compress.

You won't generally hear or read about this in courses on the ideal gas law or the kinetic theory, as these teach what is essentially a falsehood that gas molecules "never" contract and never attract each other but move freely never interacting at all.

Why such false and misleading information is so prevalent and allowed to proliferate in our educational institutions is difficult to fathom.
Attractive force is different for different gases:

https://chem.libretexts.org/Ancillary_M ... Real_Gases

https://chem.libretexts.org/Bookshelves ... lar_Forces

Cooling or compressing a gas causes the gas molecules to attract each other more and more.

The more the gas molecules are attracted to each other due to cooling or being compressed, the attractive force offsets the pressure so the pressure on the container is less.

At first you need to put pressure on the magnets to push them towards each other, but eventually they begin to attract each other more and more until the pressure on your fingers lessens and then goes away completely as the magnets snap together. But the magnets always had some attractive force. It just gets much stronger as they are brought closer together.

In the same, or a similar way, gas particles attract each other. The more their "kinetic energy" is reduced, that is; the more they are cooled or the more they are compressed (or both) the more the attractive force increases. Eventually if the molecules get close enough and the attractive force dominates the gas will liquify.

But even before liquefaction, like the magnets, the mutual attractive forces is ALWAYS present, it is just overcome by the repulsive forces at high temperature and low pressure. But as the temperature is reduced or the gas is forced to occupy a smaller area the attractive force increases and the gas contracts.

Fool wrote: ↑Fri Apr 19, 2024 10:45 am ...
100 Joules in 100 Joules of work out for an isothermal process only applies if the area under the curve goes to zero.
...

I doubt you can actually back up that statement. Or what you mean, or are trying to imply I don't know.

Do you suppose that if there is a compression stroke after an isothermal expansion 90% of the joules go back in time so instead of being transformed into work they can go to the "cold reservoir"?