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

[Fool]

For an electric motor connected to a power line:

Vin = Power line voltage.
A = Current through the motor.
R = Resistance of the motor windings.
Vloss = Voltage loss from the resistance.

The power in 'Pin' :
Pin = Vin•A

The power out 'Pout' :
Pout = Pin - Ploss

Power loss from winding resistance 'Ploss' :
Ploss = R•A^2

Efficient of the motor 'n' :
n = (Pout)/Pin

Substituting in Pout :
n = (Pin - Ploss) / (Pin)

Substituting in Pin and Ploss:
n = (Vin•A - R•A^2)/(Vin•A)

Dividing top and bottom be 'A':
n=(Vin - R•A) / Vin

R•A = Vloss , substituting into the last equation:

n = (Vin - Vloss) / Vin

Does that look familiar? How about if Vin becomes Vh, and Vloss becomes (V zero minus Vc). V zero is zero so:

n = (Vh-Vc)/Vh

The Carnot Theorem even applies to electric motors. Granted R is very low, so Vc will be close to absolute zero, unlike for heat engines running on Earth, where Tc is 300 K.

[Tom Booth]

Let Qh be Delta Qh = DQh, the heat added per cycle. 100 J

That 100J is where this starts to fall apart...it's not cycle input.

Tom seems to have a problem with the definition of Qcz, the absolute internal energy of the system at the starting point.
....

No I don't.

What I have a problem with is turning around and treating this given internal energy, the pre-existing internal energy that is neither added nor "rejected" as if it IS somehow "added" and then "rejected".

The baseline ambient energy pervading everything is not "HEAT" transfered into the engine. But you, (and the. Carnot limit generally as treated throughout all of academia) keep treating this "internal thermal energy" as if it were included as part of the "added" and then "rejected" "HEAT".

I've said this over and over now, literally for years. I don't know how my meaning is not clear.

It isn't how you define the "internal energy" it's how you treat it as you go along, (conceptually and .mathematically) as if it were heat transfered into the engine during each cycle and then "rejected" each cycle.

[matt brown]

Here is a link to a good Calculus based proof of how the Carnot Cycle equals the Carnot Theorem:

https://galileo.phys.virginia.edu/class ... Engine.htm

Awesome link that most guys should be able to follow. My only complaint is that internal energy is glossed over and that U=T per mole being constant would have been nice, especially since this allows Q=W credibility. Oh well, this is part of a series, so I'll have to review the other parts.

For the Otto Cycle, from Wikipedia:

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

This wiki link is another great read.

It seems funny to me that the same equation that leads to n=(Th-Tc)/Th , leads to the one above for Otto Cycle's compression ratio, and that everyone accepts high compression ratio equals higher efficiency, but higher temperature ratio is unbelievable? Higher temperature ratio is higher compression ratio. Same equation, same premiss.

As the compression ratio increases, the Otto Cycle efficiency gets closer to the Carnot cycle, but never quite there.

Indeed, funny to me also. Most ECE guys enjoy discarding anything/everything ICE in their windmill chasing and remain convinced that the Holy Grail is an unique heat engine.

[matt brown]

BTW Fool, I recently saw you relate entropy as a measure of usefulness (I can't remember where). Kudos for 'useful' buzz vs typical chaos, randomness, etc. The academics have chosen to add this to their dancecard whereby their gatekeeper status increases. However, entropy is nothing more than an invented statistical calc that relates relative work potential of (usually) 2 states, and then we're left to consider (compare) these 2 states. So, there's no direct comparison between 2 efficiencies or whatever. The best example is probably comparing 2 isothermal states where we know T, U, and m are constant. The larger volume state will have less pressure and more entropy since it's less useful at producing work than the lower volume state with higher pressure and...less entropy. The only entropy thing I'll go along with is that returning the larger volume state to a smaller volume (isothermally) will require "trading entropy" with something else. Entropy is nothing more than a statistical measure of PVT proportions that are largely meaningless for engine cycles.

[Fool]

Was it this post? :

viewtopic.php?p=21677&hilit=Dilution+entropy#p21677

I equated it to a dilution factor, and uselessness. The grater the Entropy, the more useless and diluted it is. It seems backwards to me. More energy equals better, more entropy equals worse. Drives me crazy, especially when it is defined as DQ/T and disorder, because, heating something up, causes the particles to behave more chaotic, is more more useful, and higher entropy and and more ordered, and lower and and and arrrrrgh! I'm total confused as to why statistical mechanics doesn't do more harm than good. Also why there is a one to one equivalence between a PV diagram and a TS diagram, or even how a TS diagram is actually a T vs DQ/T diagram. Foorratt.

Rudolf Clausius did the same thing. He coined the term 'entropy'. It was from his observation that steam going through an engine came out of the exhaust much less useful. He was the initiator of the death spiral that Caloric Theory died from.

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

In the 1850s and 1860s, German physicist Rudolf Clausius objected to the supposition that no change occurs in the working body, and gave that change a mathematical interpretation, by questioning the nature of the inherent loss of usable heat when work is done, e.g., heat produced by friction.[6] He described his observations as a dissipative use of energy, resulting in a transformation-content (Verwandlungsinhalt in German), of a thermodynamic system or working body of chemical species during a change of state.[6] That was in contrast to earlier views, based on the theories of Isaac Newton, that heat was an indestructible particle that had mass. Clausius discovered that the non-usable energy increases as steam proceeds from inlet to exhaust in a steam engine. From the prefix en-, as in 'energy', and from the Greek word τροπή [tropē], which is translated in an established lexicon as turning or change[7] and that he rendered in German as Verwandlung, a word often translated into English as transformation, in 1865 Clausius coined the name of that property as entropy.[8] The word was adopted into the English language in 1868.

Statistical Mechanics, of which Quantum Mechanics is one, came in during the 1900's, and put the final, death to Caloric Theory, and proof of Entropy Theory, and an end to any hope of breaking the 2nd law, and became standard teaching after about 1940 or 50. About when Tesla had died. It is no wonder he was a holdout of the old 'complete energy conversion' theories of the 1800's and early 1900's. Perhaps now it would make sense to Tesla too. I don't know to what level his mathematics skills were, he seems more an intuitive thinker, builder, than a theoretician. Only theory proves the Carnot Theorem. I know you know that, I just type too much. LOL

The Carnot proof, is awesome, and a educational series. It seems also to prove that, you can't understand this from one little viewpoint, proof, or source. Took me many years of education, and a whole lot more of deprogramming, self learning, and listening to other from all sides of correct, fantasy, experimentation, knowledge, ignorance, skilled, and unskilled. I'm still learning, mostly now by trying to explain things better. It is not as easy to explain as it is to understand. However, trying to explain it seems to improve one's understand.

I think I could sum up my ECE windmill chasing by a desire to have a safe self feeding wood fired electric generator. Tom appears to be headed that way too. I'm slowed down by farm, vehicle, and home, maintenance. I'd like to have reasonable efficiency too, 25% to me is reasonable. Of course more would be better. I dream.

[Fool]

Tom I haven't left you out.

No I don't.

What I have a problem with is turning around and treating this given internal energy, the pre-existing internal energy that is neither added nor "rejected" as if it IS somehow "added" and then "rejected".

The baseline ambient energy pervading everything is not "HEAT" transfered into the engine. But you, (and the. Carnot limit generally as treated throughout all of academia) keep treating this "internal thermal energy" as if it were included as part of the "added" and then "rejected" "HEAT".

I think I can work with that. If just the internal gas is looked at, it starts at T 300 K and 15 psi. If heated to 400K with 100 J, volume being held constant, it will have a pressure of:

(P1•V1)/T1 = (P2•V2)/T2

Constant volume so V2=V1:

P2 = (P1•T2)/T1= P1•(400/300)= 15•1.3333=about 20

If the maximum work out is desired, P2 must be expanded back down until it equals P1, or about 15 psi :

V3 = V2•P2•T3/P3•T2=V2•20•T3/(15•400)

That single equation has two unknowns V3, and T3 (V2 is knowable.), so the values can't be determined from this one equation.
It however can be stated that V3 is larger than V1,
so T3 must be larger than T1
because if the pressure P3 is P1,
and T3 were T1 or lower,
V3 would be lower than T1.

So T3 by nature is higher than T1.

This means that to return to T1 and V1, heat must be rejected.

The amount rejected depends on Qcz because that value is the internal pressure and temperature that the gas is being compressed, and cooled, 'to'. Plus the residual heat leftover from DQh. Kind of like, not all of DQh is getting used during the "full" expansion. The total will be 75% of DQh. Actually it would be more than 75%, as the above includes heat that would be saved in the regenerator.

Qcz is the top of the hill that a car is being driven too.

Another way of looking at it is that 15 psi needs to be compressed back to V1 from V3. That 15 psi is caused by the energy Qcz. It would get hotter when compressed if DQc were not rejected, and it would take away more work. DQc, rejected heat, is proportional to (DQh plus Qcz) over Qhz. The higher Qcz the higher the back work, and heat rejection.

[Tom Booth]

....

If the maximum work out is desired, P2 must be expanded back down until it equals P1, or about 15 psi :

In a real engine, judging by actual PV diagrams and some basic Newtonian laws of motion. At P2 heat, or if you prefer "internal energy" from the expanding gas has been converted into motion/velocity which continues to carry the piston so that pressure drops below atmospheric pressure.

It however can be stated that V3 is larger than V1,
so T3 must be larger than T1

because if the pressure P3 is P1,
and T3 were T1 or lower,
V3 would be lower than T1.

So T3 by nature is higher than T1.

These statements or assumptions are contrary even to the Carnot cycle itself which describes T3 as you've defined it (piston at BDC or full expansion, presumably) as having reached a temperature

Random reference:

In (b), the process is reversible adiabatic gas expansion. Here, the system is thermally insulated, and the gas continues to expand and work is done on the surroundings. Now the temperature is lower, Tl.

https://byjus.com/physics/carnot-engine ... ompression.

At BDC the temperature of the working fluid (in a Carnot cycle) has, through adiabatic expansion already returned to the starting temperature before heat was added

How can heat be "rejected" from a temperature of T1 INTO a temperature of T1 (same ambient starting temperature) within the fraction of a second that it takes the piston to travel from BDC back to TDC?

The Carnot cycle does not take into account the velocity and momentum carrying the piston a little further than what Carnot imagined

So the working fluid at V3 ending up at a temperature LOWER than T1 is not out of the question IMO, and at a pressure LOWER than P1 or below atmospheric pressure, is not only possible but has been recorded experimentally.

This means that to return to T1 and V1, heat must be rejected.

Based only on your various assumptions above.

The amount rejected depends on Qcz because that value is the internal pressure and temperature that the gas is being compressed, and cooled, 'to'. Plus the residual heat leftover from DQh. Kind of like, not all of DQh is getting used during the "full" expansion. The total will be 75% of DQh. Actually it would be more than 75%, as the above includes heat that would be saved in the regenerator.

Qcz is the top of the hill that a car is being driven too.

Are you redefining Qcz again?

Earlier you wrote:

Qcz is not rejected during the running. Nor is it added. It is the base energy that DQh is added to. Earlier definition discarded.

Would not the "base energy" before heat is added be the BOTTOM of the hill?

That "not all of DQh is getting used during the 'full' expansion" is neither here nor there IMO.

By simple observation it can be seen that ENOUGH heat (or internal energy) is used for the piston to return, reliably and repeatedly, within a time frame that renders "heat rejection" after BDC unlikely especially considering that in actuality, adiabatic expansion has reduced the temperature of the working fluid to T1 (According to the Carnot cycle) at BDC making "heat rejection" at best, more difficult, requiring more time (infinite time), for isothermal compression.

You don't seem to be able to clearly define or adhere to your own definition of your own made up variables. This "magic" Qcz seems to become whatever you need it to be at any given moment to produce whatever result you want.

[matt brown]

It is not as easy to explain as it is to understand. However, trying to explain it seems to improve one's understand.

Definitely easier to "get than give".

You would hit on my major entropy sniglet where entropy increases during isochoric heating. This short video shows the relationship, although I still find this somewhat odd, as if slight of hand.

https://www.youtube.com/watch?v=-9bXpv2YgFw


I found this relationship long ago via graphical analysis. How ??? I was plotting successive 'heat triangles' (isochoric input with isobaric input to common isotherms) where Q followed T and a creepy 1.38 "issue" kept coming into stuff. Hmmm, Boltzmann's constant kB=PV/TN had reared itself. I've never arrived at any explanation for how entropy increases during isochoric temperature increase vs simple explanation for entropy increase during isothermal volume increase. I've simply chosen to ignore entropy due to minor impact on lowend ECE (DIY).

I'd like to have reasonable efficiency too, 25% to me is reasonable. Of course more would be better. I dream.

I'm shooting for 35-50% ideal with 25-40% real. Even then, trashing 60-75% of input is hard to swallow as the Holy Grail.

[Tom Booth]

.... trashing 60-75% of input is hard to swallow as the Holy Grail.

Not if you count the heat in the ambient surroundings as "input".

300K (or 300 joules) ambient heat "added" for "free".

Supply another 100 joules to get Th up to 400K

Carnot efficiency is 25% of "all the heat" (from 0K to 400K).

Efficiency = (Th - Tc)/Th x 100%

At maximum, 25% of "all the heat' or 100 joules can be converted to work.

That is in perfect accord with conservation of energy.


Academia, however, for no good reason applies the 25% figure, not to "all the heat", including the 300 joules supplied by ambient but rather the meger bit we added to that.

Result:

25 Joules out of the 100 we supplied can be converted to work. While all the rest must be "rejected".

375 joules from the 100 supplied "rejected". Huh?

Or 25 Joules converted to work

75 rejected. and: The magic 300 joules "supplied' as baseline ambient heat just magically disappears out of the engine.

If there was consistency to how "fool's" Qcz was interpreted there would be no problem and we wouldn't be having this conversation.

If it is "added" for free from the environment and "rejected" then it should be included as part of the "heat" to which the 25% Carnot efficiency applies (all 400 joules or "all the heat from 0 to 400 K.)

Consistency.

[Fool]

Tom, I don't know how I can be any more consistent than the definition:

Qcz = M•Cv•Tc

You are confusing other things being equivalent to it, such as:

Qcz = M•Cv•Tc being the base, or starting point, or even the energy contained in the gas, or the internal energy, or energy at P1, V1, and T1, or the energy of M•Cv•(Tc-0 K)

The definition starts it. The equivalents follow. They are all the same thing and extremely consistent because they are all equal to :

Qcz = M•Cv•(Tc Kelvin)

You also seem to be confusing Qcz with the path to get to Qcz.

If you define a point part way up a hill, the point is the same point regardless of what it looks like coming from different directions. From higher up it will look like a point down in the valley. From lower it will look like a point way up on a hill. Regardless it will always be at the same height, as in the case of Qcz, at the same energy level. Climbing there from your current point will require or provide work accordingly to the fixed Qcz level and your current level. And will be a function of Qcz. If you are headed to it, there is no escaping it's influence.

On other words, once starting with Qcz, it will dictate returning there and climbing/(work input) or descending/(heat rejection) to get there, after, the maximum amount of work has been extracted from the expansion stroke.

In a real engine, judging by actual PV diagrams and some basic Newtonian laws of motion. At P2 heat, or if you prefer "internal energy" from the expanding gas has been converted into motion/velocity which continues to carry the piston so that pressure drops below atmospheric pressure.

Piston speed, 1/2MV^2, is the result of work out from the gas, as P3 reaches P1 the speed is at a maximum. If it passes that point the inside pressure drops below the outside pressure and the piston begins to slow. This means the working gas is reabsorbing energy from the piston's momentum.

In other words energy returned to the gas is less energy out. The maximum out is at P3 = P1. As I stated.

These statements or assumptions are contrary even to the Carnot cycle

True. I was describing a cycle that had constant volume heat addition, followed by adiabatic expansion. A lot less efficient than Carnot. Less than 25% at those temperatures.

Although the above is a valid theoretical description, let me redescribe it with a more Carnot cycle like description, LOL.

Start at Qcz, V1, P1, T1=Tc. Compress the gas, work input, (At a loss of piston momentum.), until the gas is at Th. V2, P2, T2=Th. Smallest volume.

Expand the gas while adding heat to maintain Th, (Yes from a hotter than Th source), positive work output, until P3 = P1 for maximum work out, V3, P3=P1, T3=T2=Th. T3 still equals Th.

Expand the gas further at a loss of work out. I.E., work goes back into the gass as reduced pressure. Inward piston force. Having to fight inward force by pulling out or slowing of the piston's, or flywheel's, velocity. This will be continued until T4=T1. Volume will be much larger and pressure will be much lower. V4, P4, T4=T1=Tc

Compress the gas maintaining the temperature Tc (Yes, this will require a much lower Temperature than Tc.). This is the trip back to Qcz, the starting point. The gas will heat up if DQc is not rejected. That is often called the heat of compression. This process is a work into the gas heat out/rejected stroke. It is powered by the outside air pressure being higher than the inside pressure. This energy gain is used up by the energy loss in the previous stroke, and the following stroke for a net amount of zero. This means the buffer pressure effect can be ignored. It's just as if the engine were in a vacuum, the same net zero help from outside pressure, in regard to work out only. This plus the next stroke (The same as the first stroke.) Are the "back work" that Matt calls it. This stroke returns the engine's internal energy back to Qcz, and why there is a battle against the internal pressure, associated with Qcz. Back to V1, P1, T1=Tc.

Another way to look at Qcz=M•Cv•Tc is, it is the pressure energy that is opposing the compression back to V1, P1, and, T1, from a larger volume. Qcz isn't used but, it is fought all the way back home, to the starting point. The starting point isn't at the bottom of a valley. It is 300 K up the hill. It dictates a pressure level. It is Qcz=M•Cv•Tc as well as other things are equal to it. Do not confuse the direction of definition, with equivalents.

This is a lot easier to follow on a PV diagram for an ideal Carnot, or Stirling, Cycle.

[Fool]

You would hit on my major entropy sniglet where entropy increases during isochoric heating. This short video shows the relationship, although I still find this somewhat odd, as if slight of hand.

Thanks for the link.

Yes, all mathematical proofs appear as slight of hand. Some are worse. It is the short cuts that make them harder.

That YouTube link assumes knowledge of the fact that the integral of 1/x is ln(x) and that ln(x1) - ln(x2) = ln (x1/x2). Easy if you know it. It still questions how DQ/T, becomes a function of dQ and T1/T2. And why there is no DT?

Not as bad as proving 1+1=2+2, when 2+2 = 22, one plus one Equals two too. LOL an every one knows 1+1=10. Hahaha...

3=11
100=4

I'm shooting for 35-50% ideal with 25-40% real. Even then, trashing 60-75% of input is hard to swallow as the Holy Grail.

There are some combined steam cycles that get up to those levels of efficiency. A combined cycle Stirling might be possible too. I really like the idea of a 50 cid Stirling, pressurized with cheap hydrogen, that gets 150+ hp. Could use that for my sawmill. Perhaps generating power to put back out on the line. I got lots of Wood.

I'm fairly sure 25% would blow away the wood gas, or alcohol, ICE power generating efficiencies. Plus potentially a lot longer life. I dream, and mow more lawns.

[Tom Booth]

Tom, I don't know how I can be any more consistent than the definition:

Qcz = M•Cv•Tc

You are confusing other things being equivalent to it, such as:

...

Well, before I even go through all that you've written, you first defined(?) Qcz:

The heat rejected from zero K all the way to Tc will be Qcz

Throughout this exercise you have not clearly differentiated between "heat" as energy in transit and "internal energy", or environmental ambient thermal energy.

I have no idea really how to even conceptualize "The heat rejected from zero K all the way to Tc"

That makes no sense

When I asked for clarification, instead you just threw that out.

Later you wrote:

Qcz is the internal energy of the mass of gas that is in the engine.

It is equivalent to heat that would be added if heating that mass from zero K to Tc K. It is also the amount of heat energy that would come out if cooled from Tc to zero Kelvin.

It is automatically supplied by the ambient temperature. It is not used up to run the engine.

It is the base line energy that DQh is added to, to get from Tc to Th.

If M•Cv is such that it equals one, Tc=300, then Qcz=300.

This, I think seems clear enough.

Maybe M•Cv has some meaning or implication or definition I'm not aware of or not getting.

Presumably it applies exclusively to the M mass(?) of working fluid INSIDE the engine. Cv means what? Heat capacity of the volume of working fluid inside the engine before additional heat is added and the volume expanded. Also the point or volume and heat capacity of the working fluid to which the engine is presumed to return to upon completion of a cycle. (TDC or "full compression)..

To my mind this initial "internal energy" IS NOT "heat", is NEVER "added" at any point we are concerned with and never "rejected".

It is simply the substance or mass of the working fluid along with whatever "internal energy" it may contain when the engine was first constructed. As such, to my mind it is completely "inert" and irrelevant. This so-called "heat" as you have repeatedly characterized it is effectively ZERO. The baseline before any heat is added to operate the engine.

Maybe M is "moles"? mass? Same thing basically. Right?

So, how could this "internal energy" ever be conceived in terms of:

"The heat rejected from zero K all the way to Tc"

Heat "rejected" would be heat LEAVING the engine. Going OUT to the surrounding environment or to the "cold reservoir" or wherever. But heat, once "rejected" would no longer be a part of or quality of, or "internal" energy of the working fluid.

The working fluid is inside the engine. Heat added WAS outside, heat "rejected" is outside the engine. No part of the working fluid Is INSIDE.

So on the one hand you seem to say Qcz is STRICTLY associated with the given internal energy of the working fluid, then go on to say things such as:

Qhz will become the heat added to get to Qcz

"Heat added" ?????

"Heat added" when, and to what?

There is no "heat added" to get to the working fluid starting temperature and/or starting working fluid internal energy level. Which is what Qcz is supposed to represent, so how do we "add' heat to "get to" where we already are at the start.

The "confusion" is not mine IMO, but goes all the way back to Caloric theory and the general debate and confusion regarding the true nature of heat. Is it simply a form of energy, or a quantity of joules or is it a substance?

I don't think I'm confused about anything, it is basically ALL of academia and the so-called "science" of thermodynamics that is confused, that has been confused for the past 200 years.

I'm just trying to point out and highlight the logical inconsistencies and contradictions inherent in the "Carnot limit" which is exemplified by your "Qcz" and its apparent magical properties.

It can be inside the engine. Never added and never leaves on the one hand, but then "added" and "rejected' on the other.

[Tom Booth]

...

On other words, once starting with Qcz, it will dictate returning there and climbing/(work input) or descending/(heat rejection) to get there, after, the maximum amount of work has been extracted from the expansion stroke.
....

This is an example of the illogical nonsense you seem to have so engrained in your way of thinking the "double-think" is second nature and goes completely unnoticed as if it actually makes perfect sense.

Why do we have to "return there" if that is where we "start"?

Who has to "climb" to get to the bottom?

[Tom Booth]

...

On other words, once starting with Qcz, it will dictate returning there and climbing/(work input) or descending/(heat rejection) to get there, after, the maximum amount of work has been extracted from the expansion stroke.
....

This is an example of the illogical nonsense you seem to have so engrained in your way of thinking the "double-think" is second nature and goes completely unnoticed as if it actually makes perfect sense.

Why do we have to "return there" if that is where we "start"?

Who has to "climb" to get to the bottom?

After mulling this over, to clarify a bit...

I assume, not being completely daffy, you are assuming it is (of course as a basic premise or "law") "impossible", as a presupposed "fact" that "after, the maximum amount of work has been extracted from the expansion stroke." There is necessarily or absolutely no chance in the world that there would not be.some remaining heat left over from the heat that was added.

Therefore it will take "work" compression of the gas, or heat "rejection" (or both) to complete the cycle and get back to the starting point at TDC.

This "work" or whatever "heat" or "internal energy" that you suppose needs to be put in and or "rejected" is NOT, IMO, any part of, and. NEVER belonged with or became co-mingled with or indistinguishable from your "Qcz" which is supposed to be the baseline from which the cycle starts.

You keep treating this Qcz as-if it is somehow carried along with or included in heat that may. be "rejected",.

That, to me is just faulty accounting. Completely .mathematically unsound and completely illogical.

But somehow this kind of "slight of hand" has become routine in thermodynamics for calculating so-called "Carnot efficiency".

It is transparently absurd.

[Tom Booth]

...
Piston speed, 1/2MV^2, is the result of work out from the gas, as P3 reaches P1 the speed is at a maximum. If it passes that point the inside pressure drops below the outside pressure and the piston begins to slow. This means the working gas is reabsorbing energy from the piston's momentum.
...

Says who?

BTW, I assume you mean. Well, let's just say I don't know what you mean by "as P3 reaches P1 the speed is at a maximum".

Speed of what? At what point?

Your understanding of engine mechanics and piston motion seems out of touch with reality.

Anyway, ignoring that, you go on to say: " If it (the piston) passes that point" (which it does about midway down the cylinder towards BDC), "the inside pressure drops below the outside pressure and the piston begins to slow. "

OK, so far. Sort of.

This means the working gas is reabsorbing energy from the piston's momentum.

Your physics understanding is not too keen either.

During the expansion stroke, the working fluid is doing work pushing the piston, regardless if the piston is accelerating or decelerating, as it moves toward BDC or full expansion.

But your description of the sequence of events is discombobulated.

as P3 reaches P1 the speed is at a maximum. If it passes that point the inside pressure drops below the outside pressure and the piston begins to slow.

That makes no sense.

At P1 the piston is stopped. Motionless. About to change direction. Is not P1 the starting pressure?

OK supposing the pressure drops midway to BDC. That does not change the fact that: During the expansion stroke, the working fluid is doing work pushing the piston, regardless if the piston is accelerating or decelerating, as it moves toward BDC or full expansion. To say that "the working gas is reabsorbing energy from the piston's momentum" is wrong