The Carnot efficiency problem

Discussion on Stirling or "hot air" engines (all types)
Fool
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Re: The Carnot efficiency problem

Post by Fool »

Matt, what Fool should have said:

Adiabatic processes are those with zero added or rejected heat, for any of the following: expansion, compression, constant pressure, constant volume, not necessarily all at once. They are a constant unchanging entropy process. Vertical line on a TS diagram. If and only if it is a fully reversible process.

Tom, science isn't about faith or dogma. It is about mathematics correlating from time to time with observation and the perfect theory. Definitions are just the starting point and are not just semantics. The Carnot effiency rule and Carnot and Stirling cycles are examples of perfectly proveable theories, as are the perfect circle equations and trigonometry.
From Wikipedia
From Wikipedia
Entropyandtemp.PNG (79.69 KiB) Viewed 5486 times
That PV diagram I think is for an ideal gas and is derived using PV=nRT. A real gas phase diagram would have similar attributes and principles including a vapor dome.

The red lines are isotherms the black are adiabatic lines. Even though one can move along an adiabatic to different temperatures using work input, the only way to change from one adiabatic line to another is through heat conduction from a hotter heat source or to a colder heat sink.

A cycle that produces work output or input has to change between adiabatic lines. Staying on one adiabatic line will add to zero work for any cycle so constrained.

This would be so much easier with drawing boards or paper.
Tom Booth
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Re: The Carnot efficiency problem

Post by Tom Booth »

Fool wrote: Sat Aug 05, 2023 2:29 am Matt, what Fool should have said:

Adiabatic processes are those with zero added or rejected heat, for any of the following: expansion, compression, constant pressure, constant volume, not necessarily all at once. They are a constant unchanging entropy process. Vertical line on a TS diagram. If and only if it is a fully reversible process.

Tom, science isn't about faith or dogma. It is about mathematics correlating from time to time with observation and the perfect theory. Definitions are just the starting point and are not just semantics. The Carnot effiency rule and Carnot and Stirling cycles are examples of perfectly proveable theories, as are the perfect circle equations and trigonometry.

Entropyandtemp.PNG

That PV diagram I think is for an ideal gas and is derived using PV=nRT. A real gas phase diagram would have similar attributes and principles including a vapor dome.

The red lines are isotherms the black are adiabatic lines. Even though one can move along an adiabatic to different temperatures using work input, the only way to change from one adiabatic line to another is through heat conduction from a hotter heat source or to a colder heat sink.

A cycle that produces work output or input has to change between adiabatic lines. Staying on one adiabatic line will add to zero work for any cycle so constrained.

This would be so much easier with drawing boards or paper.
In reality, as far as I know, there are no fixed adiabatic lines or paths. Adiabatic just means without heat exchange, but work transfers in and out of the system are possible and you can still call it "adiabatic"

Apparently what your chart shows (the black lines) are "entropy" or "polytropic" (constant entropy) lines.

That is, the lines represent constant entropy where the process must be "reversible" and there are no transfers of kinetic energy or any "work" is negligible ("frictionless").

In other words these lines don't represent any REAL adiabatic process in any REAL engine. Those are just IDEALIZATIONS used as a.basis for comparison with mostly processes involving mass flows, and probably not applicable to Stirling engines that have no mass flow but do have significant transfers of kinetic energy.

I've only spent a few hours using reverse image search for that graph and reading up on what it represents, so feel free (anyone) to prove me wrong or elaborate or educate, I'm just giving my general impression and overview of what I just found.

It does not appear that these constant entropy "adiabats" account for the real heat input and work output in a real engine.
matt brown
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Re: The Carnot efficiency problem

Post by matt brown »

Fool wrote: Sat Aug 05, 2023 2:29 am
Tom, science isn't about faith or dogma. It is about mathematics correlating from time to time with observation and the perfect theory. Definitions are just the starting point and are not just semantics.
Many scientific conclusions are backed by math, and thermodynamics is loaded with them. Anyone challenging these old conclusions without supporting math for their new conclusions demonstrates a failure to grasp the math behind old conclusions. Yes, the burden of proof is on the new guy.

Understanding thermo requires a math mindset. Unfortunately, the usual calculus hieroglyphics intimate most, but this mumbo-jumbo is largely only required for actual calculation and not for a working knowledge. I tend to avoid actual calcs (boring) and focus on the proportional relationship of processes and cycles via ratio analysis (aka RA) and graphical analysis (aka GA). RA and GA help reduce a wide range of thermo concepts (from simple to abstract) to a more unified level. Formal thermo education is heavy in GA (PV plots, etc) but misses a lot of RA inherent to the subject.
Fool wrote: Sat Aug 05, 2023 2:29 am This would be so much easier with drawing boards or paper.
I tried to find a descent youtube video on adiabatics, but nothing without a heavy Paki accent.
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Re: The Carnot efficiency problem

Post by Fool »

This may help:

https://energyeducation.ca/encyclopedia ... 20transfer

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

From that last link, just below the earlier PV adiabatic isothermal diagram:
An adiabat is a curve of constant entropy in a diagram.
Measured PV diagrams, often called phase diagrams, are proven science. They will have set constant entropy lines which is the same as an adiabatic line. The work is the integral of that path. An integral is the area under the path and above the zero line. That area is work generated by the process.


Processes moving to the right produce positive work output. Processes moving to the left produce negative work output, requiring a reduction in internal energy, or work input provided from outside the gas envelope.

The following is a real measured diagram for nitrogen. It even shows the liquid-vapor phase, "under the liquid-vapor dome" (for lack of a better term).
From Entropy Wikipedia.
From Entropy Wikipedia.
ST_diagram_of_N2_01.jpg (115.3 KiB) Viewed 5459 times
Fig.2 Temperature–entropy diagram of nitrogen. The red curve at the left is the melting curve. The red dome represents the two-phase region with the low-entropy side the saturated liquid and the high-entropy side the saturated gas. The black curves give the TS relation along isobars. The pressures are indicated in bar. The blue curves are isenthalps (curves of constant enthalpy). The values are indicated in blue in kJ/kg.
https://en.m.wikipedia.org/wiki/Entropy ... odynamics)

The isotropes are vertical lines. Isotrope is another way of saying adiabatic with work and is reversible.

https://en.m.wikipedia.org/wiki/Isentropic_process
In thermodynamics, an isentropic process is an idealized thermodynamic process that is both adiabatic and reversible.
Isotropic, adiabatic, and heatless-reversible, are all the same as your "expansion with work" or with a load. Just different words for the exact same process, and they are represented by a single path on a thermodynamic diagram.

The nitrogen phase diagram may as well be called a law. If you can't accept it as accurately depicting reality you won't have any chance of understanding thermodynamics. It was created from measured data, real gas, real temperatures, real volumes, real pressures and real provable mathematics.
matt brown
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Re: The Carnot efficiency problem

Post by matt brown »

Tom Booth wrote: Sat Aug 05, 2023 12:21 pm
In reality, as far as I know, there are no fixed adiabatic lines or paths. Adiabatic just means without heat exchange, but work transfers in and out of the system are possible and you can still call it "adiabatic"
These curves are the mathematical PVT relationship, however there's a minor distinction that must be made. The isotherms have the same expansion "drop" for monatomic and diatomic gasses, but adiabats have two different drops: one for diatomic (air) and one for monatomic. The difference in adiabatic drops is due to the so called gamma of the gas which is the ratio of constant volume heat (Cv) to constant pressure heat (Cp). This Cp/Cv gamma function (tho not to be confused with another gamma function) is the often seen 1.66 for monatomic and 1.4 for diatomic.

To find the corresponding pressure and temperature change for any volume change during an adiabatic process, simply take the volume ratio to the gamma power for pressure change and gamma minus one power for temperature change like this...

Using your previous 10x compression for air which has gamma=1.4 and where Vr=volume ratio, Pr=pressure ratio, and Tr=thermal ratio, this 10x adiabatic compression for air is

Vr=10
Pr=10^1.4=25.118
Tr=10^.4=2.5118

where the Vr is a decrease vs Pr and Tr are an increase.

And just to round this out, eff=1-1/Tr or about eff=.60 here (and inline theoretical ICE values)
Tom Booth wrote: Sat Aug 05, 2023 12:21 pm
Apparently what your chart shows (the black lines) are "entropy" or "polytropic" (constant entropy) lines.
I think the correct term is isentropic, but most of this poly vs isen buzz is academic job security.
Tom Booth wrote: Sat Aug 05, 2023 12:21 pm
In other words these lines don't represent any REAL adiabatic process in any REAL engine. Those are just IDEALIZATIONS used as a.basis for comparison with mostly processes involving mass flows, and probably not applicable to Stirling engines that have no mass flow but do have significant transfers of kinetic energy.

It does not appear that these constant entropy "adiabats" account for the real heat input and work output in a real engine.
More real than not. Yes, everything is an approximation, but the heat capacity of a gas is not, and ultimately this is what limits output.
Fool
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Re: The Carnot efficiency problem

Post by Fool »

Matt, hawh! Beat you by one minute. LOL

Yes isentropic. I would also call that buzz, extra painful forced learning.

Polytropic pretty much describes any process on a diagram. If it doesn't, completely follow one, it can be diced up into a bunch of smaller processes that are.

On the other hand, adiabatic processes are a very specific polytropic process where n=gamma. Reversible adiabatic, with work.

Isenthalpic processes, constant enthalpy, constant internal energy, adiabatic with zero work, throttling, fire extinguisher blasted into space, are also adiabatic and completely irreversible. That is irreversible. And don't appear to be polytropic. That then requires the use of "steam tables". The "steam" table chart, I just supplied, was for nitrogen steam/vapor/liquid/gas and gas. It has constant enthalpy lines too.
matt brown
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Re: The Carnot efficiency problem

Post by matt brown »

Fool wrote: Sat Aug 05, 2023 5:39 pm Matt, hawh! Beat you by one minute. LOL
Yikes, nailed by the arrow of time, my entropy/chaos has increased...
Fool wrote: Sat Aug 05, 2023 5:39 pm Yes isentropic. I would also call that buzz, extra painful forced learning.

Polytropic pretty much describes any process on a diagram. If it doesn't, completely follow one, it can be diced up into a bunch of smaller processes that are.
Yep, and why I mentioned academic job security.
Fool wrote: Sat Aug 05, 2023 5:39 pm That then requires the use of "steam tables". The "steam" table chart, I just supplied, was for nitrogen steam/vapor/liquid/gas and gas. It has constant enthalpy lines too.
Oh...now I recognize it (LOL) it's a Mollier chart. Coming from the steam camp, we used to dissect these charts as if seeking some magical insight (that never happened). The difference with this chart is that it highlights the supercritical region which is was turned me on to single-phase gas in the first place
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Re: The Carnot efficiency problem

Post by matt brown »

Hey Fool, did you ever study the Kalina Cycle ?

It uses a solution of 2 fluids with different boiling points for its working fluid. Since the solution boils over a range of temperatures as in distillation, more of the heat can be extracted from the source than with a pure working fluid. The same applies on the exhaust (condensing) end. This provides efficiency comparable to a Combined cycle, with less complexity. (wiki)

The struggles that Alex Kalina endured attempting to prove his scheme exemplify how even a good idea can be dismissed. I have a soft spot for this guy, since I was working on the same scheme at the same time (late '70s) but no one knew of him until he emigrated here from Ukraine. There's a few power plants that use this, but largely ignored as too complex and costly for output.


Alex Kalina.png
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matt brown
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Re: The Carnot efficiency problem

Post by matt brown »

Tom Booth wrote: Sat Aug 05, 2023 12:18 am
Whatever doesn't arrive at the "sink" (water bath) can be assumed to have escaped "elsewhere".

Nearly all these possible "loses" cannot exist as loses until AFTER the heat has already been converted to motion (work).

Potential direct loses of heat from the lower water bath can be controlled with insulation to whatever degree possible but would be difficult to quantify.
Let J=energy (heat or work equivalent), 1=initial, 2=final

My premise is that Jhot water 1 + Jcold water 1 = Jhot water 2 + Jcold water 2 + work. Granted, ambient insulation is paramount, but this setup simplifies tracking heat input and output. I'm still scheming a similar simple way to arrive at output comparison without actual work measurements. How, you may ask? Well, in theory, with constant load, when work output is 100% heat input then there's 5x as many strokes as when work output is 20% heat input. A 5x factor is hard to miss and just about right for this setup which is not time dependent, so favors a shaft counter, but optical pickup and stop watch should work. Any "partial" heating of gas at higher speeds vs lower speeds should be offset by lower output and shaft speed, so for now, I'll ignore this potential issue.

If load is only partial per anticipated max output, everything is OK, since the proportional speed difference between 1.0 vs .20 is uneffected. So, everything's cool with this setup except for one sniglet...I need a simple way to integrate a heat:work baseline into this.
Tom Booth wrote: Sat Aug 05, 2023 12:18 am Does the cold water bath actually receive any heat from the hot water bath through the engine while it is converting heat into work or "other forms of energy" (squeaks, friction etc.) that we need not be overly concerned with.
I don't see anymore issues, yet. This test scheme nixes thermal short issues that plague small scale testing and only a "flat plate" design can use this setup.
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Re: The Carnot efficiency problem

Post by Tom Booth »

matt brown wrote: Sat Aug 05, 2023 4:02 pm
Tom Booth wrote: Sat Aug 05, 2023 12:21 pm
Apparently what your chart shows (the black lines) are "entropy" or "polytropic" (constant entropy) lines.
I think the correct term is isentropic,...
Thanks. The terminology is not second nature for me. I had multiple windows open and C/P'd the term. I was looking at both because isentropic is a sub-set of polytropic.

The term Isentropic, (my understanding at this point) is a subset of adiabatic which is constrained: constant entropy, reversible and negligible work.

The chart with the "adiabats" apparently represents constant entropy/isentropic lines. A narrow subset of the adiabatic "spectrum" so to speak.

I believe what these lines represent, is what you would derive from a Rüchardt experiment

https://en.m.wikipedia.org/wiki/R%C3%BC ... experiment


My point, or the point I'm trying to make is that a Stirling engine is not a Rüchardt apparatus.

A Rüchardt tube does not have any heat input or any actual work output.

Yes there is a minimal amount of theoretically "frictionless" work, just whatever is involved in compressing and expanding the gas. There is no external load: shaft-work output to an external crankshaft, flywheel, generator etc.

So yes, in the Rüchardt experiment, the adiabatic process is 0 entropy, negligible "work", just sliding up and down the same adiabat.

A REAL adiabatic expansion or contraction in an actual engine cuts through those lines like a knife through a layer cake.
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Re: The Carnot efficiency problem

Post by Tom Booth »

If you really wanted to though, you could reproduce that type of curve in a REAL engine.

Take a Rüchardt apparatus and set the piston bouncing. Add a Ringbom type (pressure actuated) displacer and just enough heat input to compensate for friction loss.

At the bottom of each "adiabatic bounce" the pressure increases, the displacer lifts due to the pressure differential between inside the bottle and atmosphere.

Now you have a steady state Rüchardt tube.

Now, make the piston a magnet and wrap a coil around the narrow tube as a load for work output. Whenever the magnet piston passes the coil it generates electricity and lights a light bulb -actual work output.

With work out the "bounce" will decrease, so turn up the heat a little to compensate.

In this way you could increase the work output by adding more light bulbs in the circuit to increase the load, while simultaneously turning up the heat.

Heat input = Real external work output

As far as the "adiabat" is concerned, the heat input and work output cancel out.

You still have a steady state "adiabatic bounce" but now work is no longer negligible.

Heat input is mostly isochoric (constant volume) taking place only when the piston reaches bottom and is about to change direction.

The heat input in Joules above whatever was required to compensate for friction loses is directly matched by Real work output in Joules to the linear generator. Of course there will be loses in electrical transfer that will also need to be compensated.

The point is, you end up with the same, or a very similar "adiabatic" on a PV graph as the Rüchardt experiment with no heat input or "real" work output.

These graphs do not account for such simultaneous heat input along with external work output, it cancels out in the equation.
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Re: The Carnot efficiency problem

Post by Fool »

Matt, I have not studied the Kalina cycle. It is awesome that there are power plants running it. Looking at a schematic, briefly, it appears to start as a Rankin cycle with reheat. There is then a third "magical" turbine where reheat seems to come from a condenser phase change latent heat process.

The maniac that was able to think up that cycle must either be wickedly smart, or a lucky fool. I'm voting for the smart explanation. It amazes me the things people come up with.

Thanks. I had heard of it. It seems to use heat energy at a lower temperature, awesome.
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Re: The Carnot efficiency problem

Post by Tom Booth »

Figure 1a shows a schematic diagram of the KCS11. It consists of a turbine, absorber, condenser, evaporator, separator, regenerator, pump and throttling valve. In the evaporator, the ammonia–water mixture is heated up by the low-temperature heat source and then passes to the separator. In the separator, the saturated vapour part of the mixture separates from the liquid. The rich ammonia saturated vapour mixture then expands through the turbine producing power output and then it passes through the absorber. The ammonia–water solution leaves the absorber to the condenser where it is condensed and then it is pumped to increase its pressure to that corresponding to the evaporator. The hot weak ammonia–water saturated liquid mixture leaving the separator is then directed to the regenerator where it is cooled by the rich ammonia mixture flowing back to the evaporator. After the regenerator, the weak ammonia solution passes through a throttling valve to lower its pressure. The ORC consists of four components, namely the turbine, evaporator, condenser and pump, as shown in Figure 1b. In the ORC modelled in this paper, pure ammonia or R134a were used as the working fluid.
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https://academic.oup.com/ijlct/article/ ... i69/770890
Sounds/Looks like basically an ammonia absorption refrigerator with a turbine where the expansion valve would normally reside.

This is the same method used in an air cycle refrigerator. (A turbine to replace the expansion valve)

The turbine is more efficient because it can extract "work" in addition to (or instead of) the more modest cooling of the Joule Thomson expansion valve in a normal refrigerator.
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So what produces the refrigeration to allow lower temperature operation is the work output from the system.

Looks a lot like my "Stirling turbine" Tesla "Ambient heat engine" in some respects. (Expansion turbine to extract power while producing cold cold enough to absorb heat at a low i.e. ambient temperature)
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Re: The Carnot efficiency problem

Post by matt brown »

Fool wrote: Sun Aug 06, 2023 6:43 am
The maniac that was able to think up that cycle must either be wickedly smart, or a lucky fool. I'm voting for the smart explanation. It amazes me the things people come up with.
absorption_1.jpg
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This is a good diagram of small ammonia absorption setup common in RV (aka gas or 3 way reefer). Many guys are familar with this, but few know it was Einstein and 1 or 2 buddies that invented this while he was at the Swiss PTO. Note there's no mechanical pump, instead a sneaky clever thermo siphon is achieved via partial pressure from a third component: hydrogen gas. This is the reefer version that Serval made famous.

absorption_2.png
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During my steam days, I spent a few mornings studying absorption systems over breakfast. Note the difference between these 2 diagrams where the first diagram is so detailed that it's hard to see much beyond what's in front of you. However, the second diagram is a simple schematic devoid of clutter (distractions). As I was bouncing between these two diagrams, I became convinced that I was missing something. On the third day at breakfast, the lights went on...in this second diagram, simply remove both condenser and evaporator (the 2 blue parts) and insert an engine.

This is the basic Kalina scheme, and here's the nickel tour. At room temperature, 1 vol of water can absorb 700 vols of ammonia to make 2 vols of solution, but as you raise the temperature, the ammonia "cooks off". My thinking was sweet, a steam cycle without the inherent loss of the heat of vaporization. Well, technically this is correct, but on the other side of the cycle - the absorber - even if the ammonia gas is the same temperature as the "water" (weak ammonia solution) when the ammonia is absorbed, the heat of solution raises the temperature quickly and prevents further solution (solubility a function of temperature). It turns out, that the cooling required to achieve absorption back to start state in this scheme is basically the same as the heat of vaporization for a similar Rankine cycle.

What Kalina did was to manipulate the 'timing' of heat into and out of the cycle via an array of counterflow heat exchangers. My contribution to this scheme is upping the ante while nixing the nasties...forget ammonia-water and go with another combo, maybe 3 element. I spent many hrs checking my CRC and figured water-glycerol would be better baseline. I'll stop here before I bore you guys more, but during this time, my single-phase gas sidebar tookover and I rarely look back.
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Re: The Carnot efficiency problem

Post by Tom Booth »

Fool wrote: Sat Aug 05, 2023 4:01 pm.
(...)
Processes moving to the right produce positive work output. Processes moving to the left produce negative work output, requiring a reduction in internal energy, or work input provided from outside the gas envelope.

(...)
Can't always tell from just a PV diagram though.

I've been hammered over the head a dozen times with these diagrams. Over and over being told the expansion work has to. be stored in the flywheel so it can push the piston back because even with a "sink" not all the heat. can be removed, whenever I've tried to point out any "self-cooling" type observation.

Finally I read Mechanical Efficiency of Heat Engines, by Senft, James R. which was enlightening in that it showed that atmospheric pressure does the work on the return stroke.

Of course, the engine has to displace atmosphere in the first place, but THAT is WORK output, not HEAT output. I guess displacing atmosphere does work on the atmosphere adding "internal energy" to the atmosphere (?) Ala "air spring" so the work is immediately turned back into heat. Can that be called "heat rejection"? It is going to the "sink' so in a way... The net "efficiency" may work out to be very small but, since this considerable internal energy was transfered out as work there is no need to "reject" HEAT to lower the internal energy. To get the piston to return.

Funny thing though, the atmosphere puts that work right back in!

Another possibility. Real gases are attracted to each other. This seems strange to me but apparently the further apart gas molecules get the stronger there are attracted and the closer they get the .ore they repel each other.

Atmospheric air, naturally holds these forces in balance, otherwise the atmosphere would just drift off into space I suppose, or become so dense and heavy we wouldn't be able to breath.

Maybe, if the somewhat attenuated hot gas expands enough in the cylinder the gas molecules start being attracted more than they repel, which maybe makes sense if the forces were balanced to start with. Hard to know, but I don't have to tug on a connecting rod much before I start to feel a "vacuum".

At any rate, atmosphere is doing work on the return stroke and maybe the gas is actually helping due to the presumed "attraction" between particles.

My point being, work is being done on the return stroke and this is not "negative". Or is it?

The engine is continuing to run and "do work" the full 360° of rotation, so some 'work" of some sort is being done on the return stroke by either at.osphere or "attraction" of the molecules of working fluid or both. And this work, presumably results in either the cooling of the working fluid in the cylinder "pulling" the piston in or cooling of the atmospheric air in the cylinder pushing the piston in. Either way, it would seem that the power cylinder is feeling some siphoning off of energy resulting in a temperature drop in the power cylinder.

The piston is the focus of action, it's being worked on, alternately from both sides and the gas on either side of the piston is doing that work shoving the piston one way then the other.

There is, however, so a concentration of friction at that point.

If the friction can be reduced by say, using a magnetic ferrofluid coated piston maybe this cooling at the power cylinder could be more easily detected.

What's the point of all this?

Not trying to "prove" any "claim" here or "overturn 200 years of thermodynamic theory". My only aim is to observe and try to figure out how these engines really work because all I've been able to find is a lot of conflicting, mutually incompatible theories.

Would a cold power cylinder hurt or help?

Not really sure. If true, is there any way this cooling effect, if it exists, could be put to good use?

I don't know that either. Just "thinking out loud" at this point.

One more time:
Processes moving to the right produce positive work output. Processes moving to the left produce negative work output, requiring a reduction in internal energy, or work input provided from outside the gas envelope.
"Negative work output" what is that exactly?

I see an "air spring" that returns work to the engine. Seems pretty positive to me.

Is there a "reduction in internal energy"?

Well, work was done on the expansion stroke but now "moving to the left" atmosphere is returning that work, so, really, moving to the right reduced internal energy work is being returned, this contributes to the build up and "explosion" of heat at TDC...

Seems the "negative" has been turned to positive effect. The recovered energy will be used for work output with the next expansion.

"work input provided from outside the gas envelope."

Well, atmospheric pressure. That was already "paid for" though. It isn't like you have to run some auxiliary motor to keep the engine going.
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