The Reversible engine

[Fool]

https://www.tellerreport.com/news/2023- ... T6XS2.html

"The Tesla you don't know." Why did he reject Einstein's theory? And was he sick with OCD?

Carnot science doesn't disclude a piston from returning to TDC without rejecting heat. It just says if you don't absorb and or reject heat in a cycle the path will have zero energy useful for output.

Carnot Mentioned that rejecting heat is as important as absorbing heat. We now know that less energy is rejected making it look as if some heat energy was converted to work. It is during the rejection process where work is saved leaving the total work for the cycle positive. The more you save work during compression the more efficient your machine will be. We measure that saving effect by a lower temperature.

[Tom Booth]

https://www.tellerreport.com/news/2023- ... T6XS2.html

"The Tesla you don't know." Why did he reject Einstein's theory? And was he sick with OCD?

Carnot science doesn't disclude a piston from returning to TDC without rejecting heat. It just says if you don't absorb and or reject heat in a cycle the path will have zero energy useful for output.

Carnot Mentioned that rejecting heat is as important as absorbing heat. We now know that less energy is rejected making it look as if some heat energy was converted to work. It is during the rejection process where work is saved leaving the total work for the cycle positive. The more you save work during compression the more efficient your machine will be. We measure that saving effect by a lower temperature.

When someone can't argue based on the merits of an idea they resort to ad hominem character attacks. In this case Tesla bashing. Tesla was certainly "weird" in ways and sometimes wrong but his theories stand or fall on their own.

BTW the universally present "quantum foam" of "virtual particles" looks kind of like "something" to me.

So let me get this straight, in your view: "We now know that less energy is rejected making it look as if some heat energy was converted to work"

So your THEORY is, no heat is actually "converted" into work output, it's just that we SAVE some energy so it LOOKS as if that's the case.

"The more you save work during compression the more efficient your machine will be. We measure that saving effect by a lower temperature."

Your wording: "We now know that less energy is rejected making it look as if some heat energy was converted to work"

Is rather bold on your part since, as far as I know this is a theory of YOURS and yours only that you have just now pulled out of thin air.

Nevertheless, possibly valid and worth considering or testing experimentally.

A penny saved is a penny earned as they say.

So.... Heat isn't converted to work, but "less energy is rejected making it look as if some heat energy was converted to work"

Your statements I'm afraid, are self contradictory on their face, so your argument IMO is invalid.

These two side by side assertions contradict each other:

...less energy is rejected making it look as if some heat energy was converted to work. It is during the rejection process where work is saved...

Theoretically heat rejection during compression saves work. So how does it follow that if "less energy is rejected" then "work is saved".

[Fool]

Because the heat that isn't rejected has to be accounted for by increased work output. The amount of heat rejected depends on the temperature during rejection and the volume decrease. Higher temperatures higher heat rejection all the way up to T-hot where it takes as much work to compress as is gained during expansion. Remember it takes a temperature differential for the engines to run, you've said so. Bigger temperature differences bigger power, lower T-cold, lower amount of work to compress, more work saved.

You are the one that brought up Tesla. I think he was awesome, weird, but definitely awesome. Not me bashing. Just that link giving rarely discussed truth. You brought up the logical fallacy of argument from authority. Tesla was either right or wrong about it, however we can't use his authority to prove our condition.

It isn't the fact he was human and wrong that is a problem. Einstein has been known to be wrong. It is that he was wrong a lot more than followers care to admit. Virtual particles sounds like unicorns. Not that I disagree, just that I don't think you understand them any better than I, and they have not been used to prove any aether.

This is also true of your tireless attack on the second law. Prove your theory. The second law is backed by 200 years of trying to prove it wrong. The Carnot limit is proven for every engine that has been built and had it's efficiency tested. None have proven the limit wrong. Mathematics proves it correct. Let's see some GEET Engine numbers? It's a formidable battle you are waging. Make sure you have sufficient data, and peer support before concluding anything.

Yes, insulating an LTD engine makes it idle faster. That is insufficient data to challenge mainstream thermodynamics, even when adding your temperature data. Theory, mathematics, prediction, and confirmation. Repetition. And then see if you can quietly get someone to repeat the experiment. It is not cheap or easy to do good science, but one can dabble. I like your dabbling. Ease of on the Carnot bash for now. Please, for your own sake.

Where are your experimental power output numbers? All we've seen are temperature readings and rpms. Perhaps we could see it lighting an LED, single digit Watts from an 80 watt input? Why have we not at least seen that? A few magnets on the flywheel and a small coil, should light an LED, or stop it when switched on.

[Tom Booth]

Because the heat that isn't rejected has to be accounted for by increased work output. ...

OK, but, ...

Without addressing the rest of your post, this introductory statement still seems to me to contradict your previous statement:

"...less energy is rejected making it look as if some heat energy was converted to work...."

More energy "rejected" to the sink might be mistaken for heat conversion into work.

More heat rejected; more work saved.

So how in heck can "less energy" rejected look as if "some heat energy was converted to work". ???


The full paragraph reads:

Carnot Mentioned that rejecting heat is as important as absorbing heat. We now know that less energy is rejected making it look as if some heat energy was converted to work.

Carnot originally believed heat was a fluid so NONE at all was converted to work or anything else.

The first law of thermodynamics: conservation of energy, had not yet been formulated. So Carnot thought it was just fine for heat to ALL pass through an engine, just like water through a turbine.

You say: "We now know that less energy is rejected..."

Yes, less than the ALL Carnot imagined.

Less heat is rejected BECAUSE at least SOME heat is converted to work, so doesn't need to be rejected.

There is no controversy regarding SOME of the heat being converted, unless you want to propose a new theory and create a controversy.

It doesn't just "look as if" heat is converted to work.

Heat IS converted to work.

The only controversy is how much heat can be converted to work.

The Carnot efficiency formula allegedly sets a limit on how much heat is converted to work.

This Carnot limit appears to be arbitrarily based upon the temperature ratio on the kelvin scale. There seems to be no justification for this. It has never been proven experimentally or even subject to testing, (except by me apparently).

What do you mean by "making it look as if some heat energy was converted to work", maybe I'm misreading that, but it comes across like a denial that any heat is actually converted to work, that it only looks that way.

[Tom Booth]

(...)

Where are your experimental power output numbers? All we've seen are temperature readings and rpms. Perhaps we could see it lighting an LED, single digit Watts from an 80 watt input? Why have we not at least seen that? A few magnets on the flywheel and a small coil, should light an LED, or stop it when switched on.

There are two possible outputs from an engine, heat and "work".

To quote one source" :

A transfer of energy to or from a system by any means other than heat is called “work”.

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

That's not my definition. That's the accepted definition in thermodynamics generally.

Heat can be easily measured, "any other means", that could include vibrations, sound waves, air resistance, friction, stored momentum, velocity, displacement, work against gravity, air pressure etc. is way too involved and would unnecessarily complicate the experiment.

Heat transfer at the sink is relatively easy to measure with inexpensive equipment readily available to anyone.

If the temperature of the working fluid drops WITHOUT measurable "heat rejection" then the fall in temperature can be assumed to be because of "work" output.

Those are the only two choices. Or the only two possibilities.

Sure, it might be nice to measure accurately as far as possible every detail of where ALL the work output are going, but unnecessary for my purposes.

[Fool]

During a conventional discussion of heat engines it is customary to break a PV diagram into four idealized processes. That's for the ease of seeing and explaining what happens for each and the whole cycle. Real engines mimic those ideal processes, but for the most part are not all that close. So, let us just use a real PV diagram. You have provided the following.

Resize_20230522_031849_9086.jpg (175.66 KiB) Viewed 6577 times

If one starts drawing vertical, constant volume lines, the left most line, barely touching the curve, is top dead center TDC, also known as minimum volume, or V-min. The right most line is BDC, V-max.

Thanks for adding constant temperature, isothermal, lines. If a process follows an isotherm, temperature doesn't change. Please notice there are two short parts of the path that mimic isotherms. Lower left and upper right, corresponding to the end of the compression stroke and the other to the end of the expansion stroke. Any temperature change is depicted by a move off or across those lines.

One also can notice that there aren't any distinct separate processes. It is just a smooth curving loop cycling back onto itself. That is caused by the mechanisms mimicking harmonic motion.

There is no single process where we can observe heat being converted to work.

Work is displayed on this chart by a change in volume. The vertical line displaying zero work. Moving to the right positive work output. Moving to the left negative work output, or positive work input, to the working gas.

If there were adiabatic lines drawn on the chart, we could see lines of zero heat. Heat is thermal energy change, ∆Q. We will need to visualize them. Adiabatic lines are steeper than isotherms.

Entropyandtemp.PNG (79.69 KiB) Viewed 6577 times

If heat is added without work, a constant volume line will be followed. That line will cross both adiabats and isotherms. This means that heat is added and temperature changes without any work.

The loop, for the complete cycle, crosses many isotherms, adiabats, and constant volume lines. That makes it a summation of all the work (constant volume) it crosses, positive and negative, depending on the left or right direction as it crosses. And it is a summation of all the heat (adiabats) it crosses. That area of work and area of heat are equal. They do not cancel. They are equal. One is added going in, the other comes out. It takes the entire loop to measure this. There is no subprocess where heat is converted to work. Therefore heat appears to be converted to work somewhere in the whole cycle, not any single conversion process.

Also, heat increases internal energy. Higher internal energy is reflected by an increase of pressure and or volume. Pressure pushes the piston as volume increases, producing work. It's hard to say exactly where and heat is converted, so it looks as if it's converted. And I never used the word "just".

[Tom Booth]

We could debate this back and forth all day and never get anywhere I'm afraid.

The problem with saying volume changes represent work is that it doesn't. It represents volume change.

Let's say temperature times volume corresponds roughly to "internal energy" and let's imagine "internal energy" is the volume of gasoline in my cars gas tank.

If I fill up at the gas station I can measure the increase in volume of gasoline.

If I'm driving I can measure the decrease in volume due to "work" output.

If in some way my tank is being filled WHILE driving it could appear that no gas is being added and/or no gas is being converted to work.

The state of my gas tank at any one point in time has no real bearing on how fast I'm driving, how efficient the engine is or by what means a change took place between state 1 and state 2.

Between point 1 and 2 on a PV chart I could have added a whole lot of heat (gas added to the tank) and also did a whole lot of driving (gas subtracted from the tank as "work") OR

I might have added just a little gas to the tank without doing any driving. OR

I could have done a lot of driving then added fuel then did more driving

In other words, the chart only shows a series of connected dots which represent states of "internal energy" p X v not how much heat was added or how much work.went out between each dot and it does not matter how finely this is divided, even into "infinite" dots, you just then have infinite spaces between the dots where there could be any amount of heat addition of work output.

Finally there is virtually no time when heat input and work output are NOT going on simultaneously in a real heat engine.

I think the point of a PV diagram is that it is composed from "state variables" while ignoring "path variables".

Suppose you come home and find your gas tank half empty? How did that happen? You know you filled it up just yesterday and haven't driven anywhere, so what's up with that?

Well someone 'borrowed" the car.

Did they just drive 100 miles leaving the tank half empty or did they drive 2000 miles filling the tank along the way?

There is no real measure of heat input and work output in a PV diagram as far as I can see.

[Tom Booth]

..(..,)
One also can notice that there aren't any distinct separate processes. It is just a smooth curving loop cycling back onto itself. That is caused by the mechanisms mimicking harmonic motion.

???

"That is caused by the mechanisms mimicking harmonic motion".

No real idea what you mean there.

There is no single process where we can observe heat being converted to work.

Work is displayed on this chart by a change in volume

.


Though I can agree with the first part, these two statements seem vaguely contradictory and the second:
"Work is displayed on this chart by a change in volume[" is based on what?

Heat input or output could result in a change in volume.

Work input or output COULD in the absence of any simultaneous heat input or output be roughly represented by volume change.

As I understand it, when deriving one of these "ideal" PV cycles, (or a "real" PV reading for that matter) actual "work" (shaft work) input or output does not enter into the equations or the graph at all.

The "work" (in an "ideal" case), is derived from the gas law which only accounts for the work the gas does in the process of expanding itself. That is, not doing any external work at all. (i.e. having its "pebbles" removed for it)

[Fool]

You seem to accept isothermal lines on a PV diagram? No temperature represented, yet you seem to accept them.

What other lines do you accept?
Constant pressure, horizontal?
Constant volume, vertical?
Adiabatic lines, similar but steeper to the isotherms.

[Fool]

Work is depicted on a PV diagram by the bases of

Work = F•d, Force times distance.

Where pressure is the force and change in volume is the distance.

For every pebble removed by rolling it off onto a hill, of the same height preserving their potential energy, the rest of the pebbles are lifted higher. The last pebble is lifted the highest.
The total work/potential energy involved will be represented by the mass and height of the pebbles added up, and is equal to the added up area under the PV curve. 100% thermodynamic to mechanical conversion. Once. One process. 1/4 cycle. Genuine Isothermal expansion.

[Fool]

Harmonic motion is the kinematic engineering term for a motion that purely sinusoidal in time. Since a crank pin describes a circle anything hooked to it will move, more or less, sinusoidally.

A sine wave in kinematics is desirable because it's derivative is a cosine, and the derivative of a cosine is a sine. So the position in time is a smooth curve, the velocity (derivative of the position) is a smooth curve, and acceleration (derivative of the velocity) is a smooth curve. This can be continued into the mystery terms and derivatives "ping" and "jerk". Yes those are real engineering terms. There is no end to these types of trivial details.

The up shot of harmonic motion is that it smooths things off and avoids sudden banging accelerations and more. I had no idea how excellent an engines crankshaft is until I learned about them senior year in college. Kinematics is actually kind of fun. Most of the drawing fun has been taken away by electronic computers nowadays. LOL

[Tom Booth]

(...)

The loop, for the complete cycle, crosses many isotherms, adiabats, and constant volume lines. That makes it a summation of all the work (constant volume) it crosses, positive and negative, depending on the left or right direction as it crosses. And it is a summation of all the heat (adiabats) it crosses. That area of work and area of heat are equal. They do not cancel. They are equal. One is added going in, the other comes out. It takes the entire loop to measure this. There is no subprocess where heat is converted to work. Therefore heat appears to be converted to work somewhere in the whole cycle, not any single conversion process.

Also, heat increases internal energy. Higher internal energy is reflected by an increase of pressure and or volume. Pressure pushes the piston as volume increases, producing work. It's hard to say exactly where and heat is converted, so it looks as if it's converted. And I never used the word "just".

Heat is converted to work at the instant a gas particle impacts the piston and transfers energy to the piston.

Also, heat increases internal energy. Higher internal energy is reflected by an increase of pressure and or volume

This is true, but work output decreases internal energy, and lower internal energy is reflected by a decrease of pressure and or volume.

Sum these two together and you end up with zero or some fractional difference.

This simultaneous heat addition with expansion and work output with temperature reduction and volume reduction is not cyclical. It is happening continuously and instantaneously with each impact, summed together for every incremental change so that what is represented on the PV diagram is only the resultant.

If you add a joule of heat which causes the gas to expand and do a joule of work pushing the piston, then the net change on the PV diagram is zero.

Whatever volume increase took place or should have taken place due to heat addition is canceled or reversed due to the equivalent work output.

You could view it as a few molecules are heated which then strike the piston and cool back down. Overall, there is no net change in internal energy, but the volume increased slightly and the temperature dropped and the pressure dropped.

If more heated molecules did not continue striking the piston it would go right back to where it was before, pushed back by atmospheric pressure, due to the lower internal pressure after the joule of work output.

All the textbooks say that the piston can only return by isothermal heat rejection during compression, but this is observably not the case.

I'm no expert, I'm just trying to account for the "missing" work output that lowers the temperature enough to allow the piston to return rapidly, almost instantaneously, obviously NOT isothermally as there isn't time for heat to gradually conduct out instantaneously.

It is openly acknowledged that the "infinitesimal" pressure drop and temperature drop (and subsequent volume reduction) with every incremental change in volume is not being accounted for, as if somehow dividing the pie into enough sections making the volume and mass of each section negligible actually obliterates the total mass and volume of the pie altogether.

I'm saying if you sweep up all the crumbs, you will still have the total mass and volume.

Is you add up all the "infinitesimal" and supposedly inconsequential and therefore ignored or discounted pressure and temperature drops, or reductions in internal energy due to actual work output, you have enough temperature and pressure drop to account for what can clearly be observed: ... the piston returns adiabatically, (extremely rapidly) without or with very little discernable "isothermal" heat rejection.

[Tom Booth]

I should clarify, when I wrote:

"You could view it as a few molecules are heated which then strike the piston and cool back down. Overall, there is no net change in internal energy, but the volume increased slightly and the temperature dropped and the pressure dropped."

I of course meant:

You could view it as a few molecules are heated which then strike the piston and cool back down (or return to the initial temperature due to a loss of internal energy or "work" output, NOT "cooling" by heat transfer out of the engine) Overall, there is no net change in internal energy, but the volume increased slightly and the temperature dropped and the pressure dropped.

I suspect, if you add up all these "infinitesimal" supposedly unimportant reductions in temperature and pressure, each molecular impact on the piston with loss in internal energy, you could account for the piston returning and completing the cycle without a flywheel and without heat "rejection".

[Tom Booth]

Really, if you were to say that, adding 10 joules of heat energy causes the gas to expand and do ten joules of work moving the piston and increasing the volume, then

To get the piston to return you now have to remove ten joules of heat, that would be a clear violation of conservation of energy.

10 joules in as heat -> [gas] -> 10 joules out as work

Then

10 joules back out as heat <- [gas]

When the ten joules go out as work, this causes the temperature and pressure to drop and that SHOULD cause the piston to immediately return to restore the original volume.

It doesn't return immediately because some of the energy has been converted to momentum which carries the piston forward, along with the piston continuing to be bombarded by more hot gas molecules.

Ultimately the piston does return because all these incremental loses from the gas out to the piston as "work" eventually add up and the pressure drop or "vacuum" overcomes the momentum of the piston.

[matt brown]

Really, if you were to say that, adding 10 joules of heat energy causes the gas to expand and do ten joules of work moving the piston and increasing the volume, then

To get the piston to return you now have to remove ten joules of heat, that would be a clear violation of conservation of energy.

10 joules in as heat -> [gas] -> 10 joules out as work

Then

10 joules back out as heat <- [gas]

When the ten joules go out as work, this causes the temperature and pressure to drop and that SHOULD cause the piston to immediately return to restore the original volume.

Tom, why do keep trying to cheat the system when this is all VERY simple...

(1) when back pressure is zero (vacuum) all heat input during an isothermal expansion is AVAILABLE as 'shaft' work output

(2) when there's back pressure (ambient/buffer) all heat input during an isothermal expansion is available for shaft work output EXCEPT for the 'neg' work proportional to this back pressure

(3) under an ideal analysis, the Wneg of back pressure during iosthermal expansion simply reduces the Wneg req'd during similar isothermal compression process to return to start state (aka a cycle)

(4) if you add a flywheel (by option or necessity with whatever back pressure you favor) this taxes the available shaft output during each isothermal expansion process, and is returned during similar iosthermal compression process

(5) any momentum, angular velocity, or whatever other mumbo-jumbo you care to add is bogus, since ALL these 'downstream' effects originate from input, and at best, can only be returned later in cycle

(6) isobaric processes and cycles follow closely isothermal ones

(7) adiabatic processes and cycles are more unique (nonlinear PVT relationships) but otherwise similar