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

[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

To clarify some of this a little, (my understanding of what you are trying to to say)

When I said above:

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. "

There should be a question mark.

I don't think I can agree with your statement: "as P3 reaches P1 the speed is at a maximum". regardless of which way the piston is, or was traveling.

There are several ACTUAL real time recorded PV tracing posted in the forum. It might help clarify things if you could indicate where in such a PV diagram you would locate P1, P2, P3, the outside atmospheric pressure (horizontal) and so forth.

Generally, if your "P1" is atmospheric pressure or about 15psi that generally occurs at, or near TDC and BDC, at the end of the power stroke the pressure falls below atmosphere and the piston returns. At or near the end of the compression stroke (or TDC) the pressure increases crossing the line of 1ATM and again at that point the piston reverses direction to begin the power/expansion stroke.

Where you get that: "as P3 reaches P1 the speed is at a maximum".

Or that the piston slowing down during the power stroke means:

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

That is probably true on the return stroke

I suppose, maybe you could say the return stroke is "powered by" the power stroke, or the energy "stored" in the "buffer" (atmosphere) and, so...

That is returned and converted back into heat during the compression stroke. But when the pressure is at or near atmosphere, the piston is at or near a complete standstill about to reverse direction.

The engine also has a great mechanical advantage as it approaches BDC and TDC, if it has a crankshaft.and connecting rod.

[Fool]

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.

You are correct, I am wrong. Pretty amusing... me, admitting to be a fool on a day like today. LOL

The working gas is still providing positive work, gasses always push, never pull. The buffer/atmospheric gas is slowing the piston, and absorbing the energy of the piston and is in a battle with the positive work, and pressure, from the working gas. The working gas is loosing, after passing the equal pressure point.

Not all engine configurations have this point. Minimum internal pressure has to be lower than outside, or buffer pressure for that point to happen.

Senft photo

591.jpg (48.32 KiB) Viewed 177 times

For a free piston: At the Pb line at the two points where it crosses the engines path. Both points will have internal pressures equal to external pressure. At those two points piston speed will not be accelerating therefore will be at a maximum. The acceleration/deceleration direction will be reversing.

Many engine configurations, such as having a crankshaft, will have piston speed constrained by the crank and flywheel motion, as well. With such a mechanism the thing to compare is work. Zero work is being added at those points. Other places will have energy added or subtracted from the motion of the system.

I try to visualize a free piston, being connected to nothing else, it will be only the pressure differential that provides acceleration.

[Fool]

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

I have so moved on from that poorly constructed sentence. The first correction I'd like to make is :

"The heat rejected from Tc all the way to zero Kelvin will be Qcz."

Just realize one thing, heat added must come from something rejecting/releasing it, and it's the same amount. Qcz is just an amount, and it's energy, internal energy, as I've constantly defined it. It is equivalent to the amount that "could be rejected", but isn't.

The second correction is :

"The potential heat that could be rejected from Tc all the way down to zero Kelvin will be Qcz, but it isn't rejected, just potentially."

If a mass of gas, M, is at zero Kelvin, it's internal energy will be zero Joules.

If 100 J of heat are added, the temperature will be at 100 K, if Cv and M are chosen accordingly.

If an additional 200 J of heat are added, the temperature of the gas will be 300 K = Tc. And the internal Energy will be 300 J.

If that gas is then cooled back down to zero Kelvin, it will release 300 J of heat. So the described mass of gas at Tc = 300 K has a potential of releasing 300 J of heat to a sink that is at T = 0 K.

Internal energy is typically denoted as the letter 'U' or letters 'Uc'. I have chosen to call it Qcz.

It is important to understand that the example engine only absorbs 100 J more of heat bringing Qhz to 400 J, and then back to Qcz during the cycle, never expending the Qcz energy.

DQh, 1/4 of the total energy not just heat energy, is the only energy pressure pushing out, because atmospheric energy pressure is pushing back. 15 psi (inside) plus MCvTh - 15 psi (atm)

However, the return stroke requires the full return work against Qcz pressure, say 15 psi. That return stroke heats the air proportionally to where Qcz is, not zero K pressure. That compressive energy must be released as heat or it will be higher. Without heat rejection the return work will equal the power stroke work.

[Tom Booth]

In reviewing your previous post there are a few things I failed to consider or missed entirely. Often I give these posts too hasty a read and too hasty a response, as I'm more often than not pressed for time.

Your analysis involved "constant volume" which caused me some confusion as I missed that or failed to remember it as I continued reading your post.

You also incorporated the combined gas law equation which I did not recognize., so your analysis seemed an utter jumble of nonsensical values. Nonsensical if trying to imagine them applying to a "real" or at least typical engine (where volume is not constant) so I was imagining P2 taking place half way to BDC rather than at TDC and nothing really made any sense.

Anyway I need to revisit that after I make myself more familiar with the combined gas law equation and it's usage.

A few thoughts regarding your recent post here though:

....

If that gas is then cooled back down to zero Kelvin, it will release 300 J of heat. So the described mass of gas at Tc = 300 K has a potential of releasing 300 J of heat to a sink that is at T = 0 K.

With a sink at 0°K there can be 100% efficiency.

Correct?

So your statement here does not apply to an actual engine with work output does it?

At 100% efficiency all 300 joules will be converted to work. Work out = 300 joules

But heat rejected = 300 joules

Do you see why I say this Carnot efficiency business violates conservation of energy?

Do you have the same "magic" 300 joules of energy going out as work and also.being "rejected" into a "cold reservoir" at 0°K ?

Internal energy is typically denoted as the letter 'U' or letters 'Uc'. I have chosen to call it Qcz.

It is important to understand that the example engine only absorbs 100 J more of heat bringing Qhz to 400 J, and then back to Qcz during the cycle, never expending the Qcz energy.

DQh, 1/4 of the total energy not just heat energy, is the only energy pressure pushing out, because atmospheric energy pressure is pushing back. 15 psi (inside) plus MCvTh - 15 psi (atm)

However, the return stroke requires the full return work against Qcz pressure, say 15 psi. That return stroke heats the air proportionally to where Qcz is, not zero K pressure. That compressive energy must be released as heat or it will be higher. Without heat rejection the return work will equal the power stroke work.

All I can say about all that at this point is "pressure" is basically relative. It is pressure difference that moves the piston.

At the start the piston has 15 psi on the inside and 15 psi on the outside. The effective "pressure" is zero.

So your characterizing the piston as having to work against or "compress" the "full" 15 psi of Qcz is not taking account of the actual BALANCE of pressure inside and out

If Qcz is 15 psi, that is the starting pressure, which is zero relative pressure, rf no effective "back pressure" whatsoever because the buffer pressure (atmospheric pressure) is also 15 psi and the two cancel out.

[Fool]

An expansion that ends at zero K, does not have a Carnot efficiency application. Carnot efficiency applies only to a complete cycle. A single generic cycle has an indeterminate efficiency. It requires a description of the cycle. When I mention heat rejected from Tc to Zero K, it has zero expansion, so zero work, because it is constrained to zero Volume. It has zero efficiency because all the energy would be rejected as heat. It is not an engine stroke, it is a depiction of the energy Qcz if released fully as heat, no work. It is a way of depicting the value of Qcz.

Yes, it's a pressure difference that motivates an engine or it's free piston. Energy creates that pressure difference and that is reflected by temperature. A good representation of this are the three equations :

PV = nRT
Q = M•Cv•T
Q = M•Cp•T

If the process is neither constant pressure nor constant volume it becomes a combination of the second and third equations.

In other words heat addition creates a pressure difference corresponding to a Temperature and related temperature differences.

Qcz is the starting energy level. Since it is at a minimum volume, to get back to it will require compression of the internal gas. Compression means the gas will heat. Requiring work input or heat rejection or both. The maximum efficiency of the complete cycle has been proven to be lower than the Carnot Theorem predicts. This is because the back work must contend with compressing the gas back to the energy level of Qcz.

If Qcz is zero at zero Kelvin, the back work will be compressing zero pressure back to zero pressure for zero energy required and zero heat rejected. The back work energy provide by the outside will be 100% positive on the return stroke an equal and opposite on the power, expansion, forward, stroke, so will perfectly cancel and be ignorable. The atmospheric pressure has zero effect on efficiency or work output.

[Tom Booth]

... The maximum efficiency of the complete cycle has been proven to be lower than the Carnot Theorem predicts. ...

Really?

"Proven" when? On what date? By whom? Using what experimental procedure? Or, being generous, what actual mathematical proof ?

This is just going in circles. I've heard this confident unequivocal assertion of there being some concrete "proof" and have asked the same questions for over a decade. Proven when? By whom? By what experiment ?

Turns out to be empty talk with no actual data to back it up. Just the same mumbo jumbo contradictory "slight of hand".

If I disprove one argument you just change your tack, like a religious zealot convinced of "the truth" on the basis of hero worship. The great infallible Carnot. And "faith" that 200 years of past "science" could not be wrong. Nothing more.

Who is credited with this "proof"? What great scientist or mathematician or engineer ?

How many engines of what type were involved? Or what were the actual circumstances

Every REAL science discovery of the past is backed up with such factual, historical information and the experiments are all repeatable and verifiable.

[Fool]

The is no such thing as proof in experimental science. A proof, is a mathematical process. I've already given you several, and some several times over. I've even linked some. Here is another good read that is easy:

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

Experiments are used to check for compliance, verification, repeatability, reliability, contradiction, refutation, and precision.

Prove experimentally that a circle is given by the equation :

(x-h)2 + (y-k)2 = r2. Where (h,k) is the coordinates of center of the circle and r is the radius.

It can never be perfectly shown by experiment, but it can be shown accurate within a certain precision and accuracy.

However, the are many corroborating mathematical proofs that it is a perfect circle.

Since an experiment can show refutation of the theory, it is the refuters job to set up a scientifically valid, precise, and accurate experiment. And it needs to be so well thought out with theory, and orchestration that any well trained scientist would repeat the experiment.

The experiment in this case needs to have more than the following. Power out measurement using a dynamometer. Temperature on each side of the hot plate and cold plate (Minimum of four.). Power supplied measured by fuel flow, or electrical. The power out plus the heat out must equal the heat supplied.

Any experiments with less information will be met with rolling of the eyes. Even with the above data, there will be a lot of eye rolling and scepticism. Without that data the process of 200 year old theory denial and bashing will be met with vial contempt. You seem baffled by that response. I'm continuing to be patient here explaining as clearly as able.

I can see that we are reaching an difficult point, so I'd like to shift gears to discussing the processes of the heat pump in hopes that it might be a better point of view and study.

[Tom Booth]

The is no such thing as proof in experimental science. A proof, ...

That's a cop out.

A proof in experimental science is an experimental result that validates the hypothesis. There may not be absolute proof of anything but there can certainly be an abundance of compelling empirical evidence.

"Proof" is YOUR assertion, which is a very high bar.

What you've posted so far is a jumble of contradictions that make little sense at all.

Anybody can make any cockamamie assertion that nobody could possibly disprove, that doesn't make it science. Quite the opposite.

You should apply your standards to drumming up some shread of experimental evidence that actually supports the "Carnot limit" THEORY so-called "LAW" that was never experimentally established in any way, by anyone EVER in 200 years.

[Tom Booth]

The Carnot limit can be DISPROVEN experimentally very easily.

An engine has a hot inlet side at 400°K and a cold ambient sink at 300°K

That gives a .maximum "Carnot efficiency" of 25%

So of every 100 joules of heat supplied each cycle 75 joules of heat would be expected to arrive at the sink as "waste heat" being "rejected". A less efficient engine would reject more heat as "waste heat" not less.

I've done just this experiment a dozen times with the same result. Every time MUCH much less heat is found to be "rejected" at the sink. Sometimes no heat at all, sometimes the cold side is actually slightly cooler than ambient.

So I'm just an amateur experimenter, but this is a very straightforward "proof".

So show me one comparable experiment performed in the past 200 years that shows the abundance of "waste heat" at the "sink" as predicted by the Carnot limit, or any comparable experiment whatsoever that lends any kind of support at all to this so-called unbreakable universal "LAW'".

I'd love to see it.

What have you got?

Excuses.

"No such thing as proof" what a laugh.

[Tom Booth]

A proof, is a mathematical process. I've already given you several, and some several times over. I've even linked some. Here is another good read that is easy:

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

...

Your so-called mathematical "proofs" so far remind me of a dog chasing his own tail.

You put together some jumbled up nonsense that has no meaning whatsoever.

I point out some obvious errors.

You make changes or redefine your variables or make up new ones, and it goes on and on page after page.

What mathematical proof have you given that isn't currently in revision?

None that I'm aware of.

Your references are long rambling articles I've already read some I've already posted here myself.

Which, if any contain any actual "proof" of anything, in your estimation? Please point out the actual paragraph(s) or passages containing some "proof".

I have proof too. Here, read the encyclopedia Britannica, it's in there somewhere.

You just keep throwing spaghetti against the wall hoping something will stick.

[Tom Booth]

As a matter of fact, very early on, you basically proved my case for me, mathematically:

But horror of horrors, we can't have that!

If you are comfortable with:

Qh = Ql +W

Are you comfortable with rearranging the terms:

W = Qh - Ql

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

Sorry Fool, but I'm not comfortable with this entire post
(...)
Let's say Q = energy vs 'heat'. If W = Qh - Ql then work equals high energy minus low energy, and Tom's idea that ALL additional energy above Ql can become work appears correct. (...)

Since then you've just been muddying the waters, trying to come up with a different result.

Heat "ALL" becoming work, however, is not the same as overall engine efficiency.

Some of the work is immediately converted BACK into heat through friction, lost as vibration, noise and so forth

But such loses are derived from work (mechanical motion). But such loses can be minimized, resulting in very high net efficiency as well.

This is very different from a "Law of the universe" that sets a very severe mathematical limit on initial or actual heat conversion into mechanical motion, based on absolutely nothing but a ratio derived from the ∆T.

[Tom Booth]

The mathematics is not at all complex or mysterious. It's ridiculously simple.

Provide the temperature difference and the problem is already solved. We don't need any engineering skills or specialized high tech equipment to figure this out, we just need to know the weather.

What's the ambient temperature?

And the heat of our fire.

No need for any logical or mechanical acumen. We don't need to know how any engine of any kind actually works.

Just hold up a wet finger or read a thermometer.

Science? Who needs science? Who needs an experiment to prove anything? We have a thermometer!

[Fool]

I don't know why, but your last comment reminded me of a Gary Larsen comic:

https://ifunny.co/picture/i-grew-up-wit ... -4SQxWnOj9

An instant later both professor waxman and his time machine are obliterated leaving the cold blooded warm blooded dinosaur debate still unresolved.

[Tom Booth]

The giant thermometer maybe?

the_big_thermometer_maybr.jpg (467.64 KiB) Viewed 115 times

[Tom Booth]

Anyway, specifically, what we have with the "Carnot Limit on Efficiency" is a ratio. Th - Tc / Th.

What "proof" or evidence of any kind whatsoever is there that this ratio has any applicability to how much of the heat supplied in joules can be utilized by any engine whatsoever?