Why a temperature differential?

[Jack]

Yeah I might have mixed up BDC and TDC. Kinda depends on the picture of the engine I have in my head.

With BDC I meant fluid and engine volume at it's "smallest".. in the ltd engine I had in my head that means BDC hehe.

[Tom Booth]

...
With BDC I meant fluid and engine volume at it's "smallest"..

Right, that was pretty obvious. Terminologies borrowed from the IC engine world can be confusing applied to Stirlings.

Like "compression stroke", which Andrew Hall explains, in a Stirling engine is really a "contraction stroke".

https://youtu.be/SHyke4hUNOs

[Tom Booth]

And just to be clear, nobody is talking about perpetual motion or "over unity" here.

There is plenty of heat being supplied. No "free lunch"

It's just that as Hall explains, some of the heat is converted to momentum so that by BDC the pressure drops below ambient.

That momentum of the piston didn't appear out of nowhere.

As you said earlier: " I could imagine a situation where the piston, on the power stroke, goes a little further than needed using momentum".

No need to "imagine" it.

Andrew Hall, as you probably know, unlike some of the clowns in here, is a REAL authority on Stirling engines and he agrees with you 100%

Obviously this return work will reduce the "net" efficiency or external work output but that is not "heat removal" to a sink.

[Jack]

I think I now would need to be convinced as to why this wouldn't be possible. I don't think it's magic and not even against the second law.
Please try to convince me without using too many fancy words haha.

[VincentG]

Jack this is the best I can do without a white board(lol). Take a look at the specific heat values of a gas for Cv(constant volume), and Cp(constant pressure). The heat required for Cv is the minimum energy needed to start playing the game. The additional heat needed for Cp is the extra energy needed to affect the greatest change to the gas.

Note that the energy of Cp is not that much more than Cv. It's this difference between the two that you have to play with, for the most part.

So in effect what you are trying to do is remove the heat of Cv, using only the difference of Cp-Cv, at best.

[Jack]

I'm gonna need that whiteboard haha.

I'm letting heat do work on my piston. But at some point there will be some left over that isn't enough to push the piston. That leftover is the only part I want to cool by the momentum of the piston, so to stretch the fluid and give that heat a way out as work. So this little extra travel of the piston will partly be assisted by that heat and the rest by momentum, or flywheel power.

[Fool]

To sum up what you all have said:

The temperature difference is needed to produce a higher pressure during expansion, and a lower pressure during compression. Thus providing a positive work output for the complete cycle.

Many have suggested that for a single expansion stroke Th constant, isothermal from Pmax to Patt, the conversion of input heat to output work will equal for a 100%.

I contend that at that point the inside gas will be at Th and simply pushing the piston back in will require the same amount of work and Qh will be rejected to the hot plate in this isothermal compression back to the beging of the stroke for a complete cycle. Ending up with zero heat used and zero work generated. Zero zero. No free lunch.

Many have contended that if instead of isothermal an adiabatic single stroke expansion takes place, that the gas temperature will drop. It is often called adiabatic cooling to separate it from heat loss/rejection cooling.

I contend that even at the point that inside pressure equals atmospheric pressure, that the gas temperature will still be well above the Tc of the cold side. This was proven mathematically in the "Let's beat up Carnot" thread.

So the dilemma becomes, what to do with that temperature/energy? Three things come to mind:

1, Throw it out, reject it. Unfortunately done in many homebuilt engines.

2, Save it to a regenerator and reuse it after the compression stroke. We know it now as being in the Stirling cycle. Very good.

3, Expand the gas further by "pulling a vacuum". Work is put into the system. That work can come from the first part of the stroke in the form of momentum. Carnot suggested this process. We know it now as being in the Carnot cycle.

Those three things can be done for an isothermal stroke too. The first would throw away more energy. The second would require a larger regenerator. The third would require a deeper vacuum, and higher percentage of the forward work. But the isothermal stroke would generate a lot more forward work output, 100% of the heat input. They will also work if the stroke process is somewhere between isothermal and adiabatic strokes.

I contend that the work being put into the system doesn't go into the the internal gas. It goes into the atmosphere. It is stored there like a gigantic but weak air spring, changing the atmospheric pressure very very little, negligible. It is modeled as a constant pressure and temperature process. It can do this because the mass of the air moves in and out of the local area without restrictions. It acts very much like a spring.

The internal gas during this process of expansion, has a temperature decrease because it is still doing positive work, it is less work than for the negative atmospheric work, for this part of the stroke. For the first part the gas's work was larger than the atmosphere's.

The following is where you all are struggling, off the deepend, and perhaps are in the realm of magic:

A very few of you contend that, since the gas in situation 3, is now at Tc, all the "heat" is gone! Converted to work. Since it's gone, the piston can retract to the smallest volume with the push of outside air pressure, while not experiencing any pressure build up, except back to P1, and end at the cycle beginning at T1/Tc and V1 (sometimes called TDC). Waiting for the next round of heat addition. Zero heat needs to be rejected because there is zero compression. Hence no heat of compression. At least One person attributes this to 'gas contraction'.

I contend that, heat doesn't get converted to work during an expansion, especially with an adiabatic expansion. See my "Struggling With Internal Energy" thread. Internal energy is converted to work output. The heat is still there and returns during adiabatic compression. It's easy if you think about adiabatic bounce. It returns temperature and pressure when the volume returns.

Also gas doesn't contract in any of the engines we work on here. They all work in the area where real gases behave very closely to the ideal gas laws. This means for the temperature and pressures these engines produce, the gas always has positive pressure. It always pushes outwardly. The very word contraction means a pulling inward. There is no record of gases pulling inward, they always push outward. This means that reducing the volume will both increase the pressure and temperature. This always will produce the colloquial 'heat of compression' that will be rejected if placed against a colder mass such as the cold plate as accomplished by the mechanical pieces of the engines discussed here.

The return stroke is work into the gas and adiabatic temperature rise, no matter what the outside pressures are. No contraction. Internal energy converted to work in each single stroke. Not heat. Heat is the transfer of energy from one mass's internal energy to the next. It can only add or replenish internal energy. It is internal energy that does the work. Flowing heat, and heat is only flowing energy, doesn't do work. That is the reason the old Caloric waterfall theory was abandoned. They discovered heat doesn't do any work. Internal energy does. Heat and internal energy are not the same. Many people misleadingly describe it that way.

This is my explanation. I don't think you will find it anywhere. However, the data and mathematics supporting it are out there many places. Investigate if you'd like. Denial will force you to not investigate, that's for sure. Skepticism is appropriate.

In other words, the second law forced the discarding of Caloric Theory. If you discard the second law, you welcome back Caloric heat flow theory. Denial, as always, won't help. Questioning yourself will.

[Tom Booth]

...I contend that, heat doesn't get converted to work during an expansion, especially with an adiabatic expansion. .(...)Internal energy is converted to work output. The heat is still there and returns during adiabatic compression.
...

What do you mean by "The heat is still there"?

You jump down everybody else's throat if they refer to 'internal energy" as "heat", but aren't you doing the same thing? Word police.

So saying: "internal energy is converted to work output. The heat is still there" is basically a contradiction since the heat energy was transfered becoming "internal energy" so how is the heat "still there" except as internal energy. You stated: ”Internal energy is converted to work output.".

Heat energy could be added to the "internal energy" isochorically or isothermally before that same energy is converted to external work output.

How do you imagine or suppose that the "heat" portion of the internal energy is somehow "still there" after the internal energy is used for work?

It's just ridiculous.

https://www.vaia.com/en-us/textbooks/ph ... t-does-wo/

When a gas expands adiabatically (meaning no heat is transferred into or out of the system), the energy for the work done by the gas comes from the internal energy of the gas itself.

In an adiabatic process, the change in internal energy of the gas is solely responsible for providing the energy needed to do work on the surroundings. This change in internal energy can occur due to the gas expanding against a piston, for example, which converts the internal energy into work done on the surroundings.

The first law of thermodynamics, which is the principle of energy conservation, states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

ΔU = Q - W

In the case of an adiabatic process where Q = 0, all the energy for the work done by the gas comes from the internal energy of the gas itself. This internal energy can be in the form of kinetic energy of the gas molecules, potential energy due to intermolecular forces, or other forms of energy associated with the gas molecules.

So, in an adiabatic expansion, the energy comes from the internal energy of the gas, and this internal energy is converted into work done on the surroundings as the gas expands

I'm not usually a big fan of bots, but in this case the Quora Bot makes more sense than you do.

https://www.quora.com/When-a-gas-expand ... -come-from

Anyway, there is really no pure "adiabatic", isochoric, or isothermal, expansion. Expansion of a gas in an engine is a mix, just as "internal energy" is a mix, just like my bank balance is from a mix of income sources.

To say the heat input added to the internal energy somehow remains in some separate energy "account" and then "returns" aside from being silly just doesn't even make sense.

How does it "return" if it is "still there".

And that's just one paragraph out of your ramble. Full of contradiction and outright nonsense.

[Fool]

What do you mean by "The heat is still there"?

PV=nRT

If energy in the form of heat is put into a closed cylinder of gas and the piston is held stationary. The above experimentally discovered equation can be used to determine the pressure from just measuring the temperature, or vice versa.

For example:

T1=Tc=300K
T2=Th=600K
P1=100 kPa
P2=200 kPa
V1=V2=2 m^2
n=2/3 kPa m^3/K

PV=nRT
So:
nR=PV/T
nR=100•2/300=2/3

V and nR are constant:

P1V1/T1=nR= P2V2/T2

V1/V2= 1 so:
P2=P1•V1/V2•T2/T1= 100•1•600/300=200 kPa

But we knew that already by inspection. Double the temperature, double the pressure.

Now the heat entered has become internal energy. The energy was converted into increased pressure and temperature. Zero work so far, no ∆V, volume hasn't changed.

So the heat has disappeared, but is still there in the form of internal energy temperature and pressure, Th and P2. But again heat really doesn't exist. Any temperature difference between two masses can spontaneously transmit energy as heat if put into conduction. So temperature difference is important for heat flow.

So what happens now when the gas expands against a piston? First, expansion of a gas against a moving wall, such as in any of the engines we have here, is 'with work'. That work can be collected by a electric generator or just by the momentum increase of the piston. Both are a load. This allows the gas to have a maximum adiabatic temperature drop, and minimum work out, for all our piston engines. Or maxim heat absorption, and work, if isothermal or no temperature drop.

Second, the end of the expansion power stroke is at the point gas pressure decreases back to P1. It's easier to think of this as the inside pressure becomes equal to the outside pressure, and any further expansion will require pulling a vacuum. Or being at the bottom of a valley, nowhere to go but up, work.

So now for a purely adiabatic stroke, the internal energy, which came in as heat, is converted to temperature (Still higher than T1\ Tc, as proven in the "Let's beat up Carnot" thread), volume, and work. The energy, which transfered in from an outside mass of internal energy, often called heat, is now, not heat but internal energy and some work and larger volume. So the energy input as heat is still in the system in the form of a larger volume, some temperature increase and some work.

To return to the starting volume, the volume of gas must decrease. This is called compression with work, because it is accomplished by a moving wall/piston. To do this without cooling/(rejecting heat to a cold plate), the exact same amount of work must be put in. Conservation of energy dictates that the point 2 will be the stopping point where the states will be again P2, V1 or V2 (they are equal, V3 was the final stroke volume.) and T2=Th. Where again the same amount of heat can spontaneously transfer to Tc. The heat returned, or more accurately the heat potential, is again identical. All the energy that was converted was forced to return to allow the same potential for heat to flow out.

So the heat disappeared, energy was slopped around to different types, and returned to become the original form, for heat (returned but the energy never gone.) to be able to flow again.

At this point the energy can only flow spontaneously to a colder than Th mass. It can't flow back into the hotter flame that is keeping the hot plate at Th. Flame Tf to Th is an example of an irreversible energy transfer. Hot to cold dispersion of energy.

It is easier to ask for these incredibly detailed proofs than to produce them. Please be more tolerant and patient.

[Tom Booth]

That is not a "proof" it's nonsense.

As always, you conflate heat added to the system in joules with the resulting total internal energy at Th.

That is the fundamental error. Temperature is not heat and heat is not temperature.

You could add zero heat in joules and the system would still have an internal energy and a temperature.

Specifically:

So the heat has disappeared, but is still there in the form of internal energy temperature and pressure, Th and P2

Absolutely not.

The added heat has nothing to do with Th or P2 for that matter.

As if you were to spit in the ocean, and your spit becomes the entire ocean.

The ocean was already there.

[Fool]

That is not a "proof"

You are correct. A rigorous proof would be many times more complex. That was just a common language description of the science so as to allow more people to understand. A simple proof.

it's nonsense.

Off topic attack on poster and poster's style noted and disregarded.

The added heat has nothing to do with Th or P2 for that matter.

If the gas is at Tc and it is heated by Th to Th. The change in internal energy can be calculated by ∆U=MCvTh-MCvTc. The system reaches point two at P2, Th, and V1=V2.

As if you were to spit in the ocean, and your spit becomes the entire ocean.

Qh isn't going into a gigantic ocean or atmosphere. It is going into a tiny amount of gas mass, that is contained inside a tiny engine. It is sufficient to raise the gas temperature from Tc to Th. It is a linear increase. Linear is a very important part of the proof.

Tc was already there. It has been separated by cylinder walls and piston. It is now a very tiny piece of Tc. It is greatly affected by Qh.

[Jack]

I think I'm seeing where my previous reasoning was wrong.

It's a game of temperature, pressure and volume.
What I previously thought was heat converted to work is actually heat transferred to internal energy which expands the fluid. An expanded fluid in the same volume is hotter and will cool down as it expands and moves the piston. If you move the piston so much that the fluid reaches its starting temperature it will still have a bigger volume. The heat is converted to work through pressure and volume. I didn't see this before and didn't understand what you meant with the second half of the stroke. Now you have to bring the piston back and achieve the previous volume and this will create heat as you're compressing the fluid. Heat that needs to be rejected.

I still believe there might be a way around it, but I'm now thoroughly convinced that that can't be done with a piston engine.

[Tom Booth]

As if you were to spit in the ocean, and your spit becomes the entire ocean.

Qh isn't going into a gigantic ocean or atmosphere. ....

Relatively speaking, on an absolute temperature scale adding "the heat of a hand" to an LTD it is.

Put simply in practical terms; the Carnot Limit postulate asserts that for 100% efficiency, to fully utilize that infinitesimal increase in temperature provided by my hand, (a veritable drop in the ocean on an absolute temperature scale), Rather than the engine utilizing the "drop", ALL the heat, all the "internal energy" of the working fluid down to absolute zero must be completely utilized. Then and only then could we avoid having to "reject" additional heat to the 300° K surroundings.

Utter poppycock.

Completely illogical, unreasonable, irrational nonsense.

[Tom Booth]

Not to say this supposed absolute zero requirement originated with "fool".

https://youtu.be/V3nNgygrmsI

https://youtu.be/_n3Z_YBzvDQ

As seen in the above couple of videos, this is what is being taught as "proven" on a regular basis, at the university level.

Proven on what basis?

Supposedly, according to the instructor, Carnot was able to "prove mathematically"...

Well no. Carnot never put his water wheel theory of heat into mathematical form.

And... on the basis of...

an admittedly "theoretical" engine that could never actually exist or even operate.

So basically, on the basis of nothing at all other than a modern interpretation of a supposed "mathematical formula" which is really nothing more than some representation of Carnot's temperature difference heat as a waterfall idea... An unproven theory, which Carnot, BTW later himself rejected. An equation of questionable origin, where it came from seems lost in history, but certainly not Carnot himself.

And nothing more.

No actual engine. No experimental evidence. No experiments. No "proof".

And on that basis I must accept that to fully utilize the miniscule amount of heat transferring from my hand to the LTD the engine must drop its internal temperature down to absolute zero? It must consume "ALL" the heat of its "internal energy". That BTW includes the nuclear strong force. Einsteins E=mc2

How much nuclear force is there in a few thimbles full of gas

Well, I think the equivalent of maybe a few billion atomic bombs?

[MikeB]

Put simply in practical terms; the Carnot Limit postulate asserts that for 100% efficiency, to fully utilize that infinitesimal increase in temperature provided by my hand, (a veritable drop in the ocean on an absolute temperature scale), Rather than the engine utilizing the "drop", ALL the heat, all the "internal energy" of the working fluid down to absolute zero must be completely utilized.

No, that's not correct.

Carnot starts with the fuel.

To be 100% efficient as an engine, your HAND must drop to absolute zero.