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In fact I'll do it again.

Qh heat added from Tc to get to Th.

Uh total energy contained in the gas at Th
Uc total energy contained in the gas at Tc

Qh=Uh-Uc
100=400-300
Heat equals the change in internal energy.

Starting point one:
Uc=MCvTc
300=MCv300

Maximum point two:
Uh=MCvTh
400=MCv400

MCv= Mass times constant chosen so it is one.

If an engine forward stroke starts while containing Uh 400 Joules, the most work it can produce is 400 J. This will be when it gets to absolute zero Kelvin. We want a maximum. The piston returns to the starting point costing zero back work, because it is at zero Kelvin. The buffer pressure has added a total of nothing because the same pressure opposed the forward stroke, as helped the back stroke. P∆V work, out forward stroke, in backward stroke, same P∆V

Now, back at the starting volume and zero K, 300 J must be added to complete the cycle. That 300 Joules must come from the forward work. Why? Because we are only adding 100 J. 300 J more must come from somewhere to get to the 400 J second position and 300 J starting first position. Forward work is the only energy available.

If the atmosphere were expected to add the Joules, it will do it at Tc for the entire return stroke and 300 J of work will be needed to compress the gas.

If the buffer pressure force compresses it, 300 J Uc will be rejected as energy of compression to the cold sink as heat, to keep the reverse work at a minimum.

So, it starts with Uh and returns the amount Uc to complete the cycle. The maximum efficiency to start at Uc, add Qh, expand all the way to zero borrowing from the Uc energy, return the extra gained by over expanding to become a cold hole by adding back the borrowed heat Uc, during the return stroke, to complete the cycle is:

n=(Uh-Uc)/Uh

Maximum efficiency after all ideas have been tried.

Substituting in the temperature energy formulas Ut=MCvTh:

n=(MCvTh-MCvTc)/(MCvTh)

Canceling MCv top and bottom.
n=(Th-Tc)/Th

This raises the question, Why is Qc subject to the same efficiency as Uc? It's the same engine operating from the same temperatures. It should get the same efficiency even if we caculated all the work Uc could supply on the single forward stroke. Again it is the return stroke that gets it.

If it is obfuscated with buffer pressure, then the forward stroke gets it. There is no efficiency win from applying buffer pressure to a full cycle. It cancels too. Or as Matt says, there's no free lunch from buffer pressure in regard to efficiency.

This is Proof that mathematically shows the cancelations, and scientifically describes why it is an absolute maximum when operating in a full cycle from two limiting temperatures.

If you want to beat the following equation:

n=(Th-Tc)/Th

You will need to measure the temperatures and :

n=W-Qsupplied

Qsupplied is the outside energy of your energy source and not necessarily the Qh absorbed by the internal gas. It would be better to know Qh. A good indicator diagram will give that.

Measure Work, and the wattage of the heat source, and elapse time for the run.

And if you want to do good science you will run the experiment many times.

Fool wrote: ↑Sun May 26, 2024 6:14 am In fact I'll do it again.

....the most work it can produce is 400 J. This will be when it gets to absolute zero Kelvin.
...

Among many other things, your expansion to "absolute zero" is completely unrealistic nonsense.

I'm interested in real engines and real engine behavior in the real world, not all this impossible, hypothetical pie-in-the-sky number juggling.

I'm not wasting any more time on your supposed "derivations".

As far as I can see you've added nothing new to your previous failed derivations. I'm not going to keep rehashing this.

Fool wrote: ↑Sun May 26, 2024 6:14 am In fact I'll do it again.

Qh heat added from Tc to get to Th.

Not "Q" when Stirling, since this is regen heat within system (vs to or from system). Q always denotes heat between system and surroundings.

Fool wrote: ↑Sun May 26, 2024 6:14 am Uh total energy contained in the gas at Th
Uc total energy contained in the gas at Tc

Qh=Uh-Uc
100=400-300
Heat equals the change in internal energy.

This is why mixing path functions with state functions is frowned on. Firstly, a more empirical expression would be
Q=Uh-Uc, but both are bogus. Secondly, thermo is taught from the Stirling cycle simply due to constant internal energy processes which many newbies never can get their head around (due to other common heat engines).

Here's my juggernaut issue with this (and the following refers only to a strict Stirling cycle)...Wpos occurs only at Uh and Wneg only at Uc, so if input was Qh=1000 while Uh=400 and Uc=300, then 1000=400-300 ???

Fool wrote: ↑Sun May 26, 2024 6:14 am If an engine forward stroke starts while containing Uh 400 Joules, the most work it can produce is 400 J. This will be when it gets to absolute zero Kelvin. We want a maximum. The piston returns to the starting point costing zero back work, because it is at zero Kelvin. The buffer pressure has added a total of nothing because the same pressure opposed the forward stroke, as helped the back stroke. P∆V work, out forward stroke, in backward stroke, same P∆V

As a sidebar, I never cared for the long held work definition P∆V which at a glance only refers to isobaric process (there's too many qualifiers for P for me to ever consider this valid).

Fool wrote: ↑Sun May 26, 2024 6:14 am Now, back at the starting volume and zero K, 300 J must be added to complete the cycle. That 300 Joules must come from the forward work. Why? Because we are only adding 100 J. 300 J more must come from somewhere to get to the 400 J second position and 300 J starting first position. Forward work is the only energy available.

If the atmosphere were expected to add the Joules, it will do it at Tc for the entire return stroke and 300 J of work will be needed to compress the gas.

Nope...going along with typical cold hole premise, you compress the gas at 0k (another apparent free lunch) then you regen back to 300k.

Fool wrote: ↑Sun May 26, 2024 6:14 am If the buffer pressure force compresses it, 300 J Uc will be rejected as energy of compression to the cold sink as heat, to keep the reverse work at a minimum.

The cold hole camp will have no buffer pressure (vacuum) since no work is req'd for compression at 0k and zero buffer pressure won't tax expansion work. And at 0k, they tell me there's no heat of compression, so they don't sweat worrying about any heat sink below 0k akin common isothermal 'cooling'. I just love this cold hole spin...fantasy on steroids.

Fool wrote: ↑Sun May 26, 2024 6:14 am So, it starts with Uh and returns the amount Uc to complete the cycle. The maximum efficiency to start at Uc, add Qh, expand all the way to zero borrowing from the Uc energy, return the extra gained by over expanding to become a cold hole by adding back the borrowed heat Uc, during the return stroke, to complete the cycle is:

n=(Uh-Uc)/Uh

The description is a tad fuzzy, but the conclusion is correct. Gotta stop here before being timed out.

No worries Fool, not giving you a hard time, just squaring it up for the time capsule...

Understanding all this math stuff is totally unnecessary, but understanding compression cycle theory is paramount to any engine design. The basic takeaway is that an engine runs on a pressure differential and there's various ways to achieve this. Most early engines had no or minimal compression and were simple/crude schemes. Prior Otto, credible engineers avoided compression cycles, since any compression process would tax meager expansion process of the day. After Otto, most heat engines employed compression cycles that simply compress gas at a lower energy state then expand gas at a higher energy state.

The pros and cons of the compression cycle slipped past the old boys and continue to slip past many today. It was only after Otto that thermo evolved from more output to more efficiency.

I think the Phillips Company had nothing slip by. In the mid of century numbered "19", 1900's, they were building Stirling Engines and Cryo Coolers that people contributing to this forum only dream of.

Thanks for the attaboys, and thanks for the ah rats too. The attaboys are more fun. LOL

You are correct that my description isn't for any particular engine, let alone for a Stirling. I'm driving for generic, or a general proof. It seems to me that it doesn't matter how or when 400 J gets into an engine, the absolute most that can come out as work for a single stroke is 400 J. That will only occur if outside atmospheric/buffer pressure is zero. And it continues to zero Kelvin. The path will be such that the area underneath is 400 J. The mathematics works for any paths/engines if maximizing reversibility and energy out.

Since a real engine doesn't operate down to zero Kelvin, only down to 300 K, it remains operating in its ideal gas approximate range. Since it is operating in the ideal gas range it is fair to use ideal gas modeling and absolute zero Kelvin is a fair value for that model.

At the end of the forward/power stroke, since buffer pressure is zero and the internal gas pressure is zero, the return stroke to V1, the starting volume, will neither supply nor require work energy. Since it is now at V1 and zero Kelvin, 300 J of heat Uc must be put back in to get to the V1 300 J start of the he cycle. This is taken from the 400 gained from the forward stroke. 400 used, 300 returned, 100 output as work, 25% efficiency.

Adding outside atmospheric/buffer pressure just complicates the mathematics with ultimately zero gain in energy out. I know you know this. Hmmm is that sentence a double negative. LOL it has two "knows" in it. LOL it has two "nose" in it.

If a real engine starts expanding adiabatically from 400 J, with an outside pressure it will only have positive work until the inside pressure is equal to the outside pressure. This forward stroke will generate less than 100 J of work. If allowed to expand further that net work will be even less. If it doesn't expand further there is no return force to compress the gas, and the return energy must be taken from the forward energy. Ultimately the net energy gained for a complete cycle will be 25 J or less, as dictated by the temperatures.

The work contributed by the outside atmospheric/buffer pressure is isobaric. That is why it is equal and opposite. So it cancels out. Hence the P∆V premise.

Thanks for squaring things up. We agree, just get scrambled writing it, and reading it. I hope the above helps explain what I'm thinking without too many new mistakes.

Tom replied with the following in the "Heat is never hot or cold. Internal energy can be hot or cold.", thread. I thought it belonged more here, so will address it here.

Tom Booth wrote: ↑Sat Jun 01, 2024 9:26 am What I object to mostly is how (added) "heat" and (existing) "internal energy" are treated as equivalent for the purposes of the "Carnot limit" equation

η=(Th -Tc)/Th ⋅100%

Tc is the internal energy before adding heat
Th is the internal energy after adding heat

Th - Tc would then represent the heat added.

Let's say 100 joules are added
Tc = 300

η=(400 - 300)/400 ⋅100%

η=100/400 ⋅100%

η=1/4 ⋅100%

η=25%

All that really says is that the temperature was raised 25% or the internal energy was increased by 25%

To illustrate: suppose I have a gas tank that contains 300 ounces of gasoline.

I add 100 ounces making 400

Suppose my gas tank was empty and I added 100 ounces and drove 50 miles.

My mileage (efficiency) is 0.5 miles per ounce

Now suppose my tank has 300 ounces and I add 100 to make a total of 400.

I again drive 50 miles and burn 100 ounces.

My car efficiency is still 0.5 miles per ounce.

The "reserve" 300 gallons that never gets touched has zero influence on efficiency. It is how much is added vs.how much is burned or actually used that determines efficiency.

The 300 "internal energy" is balanced by an equivalent 300 external ambient energy These simply cancel out, may as well not exist. It's just the given environmental heat.

So "Carnot efficiency" is actually the percentage of heat that can be utilized of the environmental "internal energy" plus the added heat rather than just the added heat.

Why should the pre-existing "internal energy" be included along with the increase in "internal energy" that results exclusively from the added heat?

Actual efficiency is, or should be, heat added in joules vs internal energy (resulting from the added heat exclusively) converted to work in joules.

I didn't add 100+300=400 joules. I added 100.

Efficiency is (or IMO should be) a .measure of Joules added vs joules converted to work NOT joules added PLUS the pre-existing 300 joules of internal energy vs joules converted to work.

100 joules added, 100 converted = 100%

Not

100 joules added 100 converted = 25% because 100 is 25% of 400 which is joules added PLUS ambient "internal energy".

I see no justification for such a convoluted measure of efficiency.

Tom wrote: "What I object to mostly is how (added) "heat" and (existing) "internal energy" are treated as equivalent for the purposes of the "Carnot limit" equation ".

The two are not treated as equivalent. The derivation shows how the differences cancel in the calculation of the ratios.

It is similar to 4•3 is not equivalent to 3. And 4•5 is not equivalent to 5.

3/5 = (4•3)/(4•5) = 0.6 = 12/20 = 3/5 = 15/25 = 150/250 = 30/50

The fours cancel. The ratio is equivalent but the inputs aren't. 3/5 is just the fraction with the lowest denominator.

It is impossible to compare fuel tank levels with internal energy levels because it is just as easy to fill the top 1/4 of a tank as the bottom 1/4, and it is harder to fill at higher temperature than zero Kelvin, it harder to compress 14.7 psi than zero psi. That is the penalty of back work against an ambient pressure and or temperature. Your gas tank analogy didn't include any back work. You are considering the entire cycle forward work. Recompressing the gas is a penalty unique to thermodynamic engine cycles.

The only way to get 100 J of work out is to drive 100 J down hill consuming 25 J of fuel. You are now at a different level than the beginning of the cycle. That back driving will cost you 75 J to return to the station and trip starting point.

I will try to supply a mathematical proof that an ideal Stirling Cycle has an identical efficiency to the Carnot Theorem. Please give me some time.

Fool wrote: ↑Sun Jun 02, 2024 2:23 pm ....
It is impossible to compare fuel tank levels with internal energy levels because it is just as easy to fill the top 1/4 of a tank as the bottom 1/4, and it is harder to fill at higher temperature than zero Kelvin, it harder to compress 14.7 psi than zero psi. That is the penalty of back work against an ambient pressure and or temperature

I disagree with this in principle.

Before heat is added, inside pressure & temperature is equivalent to outside pressure & temperature.

They cancel out.

Forces are in balance at the start of expansion and return to a balance at the end of compression. There is no "back work" any more than there is "back work" when a scale is zeroed out and then a weight is added and removed.

The added heat is what causes an imbalance. An equivalent work output results in the piston returning to restore the original state of equilibrium.

The only "back work" is in your imagination, and that of every other Carnot limit advocate I suppose.

The only "back work" is in your imagination, and that of every other Carnot limit advocate I suppose.

Tom, even with a completely ambient pressure driven return stroke, the internal gas is still being compressed, hence the back work.

The only way to get a free ride is if internal pressure falls to a total vacuum for the entirety of the return stroke. And the only way for that to happen is reaching 0k.

I'd like to know how you think this part is avoidable.

VincentG wrote: ↑Mon Jun 03, 2024 8:08 pm

The only "back work" is in your imagination, and that of every other Carnot limit advocate I suppose.

Tom, even with a completely ambient pressure driven return stroke, the internal gas is still being compressed, hence the back work.

The only way to get a free ride is if internal pressure falls to a total vacuum for the entirety of the return stroke. And the only way for that to happen is reaching 0k.

I'd like to know how you think this part is avoidable.

It's pressure difference that matters not absolute pressure.

Reducing the internal pressure to say 10 psi is the equivalent of 14.7 - 10 or 4.7 psi of external pressure.

It's pressure difference that matters not absolute pressure.

Reducing the internal pressure to say 10 psi is the equivalent of 14.7 - 10 or 4.7 psi of external pressure.

That still leaves 10psi fighting the atmosphere on the return stroke. That's what the equation is validating.

VincentG wrote: ↑Tue Jun 04, 2024 6:37 am

It's pressure difference that matters not absolute pressure.

Reducing the internal pressure to say 10 psi is the equivalent of 14.7 - 10 or 4.7 psi of external pressure.

That still leaves 10psi fighting the atmosphere on the return stroke. That's what the equation is validating.

It's really beyond me how you guys don't seem to comprehend relative pressure.

Equal pressure, equal weight, or equal temperature on either side of anything cancels out.

How do you think a piston gets driven out on the power stroke?

You start out with 14.7 psi inside and outside the engine, on either side of the piston.

Then the inside air is heated. The pressure increases what? 1 or 2 psi on the inside?

So inside is then maybe 16 psi and outside atmosphere is still 14.7 a 1.3 psi pressure differential.

Why in bloody hell do you and "fool" think you need an "absolute vacuum" or a temperature of 0°K to reverse the process?

You don't.

With an isothermal expansion through say 3/4 of the power stroke the pressure drops below atmospheric. Adiabatic expansion the rest of the way to BDC reduces the internal pressure even more.

So you have a pressure differential of maybe 2 psi that drives the piston out during the power stroke, but you think a pressure differential of 14.7 or absolute zero internal pressure is required on the return stroke?

That makes no sense.

An internal pressure of 13.4 psi would drive the piston back to TDC with as much force as an internal pressure of 16 psi drove it out in the first place.

I don't even understand how that is not obvious or why it needs explaining.

So you have a pressure differential of maybe 2 psi that drives the piston out during the power stroke, but you think a pressure differential of 14.7 or absolute zero internal pressure is required on the return stroke?

Who said it's required?

The question is simple.

What would make more power?

1- An engine whos internal pressure drops to 10psi on the return stroke.

2- An engine whos internal pressure drops to 1psi on the return stroke.

VincentG wrote: ↑Tue Jun 04, 2024 8:43 am

So you have a pressure differential of maybe 2 psi that drives the piston out during the power stroke, but you think a pressure differential of 14.7 or absolute zero internal pressure is required on the return stroke?

Who said it's required?
...

Well, "fool" has been arguing all along that it is "impossible" for the piston to return without "rejecting" heat.

Matt Brown I think, a "nobody" previously and a few others. Most, if not all of the moderators of the Science forums.

It is a basic premise of the "Carnot Limit" and the 2nd Law of thermodynamics.

If that is not what you were saying, my apologies. Then I guess we are in agreement.

I'm not saying cooling might not be helpful at times, just that it is not necessary in order to have a running engine.

Heat can be converted to work without any necessity for heat "rejection" or heat passing through the engine to a "sink".

https://fordfestiva.com/forums/forum/cu ... -on-festys

As far as I am concerned this is still up in the air, I'll reserve judgment until I can test this myself.

All I was suggesting is that the engine that reaches zero k, or zero internal pressure will be more efficient and make more power, as the formula suggests.

Separately I'm saying that we live on Earth, so thats not going to happen.

VincentG wrote: ↑Tue Jun 04, 2024 9:52 am

https://fordfestiva.com/forums/forum/cu ... -on-festys

As far as I am concerned this is still up in the air, I'll reserve judgment until I can test this myself.

All I was suggesting is that the engine that reaches zero k, or zero internal pressure will be more efficient and make more power, as the formula suggests.

Separately I'm saying that we live on Earth, so that's not going to happen.

What's that forum link?

Anyway, I completely disagree.

Assuming by "the formula" you mean the "Carnot efficiency Limit" equation.

IMO it is completely ridiculous and contrary to observable, demonstrable experimental outcomes, and the whole nonsensical water wheel theory of heat engine operation it's based on is false.