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

Discussion on Stirling or "hot air" engines (all types)
Tom Booth
Posts: 4670
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

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

Post by Tom Booth »

Fool wrote: Tue Jan 02, 2024 1:39 pm I'm okay with heat being a form of energy passing from hotter to colder. I am not comfortable with internal energy being called heat. Something hotter has a higher temperature and higher internal energy than that same thing at a colder temperature, everything else the same.

Sorry it took so long to reply, I get busy, tired, lazy, distracted, procrastinated, plus, I like to think about your questions, and think a lot, sometimes for days. I research and think about my answers, also often for days, sometimes longer. I want to check and recheck that I'm not missing anything.

I think the question, "what is heat", is a proverbial 'good one'. I'm glad you put out an answer before I had a chance to do so. Although I think heat needs more than that description, it isn't your definition of heat that is our problem. It appears to be what follows.

Heat can be 'described' just as you have stated. Mathematics 'defines' the variable Q as heat, and it is solely to be used in equations for models. Subscripts are used to separate different processes or values of heat.

Moving on, I have a problem with your last statements. Conservation of energy is not derived from conservation of caloric, the two tend to contradict each other. Similar to how in an inelastic collisions kinetic energy and momentum appear to not both be conserved. They are, however, caloric doesn't exist. Let us forget about caloric.

First law:

Qh = Ql + W

Has zero to do with caloric theory. It reflects the observation that energy is conserved and caloric isn't. If caloric were conserved, the equation would be of the form:

Qh = Ql

Caloric/heat in would equal caloric/heat out. Work energy would then have to spontaneously form and be pushed out magically. Since this has never been observed, we conclude energy passes through and is conserved, caloric isn't (Which agrees with the lack of observed caloric.). Done talking about caloric. Leading to many energy balance equations such as the following:

Qh = Ql +W

Which means that for work out, the ejected Ql must be smaller than Qh. The process of 'converting heat to work', isn't direct. Heat must first be converted to internal energy. This is accomplished by a rise in temperature and or pressure, or both. 'Internal energy'/'pressure', pushes out the work, and at the same time pressure decreases, decreasing internal energy. That equation doesn't, by itself, disallow Ql from being zero. It does have a serious, divide by zero, problem with Qh being zero.

A side example. That is not the only equation. Example, the following:

Qh = U + Ql + W

U = Energy saved, stored inside the engine. Also called internal energy.

The left side is energy in, the right side is energy out plus energy saved. The equals symbol means the two sides must balance, similar to putting weight on either side of a balance.

The energy balance equation can be expanded even further:

Win + Qh + Uout = Uin + Ql + Wout

U can be in the form of potential, kinetic/momentum, light, or heat, chemical and perhaps others. Nuclear decay strikes me as one I left out.

Ending with:
Since the equation, Qh=Ql+W, doesn't come from caloric theory, and you said you were comfortable with it, can we move on with accepting it as the first law of kinematic thermodynamics? Please. If more clarification is desired, please explain.

I'm hoping to get to efficiency, n =W/Qh next.
There seems to have been some cross posting/editing as there are some equations in this earlier post I don't recall having seen and don't think I would have been OK with.

Starting with:

A side example. That is not the only equation. Example, the following:
Qh = U + Ql + W

U = Energy saved, stored inside the engine. Also called internal energy.
That makes no sense.

Say we add 50 joules = Qh

Or say we add 1 Joule = Qh

How could that ever be equivalent to all the internal energy - U, plus the heat rejected, plus the work output?

A typo maybe?
Fool
Posts: 1217
Joined: Sun Jul 16, 2023 9:14 am

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

Post by Fool »

Tom Booth wrote: Mon Mar 25, 2024 9:48 am Well, right off the bat I see the same conceptual difficulty with:
Let Qhz be the internal energy from heat added to the cycle all the way from zero K Kelvin to Th.
We add 100 joules at a temperature of 400°K

As stated previously "internal energy from heat ADDED to the cycle all the way from zero K Kelvin to Th" counts the baseline "internal energy" of 300 joules as if it is carried along or included in the 100 joules actually supplied as "heat".

Then later this same baseline internal energy is supposedly "rejected" though in actuality it was never added in the first place.

You already conceded previously:
Qcz is not rejected during the running. Nor is it added. It is the base energy that DQh is added to. Earlier definition discarded.
viewtopic.php?p=21686#p21686

This is the fundamental fallacy of the whole Carnot Limit so-called "LAW". It's an accounting error.

The baseline "internal energy" between 0°K and 300°K cannot be included with "heat added".

The only way that statement begins to make sense is in the context of Caloric theory where heat is considered to be a fluid that passes through the engine and temperature is the measure of a quantity of a fluid.
I see that you are challenging the Qcz term. It technically should use the U variable name. I used 'Q' because I'm only modeling the internal thermal energy. Since it can be represented as coming only from heat input, I thought it would be okay to represent it as a name similar to heat. It is just a name. U gets confused with 1/2MV^2 and Mgh and I^2R, E.T.C.... Qhz and Qcz are concerning temperature values, Kelvin scale/absolute.

If the gas is at 300 K, one doesn't need to ask how it got there. We just use it as an absolute quantity of energy.

The comparison to cooling is only to bring in absolute zero K, and have some reference to calculate the absolute energy contained in the engine at rest before even heating it to Th.

It is an absolute accounting of total energy. Sort of like having a pot of money, say 300 dollars. Someone asks to get paid the next amount you put into that pot. You put in $100. And you pull out $100, but the dollars you pull out are a mix of all the dollars you have. So instead of paying the $100 you last put in, there will be a 25~75 mix of bills that you pay out.

The person who you paid has to use the old bills for food, and the new ones for gasoline (silly rule made up for demonstration only). He would have $25 to spend on gasoline. And $75 to reject for food.

My derivation shows how absolute energy gets substituted for Delta energy. The same amount gets used by the engine but it becomes clear as to were it must be spent. 25 J on work, 75 J on back work, and or, heat transfer. The transfer of lower temperature heat, helps get more power out of the complete cycle by saving energy to complete the compression.

Hence it's often called 'the heat of compression'.

The two equations that are equal to 'n', do not imply that the numbers are equal:

Case in point, example:

n= (DQh-DQc)/DQh=(100-25)/100=0.25

n=(Qhz-Qhc)/Qhz=(400-300)/400=0.25

It's the same ratio of heat going out either way. The accounting is the same either way. You put into your bank $100, you take out $100. You have a bank account of $400, you have a balance of $300. The same $100 amount, just more information the second way. When you deposit a check, don't you ask for a printout of your balance? I do. More information, right.

The difference is that I know what Qcz is, so can use it in calculations to find 'n' and then DQc.
Tim Booth wrote:This is the fundamental fallacy of the whole Carnot Limit so-called "LAW". It's an accounting error.
Not really. There are other way to derive/prove the Carnot Theorem. I have come up with this way because the other methods require more advanced mathematics. Integral Calculus specifically. I wanted it more understandable by more people.

It wouldn't be consistent reliable science if there weren't multiple cross checks. That is why one data set won't overthrow a 200 year old theory. It will take many many plus, cross checks and theories. That is why people keep asking for additional tests on your experiment. That's why I ask for power output, input temperatures, theory, vacuum tests, oven tests, upside down tests. If we weren't trying to help we wouldn't be asking.

The best way to look at Qcz is the mathematical definition:

Qcz = M•Cv•Tc

It is the starting point of the engines cycle. And the end point. Starting point and end point must be the same for a complete cycle. It is also the point reached after DQc is releasd, regardless of how much that will be.

Heat is flowing through an engine. The flow isn't any particle, or caloric, nor does the "flow" cause the work. It is necessary to produce pressure increase and decrease. Heat in, higher pressure. Heat out lower pressure, back work. Calorics have been discarded completely in main stream science. That was accomplished about 100 years ago. Heat flows more the way light flows.

The parts of Thermodynamics that are kept are the ones that can be derived from current understandings of Thermodynamics, including but not limited to, Kinetic Theory, Statistical Theory, and Quantum Mechanical Theory. No, I don't know all those, so don't ask for those proofs. Just take comfort that if current theories don't agree, someone is working on it. From what I do know of those theories, Entropy, and Carnot, allegedly, hold up well. Kind of like Relativity Theory, both Einstein and Carnot have been picked to death, and still prevail. No matter how smart we are, we'd have zero chance of changing that. And, so far, I only see evidence of support for Carnot. Not even Matt has anything substantial, claims yes, but no working devices. Improvements perhaps. Throwing Carnot out? Not even close. All your points don't add up. My derivation is consistent with others, I'm surprised no one else has mentioned it. I suppose it is considered non ground breaking.
Fool
Posts: 1217
Joined: Sun Jul 16, 2023 9:14 am

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

Post by Fool »

I'm taking a time out. I need to study all that's been said. My derivation appears consistent. Qcz is in question, but it is not a problem as I see it. Will study what the issue is. Please be patient until I return. Thanks for working here. Your time is important, and I don't want to waste it.
matt brown
Posts: 749
Joined: Thu Feb 10, 2022 11:25 pm

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

Post by matt brown »

Fool Re- wrote: Wed Mar 20, 2024 7:36 am
Let Qh be Delta Qh = DQh, the heat added per cycle. 100 J
That 100J is where this starts to fall apart...it's not cycle input.
Fool Re- wrote: Wed Mar 20, 2024 7:36 am
W = Work output from the working gas.
W = DQh - DQc
n = efficiency = W/DQh = (DQh - DQc) / DQh
Hey Fool, don't you think this is a better starting premise...since Carnot refers to compression cycles (tho hasn't been beat by non-compression cycles) how about SIMPLY

n=(Wpos-Wneg)/Wpos, therefore
when Th=Wpos and Tc=Wneg, then
n=(Th-Tc)=Th

This eliminates U considerations, but presupposes Q=W
Fool Re- wrote: Wed Mar 20, 2024 7:36 am Basically we took relative scale Delta heat:
n=(DQh-DQc)/DQh
Converted it to absolute temperature scale heat:
n=(Qhz-Qcz)/Qhz
Then converted it to temperature calculated heat:
n=(Th-Tc)/Th
They are all the same value of n, because they are all the same heat-in heat-out ratios. They are just three different ways of obtaining the same thing.
Why wade thru the weeds when you can cut to the chase ???
matt brown
Posts: 749
Joined: Thu Feb 10, 2022 11:25 pm

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

Post by matt brown »

Fool wrote: Mon Mar 25, 2024 8:55 pm
I see that you are challenging the Qcz term. It technically should use the U variable name. I used 'Q' because I'm only modeling the internal thermal energy. Since it can be represented as coming only from heat input, I thought it would be okay to represent it as a name similar to heat. It is just a name. U gets confused with 1/2MV^2 and Mgh and I^2R, E.T.C.... Qhz and Qcz are concerning temperature values, Kelvin scale/absolute.

My derivation shows how absolute energy gets substituted for Delta energy.
This substitution is very creative, but IMO too complex for this site. The problem with proving Carnot remains proving T as a consequence vs a coincidence. However, I'd still rather "hash thru your pitch" (wink-wink) than any calculus dibble.

The amusing thing about Carnot limit is that it also carries over into non-compression cycles, open or closed, so this guy is hard to beat.

If you still toy with super-Carnot cycles, have you ever checked out low volume ratio Braytons with regen ?
Fool wrote: Mon Mar 25, 2024 8:55 pm My derivation is consistent with others, I'm surprised no one else has mentioned it. I suppose it is considered non ground breaking.
The gatekeepers and grantmeisters make their living from complicating stuff.
matt brown
Posts: 749
Joined: Thu Feb 10, 2022 11:25 pm

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

Post by matt brown »

matt brown wrote: Mon Mar 25, 2024 9:04 pm
Hey Fool, don't you think this is a better starting premise...since Carnot refers to compression cycles (tho hasn't been beat by non-compression cycles) how about SIMPLY

n=(Wpos-Wneg)/Wpos, therefore
when Th=Wpos and Tc=Wneg, then
n=(Th-Tc)=Th
edit: n=(Th-Tc)/Th

Yikes, missed that big time, but you guys likely caught it...
Tom Booth
Posts: 4670
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

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

Post by Tom Booth »

matt brown wrote: Mon Mar 25, 2024 11:59 pm
matt brown wrote: Mon Mar 25, 2024 9:04 pm
Hey Fool, don't you think this is a better starting premise...since Carnot refers to compression cycles (tho hasn't been beat by non-compression cycles) how about SIMPLY

n=(Wpos-Wneg)/Wpos, therefore
when Th=Wpos and Tc=Wneg, then
n=(Th-Tc)=Th
edit: n=(Th-Tc)/Th

Yikes, missed that big time, but you guys likely caught it...
IMO the entire Carnot efficiency "Limit" derives from the false premise that heat as a substance (caloric) goes through a heat engine like water through a turbine, carried through by some force.

The force that causes water to flow through a turbine from a high to a low level is gravity.

There is no HEAT GRAVITY or any equivalent which causes heat to "flow" from a "high" to a "low" temperature.

If you had a reservoir or water at a high elevation and another at a low elevation and open a channel for water to flow between the two, ALL the water will flow down into the lower reservoir. It is compelled to do so by gravity.

If you have a hot object and a cold object and put them together, random motion will eventually result in the two objects reaching equilibrium. The hot object will cool down and the cold object will heat up until both objects are at an intermediate warm temperature. No directional "flow". Just a gradual neutralization of the ∆T.

Heat in a gas, is merely energy or the motion (vibration) of the gas molecules.

Heat added to a gas increases the molecular vibration, resulting in expansion. The molecules move around or vibrate more vigorously and so require more room, spreading further apart. They repel each other more.

This happens on a molecule by molecules basis. There is no "flow" anywhere, and the presence or absence of any "cold" near or far has ZERO influence on the individual molecules vibration or behavior or motion.

The whole Carnot theory of heat as a fluid that "flows" from a "high" to a "low" temperature is primitive, silly, juvenile nonsense based on nothing whatsoever but ignorance regarding the actual nature of heat.
Tom Booth
Posts: 4670
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

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

Post by Tom Booth »

So what happens when heat is added to a gas in a confined space with a moveable wall (piston)?

Add enough energy to cause one gas molecules to "jiggle around" more vigorously.

This additional "jiggling", since the chamber is sealed, has nowhere to go but may transfer if molecules collided, but cannot leave the chamber causing "pressure" within the chamber.

But the only "wall" that can move is the piston.

Eventually the "hot" molecule collides with the piston and transfers it's energy to the piston and the piston moves out.

The molecule that was "hot" lost energy to the piston so the vibration of that molecule falls back to the normal, or ambient "baseline" level of vigor.

Due to the piston motion however, the volume has increased, which results in a partial "vacuum" and the piston returns.

Rapidly heat many molecules simultaneously and the effect is more pronounced, but the result is the same.

The gas is heated causing "pressure" and the piston moves, due to energy transfer from the gas molecules to the piston. The increase in volume results in a vacuum and the piston moves back.

Heat is converted to mechanical motion by the repulsive or expansive force of some confined gas molecules transferring their acquired energy to a piston, not as a consequence of the "flow" of some imaginary "fluid".
Fool
Posts: 1217
Joined: Sun Jul 16, 2023 9:14 am

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

Post by Fool »

You could be right about all that.

In regard to all of science and especially religion:
"The first principle is that you must not fool yourself and you are the easiest person to fool."
Richard P. Feynman
https://en.m.wikipedia.org/wiki/Everyth ... w_Is_Wrong


I don't know if your theory supports how heat gets from the sun to our faces through the Vacuum of space, or that the cold of Earth is cooling the sun any.

How does a heat shield work? It blocks heat from a camp fire, but it doesn't seem to block cold.

You are trying to redefine heat as vibration, that doesn't flow? So how does our agreed definition fit in. Energy in transit?

You are making the claim that heat energy isn't conserved, it gets neutralized by neutralizing cold? How do you, "unneutralize" heat.

What powers neutralization? How is heat different than Temperature? Hot? Cold? Absolute zero Kelvin? How do you define heat work energy equivalence? How do you calculate maximum efficiency? Real efficiency? What can we expect to get from two temperatures? Two pressures? Two Volumes? What happens to heat when it is neutralized and gets colder? What happens to cold? How much cold does it take to neutralize how much hot? Why does a hotter cool faster than warm?

Many of those questions can be answered, with reasonable error, with the 200+ years old Theory. Caloric or Kinetic.
Tom Booth
Posts: 4670
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

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

Post by Tom Booth »

Fool wrote: Tue Mar 26, 2024 9:39 am You could be right about all that.

In regard to all of science and especially religion:
"The first principle is that you must not fool yourself and you are the easiest person to fool."
Richard P. Feynman
https://en.m.wikipedia.org/wiki/Everyth ... w_Is_Wrong


I don't know if your theory supports how heat gets from the sun to our faces through the Vacuum of space, or that the cold of Earth is cooling the sun any.

How does a heat shield work? It blocks heat from a camp fire, but it doesn't seem to block cold.

You are trying to redefine heat as vibration, that doesn't flow? So how does our agreed definition fit in. Energy in transit?

You are making the claim that heat energy isn't conserved, it gets neutralized by neutralizing cold? How do you, "unneutralize" heat.

What powers neutralization? How is heat different than Temperature? Hot? Cold? Absolute zero Kelvin? How do you define heat work energy equivalence? How do you calculate maximum efficiency? Real efficiency? What can we expect to get from two temperatures? Two pressures? Two Volumes? What happens to heat when it is neutralized and gets colder? What happens to cold? How much cold does it take to neutralize how much hot? Why does a hotter cool faster than warm?

Many of those questions can be answered, with reasonable error, with the 200+ years old Theory. Caloric or Kinetic.
Your straw manning.

I'm not "trying to redefine heat as vibration", and yes, there is nothing new, it's just modern kinetic theory.

And no I did not say "heat" is neutralized, I said the temperature difference is neutralized.

All matter vibrates. Some vibrates more. In terms of our senses, we call that "hot" because it makes the molecules in our finger vibrate more when we touch it.

I'm not inventing any new science here, just discarding the obsolete notion of heat as something that "flows" down to a colder temperature like a fluid flowing down hill, which is just a holdover from Caloric theory.

In a chaotic mix of moving and vibrating particles, it would not be impossible for two slow molecules to collide into a hot molecule, the cold molecules transferring their energy to the hot, changing its trajectory and all kinds of other possible transfers, based on the localized motion of the particles not any distant gravitational field compelling motion in a specific "downward" direction.

Hot and cold are relative terms and when put together tend to average out over time.
Tom Booth
Posts: 4670
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

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

Post by Tom Booth »

Tesla recognized heat was energy, but bought into the idea that heat was "compelled" to "flow" into a "cold hole", so likely wasted much of his life trying to create a "cold hole" so he could just sit back and let the heat flow "down", then extract "free energy" by simply dropping a heat engine down to intercept the "flow" like a water wheel.

Well, he was half right. Heat is mostly just a FORM of energy that can be converted. But the conversion requires considerable additional engineering beyond simply "digging a cold hole" so that the heat will be "compelled to flow in".

Compelled by what?

The moving, vibrating particles are meandering every which way and tend to be rather uncooperative in my experience,... experimentally that is.
Goofy
Posts: 37
Joined: Sun Feb 13, 2022 3:06 am
Location: Denmark

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

Post by Goofy »

Maxwell`s Demon.png
Maxwell`s Demon.png (272.79 KiB) Viewed 5565 times
Like the picture, we perhaps always imagine the gas molecule's hurling around, bumping into each other or into walls or piston tops.

But then what's in between the molecules' ? ;-)

I imagine the container or cylinder is "packed" with molecule`s. They are touching each other and cannot move or fly around as so.
But I believe they are vibrating somehow related to their temperature and the pressure.

I think they act`s more like a Newton`s cradle. The ball`s in-between don`t move.

If you think of an ocean wave, the water molecule don`t travel from cost to cost, but they push on one to the next.
The wave is moving, not the water.

In an electric wire, the electron's don`t "travel" or flow through the wire from the utility plant to your lightbulb. They make
a kind of push on their neighbors, created by a potential difference.

In a soundwave, the air molecules don`t travel anywhere either. It`s a pressure wave moving by speed of app. 343 m/sec.

Nikola Tesla compared light, to "a soundwave in the ether"

So where does that leave us, as to find out how heat moves/travels ?

The 3 classic ways are conduction, convection, and radiation.

I compare conduction in perhaps a copper rod, to the "flow" of electrons in a electric wire.
I compare convection between perhaps a hot plate with air around, like a ocean wave.
I compare radiation from perhaps a candlelight, as to a soundwave.

But we also accept, that light can both be particles and waves.
- So a humble candlelight hurls out particles/photons at a speed of 300.000 km/sec. ?

BR
Petter
Tom Booth
Posts: 4670
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

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

Post by Tom Booth »

Goofy wrote: Wed Mar 27, 2024 2:35 am Maxwell`s Demon.png
Like the picture, we perhaps always imagine the gas molecule's hurling around, bumping into each other or into walls or piston tops.

But then what's in between the molecules' ? ;-)

I imagine the container or cylinder is "packed" with molecule`s. They are touching each other and cannot move or fly around as so.
But I believe they are vibrating somehow related to their temperature and the pressure.

I think they act`s more like a Newton`s cradle. The ball`s in-between don`t move.

...
This is certainly true for solids and liquids. With gases, I think it depends on pressure and temperature.

How much "empty space" exists between each gas particles before it collides into another gas particles is called the "free path". It varies a great deal with pressure and temperature, the type of gas, the size of the molecules, and so forth.

This seems to be well established to a high degree of mathematical precision.

At low pressure and high temperature gas particles are largely flying free, independent of one another.

In an engine you have both extremes.

It suddenly occurs to me

During high compression of a gas at low temperature, you have the Newton's Cradle effect. With low compression and high temperature, more like the swarm of bees.

Convection does not rely on a Newton's Cradle type transfer of heat Convection is the boyancy of the "swarm of bees" in the air rising above the "newtons cradle" clumping around the queen down on the ground.

Electromagnetic type, infrared transfer?

I'm not really sure to what degree that might play a role in a gas.

If the heat exchanger is red hot or if the engine is a transparent "solar" type.

Infrared light traveling as a "wave" through a gas in an engine, I imagine, probably has little direct effect as far as expanding and contracting the gas.

There are many, many variables involved, so, in talking about "Stirling engines" there are so many different types, pressure ranges, working fluid choices etc. etc. along with rapidly changing conditions, there are usually exceptions to any broad generalizations.
Fool
Posts: 1217
Joined: Sun Jul 16, 2023 9:14 am

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

Post by Fool »

matt brown wrote: Mon Mar 25, 2024 9:04 pm
Fool Re- wrote: Wed Mar 20, 2024 7:36 am
Let Qh be Delta Qh = DQh, the heat added per cycle. 100 J
That 100J is where this starts to fall apart...it's not cycle input.
Tom seems to have a problem with the definition of Qcz, the absolute internal energy of the system at the starting point. Defined as:

Qcz = M•Cv•Tc

I'm not sure why? That is a standard way of defining a variable, and a standard specific heat equation.

DQh could be heat per one cycle, no reason it can't be. Just a choice of engine volume, gas, and atmospheric pressure. Although I see your point that DQh could be the amount of heat per one hundred cycles, the 100 multiple would cancel out the same as the M and Cv do. I hope that I've understood your point?
matt brown wrote: Mon Mar 25, 2024 9:04 pmHey Fool, don't you think this is a better starting premise...since Carnot refers to compression cycles (tho hasn't been beat by non-compression cycles) how about SIMPLY

n=(Wpos-Wneg)/Wpos, therefore
when Th=Wpos and Tc=Wneg, then
n=(Th-Tc)=Th

This eliminates U considerations, but presupposes Q=W
Yes I do. Good one. You sold me. The Carnot Theorem is now proven.

Now we need to get Tom to agree that there is such a thing as work positive and negative. Positive, work out of the working gas, expansion. Negative, work into the gas, compression.

Carnot was comparing the work into and out of the gas, and the heat in and out to produce those two strokes. Not heat flowing by. It takes heat in to produce expansion. It takes heat out to allow compression. Not doing that makes work in and out cancel each other, for an adiabatic bounce zero work out process.

To get work out, heat must be added. The volume and temperature will now be larger. If it weren't hotter, the volume wouldn't be bigger at the end of expansion.

To cycle back to the beginning, heat must be removed, and work must be input, to the internal gas. Negative work. If not cooled, negative work will be larger. Cooling reduces the back work, so there can be a total work out for the complete cycle. No escape for any cycle. As proven by Carnot, Clausius and Kelvin, and many many others since them, except Tesla and many other home experimenters.
matt brown wrote: Mon Mar 25, 2024 9:04 pmWhy wade thru the weeds when you can cut to the chase ???
Mathematical proofs are funny that way. Ain't they. And a pain too. Proofs depend on the recipient more than the provider. Sometimes the recipient needs to dig-in to understand, a process that is difficult.
matt brown wrote:If you still toy with super-Carnot cycles, have you ever checked out low volume ratio Braytons with regen ?
I have not. Perhaps I'll give it a gander one day. Thanks for the tip.
matt brown wrote:The gatekeepers and grantmeisters make their living from complicating stuff.
Ain't that the truth. If you can't dazel some one with brilliance, baffle them with bull shit.

By the way, didn't catch your typo. I guess I'm a poor proof reader. Thanks for the help here.
Fool
Posts: 1217
Joined: Sun Jul 16, 2023 9:14 am

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

Post by Fool »

Here is a link to a good Calculus based proof of how the Carnot Cycle equals the Carnot Theorem:

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

Scroll down for the proof. It also has a good history point leading up to the proof.

I'd quote some here but the special symbols, for integrations, seem to be lost in the cutting and pasting. Bummer.

The proof only applies to the Carnot cycle. A similar, slightly simpler proof could be given for the Stirring cycle. The basic difference is the adiabatic parts for the Carnot proof would be replaced by the zero work zero Delta Volume process of heat removal, by the regenerator displacer stroke.

For the Otto Cycle, from Wikipedia:

https://en.m.wikipedia.org/wiki/Otto_cycle
Equation 6:
n = 1 - [1/(r^(gama-1))]

From analyzing equation 6 it is evident that the Otto cycle efficiency depends directly upon the compression ratio 'r'.
Gama = Cp/Cv about 1.4 for air.
It seems funny to me that the same equation that leads to n=(Th-Tc)/Th , leads to the one above for Otto Cycle's compression ratio, and that everyone accepts high compression ratio equals higher efficiency, but higher temperature ratio is unbelievable? Higher temperature ratio is higher compression ratio. Same equation, same premiss.

As the compression ratio increases, the Otto Cycle efficiency gets closer to the Carnot cycle, but never quite there.
Post Reply