The Carnot efficiency problem

Discussion on Stirling or "hot air" engines (all types)
matt brown
Posts: 755
Joined: Thu Feb 10, 2022 11:25 pm

Re: The Carnot efficiency problem

Post by matt brown »

Tom Booth wrote: Mon Sep 04, 2023 5:17 pm
matt brown wrote: Mon Sep 04, 2023 2:54 pm (...)
Turning this Carnot limit into a mystery is ridiculous, since it's only a simple mathematical derivation of how PV=mRT plays out.

(..)
How so?

Who says it's a mystery?

Mathematically the carnot supposed limit is nothing more than a ratio derived from a temperature difference.
Note that Carnot eff. is Tmax and Tmin CYCLE ratio for "isothermal" Carnot, Stirling and Ericsson. However, Carnot eff. for "adiabatic" Otto and Brayton is Tmax and Tmin ratio of their adiabatic PROCESS, not Tmax and Tmin ratio of their cycle.

So, an Otto (or Brayton) with 6:1 adiabatic compression will have Carnot=.50 since 6:1 air goes from 300k to 600k. The adiabatic compression ratio locks in Carnot eff. for these 2 cycles (think ICE) and however much or little input above 600k (this ex) effects output only, not Carnot eff.

The jerk knee takeaway is that the Carnot limit for isothermal cycles is easy enough for grade school kids to memorize. Meanwhile, the adiabatic cycles are second fiddle with far less Carnot eff. (usually) but still allow the typical max Carnot limit dribble to proceed. Unfortunately, Carnot eff. for adiabatic cycles requires more work...compression ratio to "gamma" exponent, mono vs diatomic, and 2 versions whether seeking deltaT or deltaP. A few buttons on most calculators and you're all set. I keep a 'cheat sheet' with adiabatic PVT values indexed to volume ratios, one list for monatomic and another for diatomic.

The only mystery here is that few know the mathematical derivation from PV=mRT to Carnot buzz. It's just a little busywork, no calculus or anything major (geez, I'm too lazy for that stuff anymore).
Tom Booth
Posts: 4727
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

Re: The Carnot efficiency problem

Post by Tom Booth »

matt brown wrote: Mon Sep 04, 2023 6:43 pm
Tom Booth wrote: Mon Sep 04, 2023 5:17 pm
matt brown wrote: Mon Sep 04, 2023 2:54 pm (...)
Turning this Carnot limit into a mystery is ridiculous, since it's only a simple mathematical derivation of how PV=mRT plays out.

(..)
How so?

The only mystery here is that few know the mathematical derivation from PV=mRT to Carnot buzz. It's just a little busywork, no calculus or anything major (geez, I'm too lazy for that stuff anymore).
Still waiting.

Maybe a reference?

So how "few" know about this "simple derivation"?

Just you?
matt brown
Posts: 755
Joined: Thu Feb 10, 2022 11:25 pm

Re: The Carnot efficiency problem

Post by matt brown »

Tom Booth wrote: Mon Sep 04, 2023 5:11 pm BTW, where did this idea that an "Ideal" Stirling cycle is ALL isothermal expansion (on the expansion stroke) ?

The term "isothermal", I believe, originated with Clausius about 30 years or so after Stirling got his patent. Carnot was skeptical that hot air engines could even exist so not likely he ever said such a thing.
Most of our current Stirling buzz dates to Schmidt analysis of 1871. Schmidt is famous for his "closing the cycle" analysis and was aware of various imperfections like instantaneous equal pressure thruout engine cycle, etc. In the early 1840s, Clapeyron was exploring Carnot's cycle concept while all the legendary thermo gurus were debating the Stirling. Carnot's main thing was his cycle concept, but suggesting a compression cycle with Wneg was scoffed at until Otto had his Eureka moment. Otto changed everything, and most previous thermo was discarded as wrong, misleading or obsolete. Then Diesel came along and sealed the deal, no looking back. A lot of what happened is lost to history, except what is frozen in time at the USPTO and conveniently dated.

It really doesn't matter if a Stirling has ideal isothermal expansion, what matters is knowing what ideal PVT values are for an isothermal expansion vs ideal PVT values for an adiabatic expansion, then we know that reality is somewhere in between. Any closed cycle requires one to "close the cycle" (valid PV plot) or the cycle will stall, but any open cycle cheats this requirement.
Tom Booth
Posts: 4727
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

Re: The Carnot efficiency problem

Post by Tom Booth »

Ummm... ok well;

Here's one problem that seems to me a blatant contradiction

Take that "Carnot cycle":

(1) Process 1 – 2: Isothermal expansion at TH.
(2) Process 2 – 3: Adiabatic expansion (Q=0).
(3) Process 3 – 4: Isothermal compression at TC .(4) Process 4 – 1: Adiabatic compression (Q =0).

The site here says:
"By using the ideal gas equation (pV = nRT), the fact that W = Q for isothermal processes and the fact that pV()=constant for adiabatic ideal gas processes it can be shown that, Qh/Qc=Th/Tc
Therefore,...."
https://www.physics.louisville.edu/clda ... cycle.html

If I put in 10,000 joules for Q at let's say 370K and " W = Q for isothermal processes " then Q out is

e=W/Qh

So if W= Q then e=10,000/10,000

Or e=1

Well my hot side is 370k and my cold side is 270k

Above it says: " it can be shown that, Qh/Qc=Th/Tc ..."

Really? Shown how exactly?

Qh is 10,000 joules

If W=Q then that 10,000 joules went out as work in the isothermal expansion right? So Qc is zero

10,000/0 does not equal 370/270

BTW, just generally:
QH is the heat transfer interaction with the hot reservoir and QC is the heat transfer interaction with the cold reservoir
I assume that holds true here.
Tom Booth
Posts: 4727
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

Re: The Carnot efficiency problem

Post by Tom Booth »

So....

To me, this statement, which I've actually seen many times: "" it can be shown that, Qh/Qc=Th/Tc ..." in many places, has no actual basis.

It looks like circular reasoning where the Carnot efficiency limit assumption, and, apparently, that is all it is, an assumption (that the temperature difference limits efficiency via "the height of the fall" i.e. heat 'automatically" "flows down" high to low temperature: from hot to cold a.k.a. obsolete Caloric theory) is adopted as an axiom to draw the conclusion that the unproven axiom itself has some basis or foundation.

To simply SAY "It can be shown" that 1 - 1 > 0 by (some arbitrary fraction) does not actually show or prove anything.

So if I put in 10,000 joules of heat and do 10,000 joules of work then 1/0=270/370

Why?

Because "it can be shown" that 1/0=Qh/Qc

So... how does that follow exactly?

Because otherwise you would be able to exceed the Carnot limit, and the Carnot limit says you cannot exceed the Carnot limit, so that's why you can't exceed the Carnot limit, because the Carnot limit says you cannot exceed the Carnot limit because the Carnot limit says you can't, therefore it can be shown that you can't because....etc. etc.

Circular reasoning.

So how exactly can it be shown that my 10,000 joules of heat input equals 370°K ?

And my zero waste heat after doing 10,000 joules of work equals 270°K ?

The whole proposition doesn't even make sense.
matt brown
Posts: 755
Joined: Thu Feb 10, 2022 11:25 pm

Re: The Carnot efficiency problem

Post by matt brown »

Tom Booth wrote: Tue Sep 05, 2023 4:04 am
The whole proposition doesn't even make sense.
That link is very poorly written. I also found it, prior your post.

eff_1.png
eff_1.png (104.58 KiB) Viewed 11687 times

Rewording the last couple lines, it should read: work equals source heat minus sink heat where source heat is heat entering at Thot during isothermal process AB and sink heat is heat leaving at Tcold during isothermal process CD.


eff_2.png
eff_2.png (17.69 KiB) Viewed 11687 times

It then continues with by using:
(1) the ideal gas equation PV=nRT
(2) the fact that W=Q for isothermal processes
(3) and the fact that PV^y=constant

IT CAN BE SHOWN that source input/sink output=Thot/Tcold

That PV^y=constant is where y represents the Greek lower case gamma symbol and means the Cp/Cv ratio of specific heats (1.4 for diatomic air, 1.66 for monatomic helium).

Unfortunately, the "it can be shown" part is the derivation. The problem here is that they're using a Carnot cycle instead of a Stirling cycle, and the Carnot cycle has those squirrelly adiabats that complicate stuff vs simple Stirling isochors (note how sketchy the volume demarcations are in this Carnot PV).

eff_3.png
eff_3.png (25.75 KiB) Viewed 11681 times


They then continue to their trusty conclusion, but it comes off as slight of hand.

The PV=nRT to Carnot Eq. is only for those who need verification but haven't heard of internal energy. Once you grasp that internal energy is linear T, lots of this stuff should be nearly obvious. Again, all this Carnot buzz mainly relates to compression cycles, but no one yet has beat this limit with any other heat engine cycle. If you want to mess with non compression cycles (Lenoir or whatever) you're in for a much longer ride...
matt brown
Posts: 755
Joined: Thu Feb 10, 2022 11:25 pm

Re: The Carnot efficiency problem

Post by matt brown »

Something else to consider (and I don't recall anyone mentioning) is that any engine without cooling and compression can't be a Stirling or Ericsson, or Otto, or Brayton...
Tom Booth
Posts: 4727
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

Re: The Carnot efficiency problem

Post by Tom Booth »

Not sure where your going with this. The "simple derivation" seems to have become a bit complicated.

A few posts back on this same thread though, you wrote:
During the isothermal expansion, ALL the source heat becomes work, ... adding 10kJ source heat will produce 10kJ work during expansion, ...Great, all heat became work aka 100% efficiency.
https://www.stirlingengineforum.com/vie ... %25#p20307

I take this to be equivalent to the statement on the website being referenced:

W = Q for isothermal processes

According to the first law of thermodynamics, aka conservation of energy:

e=W/Qh

So

e=10,000/10,000=1

So, the second law contradicts the first.

Logically and mathematically, Qc can't be anything other than zero.

Anyway, I could really care less if an "Ideal" Stirling cycle is supposed to be ALL isothermal expansion according to whomever long dead theoretician . The observable REALITY is, the expansion in any real engine is not "reversible", is also ADIABATIC which does work at the expense of "internal energy" which in a real PV tracing of a real Stirling shows a sharp drop in pressure below atmospheric at BDC all the way back to very near TDC.

When theory clearly contradicts experimental observation I really don't know why anyone would consider it "foolish" to think maybe the theory needs to be re-evaluated rather than adhering doggedly to theory.

At TDC before heat was added, internal energy was whatever X.

10,000 joules were added.

10,000 joules were converted to work.

The piston is now at BDC but with the same internal energy it hat at TDC.

That is the equivalent of a stretched spring, it's going to return.

If I drive a car up a hill in opposition to gravity, it has gained the "potential energy" necessary to roll back down. Nothing more has to be added or removed.

If the gas expanded moving the piston in opposition to atmospheric pressure to BDC the piston is then in a new position where it already has the "potential energy" to return to TDC.

Somehow Kelvin or Clausius or.whomever thought heat was different from all other energy in the universe and follows some special laws that exempts it from conservation of energy. Personally I don't buy it. Sorry but it just doesn't wash. Experimentally it falls flat.

Don't blame me just because I'm apparently the only one to do any significant heat engine experiments in the past 200 years.

This whole Carnot Limit so-called "LAW" was just swallowed hook line and sinker and never challenged or tested or doubted.

Well I take that back. It's been challenged and doubted a plenty but God help the sorry soul with any common sense to come along and do so.

The rabid relentless 2nd law enforcers will come out of the woodwork to harass and ruin the poor sap and never give up until he recants or drops dead.

Funny thing is, all this theory attributed to Carnot.

Carnot himself threw it out as evidenced by his unpublished notebooks.

Kelvin resurrected the Caloric theory after Carnot had actually already rejected it.
Tom Booth
Posts: 4727
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

Re: The Carnot efficiency problem

Post by Tom Booth »

matt brown wrote: Tue Aug 08, 2023 12:22 am
Tom Booth wrote: Mon Aug 07, 2023 11:05 pm
Ummm.,. A little too cryptic. No real idea what you are saying here.
A lot of people are familiar with the FRB, but these goons only print the moola (burrrr) and regulate domestic credit. Meanwhile, the 800 lb. gorilla on the world stage is the petrodollar which the Fed has no direct control over. It's not money, but credit, denominated in US dollars. Here's the 'problem' which the rich and Uncle Sugar enjoy...since most commodities are dominated in US dollars, you need DOLLARS to buy this stuff and there's only 2 ways to get these dollars (1) trade something to the US for the dollars, or (2) 'borrow' the US dollars. The 1st method forces countries to discount exports to the US to get the desperate dollars they need for commodities they lack. The 2nd method is more sinister, a credit system created out of thin air with a massive debt that settles in dollars or default. Either way, guess what the #1 thing is that everyone is scrounging dollars to buy? Yep, it's oil.

Haven't you heard, this dollar hegemony is what the US military enforces. Nix the oil dependency (or the oil trade in dollars) and you nix the dollar...
Maybe a good reason to perpetuate this "Carnot Limit" nonsense through the educational system you think Matt? Or what exactly are you trying to say here?
Tom Booth
Posts: 4727
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

Re: The Carnot efficiency problem

Post by Tom Booth »

matt brown wrote: Mon Sep 04, 2023 6:12 pm
Tom Booth wrote: Mon Sep 04, 2023 4:41 pm Honestly, I find this explanation laughable Matt, no offense, but it makes no sense.

All you said, in regard to my actual question is:
"the reason why 0k would be required for 100% efficiency is simply that to compress any gas back to its original volume IN A CYCLE would require Wneg=0 aka zero internal energy to resist the work of compression."
That's an affirmation of an opinion not an explanation.

The assertion has no rational basis.

Maybe you could explain it another way because just saying "IN A CYCLE" proves nothing at all.

So why should a Stirling engine have to go to absolute zero to get the piston to return one cycle, when quite obviously and observably the piston returns to TDC by the same atmospheric pressure it had to work against to get to BDC in the real world all the time, quite easily.
In a cycle, everything must return to the start state, so any expansion would require a similar compression. The work of compression aka Wneg must come from somewhere whether it's from partial expansion energy stored in a flywheel or buffer/ambient pressure that was Wneg during expansion. Either way, this Wneg is proportional Wpos via PV=mRT where Wneg/Wpos=eff. The only way to have compression Wneg=0 is at 0k. Your static analogies are fine for lifting and dropping rocks, but don't cut it in thermo since there's kinetic energy at play (except at 0k).

You act as if LTDs are magical...who cares if some toy can run without a flywheeel or whatever. As I've said before, Senft didn't do us any favors with this 'educational' tinker-toy, he just ushered in endless distraction. I suspect that these LTD toys are nothing near a "Stirling", but approximate the three legged Lenoir cycle...yeah, no Wneg or physical heat sink req'd. They simply suck in a little air, mix it with some other air, semi-isolate during heating, then let this air slightly expand and leak a little to ambient, then repeat.

The Lenoir cycle is non compression with Cv input, adiabatic expansion, and isobaric 'cooling' (ambient exhaust). Yep, the same as Otto-Langden and even without Wneg of compression they suck compared to common compression cycles. Why ??? Fool supplied the 'magic' answer a while back - isothermal input is the most efficient, and processes are path dependent. Different paths between the same points (PV plots) do NOT have the same values.

If so concerned with real this and that, why worry about tinker-toys, I hadn't noticed they're real.
Seems I missed this post.

You seem to have, what to me anyway, is a strange way of looking at efficiency in a heat engine.

I think of it as heat converted into work.

Heat goes in, expands the gas and that pushes the piston. That is heat being converted into work regardless of what happens afterwards.

You have this abstract mathematical conception that this conversion of heat into work that was just accomplished can somehow be nullified by negative work in the opposite direction. Or if twice as much work is done in one direction that work nullifies itself.
The work of compression aka Wneg must come from somewhere whether it's from partial expansion energy stored in a flywheel or buffer/ambient pressure that was Wneg during expansion
Yes, that work "must come from somewhere".and it most certainly does. From the heat that was converted into work.

So what if that heat was converted during expansion? The contraction stroke does not literally erase the work that was already accomplished or stored as potential energy in an air spring or as displaced atmospheric pressure.

Yes I can agree that if, let's say I have five people that need to get to work at the top of a hill but I only have a two seater car.

To conserve on gas I drive one person up the hill at a time and let the car roll back down the hill in neutral with the engine shut off.

If I recon the "efficiency" of my car engine only by the distance traveled up the hill, (up the hill is Wpos and down the hill Wneg) well sure, my car ends up back where it started at the bottom of the hill after each trip. Not a terrifically efficient car engine there bub.

Regardless of the abstract mathematics used to calculate the efficiency, in REALITY, gasoline was still converted into work and this engines power output got five people up to the top of a steep hill and all the return trips back down the hill were with no additional fuel consumption.

Why am I "concerned with real this and that" ???

So now a model engine is not "real" only the mathematical abstractions in your mind or that you calculate on paper are "real".

I think most people are concerned with the real practical use to which an engine can be put, like getting up a hill to go to work.

Your Wneg/Wpos=efficiency does not REALLY negate the fact that heat was ACTUALLY converted to work and that very REAL work was accomplished in the process.
Tom Booth
Posts: 4727
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

Re: The Carnot efficiency problem

Post by Tom Booth »

Just recently I've been researching what might have been posted recently about the idea of using a heat pump to deliver heat to power a Stirling engine and I came across this interesting statement:
The system proposed is entirely workable and would provide "Free Energy" from the environment until the end of the World as we know it.
The Oil companies do not want you to know this and so post disinformation about free energy.
https://www.stirlingengine.com/forums/v ... 3870#p3031

Usually I just dismiss such statements with a chuckle, but lately I'm beginning to wonder.
Tom Booth
Posts: 4727
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

Re: The Carnot efficiency problem

Post by Tom Booth »

matt brown wrote: Mon Sep 04, 2023 6:12 pm
Tom Booth wrote: Mon Sep 04, 2023 4:41 pm Honestly, I find this explanation laughable Matt, no offense, but it makes no sense.

All you said, in regard to my actual question is:
"the reason why 0k would be required for 100% efficiency is simply that to compress any gas back to its original volume IN A CYCLE would require Wneg=0 aka zero internal energy to resist the work of compression."
That's an affirmation of an opinion not an explanation.

The assertion has no rational basis.

Maybe you could explain it another way because just saying "IN A CYCLE" proves nothing at all.

So why should a Stirling engine have to go to absolute zero to get the piston to return one cycle, when quite obviously and observably the piston returns to TDC by the same atmospheric pressure it had to work against to get to BDC in the real world all the time, quite easily.
In a cycle, everything must return to the start state, so any expansion would require a similar compression. The work of compression aka Wneg must come from somewhere whether it's from partial expansion energy stored in a flywheel or buffer/ambient pressure that was Wneg during expansion. Either way, this Wneg is proportional Wpos via PV=mRT where Wneg/Wpos=eff. The only way to have compression Wneg=0 is at 0k. Your static analogies are fine for lifting and dropping rocks, but don't cut it in thermo since there's kinetic energy at play (except at 0k).

You act as if LTDs are magical...who cares if some toy can run without a flywheeel or whatever. As I've said before, Senft didn't do us any favors with this 'educational' tinker-toy, he just ushered in endless distraction. I suspect that these LTD toys are nothing near a "Stirling", but approximate the three legged Lenoir cycle...yeah, no Wneg or physical heat sink req'd. They simply suck in a little air, mix it with some other air, semi-isolate during heating, then let this air slightly expand and leak a little to ambient, then repeat.
(...)
If so concerned with real this and that, why worry about tinker-toys, I hadn't noticed they're real.
The only way to have compression Wneg=0 is at 0k.
Do you understand the concept of a balance of pressure?

We live in an atmosphere with a pressure that can easily flatten a steel drum we don't implode because there is internal pressure in our cells pushing outward

A displacer moves freely in a pressurized Stirling engine, no matter what the pressure, because there is an equal balance of pressure on both sides, or top and bottom.

You have almost the same situation with the piston. There is pressure on both sides equally balanced until a little heat is added then the piston moves, does work and energy goes out and the piston returns. There is a SLIGHT rise and fall in pressure inside the engine that allows the piston to move.

So where does all this "absolute zero" is necessary for compression come from? It's crap. Complete delusional nonsense.
Senft didn't do us any favors with this 'educational' tinker-toy, he just ushered in endless distraction. I suspect that these LTD toys are nothing near a "Stirling",
To be honest Matt, I don't know why you spend so much time on a STIRLING and HOT AIR ENGINE forum doing little else other than running down Stirling engines and trying to point out how impossibly inefficient they are and how great internal combustion (Otto) engines are.

Senft, IMO showed that temperature difference is not a limitation. If you can provide more surface area you can gat more joules across the heat exchanger, so a Low Temperature Difference Stirling engine can produce just as much power as a high temperature difference Stirling engine.

That the ∆T has any bearing on efficiency is hogwash. What matters is the actual joules entering into a given volume of working fluid.

You can have a long skinny piston and cylinder heated at the end to a high temperature, or you can have a wide flat cylinder with much more surface area at a much lower temperature and still have the same quantity of actual heat in joules crossing the heat exchanger and expending the working fluid.

That, to me, was a big leap forward, and proof in itself that the "Carnot limit" based on the temperature spread is pure hooey.

I honestly sometimes find it hard to believe that you actually truly believe the things that you say.

But then again, the same preposterous "absolute zero" poppycock is everywhere. It's textbook.

Completely ridiculous just the same. I can only attribute that to a nearly complete lack of common sense within academia.
Tom Booth
Posts: 4727
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

Re: The Carnot efficiency problem

Post by Tom Booth »

This engine is not LTD.

A propane torch burns at around 2000°F


https://youtu.be/LG09AXAjpio?si=wlkdiYp3Z1TsKErn


Where is this "absolute zero"?

Where is there even any cooling? The whole thing is non-heat conducting and covered with insulation, or actually being heated to 2000°F with a torch.

The piston returns about 20 times a second (I think probably more). No absolute zero in sight.

The piston will return to TDC even with a 30 gram weight hanging off it.


https://youtu.be/e-7DFp_B0y4?si=D8XRaiIAFK8iEhaT


The thing is being heated to 2000°F and you say it can't do what it is observably doing without being cooled to absolute zero.
Tom Booth
Posts: 4727
Joined: Tue Nov 07, 2006 2:03 am
Location: Fort Plain New York USA
Contact:

Re: The Carnot efficiency problem

Post by Tom Booth »

Tom Booth wrote: Tue Sep 05, 2023 4:04 am (...)

So how exactly can it be shown that my 10,000 joules of heat input equals 370°K ?

And my zero waste heat after doing 10,000 joules of work equals 270°K ?

The whole proposition doesn't even make sense.
This actually DOES make perfect sense, it is just the current academic interpretation that is nonsensical.

If I have a ∆T of 370° F / 270° F and add some arbitrary number of joules at 370° F and the engine really is 100% efficient so all of the joules are "consumed" or converted into work then the cold side temperature will remain unchanged so no heat whatsoever is "rejected".

There will be no heat transfer between the 270°F cold side heat exchanger or cold plate and the 270°F ambient surroundings.

It makes sense that with literal 100% efficiency this is actually the case.

But somewhere along the lines down through history this common sense result alarmed somebody, because if you flip the engine upside down and run it on ice and have no heat rejection to the sink (ice) then you could have perpetual motion, so, that would mean Tesla was right!

Look guys, we need to fix this, we can't take the chance of another Tesla coming along and figuring this out, so get busy and rewrite all the textbooks.

What if we take the real Carnot efficiency ratio and reapply it to the input heat itself?

Well we could do that, but then it would make efficiencies appear to be ridiculously low, I don't think anybody would actually fall for that.
Bumpkin
Posts: 282
Joined: Fri Mar 02, 2012 1:42 pm

Re: The Carnot efficiency problem

Post by Bumpkin »

A lot of folks gave us a great foundation, but I can’t believe any designer would be worse off for never having heard of Carnot or some others. For instance — Stirling didn’t need to think of heat as molecular motion instead of “caloric” to understand its affects. He didn’t need Boyle’s law to understand the relationship between absolute temperature and air volume/pressure. He didn’t need to know that air isn’t an “ideal” gas. (It’s close enough.) He didn’t need to know the first or second so-called laws. All he needed was an open mind and respect for cause and affect. Lack of that respect is, of course why we need laws, but if you have enough respect you can tactfully wait until nobody’s watching and do what you want. :big smile:

Bumpkin
Post Reply