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

Discussion on Stirling or "hot air" engines (all types)
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

Tom, I'd say you are proceeding towards an ethical dilemma here. You can dig into learning mathematics and thermodynamics and engineering, or continue misusing logic and science by staying ill informed. The ill informed route is more entertaining for most of the public so pays more. Look at Bill Nye, and he is more informed than most. I never followed Nye much.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Thu Apr 18, 2024 12:39 pm
Tom Booth wrote:Anything?

I'd love to see it.

So far, in ten years I've seen nothing.

Please! Please! Please...

Show me the "reverification".
You've seen plenty. ...
As usual, empty handed. Nothing but blah blah blah...

If you had any concrete evidence of the validation or "revalidation" of the Carnot efficiency limit equation you could easily produce some reference here and now for everybody to see and prove me wrong.
What you should be looking for is why your data is wrong
I think what you should be looking at is why your repackaged renamed 200 year old caloric theory is wrong. Call it "Kinematic Theory" or whatever you like it's still just the same old obsolete water wheel nonsense.
"Calorics are just carriers. They carry different quantities of heat as specified by the temperature. They go in the top at Th and come out the bottom at Tc. In the example I keep going back to. 400K and 300K, they have only 80% of the energy left. Same as water going through a waterwheel, height in 100' height out 80'. For 80% of potential left to run down hill. 20% comes out as work.
Your simply saying so and claiming there is "verification" doesn't make it true.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

VincentG wrote: Thu Apr 18, 2024 9:59 am Maybe it would be helpful to draw a distinction between passing "heat" though an engine and passing "energy" through an engine. And yes I know there's no such thing as heat. What I mean is a distinction between passing total temperature through v. passing total energy through?
What appears to be the case VincentG is unlike water flowing in a river channel downhill by the EXTERNAL force of gravity, or at least some gravitational "mutual attraction" that causes water to "flow' in a specific direction. Supplying heat to a gas in an engine simply causes expansion of the gas. Like blowing up a balloon. There is no "flow" in any particular direction. No "gravity", or anything like gravity. No "channel". No river. Just a stationary expanding balloon or expanding gas.

It is well known, that when a gas expands doing work such as driving a piston it will cool and contract.

A Stirling engine appears to be a method for causing a gas to oscillate, alternatively heated/expanded then outputting "work" as mechanical motion and contracting.

For such an oscillating cycle, the expansion needs to be rather rapid. If too slow the gas will simply absorb additional heat from the surroundings rather than cooling and contracting after outputting energy.

To initiate an oscillation a ∆T appears to be necessary so the gas has a starting low energy state to expand from and return to, but there is no "flow" of heat through the engine traveling towards some "cold reservoir".

Those are my conclusions, which anyone is welcome to take or leave.

That's based on several years of my own observations and experiments.

I have nothing to prove and no ax to grind, just drawing my own conclusions from observation and experiment without a lot of preconceived expectations and with nothing to loose or gain

I'm not saying these engines can't be improved, far from it, but I think to make any real improvements it is first necessary to understand how they actually work, or rely on chance, or trial and error, but it helps to have some inkling of an idea what's actually going on.

All I do know is the Carnot water wheel "flow through" theory is quite obviously flawed. It does not hold up to experiment.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Yes, "energy" in a sense, goes "through" the engine 100%

It goes in as "heat".

"Heat" is, of course, a term applied to a human perception. The "heat" going in is a form of energy. The heat is taken in by the gas causing it to expand and put pressure on the piston. When the piston moves, the energy in the expanding gas is transfered to the piston.

That is, on a molecular level, like a pool ball, or many pool balls striking a much larger bowling ball. The pool balls (gas molecules) loose energy and the bowling ball (piston) gains energy.

The only "flow" or transfer of energy, ideally, is between the expanding gas and the piston.

If the engine body is, say, epoxy, which is non heat conducting, then it will absorb less of the heat from the gas and more will be available to transfer to the piston. In this way, heat "flows" or enters into the gas and "work" or mechanical motion is produced.

Mechanical motion of the engine is, of course, NOT HEAT, or no longer "sensible" heat. The energy that went in as heat goes out as mechanical motion.

So, sure, in a sense, or in a manner of speaking energy "flows" through the engine, but is transformed in the process. Goes in as heat, goes out as "work".

The "Carnot Limit" allegedly puts a cap on exactly how much "heat' expanding the gas can be transformed into work. A ratio based on the temperature difference, based on this notion that heat as a "fluid" "FALLS" down from hot to cold like a waterfall.

A compelling idea that is easily visualized and captures the imagination. It has captured people's imagination for 200 years. Be that as it may, its juvenile nonsense comparable to saying babies are delivered by the stork. It's a fairy tale.
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

Fool wrote: Mon Mar 25, 2024 9:13 am The following is some improvements to the original post, not a direct Quote:
Fool Re- wrote: Wed Mar 20, 2024 7:36 am Question: How does Qh an Qc become Q's all the way to Zero, and then, Th and Tc?

Answer:

Setting of parameters, mathematical definitions, and descriptions:
Let Th be the temperature of the hot plate/reservoir.400K
Let Tc be the temperature of the cold plate. 300 K

Let Qh be Delta Qh = DQh, the heat added per cycle. 100 J

Qc then becomes DQc the heat rejected per cycle. Minimize this to zero if possible. Determined by Derivation and Calculation once 'n' is known.

Let Qhz be the internal energy from heat added to the cycle all the way from zero K Kelvin to Th.

Let Qcz be the heat added from zero K all the way to Tc. That is true but confusing, so, Qcz is the internal energy of the engine's gas mass at Tc. It is the base energy. It neither used, nor rejected. It is the absolute energy level at Tc contained the the engine gas mass.

DQh is added to Qcz, to get to Qhz at Th. Qcz is the temperature determined energy content of the engine's gas mass at Tc.

Qhz will become the heat at Qcz plus the extra heat added, or Qcz + DQh to get to Th.

Mathematical definitions follow, also true:
Qcz = M•Cv•Tc
Qhz = M•Cv•Th = Qcz + DQh

Noted equation:
Qhz-Qcz = DQh = DT• Cv•M (Noted equation)

Where:
DT = Delta T = Th-Tc
M = Mass
Cv = specific heat.

Where M is the mass of the working gas contained in the engine, and Cv the heat capacity. Cv example units: Joules per Kelvin per Kilogram of gas contained in the engine, or J/(K•Kg). M is chosen, in collaboration with Cv, to have a one to one relationship between heat energy and Temperature. M•Cv=1. The derivation here will show that it cancels out later. So the actual numbers won't matter, "works for all".

Starting with the first law, "Energy is conserved", and the definition of efficiency 'n':

W = Work output from the working gas.
W = DQh - DQc
n = efficiency = W/DQh = (DQh - DQc) / DQh

Begining of derivation:

Taking the above 'noted' equation "Qhz = Qcz + DQh" and subtracting Qcz from both sides:
Qhz = Qcz + DQh (Noted equation)
Qhz-Qcz = Qcz-Qcz + DQh

Canceling and rearranging the terms gives:
DQh = Qhz -Qcz (DQh equation)

Multiplying both sides by n:
n•DQh = n•(Qhz-Qcz)

Dividing both sides by DQh:
n = n•(Qhz - Qcz) / DQh (#1)

The term 'n' applies to the engine regardless of 'Qcz' base heat amount or reference point. It will have the same efficiency burning the same energy at the same temperatures but with different reference points. That allows us to look at the total energy in the system. So Qhz should have the same relation ship to Qcz that DQh has to DQc, in other words, 'n' is the same. Basically this says that Qcz can be calculated from 'n' and Qhz,
or:
Qcz = Qhz•(1-n)

Substituting that equivalence for Qcz into the 'DQh' equation:
DQh = Qhz - Qcz (DQh equation)
Qcz = Qhz•(1-n)

Substituting and distributing:
DQh = Qhz - Qhz(1-n)
DQh = Qhz - Qhz + n•Qhz

Subtracting:
DQh = n•Qhz (#2)

Combining #1 and #2
n = n•(Qhz - Qcz) / DQh (#1)
DQh = n•Qhz (#2)

#2 into #1:
n = n•(Qhz - Qcz) / n•Qhz

The n's on the right side, of the equals sign, cancel becoming one, and rewriting:
n = (Qhz - Qcz) / Qhz (#3)

Equation #3 shows that DQh and DQc have now become Qhz and Qcz. In retrospect, it seems logical that the efficiency profile should be the same regardless of absolute scale. Substituting in Qhz and Qcz straight across for DQh and DQc makes sense, and would have been faster. That was the mathematical derivation/proof. Note also, that it tends to maximize efficiency by equating DQc with Zero. DQc is not zero, it is just the amount of inevitable heat rejection and with maximum work out.

Now using the equations above for Qhz and Qcz:
Qhz = M•Cv•Th (The equations above)
Qcz = M•Cv•Tc (The equations above)

Substituting the above two lines into equation #3:
n = (Qhz - Qcz) / Qhz (#3)
Here:
n = (M•Cv•Th - M•Cv•Tc) / (M•Cv•Th)

Rearranging and removing the distributed 'M•Cv':
n=M•Cv•(Th-Tc) / (M•Cv•Th)

Canceling 'M•Cv' top and bottom because they become one:

n=(Th-Tc)/Th <<<The final solution.

That shows the logical step by step progression for the mathematical derivation or proof.

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.

It's logical to think that higher temperature differences produce higher pressure differences, that make greater power to weight and size ratios, ease of construction, operation, and efficiency increases.

1/10 degree temperature, would be like trying to move a piston with 1/10 of a psi. How can that possibly be as efficient as trying to move a smaller piston with 100 psi.

It would take a piston 1000 times larger in area to develop the same force. Power and efficiency sucking bad bulk would be the down fall. I hope this makes sense in regard to the Carnot Theorem.
Hopefully that is a little easier to follow. Whomever said this was easy, or simple, has my complete disagreement. LOL

And it still probably has little errors. I'm hoping the little errors we've found are it. I don't think they spoil the overall proof, but a proof isn't a proof until all the errors are eliminated. Your help has been much appreciated.
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

The Carnot theorem is not a cap. It caps nothing. It was discovered that nature has a way of utilizing heat in a cycle to produce work, heating and cooling systems have been used. A cycle is described as a starting point, and a series of processes that eventually return back to the starting point. Carnot described a cycle running between two isotherms that has a maximum efficiency of :

n = (Th-Tc)/Th

No one has been able to beat it. Not even Tom Booth. You all are welcome to try. Please document it with verified power output measurements.from a reputable impartial measuring service.
matt brown
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Tom Booth wrote: Thu Apr 18, 2024 3:01 pm
It is well known, that when a gas expands doing work such as driving a piston it will cool and contract.
Way off..when a gas expands doing work such as driving a piston it MAY cool (depending upon process) and pressure MAY decline (depending upon process) but a gas never "contracts".
Tom Booth wrote: Thu Apr 18, 2024 3:01 pm A Stirling engine appears to be a method for causing a gas to oscillate, alternatively heated/expanded then outputting "work" as mechanical motion and contracting.
So, a Stirling engine is special and operates on magic vs common ICE. Heck, I've never heard anyone refer to piston ICE as "oscillating" (only reciprocating).
Tom Booth wrote: Thu Apr 18, 2024 3:01 pm For such an oscillating cycle, the expansion needs to be rather rapid. If too slow the gas will simply absorb additional heat from the surroundings rather than cooling and contracting after outputting energy.

To initiate an oscillation a ∆T appears to be necessary so the gas has a starting low energy state to expand from and return to, but there is no "flow" of heat through the engine traveling towards some "cold reservoir".
Hmmm, it appears you think your LTD is similar a gas spring with heat added prior expansion, and when all this heat turns into work, no heat sink is required. But wait there's more...only fast adiabatic expansion insures pressure drop below ambient, whereby ambient compression is possible.
Tom Booth wrote: Thu Apr 18, 2024 3:01 pm I'm not saying these engines can't be improved, far from it, but I think to make any real improvements it is first necessary to understand how they actually work, or rely on chance, or trial and error, but it helps to have some inkling of an idea what's actually going on.
Some of us have more than an inkling of what's going on. The giveaway is "ambient compression" aka vacuum engine vs atmospheric engine. Any cycle with ambient compression will have severely restricted PVT values that require years of PV analysis to grasp.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Fool wrote: Thu Apr 18, 2024 8:10 pm The Carnot theorem is not a cap. It caps nothing. It was discovered that nature has a way of utilizing heat in a cycle to produce work, heating and cooling systems have been used. A cycle is described as a starting point, and a series of processes that eventually return back to the starting point. Carnot described a cycle running between two isotherms that has a maximum efficiency of :

n = (Th-Tc)/Th

No one has been able to beat it. Not even Tom Booth. You all are welcome to try. Please document it with verified power output measurements.from a reputable impartial measuring service.
The Carnot theorem is far more extensive than most realize. Although intended as a single-phase gas limit, many know that it extends to multi-phase cycles, but this is usually considered mere chance. Well Fool, here's one for you...did you ever notice that Carnot wins with cascading cycles ? Yep, a mercury-water-alcohol-freon "Rankine" cascade has the same Carnot eff thru cascade T range as single system thru same T range. But, if similar cascade could sink to liquid fuel that supplies initial cascade source, then we might win.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Tom Booth wrote: Thu Apr 18, 2024 6:45 pm
The "Carnot Limit" allegedly puts a cap on exactly how much "heat' expanding the gas can be transformed into work. A ratio based on the temperature difference, based on this notion that heat as a "fluid" "FALLS" down from hot to cold like a waterfall.
The Carnot theorem is only EXPRESSED via T for simplicity. If you get off this T bandwagon and onto the W bandwagon, you'll see just how lame this Carnot BS is.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

This again? fool?

First off, it looks like the same old merry-go-round
Let Qhz be the internal energy from heat added to the cycle all the way from zero K Kelvin to Th.
Can you unpack that sentence please?

How is "internal energy" from "heat added"?

You said the heat added (DQc, delta Qc, or Qc, why so many designations for the same thing/quantity anyway) is 100 joules per cycle.

Now you appear to be drawing an equivalence between the 100 joules "added" and "heat added to the cycle all the way from zero Kelvin to Th." And attach yet another designation to represent this same 100 joules.

Your definition of this Qhz makes no sense. It could mean anything, or nothing. What concept or idea of "heat" issues forth such a string of words as these:


"heat added to the cycle all the way from zero Kelvin to Th."


Are we supposed to start anew at absolute zero and start "adding" heat each and every cycle? How, when ambient is 300°K?

How is this heat represented by Qhz "added".

Do we begin each engine cycle at absolute zero and start adding heat? Or do we begin each cycle at 300°K ?

Qc is 300°k. Presumably that is the ambient temperature. Each cycle then would start and end at 300°K not absolute zero

No heat is ever "added" starting "all the way from zero Kelvin".

Each cycle begins and ends at 300°K

You seem to be suggesting that adding 100 joules to bring the temperature from 300°K to 400° K is somehow REALLY adding 400 joules "all the way" from absolute zero to 400°K (assuming 1° rise in temperature requires the addition of 1 joule of heat to the working fluid inside the engine.)

If that is what you mean, then you need to clarify how 100 joules added each cycle really amounts to 400 joules "added".

You do realize to "add" some specific quantity of heat it must have not already have been there before being "added'.

You can't have a substance at 300° K and then "add" the same "internal energy" that is already there by adding some other 100 joules of energy, to total 400 joules and then claim you "added" all 400 joules by adding the 100 joules.

Do you imagine each Joule is itself a column of heat that extends all the way down to absolute zero?

Well, in that case you have still only "added" the additional 100 columns of heat to the 300 columns of heat you did not need to add because they were already there.

So, 100 joules were added. Period.

So Qhz = DQh = delta Qh = 100 joules "all the way to 0°K" or not, the heat "added" is 100 joules.

300 joules is the "internal energy" at the start of each cycle and 400 joules is the "internal energy" after adding 100 additional Columbus of joules that somehow extend from absolute zero to 400°K

If that does not represent your intended meaning please explain. Preferably in a way that makes sense in this universe.

Generally in this universe, adding 100 joules is not the equivalent of adding 400 joules.
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

All that you just posted is hog wash.

Qcz is the internal energy of the gas at Tc. No mystery about it. It is the exact same energy that would be added if starting at zero Kelvin and adding heat until you get to Tc. It is also the same amount of energy that would be rejected if the gas were cooled to zero Kelvin. It is the free internal energy contained in the engine obtained from the atmosphere at the starting point. The engine isn't starting at zero internal energy. It's starting at Qcz. All those are the same value and definition. No mystery about them.

Qhz = Qcz + DQh or 400=300+100 or Qhz = M•Cv•Th = M•Cv•Tc + DQh

All those are the same. No mystery about them.

Everything has an accurate and logically solid mathematical definition.

It follows the law of superposition.

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

Post by Tom Booth »

Fool wrote: Thu Apr 18, 2024 9:41 pm All that you just posted is hog wash.
No, the way you phrase your statements is at best ambiguous and needs clarification.

Please try to make sense. Definitions of mathematical variables require precise definitions not slippery, meaningless nonsensical definitions that could mean anything and later on morph and change into even more confused meaningless nonsense.

Don't blame me if you can't formulate a sentence that makes sense.

Let Qhz be the internal energy from heat added to the cycle all the way from zero K Kelvin to Th.

Makes no sense.
Qcz is the internal energy of the gas at Tc. No mystery about it. It is the exact same energy that would be added if starting at zero Kelvin and adding heat until you get to Tc. It is also the same amount of energy that would be rejected if the gas were cooled to zero Kelvin.
Well, that is not the same thing.

"Internal energy from heat added" is not the same thing as "energy that WOULD be added IF starting at zero Kelvin"

We, though, are NOT starting at zero K we are starting at 300 k. So why introduce values that are not at all applicable to the situation?

Of course, later you will use this intentional ambiguity to try and prove the Qh Qc = Th Tc supposed equivalence of the Carnot limit nonsense, we've been down this road before.
It is the free internal energy contained in the engine obtained from the atmosphere at the starting point.
Again you are implying this Qcz is "added" or now "obtained" at "the starting point".

At what starting point? Your original definition was "added to the cycle".

Cycles repeat.

Nothing is "obtained". Nothing is "added". So why use such misleading terminology?
The engine isn't starting at zero internal energy. It's starting at Qcz.
Great.

So why do you continually use these phrases "added' and "obtained" about a value that is not added or obtained?

Because later you will try and claim that this Qcz is "rejected" which of course is not true, but MUST be true for the Carnot Limit to be true, so you introduce the ambiguity early.
All those are the same value and definition. No mystery about them.

Qhz = Qcz + DQh or 400=300+100 or Qhz = M•Cv•Th = M•Cv•Tc + DQh

All those are the same. No mystery about them.

Everything has an accurate and logically solid mathematical definition.

It follows the law of superposition.

Matt yes. Thanks.
LOL.

Superposition it is now. LOL.

Like Schrodinger's Cat?

So your Qcz represents heat that is not inside the engine nor outside the engine but both? Or neither, until you call it into existence. The magical Qcz strikes again.

So quantum mechanics it is now?

Will you also be utilizing quantum teleportation? The many worlds hypothesis?

Qcz, I suppose exists as only a probability wave so Qcz does not actually exist inside or outside the engine until someone looks, then it is either or neither of both, as necessary to have the Carnot Limit formula work out.

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

Post by Tom Booth »

Fool wrote: Thu Apr 18, 2024 7:48 pm ...
Qc then becomes DQc the heat rejected per cycle. Minimize this to zero if possible. Determined by Derivation and Calculation once 'n' is known.
...
To be clear, your 'n' I suppose represents η, Greek letter eta, The symbol for Carnot efficiency ?

Do we not already know the Carnot efficiency ?

Tc = 300, Th = 400, η = 25%

Of 100 joules 25 Joules can (at best, according to the Carnot limit) be converted to "work".

So why play ignorant?

With 100 joules input, we already know Qc/DQc will be 75 joules

Qc or DQc is 75 joules

Yes? No?
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Tom Booth wrote: Thu Apr 18, 2024 10:57 pm
No, the way you phrase your statements is at best ambiguous and needs clarification.

"Internal energy from heat added" is not the same thing as "energy that WOULD be added IF starting at zero Kelvin"

Great.

So why do you continually use these phrases "added' and "obtained" about a value that is not added or obtained?
Fool, Tom is right #1, you're "switching it up" as the kids say.
Tom Booth wrote: Thu Apr 18, 2024 10:57 pm Because later you will try and claim that this Qcz is "rejected" which of course is not true, but MUST be true for the Carnot Limit to be true, so you introduce the ambiguity early.
Tom is right #2, but now I see what's going on here, Fool is working this out as he goes where writing forces one to think.

Tom - he's trying to prove Carnot from a novel approach (a work in progress)

Fool - indeed unique, don't stop, maybe someone beyond this forum will notice. I only got this after my recent zero-zero comment, a buzz I haven't used in years. If I have your pitch correct, summing from zero appears possible for a Stirling cycle with isotherms, but remains challenging for adiabatic cycles. Kinda reminds me of the buffer pressure issue where

shaft work out during expansion = Wpos - Wbuffer and shaft work in during compression = Wneg - Wbuffer

I'll be thinking about your approach myself...
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Well, fool, after going through your derivation I find no particular MATHEMATICAL flaws.

If you have a temperature difference at the hot plate of 400°K for example, and a cold plate at 300°K you have a ratio on the absolute temperature scale.

In this case 25%

300°K is 75% of 400°K.

It's undeniable the ratio exists. You can carry out all kinds of mathematical transformations and permutations.

Of the total 400 joules only 25% are "available".

That is, 100 joules supplied is 25% of 400 joules.

300 joules are "unavailable"

You flip and toss these values around over and over and derive the same ratio over and over in a variety of permutations and, well. The math itself seems fine.

I don't detect any mistakes.

But I don't agree that this temperature ratio, however you scramble it up or rewrite it represents "efficiency"

It's still just the temperature difference.

You cleverly made the ∆T equivalent to joules supplied, 100 degrees between 300 and 400. 100 joules added per cycle.

What you have done in essence is to put an "η" in front of a temperature ratio and then rearranged the equation.

The heat supplied is 100 joules which is 25% of 400 joules.

Yes, on an absolute scale only 25% of the total 400 joules is available to be converted to work. ,(i.e. the 100 joules actually supplied).

Because you've put that in front of an equal sign with a η efficiency symbol on the other side doesn't justify making the conclusion that this now proves only 25% of the 100 joules actually supplied, or 25 joules can be converted to work.

You've simply assigned an efficiency symbol to a temperature ratio in accordance with the Carnot theory.

That assignment is based on the Carnot theory. If the Carnot theory is valid the assignment is valid. If the Carnot theory is invalid the assignment is invalid.

It's basically like saying that because my gas tank is only 1/4 full the efficiency of my car engine is only 25%

Well, in a sense, that is true. The engine can only utilize 25% of the fuel tanks full capacity.

That however has no real bearing on actual engine efficiency which is mpg.
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