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

[Tom Booth]

Experimentally, the heat "rejected" at the door so to speak, or rejected or perhaps reflected away on the hot side, so as to not ever find admission into the engine, makes more sense than the "excess" heat is "rejected" by being sent through the engine to the cold side.

In these experiments running LTD engines on ice for example:

https://youtu.be/-7zntz8kwIk?si=6bWhJ8X9fMdRKzoq


Compared with a "control", the ice under a running engine consistently took longer to melt than an identical cup of ice just sitting,wrapped in insulation, under the same engine NOT running.

Logically, to my mind anyway, if heat was being "rejected" by passing through the engine and into the ice, the ice would likely melt more quickly, given that the running engine is actively circulating the working fluid between the hot and cold sides transporting heat to be "rejected" into the ice in the process.

If, on the other hand the heat is being "rejected" in the sense of never being admitted into the engine in the first place, then the engine is actively blocking or preventing heat from getting through to the ice. In addition, heat that is admitted, limited by the heat capacity of the working fluid, is converted to mechanical "work" which, though producing some friction in the power cylinder and flywheel and bearings and such, is all occuring above the working fluid, up above the engine body in the ambient air.

Naturally, with all this heat being "rejected" and/or re-emerging on the HOT side, the cold side tends to remain cold longer

Whatever the case, the fact of the matter is, the ice took significantly longer to melt when used to run the engine, time after time. Sometimes using solid blocks of ice, sometimes using ice cubes floating in ice water.

In that particular experiment in the above video the engine ran for 33 hours on a cup of ice before the ice completely melted. With the engine non-operational as a "control" the ice melted completely in just 28 hours.

[Tom Booth]

....heat "rejected" ....so as to not ever find admission into the engine, makes more sense than the "excess" heat is "rejected" by being sent through the engine to the cold side....
...

Thinking about this a bit more, if heat "rejection" from the engine, in terms of the efficiency limit, has anything to do with the heat capacity, how could it be any other way.

"Capacity", by definition, really, means a limitation on the quantity that could be admitted, or carried or contained.

A bus for example, let's say, has a capacity of 35 passengers.

That limited capacity doesn't mean the bus can let 100 people on, carry 35 to the destination and dump the other 65 off as excess passengers someplace beyond the destination.

The examples, of course could be multiplied almost endlessly, buses, planes, ships, passengers vehicles, cargo bays, fuel tanks, etc. etc.

the maximum amount or number that can be received or contained

https://www.dictionary.com/browse/capacity

So, if you have, say, boiling water in a "perfect" Dewar and set an engine on top with a working fluid with a heat capacity of "C", and the engine body is otherwise perfectly non-heat conducting, heat below the engine inside the thermos bottle will just have to wait it's turn.

The working fluid could not carry excess heat to be carried through and deposited on the other side.

Suppose you have some loiterers on the bus that take up the capacity but never get on or off.

Well, again, by the modern definition, they would not constitute "heat" which is only that quantity of energy actually received or transfered.

So, the "efficiency" per trip or cycle could be limited by any "latent heat" or the existing ambient in thermal equilibrium everywhere, but the "efficiency" per Dewar full of "heat" would still be 100%. Although the effective carrying capacity is diminished due to the vagrants occupying seats on the bus or train, the passengers waiting at the station or bus stop for an empty seat are never "wasted". They aren't ever allowed on until an empty seat becomes available.

Seats become available when some heat/energy leaves as "work". There are no additional seats, or there is no additional capacity for carrying "waste heat".

"Carnot efficiency" as usually interpreted counts the vagrants living on the bus as "waste heat" being taken on, carried and then dumped. But this ambient energy, in general thermal equilibrium doesn't get on or off. It is not "heat". It just cancels out.

Interestingly, the heat capacities of Helium and Hydrogen, even unpressurized are respectively 5x and about15x greater than air.

https://www.engineeringtoolbox.com/amp/ ... d_159.html

[Tom Booth]

Honestly, watching some of these professors expounding upon Carnot efficiency and the second law reminds me of the performance of a stage magician trying to wow the audience by pulling a Rabbit out of a hat.

Watch this folks, I just put 800 joules INTO the hat, now to get that 800 joules back out to transform it into work, I need to remove FOUR THOUSAND Joules to bring the temperature all the way down to ABSOLUTE ZERO!!!

Isn't that AMAZING!!!!!

[Fool]

Question: How does Qh an Qc become Q's all the way to Zero, and then, Th and Tc?

Answer:

Let Th be the temperature of the hot plate/reservoir.
Let Tc be the temperature of the cold plate.

Let Qh be Delta Qh = DQh, the same definition and value, the heat added per cycle.

Qc then becomes DQc the heat rejected per cycle. Minimize this to zero if possible, for 100%.

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

The heat rejected from zero K all the way to Tc will be Qcz

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

Qhz = M•Cv•Th = Qcz + DQh
Qcz = M•Cv•Tc

Note: Qhz-Qcz = DQh-DQc

Where M is the mass of the working gas, and Cv the heat capacity. Cv example units: Joules per Kelvin per Kilogram, or J/(K•Kg)

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

Taking the above equation Qhz = Qcz + DQh and subtracting Qcz from both sides and rearranging to get:
DQh = Qhz -Qcz
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 reference point. It will have the same efficiency burning the same energy at the same temperatures but different units. That allows us to look at the total energy in the system, Delta Qh added to ambient. 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
Qcz = Qhz•(1-n)

Substituting that into the above equation for Qcz:
Qhz = Qcz + DQh
Substituting and distributing:
Qhz = Qhz(1-n) + DQh
Qhz = Qhz - n•Qhz + DQh
Subtracting Qhz - n•Qhz from both sides:
Qhz - (Qhz - n•Qhz) = DQh
Subtracting:
n•Qhz = DQh
Rearranging:

DQh = n•Qhz #2

Combining #1 and #2

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

n = n•(Qhz - Qcz) / n•Qhz
The n's on the right side cancel becoming one, 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 makes sense, and would have been faster. Note also, that it tends to maximize efficiency by equating DQc with Zero. It's not zero is just the amount left over after inevitable heat rejection and maximum work.

Now the equations above for Qhz and Qcz:

Qhz = M•Cv•Th
Qcz = M•Cv•Tc
Substituting into equation #3:

n = (Qhz - Qcz) / Qhz #3

n = (M•Cv•Th - M•Cv•Tc) / (M•Cv•Th)
Rearranging 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 progression of the mathematical proof.

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 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 is enough said about the Carnot Theorem.

[Fool]

Watch this folks, I just put 800 joules INTO the hat, now to get that 800 joules back out to transform it into work, I need to remove FOUR THOUSAND Joules to bring the temperature all the way down to ABSOLUTE ZERO!!!

That's not how it works. If you start at zero Kelvin, and add 800 Joules, you can get all 800 out as work, because rejecting zero heat at Zero Kelvin and zero pressure as the cycle returns back to the starting point uses zero energy.

A heat pump operating at that temperature, Tc=Zero K, will have a COP no better than one. And probably worse.

If you start at 300 Kelvin, you will only be able to convert a fraction of the 800 Joules to work. The rest, for 100% energy tally, will be rejected at the expense of work to the internal gas by compression during the return stroke against the pressure of the return stroke caused by the gas being at 300 Kelvin instead of zero, and as heat to cool that compression.

You will get more energy out than your 800 Joules, if you start at 300K and cheat by cooling below your Tc with some magical Tcolder than Tc imaginary cold hole. It will take more energy, than you save in work, to make a cold hole. The Carnot theorem and Entropy were in their infancy and not fully accepted until about the 1940's. Do not fault those working before this date, and do not consider their contrary words to be valid. They couldn't have known.

[Tom Booth]

Question: How does Qh an Qc become Q's all the way to Zero, and then, Th and Tc?

Answer:

....

About twenty paragraphs of equations later

I hope this is enough said about the Carnot Theorem.

Sorry, but I'll have to wait until I have a few hours to go through that whole tangled mess of mathematical substitutions and all to even figure out what you're doing nevermind formulate a response.

At this point seems like obfuscation to me. To complicated to argue with.

In the mean time, here is a simple textbook example: (among many to be found online).

Question
A Carnot engine absorbs 1000 J of heat from a reservoir at 127C and rejects 600 J of heat during each cycle.
Calculate

(a) The efficiency of the engine
(b) The temperature of the sink
(c) The amount of the useful work done during each cycle.

https://www.toppr.com/ask/question/a-ca ... d-rejects/

Very simple, uncomplicated and straightforward, as are all such example problems. So why does your dissertation on the subject require 20 paragraphs?

Anyway, I'll go over it with a fine tooth comb and a magnifying glass when I have some time to kill.

[Tom Booth]

Watch this folks, I just put 800 joules INTO the hat, now to get that 800 joules back out to transform it into work, I need to remove FOUR THOUSAND Joules to bring the temperature all the way down to ABSOLUTE ZERO!!!

That's not how it works. If you start at zero Kelvin, and add 800 Joules, you can get all 800 out as work, because rejecting zero heat at Zero Kelvin and zero pressure as the cycle returns back to the starting point uses zero energy.

A heat pump operating at that temperature, Tc=Zero K, will have a COP no better than one. And probably worse.

If you start at 300 Kelvin, you will only be able to convert a fraction of the 800 Joules to work. The rest, for 100% energy tally, will be rejected ...

Well, I was making up numbers off the top of my head, but...

Take a real-ish world example

Suppose I have an "Ultra" LTD that runs on a 0.5° K ∆T

Starting at 300°K ambient, I heat water with about 1000 Joules bringing it up to 301°K

(It takes 989.89 joules to heat 1 cup of water 1°K, say 1000 for simplicity sake)

For 100% efficiency or, in other words, to use ALL of that 1000 (or 989.89) joules, how many joules would have to be "rejected"? (To result in a drop in temperature to 0°K)

You say "The rest, for 100% energy tally, will be rejected ..."

The rest of the 100% energy tally...

That would be ALL the heat in the cup of water? Presumably?

So 989.89 joules per degree, from 300°K down to 0°K or 300 x 989.89 = ?

Well, roughly 300,000 joules?

IMO the whole idea you need to go to ABSOLUTE ZERO to use however many joules you actually supply is completely ridiculous.

In any reasonable way of talking about efficiency, IMO, if you could get ALL the heat actually supplied to go into the engine for complete utilization at 100% efficiency that would bring the temperature of the water down 1° to where it started at 300°K

You would have used ALL 989.89 joules not 296,964 joules to bring the water down to 0°K

[Tom Booth]

Watch this folks, I just put 800 joules INTO the hat, now to get that 800 joules back out to transform it into work, I need to remove FOUR THOUSAND Joules to bring the temperature all the way down to ABSOLUTE ZERO!!!

That's not how it works. If you start at zero Kelvin, and add 800 Joules, you can get all 800 out as work, because rejecting zero heat at Zero Kelvin ...

This is another rather irrational aspect of the so-called Carnot theory.

At zero K on the cold side there could, theoretically, be 100% efficiency regardless of the temperature of the hot side. So NO HEAT WHATSOEVER would be left to "FLOW" to the sink. It would ALL be converted to work. Supposedly

Setting aside the fact that no known gas can exist as a gas at zero K, if a heat engine can theoretically be 100% efficient with NO heat "rejection", NO FLOW out to the "cold reservoir" then according to this theory, no "sink" is actually necessary.

Originally, heat as a fluid or "caloric" would ALL flow to the "cold reservoir" even at 0°

There was no concept of heat being converted to some other form of energy.

If a "flow" through the engine, between hot and cold is necessary than how could a 100% efficient engine run at all while, as you say: "rejecting zero heat at Zero Kelvin..."?

As I said, modern thermodynamics is a patchwork. It has no inner consistency. It's a hodge podge of conflicting theories and models developed and revised over and over. It has lost touch with reality, having no empirical basis grounded in actual experiment.

[Goofy]

Tom, I really admire your patience with this "Fool".

But like Don Quixote, it is . . . .

In my (practical) world, efficiency is the relation between what you put in, versus what you get out.

Please, read again : What YOU put in !

So if the input is 1 Kelvin, where does "the rest" of the heat comes from, Fool ?

Perhaps the only practical thing you do, is to draw your chair out before sitting down in front of your desk ?

Allow me to quote Nikola Tesla :

“Today’s scientists have substituted mathematics for experiments, and they wander off through equation after equation, and eventually build a structure which has no relation to reality. ”

BR
Petter

[Tom Booth]

Tom, I really admire your patience with this "Fool".

But like Don Quixote, it is . . . .

In my (practical) world, efficiency is the relation between what you put in, versus what you get out.

Please, read again : What YOU put in !

So if the input is 1 Kelvin, where does "the rest" of the heat comes from, Fool ?

Perhaps the only practical thing you do, is to draw your chair out before sitting down in front of your desk ?

Allow me to quote Nikola Tesla :

“Today’s scientists have substituted mathematics for experiments, and they wander off through equation after equation, and eventually build a structure which has no relation to reality. ”

BR
Petter

Thank, but assuming you and I are in agreement, by all appearances, we're in the minority.

I'm grateful for "fools" patience as well in trying to bridge the gap. His point of view is apparently shared by the entire body of academia and the science community at large.

I also don't necessarily assume I'm in the right. I assume it must be me. Some lack of understanding on my part, because how could the accumulated wisdom of 200 years of scientific thinking be wrong and some uneducated grease monkey, such as myself, have a better grasp on reality than the entire staff in the science department of Harvard, MIT, CalTech, NASA, etc. etc. combined?

So I humbly sit at the feet of this representative of those more well informed and better educated than myself, hoping for some glimpse of truth, or on the other hand, perhaps it will finally be acknowledged that I'm right, but I don't really expect that to ever happen.

[Tom Booth]

Question: How does Qh an Qc become Q's all the way to Zero, and then, Th and Tc?

Answer:

Let Th be the temperature of the hot plate/reservoir.
Let Tc be the temperature of the cold plate.

Let Qh be Delta Qh = DQh, the same definition and value, the heat added per cycle.

Qc then becomes DQc the heat rejected per cycle. Minimize this to zero if possible, for 100%.

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

The heat rejected from zero K all the way to Tc will be Qcz

....

I'm trying to navigate this, and got this far. These statements, however:

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

The heat rejected from zero K all the way to Tc will be Qcz

Present some conceptual difficulties

"Heat added to the cycle..."

And

"All the way from zero K Kelvin to Th..."

Only makes sense if Tc is 0K to begin with, which is generally recognized as an impossibility, so not a valid real world example, so why is this even up for discussion?

The "internal energy from heat added to the cycle" does, or SHOULD in my opinion, NOT include the "internal energy" already present before actual "heat" was added.

This goes back to "what is your definition of heat" asked at the start of the thread:

viewtopic.php?p=21059#p21059

You agreed, I think, to the definition I cited:

Heat:

energy that is transferred from one body to another as the result of a difference in temperature.

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

Makes sense how?

The internal energy, "all the way from zero Kelvin to Th" was NOT all "ADDED" at all!

Some few joules are supplied each cycle to bring the temperature of the working fluid up from Tc which is generally, in most cases around 300°K

Everything below 300°K down to 0°K was "internal energy" already present. Not "added".

Internal energy that was not added is not "heat" by our agreed upon definition:


"energy that is transferred from one body to another as the result of a difference in temperature"

Not one joule of "internal energy" below ambient, Tc, (or generally, around 300°K) was transfered during any cycle.

All the "heat" below Tc was already ubiquitous, already present, not added, not transfered or supplied at all, except perhaps by the Sun eons ago to warm the surface of the earth before our engine or any of us even existed.

We have the same conceptual issue with the next sentence:

The heat rejected from zero K all the way to Tc will be Qcz

In what sense could this "heat" which was not supplied, and so does not even meet our definition of heat be "rejected" by the engine?

The "heat" from 0°K UP TO Tc is where we generally begin BEFORE heat is supplied to raise the temperature to Th. Correct?

This is the crux of my issue or complaint.

According to this Carnot Limit rationale, starting at 300°k and adding or supplying 1000 Joules to bring the EXISTING temperature of the working fluid up a few degrees, suddenly we have:

Qh or the actual HEAT by our accepted definition, being the equivalent of "the internal energy from heat added (sic) to the cycle all the way from zero K Kelvin to Th"

I'd like to emphasize that this illogic is, incredibly, to me anyway, accepted and taught throughout all of academia and in all courses on thermodynamics. Not some nonsense dreamt up by "fool", all by himself. How could this silliness be perpetuated without anyone besides me, and maybe "goofy" noticing how silly it actually is?

You've added a "z" to produce:

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

Not sure how adding a "z" to Qh fixes anything at this point, but I shall press on and hopefully find some logical conclusion in all this.

So far it just looks like a repackaging of the same old snake oil.

[Tom Booth]

Qhz will become the heat added to get to Qcz plus the extra heat added or DQh to get to Th

Same issue, though this seems to be making less and less sense as things progress.

How do you figure heat being "added" in order to "get to" Qcz.

Qcz you defined as "The heat rejected from zero K all the way to Tc"

There is no "heat" being "added" to get to Tc.

Tc is the temperature of the "cold reservoir". It's a given. Pre-existing. Ubiquitous energy. Not anything anyone added. Not heat supplied. By our accepted definition, not "heat" at all.

Or are you using some other definition of heat. Or maybe you can explain how "heat" between 0°K and our "cold reservoir" was "added" to get to Tc, or in what way am I confused?

[Fool]

Qcz is the internal energy of the mass of gas that is in the engine.

It is equivalent to heat that would be added if heating that mass from zero K to Tc K. It is also the amount of heat energy that would come out if cooled from Tc to zero Kelvin.

It is automatically supplied by the ambient temperature. It is not used up to run the engine.

It is the base line energy that DQh is added to, to get from Tc to Th.

If M•Cv is such that it equals one, Tc=300, then Qcz=300.

If DQh=100, then Qhz=400

n=0.25
DQc=75
W=25
Th=400

This will be a fairly large engine, not huge.

Experiment with other numbers.

[matt brown]

Good catch, Tom. Fool has various inconsistencies and bombed when he brought internal energy into it. I get what Fool is attempting (again) but this is NOT slick nor coherent (homebrew multi-substitution). Since thermo is loaded with ratios, it's very easy to prove Carnot from examples, and an ideal Stirling Cycle is ripe for this with minimal thermo knowledge.

My favorite proof is once we know that when T is Kelvin AND...

internal energy is linear T per mole regardless of PV values !!!

an ideal 300-600k Stirling Cycle has

(1) 600k isothermal expansion Wpos at constant internal energy
(2) 300k isothermal compression Wneg at constant internal energy
(3) internal energy during expansion is 2x internal energy during compression
(4) thus, Wpos is 2x Wneg, whereby Wnet=Wneg/Wpos=.50 a la Carnot (duh)
(5) but the exact location of Wpos vs Wneg will vary due to charge vs buffer pressure where the best mental model is zero buffer pressure (vacuum)

These work ratios will follow thru any Stirling Cycle and similar Ericsson Cycle due to isothermal processes. However, both cycles only follow Carnot when regen is included, otherwise both have less efficiency and further prove Carnot "Theorem" is a limit, not an equation.

The 2 common adiabatic cycles, Otto and Brayton, require more care to 'prove' but at least the adiabatic processes don't include any input that requires distinct quantification...both follow the PV=T relationship, but require gamma function to complete PVT values. Adiabatic cycles have Wpos and Wneg T gradients, so less efficient than isothermal cycles within the same thermal cycle range.

This is not rocket science and requires grasping only ONE internal energy relationship. This is why I say Carnot is chiseled in stone for common compression cycles.

[Fool]

Matt, you said the same thing I just said, only you used giant steps. I was attempting to use small enough steps that it is easy to follow, and is generic for all numbers. But thanks for the help. Please point out any specific typos/errors as I strive to be accurate.