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

Discussion on Stirling or "hot air" engines (all types)
VincentG
Posts: 629
Joined: Tue Feb 21, 2023 3:05 pm

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

Post by VincentG »

Apples and oranges...the system vs surroundings pressure differential (charge vs buffer) effects work per process (expansion vs compression) while the system vs surroundings temperature differential effects efficiency per process.

Point is, with no atmosphere to buffer piston, what is absolute zero anyway? There's no more surroundings to affect the engine externally. Just an engine floating in the either powered by sunlight and cooled by IR heat radiation into space.
matt brown
Posts: 488
Joined: Thu Feb 10, 2022 11:25 pm

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

Post by matt brown »

VincentG wrote: Sat Mar 23, 2024 3:29 pm
Apples and oranges...the system vs surroundings pressure differential (charge vs buffer) effects work per process (expansion vs compression) while the system vs surroundings temperature differential effects efficiency per process.

Point is, with no atmosphere to buffer piston, what is absolute zero anyway? There's no more surroundings to affect the engine externally. Just an engine floating in the either powered by sunlight and cooled by IR heat radiation into space.
If ICE style configuration, the buffer space is the crankcase and could be a vacuum ;-)
Tom Booth
Posts: 3313
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: Sat Mar 23, 2024 2:43 pm
Tom Booth wrote: Sat Mar 23, 2024 2:15 pm
The way the Carnot "efficiency" is being calculated and interpreted is nonsense.

It is NOTHING but the temperature difference.

Temperature has nothing to do with the QUANTITY of heat or efficiency.

All it is is the amount of "fuel" available.
I'll willing to go along with Carnot buzz under certain conditions where I'd describe Carnot as a coincidence. However, Fool is trying to convince us Carnot is a consequence by qualifying energy into temperature whereby Carnot can be quantified via temperature. Indeed, a slippery slope which I also consider 'slight of hand'...
My problem is not with the 2nd Law statement that you can't ever achieve 100% conversion efficiency.

There will always be some friction, vibration, noise, loses here, there everywhere. All we can do is minimize these loses. We are getting better and better at that with magnetic levitating bearings, non-expanding alloys for incredibly close no-contact pistons and cylinders, flexure bearings, air bearing, Ferrofluid bearings, highly non-heat conducting/insulating materials etc. etc.

But the Carnot Limit is a different story.

It doesn't say you can't have 100% efficiency. It claims there is an extremely strict mathematical or actually a kind of physical barrier, a line in the sand, limiting efficiency, an ARBITRARY line that can never be crossed SEVERELY limiting efficiency to a truly paltry value, completely dependent on temperature difference and ONLY temperature difference with ZERO empirical evidence, no experimental backing whatsoever, no sound mathematical basis, as seen by the endless tail chasing and redefining of variable and never ending escalation in complexity and obfuscation with no resolution.

There is a plethora of experimental evidence which is at worst inconclusive, failing to provide support for this "LAW" and at best, disproves it completely. (Depending on your proclivities, in terms of outcome).

The whole thing is admittedly based on an obsolete, verifiably fallacious theory of heat.

If the "Carnot Limit" calculation reveals that an engine has at best 10% efficiency based on temperature difference only, but experimentally there is no temperature rise at the interface between the cold side of the engine and the "cold reservoir", where, pray tell, is the other 90% of the heat going?

Fool said a few posts back, if there is no temperature difference there can be no transfer of heat. So what are we to conclude when the cold side of the engine is the same temperature as the ambient surroundings?

Can there be any "waste heat" leaving the engine at that point?
Fool wrote: Sat Mar 23, 2024 9:31 am Qcz is not rejected during the running. Nor is it added. It is the base energy that DQh is added to. Earlier definition discarded.
Cheers!

But is it discarded or isn't it?

Now you say Qcz hasn't changed. You just need to clarify your explanations or descriptions or something. Not discarded at all.

I say baloney.

I've been getting the runaround on this topic for over a decade. I've been banned from all the science and physics forums for simply posting videos of experiments and asking questions.

Maybe someone could explain how a toy engine with a thermal efficiency of maybe 21% rather than the Carnot Limit of 20% constitutes the dreaded "perpetual motion" making discussion on the results of an experiment forbidden to the scientific community at large.

Here I posted videos of two different but very real heat engines, a drinking bird and an "ultra LTD" Stirling.

Post in thread 'Assumed violation of physics - Heat vs. Work' https://www.physicsforums.com/threads/a ... st-4247477

The discussion was immediately shut down.

Thank you "Fool" for at least being willing to address the subject with some serious and sustained consideration rather than the sort of avoidance displayed on the various Classical Physics and Science forums.
Tom Booth
Posts: 3313
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: Sat Mar 23, 2024 1:28 pm VincentG,

The real test would be to get any engine without a crankshaft, or spring, or buffer pressure, or double acting piston and cylinder, to run in a vacuum. The cylinder will become a gun barrel every time spitting the piston out the one end. That will happen regardless of the pressure in the cylinder, or whether it is heated or cooled. There is no such thing as contraction in gas or liquid. Only solids can produce a pulling force, like springs.

Liquids can reduce in size, but they turn to gas in the space of a vacuum and it is the outside pressure that causes the volume to decrease.

In liquid systems engineers call it cavitation, boiling of the liquid from too low a pressure. Hydraulic systems never work on suction. At least not well.

Gasses expand to fill their container. This is called pressure. Cooling them, heat rejection or addition, doesn't cause volume change, it only reduces pressure, or increases it. The volume change has to come from outside, either mechanical force, or a higher outside pressure, or lower. We just view it from different perspectives/pressures.
I think all of this is simply your opinions education, indoctrination talking.

You have been "educated" to believe that gasses always "expand to fill their container".

Earths atmosphere is a gas. Outside the atmosphere is the vacuum of empty space.

All the planets, galaxies, celestial bodies and systems of all sorts are supposed to have formed out of gases. The mutual attraction of gas particles holds the entire universe together. That gases always expand to fill their container is theory taught in cloistered halls by professors with thick rimmed glasses.

I find it puzzling myself, though, why the piston doesn't just fly out of the cylinder like a bullet in a free piston engine with no flywheel to persuade the piston to return while the gas inside the engine is continually heated by means of a high temperature flame.

Well, "heat rejection" is supposedly what makes this return possible.

That's a pretty good theory. A reasonable and logical conclusion. I accepted it myself, taking it for granted for many years.

But to test this, I ran my engine free piston without a flywheel and with all points of possible "heat rejection" made of non-heat conducting material and/or insulated, while heating the engine evenly across the entire length of the hot chamber to assure myself that there were no cold spots.

So, under such controlled test conditions, why should the piston continue to return at all? Why doesn't the piston just pop out the end of the cylinder?

https://youtu.be/LG09AXAjpio?si=8l3zSRjf65RSnbz_

The return will happen even when opposed by a weight hanging on the piston.

https://youtu.be/e-7DFp_B0y4?si=N6hLNRAvRpAb7dBO

This, IMO, appears to be a very clear example of a kind of oscillating "spring" action of some kind.

It seems clear to me that "something else" causes the piston to return. Something other than "heat rejection".

Additionally. Heat rejection by conduction is a relatively slow process but the oscillations are very rapid.
Fool
Posts: 238
Joined: Sun Jul 16, 2023 9:14 am

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

Post by Fool »

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.
Tom Booth
Posts: 3313
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 »

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.
Fool
Posts: 238
Joined: Sun Jul 16, 2023 9:14 am

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

Post by Fool »

Tom Booth wrote:Maybe someone could explain how a toy engine with a thermal efficiency of maybe 21% rather than the Carnot Limit of 20% constitutes the dreaded "perpetual motion" making discussion on the results of an experiment forbidden to the scientific community at large.
That is a tall order. It is my humble opinion that if someone were to build an engine with the slim margin of 1% better than a Carnot, another engine running in reverse as a heat pump, would be so inefficient that perpetual motion would not be possible. Like a flywheel spinning up with one percent higher energy, hooked to a 98% or worse generator. It would just loose power. Plus, a mythical engine performing one percent better than a mythical Carnot engine, would be very hard to test to get any imperial data backing my opinion. Go figure?

Nevertheless, theoretically, a 1% over Carnot engine could theoretically be attached to a perfect Carnot heat pump and recycle 101% of the energy back into the hot reservoir out of the cold reservoir. Allowing "theoretical" perpetual motion.

That demonstrates that it will be impossible to have an over Carnot engine, but not "Why"?

Why is quite a bit more difficult.

Take two heat reservoirs Th and Tc, assume they are isolated large masses. Allow them to conduct heat to an engine that is connected to heat pump that is connected to the two reservoirs. Everyone knows that if Tc and Th are closer together the engine will produce less power, less heat will poured into the engine. Less heat in, less work out. The reverse is true of the heat pump it will take more energy to pump heat up into Th from Tc if the difference is greater.

So looking at the idea that heat won't flow between things at the same temperature, the max temperature of the gas in the engine will be significantly lower, for a real engine, than Th. The power out will be significantly lower as a consequence.

This means that the input power to the heat pump will be lower. In addition the heat will not move out of the heat pump unless the gas in the heat pump is significantly hotter than Th. This will require more work energy input to move the same heat to the hot reservoir, further decreasing the effect.

The cold sides will have similar degradation. The engine will need a higher temperature than Tc, and the heat pump will have lower than Tc. Effectively reducing engine work, and increasing needed work to the heat pump.

All that working together means engine efficiency will be much lower than the Carnot Limit. And heat pump COP will be much lower than the Carnot Limit. Usually by more more than half. It rapidly becomes a lose lose situation.

When and if you get an engine or heat pump with measured efficiency greater than Carnot, I'd love to talk. Until then we'll have to resort to talking of 200+ year old Thermodynamics Theory. And engine building.
Tom Booth
Posts: 3313
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: Mon Mar 25, 2024 10:16 am
Tom Booth wrote:Maybe someone could explain how a toy engine with a thermal efficiency of maybe 21% rather than the Carnot Limit of 20% constitutes the dreaded "perpetual motion" making discussion on the results of an experiment forbidden to the scientific community at large.
...

Nevertheless, theoretically, a 1% over Carnot engine could theoretically be attached to a perfect Carnot heat pump and recycle 101% of the energy back into the hot reservoir out of the cold reservoir. Allowing "theoretical" perpetual motion.

...
How so?

Take my LTD running on boiling hot water.

Say ambient temperature is 300°K

Just under boiling water is about 370°K

Carnot efficiency at that ∆T is about 19%

So let's say we supply 70 joules per cycle at 370°K

The engine converts 13.3 joules to work and "rejects" 56.7 joules.

That's a Carnot engine or Carnot efficiency. The maximum efficiency possible, supposedly.

So let's imagine we have an "impossible" engine that is 25% efficient at the same ∆T of 300°K Tc to 370°K Th.

Again we supply 70 joules of heat.

25% of 70 joules is 17.5 joules converted to work and 52.5 joules "rejected".

How does your "perfect Carnot heat pump" retrieve 101% of the energy while being driven by an engine that can only utilize, at best; 17.5 out of every 70 joules supplied?
Fool
Posts: 238
Joined: Sun Jul 16, 2023 9:14 am

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

Post by Fool »

A perfect Carnot Engine:

n =(370-300)/370 = about 0.19

A perfect Carnot heat pump:

COP=370/(370-300) = about 5.3

70 Joules in your engine turns into about 17.5

17.5 Joules into a Carnot Heat pump:
17.5 x 5.3 = 92.75 J

92.75 / 70 = about 1.325

Or 132.5%
Tom Booth
Posts: 3313
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: Mon Mar 25, 2024 3:13 pm A perfect Carnot Engine:

n =(370-300)/370 = about 0.19

A perfect Carnot heat pump:

COP=370/(370-300) = about 5.3

70 Joules in your engine turns into about 17.5

17.5 Joules into a Carnot Heat pump:
17.5 x 5.3 = 92.75 J

92.75 / 70 = about 1.325

Or 132.5%
Well, that's interesting.

Suppose instead of trying to draw heat out of the "sink" what would be the COP of a "perfect Carnot heat pump" if just MOVING heat from one place to another within the "hot reservoir"?

That is "refrigerating" the ambient environment, moving the heat and dumping it into a heat engine that's "running on ice"

Maximum COP cooling = Qc/Qh-Qc = Tc / Th – Tc

300/(300-300) = 300

Is that right?

17.5 joules × 300 = 5,250 joules moved for every 17.5 consumed by the heat engine ?

I mean that was basically Tesla's proposition.

Run the heat engine on atmospheric heat at ambient temperature, using a refrigerator as a "fuel pump".
Tom Booth
Posts: 3313
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 »

Anyway, I'm more interested in your response to this post, did you miss it?

viewtopic.php?p=21729#p21729
Fool
Posts: 238
Joined: Sun Jul 16, 2023 9:14 am

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

Post by Fool »

300/(300-300)= 300/0 = infinity

(300-300)/300 = 0/300 = 0

COP x n = Infinity x 0 = unknown. Must use limits as Tc goes to Th.

Let Tc go to Th by subtracting dt from Th to get Tc, in other words:
Let Tc=Th-dt as dt goes to zero.

Substituting it into the equation for COP x n:
(Th-Th+dt)/Th x Th/(Th-Th+dt) as dt goes to zero

All the Th's cancel:
dt/Th x Th/(dt)
dt/1x1/dt
For:
n•COP=dt/dt

And the dt's cancel:
n•COP=dt/dt = one

No need to take the limit as the dt's have all cancelled out too.
n•COP=1

Sorry, no better than one, in a "perfect" case, theoretically.

If everything is 300, zero heat transfer. Zero heat transfered equals both, zero work out, and zero heat pumped.

Just understand, that for real engines and heat pumps, there will always be some lag behind the "one" reservoir, so:

300 will become 290 for the engine, And 310 for the heat pump, both to pickup heat.

The engine will be then be running between 290 and some even lower temp. Must be a difference, remember. Say 250. Because too, remember, expansive heat drop.

The heat pump will have to run between 310 and 350. But both will only have 300. So half their cycle will fail, so the whole cycle will fail.

So the two together will fail because there will be no middle ground.

One temperature operation is called adiabatic bounce. Zero net work.

Adiabatic bounce is a lose lose situation. At the best, a bouncer would barely break even. No one has ever seen otherwise, not even close.

If they ever went over unity we'd all have them running in our basement or attic now. Unless they cost too much like a drinking bird Rankin cycle machine.

There have been many charlatans scamming people and followers, but no honest demonstrations. Lots of people thinking they might one day get it to work, but never doing so. It has driven many into declining health while trying.

Even a few patent office people have been fooled. That's why there are patents for unworkable machines.

If a single 300 source is used, and an internal heat mass. The heat pump could pull the temperature of the mass down to say 250, by operating at an internal heat pump gas temperature range of 240 and 310.

The engine would then be working at 290 and 260.

n = 290-260/290 = approximately 0.103

COP = 310/(310-240) = approximately 4.43

n x COP = 0.103 x 4.43 = approximately 0.456

Or, more than a 50% loss, with zero extra work out of it to show for the trouble. You are certainly welcome to try it.

Tying a heat pump and heat engine together to "recycle" energy is a lose lose situation. It's the same as hooking two electric motors together. Calling one a generator will not help any. They are the same thing. Twice the I^2R losses, zero extra work to output.
Tom Booth
Posts: 3313
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: Mon Mar 25, 2024 7:06 pm ...
If everything is 300, zero heat transfer. Zero heat transfered equals both, zero work out, and zero heat pumped.
...
That's funny. My refrigerator, before plugging it in is in a 300°K environment. Everything is 300°K

Do you think if I leave the refrigerator doors off and run the refrigerator it will not pump any heat?
Tom Booth
Posts: 3313
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 »

Anyway back to the analysis.

You wrote:
Fool wrote: Mon Mar 25, 2024 9:13 am ...

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.
This is no different than before.

Not long ago you said:
Fool wrote: Sat Mar 23, 2024 9:31 am Qcz is not rejected during the running. Nor is it added. It is the base energy that DQh is added to. Earlier definition discarded.
Now you have snuck that back in again.

"Let Qcz be the heat added from zero K all the way to Tc"

"base energy" is not "heat" transfered. Nor is it "heat added".

Temperature is not a measure of a quantity of heat.

The "baseline" ambient energy already exists within the working fluid before the engine starts. Before the engine is heated up. It is never added, so how can it be "rejected".

You keep treating this baseline energy as if it is being "added" and then "rejected".

Technically, as heat is defined in modern science generally, the baseline energy is not "heat" since it is not:
energy that is transferred from one body to another as the result of a difference in temperature.
viewtopic.php?p=21084#p21084

Do you agree with the Britannica definition?

Back to square one:

"How do you define heat?"

viewtopic.php?p=21059#p21059
Tom Booth
Posts: 3313
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 »

Here you said:

viewtopic.php?p=21109#p21109
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.
But you habitually refer to this baseline "internal energy" as "heat added" and "heat rejected".

How can "heat" or internal thermal energy that is never added to and never leaves the working fluid be conceptualized and treated mathematically AS IF it were "added" and later "rejected".
Post Reply