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Re: Could Both Carnot and Tom be Correct?
Posted: Thu Aug 22, 2024 6:03 pm
by Tom Booth
So what does that mean if the engine is "running on ice"?
Suppose your ice is at 260°K
You supply 300°K ambient heat.
The engine takes in, NOT 300 units of heat, but 40. The difference between 260 and 300
The engine uses up those 40 units of energy, converting it to work output, then the engine takes in another 40 from the 300°K surroundings.
OK so, the engine has "rejected" 260 units of heat.
Do we care?
The engine only has a "Carnot efficiency" of about 12%
Do we care?
No heat at all is being "rejected" into the ice, the heat is being "rejected" before it goes into the engine to make up the difference. So no refrigeration is required to keep the ice cold or to remove the heat "flowing" in.
All of the 40 units of energy is converted to work. And this continues... No heat is "rejected" into the ice. That "heat" is "rejected" before entering the engine at all. The ambient heat only makes up the 40 difference. The ice remains at 260°K a little below freezing, as long as it is otherwise perfectly insulated, no heat is getting to it through the engine.
Re: Could Both Carnot and Tom be Correct?
Posted: Thu Aug 22, 2024 10:02 pm
by Tom Booth
The problem with considering "heat" being "rejected" that doesn't enter the engine is that is really entirely meaningless.
Temperature is not heat.
Part of a temperature reading is not part of a quantity of heat.
You could have hot water in a heat proof insulated dewar so no heat escapes at 375°K and another metal can that radiates heat easily.
Each container starts out the same. Transfers equal amounts of heat to the engine but the metal can cools off or "rejects" more heat to ambient sooner
What has that got to do with the actual efficiency of the engine itself?
Estimating efficiency of an engine based on heat "rejected" before it enters the engine is meaningless nonsense.
So really, the whole Carnot limit theory is just obsolete and meaningless.
Temperature is not heat. A ∆T is not a determinant of engine efficiency in terms of how heat actually entering the engine is utilized.
Re: Could Both Carnot and Tom be Correct?
Posted: Fri Aug 23, 2024 5:57 am
by Fool
Tom is so tough that when there is no one around to fight with, he starts a fight with himself by baffling himself with his own bogus banter.
Tom Booth wrote:Starting at, for example, 300°K ambient
Raising T-hot to 375°K is an increase of 20%
375 is 20% higher than 300.
Subtract 20% from 375 leaves 300.
Last I checked, 20% of 300 was 60.
300•0.20=60.
75 is 20% of 375
375•0.20=75
So 300 is 20% lower than 375. Or 300 is 80% of 275. 375 is not 20% higher than 300.
360 is 20% higher than 300.
And as Tom repetitively states, the rest is too convoluted to comment. Learn better mathematics.
The trouble with you, Tom, is you often do simple math incorrectly. This limits any ability you have to understand how to correctly use Carnot"s Theorem, or any other science.
You have an inability to logically discuss these principles. I've simplified the math as best as anyone could. You still resort to your technique of fallacious rantings. Sorry, you aren't worth the time and suffering to straighten you out.
Go learn some correct math. Hint temperature ratios being the same as heat ratios doesn't not mean that heat and temperature are equivalent. But as you said it, 'how do you measure heat without measuring temperature '. The difference between temperature and heat cancels out from the top and bottom, of the Carnot theorem, leaving the same ratio intact. Shown by simple mathematics that is apparently above your head.
Furthermore, ratios are dimensionless numbers, if you even have a clue what that means.
Temperature difference is the thermodynamic pressure that drives the gas pressure differential that powers these engines. The lower the difference the lower the power potential per engine size.
.
Re: Could Both Carnot and Tom be Correct?
Posted: Fri Aug 23, 2024 6:02 am
by Fool
VincentG wrote: ↑Thu Aug 22, 2024 9:21 am
How can temperature be equated with energy transfered?
Because the gas has a known heat capacity.
Quite rightly so. Heat capacity, as you just pointed out, is the bridge between temperature change ∆T, and energy difference ∆U. Q=∆T•C .
Thanks.
Re: Could Both Carnot and Tom be Correct?
Posted: Fri Aug 23, 2024 7:59 am
by Tom Booth
Fool wrote: ↑Fri Aug 23, 2024 5:57 am
Tom is so tough that when there is no one around to fight with, he starts a fight with himself by baffling himself with his own bogus banter.
Tom Booth wrote:Starting at, for example, 300°K ambient
Raising T-hot to 375°K is an increase of 20%
375 is 20% higher than 300.
Subtract 20% from 375 leaves 300.
Last I checked, 20% of 300 was 60.
300•0.20=60.
75 is 20% of 375
375•0.20=75
So 300 is 20% lower than 375. Or 300 is 80% of 275. 375 is not 20% higher than 300.
360 is 20% higher than 300.
And as Tom repetitively states, the rest is too convoluted to comment. Learn better mathematics.
The trouble with you, Tom, is you often do simple math incorrectly. This limits any ability you have to understand how to correctly use Carnot"s Theorem, or any other science.
You have an inability to logically discuss these principles. I've simplified the math as best as anyone could. You still resort to your technique of fallacious rantings. Sorry, you aren't worth the time and suffering to straighten you out.
Go learn some correct math. Hint temperature ratios being the same as heat ratios doesn't not mean that heat and temperature are equivalent. But as you said it, 'how do you measure heat without measuring temperature '. The difference between temperature and heat cancels out from the top and bottom, of the Carnot theorem, leaving the same ratio intact. Shown by simple mathematics that is apparently above your head.
Furthermore, ratios are dimensionless numbers, if you even have a clue what that means.
Temperature difference is the thermodynamic pressure that drives the gas pressure differential that powers these engines. The lower the difference the lower the power potential per engine size.
.
Given the ∆T between 300° to 375°
375 is 20% higher than 300
Moron.
Your the one who needs to learn how to do simple mathematics. You contradict yourself.
Last I checked, 20% of 300 was 60.
300•0.20=60.
Naturally, but a single temperature is not a ∆T is it, and Th in my example is fixed at 375 not 360, obviously to anyone who isn't an idiot.
Re: Could Both Carnot and Tom be Correct?
Posted: Fri Aug 23, 2024 8:35 am
by Tom Booth
Tom Booth wrote: ↑Thu Aug 22, 2024 9:15 am
Let's say, for example,...
With a typical ∆T of 300° ambient and 375° heat input in Kelvin.
...Th-Tc/Th yields 20% efficiency...
Because 375°K is approximately the temperature at which water boils, which is approximately Th in the majority of my experiments.
Not exactly, of course, but it makes for nice round numbers as an example.
Re: Could Both Carnot and Tom be Correct?
Posted: Fri Aug 23, 2024 10:27 am
by Fool
"375 is 20% higher than 300
Moron."
You will never get simple arithmetic correct if you keep insisting on you fallacy.
Re: Could Both Carnot and Tom be Correct?
Posted: Fri Aug 23, 2024 2:01 pm
by Fool
Dream on.
Re: Could Both Carnot and Tom be Correct?
Posted: Fri Aug 23, 2024 2:04 pm
by Fool
Yes. It's 25% higher.
300x1.25=375
300x1.20=360
You are starting at 300. And adding a percentage.
Re: Could Both Carnot and Tom be Correct?
Posted: Fri Aug 23, 2024 2:35 pm
by Tom Booth
Fool wrote: ↑Fri Aug 23, 2024 2:04 pm
Yes. It's 25% higher.
300x1.25=375
300x1.20=360
You are starting at 300. And adding a percentage.
Your contradicting yourself. You said above:
"So 300 is 20% lower than 375"
Carnot efficiency is calculated based on T-hot. The ∆T between 375 and 300 is 20% of T-hot as per the Carnot efficiency formula.
That is, 75 is 20% of 375.
People who believe in the Carnot limit do so because they are too bad at math to understand what ridiculous simplistic nonsense it's based on.
Only a moron like "fool" could take it seriously. A 20% ∆T is 20% Carnot efficiency. Nothing more or less than the temperature difference or "Fall of caloric"
To "fall" down you have to start at T-hot, obviously, not climb up from T-cold.
Your so deeply enamored with it but can't even do the math.
Re: Could Both Carnot and Tom be Correct?
Posted: Fri Aug 23, 2024 10:56 pm
by Fool
Tom Booth wrote:Starting at, for example, 300°K ambient
Raising T-hot to 375°K is an increase of 20%
375 is 20% higher than 300.
Subtract 20% from 375 leaves 300.
"375 is 20% higher than 300. "
That is wrong.
"Subtract 20% from 375 leaves 300. "
That is correct.
375 is 25% higher than 300. Correct.
300 is 20% lower than 375. Correct.
75 is 20% of 375. Correct.
75 is 25% of 300. Correct.
The ratio (375-300)/375 is 0.20, and multiplying it buy 100 gives 20%.
Re: Could Both Carnot and Tom be Correct?
Posted: Fri Aug 23, 2024 10:57 pm
by Fool
Shazam.
Re: Could Both Carnot and Tom be Correct?
Posted: Fri Aug 23, 2024 11:48 pm
by Tom Booth
Fool wrote: ↑Fri Aug 23, 2024 10:56 pm
Tom Booth wrote:Starting at, for example, 300°K ambient
Raising T-hot to 375°K is an increase of 20%
375 is 20% higher than 300.
Subtract 20% from 375 leaves 300.
"375 is 20% higher than 300. "
That is wrong.
"Subtract 20% from 375 leaves 300. "
That is correct.
375 is 25% higher than 300. Correct.
300 is 20% lower than 375. Correct.
75 is 20% of 375. Correct.
75 is 25% of 300. Correct.
The ratio (375-300)/375 is 0.20, and multiplying it buy 100 gives 20%.
Well, I may have worded it in a way that caused you to misunderstand, but in the context of ,Carnot efficiency 375°K down to 300°K is a 20% "fall" giving a 20% so-called Carnot efficiency.
Between 300°K and 375°K is a 20% difference when calculated from the highest temperature.
The highest temperature in this case is 375° not 360° that is a given.
If you are climbing a hill to a mark that is 375 feet up and stop at 300 feet to rest, what percentage of the distance to the 375 mark do you have left to climb?
It is not "wrong" to say going up from the 300 mark to the 375 mark is 20% higher in terms of the 375 high mark.
Sure, 60 is 20% of 300, so what?
300°K does not represent a temperature difference.
The 75° ∆T is 20% of the temperature reached. You can play dumb and pretend like you have proven I can't do simple mathematics, but the fact is, you are still a demented moron with a complete lack of common sense.
At best you've proven that your incapable of basic reasoning ability and can't figure out the meaning of a sentence from the context, or you can and just like argument and playing mind games.
If you move from 300 to 375 you have proceeded 20% the total distance. That is not wrong and the meaning should have been obvious from the context.
You have an obsession about trying to prove Tom Booth wrong and just can't pass up any opportunity regardless if it only demonstrates what a complete moron you are and what your agenda here is.
Re: Could Both Carnot and Tom be Correct?
Posted: Sat Aug 24, 2024 5:06 am
by sarimshaikh
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Re: Could Both Carnot and Tom be Correct?
Posted: Sat Aug 24, 2024 7:07 am
by Tom Booth
Fool wrote: ↑Fri Aug 23, 2024 10:56 pm
[
"375 is 20% higher than 300. "
That is wrong.
(...)
375 is 25% higher than 300. Correct.
300 is 20% lower than 375. Correct.
75 is 20% of 375. Correct.
75 is 25% of 300. Correct.
The ratio (375-300)/375 is 0.20, and multiplying it buy 100 gives 20%.
375 is not 25% higher than 300 on a scale from 300 to 375.
300 is 80% of 375
What then is the remainder?
Sorry but 375
is 20% higher than 300, on a scale of measure between 0 and 375.
Your logic is flawed bot. Who programmed you? They should be fired.