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Nikola Tesla Vs. The Second Law Of Thermodynamics


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Posted (edited)

459 + 212+1(for zero) = 672 (F measuring units)

 

672  X  0.189 = 127 F measuring units

672  X  0.811 = 545 F measuring units

 

545+127=672

 

So-called "Carnot efficiency" is exactly nothing more and nothing less than the temperature difference.

 

In other words the "available heat" above equilibrium, or above the ambient baseline.

 

Carnot efficiency is really just a measure of the amount of heat available for conversion to work. In this case 18.9% of the imagined total "caloric" (contained in the steam).

Edited by TomBooth
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Carnot efficiency is the maximum theoretical efficiency of a heat engine operating between ANY two temperatures.  That would be for a "perfect" engine without any loses to friction, heat leakage,bad s

I ordered this digital thermometer a few days ago. It comes with four probes for taking simultaneous readings. It doesn't record data to the cloud or have software to build charts and plug into the co

I said "If", hypothetically.   Just supposing that the engine could utilize that additional 7 joules otherwise rejected to the sink. That would only bring the temperature back down to equilibrium with

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To be fair Ocean,you speak of hijacking but the thread is hard to follow outside basic examples. A black hole is a heat engine and so to discuss the roles of a heat engine, we sometimes resort to the most simplest of cases. To me a black hole is quite simple theoretically, though we have strange threads willing to accept mathematical spaghetti concerning whether things can cross an event horizon. Let's be certain, you yourself, after me, admitted there is no breakdown of Carnots system, which begs the question of the motive behind further discussion. The case of absolute zero as has been discussed many times is an unobtainable limit of experimental prowess. We know there are no perfect insulators in physics as far as the mind may take us.

Edited by Dubbelosix
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One thing is good, at least the OP recognizes the importance behind analogies which certainly helps anyone's understanding. I remember for agood year you would not accept my rejection of a system at absolute zero.... It seems since you have waned from this, which is also good. So long as we have adiabatic irreversible dynamics, the thermodynamic laws stand classically.

Edited by Dubbelosix
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Maybe redo that scale comparison a bit:

 

 

 

0 _ 10 _ 20 .... 300 _ 310 _ 320 _ 330 _ 340 _ 350 _ 360 _ 370 _ 380..(Kelvin)

                             ^                                                                      ^

                          302K                                                               373K

-459F                  85F                                                                212F

 

<------81.1%-------><------------------------18.9%--------------------->

<---------------------------------100%------------------------------------->

 

 

Well OK,  :rolleyes:  but if you are not using 0 F as your reference, you are not really using the Fahrenheit scale. What you are doing is using degrees F referenced to 0 K. That is not a standard usage and can easily lead to confusion.

 

Personally, I find it much simpler to take the ratio of (373-300) K / 373 K than the ratio of (212-85) F / [ 212- (-459.67) ]F. Normally I might say “to each his own” but if we did that in science it would lead to chaos.

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I think your argument contains a logical inconsistency. You start out with "if one cup of water at 373 K cools down to ambient 300 K, it releases about 72,500 Joules of heat energy." <--- That is the 20%

 

The 80% is all the remaining potential "heat" from 300K down to absolute zero that is supposed to be rejected to the sink.

 

In a steam engine, I suppose this makes some sense because all that "latent heat" if it can be called that, is actually contained in the molecules of H2O that actually do pass through the engine. The unused "heat" or internal molecular kinetic energy IS carried along through the cylinder and goes out through the valves and is "rejected" or exhausted

 

In a Stirling engine, though, no steam or air or other substance passes through only the 20% of additional added heat above the ambient baseline is in any way utilized.

 

Even if I take a bucket of water that represents 20% of a bathtub full of water and elevate it to pour through a turbine, and the entire bucketfull returns to the bathtub "sink", the 80% remaining in the tub did not pass through the engine.

 

In a Stirling engine even the 20% does not return to the tub because it is really energy that is converted, so never returns to the tub.

 

The other 80% never left it.

 

 

Yes, you’re right. The max theoretical Carnot efficiency is 20% and that is the 72,500 joules of heat energy released by one cup of water falling in T from 373 K to 300 K. What I am saying is that all of that energy cannot be converted to useful work to run the engine. There must be some heat lost from the cylinder when the gas expands. For no particular reason, I stayed with 20 % system thermal efficiency, which is rather low, and I should have explained what I was doing.

 

However, even if I were to go with a thermal efficiency, of the overall system, being half of the theoretical limit, then 50 % of the 6.7 Watts of continuous power is converted to mechanical work and the other 50 % is dissipated as waste heat.  (Actually, everything eventually winds up as heat, including the mechanical energy, as energy is conserved). Now, instead of 5 W being dissipated by the heat sink, there is only a 3.5 W dissipation.

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To be fair Ocean,you speak of hijacking but the thread is hard to follow outside basic examples. A black hole is a heat engine and so to discuss the roles of a heat engine, we sometimes resort to the most simplest of cases. To me a black hole is quite simple theoretically, though we have strange threads willing to accept mathematical spaghetti concerning whether things can cross an event horizon. Let's be certain, you yourself, after me, admitted there is no breakdown of Carnots system, which begs the question of the motive behind further discussion. The case of absolute zero as has been discussed many times is an unobtainable limit of experimental prowess. We know there are no perfect insulators in physics as far as the mind may take us.

 

I tell you what, you go ahead and hijack this thread, talk about black holes, event horizons and plug your nonsensical "paper" all you want to. You can even claim that -1 is the smallest negative number and zero is not a number. I don't give a fck what you do. It is a shame that an asshat like you has to come along and ruin an otherwise interesting thread with your nonsense, but I am not interested in your BS, so I withdraw from this discussion.

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As the graph apparently does not render consistently on different computers/browsers, I took a screenshot:

 

post-96374-0-34500000-1597766645_thumb.jpg

 

LOL

 

But seriously, what temperature is a black hole anyway?

 

Also, air is no longer air, but condenses into a liquid, or various liquids, at very low temperatures, so in a "Hot air engine" does efficiency below about 80K have any meaning?

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Posted (edited)

Well OK,  :rolleyes:  but if you are not using 0 F as your reference, you are not really using the Fahrenheit scale. What you are doing is using degrees F referenced to 0 K. That is not a standard usage and can easily lead to confusion.

 

Personally, I find it much simpler to take the ratio of (373-300) K / 373 K than the ratio of (212-85) F / [ 212- (-459.67) ]F. Normally I might say “to each his own” but if we did that in science it would lead to chaos.

 

I"m just trying to get across that the "Carnot efficiency" IS THE SAME THING as the temperature difference. Not, in this example, 18.9% of the temperature difference. Which would be 18.9% of 18.9% which is like 3.6% or something?

 

But "Carnot efficiency" generally is supposed to mean that the 81% down to absolute zero goes through the engine to the sink, not 15.3% (18.9 - 3.6) or whatever.

 

I know it is an unusual way of measuring on the Fahrenheit  scale (but not that unusual), but I'm just trying to reverse obfuscate the fact that Carnot efficiency is really nothing more than the temperature difference.

 

In a steam engine, I guess it works out, that was all Carnot dealt with, because steam carries latent heat, but for a Stirling engine, there is no material substance to carry latent heat through the engine. All that "unavailable" heat should, IMO, for a Stirling engine, just be ignored.

 

Any scale is just a measuring stick, marked off in arbitrary units. -459 Fahrenheit is perfectly legitimate. (or -459.67 to be exact, I guess)

Edited by TomBooth
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Posted (edited)

I tell you what, you go ahead and hijack this thread, talk about black holes, event horizons and plug your nonsensical "paper" all you want to. You can even claim that -1 is the smallest negative number and zero is not a number. I don't give a fck what you do. It is a shame that an asshat like you has to come along and ruin an otherwise interesting thread with your nonsense, but I am not interested in your BS, so I withdraw from this discussion.

I'm going to go out and buy some 9 volt batteries for the thermometer. I'll be trying different experiments and different readings and probably making various modifications. Dubbelosix's posts, I tend to agree, seem to be self promotion of his paper, which contributes little to the discussion, as far as I can see, so far, other than the fact? that a black hole is a heat engine (I guess). But I know next to nothing on the subject (of black holes), so I'll reserve judgement.

 

I very much value your input, so I do hope you will continue to contribute to the thread.

Edited by TomBooth
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Posted (edited)

Yes, you’re right. The max theoretical Carnot efficiency is 20% and that is the 72,500 joules of heat energy released by one cup of water falling in T from 373 K to 300 K. What I am saying is that all of that energy cannot be converted to useful work to run the engine. There must be some heat lost from the cylinder when the gas expands. For no particular reason, I stayed with 20 % system thermal efficiency, which is rather low, and I should have explained what I was doing.

 

However, even if I were to go with a thermal efficiency, of the overall system, being half of the theoretical limit, then 50 % of the 6.7 Watts of continuous power is converted to mechanical work and the other 50 % is dissipated as waste heat.  (Actually, everything eventually winds up as heat, including the mechanical energy, as energy is conserved). Now, instead of 5 W being dissipated by the heat sink, there is only a 3.5 W dissipation.

Regarding the highlighted statement.

 

This is ultimately true, of course, but, suppose the engine is coupled to an electric generator, connected to the grid, sending current across the country to power someone's electric range. The heat reappears in the stoves heating element to cook someones food, miles away. If the engine was running on ice, that heat is not really going to the "sink" (the ice), it is far far away from the sink.

 

With a little Stirling engine running on a cup of ice, this is mostly also true of heat from friction. The engine is running in the open air above the ice and heat from friction, from the gears and the flywheel and such is carried away by the surrounding air, which is the source of heat, not the ice. The heat from friction actually, if anything contributes to the temperature differential by warming the surrounding air.

 

True "no insulation is perfect" but with something like a Dewar within a Dewar perhaps, we could get pretty darn close.

Edited by TomBooth
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Posted (edited)

A black hole can nearly any temperature depending on size.

 

 

https://youtu.be/LRtvGABJC2Q

 

If nothing else, I find it very interesting that a virtual particle pair can be separated, so that 1/2 of the virtual particle pair becomes a real particle in our universe.. So I learned something new today, And, black holes are "evaporating". That's two things.

Edited by TomBooth
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Posted (edited)

The thermometer seems fairly accurate, as far as I can tell, out of the box, with all four probes plugged in and just measuring ambient room temperature.

 

There is some fluctuation, within 0.5 degrees F, possibly due to drafts, body heat? But close enough for the purpose at hand I think. It is factory calibrated, so I don't think I'll need to recalibrate or anything.

 

The three photos were taken a few seconds apart. The probes are just dangling off the edge of the table in the open air

post-96374-0-17560100-1597776131_thumb.jpg

post-96374-0-56141700-1597776155_thumb.jpg

post-96374-0-50073300-1597776183_thumb.jpg

Edited by TomBooth
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If nothing else, I find it very interesting that a virtual particle pair can be separated, so that 1/2 of the virtual particle pair becomes a real particle in our universe.. So I learned something new today, And, black holes are "evaporating". That's two things.

Whatever the cause. Hawking evaporation is a theory of how a heat engine in nature gives up its thermal abilities to the environment. Believe me when I say, it's not entirely hypothetical since we have observed analogue radiation from analogue black holes. Edited by Dubbelosix
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