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Author Topic: 2nd law of thermodynamics  (Read 10452 times)

Offline PaulLowrance

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2nd law of thermodynamics
« on: November 10, 2006, 11:55:23 PM »
Any EE knows that an enclosed box at room temperature containing air, wire, and an appropriate load can generate so-called "free energy" 24 hours a day, 365 days a year.

The amount of continuous power for a single object is:

P = 4 K T dF

K is Boltzmann constant
T is temperature in Kelvin
df is bandwidth.

@ room temperature say 1 GHz dF we get just 17 pW.  Not much power, but such an object could be a nano particle.  One trillion of such objects generates 17 watts. Increase the bandwidth 1 THz and we get 17 kilowatts.

Quote taken from WikiPedia.org:
"Since any thermodynamic engine requires such a temperature difference, it follows that no useful work can be derived from an isolated system in equilibrium; there must always be an external energy source and a cold sink. The second law is often invoked as the reason why perpetual motion machines cannot exist."

^ Incorrect ^

I think the Philosopher / Physicist P.W. Bridgman, (1941) said it best "There are almost as many formulations of the second law as there have been discussions of it."


Regards,
Paul

Free Energy | searching for free energy and discussing free energy

2nd law of thermodynamics
« on: November 10, 2006, 11:55:23 PM »

Offline PaulLowrance

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Re: 2nd law of thermodynamics
« Reply #1 on: November 13, 2006, 06:01:54 PM »
The following is a specific example that breaks the 2nd law of thermodynamics.

Carbon resistors generate more thermal noise than Metal film resistors. Think about the difference between the following two experiments :

Experiment #1:
Carbon resistor in series with an antenna.

Experiment #2:
Metal film resistor in series with an antenna.

Which experiment generates more radiation? Answer: Experiment #1

That's why some surfaces are cooler or hotter than other surfaces.

Experiment #1 becomes colder than experiment #2.  Place both experiments in a vacuum jar and you'll begin to see a large difference in temperatures.


The 2nd Law of Thermodynamics is incorrect.



Regards,
Paul Lowrance

Offline PaulLowrance

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Re: 2nd law of thermodynamics
« Reply #2 on: November 13, 2006, 08:11:53 PM »
Here's an exact example:

Experiment A:
Source Resistance: R
Antenna radiation resistance: Rr
RMS thermal noise: Va
RMS current: Va / (R + Rr)
Radiated power: I^2 Rr = (Va / (R + Rr))^2 * Rr

Experiment B:
Source Resistance: R
Antenna radiation resistance: Rr
RMS thermal noise: Va * 1.1
RMS current: Va * 1.1 / (R + Rr)
Radiated power: I^2 Rr = (Va * 1.1 / (R + Rr))^2 * Rr

Experiment B radiates more power.  Experiment B will be cooler than experiment A.

Lets simply. Neither experiment A or B have a power source except thermal noise. Experiment B radiates more power. It is a very simple circuit. Over time, more energy is leaving experiment B than experiment A. Therefore experiment B will be colder than experiment A.


Regards,
Paul Lowrance

Offline PaulLowrance

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Re: 2nd law of thermodynamics
« Reply #3 on: November 27, 2006, 07:08:44 PM »
Here's another experiment that is extremely straightforward and simple.

We know that thermo noise has no theoretical upper crest limit. Normally we refer to noise in terms of root mean square. When studying real thermo noise we see that given enough time the noise will eventually drift to a higher crest. The experiment is simple. Connect one resistor in series with an LED. That is it. To save yourself a lot of time you should pick a high frequency LED as used in GHz optics. This will provide a lot of bandwidth, which is what you want given voltage thermo noise is (4 K T R B)^0.5, where B = Bandwidth.  Also you want to pick a resistor that matches the LED for an optimum effect. Also it doesn't hurt if the resistor is a noisy one such as carbon composite and as small as possible. Smaller carbon composite resistors generate more noise. Of course the *extra* noise is 1/f. This results in a resistor with real noise. When then voltage noise crest overcomes the LED's forward voltage then the LED will emit photon(s).  Also note the LED emits photons far below the forward voltage. So in that sense, it is possible the LED will emit an occasional photon even when the noise voltage is far below the LED's forward voltage.

Now the question is, "Where do we aim the photons?"  Note the above experiment is in an isolated system. We have two experiments. Experiment #1, the resistor absorbs the photons.  Experiment #2, the LED absorbs its own photons (we coat the LED with opaque material).  The main difference between the two experiments is the resistor in experiment #2 is colder than the resistor in experiment #1.


Note, the above experiments could require vast amounts of time, depending on the exact parts used in such experiments. Given enough time, the noise crest will reach the LED's forward voltage. For those who are less patient, it is possible you will see some photons emitted even below the LED's forward voltage.

Regards,
Paul Lowrance

Offline Nabo00o

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Re: 2nd law of thermodynamics
« Reply #4 on: May 08, 2010, 05:50:01 PM »
Hey Paul, I know this is a terribly old tread and this an awful late post ;D
But, I am really interested in this thermal noise, and I think I will create a new tread in the heat to mechanical section. What I was really interested in was the concrete numbers on how much radiation in wattage a given temperature will give, both in a certain wavelength and in the complete spectrum.

And you are absolutely right, thermodynamics has a fundamental problem with heat related energy, namely electromagnetic emission.

Julian

Free Energy | searching for free energy and discussing free energy

Re: 2nd law of thermodynamics
« Reply #4 on: May 08, 2010, 05:50:01 PM »
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Offline mscoffman

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Re: 2nd law of thermodynamics
« Reply #5 on: May 08, 2010, 07:45:03 PM »
@juilian

I think Paul_L no longer visits this site.
You may want to contact him directly at his blog site;

http://www.globalfreeenergy.info/

What you are referring to is called the Planck Thermal radiation spectrum.
It turns out that random latent heat (molecules bumping into one
another) comes into equilibrium with thermal radiative heat which
is EMF, alternately known as IR radiation. By an equation of what
is called the black-body radiation spectrum. It's important to note
that the earth has a radiation spectrum that differs slightly from
black-body matter because it is internally driven by energetic
processes which color its output. My feeling is that Brownian motion
of molecules in matter forms loops combined with the electrically
bipolar nature of matter which automatically converts to emf by
way of Maxwell's theorem. Matter both transmits and absorbs this
EMF. The bidirectional nature of random process is what is used
to justify leaving it out of theoretical statistical discussion involving
latent heat for example; in the Carnot perfect gas cycle.

Here is a link that explores this radiation in detail;

http://en.wikipedia.org/wiki/Black_body

which implies:
http://en.wikipedia.org/wiki/Dark_current_(physics)

Which creates the forward voltage pedestal in semiconductor
diode junctions.

:S:MarkSCoffman

Offline Nabo00o

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Re: 2nd law of thermodynamics
« Reply #6 on: May 08, 2010, 11:37:53 PM »
Thanks for the information Mark!

And yes I loved that link, a really informative, in depth wiki page, not as usual since its not their style, but very useful for people who want to learn a whole bunch.

I also like the thought of creating almost 100% black artificial "black body", by letting radiation enter a cavity with black internal walls. Since it will reflect back and forth most of the time, almost everything will be adsorbed and only re-emitted by emission.


But I think radiative equilibrium can be violated....
Julian

Free Energy | searching for free energy and discussing free energy

Re: 2nd law of thermodynamics
« Reply #6 on: May 08, 2010, 11:37:53 PM »
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