Re: 2nd law of thermodynamics in question

In message <1163444775.906688.99910@xxxxxxxxxxxxxxxxxxxxxxxxxxx>, Paul <softwarelabus@xxxxxxxxx> writes
Richard Herring wrote:
In message <1163436256.228189.76090@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, Paul
<softwarelabus@xxxxxxxxx> writes
>jambaugh wrote:
>> Paul wrote:
>> > Any true 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 from thermal noise.
>> [snip]
>>
>> But no EE true or otherwise could build such a device. Your load being
>> themal will on average put as much energy back into the circuit as it
>> takes out. You can't prevent this without a perfect rectifier i.e. a
>> version of Maxwell's demon.
>
>
>Easy for deep thinkers. 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 >[previously corrected!]
>
Sure about that? ;-)

Yes, I am sure about that. Experiment #1. Did you see my typo
correction within minutes after posting? Perhaps you want to be a j---.

No, it should be obvious that when I posted I had not seen your correction.

>Think about it any you might figure out why some surfaces are cooler or
>hotter than other surfaces.
>
>In case you missed it, 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.
>
Experiment #1:
Glass of hot water standing in a bowl cold water.

Experiment #2:
Expanded polystyrene cup of hot water standing in a bowl of cold water.

Which experiment generates more heat flow? Answer: Experiment #1.

Think about it and you might figure out why some surfaces take longer
than others to reach equilibrium with their surroundings.

In case you missed it, the inner beaker in experiment #1 cools more
quickly than in experiment #2.


>The 2nd Law of Thermodynamics is incorrect.

Not from anything you've posted here.

You are incorrect. The answers just for you Richard -->

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

Doesn't take a genius, Richard, to understand experiment B radiates
more power.

And where did I suggest otherwise?

Doesn't take a genius, Paul, to understand my experiment B has a greater power flux than A.

Experiment B will be cooler than experiment A.

Until it reaches thermal equilibrium.

Experiment B tends towards equilibrium faster than experiment A. So?

Very simple
stuff Richard. That is direct proof the 2nd Law of Thermodynamics is
incorrect.

And that's what we call a non sequitur. Since, as you pointed out, there are many formulations of the 2nd Law, perhaps you should clarify matters by posting the one you are using, and then explain exactly which part of it you think is contradicted.

My cooling-water examples are exactly analogous to yours. Do you think _they_ contradict the 2nd Law?


Lets simply this for you Richard. Neither experiment A or B have a
power source except thermal noise.

Agreed.

Experiment B radiates more power.

Agreed.

If
you disagree then show the error in the math. 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.

Until it reaches equilibrium.

So?

Do I have
to draw you a picture?

Maybe you should draw yourself one. You need to include a heat source, a heat sink, and a thermal resistance. You can model it as an electric circuit, with capacitors for the source and sink, a resistor for the thermal resistance, voltage for temperature and charge for heat.

>Richard Herring wrote:
>> > P = 4 K T dF
>> >
>> > K is Boltzmann constant
>> > T is temperature in Kelvin
>> > df is bandwidth.
>> That 4 shouldn't be there.
>
>You are assuming the load generates ZERO noise.

No, I'm not. I'm just taking account of the _direction_ of power
transfer.

> Most loads, except a
>few such as radiation resistance, generate thermal noise.

Anything that turns electrical energy into heat, in fact.
Fluctuation-dissipation theorem.

So Richard Herring believes radiation resistance generates thermal
noise?

Strawman. Since when did radiation resistance turn electrical energy into *heat* ?

But you can certainly treat the radiation resistance as a noise source, if the antenna is coupled to other antennas being driven by thermal sources. Does the name Jansky mean nothing to you?

Of course, the noise is attenuated by the transmission loss between the antennas, so its effective temperature is usually very low.

>I provided a
>broad example by stating "appropriate load." If both the source and
>load generate thermal noise then the power across the load is 4 K T dF,
>but if the load generates no noise then the power across the load is K
>T dF.

I don't know what you think you mean by the power "across" a load. It's
either "into" or "out of".

"Power across a load" means the consumed power as measured across a
load.

Semantic confusion. You measure the voltage "across" the load, but measure (or infer) the current "through" it, and the sign of their product is significant.

--
Richard Herring
.

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