Selfrunning cold electricity circuit from Dr.Stiffler

[nul-points]

Hi Sandy,

Wish you a new joyful year, Sorry if I offended you,

hi Elias

you're very gracious to apologise but you have not offended me

i've been priviliged to meet and/or study writings of some honourable engineers: over 30 years ago when i started work as a graduate engineer i was in the next lab to Eric Laithwaite - an inspiration to inquiring minds

when i was introduced to him he was testing one of his prototype MAGLEV systems, using it to fire pieces of steel lamination down the length of the track and into the wall at the far end of the room! he later became more famous for demonstrating anomalous OU-like gravity-altering behaviour of gyroscopes - something the 'Established' classical-engineering community couldn't accept and for which they never forgave him

i have also been inspired by the writing, research and patents of Harold Aspden (someone who is older even than me!) who has been involved in the proven OU/cooling behaviour of dissimilar metal junctions in cross-EM fields and who also produced some research in conjunction with Robert Adams (creator of the Adams Motor)

these are people who had classical engineering training and were not afraid to use it to good effect investigating anomalous behaviour in a variety of engineering disciplines

but I certainly think that we all must respect Dr Stiffler here, as for sure he is here to help us, and a very honorable man. 

i have no experience of Dr Stiffler so i could not comment on how honorable he is

all the best to you all with your research!
sandy

[Mr.Entropy]

In the mean time could you review my method of determining how much classical energy is being harvested with the DIAC collection approach?

Sure:

My method is to merely compare energy (in mJ) between what the storage capacitor dumps compared to how much energy was inputted in the same period of time.

Yes, that's right, but:

The capacitor dumps every 135 mS. The start voltage is 32 Volts the residual voltage is 22.44 Volts. This is about aÃ,  9 Volt differential. With a 200 uF capacitor this means a loss ofÃ,  5.22 mJ.

The energy stored in a capacitor of C farads with V volts across the terminals is C*V*V/2.Ã,  Because the voltage is squared, you can't calculate an energy differential from a voltage differential alone -- you need to use the two voltages:

E1 = 0.0002 * 32 * 32 / 2 = 0.1024 J
E2 = 0.0002 * 22.44 * 22.44 / 2 = 0.0504 J
E1 - E2 = 0.052 J

Since you get this 0.052J out every 0.132s, power

Pout = 0.052J/0.132s = 0.394W

In the same time period of 135 mS the Power supply is running at 15.06 Volts at an average current of 65 mA. This calculates to an energy input ofÃ,  132 mJ.

That's Pin = 15.06V *0.065A = 0.979W, which is more than the 0.75W you quoted above, giving a COP of 0.394W / 0.979W = 0.402.Ã,  Still quite reasonable.

Cheers,

Mr. Entropy

[Mr.Entropy]

Okay, how are you accounting for the duty cycle not being 50/50? Would it not be the same in the general sense as that of a squarewave calculation?

You don't need to account for the duty cycle in this calculation.

The duty cycle compensation you refer to arises from the calculation of average power using voltage and impedance.  Instantaneous power is V*V/R, and average power is:

P = \integral (V*V/R)dt / (t1-t0)

What we call the RMS voltage is:

Vrms = sqrt( \integral (V*V)dt / (t1-t0) )

And that makes:

P = Vrms*Vrms / R

as long as R is constant.  Vrms for an on-off waveform is:

sqrt( (Von * Von * Ton) / (Ton + Toff) )

= Von * sqrt(Ton/(Ton + Toff)).


But you're not calculating average power using voltage and resistance.  You're using energy and time, dumping Jc joules every (Ton+Toff) seconds, so you can just divide.  If you were calculating using voltage and resistance, you couldn't use the on-off duty cycle calculation, because you wouldn't be getting an on-off waveform out of the capacitor.  You'd have to do the integration.  If Ton << Toff, you'd find that:

\integral (V*V/R)dt = Jc

Note that the simple method of using the capacitor energy differential, Jc, neglects any energy delivered by the circuit during the discharge cycle.  Since you have Ton / T = 0.2, you might expect your answer to be short by 20% or so -- exactly how much depends on many things.  This error would be reduced if you made the duty cycle shorter.

Cheers,

Mr. Entropy

[Spokane1]

Dear Non-Funded Researchers

Please accept my apologies for the questions about the anomalous 7 degree temperature drop.

The instrument I was using to measure the transistor temperature proved to be defective and was reversing the readings. What I thought was a lower temperature was really a higher temperature. Thus the return to normal reading when the circuit was shut down or the probe moved off the transistor. This really got me confused.  Fortunately I had a back up meter at the barns that quickly displayed the error. I was reluctant to touch the transistor directly because I thought it might push it into an unstable condition.

My inexperience with temperature probes is showing.

This only shows me how easy it is to make gross mistakes and how much we rely on single instrument readings.

Well, I shall continue on.

Spokane1

[derricka]

@Spokane1

I too, have never heard of an entire transistor  cooling below ambient. In most cases, ohmic (resistance) heating would swamp any localized cooling. Transistors are normally expected to heat above ambient, often needing a heatsink to prevent overheating.  While there would certainly be a small  (Peltier) temperature difference between the PN junctions on the transistor die, the structural design of most transistors is intended to maximize the shedding of heat by means of thermal transfer to its case and leads.  In your scenario, If the Peltier effect is taking place, then 1 or more of the transistors leads would be getting quite warm or even hot. If the transistor case AND all leads are below ambient temperature, then something truly weird is going on! (According to traditional thermodynamics, heat can be moved, but not destroyed.)   Perhaps an infrared camera with a macro lens, if available from a lab or building inspector, would reveal the hot and cold spots in your circuit.


In the Peltier effect, when electrons flow from a region of high density to a region of lower density (like in a PN Junction), they expand and cool (like the freon gas in a refrigerator). The other side of the junction gets hotter.  The Seebeck effect is the mirror image of this, where a heat differential creates a voltage. Both of these effects are really just different expressions of the thermoelectric effect.