D14, Meyersova najnovšia aktualizácia, december 2007

[sano]

V tejto aktualizácii, p. Kelly popisuje spôsob ako ladiť trúbky v Meyersovom vyvíjači. Čo sa deje? Dávajú skoro všetko čo doteraz tajili. Nevie z Vás niekto?

[sano]

Uvádzam adresu fóra, kde sa dá konzultovať problém ku Myersovmu vyvíjaču.

http://www.overunity.com/index.php?topic=3079

[Návštěvník]

strojový překlad

[la]

Zdravim.Velmi pěkné.

[sano]

Najnovšia aktualizácia Meyersovho vyvíjača marec 08. Od stránky 101 až do konca kapitoly. Prepáčte, v originály je spoločne s Bobom Boysom.
Archív, z ktorého som doteraz čerpal na panaceauniversity.com, už prestal fungovať. Dúfam, že ostane aspoň toto.
http://www.free-energy-info.co.uk/

[martinq]

Kód: Vybrat vše

http://www.hotlinkfiles.com/browse/srawofni/33693

je nutno se zaregistrovat a kliknout v době kdy není překročení download ...


seznam souborů:

[sano]

[Vo]:Re: Sample numbers for resonant AC electrolysis
Horace Heffner
Sat, 01 Sep 2007 15:54:10 -0700




On Jun 29, 1999, at 11:42 PM, Horace Heffner wrote:

Reviewing the resonant circuit proposed for use with an AC electrolysis cell:

I1
--------------- V1
| |
| -------------
| | |
| | C1
AC L1 | I2
| | |
| | I3 R1
| | |
| -------------
| |
--------------- Ground

AC - AC source
L1 - Inductor
C1 - The cell capacitance
R1 - The net cell resistance
I1 - Input current (rms)
I2 - Cell current (rms)
I3 - Inductor current (rms)
V1 - supply voltage = cell voltage
Xl - Reactance of L1
Xc - Reactance of C2

Fig. 1 - Resonant circuit for AC electrolysis cell

When the operating frequency is at the resonant frequency for the tank
circuit L1, C1, R1, the net impedence of the tank circuitis maximum to the
AC source, thus the current through the cell I2 is at a maximum with
respect to the input current I1. In fact,

I2 = Q * I1 = I3

Where Q is given by:

Q = Xl/R1

and is a measure of the sharpness of the resonance peak. Since values of Q
over 100 are not uncommon in ordinary resonance circuits, this is
fascinating, and hints at ou behaviour all by itself, assuming the cell can be made efficient enough that heat from ambient becomes a large contributor
to the splitting reaction.

The cell is portrayed as a capacitor C1, though it is known that its makeup is really more in the form of a lattice where elements connecting nodal points consist of a resistor and capacitor in parallel. In addition, the cell is capacitively linked, so there is a set of capacitances around the electrolyte as well. A simplified model of the cell might look like the
following:



|
|
C2
|
-----------------
| |
| |
C4 R2
| |
| |
-----------------
|
|
C3
|

Fig. 2 - Simplified circuit for
capacitively linked cell

Here C2 and C3 are the capacitive interface to the electrolyte, either an insulator around the electordes, the tank walls, or a combination. The
elecrolyte, or slurry, is represented by C4 and R2 in parallel. For
purposes of simplifying the analysis, the electrolyte resistance R2 can be
rolled into the tank resistance R1, and we also have for the series
capacitances:

1/C1 = 1/(C2 + C3 + C4)

In other words, if R2 is large, if the cell walls and electrolyte all have the same dielectric constant, then the cell is equivalent to a capacitor the width of the cell, including electrolyte and plate insulators. If R2
is very small, then:

1/C1 = 1/(C2 + C3)

and we can treat the two insulators as an equivalent insulator as thick as
the sum of the two.

It is especially of interest that, as the cell area gets bigger, R2 drops,
yet the capacitance C1 increases, thus Q increases. This has the
appearance of some kind of economy of scale.

Now for some sample design numbers.

ASSUMPTIONS

10 cm x 10 cm plates (plate size)

V1 = 20 kV (resonant operating voltage)

Ke = 50 (dielectric constant of ceramic)

Ks = 1000 V/mil = 3.94 x 10^5 V/cm (dielectric strength)

F0 = 16,000 Hz (resonant frequency)

R1 = cell resistance = 1 ohm

CALCULATIONS

Total plate thickness D1 = V1/Ks = (2x10^5 V)/(3.94 x 10^5 V/cm) = 0.508 cm

Using 50 percent margin, D1 = .762 cm

Area A = 100 cm^2

The ratio A/d1 = (100 cm^2)/(0.762 cm) = 131

Capacitance C1 = Ke (A/d1) (8.85 x 10^-12 F) = 5.80x10^-8 F

Impedence of capacitor Xc = 1/(2 Pi F0 C1) = 1/(2 Pi (16,000) (5.80x10^-8 F))

Xc = 172 ohms

Current through cell I2 = V1/Xc = (20,000 V)/(172 ohms) = 116 amps

Knowing energy of inducatance is equal to energy of capacitance we have:

L1/C1 = (V1/Ic)^2 = (20000/116)^2 ohms^2, so we have

inductance L1 = (2.97x10^4)(C1) H = (2.97x10^4)(5.80x10^-8) H

L1 = 1.72 mH

Inductive reactance Xl = 2 Pi F0 L1 = 2 Pi (16000) (1.72x10^-3) ohms

Xl = 173 ohms

Q = Xl/R1 = 173/1 = 173

Thus we have the driving current:

I1 = I2/Q = (116 A)/173 = 0.67 A

And driving power:

Pd = I1 * V1 = (0.67 A) (20,000 V) = 13.4 kW

Apparent power:

Pa = I2 * V1 = (116 A) (20,000 V) = 2.32 MW


Horace Heffner
http://www.mtaonline.net/~hheffner/





[Vo]:Re: Sample numbers for resonant AC electrolysis Horace Heffner
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[sano]

http://www.alexpetty.com/2010/11/17/mey ... mment-2231

malo by tam byť všetko. Autor je skúsený konštruktér vodíkových vyvíjačov. Jeho replika bola overená viacerými členmi na overnunity.