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/
