Notes continued from:
Dielectric Properties of Water and Water/Ethylene Glycol mixtures for use in Pulse Power
System Design by M, Zahn et al Vol 79 no.9 september 1986
previous url address was: http://www.rle.mit.edu/cehv/documents/35-Proc.IEEE.pdf:
symbols used in coaxial cables and in this paper can be seen here:http://www.rfcafe.com/references/electrical/coax.htm
you'll need to know what the symbols mean in order to understand this paper.
Page 1186, starts analysis of a water capacitor in the from of a coaxial line,
stans capacitor can be viewed as a coaxial line with water being the dielectric
between two metal conductors. So the information in this section should
be beneficial to understanding stans tubes. I would think it would be a necessity
to have some understanding of this in order to understand stans system.
The results of this section will apply to any transmission line form used in a
pulse forming network, they mention other types: stripline, triax, triplate
and coax. Fortunately the detailed maths they discuss is for coaxial line
which matches stans tubes ( or injector), though it does open the possibility
that other transmission line geometries might also be useable, I'm not familiar
with those other geometries ( stripline/triax/triplate), though triplate might
well be the form used by stephen meyer in his patent .
Just another thought: is stan using an external resistor as a load, its value
being such that the energy released is reflected back into the water capacitor?
Stans system as far as I can tell is not trying to maximise the amount of energy
that can be stored in the water capacitor but rather he is trying to maximise
efficiency of the conversion process.
For a coaxial line the maximum electric field strength is at the inner electrode.
( noted in the paper)
There are numerous formulas given that relate to stans tubes as a coaxial
transmission line and water capacitor.
Many of the formulas given should be useable, for example, equation 2.3 ( page 1186),
gives calculation of the line length ( tube length) for a given pulse time.
In the formula they use the round trip time, so the answer ( if you use the
formula 2.3) is divided by 2, take out that '2' from the formula!
So for a 1 microsecond charge time, the line length would be 16.8 metres, this is given
result shown in this research paper for water. So for a given charge time there is an ideal
length that I'm guessing maximises the efficiency of charging time. You could regard this
as one type or from of resonance.
( just another thought: is it possible stan uses the tubes in series in order to increase the
total length of the tubes? so although there is 9 lots of 12cm tubes ( approx), in series they
act like a long single tube and make it possible to match the pulse time with the ideal length
of the equivalent coaxial cable or at least to get some harmonic length of the resonant length?)
So lets put some numbers into the equation ( sorry I cant cut/paste the formula or use the
symbols etc), I'll try work out a way to do that later.
The time for the electrical wave to travel along the conductors ( arranged in coax geometry as
in stans tubes) for a one way trip (from one end of the tube to the other?) is:
time = sqrt ( epsilon * mu )
for round trip ( I'm not sure if this means reflected at the end of tubes/coax, or if it means
wave goes along the positive conductor then comes back down the negative conductor?)
time = 2 * sqrt ( epsilon * mu )
which gives equation 2.3 which is:
length = 1/2 * time / ( sqrt ( epsilon * mu ) ( equation 2.3)
which simplifies to:
length = 16770510 * time ( I've substituted epsilon = 80 )( answer in metres)
So it turns out thats a pretty simple formula to find the resonant length of stans tubes
for a given pulse time)
lets say the pulse on time ( charge time of the water capacitor) is 1 microsecond then
put 0.000001 seconds into the formula
length = 16770510 * 0.000001 = 16.8metres ( rounding off)
lets put in 10 microseconds we get 168metres
lets put in 0.1microseconds we get 1.68metres
lets say stan was using a charge time of 0.1microseconds or a resonant length of 1.68 metres,
lets divide that by 9 tubes in series and we get 0.19metres for resonant length.
Lets work backwards, lets say tube length of 12cm ( 0.12m), and 9 tubes in series gives
total length of 1.08metres for resonant length ( not including wires between tubes)
rearranging equation 2.3 for time gives:
time = length / 16770510
time = 1.08 / 16770510 =0.064 microseconds or 64nanoseconds
That seems like a very short pulse time, rather than using the charge time, since its step charging
in stans system, lets say this represents the pulse on time. Thats assuming the tubes in series you can
add them as one long coax cable.
Since this time is too short for stans frequencies, I would think that the tube length is something like
a 1/2 wave or 1/4 wave or 1/8 or 1/16 harmonic length. I've used a value of 80 for permittivity I think stan
used 79 but it wont make much difference to the answer.
Lets try with epsilon = 79
formula 2.3 becomes:
length = 16876319 * time
lets say 10khz pulse time is used, then on time is reciprocal which is 100microseconds ( 0.0001sec) ( assuming
the freq reciprocal represents the on time?)
using formula above harmonic length is :
l = 16876319 * 0.0001 = 1687m or 1.687km yikes!
no resonance there, even using the 9 cells in series as a long coax its still nowhere in the ball park!
I previously did some calculations ( using alternate method) gave some similar results but seems confirmed
using different method using formula in this research paper.
So perhaps the efficiency of matching the tube length to the pulse time is not particularly important in stan
system? In the injector being much shorter than the tubes matching the voltage zone as a coax cable arrangement
also seems not relevant.......seems that way to me?