At 11:46 PM 7/3/2012, Finlay MacNab wrote:
Sorry, I fail to see why the voltage drop is 3kv across the acrylic
layer. Why is that exactly?
There are three regions involved, between the plates that are
connected to a high voltage supply, 6 KV.
There is the first cell wall, 1/16 inch (1.6 mm) thick, made of
acrylic plastic. Call this R1.
There is about an inch (25 mm) of electrolyte, which is not pure
water, but which has an "electrolyte" dissolved in it, lithium
chloride. Call this R2.
There is the opposite cell wall, the same as the first. Call this R3.
The resistances of R1 and R3 are roughly 1.6 x 10^14 ohms each.
That's a roughly calculated value from the properties of acrylic.
The resistance of the electrolyte, R2, is on the order of 100 ohms.
Easy to measure, routinely measured, voltages and currents are known.
Consider these three resistances in series, with 6 KV across the
assembly. Use Ohm's law to calculate the voltage across each resistance.
You will find that the voltage is equally divided between the two
plates, at about 3 KV per plate. The voltage across the electrolyte is low.
The current would be 6000/(3.2 x 10^14) or about 2 x 10^-11 amps,
that's 20 picoamps. The voltage across R2 would be 2 nanovolts. To
measure this if there were no other activity in the system would be
quite difficult. Microvolts are bad enough. But maybe you could do it.
However, there is electrolytic current added to the electrolyte by
the experiment. It might be a current at initial plating on the order
of a milliamp, the voltage would be under two volts at first. The
noise in the power supply would be well above the level of voltage
from the HV source.
Current flow through an electrolyte is complex, as you know. But we
don't need to go into that complexity. At steady state, DC, the
electrolyte will behave as a somewhat noisy resistor. (And at this
stage of electrolysis, the noise would be low, it gets noisier,
later, when deuterium gas is being evolved.) There is a parallel
capacitance, but it has practically no effect.
Bottom line: in his basic thesis, Rich is correct. The external
plates with a high voltage on them can be expected to have no effect
on the electrolytic activity. It looks like the SPAWAR team simply
overlooked this consideration, we could do a whole study on the
psychology of cold fusion; suffice it to say that this was a human
error, and an understandable one. What is surprising is that this
made it past peer review.
If the claimed experimental result were verified, we'd have to start
to look for some flaw in this argument. However, reading the paper, I
don't see that the result is clearly established even in the paper.
It's asserted without showing the basis of the analysis. It appears
to be subjective.
Now, these researchers had looked at a lot of cathodes. Variation in
cathode appearance can be great, depending on very subtle conditions
that are difficult to control. This is the big problem with the
electrochemical approach to cold fusion, it's extraordinarily
difficult to control the conditions.
However, how important is Rich's objection? In another post today,
Rich speculates about all kinds of fantastic phenomena that he thinks
might happen if the high voltage leaks through the plastic. I suspect
that he links this in his mind to some of the reported phenomena, but
he's made a huge error himself. He thinks, it seems, that the use of
an external high voltage field is common, such that it could explain
effects reported. No, that was pretty much an isolated experiment.
SPAWAR did not continue to use an HV field. This published paper was
simply a report of something that seemed anomalous to them, an effect
of an external electric field on codeposition morphology. It's a
hiccup in an avalanche of findings.
There is no leakage through the plastic. This plastic is not riddled
with "ionized radiation tracks." (It would be murky, not clear, and
those tracks would not stay ionized, Rich has confused the ionization
which is caused by charged particle passage, which disrupts the
plastic structure, with some sort of permanent ionization which would
facilitate current flow. No, that doesn't happen. The ionization will
resolve itself rapidly; after all, the plastic does conduct. What is
left is simply disrupted plastic. Same material as before. Same resistance.
(Rich is talking about background radiation ionization, accumulated
after the polymerization of the plastic. This would accumulate very
slowly, so even if it takes days or weeks for ionization to resolve
(which I doubt), it would nevertheless resolve. Experimental fact:
acrylic is an excellent insulator, and it stays that way for a long
time. You can bet your life on it, and these experimenters did, every
time they touched any part of that cell with the HV turned on. They
may have avoided that, and it is *this* effect that might explain a
morphological difference in plating, if indeed there was one!)
There is a nice picture on Wikipedia of a block of acrylic where the
breakdown voltage of the plastic has been exceeded and there was a discharge.
http://en.wikipedia.org/wiki/File:PlanePair2.jpg
(this is a different situation, a charge was built up *in the
plastic* through e-beam irradiation. But you can see the effect of a
discharge, damaging the plastic.)
3 KV across 1.6 mm of plastic is well below the breakdown voltage,
which is about 17 KV per mm. That "breakdown voltage" is a
specification that is used for safety calculations. It is always very
conservative. I would not expect to see an actual breakdown until
well above the specified voltage.
