At 12:10 AM 3/12/2010, Rich Murray wrote:
1. Perhaps Earthtech and others like Ludwik Kowalski would like to
join Lomax in developing a simple, low-cost standard version of the
SPAWAR co-deposition cell.
Common in the history of CF is lack of exact replication
of these always surprisingly complex devices,
confounding any resolution of the fundamental issue of
clearly confirming nuclear reactions.
I announced the kit concept here at the end of August last year, and,
September 3, started a yahoogroup to help develop
ideas,
http://tech.groups.yahoo.com/group/coldfusionproject/messages. Most
of the discussion has, however been here and in private
correspondence that led me to come up with specific details of
experimental design. I was advised by Steve Krivit to avoid "design
by committee," and, since I'm putting my own time and funds into what
I personally do, I've been making decisions on my own responsibility.
However, the "kit company," which is just continued operation of
Lomax Design Associates, will attempt to assist anyone who wants to
work on similar concepts, that is, standardized experimental kits,
the basis of which is clear and highly specific design, where every
detail is covered.
The basis of the kits I'm preparing is the Galileo protocol, using a
gold cathode wire, and with quantities reduced proportionally to wire
length, which will be roughly half that of what was used (and not
precisely characterized, unfortunately) in the Galileo project. It
looks like I'll be starting with LR_115 detectors, for practical reasons.
2. [ " LR-115 is a simpler material for this purpose;
the active layer is 6 microns of cellulose acetate, red in color.
Radiation damage to the LR-115 causes, then, after etch,
a hole to appear completely through the red layer,
which is carried on 100 microns of polyester,
the hole appears as a bright dot.
It's much easier to read,
and it will read higher track density than CR-39, allegedly." ]
Video imaging chips are cheap,
with megapixel resolutions at above 30 fps.
Sure. However, I bought a Celstron microscope, and I'll be modifying
the stage to hold a cell (with the microscope laid down, so the
microscope is looking at an upright cell -- or the contents would
spill! -- through the side, to where the cathode wire will be. It now
looks like, for the first runs, the cathode wire will be against the
edge of the cell, with the SSNTD being on the outside. I will have
some of the cathode wire visible, extending beyond the edge of the
SSNTD. It is possible that bright spots would be visible through the
SSNTD (it would definitely be possible with CR-39, that's one reason
why I want to move to CR-39 when I can, it will be enough for a first
pass to see if I can see anything at all on the active cathode,
particularly in the dark. I do not know if the response time of the
digital camera that is in the Celstron microscope will be fast
enough. I'd be happer, probably, with a regular film camera
accumulating light the way that the SSNTD accumulates tracks. But
maybe the digital camera will be fast enough.
For me, *simple* is very important.
If a thin radiation sensitive phospher screen is put on
a horizontal video chip, then a LR-115 SSNTD film on top of
that, and a 1 micron gold cathode film,
[ or a gold screen cathode with a mylar layer below it
to shield the LR-115 film ]
with a clear box on all four sides,
then a 1 cc cell could be cheaply made, allowing the device
to record the cumulative spatial location of pits on the LR-115,
and real time recording of visible light and IR (from hot spots),
to give both high time resolution of both hot spot and
radiation events -- which could be shared real-time with
the world on the Net.
LR-115 doesn't show radiation evidence until developed by etching,
same as CR-39. However, if the spots are bright enough, yes, they
would show through the red cellulose acetate. CR-39, in the end, will
be better, if the right material is obtained. Landauer isn't easy to
buy from, and they seem to have only 1/16 in dosimeter chips, which
are way too thick, I'd say. There is a project for someone to work
on: obtaining good CR-39 that is thin and characterizing it as to
background radiation and response.
The kind of detector sandwich described is certainly of interest, but
is more complex than what I can take on at this time. Note that, as
described, the active surface of the gold would be away from the
phosphor screen. Horace Heffner, here, suggested another approach,
which would have a photofabricated cathode with holes in it too small
for the liquid to escape through at the pressures involved. The
active surface would be viewed edge-on, light, including EUV that
some theorize may be present, could pass through the holes. Again,
great ideas and some might even be attempted later, but too complex.
The Galileo protocol was very simple, and I'm adding to it very cheap
and very simple monitoring.
Besides looking through a microscope -- which is a USB video camera
out-of-the-box -- I'll be monitoring the output of one or two
piezoelectric sensors attached to the cell. SPAWAR actually
constructed an electrode by plating a piezo sensor with silver, I
think it was. Again, I'm not doing that, I want to use a simple wire
cathode, so I won't be as sensitive as the SPAWAR configuration.
But I should do as well at detecting neutrons as they did, or better.
I have some Boron-10 neutron converter material, and will be, in
addition to checking for proton-knock on tracks, looking more
directly for neutrons as converted by Boron-10 to alphas.
The problem with using direct electronic radiation detectors is that
they are far more subject to noise, and placing the actual detector
surface very close to the cathode is difficult. That's why SSNTDs
are used. In this case, there will be, as with the SPAWAR wet
experiments, 1/16 inch of plastic between the cathode and the
detector surface, the one important for neutrons. Which is, in the
wet configuration, the "back side" of the chip. In mine, dry
configuration, it will be the front side of an LR-115 chip. To put
this in perspective, though, there may be, in a small area, maybe one
neutron-caused track per minute. For a lot of minutes... that's quite
a low flux, but it's way above background here because we are talking
about that flux appearing in a very small area, smaller than any
standard electronic radiation detector. But maybe someone knows of
something that could be used. I'd sure like to know how the radiation
varies with time and not just have the integrated result. I'm
thinking of replaing the detector chip at times during the
experiment, that's possible with the dry configuration.
A cheap webcam close above the cell can focus down
with about 10X magnification, to record the actual Pd
deposits from electrolysis, and any flashes of light from
micropit formation.
Such a multifunctional, low-cost device might be marketable for
many other types of research.
Sure. My Celstron microscope was $200, but it's way cool all by
itself. http://www.celestron.com/c3/product.php?CatID=81&ProdID=516
I also use it for inspecting etched SSNTDs.
3. The topic of micropit formation from recombination
of O2 and H2 on the Pd and gold cathode deserves research
in its own right, as well as being a factor that has to be
studied as an obvious major alternative to nuclear
reactions. At the end of this post, I offer an 1998 estimate
of the available chemical energy density.
I'm going to say right off that I'm very skeptical about serious
recombination taking place in these cells at the cathode. Gas is
being evolved from the cathode, under the surface of the D20,
deuterium gas from the cathode, oxygen from the anode. While there
would certainly be dissolved oxygan in the solution, substantial
presence of oxygen bubbles at the cathode, the important site for
"micropits" will be low, because the gas flow will be away from the
cathode. If a tiny bubble of oxygen manages to mix with a tiny bubble
of deuterium, this will produce a tiny bubble of explosive misture,
perhaps. Something would have to ignite it, deuterium and oxygen
don't spontaneously burn. Okay, it ignites. A tiny microbubble
explodes, releasing how much heat? Looking at total heat that might
be evolved over a relatively large surface doensn't explain how this
heat could be sufficiently concentrated, below the surface of a
liquid with low boiling point, to melt palladium (the issue here, not gold).
Basically, it's such a stretch that I'd prefer to hypothesize ... low
level low energy nuclear reactions. Matter of taste, I suppose. I
think the neutrons might tall us something, likewise the apparent
nuclear transformatinos that have been so often reported, and, killer
result, when excess heat is being measured accurately, and helium
captured and measured, helium correlated with excess heat has been
found by multiple groups, in range to be explained by nucleosynthesis
of helium from deuterium. Some don't like to call that fusion, but a
rose by any other name....
What's so important about heat/helium correlation is that in these
experiments, when they don't find excess heat, they don't find
helium. When they find excess heat, the more excess heat they find,
the more helium they find, with high reliability. Rich, I'm sure you
realize the implications. You can impeach excess heat, just claim
calorimetry error or Shanahan's calibration constant shift or failure
to stir or the imagination is unlimited. You can impeach helium
results because even the best results are not very far above ambient
air, so maybe there was leakage. But why, accross many experiments,
would the two results be correlated?
This is the smoking gun that was supposedly demanded in 1989: fusion
product, where is it? In amounts that explain the heat? It was there
early on, Miles' work was published, when, 1994 or so? Yes. Fusion
without neutrons or gammas. Supposedly impossible. But not
impossible, that was simply poverty of the imagination. Imagine 4
deuterons finding themselves in a tetrahedral configuration confined
in the lattice. Perhaps they are actually deuterium molecules, which
means they have their electrons bound with them. They form a
Bose-Enistein condensated, which is predicted, from quantum field
theory, to collapse and fuse within a femtosecond. Getting four
deuterons in that space is very difficult, the space doesn't have the
energy to confine them for long, it would rupture. Probably it's very
lucky that it's very difficult, or else some labs might have
vaporized. That's only one theory, certainly unproven, but by no
means "impossible" on the face. What would happen if there were
double molecular confinement? A single molecule would be rare and
would do nothing, the molecule, after a time, would dissociate,
losing its electrons to the lattice. Two would be more rare, but if
Takahashi's math is correct -- and nobody has yet even attempted to
refute it as far as I've seen, and he's been published and cited,
it's one of the more notable extant theories -- two wouldn't last
very long! And then, a Be-8 nucleus is formed, with four electrons
that probably don't stick around. The Be-8 nucleus is highly excited
and radiates energy as photons for the period until it decays, as I
recall the half-life is also about a femtosecond. It forms two alpha
particles carrying at least about 90 keV, which represents the Be-8
ground state decay. If it decays faster, they may have more energy,
theoretical maximum 23.8 MeV each. But practically all of the energy
released ends up as heat. And that's what's found experimentally,
once one factors in that some of the helium won't be detected unless
extraordinary measures are taken.
Sure, hold on to the idea of chemical recombination. But don't be so
optimistic in your calculations of how much energy could be released
by chemical recombination, without having any mechanism to explain
how that much energy could become concentrated, underwater, in a very
small space. An exploding bubble could only release as much energy as
the chemical energy of the bubble itself. You were looking at older
work, possibly with larger pits. The pits I've seen in SPAWAR
publications seem to be around 10 microns across. To concentrate its
chemical energy into melting the palladium, the bubble could be no
larger. How much energy is contained in a 10 micron bubble of
deuterium/oxygen, even assuming somehow there was optimum mixture?
The optimistic analysis seems to assume that all the oxygen and
deuterium liberated in the run are available for these explosions,
whereas by far most of the oxygen escapes the cell and most of the
deuterium is initially absorbed, as plating proceeds, and then,
later, with overcurrent, most is evolved and escapes. It would be
possible in a more complex experiment to capture and burn this
escaping deuterium to recover heavy water, and recombination cells do
this, and if the recombiner is a fuel cell, recombination rate could
be continuously measured.
Bottom line, I doubt that much oxygen is available at the cathode.
Rich wrote this:
Now, the results are the same if we have one 0.1 cm3 O2 bubble,
or a million bubbles of size 10E-7 cm3, spread out randomly over
the 30 day run, about 2-3 event/sec, creating the same total of
10 mg melted Au.
That's preposterous, Rich. One large O2 bubble will create a maximum
amount of heat in a very short time, if it reacts with the deuterium.
Maybe that could do something, though I'm still skeptical, how you
would channel the heat down to the palladium isn't stated. Most
likely it would break the cell, without melting anything at all.
Little firecracker. But a million tiny bubbles, "spread out randomly"
would each generate so little heat that *in the environment,* which
is, after all, very intimately water-cooled, it would only raise
overall temperature and probalby no melting effect at all would be visbile.
If I had time, I'd take the energy from a 10 micron bubble at
atmospheric pressure and see how hot that volume of palladium would
get on ignition. That would be a temperature that would be reached
only if energy transfer were 100% efficient. All the calculations of
total energy into the cell are basically hand-waving. Most of the
energy going into open cells ends up as oxygen in the atmosphere and
only the early deuterium is retained, later deuterium escapes from
the cell and the increased electroysis currents are merely to create
disequilibrium in the lattice, deuterium flow. Until codep is
complete, almost all deuterium is immediately absorbed, it would
seem. Oxygan can't mix with the absorbed deuterium, not to any
significant degree, unlike deuterium, it's not soluble in palladium.
So all that is available for chemical reaction with oxygen is a
little deuterium at the surface or in the solution, and likewise only
a little of the oxygen sticks around. Later, when deuterium is
bubbling out, almost all of it is escaping, for sure. In an open
cell. In a closed cell, there will be excess oxygen in the solution
and atmosphere, corresponding to the absorbed deuterium, but if
pressure is being kept low, i.e., the recombiner is working, that
energy is recovered, only the relatively low concentration of oxygen
from the sequestered deuterium would be floating around.
(I don't know details about that, and I worry about it. Basically,
I'd expect oxygen pressure build-up in a closed cell, corresponding
to the absorbed deuterium not available for recombination, but my
impression was that it doesn't. What do they do with the oxygen? If
the oxygen corresponding to the absorbed deuterium were free in the
cell, I might expect the pressure to be high. That deuterium in the
palladium is at very high effective pressure, -- like deuterium
metal! -- but perhaps the volume of palladium deuteride is low
enough, compared to cell volume, that the oxygen pressure remains pretty low.)
But these are open cells, and much of what I've seen pit photography
on has been open cells, co-dep cells are usually open, if I'm correct.
I'll say one more thing before I leave Rich's text below with his
calculations. I've thought of putting in something to segregate the
oxygen and deuterium. I'd like to channel them out and run a fuel
cell with them, there are very cheap fuel cells. This would provide,
from the current generated, a direct measure of evolved deuterium,
which, when compared with integrated current input, would provide a
measure of loading. And would recover some of the deuterium oxide,
which is the most expensive thing in the cell. However, "very cheap
fuel cell" is probably about as expensive as everything else in the
cell. I want to keep cell cost below $100 at "retail" pricing.
Actual cost to experimenters might well be lower if there are
donations to subsidize kits. But I wanted to set this up so that it's
totally self-supporting requiring no donors other than those who
directly donate to experimenters. I have received a substantial
donation, and an interest free-loan, that has covered a good chunk of
my costs so far, maybe half. I doing this on a shoestring, and that's
probably good, because the cells *must* be cheap, and cheap means
that many of them can be run. The design that I sell as kits will be
fully documented. No trade secrets, really, except maybe some
production methods that simply save a little time. I.e., I might make
a hundred cells with the work you might put into a few. But
hopefully, if you follow the plans that will be available, the only
difference between a "fully independent replication" that you might
do following the designs would be exact material batches, which,
hopefully, won't make a difference with codeposition though it
certainly made a difference with sold palladium Fleischmann cells.
To emphasize, the primary experimental measurement in the planned
experiments is of neutron radiation, as seen through secondary tracks
produced in SSNTDs. I believe that the primary reaction likely
produces alpha radiation, but the range of energies produced is quite
unclear and measurement of that radiation is tricky because if you
put the detector up close and personal with the cathode, there is the
possibility of chemical damage, and if you back it off, perhaps with
a mylar film, you block most of the alphas. But a layer or mylar or
other plastic will, with neutrons, generate knock-on protons and
other kinds of tracks. Get lucky, some triple-tracks from C-12 breakup.
So why neutrons? Aren't I expecting that the primary reaction doesn't
produce neutrons? Well, the primary reaction produces some hot
reaction products, which could cause ordinary fusion with the handly
deuterium sitting around in abundance, and the primary reaction isn't
the only thing going on. As a possible example, the BEC Takahashi
theorizes is neutrally charged and might fuse with stuff. Some
tritium is apparently produced. Neutrons, it appears, are produced in
secondary reactions at low rates. But, being penetrating, can be
detected where alphas can't.
Besides, neutrons are sexy. Don't you think so?
(No more original text below)
Microbubbles of O2 reacting with the D in 1 to 1 ratio absorption in
rough complex Pd or Au surfaces will generate enough heat to melt
the Pd, thus creating complex foamy microstructures.
Ohmori little lily theory details: Rich Murray 1998.06.17
June 17, 1998
Hello all, The report in May, 1998 Fusion Technology by Ohmori,
Mizuno, and Enyo describes 7 to 30 day runs at 1 to 3 A
on 2.5 to 5 cm2 Au electrodes in 0.5 M Na2CO3 and Na2SO4
H2O electrolyte, from a Pt anode.
producing after a few days up to ~1 mg mostly Au precipitates, and
leaving myriad little lily volcano-like or ear-like foam structures on
scraped (rough) sites on the Au, as large as 20 microns wide and 30
deep, with detected Pt, Pd, Ni, Os, and Ti, and other elements, with
claimed isotopic ratio anomalies.
I am disputing their claim that the precipitates and spots are evidence
of low energy nuclear transmutations, and suggesting a chemical
reaction theory, namely that the most abundant and obvious and
reactive chemicals present, naturally enough, H2 and O2, are
recombined at the cathode.
I don't know how much the Au will load with H2.
However, Pt, Pd, Ni, Os, and Ti will naturally be electrodeposited as
concentrations at any tiny rough spots, and then will both load with H
and catalyze the swift reaction of that H with any tiny O2 bubbles that
are also attracted from the anode to attach to the rough spot.
The bubble and the spot will heat up quickly, so quickly that there is
little time for heat loss by radiation , conduction, or convection
at the Au-H2O interface.
As the Au heats and softens, the contained H will build up pressure
and expand it like popcorn, creating a popped blister of frozen foam,
expelling some of the metal, and leaving the impressively ugly little
lily vocanos.
The process would tend to reoccur at the thus even rougher spot,
building up a cluster of lilies of various sizes, as is shown in Ohmori's
dramatic images.
I will calculate the details for a 0.1 cm3 amount of O2.
Au melts at 1063 degrees C, 1336 degrees K.
The molar specific heat Cm = 26.9 J/mol degC.
For Au, 197 g/mol 5.08X10E-3 mol/g 19.32 g/cm3
9.81X10E-2 mol/cm3 10.2 cm3/mol
To heat from 27 to 1063 deg C, a delta of 1036 deg C,
takes heat (1036 deg C)(26.9 J/mol) = 2.79X10E4 J/mol,
and to melt takes
1.27X10E4 J/mol, known as the molar heat of fusion.
These conveniently add up to 4.06X10E4 J/mol,
or 40.6 KJ/mol to heat and melt the Au
That certainly sounds like a lot!
Now, we get the moles of O2 in the 0.1 cm3 O2:
n = PV/RT =
(1 atm X 10-4 L)/(8.2X10E-2 atm L/degK mol)X(300 deg K) =
4.065X10E-6 mol O2. That's not very much.
We know that one mole O2 reacts with 2 moles H2, and may as well
assume with 50% loading that the H2 is held within 4 moles of Au.
The reaction is 2 H2 (g) + O2 (g) -> 2H2O (g), and the enthalpy is
2 X 241.8 KJ/mol = 483.6 KJ/mol.
So the enthalpy released is
Ec = (4.065X10E-6 mol)X(483.6 KJ/mol)
= 1.97X10E-3 KJ = 1.97 J.
Now, 2 J is the energy from 1 A at 1 V for 2 sec.
Note: this is the range that heats W to incandescence in a flashlight.
The moles of Au heated and melted by this heat are
Nm = (1.97X10E-3 KJ)/(40.6 KJ/mol) = 4.85X10E-5 mol
and the volume of Au melted is
Vm = (4.85X10E-5 mol)X(10.2 cm3/mol)
= 4.95X10E-4 cm3, which, assuming for convenience a cube,
has a width .791 mm, and
mass Mm = (4.85X10E-5 mol)X(197 g/mol) = 9.56 mg,
or ten times the maximum precipitates found by Ohmori
after 30 days of electrolysis at up to 3 A and a few volts,
an input energy for 2.592X10E6 sec, if at 5 V and 3 A,
of 38,880,000 J. So the 2 J to create 10 mg of melted Au
is a most minute fraction of the available input energy.
Now, the results are the same if we have one 0.1 cm3 O2 bubble,
or a million bubbles of size 10E-7 cm3, spread out randomly over
the 30 day run, about 2-3 event/sec, creating the same total of
10 mg melted Au.
These million bubbles would as little cubes have widths
.004641 cm = 46.4 micron, about the right size for our little lilies.
Each of these events would have an average energy of 2X10E-6 J.
It should be possible to detect IR, visible, and UV radiation, and
acoustic signals, about 2-3 event/sec.
Another test would be to use an anode which does not contribute
Pt, Pd, Ni, Os, and Ti, and in contrast, to use an anode
enriched in these metals.
Also, a barrier could be used to prevent O2 bubbles from
reaching the cathode from the anode, and in contrast,
positioning the anode to maximize O2 bubble transfer.
_____________________________________________________
Rich Murray, MA
Boston University Graduate School 1967 psychology,
BS MIT 1964, history and physics,
1943 Otowi Road, Santa Fe, New Mexico 87505
505-501-2298 rmfor...@comcast.net
http://groups.yahoo.com/group/AstroDeep/messages
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group with 142 members, 1,588 posts in a public archive
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_____________________________________________________
