On Feb 7, 2010, at 4:42 AM, Michel Jullian wrote:
2010/2/7 Horace Heffner <hheff...@mtaonline.net>:
Two things to consider: (1) reversing the current *does*
"dissolve" the Pd
surface,
True, but extremely slowly I believe. A Pd anode is known to dissolve
relatively fast in acidic electrolytes such as D2SO4, but I don't
think that's what they used. It is doubtful whether they reverted the
current long enough to dissolve more than a few atomic layers.
I think the experimenters were competent. They knew what they were
doing.
Using a Faraday constant of 96,485 C/mol, and (conservatively) a
valence of 4, n for moles produced, I for current = .2 A, t for time
= 1 s, we get:
n = I * t / (96,485 C/mol * 4)
n = (0.2 A)*(1 sec) / (385940 C/mol) = 5.182x10^-7 mol
This means that at 200 mA/cm^2, 5.182x10^-7 mol/s is removed, or
3.12x10^17 atoms per second.
We also have for Pd: (12.38 g/cm^3)/(106.42 g/mol) = 0.1163 mol/cm^3
= 7.006x10^22 atoms/cm^3. The atomic volume is 1.427x10^-23 cm^3, and
the atomic dimension is 2.426x10^-8 cm. The amount of Pd removed per
second is (3.12x10^17 atoms per second) * (1.427x10^-23 cm^3 per
atom) = 4.45x10^-6 cm/s, or 445 angstroms per second. The number of
layers of atoms removed is (4.45x10^-6 cm/s)/(2.426x10^-8 cm) = 183/s.
If this is correct (highly suspect! 8^), then at a current density of
200 mA/cm^2 we have a thickness of 183 atoms removed per second, or
445 angstroms per second.
and (2) previous work has shown that helium production takes place
near but below the surface (order of microns),
while tritium production
tends to take place on or very close to the surface (within a few
atomic
widths).
I guess you mean they are *found* there, couldn't they be both
produced on the surface, only with more kinetic energy in the helium
nuclei (alphas) than in the tritium nuclei for some reason, so that
the helium is implanted more deeply? I find the idea of two different
nuclear reaction sites producing different products a bit unlikely.
No, most of the 4He reactions occur sub-surface. What do you think
produces a "volcano"? A surface reaction? The typical 4He produced
by CF does not have MeV kinetic energy, and is not surface produced.
If it were there would be massive alpha counts. There is not
sufficient kinetic energy to push alphas that deep into the Pd.
This has been a classic problem with CF, converting the process
into a bulk effect instead of a surface effect for all practical
purposes.
Maybe it's just not possible, because you can't make large D fluxes
collide head-on
Head on collisions, i.e. kinetics, can not possibly account for cold
fusion.
in the bulk, this can only happen at a significant
scale on the surface (desorbing vs incident fluxes). In the bulk, it
seems to me the deuterons just push and follow each other down the
lattice's concentration gradients, and never really collide hard.
Also, if Bose Einstein Condensates are involved, they requires cold
bosons for their formation. Head-on collisions may be a plausible
mechanism for deuteron kinetic energy removal.
This would only be the case if the collisions were almost all totally
inelastic. The only way that can happen is if they are fusions.
Michel
On Feb 7, 2010, at 2:58 AM, Michel Jullian wrote:
2010/2/2 Abd ul-Rahman Lomax <a...@loma xdesi gn.com>:
...
A single
SRI experiment has been published that made strong efforts to
recover all
the helium, and it came up with, as I recall, about 25 MeV.
That experiment was discussed in the paper submitted by Hagelstein,
McKubre et al to the DOE in 2004:
http://www.lenr-canr.org/acrobat/Hagelsteinnewphysica.pdf
They flushed helium out by simply desorbing and reabsorbing
deuterium
several times, by varying the cell current, which they reversed
in the
end to get all the D out.
It seems to me that if they actually managed to extract all the
helium
this way, which their resulting Q value suggests (104±10 % of 23.8
MeV), the reaction can't possibly happen in the bulk. Not even
subsurface. It has to happen exactly on the surface, with some
(about
half) of the produced helium nuclei going slightly subsurface. If
the
reaction itself was subsurface, surely about half of the produced
helium couldn't be recovered without more radical means such as the
one you suggested below.
...
2. Recovery of *all* the helium -- except perhaps for minor and
unavoidable
leakage, which should, of course, be kept as small as possible.
What
occurs
to me is to dissolve the cathode.
This seems a good idea.
I forget the best acid to use, but I do
know that palladium can be dissolved.
As I recall, Aqua Regia is the best for Pd.
Michel
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
