Hi Horace, sorry for the late response, my comments below. 2010/2/7 Horace Heffner <hheff...@mtaonline.net>: > > 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.
This would be correct if palladium, when driven as an anode, did dissolve in an alkaline electrolyte (they classically used LiOD in that M4 experiment, according to their original report at http://newenergytimes.com/v2/archives/1998epri/TR-107843-V1.PDF , thanks to Steve Krivit for the link), which it doesn't, see the Pd/H2O Pourbaix diagram at http://www.platinummetalsreview.com/jmpgm/data/datasheet.do?record=532&database=cesdatabase which shows that such corrosion only occurs in an acidic electrolyte (pH <3). >>> 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 "volcanos" you mention could also be "impact craters" produced by a local chain reaction on the surface. > 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. You may well have a point here. A ref for those deep alphas would be welcome BTW. >>> 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. Not alone I agree, it's more subtle than that, but the Ds do have to meet don't they? I submit that the Ds following/pushing each other down the lattice corridors like fish in a fish swarm have no reasons to experience frequent close encounters. >> 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. Good point, although the combined effect of their respective "colleagues" pushing from behind could conceivably result in many of the collisions being inelastic. In any case, surface or subsurface, we certainly all agree that something special occurs in the surface region, so the surface plays a determinant role in CF. Maybe we could collaboratively establish a list of what we know is special about the surface, here are a few items for a start: a/ only place where frequent D encounters are possible (as mentioned above) b/ adsorption heat is higher than absorption heat, i.e. the trapping potential for Ds is deeper on the surface than in the bulk (probably due to the surface Pds having dangling bonds) c/ place where the highest electron densities occur (due to b, can be enhanced by cathodic conditions), i.e. where electron screening is the most significant, i.e. where D encounters can be the closest d/ place where the Pd atoms are the most mobile e/ boundary region for the lattice wave functions (whatever this can imply) Comments/corrections/additional items welcome. Michel > 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/ > > > > >
