In reply to Mike Carrell's message of Sat, 25 Oct 2008 13:30:19 -0400: Hi, [snip] >This is unfortunate given that He+ is also a catalyst. > >MC: But He is not a catalyst, it used as a chemically inert heat transfer >medium. When the reaction fires, undoubtedly some He will be ionized and the >H atoms around, may contribute to the energy yield. That is not unfortunate, >it is just a sideshow.
The problem is that with He present, and given the temperature at which the reaction kicks in, it may not be a sideshow, it may the main game. e.g. suppose that the only role played by the NaH is to supply H atoms, and that the actual Hydrino forming reaction is the customary:- H + He+ -> H[1/3] + He2+ 54.4 eV ? The energy released by the reaction would create extra He+ from H. How could we tell the difference between this and the mechanism that Mills is proposing? As I see it, the only way is to remove the He altogether and just use H2 itself as the heat transfer medium. >> >>The reactions involved in the test cell are complex, and discussed on pp >>10-12, equations 23-35. The next-to-bottom paragraph of p11 is specially >>interesting. >> >>NaH apparently qulaifies as a catalyst because heating can intiate a >>reaction resulting in H[1/3] which is a hydrino catalyst. > >That is secondary. The primary reason it qualifies as a catalyst is that the >sum >of the three components of the dissociation energy into the specified >components >adds to 54.35 eV, which is a close match for 54.4 eV. > >Ah, but as a compound those electrons are in place. The riddle here is that >Na in a compound does not appear to manifest the required energy hole. The entire molecule manifests the energy hole, not a particular atom within the molecule, and even then it only does so if the molecule breaks up in a certain way. >The >molecule may thermally dissociate, with the H taking back it electron. Where >is the energy to ionize the Na as it separates from the H? All the energy comes from formation of the Hydrino (108.8 eV worth). >If Na can act as >a catalyst during the separation with only thermal energy, No, the Na is not the catalyst, the entire molecule is the catalyst. >then the >"resonant raansfer" phenomenon as used/described by Mills apparently has new >aspects. The only *new* aspect is that in this case it is an entire molecule rather than an atom or an ion that is acting as the catalyst. >Ignoring this detail, and regarding the H[1/3] rpoduct of the >reaction, then a 'conventional' hydrino catalyst has appeared and can act >with any H around. Agreed, and Mills even makes use of this to explain the preponderance of H[1/4]. (If you ask me that's a little far fetched. Disproportionation reactions should also produce species other than H[1/4], yet his experiments seem to show nothing below this. He explains this with a multipole radiation theory, which IMO is a bit weak. IOW if his explanation were valid, then I would still expect a few examples of e.g. H[1/6]. BTW if anyone's interested I have my own explanation for the preponderance of H[1/4]. > >>It still is not >>clear to me where the 54.35 eV for ionizing Na to catalyze H comes from. From the Hydrino as it forms. Many on the list seem to be trying to split the process up in an attempt to get a handle on it. (i.e break the molecule into atoms, then ionize the atom(s), then look for catalysts. The very act of attempting to do this "simplification" is responsible for the difficulty in understanding (because the pieces are not the catalysts - with the possible exception of the H itself). The original hole molecule is the catalyst. The hole molecule absorbs 54.35 eV when it breaks up a certain way. Since 54.35 is very close to 54.4 and thus a multiple of 27.2, it is a Mills catalyst. [snip] >To make a long story short, when the Hydrino forms, part of the energy >released >is stored in chemical form (Na++ etc.) and part is released directly to the >environment. The part stored in chemical form is then shortly (and >separately) >also released to the environment as per equ. 24. > >MC: That helps a bit, Robin, but where does the 54.45 eV come from? The >thermal input from the heater does not seem enough, and there is no >ionization field as in the microwave cell. Yet the DSC plot clearly shows >something happening. The 54.45 is not a "trigger", it's simply what is left over when you subtract 54.35 from 108.8. IOW it also comes from the shrinkage. IOW when the shrinkage reaction takes place, 54.35 eV of the 108.8 is used to break up the NaH in that "special" way, and the remaining 54.45 eV appears as e.g. kinetic energy (or possibly as UV). I suspect everyone is wondering "so what triggers the reaction" (I certainly am ;) [snip] Regards, Robin van Spaandonk <[EMAIL PROTECTED]>
