In reply to Eric Walker's message of Sat, 24 Nov 2012 15:21:23 -0800: Hi Eric, [snip] >On Nov 24, 2012, at 14:47, [email protected] wrote: > >> Pointless line of thought, and no it wouldn't be elastic. (The ball in the >> pinball machine heats up the glue). ;) > >I'm sure you're right. I was just thinking that if the majority of the heat >energy was confined to the D nuclei, then dissipation of energy from the >system would not be too great, and there would be occasion for a large number >of collisions with varying levels of energy -- I think you were saying that a >large number of collisions would be needed to get a fusion.
Not just a large number of collisions. Ron's theory is specifically about D nuclei that approach a Pd nucleus to within about 100 fm (i.e. the size of the K shell). So it's not just about collisions, it's about two D nuclei being at a distance of 100 fm from a Pd nucleus at the same time. (Mind you I think the chances of this are about on a par with a snowflakes hope in hell ;) (Though I suspect that a head on collision between two 20 keV D nuclei in free space would be almost as useful). Note that only the 20 keV D nuclei can get that close, and they have to get lucky and penetrate to the K shell before they lose too much energy ionizing Pd valence electrons. The question that needs to be answered is: If a 23 MeV alpha from the fusion reaction creates 100(?) 20 keV deuterons, then what is the chance that two of them will arrive at the K shell of the same Pd atom concurrently? (They don't hang around for very long.) > >If the collisions are inelastic, then I understand there will be energy >transferred to the lattice. But the size of the substrate atoms is large >compared to the deuterons, so perhaps the treansfer of energy would be >minimal? A deuteron is a charged particle, and as it passes through another atom, it disrupts the (mostly valence) electrons of the atom it's passing through, and ionizes the atom to varying extents. This costs energy which comes from the kinetic energy of the passing charged particle (in this case a deuteron). Consequently, every atom along the way that gets ionized decreases the energy of the passing deuteron. BTW the same thing happens to the fast alpha from the fusion reaction. Later these ionized atoms will retrieve free electrons and release photons of various wavelengths. Thus the energy of the reaction is spread across many atoms in the lattice as "chemical" energy which is ultimately converted (degraded) into heat. BTW2 This may also tie in with Paul Brown's device. If the ionized electrons have kinetic energy of their own that is well in excess of the ionization energy, and a strong magnetic field is in place, then you may get cyclotron radiation as the electrons "head for home", i.e. back toward a positively charged ion. If you can find a way to synchronize this cyclotron radiation and capture it, then you have a means of converting fast particle energy into electric power, without using heat as an intermediary. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
