2009/10/30 Abd ul-Rahman Lomax <[email protected]>: >> Palladium hydride formation is exothermic >> (it releases around 11 kJ/mol heat for a loading factor of 0.8), so a >> hot cathode is to be expected, correct me someone if I am wrong. > > I think you are right, but I also think that with a codeposition cell, that > energy is very slowly released. Good point, though. That would be a source > of heat at the cathode. Odd that it would be in small spots, with a fully > loaded cathode with no more buildup of deuterium.
It would only be odd if, at steady state, deuterium stopped being absorbed everywhere on the cathode. But what happens is that due to the complex geometry of the surface, and to the presence of plastic behind the wire, some spots absorb a steady flow (thus releasing steady heat) while other spots desorb a steady flow (thus absorbing steady heat), and total absorption rate is equal to total desorption rate. So hot spots are still to be expected at steady state, I am afraid. My suggestion: like the SPAWAR people have done before you, forget about excess heat, hot spots, or other expensive to detect and ambiguous signatures, and concentrate on nuclear tracks. This will help make your kit cheaper i.e. more widely diffused, and the results less disputable... except by CF professionals who want to defend their favorite alternative LENR detection method of course, but the CF professionals are hardly the people you want to convince are they? Michel
