It really would be interesting to have no increase entropy—no excess heat—associated with a reaction like Claytor describes.
Bob Cook From: Frank Znidarsic Sent: Friday, July 10, 2015 9:50 AM To: [email protected] Subject: Re: [Vo]:Re: LENR-forum: Claytor generates increased tritium with Brillouin technique The question how the mass loss associated with a fusion reaction is distributed as heat still seems unanswered. Does Claytor have any ideas? Requires a force with a longer range than the Coulombic. -----Original Message----- From: Jones Beene <[email protected]> To: vortex-l <[email protected]> Sent: Fri, Jul 10, 2015 10:05 am Subject: RE: [Vo]:Re: LENR-forum: Claytor generates increased tritium with Brillouin technique From: Bob Cook Ø Ø The question how the mass loss associated with a fusion reaction is distributed as heat still seems unanswered. Does Claytor have any ideas? There probably is little excess heat in this type of experiment. Almost every known neutron-free reaction which produces tritium is endothermic. Normal deuterium fusion should produce copious neutrons plus tritium, as Claytor acknowledges. He has recently started looking for thermal gain, according to the video, but will likely not see it without adding lithium-6 and making neutrons from changes in the parameters. Of course, the hot deuteron fusion reaction itself would work to fuse to tritium and produce thermal gain, as does the Farnworth Fusor which is not hot or cold but somewhere in between - but this is NOT what is happening for Claytor, or else both helium-3 and neutrons would be evident (not to mention the super-hot neutron at 14 MeV). Obviously, this must be a unique reaction – and as such will invoke other possibilities. The Lipinski model, mentioned here recently provides one theory (but it too produces neutrons). Bottom line - since Claytor has looked for neutrons and seen none (few) – even with Li-6, he is unlikely to see excess heat and tritium together. This lack of net thermal gain is also a function of the inefficiency of needing large pulse discharges. The most logical possibility for explaining this situation (if Mills is partly correct) is that tritium formation requires a fractional species of deuterium (dense deuterium or pychno) to be made in situ - which reaction has a lower threshold and does not have the normal He-3 and neutron reaction pathways. This is completely logical, since the other reaction branches (aside from tritium) are more energetic than a tritium-only branch - and it follows that only the least energetic branch would be seen at lower input energy. Jones
