It would not be fair to criticize the Lugano team for not measuring the gas in their experiment. Rossi lent them a reactor that was not designed for sampling the product gas. After the experiment, the seal had to be "broken open" (there are various descriptions of how this happened). To have measured the gas left in the reactor at the end, the whole reactor would have to be placed in a large ulta-high vacuum system with mechanical feedthru attachments designed to break the seal on the reactor while inside the UHV. This would have been a large undertaking. Having viewed their setup, it was probably considerably out of their budget and scope.
On the other hand, the Parkhomov-like experiments, particularly with the way MFMP has modified the seals, is well suited for gas sampling. It is highly desirable to sample the gas while the reactor is still hot and before much of the hydrogen (isotope) gas can be re-absorbed into the metal as LiH. This is in my experimental plan. Historically, credibly measuring He in the electrolytic PdD experiments was hard - you have to prove it could not have come from atmospheric contamination. When the experiment is performed in high alumina reactor tubes, as Alan Goldwater has shown, the gas pressure can be quite high -200-600 PSI. When sampled, the resulting pressure in the sample vessel will be less than in the reaction tube due to the volume of the sample container, but the sample pressure could easily be in the 30-100 PSI range. Finding a measurable percentage of the sample gas to be He could confidently be determined to be a reaction product rather than contamination due to the higher pressure of the sample container which could not have been produced from atmospheric contamination. Bob Higgins On Wed, Apr 8, 2015 at 8:44 PM, Jones Beene <jone...@pacbell.net> wrote: > Well on second look, at day 32, the internal helium pressure at 1200 C is > about 2000 psi if indeed the Lugano excess heat calculation was correct (it > wasn’t) which could arguably have been tolerated by the reactor. About 0.03 > moles of helium would have been produced at 8 MeV per atom to give the 1.5 > MW-hrs of dissipated excess heat, but as we know the Lugano excess heat > calculation was grossly inflated by the incompetence of the Levi team. > > > > If the COP was closer to 1.5 as I suspect, then there would have been far > less internal pressure from the accumulated helium – if lithium fusion was > responsible. Thus, lithium fusion is not ruled out by pressure > considerations. (but is ruled out by lack of gammas) > > > > *From:* Jones Beene > > > > Blaze- Disregard previous numbers. I’ll try to calculate the internal > pressure at day 30 another way. The point remains that if lithium fusion is > responsible for the gain, lots of helium needs to have been produced and > the reactor probably could not have tolerated the pressure. >
