If one is working with a quadrapole mass spec, and especially a small one like an RGA it will be impossible to devolve the peaks of 4He and D2. Only by being certain that little D2 is present by trapping it in a cold or getter trap on the way to the mass spec can one ever be certain that the sample is 4He instead of D2. The practice is clearly informative as one learns the operation of the RGA with and without the cold trap.
Don’t be fooled by imaginary methods with such close masses. No one will ever believe 4He in such an instrument without proof that the D2 signal is suppressed. From: Jürg Wyttenbach <ju...@datamart.ch> Sent: Monday, July 15, 2019 3:34 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:If Mizuno is correct, this design is likely tobetheprecursor to all future devices For 100kw/h about 1.2mg of deuterium are needed. If pressure is lower then the relative density of D (D2 gas) increases, somewhere between 0.15 & 0.45g/l.The inventory is given by Ni/pd surface bound D, the volume (15l) of the reactor and the pressure factor (=0.003 for 300Pa) . But Mizuno recommends to always let the bottle attached and of course he did feed additional deuterium if needed. Holmid does not yet produce any energy. First he must avoid to mainly produce positive muons... Jürg Am 15.07.19 um 15:56 schrieb JonesBeene: Reality Check. Surprisingly, nuclear fusion of deuterium into helium seems NOT sufficiently energetic to account for the Mizuno claim of heating his home. Mass is apparently being converted into energy, but how? And what are the ramifications of such a low reactor inventory of deuterium gas? The main contenders for excess energy production would be: 1) D+D -> He 2) Deflation of electrons – i.e. the Millsean approach 3) Disintegration of deuterons into muons – Holmlid’s theory - which is far more energetic than fusion in terms of entropy per unit of mass 4) Sequential Coulomb explosions from cluster formation –hypothesis from Hora, Miley etc. 5) Any combination or permutation of the above If fusion of D into He is your choice - then one gram of fused deuterium yields 10^12J (one terajoule)of energy, but when based on the low operating pressure of 100-300 Pa (100 Pa = .001 bar) and the need for low metal loading, as stated in his paper - that set of factors represents a tiny fuel inventory, such that when completely fused into helium would generate about 278 kilowatt hours of equivalent heat. If Mizuno was producing close to 3 kW continuous to heat his house in a Sapporo winter, he could run it for only about 100 hours without a refill if the gain was from fusion and the inventory was at the low end of his specs. At any rate, if the gain was from nuclear fusion only - then almost all of the deuterium would be converted, and the helium ash should be easily measurable. There should be no need for a cold trap to increase the helium ratio – the residual gas after less than a week should be almost all helium, no? Even if these calculations are off by a large factor, the helium content should be obvious. IOW – in the naïve assessment of the breakthrough claim of Mizuno – specifically the heating of his home – after 100 hours or so of operation, there should be a whopping milligram of helium and little deuterium in the reactor to measure. In contrast – Holmlid’s theory proposes deuteron disintegration (with inadvertent fusion). His theory suggests that about 4 GeV of mass-energy per every two atoms of deuterium lost could be converted into energy. This is about 150 times MORE potential energy per unit of mass (converted into energy) than can be derived from fusion into helium. On the surface, then – fusion of deuterium into helium appears to be too weak a reaction to account for the Mizuno claims of heating his home, and only the Holmlid effect would have an adequate output. Why isn’t the Holmlid effect the favored hypothesis? Jones -- Jürg Wyttenbach Bifangstr.22 8910 Affoltern a.A. 044 760 14 18 079 246 36 06
