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
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