I should add that Martin Fleischmann also thinks
highly of Mengoli. He considers him one of the
world's top electrochemists. Mengoli retired several years ago.
At issue has arisen in the discussion of this
paper and in my earlier message "Rough comparison
of cold fusion Pd to UO2." The question is: to
what part of the cell should you normalize the
energy? The cathode? Or all of the parts on
inside the calorimeter walls? Or, at the other
extreme, do you only count the deuterium that is
consumed, which can only be estimated by measuring the helium produced?
As Stephen A. Lawrence pointed out, "the
palladium isn't the fuel, it's just a catalyst."
That is true. Or at least, we hope that is true.
Good arguments can be made for various way of
normalization, and the answer comes out very
different if you include lots of cell materials.
As a guide to scientific theory, obviously the
only thing you want to account for is the
deuterium, since that appears to be the only
thing which is actually consumed. But looking at
it as an engineering problem, it seems to me you
should account for the palladium because that is
what actually gets hot and will probably need to
be replaced sooner than any other component. It
is probably the limiting factor for performance.
Cathode materials are the main focus of research.
Mengoli and most others go with normalizing
against the cathode. He uses that comparison in
the caption and discussion for experiment 3, which I quoted yesterday.
Actually, with 1.4 MJ of output from a gadget the
size of the Mengoli's cell, it doesn't matter
what you normalize against! It has to be nuclear
reaction. But they and most other authors focus
on normalizing against the cathode. I go with the
cathode as well, or perhaps the cathode plus the
deuterium present in the cell. I mean the
deuterium only: the D2 gas in a gas loaded cell,
and with D2O, the deuterium only and not the
oxygen. As I see it, the oxygen in the heavy
water, the anode, the wires going to the
electrodes, and the other cell components
resemble the components of a fission reactor core
that do not participate directly in the reaction,
such as the zirconium cladding on the fuel rods,
and the moderator rods. You have to have them,
but only the uranium in the UO2 gets hot. It is
the site of the reaction, just as the cathode is
the site of the cold fusion fusion reaction.
You might count the oxygen in the UO2 as a
passive component, but I think that stretches it.
What I am thinking is that as a practical matter
-- or an engineering matter -- with a fission
reactor you have to remove and dispose of the
"spent fuel" UO2, as one thing. (Dispose of it,
or recycle it.) The oxygen doesn't actually
participate in the nuclear reaction, but it is
inseparable from the uranium, whereas the other
reactor components can be replaced on a different
schedule. Along the same lines, with a cold
fusion cell you will have to remove and recycle
the Pd cathode and remove the tritium and helium.
The Pd is somewhat analogous to the oxygen in
UO2: it is an inseparable part of the fuel that undergoes a nuclear reaction.
The palladium is also a useful comparison because
it is highly loaded and the number of deuterium
atoms that are available to participate in the
reaction at any given moment is roughly
equivalent to the number of palladium atoms, or
perhaps 70% of that number, because the cathode
will be unevenly loaded. The deuterons swap in
and out: new ones continually migrate in other
ones go out, but the total available fuel at any
moment is comparable to the number of palladium
atoms. I mean it is not a million times less or
10 times more than that, although obviously, the
number of deuterons that actually undergo fusion
in one day is many orders of magnitude fewer than
those available. Roughly 10,000 fewer, as Mengoli
shows in his discussion of heat after death, which is interesting.
After electrolysis stops, and heat after death
begins, the number of deuterons available to
undergo the reaction then becomes roughly equal
to the number of palladium atoms, so normalizing
against this number becomes more valid, or
scientifically meaningful, you might say. Mengoli
et al. address this issue in section 4.4
"After-effect" (which is what they call heat after death):
"4.4. After-effect
Power output in the absence of any power input is
the most remarkable instance of anomalous heat
generation obtained in this work. This
'after-effect' may be explained by the comments
made above for current: in o.c. [open circuit,
zero input power] conditions, 'confinement' by
the current vanishes and the 'activated'
Beta-deuteride phase finds itself in strong thermodynamic disequilibrium . . .
A second question concerns the nature of the
process sustaining the power output in o.c.
conditions for such long a time. Considering, for
instance, the data of Table 3 (exp. 3), there is
clearly no reasonable chemical explanation.
However, if heat is generated by D + D --->
4He+24 MeV fusion, steady power output of 1 W
involves the consumption of ~ 4.5 × 10E16 atoms
[per day]; but the sample used in exp. 3, loaded
to D/Pd = 0.8, as expected originally, contained
~5.7 x 10E24 atoms, which means that heat output
could indeed go on for a long time."
- Jed
