From: Dieter Britz <BRITZ@kemi.aau.dk>
Subject: RE: More on deuterium absorption
Reply-To: Dieter Britz <BRITZ@kemi.aau.dk>
Organization: Sci.physics.fusion/Mail Gateway
Date: Mon, 6 Apr 1992 13:29:36 GMT


Originally-From: ames!FNALD.FNAL.GOV!DROEGE

>For the 0.1 cc pd sample we will be using, we will be able to measure the net
>energy to about 20 J over the four hour charging time.  This should give us
>a few percent energy measurement.  I can't believe that this is a constant
>35 kJ per mole over the whole range because it is supposed to go from
>exothermic to encothermic at .9 D/Pd.  So the problem with using this as a
>measure of loading is that I don't know what the energy vs D/Pd curve looks
>like.  I have ordered the Divisek paper (Fermilab has nothing useful for
>cold fusion), perhaps it will help.

I don't think it will help you much, and I doubt that anyone knows. If anyone,
it would be the Flanagan et al lot, whose work is familiar to Dutton, and he
would have said something more about it than the qualitative guesses he has
made. You might be on your own here, Tom. I don't think there is agreement
about what happens at loadings beyond 0.8. Does the extra D really go into
vacant (but reluctant) lattice sites, or maybe into voids formed at this
stage, as compressed D2 gas? If the former, this might be a gradual admixture
of a new phase with its own definite thermodynamics; if the latter, what?

>In earlier experiments we used a third method to measure D/Pd by using a
>four point Kelvin measurement of the cathode.  This has the problem that
>the cell current is of the same order of magnitude as the measurement current
>and adds a lot of noise to the measurement.  So I have give up on resistance
>even though it is interesting.  Getting all those leads in the cell is a pain.

Also, the curve goes through a maximum at 0.8, and calibrations that I have
seen do not go much beyond that, so I don't know how you'd get a loading from
it if you are in fact beyond it. Strictly speaking, the measurement would be
ambiguous anyway - McMubre+ use a decrease in resistance as evidence of
increased loadings, but it could - in principle - also be a decrease to <0.8.


>To answer Tod about the importance of symmetry, I don't know either.  I just

Here is one theoretical possibility: if we assume that at high current
densities, you get high effective loading pressure (FPH's enormous fugacity),
then an unsymmetrical electrode arrangement might be unfavourable, because
there would be different current densities at different places on the Pd. So
in one place, you would be super-loading D into it, but in another, you might
be letting it out again. In other words, there would be transport within the
Pd, nullifying high loadings at a few points. If you have an even and high
current density, as you'd get with a symmetric setup, this can't happen.
Symmetry is possible for only two geometries: a spherical cathode, not
shielded anywhere (not possible), or a cyclindrical rod, surrounded by a
cyclindrical anode (but a spiral would be OK, at a sufficient distance of a
few cathode diameters), with the important proviso that the top and bottom of
the cathode are flush with the cell top and bottom and not electrochemically
accessible. I'd say that square cathodes would give you very uneven current
density distributions.
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Dieter Britz alias britz@kemi.aau.dk
Kemisk Institut, Aarhus Universitet, 8000 Aarhus C, Denmark.
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