Michel Jullian meant that the built-in built-in resistance heater
might go up to 3 kW. There is no other input power. The heater is
only needed to bring the temperature up to the temperature at which
the Ni reacts. I guess that would be the temperature at which it
readily absorbs hydrogen.
I do not think a heater requires any kind of fancy AC. It would be DC.
I assume they are using the same basic technique they have been doing
all these years, only using finely divided Ni instead of an Ni rod
with mysterious surface characteristics that no one else can
replicate. Here is a long paper describing their previous experiments:
<http://www.lenr-canr.org/acrobat/FocardiSlargeexces.pdf>http://www.lenr-canr.org/acrobat/FocardiSlargeexces.pdf
Here is a well-known paper that casts doubt on Focard's calorimetry,
but only method A described in the current paper, not B or C:
<http://www.lenr-canr.org/acrobat/CerronZebainvestigat.pdf>http://www.lenr-canr.org/acrobat/CerronZebainvestigat.pdf
If these authors are right, it shows that you can make a large
mistake with technique A. That doesn't surprise me.
Anyway, this 80 W strikes me as odd, but that may only be a function
of my ignorance of this technique, and the lack of detail in the
paper. But what does this 80 W mean?
Maybe this means it takes only about 80 W to bring it up to the
operating temperature. That would mean the cell is well insulated. In
that case, how do they keep from drastically from overheating when it
produces 3 kW?
Or, maybe this means it takes 3 kW from the heater to bring the cell
up the recommended operating temperature, but after the reaction
starts up they can reduce the input power down to 80 W and maintain
the high temperature. That would be a dandy way to do the experiment.
You might say it is self-calibrating, making it difficult to argue
that the input power is causing a false reading. However, if this is
what is happening, it raises a huge question. An elephant-in-the-room
sized question. Why not insulate the cell a little more, and dial the
input power all the way back to zero? In other words, why not make
the thing fully self-sustaining?!? That would eliminate any question
about input power. Why is there any input to output ratio in this paper at all?
Frankly, the whole thing is a confounded mystery to me. But as I
said, I have been advised to reserve judgement and await developments
because it may be better than it looks.
- Jed
