On Sep 4, 2007, at 5:47 AM, Michel Jullian wrote:
----- Original Message -----
From: "Horace Heffner" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Tuesday, September 04, 2007 12:55 AM
Subject: Re: [Vo]:Re: Towards verification of BG claims
On Sep 3, 2007, at 2:15 PM, Michel Jullian wrote:
Oh I see, I thought you meant the porous structure supported a
continuous (waterproof) Pd foil, in fact the mesh is made of Pd or
is Pd plated right?
The immediate surface layers could be sintered Pd granules, but the
back more granular layers could be any metal I think.
Here I should say any conductive material with very small holes in it
can be be co-deposited with an effective metal for hydrogen
generation, which is thus probably a hydride forming metal, provided
the metals won't separate when the hydride forming metal is hydrogen
saturated. The main purpose is to generate the hydrogen in locations
where it will be sucked into the cathode.
Independently of the cathode material, the idea of flowing
electrolyte through a porous electrode in order to maximize the
ratio of active (bubble free) area to total area seems good, all
the more so that it also increases the total area! In fact both
electrodes could be made thus, and one could pump electrolyte into
the interelectrode gap so the anode surface would be bubble free
too, which would further maximize achievable electrolysis current.
Has this ever been tried BTW?
I think so, but I can't find the paper. I think it was done by these
folks:
http://www.qsinano.com/white_papers/Water%20Electrolysis%20April%
2007.pdf
Interesting paper, but their Appendix 1 derivation of unity
efficiency cell voltage 1.482 V takes them nine complicated steps.
I posted a much simpler derivation here some time ago, it went like
this:
At 10^5 Pa and 25°C the endoenergetic overall electrolysis
reaction H2O(l) -> 0.5 O2(g) + H2(g) consumes 285.83 kJ/mol_H2O
(consumed energy = enthalpy change of the reaction = formation
enthalpy of H2O since O2(g) and H2(g) being in their reference
states have zero formation enthalpy). Per H2O molecule that's E =
285830/6.02e23 = 4.748e-19 J
Efficiency is unity when E is exactly equal to the consumed
electric energy per H2O, which is equal to the supply voltage V
times the charge 2*e (two electrons circulated per molecule), so:
V = 4.748e-19 / (2 * 1.602e-19) = 1.482 V
Any electrolysis at a lower voltage would have to borrow energy
from the environment, I don't know if that's possible ;
It is more than possible. Managing the thermal energy contribution
is essential to high efficiency industrial electrolyzer operating
design and a major component of the energy.
Something you might want to think about regarding back loading at a
somewhat lesser negative potential V<v<0 than the front loading
negative potential V
Why negative, relative to what?
The anode of course. In electrochemistry the cathode is commonly
referred to as negative, the anode positive.
is the fact that most all CF experiments, due to
electrolyte currents and resistances, produce a range of potentials
across the cathode surfaces - especially the fluidized bed
experiments using beads. Nothing repeatable resulted from these
experiments AFAIK.
I don't see how these are relevant to my back-loading + front-
deloading & -electrolysis scheme, which doesn't mean they aren't,
kindly explain.
First let me be clear that whenever I talk about the front side that
is the side from which the loading occurs. The hydrogen moves on
average from the front side to the back side. Somewhere mid-stream
in the discussion you seem to have decided to reverse this meaning.
Maybe it was for your discussion in other groups.
To obtain backside de-loading while simultaneously obtaining hydrogen
movement toward the back side from the electrolyte, the back side
must in fact be a cathode, and thus it is negative, but not equal in
negativity to the front side, otherwise it would in fact be the front
side, and the front side the back. Of course an electrode can be run
using AC or AC superimposed over DC, but that is not the scheme you
have been promoting. So, the back side operates at v, the front side
at V, where, V<v<0. This situation is common in ordinary
electrolysis cells. Due to electrolysis current in the resistive
electrolyte, potentials vary across electrode surfaces. In fluidized
beds the potential through the cathode material the surface potential
can vary all the way from V to 0. Hydrogen literally loads on one
side and de-loads on the other, or loads in one area while de-loading
in another, in all cases diffusing through the cathode.
Horace Heffner
http://www.mtaonline.net/~hheffner/
