----- 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. > > >> >> 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 ; note H2O formation enthalpy already includes the P*V contribution from the environment, which must be given back when pushing away the atmosphere to make room for the evolved gases. >> Now with a palladium (or nickel) porous cathode, one could still >> implement the backloading scheme (second, higher voltage anode on >> cathode venting side) which would optimize my fusion hypothesis, >> the "overfaradaic" front bubbling would still be taken away by the >> flowing electrolyte wouldn't it? > > The idea is to get the flowing electrolyte to carry the evolved > bubbles into and through the electrode holes, i.e. the equivalent to > a mesh screen hole. > >> >> Mmmm I am afraid it wouldn't work, the backloaded hydrogen would >> find it easier to leak within the thickness of the mesh, in the no >> man's land between the front and the back, where no electrolysis >> pressure holds it inside the metal, don't you think? > > 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? > 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. Michel
