Project UFO/Project UFO (Disk 3 of 6).adf/FUSION.CHE

Text File | 1980-01-09 | 10.3 KB | 223 lines

Article 22 of sci.chem:
Path: dasys1!cucard!rocky8!cmcl2!rutgers!gatech!gitpyr![e-mail]
From: [e-mail] (Poppa Smurf)
Newsgroups: sci.chem
Subject: Why is current needed to fuse deuterium in palladium?
Keywords: Fusion, deuterium, palladium, electrochemical Sorry for the previous mangled posting.
     It is well known that both platinum and palladium catalyze reactions
Message-ID: <[e-mail]>
Date: 31 Mar 89 02:37:13 GMT
Sender: [e-mail]
Reply-To: [e-mail] (Poppa Smurf)
Organization: Georgia Institute of Technology
Lines: 29
Posted: Thu Mar 30 21:37:13 1989

has always been explained by postulating that molecular hydrogen sticks to
crystal sites on the metal and is "stretched" in a way that loosens the
molecular bonds and causes the hydrogen to react as if it were a free
atom instead of a bound diatomic molecule.

     Apparently the groups which claim to have observed fusion in electro-
chemical cells containing deuterium oxide are seeing the opposite kind
of phenomena.  The reaction within the metal must be compressing rather than
loosening the deuterium atoms or else they could not possibly get close
enough to undergo fusion.

     Okay, fair enough - one is a surface process, the other operates at
depth.  What is the purpose of electrochemical induction in terms of
mechanism?  Why is it required, apparently, to initiate fusion?  It would
seem to me that simply pressurizing a passivated metal cell in which
a large amount of palladium (probably in thin spiral layers) were present
would induce fusion as the gas diffused into the palladium.

     Correct me if I'm wrong, but the primary electrode process which
occurs at the palladium electrode is formation of D2 from D2O is it not?
Would not a pressurized gas cell present the surface with the same
concentration of D2 as a cell full of electrolyzed D2O would?  You can't
diffuse deuterium into the body of the palladium electrode any faster
than the concentration gradient and diffusion constant will allow.  So long
as the surface is saturated with deuterium you should be diffusing it in
at the higest rate, so why is the current required?

                               Steve Fischer
                               Georgia Tech


Article 28 of sci.chem:
Path: dasys1!cucard!rocky8!cmcl2!rutgers!mailrus!tut.cis.ohio-state.edu!cs.utexas.edu!utastro!bigtex!pmafire!mike
From: [e-mail] (mike caldwell)
Newsgroups: sci.chem
Subject: Re: Why is current needed to fuse deuterium in palladium?
Summary: diffusion not just concentration
Keywords: Fusion, deuterium, palladium, electrochemical
Message-ID: <[e-mail]>
Date: 31 Mar 89 16:19:11 GMT
References: <[e-mail]>
Reply-To: [e-mail] (mike caldwell)
Organization: WINCO, INEL, Idaho
Lines: 43
Posted: Fri Mar 31 11:19:11 1989

In article <[e-mail]> [e-mail] (Poppa Smurf) writes:
>     Okay, fair enough - one is a surface process, the other operates at
>depth.  What is the purpose of electrochemical induction in terms of
>mechanism?  Why is it required, apparently, to initiate fusion?  It would
>seem to me that simply pressurizing a passivated metal cell in which
>a large amount of palladium (probably in thin spiral layers) were present
>would induce fusion as the gas diffused into the palladium.
>
>     Correct me if I'm wrong, but the primary electrode process which
>occurs at the palladium electrode is formation of D2 from D2O is it not?
>Would not a pressurized gas cell present the surface with the same
>concentration of D2 as a cell full of electrolyzed D2O would?  You can't
>diffuse deuterium into the body of the palladium electrode any faster
>than the concentration gradient and diffusion constant will allow.  So long
>as the surface is saturated with deuterium you should be diffusing it in
>at the higest rate, so why is the current required?
>
>                               Steve Fischer
>                               Georgia Tech

I might be misunderstanding what you said, but the chemical potential
gradient is what drives diffusion.  In most cases, the chemical
potential can be reduced to concentration.  When this is the case, the
diffusion is called ordinary diffusion.  Other gradients, such as
pressure in reverse osmosis, can cause diffusion against the pressure
gradient.  Another possibility to enhance diffusion is electrical
forces.

For ordinary diffusion, the diffusion rate limits of the mass flux.  By
using an electrical gradient in an ionic solution, you increase the mass
flux.  Otherwise, commercial electrochemical cells wouldn't be used.
You could save the huge electrical bills and just stick the cathode and
electrode in the solution and let it run at the limiting rate.

I don't know if that sheds some light on the last question (I hope so).
Of course, the real question is the first set and I don't have the
answers.  If I did, the state of Utah would be trying to force me to
talk a $5 million grant.


--
Mike Caldwell ([e-mail])
Paths: ...uunet!pmafire!mike | ...!ucdavis!egg-id!pmafire!mike


Article 30 of sci.chem:
Path: dasys1!cucard!rocky8!cmcl2!rutgers!mailrus!csd4.milw.wisc.edu!lll-winken!uunet!portal!cup.portal.com!James_J_Kowalczyk
From: [e-mail]
Newsgroups: sci.chem
Subject: Re: Why is current needed to fuse deuterium in palladium?
Message-ID: <[e-mail]>
Date: 1 Apr 89 07:11:43 GMT
References: <[e-mail]>
Organization: The Portal System (TM)
Lines: 21
Posted: Sat Apr  1 02:11:43 1989

[e-mail] (Poppa Smurf) writes:

>depth.  What is the purpose of electrochemical induction in terms of
>mechanism?  Why is it required, apparently, to initiate fusion?  It would
>seem to me that simply pressurizing a passivated metal cell in which
>a large amount of palladium (probably in thin spiral layers) were present
>would induce fusion as the gas diffused into the palladium.

I went to a seminar by Stanley Pons at the Univ. of Utah today, and
the way he explained it was that in the electrochemical cell, the palladium
cathode has a potential gradient of 800 mV from the center to the surface
of the metal (in their system, at least).  He said that when the D2
has saturated the metal, there are 0.7-1.0 D atoms per Pd atom.  Thus,
the 800 mV potential is the equivalent of 10^27 atmospheres (ten to the
twenty seventh power).  This is a very large effective pressure, and is
likely responsible for any fusion.

Jim Kowalczyk

[e-mail]
[e-mail]


Article 21 of sci.chem:
Path: dasys1!cucard!rocky8!cmcl2!lanl!hc!lll-winken!uunet!jarthur!hvr
From: [e-mail] (Hal Van Ryswyk)
Newsgroups: sci.chem
Subject: RE: Fusion
Summary: Solid state chemistry of deuterium on palladium
Keywords: fusion, palladium, deuterium, adsorption
Message-ID: <[e-mail]>
Date: 29 Mar 89 17:36:01 GMT
Reply-To: [e-mail] (Hal Van Ryswyk)
Organization: Harvey Mudd College, Claremont, CA
Lines: 38
Posted: Wed Mar 29 12:36:01 1989


To fill in a few details about the chemistry of the "Utah Reactor":

A palladium surface catalyzes the breakdown of molecular hydrogen to
elemental hydrogen.  The elemental hydrogen can then diffuse into the
palladium lattice, reaching concentrations in the high atomic-percent
region.  In some instances I've seen figures as high as one H for every
Pd.  Deuterium behaves in the same fashion--molecular deuterium on the
surface is broken down into atomic species which are then free to enter
the lattice.  Palladium is unique in this regard with respect to the
elements.  Platinum will catalyze the same process, but at much lower
concentrations of included hydrogen.  The process is reversible; anything
which scavenges atomic hydrogen at the metal's surface will draw the
hydrogen out of the system.

The "dissolved" hydrogen (or deuterium) can be thought of as a formal
palladium hydride.  The implications for long-term use are not good:
sustained use of a palladium catalyst in this fashion leads to metal
fatigue, embrittling the metal lattice until the palladium flakes off.

Certain other solids are used as "hydrogen sponges," most notably
tungsten-containing hydrides.  Once again, these systems are capable
of dissolving large amounts of hydrogen, often with better long-term
metallurgical properties.

An article in yesterday's LA Times stated basically that the process
observed in the Utah reactor is a "new type of fusion," one that does
not produce as many neutrons as expected.  If they are seeing 4 W of
excess power generated, then one would expect on the order of 4 billion
neutrons per second to be produced.  The Times article hinted that they
see no such neutron flux.  Although I'm waiting anxously for the Nature
article in May, I wonder if they have taken into consideration the
thermodynamics of the catalytic process on the metal's surface?  A back-
of-the-envelope calculation leads me to believe that this is in the right
ballpark...

If anyone has a more reliable source than the popular press,
I'd love to hear what they have to say!


Article 26 of sci.chem:
Path: dasys1!cucard!rocky8!cmcl2!rutgers!ukma!mailrus!jarvis.csri.toronto.edu!me!ecf!atwood
From: [e-mail] (Robert C Atwood)
Newsgroups: sci.chem
Subject: Re: Fusion
Keywords: fusion, palladium, deuterium, adsorption
Message-ID: <[e-mail]>
Date: 31 Mar 89 04:50:48 GMT
References: <[e-mail]>
Reply-To: [e-mail] (Robert C Atwood)
Organization: Engineering Computing Facility, University of Toronto
Lines: 10
Posted: Thu Mar 30 23:50:48 1989

According to a talk I just came from, Dr Paul Chu of superconductor fame,
who has also worked with palladium hydride (superconducting properties thereof)
has tried and failed to duplicate the rt fusion experiment.  He indicates he
doesn't disbelieve it ... yet.  However, he quoted previous experiments
studying the possible metallic interaction of hydrogen atoms in the lattice
which showed that the h atoms are not quite close enough to be metallic.
So you wouldn't expect them to be close enough to fuse.  The metallic distance
is of order 10 -8 cm i believe and the fusion distance is of order 10 -13
robert
[e-mail]