Article 22 of sci.chem Subject Why is current needed to fuse deuterium in palladium? Keywo

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Article 22 of sci.chem: Path: dasys1!cucard!rocky8!cmcl2!rutgers!gatech!gitpyr!steve%revolver@gatech.edu From: steve%revolver@gatech.edu (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: <7783@pyr.gatech.EDU> Date: 31 Mar 89 02:37:13 GMT Sender: news@pyr.gatech.EDU Reply-To: steve%revolver@gatech.edu (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: mike@pmafire.UUCP (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: <609@pmafire.UUCP> Date: 31 Mar 89 16:19:11 GMT References: <7783@pyr.gatech.EDU> Reply-To: mike@pmafire.UUCP (mike caldwell) Organization: WINCO, INEL, Idaho Lines: 43 Posted: Fri Mar 31 11:19:11 1989 In article <7783@pyr.gatech.EDU> steve%revolver@gatech.edu (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 (mike@pmafire.UUCP) 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: James_J_Kowalczyk@cup.portal.com Newsgroups: sci.chem Subject: Re: Why is current needed to fuse deuterium in palladium? Message-ID: <16535@cup.portal.com> Date: 1 Apr 89 07:11:43 GMT References: <7783@pyr.gatech.EDU> Organization: The Portal System (TM) Lines: 21 Posted: Sat Apr 1 02:11:43 1989 steve%revolver@gatech.edu (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 James_J_Kowalczyk@cup.portal.com Kowalczyk@chemistry.utah.edu Article 21 of sci.chem: Path: dasys1!cucard!rocky8!cmcl2!lanl!hc!lll-winken!uunet!jarthur!hvr From: hvr@jarthur.Claremont.EDU (Hal Van Ryswyk) Newsgroups: sci.chem Subject: RE: Fusion Summary: Solid state chemistry of deuterium on palladium Keywords: fusion, palladium, deuterium, adsorption Message-ID: <622@jarthur.Claremont.EDU> Date: 29 Mar 89 17:36:01 GMT Reply-To: hvr@jarthur.UUCP (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: atwood@ecf.toronto.edu (Robert C Atwood) Newsgroups: sci.chem Subject: Re: Fusion Keywords: fusion, palladium, deuterium, adsorption Message-ID: <821@mv03.ecf.toronto.edu> Date: 31 Mar 89 04:50:48 GMT References: <622@jarthur.Claremont.EDU> Reply-To: atwood@mv03.ecf.UUCP (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 atwood@ecf.toronto.edu

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