Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to Edmund Storms's message of Thu, 14 Mar 2013 22:04:51 -0600: Hi Ed, [snip] Robin, according to my tables, the mass of a bare d is 2.014101778, which is the value I used. I don't know where you got the idea an electron is involved. These are nuclear reactions. Ed Your table is correct, however your interpretation of it is not. The mass quoted is that of a D atom, i.e. the nucleus plus it's electron. (See http://atom.kaeri.re.kr/ton/nuc1.html for atomic masses.) Most publicly available tables provide atomic masses. That's because the mass is determined using a mass spectrometer which in turn measures the mass of ions with a single positive charge, then the mass of an electron is added to yield that of the whole atom. It's difficult to create completely ionized nuclei, to allow measurement of the actual *nuclear* mass, and it gets more difficult as the atomic number increases. Hence the approach used, and the form of the tables. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to James Bowery's message of Fri, 15 Mar 2013 00:56:35 -0500: Hi, [snip] I don't think even Jones suggested D = H + n. His words: Neutrons are 'stripped' from the deuterium ...and so they are, but only when either the neutron or the proton is immediately absorbed by another nucleus. I doubt he intended to imply that the result of said reaction would be both a free neutron and a free proton. IOW at least one of the two needs to be absorbed by a target nucleus for this reaction to occur. You can borrow 2.2 MeV from the Heisenberg bank, but you only get a *very* short term loan. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Robin, that is not my understanding. The values are from GE nuclear Energy 15 Edition that give the mass of the nucleus. The mass is not only obtained using a mass spectrometer. It is obtained by IUPAC using a complex evaluation based on nuclear decay and energy measurements as well. The mass spectrometer can not give the number of significant figures to which these values are given. Ed On Mar 15, 2013, at 2:34 PM, mix...@bigpond.com wrote: In reply to Edmund Storms's message of Thu, 14 Mar 2013 22:04:51 -0600: Hi Ed, [snip] Robin, according to my tables, the mass of a bare d is 2.014101778, which is the value I used. I don't know where you got the idea an electron is involved. These are nuclear reactions. Ed Your table is correct, however your interpretation of it is not. The mass quoted is that of a D atom, i.e. the nucleus plus it's electron. (See http://atom.kaeri.re.kr/ton/nuc1.html for atomic masses.) Most publicly available tables provide atomic masses. That's because the mass is determined using a mass spectrometer which in turn measures the mass of ions with a single positive charge, then the mass of an electron is added to yield that of the whole atom. It's difficult to create completely ionized nuclei, to allow measurement of the actual *nuclear* mass, and it gets more difficult as the atomic number increases. Hence the approach used, and the form of the tables. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
And you get your short term loan and it pays you back with interest. :-) Dave -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Fri, Mar 15, 2013 4:39 pm Subject: Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine? In reply to James Bowery's message of Fri, 15 Mar 2013 00:56:35 -0500: Hi, [snip] I don't think even Jones suggested D = H + n. His words: Neutrons are 'stripped' from the deuterium ...and so they are, but only when either the neutron or the proton is immediately absorbed by another nucleus. I doubt he intended to imply that the result of said reaction would be both a free neutron and a free proton. IOW at least one of the two needs to be absorbed by a target nucleus for this reaction to occur. You can borrow 2.2 MeV from the Heisenberg bank, but you only get a *very* short term loan. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to Edmund Storms's message of Fri, 15 Mar 2013 14:55:45 -0600: Hi Ed, [snip] Robin, that is not my understanding. The values are from GE nuclear Energy 15 Edition that give the mass of the nucleus. The mass is not only obtained using a mass spectrometer. It is obtained by IUPAC using a complex evaluation based on nuclear decay and energy measurements as well. The mass spectrometer can not give the number of significant figures to which these values are given. Ed See http://physics.nist.gov/cgi-bin/cuu/Value?mdu. (Note that the mass given differs from that which you provided by 1 electron mass.) Furthermore, from http://atom.kaeri.re.kr/ton/nuc1.html you will see that the binding energy of Deuterium is 2224.573 +- 0.002 keV (i.e. 2.2 MeV). The binding energy is of course the energy release when the nucleus is created from free particles. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
On Fri, Mar 15, 2013 at 3:39 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 00:56:35 -0500: Hi, [snip] I don't think even Jones suggested D = H + n. His words: Neutrons are 'stripped' from the deuterium ...and so they are, but only when either the neutron or the proton is immediately absorbed by another nucleus. I doubt he intended to imply that the result of said reaction would be both a free neutron and a free proton. IOW at least one of the two needs to be absorbed by a target nucleus for this reaction to occur. You can borrow 2.2 MeV from the Heisenberg bank, but you only get a *very* short term loan. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html How could such a short-lived neutron could get to the engine block? -- Forwarded message -- From: Jones Beene jone...@pacbell.net Date: Thu, Mar 14, 2013 at 1:48 PM Subject: RE: [Vo]:RE: [Vo]:chlorine - hydrogen ion explosion To: vortex-l@eskimo.com No – I am saying that if he did use D for the gain, the engine would become strongly radioactive in a short time from neutron activation of the engine block and pistons.
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to James Bowery's message of Fri, 15 Mar 2013 17:02:49 -0500: Hi, [snip] On Fri, Mar 15, 2013 at 3:39 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 00:56:35 -0500: Hi, [snip] I don't think even Jones suggested D = H + n. His words: Neutrons are 'stripped' from the deuterium ...and so they are, but only when either the neutron or the proton is immediately absorbed by another nucleus. I doubt he intended to imply that the result of said reaction would be both a free neutron and a free proton. IOW at least one of the two needs to be absorbed by a target nucleus for this reaction to occur. You can borrow 2.2 MeV from the Heisenberg bank, but you only get a *very* short term loan. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html How could such a short-lived neutron could get to the engine block? ..the reaction happens at or in the wall of the engine block. e.g. D + 56Fe = 57Fe + H + 5.4 MeV, or D + 58Fe = 59Fe (radioactive) + H + 4.3 MeV. BTW this is interesting because there is actually only a small amount of 58Fe in natural iron. Adding a neutron to either of the more common isotopes 56Fe or 57Fe yields a stable isotope in both cases. There are of course other elements present in steel in small amounts, and adding neutrons to these can also create radioactive isotopes. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
On Fri, Mar 15, 2013 at 5:17 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 17:02:49 -0500: Hi, [snip] On Fri, Mar 15, 2013 at 3:39 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Fri, 15 Mar 2013 00:56:35 -0500: Hi, [snip] I don't think even Jones suggested D = H + n. His words: Neutrons are 'stripped' from the deuterium ...and so they are, but only when either the neutron or the proton is immediately absorbed by another nucleus. I doubt he intended to imply that the result of said reaction would be both a free neutron and a free proton. IOW at least one of the two needs to be absorbed by a target nucleus for this reaction to occur. You can borrow 2.2 MeV from the Heisenberg bank, but you only get a *very* short term loan. ;) Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html How could such a short-lived neutron could get to the engine block? ..the reaction happens at or in the wall of the engine block. e.g. D + 56Fe = 57Fe + H + 5.4 MeV, or D + 58Fe = 59Fe (radioactive) + H + 4.3 MeV. BTW this is interesting because there is actually only a small amount of 58Fe in natural iron. Adding a neutron to either of the more common isotopes 56Fe or 57Fe yields a stable isotope in both cases. There are of course other elements present in steel in small amounts, and adding neutrons to these can also create radioactive isotopes. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html But now you're denying the original premise Jones put forth which was that the chemical energy from reactions involving neutron transfer from D were uniquely energetic as in the formation of hydrogen chloride.
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Thanks Robin, I see the problem. In this case, both the d and the p mass I used contains the electron. As a result, the energy change I calculated is correct because the electron mass cancels out. Nevertheless, this is not what Jones was claiming, so the value is not relevant. Ed On Mar 15, 2013, at 3:31 PM, mix...@bigpond.com wrote: In reply to Edmund Storms's message of Fri, 15 Mar 2013 14:55:45 -0600: Hi Ed, [snip] Robin, that is not my understanding. The values are from GE nuclear Energy 15 Edition that give the mass of the nucleus. The mass is not only obtained using a mass spectrometer. It is obtained by IUPAC using a complex evaluation based on nuclear decay and energy measurements as well. The mass spectrometer can not give the number of significant figures to which these values are given. Ed See http://physics.nist.gov/cgi-bin/cuu/Value?mdu. (Note that the mass given differs from that which you provided by 1 electron mass.) Furthermore, from http://atom.kaeri.re.kr/ton/nuc1.html you will see that the binding energy of Deuterium is 2224.573 +- 0.002 keV (i.e. 2.2 MeV). The binding energy is of course the energy release when the nucleus is created from free particles. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html Atomic Weights and Isotopic Compositions for Hydrogen IsotopeRelative Atomic Mass Isotopic Composition Standard Atomic WeightNotes 1 H 1 1.007 825 032 07(10) 0.999 885(70) 1.007 94(7) g,m,r,b,w D2 2.014 101 777 8(4) 0.000 115(70) T3 3.016 049 2777(25)
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to James Bowery's message of Fri, 15 Mar 2013 17:20:21 -0500: Hi, [snip] How could such a short-lived neutron could get to the engine block? ..the reaction happens at or in the wall of the engine block. e.g. D + 56Fe = 57Fe + H + 5.4 MeV, or D + 58Fe = 59Fe (radioactive) + H + 4.3 MeV. BTW this is interesting because there is actually only a small amount of 58Fe in natural iron. Adding a neutron to either of the more common isotopes 56Fe or 57Fe yields a stable isotope in both cases. There are of course other elements present in steel in small amounts, and adding neutrons to these can also create radioactive isotopes. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html But now you're denying the original premise Jones put forth which was that the chemical energy from reactions involving neutron transfer from D were uniquely energetic as in the formation of hydrogen chloride. I take Jones' original statement to mean that the formation of HCl (DCl?) is the only chemical reaction known to result in nuclear reactions. If Hydrino formation is the actual mechanism involved, then the explanation is simple:- The chemical chain reaction which creates DCl results in the production of lots of free D atoms and also DCl molecules (the latter being a Mills catalyst). These two in combination can create some highly shrunken D or D2, which then could at least in theory, react with the metals in the cylinder wall, as described here above. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Robin, Yes - in the O-P effect, as you say - one or the other nucleons of D is absorbed by a target nucleus; and in the Fusor, it usually is the neutron. When the neutron is absorbed the net energy is ~20% less even though the mass of reactants cannot explain that difference, and the net energy should go the other way. O-P is a lower net energy reaction than thermonuclear and the explanation offered by Oppie is negative kinetic energy. Kim of Purdue has a couple of papers out on the branching ratio energies. To get both a free neutron and a free proton, when deuterium is the only reactant there must be two reactions: D+D - He-3 + n and D+D - Tritium + p. However, since you have both a free proton and a neutron in the end, there is little way to be certain how they got there . other than theory. The reason that Fusor enthusiasts can be fairly certain that this is warm and not exactly hot fusion as seen in a Tokomak besides the low input energy is the branching ratio is different, the energy per fusion event is less, and the plasma is comparatively cold. Comparatively less tritium is seen, and significantly less net energy per fusion event occurs than if the branching was the hot since the tritium reaction is more energetic by 700 keV and it is suppressed. In effect, that's a lot of negative kinetic energy due to tunneling, no? Jones Neutrons are 'stripped' from the deuterium...and so they are, but only when either the neutron or the proton is immediately absorbed by another nucleus. I doubt he intended to imply that the result of said reaction would be both a free neutron and a free proton. IOW at least one of the two needs to be absorbed by a target nucleus for this reaction to occur. You can borrow 2.2 MeV from the Heisenberg bank, but you only get a *very* short term loan. ;)
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to Jones Beene's message of Fri, 15 Mar 2013 16:39:32 -0700: Hi Jones, Robin, Yes - in the O-P effect, as you say - one or the other nucleons of D is absorbed by a target nucleus; and in the Fusor, it usually is the neutron. When the neutron is absorbed the net energy is ~20% less even though the mass of reactants cannot explain that difference, and the net energy should go the other way. O-P is a lower net energy reaction than thermonuclear and the explanation offered by Oppie is negative kinetic energy. Kim of Purdue has a couple of papers out on the branching ratio energies. I don't suppose you have URL for any of Kim's papers? To get both a free neutron and a free proton, when deuterium is the only reactant there must be two reactions: D+D - He-3 + n and D+D - Tritium + p. However, since you have both a free proton and a neutron in the end, there is little way to be certain how they got there . other than theory. True, but you have consumed 4 D's (not 1) to get 1 proton and 1 neutron. The reason that Fusor enthusiasts can be fairly certain that this is warm and not exactly hot fusion as seen in a Tokomak besides the low input energy is the branching ratio is different, the energy per fusion event is less, and the plasma is comparatively cold. Comparatively less tritium is seen, This would appear to be inconsistent with the statement at the top: and in the Fusor, it usually is the neutron. If it were usually the neutron, then the product would be D + n = T. IOW one would expect more T than in the hot fusion branching ratio, not less. and significantly less net energy per fusion event occurs than if the branching was the hot since the tritium reaction is more energetic by 700 keV and it is suppressed. Please explain. Both reactions are energy positive, so how can either be suppressed? In order for energy loss to be the cause of suppression of the reaction that produces the least energy (i.e. the 3He reaction), (and thus retains the most) , one would need to lose the usual energy of that reaction i.e. 3.27 MeV, which would hint very strongly at a complete misunderstanding of the processes involved, or even the reaction taking place. Note that suppression of a reaction can also have reasons other than a lack of energy, so it may not be correct to claim that because the reaction appears to be suppressed, that the implication is an energy shortage. I get the strong impression here that what we are dealing with is another set of results for which someone, somewhere along the way, has come up with the wrong explanation. In order to sort this out, it is necessary to see the original experimental results. Can you point the way? [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Try this one: http://www.fulviofrisone.com/attachments/article/375/Cross%20Section%20for%2 0Cold%20Deuterium-Deuterium%20Fusion.pdf there is a newer one, but the citation is eluding me. Fusion Technology apparently does not like for their articles to appear online without a cash exchange taking place first. -Original Message- From: mix...@bigpond.com In reply to Jones Beene's message of Fri, 15 Mar 2013 16:39:32 -0700: Hi Jones, Robin, Yes - in the O-P effect, as you say - one or the other nucleons of D is absorbed by a target nucleus; and in the Fusor, it usually is the neutron. When the neutron is absorbed the net energy is ~20% less even though the mass of reactants cannot explain that difference, and the net energy should go the other way. O-P is a lower net energy reaction than thermonuclear and the explanation offered by Oppie is negative kinetic energy. Kim of Purdue has a couple of papers out on the branching ratio energies. I don't suppose you have URL for any of Kim's papers? To get both a free neutron and a free proton, when deuterium is the only reactant there must be two reactions: D+D - He-3 + n and D+D - Tritium + p. However, since you have both a free proton and a neutron in the end, there is little way to be certain how they got there . other than theory. True, but you have consumed 4 D's (not 1) to get 1 proton and 1 neutron. The reason that Fusor enthusiasts can be fairly certain that this is warm and not exactly hot fusion as seen in a Tokomak besides the low input energy is the branching ratio is different, the energy per fusion event is less, and the plasma is comparatively cold. Comparatively less tritium is seen, This would appear to be inconsistent with the statement at the top: and in the Fusor, it usually is the neutron. If it were usually the neutron, then the product would be D + n = T. IOW one would expect more T than in the hot fusion branching ratio, not less. and significantly less net energy per fusion event occurs than if the branching was the hot since the tritium reaction is more energetic by 700 keV and it is suppressed. Please explain. Both reactions are energy positive, so how can either be suppressed? In order for energy loss to be the cause of suppression of the reaction that produces the least energy (i.e. the 3He reaction), (and thus retains the most) , one would need to lose the usual energy of that reaction i.e. 3.27 MeV, which would hint very strongly at a complete misunderstanding of the processes involved, or even the reaction taking place. Note that suppression of a reaction can also have reasons other than a lack of energy, so it may not be correct to claim that because the reaction appears to be suppressed, that the implication is an energy shortage. I get the strong impression here that what we are dealing with is another set of results for which someone, somewhere along the way, has come up with the wrong explanation. In order to sort this out, it is necessary to see the original experimental results. Can you point the way? [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to Jones Beene's message of Fri, 15 Mar 2013 18:09:38 -0700: Hi, [snip] Try this one: http://www.fulviofrisone.com/attachments/article/375/Cross%20Section%20for%2 0Cold%20Deuterium-Deuterium%20Fusion.pdf This is an interesting paper, but I have a few problems with it. 1) Kim apparently proposes using the inverse reactions p + 3He = D + D and n + T = D + D at energies of 0.3 to 4 keV to determine the cross section of the normal reaction, however at the energies proposed, the reverse reaction won't happen at all, because you need at least 3+ MeV for the first, and 4+ MeV for the second. (Unless I misunderstood the intent.) 2) The whole edifice is based on the assumption that during cluster fusion, the energy of the cluster is evenly distributed across all the molecules in the cluster. However this is may well not be the case. It's possible that the molecules at the leading edge get more than their fair share during the impact. The mechanism being the same as when football fans get crushed to death in a stadium riot. Those at the back each provide a small force, which gets added to as it moves forward, until it is received by those at the front, who have nowhere to go, and find themselves between a rock and a hard place. 3) Resonance at low energy is offered as a possible explanation, and it may well be, but there are alternatives, Hydrinos being not least among them. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Jim, why assume the neutron is stripped from the D? This requires 1.7 MeV/event. Where does this amount of energy come from? We know that fractofusion occurs when D is present and this produces neutrons. An explosive reaction would certainly create cracks in the container that could cause this version of hot fusion. Ed On Mar 14, 2013, at 3:21 PM, James Bowery wrote: On Thu, Mar 14, 2013 at 10:49 AM, Jones Beene jone...@pacbell.net wrote: The chlorine-hydrogen photoactivated reaction is the only chemical reaction which is known to produce nuclear reactions (when deuterium is used in place of hydrogen). Neutrons are “stripped” from the deuterium in that case. Normal water is 0.02% D2O, so can't we expect: 2014101.77812 + 15994914.61957 = 16999131.75650 + 1007825.03223 + 2059.609uamu energy D + O16 = O17 + H + 2059.609uamu energy in appropriately dilute amounts?
RE: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
From: Edmund Storms Jim, why assume the neutron is stripped from the D? This requires 1.7 MeV/event. No it doesn't - that number is way off - 3 orders of magnitude off. Neutron stripping occurs as low as 10 keV. See Tom Ligon's IE article. The Farnsworth Fusor is documented proof of large neutron production at keV energy levels. Here is another version of Ligon's article. One can argue the point of whether this is due to Boltzmann's tail of the energy distribution or Oppenheimer-Philips stripping, but one must accept that it is far removed from 1.7 MeV/event. http://www.fusor.net/newbie/files/Ligon-QED-IE.pdf Jones
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Jones, I assume you accept that E=mc2 and that if the mass of a reaction changes, the energy has to come from somewhere. Here is the mass change D = 2.014101778 H= 1.00727647 n= 1.0086649 The gain in mass is D-n= p -0.001839592 which = 1.713569649 MeV has to be added to provide the increased mass of the resulting p. The Farnsworth Fusor is producing hot fusion, which generates energy. The only issue is what amount of energy must be applied to overcome the Coulomb barrier, after which energy is released. That amount to get over the barrier can be a few kev to cause a little hot fusion. In the case of neutron stripping, energy must be ADDED to the system to produce the result. On the other hand, if you assume that the chemical reaction is creating hot fusion in the gas, then you must assume that each D has been given a 10 keV as kinetic energy as a result of the chemical reaction. That is not possible because to make any energy the DCl molecule has to form, which can not have the required kinetic energy simply based on momentum considerations. Ed On Mar 14, 2013, at 4:09 PM, Jones Beene wrote: From: Edmund Storms Jim, why assume the neutron is stripped from the D? This requires 1.7 MeV/event. No it doesn’t - that number is way off - 3 orders of magnitude off. Neutron stripping occurs as low as 10 keV. See Tom Ligon’s IE article. The Farnsworth Fusor is documented proof of large neutron production at keV energy levels. Here is another version of Ligon’s article. One can argue the point of whether this is due to Boltzmann’s tail of the energy distribution or Oppenheimer-Philips stripping, but one must accept that it is far removed from 1.7 MeV/event. http://www.fusor.net/newbie/files/Ligon-QED-IE.pdf Jones
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
I was assuming the phenomenon reported by Jones Beene (without citation) was real. A citation of neutrons being produced by H+Cl =HCl is now in order isn't it? Moreover, endothermic D = H + n plausibly produces cold neutrons whereas fractofusion produces hot neutrons, doesn't it? On Thu, Mar 14, 2013 at 4:29 PM, Edmund Storms stor...@ix.netcom.comwrote: Jim, why assume the neutron is stripped from the D? This requires 1.7 MeV/event. Where does this amount of energy come from? We know that fractofusion occurs when D is present and this produces neutrons. An explosive reaction would certainly create cracks in the container that could cause this version of hot fusion. Ed On Mar 14, 2013, at 3:21 PM, James Bowery wrote: On Thu, Mar 14, 2013 at 10:49 AM, Jones Beene jone...@pacbell.net wrote: The chlorine-hydrogen photoactivated reaction is the only chemical reaction which is known to produce nuclear reactions (when deuterium is used in place of hydrogen). Neutrons are “stripped” from the deuterium in that case. Normal water is 0.02% D2O, so can't we expect: 2014101.77812 + 15994914.61957 = 16999131.75650 + 1007825.03223 + 2059.609uamu energy D + O16 = O17 + H + 2059.609uamu energy in appropriately dilute amounts?
RE: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
From: Edmund Storms Jones, I assume you accept that E=mc2 and that if the mass of a reaction changes, the energy has to come from somewhere. Here is the mass change D = 2.014101778 H= 1.00727647 n= 1.0086649 The gain in mass is D-n= p You are making an incorrect assumption. The O-P effect (i.e. stripping) is not thermonuclear, it is quantum mechanical - in effect a tunneling reaction. Quantum tunneling is one of Oppenheimer's claims to fame. In the Fusor, the transmuted nucleus is left in an energy state as if it had fused with a neutron of negative kinetic energy, so there far less mass change than the thermonuclear reaction. The Fusor can be called warm fusion not hot, since the threshold energy for thermonuclear reaction is never attained. Jones
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Jones do you have a cite for neutrons being produced from an H+Cl system? On Thu, Mar 14, 2013 at 6:47 PM, Jones Beene jone...@pacbell.net wrote: ** ** ** ** *From:* Edmund Storms ** ** Jones, I assume you accept that E=mc2 and that if the mass of a reaction changes, the energy has to come from somewhere. Here is the mass change ** ** D = 2.014101778 H= 1.00727647 n= 1.0086649 ** ** The gain in mass is D-n= p ** ** ** ** You are making an incorrect assumption. The O-P effect (i.e. “stripping”) is not thermonuclear, it is quantum mechanical - in effect a tunneling reaction. Quantum tunneling is one of Oppenheimer’s claims to fame. ** ** In the Fusor, the transmuted nucleus is left in an energy state as if it had fused with a neutron of negative kinetic energy, so there far less mass change than the thermonuclear reaction. The Fusor can be called “warm fusion” not hot, since the threshold energy for thermonuclear reaction is never attained. ** ** Jones ** **
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Ed On Mar 14, 2013, at 5:47 PM, Jones Beene wrote: From: Edmund Storms Jones, I assume you accept that E=mc2 and that if the mass of a reaction changes, the energy has to come from somewhere. Here is the mass change D = 2.014101778 H= 1.00727647 n= 1.0086649 The gain in mass is D-n= p You are making an incorrect assumption. The O-P effect (i.e. “stripping”) is not thermonuclear, it is quantum mechanical - in effect a tunneling reaction. Quantum tunneling is one of Oppenheimer’s claims to fame. OK Jones, then were does the mass come from? No matter what you call the process, the energy MUST be conserved. This reaction requires energy be added to create the mass of the product. Where does this energy come from? In the Fusor, the transmuted nucleus is left in an energy state as if it had fused with a neutron of negative kinetic energy, so there far less mass change than the thermonuclear reaction. The Fusor can be called “warm fusion” not hot, since the threshold energy for thermonuclear reaction is never attained. The only issue here is how the barrier is overcome, because once this happens, energy is created by the normal hot fusion reaction, i.e. the combined nucleus fragments into the observed particles which includes neutrons. Why suggest some magic condition like negative energy. The process is very simple. The two D are given enough energy to surmount the barrier. The Fusor simply does this in an efficient way. Ed Jones
RE: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
From: Edmund Storms Here is the mass change D = 2.014101778 H= 1.00727647 n= 1.0086649 The gain in mass is D-n= p You are making an incorrect assumption. The O-P effect (i.e. stripping) is not thermonuclear, it is quantum mechanical - in effect a tunneling reaction. Quantum tunneling is one of Oppenheimer's claims to fame. OK Jones, then were does the mass come from? No matter what you call the process, the energy MUST be conserved. This reaction requires energy be added to create the mass of the product. Where does this energy come from? Yes, mass-energy is conserved but we are talking about deuterium being converted into something else (tritium or He3)- so there is NOT necessarily a non-conserved mass of anything, since there is always the neutrino wild card. That, essentially, is the crux of your incorrect assumption. In the Fusor, the transmuted nucleus is left in an energy state as if it had fused with a neutron of negative kinetic energy, so there far less mass change than the thermonuclear reaction. The Fusor can be called warm fusion not hot, since the threshold energy for thermonuclear reaction is never attained. The only issue here is how the barrier is overcome, because once this happens, energy is created by the normal hot fusion reaction, i.e. the combined nucleus fragments into the observed particles which includes neutrons. That is what you seem to be missing in all of this. It is not hot fusion but CoE does apply. In the O-P reaction, the Coulomb barrier is overcome when two deuterons approach each other with the neutron end of each facing the other - i.e. being geometrically ahead of the proton end. The 1.7 MeV barrier is effectively lowered to about 10 keV. Why suggest some magic condition like negative energy. Robert Oppenheimer and Melba Philips suggested this. Who am I, or you, to suggest otherwise? The process is very simple. The two D are given enough energy to surmount the barrier. The Fusor simply does this in an efficient way. No, the Fusor never gets close to doing this at all, without QM. The energy to surmount the barrier is reduced by a similar amount to the deficit in net energy transfer. Once again, we appear to be seeing experts in one field who do not understand the full implications of QM and nuclear tunneling - and refuse to believe that energy on the quantum scale can be borrowed for a few femtoseconds before it is repaid. There is no 1.7 MeV threshold and there is corresponding mass change. In QM tunneling, the energy barrier for fusion is reduced and the excess energy is likewise reduced. Jones
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
On Thu, Mar 14, 2013 at 4:21 PM, James Bowery jabow...@gmail.com wrote:
On Thu, Mar 14, 2013 at 10:49 AM, Jones Beene jone...@pacbell.net wrote:
The chlorine-hydrogen photoactivated reaction is the only chemical
reaction which is known to produce nuclear reactions (when deuterium is
used in place of hydrogen). Neutrons are “stripped” from the deuterium in
that case.
Normal water is 0.02% D2O, so can't we expect:
2014101.77812 + 15994914.61957 = 16999131.75650 + 1007825.03223 +
2059.609uamu energy
D + O16 = O17 + H + 2059.609uamu energy
in appropriately dilute amounts?
Still awaiting the cite from Jones about H+Cl = HCl producing neutrons.
Meanwhile, here's a wrap-up of the arithmetic for this explanation for the
Papp engine's energy source
Starting with the energy of a molar reaction:
(6.0221415e+23*2059.609udalton*c^2)?J
(6.0221415E23 * [2059.609 * {micro*dalton}]) * (speed_of_light^2) ? joule
= 1.8510799E11 J
Now we take 100hp as the Papp engine output and as how moles per hour of
deuterium it would consume:
(6.0221415e+23*2059.609udalton*c^2)/mole;100hp?mole/hour
([{6.0221415E23 * (2059.609 * [micro*dalton])} * {speed_of_light^2}] /
mole)^-1* (100 * horsepower) ? mole / hour
= 0.0014502439 mole/hour
We take that and convert that to grams of deuterium:
0.0014502439 mole;2g/mole?g
(0.0014502439 * mole) * ([2 * gramm] / mole) ? gramm
= 0.0029004878 g
And since we know that deuterium is 0.0156% of hydrogen and hydrogen is 1/8
the mass of water we can ask how much water per hour is consumed per hour
by the Papp engine running at 100hp:
8*0.0029004878 g/0.000156;1kg/l?l
([8 * {0.0029004878 * gramm}] / 0.000156) * ([1 * {kilo*gramm}] / liter)^-1
? liter
= 0.14874296 l
or about a half cup of water per hour.
Of course, the power level would decrease as the deuterium is burned up and
is therefore more dilute.
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
On Mar 14, 2013, at 7:05 PM, Jones Beene wrote: From: Edmund Storms Here is the mass change D = 2.014101778 H= 1.00727647 n= 1.0086649 The gain in mass is D-n= p You are making an incorrect assumption. The O-P effect (i.e. “stripping”) is not thermonuclear, it is quantum mechanical - in effect a tunneling reaction. Quantum tunneling is one of Oppenheimer’s claims to fame. OK Jones, then were does the mass come from? No matter what you call the process, the energy MUST be conserved. This reaction requires energy be added to create the mass of the product. Where does this energy come from? Yes, mass-energy is conserved but we are talking about deuterium being converted into something else (tritium or He3)– so there is NOT necessarily a non-conserved mass of anything, since there is always the neutrino “wild card”. That, essentially, is the crux of your incorrect assumption. I'm making no assumption. I'm simply applying conservation of energy. If instead of the D= n +P reaction, you propose the normal hot fusion reaction, then of course the situation changes. When two D come together with enough energy, the nuclei combine and then explodes into tritium + p and He3 + n. This is the normal hot fusion reaction that generates energy. That is not a neutron stripping reaction. In the Fusor, the transmuted nucleus is left in an energy state as if it had fused with a neutron of negative kinetic energy, so there far less mass change than the thermonuclear reaction. The Fusor can be called “warm fusion” not hot, since the threshold energy for thermonuclear reaction is never attained. The only issue here is how the barrier is overcome, because once this happens, energy is created by the normal hot fusion reaction, i.e. the combined nucleus fragments into the observed particles which includes neutrons. That is what you seem to be missing in all of this. It is not hot fusion but CoE does apply. In the O-P reaction, the Coulomb barrier is overcome when two deuterons approach each other with the neutron end of each facing the other – i.e. being geometrically ahead of the proton end. The 1.7 MeV barrier is effectively lowered to about 10 keV. Yes, the barrier is lowered and the expected fusion reaction occurs. That was not the original subject. Why suggest some magic condition like negative energy. Robert Oppenheimer and Melba Philips suggested this. Who am I, or you, to suggest otherwise? The process is very simple. The two D are given enough energy to surmount the barrier. The Fusor simply does this in an efficient way. No, the Fusor never gets close to doing this at all, without QM. The energy to surmount the barrier is reduced by a similar amount to the deficit in net energy transfer. Once again, we appear to be seeing experts in one field who do not understand the full implications of QM and nuclear tunneling - and refuse to believe that energy on the quantum scale can be “borrowed” for a few femtoseconds before it is repaid. I know about tunneling. It is simply a way of saying that the expected barrier is lowered by some process. You can describe the process using QM if you want. Or you can propose that the orientation of the two d is important or, if the d are in a material, the electron concentration is important. Or you can imagine borrowed energy. These are all assumptions used to explain what is observed. It has nothing to do with the initial subject. Ed There is no 1.7 MeV threshold and there is corresponding mass change. In QM tunneling, the energy barrier for fusion is reduced and the excess energy is likewise reduced. Jones
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to Edmund Storms's message of Thu, 14 Mar 2013 16:31:19 -0600: Hi Ed, [snip] Jones, I assume you accept that E=mc2 and that if the mass of a reaction changes, the energy has to come from somewhere. Here is the mass change D = 2.014101778 H= 1.00727647 n= 1.0086649 The gain in mass is D-n= p -0.001839592 which = 1.713569649 MeV has to be added to provide the increased mass of the resulting p. You have used the mass of a bare proton and neutron, but the atomic mass of D (i.e. including the electron). The actual reaction energy is therefore .511 MeV larger, (since really only a D nucleus is produced, not atomic D), i.e. 1.7 + 0.5 = 2.2 MeV. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
RE: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
There are obvious reasons why you will have to present a need to know for this information. You can start your search with the inventor of the Mark 1 and 2 triggers, referred to here. http://pubs.acs.org/cen/priestley/recipients/1972kistiakowsky.html From: James Bowery Still awaiting the cite from Jones about H+Cl = HCl producing neutrons.
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to Jones Beene's message of Thu, 14 Mar 2013 16:47:50 -0700: Hi, You are making an incorrect assumption. The O-P effect (i.e. stripping) is not thermonuclear, it is quantum mechanical - in effect a tunneling reaction. Quantum tunneling is one of Oppenheimer's claims to fame. To be fair Jones, you were not exactly clear in your original post. The bottom line is that stripping can only happen when one half or the other of the D nucleus is absorbed by the target nucleus. Part of the energy liberated by the absorption (usually around 3-8 MeV) is used to pry apart the original D nucleus. It could also be seen as absorption of the entire D nucleus, followed by emission of either a proton or a neutron, though I doubt that's the way it works. More likely that the neutron migrates to the target nucleus (most of the time), and at least part of the energy liberated by this event is carried away by the now lone proton. In the Fusor, the transmuted nucleus is left in an energy state as if it had fused with a neutron of negative kinetic energy, so there far less mass change than the thermonuclear reaction. One would expect there to be 2.2 MeV less energy than from the absorption of a bare neutron, since 2.2 MeV is required to break apart the D nucleus. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Jones, Are you saying that there are two reactions taking place in this situation where the final product results in the release of energy? I agree with Ed if the end products are a neutron and proton that are now unconnected. Perhaps it is possible to borrow energy for a short period of time with a quantum tunneling effect, but it must be repaid soon afterwards. Please explain when that happens. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Thu, Mar 14, 2013 9:06 pm Subject: RE: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine? From:Edmund Storms Here is themass change D = 2.014101778 H= 1.00727647 n= 1.0086649 The gain inmass is D-n= p You are making an incorrect assumption. The O-P effect (i.e.“stripping”) is not thermonuclear, it is quantum mechanical - ineffect a tunneling reaction. Quantum tunneling is one of Oppenheimer’sclaims to fame. OK Jones, then were does the mass come from? Nomatter what you call the process, the energy MUST be conserved. Thisreaction requires energy be added to create the mass of the product. Where doesthis energy come from? Yes, mass-energy isconserved but we are talking about deuterium being converted into somethingelse (tritium or He3)– so there is NOT necessarily a non-conserved massof anything, since there is always the neutrino “wild card”. That, essentially,is the crux of your incorrect assumption. In the Fusor, the transmuted nucleus is left in an energy state asif it had fused with a neutron of negative kinetic energy, so there far lessmass change than the thermonuclear reaction. The Fusor can be called“warm fusion” not hot, since the threshold energy for thermonuclearreaction is never attained. The only issue here is how the barrier is overcome,because once this happens, energy is created by the normal hot fusion reaction,i.e. the combined nucleus fragments into the observed particles which includesneutrons. That is what you seem tobe missing in all of this. It is not hot fusion but CoE does apply. In the O-Preaction, the Coulomb barrier is overcome when two deuterons approach eachother with the neutron end of each facing the other – i.e. being geometricallyahead of the proton end. The 1.7 MeV barrier is effectively lowered to about 10keV. Why suggest some magic condition like negative energy. Robert Oppenheimer andMelba Philips suggested this. Who am I, or you, to suggest otherwise? The process is very simple. The two D are given enoughenergy to surmount the barrier. The Fusor simply does this in an efficient way. No, the Fusor never getsclose to doing this at all, without QM. The energy to surmount the barrier isreduced by a similar amount to the deficit in net energy transfer. Once again, we appear tobe seeing experts in one field who do not understand the full implications ofQM and nuclear tunneling - and refuse to believe that energy on the quantumscale can be “borrowed” for a few femtoseconds before it is repaid. There is no 1.7 MeVthreshold and there is corresponding mass change. In QM tunneling, the energybarrier for fusion is reduced and the excess energy is likewise reduced. Jones
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Robin, Why is the energy required to break apart the D 2.2 MeV? Ed calculated 1.7 MeV by calculating the mass difference which seemed correct. I would assume that there is no charge change taking place which involves an electron since the same number of protons are present in both the initial and final products. Could you explain your reasoning? Dave -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Thu, Mar 14, 2013 10:58 pm Subject: Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine? In reply to Jones Beene's message of Thu, 14 Mar 2013 16:47:50 -0700: Hi, You are making an incorrect assumption. The O-P effect (i.e. stripping) is not thermonuclear, it is quantum mechanical - in effect a tunneling reaction. Quantum tunneling is one of Oppenheimer's claims to fame. To be fair Jones, you were not exactly clear in your original post. The bottom line is that stripping can only happen when one half or the other of the D nucleus is absorbed by the target nucleus. Part of the energy liberated by the absorption (usually around 3-8 MeV) is used to pry apart the original D nucleus. It could also be seen as absorption of the entire D nucleus, followed by emission of either a proton or a neutron, though I doubt that's the way it works. More likely that the neutron migrates to the target nucleus (most of the time), and at least part of the energy liberated by this event is carried away by the now lone proton. In the Fusor, the transmuted nucleus is left in an energy state as if it had fused with a neutron of negative kinetic energy, so there far less mass change than the thermonuclear reaction. One would expect there to be 2.2 MeV less energy than from the absorption of a bare neutron, since 2.2 MeV is required to break apart the D nucleus. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to Edmund Storms's message of Thu, 14 Mar 2013 20:38:22 -0600: Hi, [snip] I'm making no assumption. I'm simply applying conservation of energy. If instead of the D= n +P reaction, you propose the normal hot fusion reaction, then of course the situation changes. When two D come together with enough energy, the nuclei combine and then explodes into tritium + p and He3 + n. This is the normal hot fusion reaction that generates energy. That is not a neutron stripping reaction. Actually, the hot fusion reaction you propose here probably never happens in reality, except perhaps in particle accelerators. Far more likely is that all real life fusion reactions are stripping reactions, i.e. D + D = T + p (where a neutron migrates from one D nucleus to the other leaving a proton behind), or D + D = He3 + n (where a proton migrates to the other D, leaving a neutron behind). Perhaps coincidentally, the concept of nuclear shielding of the ZPF that I mentioned previously, may help to explain this. As two nuclei get closer together, they start to shield one another. The shielding from a large nucleus would be greater than that from a small nucleus. When a D approaches another nucleus, the neutron is shielded on one side by a large nucleus, and on the other side by a mere proton, so at some approach distance, the shielding from the larger nucleus will exceed that from the proton, and the neutron will pushed toward the larger nucleus. The proton would too, but is repelled by the electric charge on the larger nucleus, hence gets pushed away. [snip] That is what you seem to be missing in all of this. It is not hot fusion but CoE does apply. In the O-P reaction, the Coulomb barrier is overcome when two deuterons approach each other with the neutron end of each facing the other i.e. being geometrically ahead of the proton end. The 1.7 MeV barrier is effectively lowered to about 10 keV. Actually the barrier has nothing to do with the binding energy of the D nucleus. It is purely electrostatic repulsion, hence the use of 1.7 MeV barrier is a result of confusion across posts. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to David Roberson's message of Thu, 14 Mar 2013 23:20:35 -0400 (EDT): Hi, [snip] Robin, Why is the energy required to break apart the D 2.2 MeV? Ed calculated 1.7 MeV by calculating the mass difference which seemed correct. I would assume that there is no charge change taking place which involves an electron since the same number of protons are present in both the initial and final products. Could you explain your reasoning? Please see my earlier post, where Ed's calculation is corrected. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
In reply to James Bowery's message of Thu, 14 Mar 2013 18:26:49 -0500: Hi, [snip] I was assuming the phenomenon reported by Jones Beene (without citation) was real. A citation of neutrons being produced by H+Cl =HCl is now in order isn't it? Moreover, endothermic D = H + n plausibly produces cold neutrons whereas fractofusion produces hot neutrons, doesn't it? I don't think even Jones suggested D = H + n. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Ah, I see what you refer to now. This calculation is also the famous one associated with the PF gamma ray energy error that caused them so much difficulty. Dave -Original Message- From: mixent mix...@bigpond.com To: vortex-l vortex-l@eskimo.com Sent: Thu, Mar 14, 2013 11:27 pm Subject: Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine? In reply to David Roberson's message of Thu, 14 Mar 2013 23:20:35 -0400 (EDT): Hi, [snip] Robin, Why is the energy required to break apart the D 2.2 MeV? Ed calculated 1.7 MeV by calculating the mass difference which seemed correct. I would assume that there is no charge change taking place which involves an electron since the same number of protons are present in both the initial and final products. Could you explain your reasoning? Please see my earlier post, where Ed's calculation is corrected. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
My only need to know is so that I can decide whether there is a chemical source of cold neutrons and thereby take seriously the arithmetic I have shown. If so, it _does_ explain a plausible source of energy for the Papp engine. On Thu, Mar 14, 2013 at 9:47 PM, Jones Beene jone...@pacbell.net wrote: There are obvious reasons why you will have to present a “need to know” for this information. ** ** You can start your search with the inventor of the Mark 1 and 2 triggers, referred to here. ** ** http://pubs.acs.org/cen/priestley/recipients/1972kistiakowsky.html ** ** ** ** ** ** *From:* James Bowery Still awaiting the cite from Jones about H+Cl = HCl producing neutrons.
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Robin, according to my tables, the mass of a bare d is 2.014101778, which is the value I used. I don't know where you got the idea an electron is involved. These are nuclear reactions. Ed On Mar 14, 2013, at 9:27 PM, mix...@bigpond.com wrote: In reply to David Roberson's message of Thu, 14 Mar 2013 23:20:35 -0400 (EDT): Hi, [snip] Robin, Why is the energy required to break apart the D 2.2 MeV? Ed calculated 1.7 MeV by calculating the mass difference which seemed correct. I would assume that there is no charge change taking place which involves an electron since the same number of protons are present in both the initial and final products. Could you explain your reasoning? Please see my earlier post, where Ed's calculation is corrected. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
On Thu, Mar 14, 2013 at 6:05 PM, Jones Beene jone...@pacbell.net wrote: That is what you seem to be missing in all of this. It is not hot fusion but CoE does apply. In the O-P reaction, the Coulomb barrier is overcome when two deuterons approach each other with the neutron end of each facing the other – i.e. being geometrically ahead of the proton end. The 1.7 MeV barrier is effectively lowered to about 10 keV. I wonder how this affects Ron Maimon's proposal -- as the two deuterons approach the palladium nucleus and reach the classical turning point, the proton ends would be oriented both away from the positively charged palladium nucleus and away from the other deuterium nucleus. The neutrons would be right in the middle of it all. Eric
RE: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
Dave, If we are talking about the Farnsworth Fusor, which is the way the thread evolved, then the Fusor is fueled with the same deuterium as palladium cold fusion but NO helium-4 is seen. The reaction is going to either Helium-3 and a 2.4 MeV neutron or Tritium and a fast proton. Both of those secondary nuclei will further react. There is a different branching ratio than hot fusion and there are few gammas– another sign that this is not hot fusion. The 2.4 MeV neutron is characteristic of hot fusion. In O-P, which is accurate for a fraction of the Fusor’s reactions, the lowered threshold translates into less net energy than hot fusion since the stripped neutron acts as if has negative kinetic energy (according to O-P not me). Most of that faction of fusion goes to an emitted proton. Some of the neutrons which are seen from Fusors are believed to be spallation neutrons from the fast proton interacting with the tungsten of the cathode and some are the characteristic 2.4 MeV neutrons from the He-3 reaction. In any event, the plasma remains “warm” and too cool to emit gammas, so it cannot be typical hot fusion but more like a hybrid. Even a neon transformer provides sufficient voltage. In QM tunneling, energy can be “borrowed” to accomplish fusion and immediately repaid to balance the books. This should not be in dispute. Unfortunately, QM reactions are low in probability and the Fusor is almost impossible to scale up to breakeven. That is the tradeoff. We will not solve the energy dilemma with a Fusor unless dozens are used as a neutron source for subcritical fission – which has been proposed. From: David Roberson Jones, Are you saying that there are two reactions taking place in this situation where the final product results in the release of energy? I agree with Ed if the end products are a neutron and proton that are now unconnected. Perhaps it is possible to borrow energy for a short period of time with a quantum tunneling effect, but it must be repaid soon afterwards. Please explain when that happens. Dave From: Edmund Storms Here is the mass change D = 2.014101778 H= 1.00727647 n= 1.0086649 The gain in mass is D-n= p You are making an incorrect assumption. The O-P effect (i.e. “stripping”) is not thermonuclear, it is quantum mechanical - in effect a tunneling reaction. Quantum tunneling is one of Oppenheimer’s claims to fame. OK Jones, then were does the mass come from? No matter what you call the process, the energy MUST be conserved. This reaction requires energy be added to create the mass of the product. Where does this energy come from? Yes, mass-energy is conserved but we are talking about deuterium being converted into something else (tritium or He3)– so there is NOT necessarily a non-conserved mass of anything, since there is always the neutrino “wild card”. That, essentially, is the crux of your incorrect assumption. In the Fusor, the transmuted nucleus is left in an energy state as if it had fused with a neutron of negative kinetic energy, so there far less mass change than the thermonuclear reaction. The Fusor can be called “warm fusion” not hot, since the threshold energy for thermonuclear reaction is never attained. The only issue here is how the barrier is overcome, because once this happens, energy is created by the normal hot fusion reaction, i.e. the combined nucleus fragments into the observed particles which includes neutrons. That is what you seem to be missing in all of this. It is not hot fusion but CoE does apply. In the O-P reaction, the Coulomb barrier is overcome when two deuterons approach each other with the neutron end of each facing the other – i.e. being geometrically ahead of the proton end. The 1.7 MeV barrier is effectively lowered to about 10 keV. Why suggest some magic condition like negative energy. Robert Oppenheimer and Melba Philips suggested this. Who am I, or you, to suggest otherwise? The process is very simple. The two D are given enough energy to surmount the barrier. The Fusor simply does this in an efficient way. No, the Fusor never gets close to doing this at all, without QM. The energy to surmount the barrier is reduced by a similar amount to the deficit in net energy transfer. Once again, we appear to be seeing experts in one field who do not understand the full implications of QM and nuclear tunneling - and refuse to believe that energy on the quantum scale can be “borrowed” for a few femtoseconds before it is repaid. There is no 1.7 MeV threshold and there is corresponding mass change. In QM tunneling, the energy barrier for fusion is reduced and the excess energy is likewise reduced. Jones
Re: [Vo]:Dilute D2O Cold Neutron Capture In Papp Engine?
On Thu, Mar 14, 2013 at 10:29 PM, mix...@bigpond.com wrote: In reply to James Bowery's message of Thu, 14 Mar 2013 18:26:49 -0500: Hi, [snip] I was assuming the phenomenon reported by Jones Beene (without citation) was real. A citation of neutrons being produced by H+Cl =HCl is now in order isn't it? Moreover, endothermic D = H + n plausibly produces cold neutrons whereas fractofusion produces hot neutrons, doesn't it? I don't think even Jones suggested D = H + n. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html His words: Neutrons are 'stripped' from the deuterium
