Re: [Vo]:List of Rossi 18-hour test parameters
On Feb 23, 2011, at 5:47 PM, mix...@bigpond.com wrote: In reply to Horace Heffner's message of Tue, 22 Feb 2011 13:35:03 -0900: Hi, [snip] This 270kWh per 0.4 g if hydrogen is obviously well beyond chemical if the consumables actually are H and Ni. The energy E per H is: E = (270kwh) /(0.4 g * Na / (1.00797 gm/mol)) = 2.54x10^4 eV / H E = 25.4 keV per atom of H. This is about 2.5 times the ionization energy of the innermost electron of Ni. This is well under expected conventional weak reaction energies feasible between protons and Ni, but not out of the range of feasibility for hydrino reactions, or deflation fusion reactions. ..we also don't know how much of the H remained in the Ni after the reaction was finished. Yes, very true. The 25.4 keV is a *minimum* energy per hydrogen atom. However, if 30% of the Ni was converted to Cu, or even if readily observable quantities of new elements were created, then we have to expect much or even most of the hydrogen was consumed. Something doesn't add up here. There should have been a very observable drop in hydrogen pressure, because the hydrogen was shut off after initial loading. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:List of Rossi 18-hour test parameters
Robin, I don't understand- excuse where is the pressure of hydrogen measured? It is adsorbed absorbed in the nanometric nickel, the temperature increases there up to say 400 C- I don't think the reactor has a manometer on it. Peter On Thu, Feb 24, 2011 at 10:39 PM, Horace Heffner hheff...@mtaonline.netwrote: On Feb 23, 2011, at 5:47 PM, mix...@bigpond.com wrote: In reply to Horace Heffner's message of Tue, 22 Feb 2011 13:35:03 -0900: Hi, [snip] This 270kWh per 0.4 g if hydrogen is obviously well beyond chemical if the consumables actually are H and Ni. The energy E per H is: E = (270kwh) /(0.4 g * Na / (1.00797 gm/mol)) = 2.54x10^4 eV / H E = 25.4 keV per atom of H. This is about 2.5 times the ionization energy of the innermost electron of Ni. This is well under expected conventional weak reaction energies feasible between protons and Ni, but not out of the range of feasibility for hydrino reactions, or deflation fusion reactions. ..we also don't know how much of the H remained in the Ni after the reaction was finished. Yes, very true. The 25.4 keV is a *minimum* energy per hydrogen atom. However, if 30% of the Ni was converted to Cu, or even if readily observable quantities of new elements were created, then we have to expect much or even most of the hydrogen was consumed. Something doesn't add up here. There should have been a very observable drop in hydrogen pressure, because the hydrogen was shut off after initial loading. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
Re: [Vo]:List of Rossi 18-hour test parameters
In reply to Horace Heffner's message of Thu, 24 Feb 2011 11:39:36 -0900: Hi, [snip] ..we also don't know how much of the H remained in the Ni after the reaction was finished. Yes, very true. The 25.4 keV is a *minimum* energy per hydrogen atom. However, if 30% of the Ni was converted to Cu, or even if readily observable quantities of new elements were created, then we have to expect much or even most of the hydrogen was consumed. Something doesn't add up here. There should have been a very observable drop in hydrogen pressure, because the hydrogen was shut off after initial loading. Two different experiments. The Copper conversion is a report from Rossi about an earlier run. We don't what if anything was created/transmuted in the run where 0.4 gm H2 was consumed, so there isn't necessarily a conflict. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: [Vo]:List of Rossi 18-hour test parameters
In reply to Peter Gluck's message of Thu, 24 Feb 2011 22:48:52 +0200: Hi, [snip] Robin, I don't understand- excuse where is the pressure of hydrogen measured? It is adsorbed absorbed in the nanometric nickel, the temperature increases there up to say 400 C- I don't think the reactor has a manometer on it. Peter Was it measured at all? Does it matter? The calculations are based on the mass change, presumably of the Hydrogen bottle, so it's a measure of the H2 that went into the device, however just because Hydrogen went into the device, that doesn't necessarily mean that it underwent a nuclear reaction. Some (most?) is sure to have been left in the Ni as Ni hydride (/or Hydrinos? ;) On Thu, Feb 24, 2011 at 10:39 PM, Horace Heffner hheff...@mtaonline.netwrote: On Feb 23, 2011, at 5:47 PM, mix...@bigpond.com wrote: In reply to Horace Heffner's message of Tue, 22 Feb 2011 13:35:03 -0900: Hi, [snip] This 270kWh per 0.4 g if hydrogen is obviously well beyond chemical if the consumables actually are H and Ni. The energy E per H is: E = (270kwh) /(0.4 g * Na / (1.00797 gm/mol)) = 2.54x10^4 eV / H E = 25.4 keV per atom of H. This is about 2.5 times the ionization energy of the innermost electron of Ni. This is well under expected conventional weak reaction energies feasible between protons and Ni, but not out of the range of feasibility for hydrino reactions, or deflation fusion reactions. ..we also don't know how much of the H remained in the Ni after the reaction was finished. Yes, very true. The 25.4 keV is a *minimum* energy per hydrogen atom. However, if 30% of the Ni was converted to Cu, or even if readily observable quantities of new elements were created, then we have to expect much or even most of the hydrogen was consumed. Something doesn't add up here. There should have been a very observable drop in hydrogen pressure, because the hydrogen was shut off after initial loading. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: [Vo]:List of Rossi 18-hour test parameters
On Feb 24, 2011, at 12:19 PM, mix...@bigpond.com wrote: In reply to Horace Heffner's message of Thu, 24 Feb 2011 11:39:36 -0900: Hi, [snip] ..we also don't know how much of the H remained in the Ni after the reaction was finished. Yes, very true. The 25.4 keV is a *minimum* energy per hydrogen atom. However, if 30% of the Ni was converted to Cu, or even if readily observable quantities of new elements were created, then we have to expect much or even most of the hydrogen was consumed. Something doesn't add up here. There should have been a very observable drop in hydrogen pressure, because the hydrogen was shut off after initial loading. Two different experiments. The Copper conversion is a report from Rossi about an earlier run. We don't what if anything was created/transmuted in the run where 0.4 gm H2 was consumed, so there isn't necessarily a conflict. Yes, right. I keep blurring or confusing the lies between the various tests and the patent itself. I don't even know if the Ni container was sealed. And, as Peter pointed out, there was no pressure gage used: On Feb 24, 2011, at 11:48 AM, Peter Gluck wrote: Robin, I don't understand- excuse where is the pressure of hydrogen measured? It is adsorbed absorbed in the nanometric nickel, the temperature increases there up to say 400 C- I don't think the reactor has a manometer on it. Peter Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:List of Rossi 18-hour test parameters
In reply to Horace Heffner's message of Tue, 22 Feb 2011 13:35:03 -0900: Hi, [snip] This 270kWh per 0.4 g if hydrogen is obviously well beyond chemical if the consumables actually are H and Ni. The energy E per H is: E = (270kwh) /(0.4 g * Na / (1.00797 gm/mol)) = 2.54x10^4 eV / H E = 25.4 keV per atom of H. This is about 2.5 times the ionization energy of the innermost electron of Ni. This is well under expected conventional weak reaction energies feasible between protons and Ni, but not out of the range of feasibility for hydrino reactions, or deflation fusion reactions. ..we also don't know how much of the H remained in the Ni after the reaction was finished. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
[Vo]:List of Rossi 18-hour test parameters
A source close to the recent 18-hour test of the Rossi device gave me the following figures. These are approximations. Flow rate: 3,000 L/h = 833 ml/s. Input temperature: 15°C Output temperature ~20°C Input power from control electronics: variable, average 80 W, closer to 20 W for 6 hours Notes from Jed 5°C temperature difference * 833 ml = 4,165 cal/s = 17,493 W 3,000 L/h seems like a lot but it is 793 gallons/h, which is how much a medium-sized $120 ornamental pond pump produces. Peter I think it would have been better to throttle back the flow rate somewhat. 15°C is probably tap water temperature. A 5°C temperature difference can easily be measured with confidence. The control electronics input of ~80 W is in line with what was reported for tests before Jan. 14. Input was high on that day because something went wrong with the controls, with cracked welding as described in the Levi report. - Jed
Re: [Vo]:List of Rossi 18-hour test parameters
More notes I do not know if they used a pump, or simply let the water flow from the tap. I have used both methods at various times, and so has Dennis Cravens, although not for such a large flow rate. They said they checked the flow rate several times which I assume means it was measured manually, with a bucket and stop watch. You might think that the flow rate would fluctuate significantly over 18 hours, but in my experience, using either tap water pressure or something like a 700 gallon/hour (gph) pond pump, the flow rate is quite stable over many hours. With a pond pump, you can use a small plastic throttle to set a lower flow rate. It stays constant longer than you might think. I have tested this out of curiosity. With an actual outdoor pond, it will change gradually over a week, as gunk accumulates in the pump. I do not have a 700 gph pond pump. I have a 170 gph circulation pump, and also a 1/3 HP sump pump that I think is rated 25 gpm, about twice the flow of the Rossi test. As far as I can see, the only likely error with this setup would be measuring the temperature too close the energy source within the gadget. Based on the photo, McKubre thought the outlet thermocouple was too close to the likely source of energy. As I mentioned, the NRL 10 kW test bed system has much better arrangement of temperature sensors and flow meters. However, with input power of only ~80 W and a flow rate of 833 ml/s, without excess heat the temperature difference would be 0.02°C. I doubt you could detect that with this arrangement. The difference between 0.02°C and ~5°C is gigantic. - Jed
Re: [Vo]:List of Rossi 18-hour test parameters
Here is some additional info on the 18-hour test. I do not think I will add this to the News section. It can wait for a paper from Levi. This may have been reported here by Cousin Peter: Approximately 0.4 g of hydrogen was consumed in 18 hours. This is based on what sounds like a crude estimate to me: measuring the weight of the hydrogen tank before and after the test with the electronic weight scale. The weight scale has a margin of error of 0.1 gram. They measured a 0.3 g difference and they assume it was actually closer to ~0.4. Total energy production was ~1,037 MJ. This seems like much less than you get from a fusion reaction with 0.4 g of hydrogen. Hydrogen fusion yields 1.35 * 10E7 per kilogram says this source, Table 1: http://gltrs.grc.nasa.gov/reports/1996/TM-107030.pdf So for 0.4 g that would be 54,000 MJ. This is ~1000 MJ, so it is off by a factor of 54. I guess that isn't such a big difference given the crudeness of these measurements. My guess is that hydrogen leaking or absorbing into the materials far outweighs the hydrogen consumed by the reaction. Unless . . . UNLESS! . . . I don't know . . . unless Mills is right? Or the W-L theory is right? It ain't my bailiwick. The experts in theory such as Krivit can hash this out. - Jed
Re: [Vo]:List of Rossi 18-hour test parameters
On Feb 22, 2011, at 11:34 AM, Jed Rothwell wrote: Here is some additional info on the 18-hour test. I do not think I will add this to the News section. It can wait for a paper from Levi. This may have been reported here by Cousin Peter: Approximately 0.4 g of hydrogen was consumed in 18 hours. This is based on what sounds like a crude estimate to me: measuring the weight of the hydrogen tank before and after the test with the electronic weight scale. The weight scale has a margin of error of 0.1 gram. They measured a 0.3 g difference and they assume it was actually closer to ~0.4. Total energy production was ~1,037 MJ. This seems like much less than you get from a fusion reaction with 0.4 g of hydrogen. Hydrogen fusion yields 1.35 * 10E7 per kilogram says this source, Table 1: http://gltrs.grc.nasa.gov/reports/1996/TM-107030.pdf So for 0.4 g that would be 54,000 MJ. This is ~1000 MJ, so it is off by a factor of 54. I guess that isn't such a big difference given the crudeness of these measurements. My guess is that hydrogen leaking or absorbing into the materials far outweighs the hydrogen consumed by the reaction. Unless . . . UNLESS! . . . I don't know . . . unless Mills is right? Or the W-L theory is right? It ain't my bailiwick. The experts in theory such as Krivit can hash this out. - Jed This 270kWh per 0.4 g if hydrogen is obviously well beyond chemical if the consumables actually are H and Ni. The energy E per H is: E = (270kwh) /(0.4 g * Na / (1.00797 gm/mol)) = 2.54x10^4 eV / H E = 25.4 keV per atom of H. This is about 2.5 times the ionization energy of the innermost electron of Ni. This is well under expected conventional weak reaction energies feasible between protons and Ni, but not out of the range of feasibility for hydrino reactions, or deflation fusion reactions. Deflation fusion reactions which do not involve the weak force can trigger shuffles between electron quantum levels post reaction, due to the post fusion reaction electron escape, and thus radiate a significant amount of x-ray and EUV energy. Here are some candidate Ni + H deflation fusion reactions, not involving the weak force, all of which show a net initial energy deficit, but positive net reaction energy, thus making strong force reactions feasible which generate x- rays and EUV: 58Ni28 + 2 p* -- 32S16 + 28Si14 + 1.859 MeV [-15.209 MeV] (H_Ni:1) 60Ni28 + 2 p* -- 32S16 + 30Si14 + 00.554 MeV [-16.327 MeV] (H_Ni:2) 60Ni28 + 2 p* -- 34S16 + 28Si14 + 1.530 MeV [-15.351 MeV] (H_Ni:3) 60Ni28 + 2 p* -- 50Cr24 + 12C6 + 00.365 MeV [-16.516 MeV] (H_Ni:4) 60Ni28 + 2 p* -- 58Ni28 + 4He2 + 7.909 MeV [-8.973 MeV] (H_Ni:5) 61Ni28 + 2 p* -- 33S16 + 30Si14 + 1.376 MeV [-15.416 MeV] (H_Ni:6) 61Ni28 + 2 p* -- 34S16 + 29Si14 + 2.184 MeV [-14.608 MeV] (H_Ni:7) 61Ni28 + 2 p* -- 47Ti22 + 16O8 + 00.026 MeV [-16.765 MeV] (H_Ni:8) 62Ni28 + p* -- 59Co27 + 4He2 + 00.346 MeV [-7.760 MeV] (H_Ni:9) 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (H_Ni:10) 62Ni28 + 2 p* -- 34S16 + 30Si14 + 2.197 MeV [-14.507 MeV] (H_Ni:11) 62Ni28 + 2 p* -- 48Ti22 + 16O8 + 1.057 MeV [-15.647 MeV] (H_Ni:12) 62Ni28 + 2 p* -- 52Cr24 + 12C6 + 3.249 MeV [-13.455 MeV] (H_Ni:13) 62Ni28 + 2 p* -- 60Ni28 + 4He2 + 9.879 MeV [-6.825 MeV] (H_Ni:14) 62Ni28 + 2 p* -- 63Cu29 + 1H1 + 6.122 MeV [-10.582 MeV] (H_Ni:15) 62Ni28 + 2 p* -- 64Zn30 + 13.835 MeV [-2.869 MeV] (H_Ni:16) 64Ni28 + p* -- 65Cu29 + 7.453 MeV [-0.569 MeV] (H_Ni:17) 64Ni28 + 2 p* -- 36S16 + 30Si14 + 2.576 MeV [-13.958 MeV] (H_Ni:18) 64Ni28 + 2 p* -- 50Ti22 + 16O8 + 3.642 MeV [-12.891 MeV] (H_Ni:19) 64Ni28 + 2 p* -- 54Cr24 + 12C6 + 4.411 MeV [-12.122 MeV] (H_Ni:20) 64Ni28 + 2 p* -- 62Ni28 + 4He2 + 11.800 MeV [-4.734 MeV] (H_Ni:21) 64Ni28 + 2 p* -- 65Cu29 + 1H1 + 7.453 MeV [-9.080 MeV] (H_Ni:22) 64Ni28 + 2 p* -- 66Zn30 + 16.378 MeV [-0.155 MeV] (H_Ni:23) taken from: http://www.mtaonline.net/~hheffner/RptH http://www.mtaonline.net/~hheffner/dfRpt Fusion Product Chart for Ni + n p reactions Relative Percent Abs. 0 10 20 30 40 50 60 70 80 90 100 Z Percent El.||||||||||| 1 3.142H |*** 2 10.106 He |* 6 1.019C |* 8 00.489 O |* 14 00.804 Si |* 16 00.804 S |* 22 00.489 Ti |* 24 1.019 Cr |* 27 00.068 Co |* 28 10.038 Ni |* 29 56.507 Cu |** 30 15.517 Zn |** ||||||||||| 0 10 20 30 40 50 60 7080 90 100 The above chart is merely a very approximate visual aid to show feasible reaction product probabilities by a rule of thumb estimate. Copper is visualized as a most likely product. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:List of Rossi 18-hour test parameters
Horace Heffner wrote: The above chart is merely a very approximate visual aid to show feasible reaction product probabilities by a rule of thumb estimate. Copper is visualized as a most likely product. Izzatso? So you think the reports of copper can be explained by your theory? - Jed
Re: [Vo]:List of Rossi 18-hour test parameters
On Feb 22, 2011, at 2:11 PM, Jed Rothwell wrote: Horace Heffner wrote: The above chart is merely a very approximate visual aid to show feasible reaction product probabilities by a rule of thumb estimate. Copper is visualized as a most likely product. Izzatso? So you think the reports of copper can be explained by your theory? - Jed Not 30% of *actual* copper from Ni, as I posted earlier. The predominant element from ordinary LENR is copper, but only from the lesser abundance Ni isotopes. If a 30% conversion to copper is actually observed (which seems questionable at this point), then 58Ni and/or 60Ni must be involved. This and an alternative explanation (however tenuously speculative) for high *apparent* copper percentages were posted here earlier and appended below. I should also note the feasibility of deflation fusion reactions with hypernuclei, which could have unanticipated outcomes. It is known hypernuclei can support (bind to) up to two sigma+ or lambdas. A sigma+ can decay into an ordinary proton (plus other stuff), so this could provide a direct pathway to ordinary copper. On Jan 25, 2011, at 1:40 PM, Horace Heffner wrote: If the experiment is not a boondoggle, and there was actually observed by Rossi a 30% conversion of *all* the Ni to Cu, then it could simply be the copper is not really copper. It would then seem necessary 58Ni must be involved. I showed some potential strange reactions earlier: http://www.mail-archive.com/vortex-l@eskimo.com/msg41755.html You can see that one of them, as a subreaction: p (938.27 MeV/c2) + e - sigma+ (1189.3 MeV/c2) + K0 ( 497.6 MeV/ c2) + e would replace a proton in the new copper with a sigma+. The resulting hyperon copper [copper hypernuclei] would be chemically indistinguishable from copper. I have no idea how long such material might be stable, or what the trigger energy would be to force decay kinetically if it is otherwise stable. Trigger energies for light hyperons [hypernuclei], like helium, are very low, on the order of 20 kEv. As I noted earlier, the following reactions work fine creating ordinary Ni in the deflation fusion process: 62Ni28 + p* -- 63Cu29 + 6.122 MeV [-1.984 MeV] (B_Ni:28) 64Ni28 + p* -- 65Cu29 + 7.453 MeV [-0.569 MeV] (B_Ni:60) However, if 30% quantities of Cu are actually found, then some 58Ni must be transmuted to non-radioactive copper. We know 59Cu is radioactive. We don't know if 59Cu with a sigma+ replacing a proton is stable, or quasi-stable. Note also, that the neutral lambda0 reactions can both create transmuted Ni which appears to have added neutrons . This could happen numerous times per Ni. In this way 59Ni , 60Ni, 61Ni, and 62Ni hyperons [hypernuclei] containing lambda0 particles could be created. These could then be transmuted by an ordinary transmutation, or a sigma+ creating transmutation, to produce what appears to the eye to be normal Cu, but which is not. A sample from Rossi's device showing in mass spectroscopy an unusual amount of 59Ni, and no signs of EC, would be an indictor this is happening. All this is extremely speculative, especially given that we know almost nothing about Rossi's device. I do find it worrisome that the gamma counts were irregular as the counter was moved about by hand. If strange quark reactions are taking place in the device, then the signature would be K0_long particles, which in part decay into positrons. They would act like neutral neutrons close to experiment, and then can decay a meter or more away from the experiment, endangering the operators. The gamma counts might actually increase with distance up to a meter away from the experiment, if the K0's are normal, further if their low excitation energy permits a longer half-life. One thing I do now feel fairly confidant is possible, that apparently no else believes is possible, is that strange pairs exist, are created from the vacuum, within protons and neutrons, and that high mass deflated electrons, if they exist, can catalyze virtual strange quark separation into real independent quarks resident in separate fissioned particles. If this is truly feasible and safely engineerable, then infinite Isp drives are feasible, as is light speed travel, as well as an infinite source of energy. On Jan 21, 2011, at 4:23 PM, Horace Heffner wrote: On Jan 21, 2011, at 8:31 AM, Peter Gluck wrote: That device working for 6 months has produced approx. 50,000 kWhours heat. Can this be explained by the reaction of transmutation of Ni to Cu? Considering first 300 grams of nichel...? Rossi can tell how much Ni is uesd - if he will. Am important rough energy balance anyway. Peter There are some very fundamental issues, and mysteries involved. The fundamental questions relate to exactly what reactions are involved. Some do not produce copper, so the new copper content only
