[The Rossi reactor] Bologna, January 14, 2011
by Jed Rothwell The experiment has been underway at U. Bologna since mid-December 2010. It has been done several times. Several professors with expertise in related subjects such as calorimetry are involved. LIST OF MAIN EQUIPMENT IN EXPERIMENT A hydrogen tank mounted on a weight scale which is accurate to 0.1 g 10 liter tank reservoir, which is refilled as needed during the run Displacement pump Tube from pump to Rossi device (The Rossi device is known as an "ECat") Outlet tube from the Rossi device, which emits hot water or steam Thermocouples in the reservoir, ambient air and the outlet tube An HD37AB1347 IAQ Monitor (Delta Ohm) to measure the relative humidity of the steam. This is to confirm that it is dry steam; that is, steam only, with no water droplets. Alternating-current heater used to bring the Rossi device up the working temperature METHOD The reservoir water temperature is measured at 13°C, ambient air at 23°C. The heater is set to about 1000 W to heat up the Rossi device. Hydrogen is admitted to the Rossi device. The displacement pump is turned on, injecting water into the Rossi device at 292 ml/min. The water comes out as warm water at first, then as a mixture of steam and water, and finally after about 30 minutes, as dry steam. This is confirmed with the relative humidity meter. As the device heats up, heater power is reduced to around 400 W. RESULTS The test run on January 14 lasted for 1 hour. After the first 30 minutes the outlet flow became dry steam. The enthalpy during this last 30 minutes can be computed very simply, based on the heat capacity of water (4.2 kJ/kgK) and heat of vaporization of water (2260 kJ/kg): Mass of water 8.8 kg Temperature change 87°C Energy to bring water to 100°C: 87°C*4.2*8.8 kg = 3,216 kJ Energy to vaporize 10 kg of water: 2260*8.8 = 19,888 kJ Total: 23,107 kJ Duration 30 minutes = 1800 seconds Power 12,837 W, minus auxiliary power ~12 kW There were two potential ways in which input power might have been measured incorrectly: heater power, and the hydrogen, which might have burned if air had been present in the cell. The heater power was measured at 400 W. It could not have been much higher that this, because it is plugged into an ordinary wall outlet. Even if a wall socket could supply 12 kW, the heater electric wire would burn. During the test runs the weight of the hydrogen tank did not measurably decrease, so less than 0.1 g of hydrogen was consumed. 0.1 g of hydrogen is 0.1 mole, which makes 0.05 mole of water. The heat of formation of water is 286 kJ/mole, so if the hydrogen had been burned it would have produced less than 14.3 kJ. I uploaded that to the News section. I was tempted to add: "Hey, Richard Garwin: here's your cuppa tea, big guy!" I will soon upload a more detailed description by Mike Melich, and I hope I can add Prof. Levi's report. I think it is all but certain these results are real. They cannot be a mistake, and fraud seems unlikely to me. - Jed --------- Jones wrote: Here is the website of the company founded by Andrea Rossi and others a few years ago. This company funded and owns the technology in question. http://www.lti-global.com/index.php However, apparently there has been some kind of falling-out with Rossi, and as you can see there is no mention of any of this on the website. It seems he is being marginalized. The company has changed focus to so-called "clean-coal". Sad. They have no comment about Rossi, who was operating out of a different branch (New Hampshire). They have large DARPA grants, unrelated to the LENR cell, and do not want to compromise those. You may or may not agree, but it is clear to me that this drama in Bologna was hastily staged, not ready for prime-time, and will end up being a disaster for Rossi and LENR in general - when all of the details emerge. First off, he will sell not a single unit in the USA without an NRC license, which is complicated, costly and takes years. As for Europe, where the need for inexpensive energy is greater, who knows? The best thing that could happen, IMHO, is that the Italian military, their Pentagon equivalent, will take over the program and work something out with LTI as to the IP. Jones ----------- Jed wrote: I revised the H2 flow measurement part already. The first report I will upload today is by Melich. This week or next we should have one by Prof. Levi. These people are busy, which is why it took so long for them to give my report the once-over, and even they overlooked the part about weighing the H2 bottle. That is what they told me -- I have the handwritten notes, but it is clearly wrong. The part about the electric wires I observed myself, from the video and photos. It is just a "reality check" observation. I would like to know more about how the steam was condensed. They must have flushed it out of the room, down a drain. Otherwise they would end up with a very hot large bucket of water. Someone here correctly pointed out that venting that much steam into the room would make it like a steam bath. A medium size water heater at Home Depot is 5.5 kW. A sauna takes about 6 kW, I think. Jed ----------- Stephen wrote: Jones, I often disagree with you, but this time I have to say your suspicions ring a chord. Something doesn't smell right here. Please check me on this, because I'm not sure I've got it right. And feel free to yell at me; I realize I'm going kind of far on not much evidence. * The basic work for this process was done by other (clearly legitimate) researchers some years back, but the usual problems with reproducibility etc dogged them. * Rossi, apparently building on that earlier work, has found the Holy Grail: He can produce large quantities of high grade heat on demand with a reproducible process. /Everybody in the field wants this. Heck, almost everybody on Earth who knows anything about energy wants this./ * When answering questions about this, Rossi seems to downplay the truly earthshaking nature of this work, saying it's something that works, doesn't matter how, and he wants to sell heaters. (I _/think/_ I got that right -- from a post of Jed's but I don't have it in front of me right now.) /I have the impression that he never talks about how he's leapfrogged everybody and how fabulous this result is ... is that right? * I don't see any acknowledgment in Rossi's comments of the serious difficulties which may arise in attempting to go straight to a salable product. /This all sounds so much like what we've seen of perpetual motion machine vendors ... they downplay the earth-shattering theoretical aspects and talk about how they're just going to sell devices./ * Nobody knows the details of the process except Rossi. * Rossi is keeping the secret ingredient secret so nobody can steal his work. Failure to reveal all has interfered with getting a patent. It has also made it impossible for anyone to attempt a replication. /If he says he's going to reveal it at the end of the patent process, that means another two years before he tells anyone what was in the box -- if I understood what I read on Vortex. And until someone has enough information to attempt a replication, there's no solid way to test his claims./ * Rossi, unlike the earlier workers, is apparently not a trained physicist or electrochemist. In fact, from what I've read here, it's not clear what his degree is in, or if he's got one. By all means yell at me if I've got this wrong, but if I've got it right, it's an important point -- outsiders can make breakthroughs, and dishonest outsiders can make breakthroughs too, but darn, it's /rare/. * Rossi has a past which includes possible con-artist work. If this isn't a huge red flag I don't know the meaning of the word "red flag". It seems to me there's just one piece missing from the puzzle: Is there a financial incentive for this demo? In short, */are there investors in the background?/* If there are, then there's a financial incentive for Rossi to produce a convincing demo, and in that case I'd say /hold onto your wallet/. Now, Jed has said some interesting things about this: * "They should not try to sell practical devices at this stage. Sooner or later they will run into huge problems..." If Rossi really has the Grail, then Jed's comment is presumably correct, and this isn't the best approach. But if Rossi is faking it, then a black-box demo is /exactly/ what he needs to do in order to keep investment dollars coming in. * "... I did not get the sense he is trying to scam someone, or hide something ... " "I have talked to many researchers and inventors who seemed much less honest." Of course. If Rossi's not on the up-and-up, then one thing's sure: He's really good at fooling people. Good con artists may /seem/ totally honest. Honest people are often not as careful to /appear/ honest as dishonest ones! * "I do not follow the work of Mills closely, but I am not aware that he has demonstrated a heat-producing device as impressive as this, or on such a large scale. Needless to say, no other cold fusion researcher has come close." Yes, indeed, Rossi hit a home run first crack out of the box. It's as though Edison demonstrated a 1000 watt mercury vapor floodlight as his first lightbulb. Is it too good to be true? * "I cannot think of any way this result could be faked." Right -- Rossi's good. But there's a black box in the middle. As far as I know, nobody who watched the Statue of Liberty disappear a few years back caught on to how it was done. In summary, I really, really, really don't like "black box" demonstrations in an area where everybody is desperate for a solution. -------- On 01/17/2011 12:52 PM, Jones Beene wrote: Stephen - There are a few other details in the "big picture" that are not common knowledge, but should be mentioned. This is not really a breakthrough in one sense, but that all depends on how "public" you think the demo is/was. After all, it did show up on the internet. Does that make you believe it was really an open demo? Stephen wrote: That's a semantic question, and it depends on what you mean by "open". But in any case a demo (of any sort) by a single researcher proves nothing, unless you are convinced of that researcher's honesty. 10 kW output is too large to be an error. Either it's real or it's faked. If Ed Storms demonstrated a 10 kW reactor, it would be Game Over and Our Side Won, because I'm sure he would never fake anything. If David Copperfield demonstrated a 10 kW reactor I would be convinced it was a fake, and I'd be extremely amused that I couldn't see how he'd done it. --------- I was going to mention this before I saw Peter's message, but he beat me to it. On 01/17/2011 11:14 AM, P.J van Noorden wrote: "Hello Jed, How do we know that all the water ( 8.8 l) evaporated? Was the Rossi device weighted before and after the test? The diameter of the device is about 10 cm, so there could still be a few liters inside after the experiment." This is another example of the disastrous consequences of depending on a "black box" test. The stuff coming out could have been dry steam, or it could have been hot air. In fact, unless the "dry steam" was recondensed and the water which resulted was weighed, all we know for sure is that Rossi has demonstrated a device which made some quantity of water /vanish/. The person presenting the demonstration -- Rossi -- claims he turned it into steam. What proof is there of that? With a single demonstration, in which only one researcher knows what's inside the box, unless you have rock solid confidence in that researcher, you should take /nothing/ for granted. Once again, this is also probably not the "trick". In fact, I don't know what the "trick" might be; chances are, if there's a "trick", it's something far cleverer than any idea we'll come up with here. But without solid evidence to the contrary, there is no way to prove that there is no "trick". Without full disclosure and independent replication there is no "solid evidence". -------- Stephen A. Lawrence wrote: If the person doing the demonstration is not honest, you cannot take /anything/ for granted. And that is why issues with Rossi's background are so important. Jed wrote: If all of the people doing the demonstration are dishonest then you cannot take anything for granted. If Rossi alone is a crook that would make no difference. The calorimetry was designed by the others and the instruments are their property. Rossi cannot fool a thermocouple or a power meter. To engineer the 54 A wall socket, Rossi would have to go to the lab secretly and rewire the place, and then substitute a superconducting wire for the heater power supplies so that the wire does not burn up, and then he would have to replace professors power meter with one that looks exactly identical but gives the wrong values. That sort of thing might happen in a pulp thriller or James Bond movie, but not in real life. This kind of scenario falls in the "rats drinking water in Mizuno's lab" category. Regarding the quality of the steam, if it is dry that makes the computation simple. If it is wet that reduces the excess enthalpy somewhat, but it does not eliminate it. Assuming the heater is at 400 W, that's 400 W * 60 s = 24,000 J/min, or 5,714 calories. The flow rate is 292 ml/min so the water temperature would rise 20°C, to 33°C. Not even close to boiling, wet or dry. The outlet temperature was measured at 101°C, by the way. Let me add that fact to the description in the News section . . . I have encountered genuine energy scams and incompetent researchers. It is obvious they are wrong. They do not begin to fool me, and it is inconceivable they would full experienced professors who have been doing calorimetry and electrical measurements for 50 years. As I said, when people who propose the hypothesis that this might be a scam or a trick, I think it is incumbent upon them to explain how this trick might work. All hypothesis must be rigorously supported. This is a simple physics experiment, albeit one with a black box in the middle. There are some complicated cold fusion experiments with iffy results that might be faked, or at least "shaded." Some are shaded, by wishful thinking. This is not among them. The laws of physics are well defined in this case. I do not see how it could be something like a staged magic trick. Such tricks fool the eye, in any case. They never fool instruments. Penn and Teller cannot change the values displayed by a power meter. - Jed --------- On Sun, Jan 16, 2011 at 8:50 PM, Jones Beene wrote: Sell reactors to whom? Terry wrote: Iran? Stuxnet free. --------- On Sun, Jan 16, 2011 at 5:42 PM, <mix...@bigpond.com> wrote: Who says you are trying to ionize the Hydrogen? Terry wrote: Not exactly accurate. If the electron capture concept is correct, then only dissociation is necessary. Further reading on my part shows that dissociation can occur on metal surfaces. I think Horace wrote on this. I'm looking over his "stuff". Normally, we would have heard from Horace. I hope he is okay. --------- Hi All, 1-17-11 This discussion of the Rossi reactor is fascinating. There was an extensive discussion in 1997 of electron capture. I'm enclosing a few excerpts below. Jack Smith ----------- Date: Wed, 19 Nov 1997 On Tue, 18 Nov 1997, Mitchell Swartz wrote: Martin, your reading further will note that the calculation does NOT work on other atoms with that accuracy. Martin Sevior <msev...@axnd05.cern.ch> wrote: You snipped a large part of my answer that described why QED is widely accepted ... Classical Electrodynamics certainly does not predict pair production. Positrons were the specific great prediction of Dirac in 1928 in his formulation of Relativisitic Quantum Mechanics. This was later expanded into QED by Feynman, Schwinger and Tommonaga for which they were awarded the Nobel prize. Amongst other things, QED calculations show that the Lamb shift occurs. QED is in no danger of being over-turned. It is absolutely mainstream and is widely taught in Graduate Courses in Universities throughout the world ... On Tue, 18 Nov 1997, Mitchell Swartz wrote: Martin, Pair production (requiring an interaction of an electromagnetic photon of greater than about two rest energies, and triplet production (conservation of energy and momentrum from a collision with an electron) DO exist. These are mainstream and well known to radiation physics. Conservation of energy and momentum is involved. Matter appearing from pure vacuum has no basis, because ... of no conservation of energy and momentum. Martin Sevior <msev...@axnd05.cern.ch> wrote: But this is just what Quantum Field Theory says happens. The effect is real, as measured by the Lamb shift, the Casimir attraction and in the precisely predicted value of the magnetic moments of the electron and muon. On Tue, 18 Nov 1997, Mitchell Swartz wrote: Martin, sorry that my requirement for conservation of energy stands (until proven otherwise), but that is the belief of mainstream science. Martin Sevior <msev...@axnd05.cern.ch> wrote: Mitch don't teach Grandma to suck eggs. This is absolutely my field. If I walk out of office and ask a random sample of the 5,000 High Energy Physicists who spend time here at CERN, do think there is Zero Point Energy in the Vacuum? 99% of them will say "yes". After I explain Hal Putoff's example of the plates colliding due to Casimir forces, 90% of the last 1% will agree. That to me shows that ZPE in the Vacuum is mainstream in the correct meaning of the word. I suggest you try to understand what I have written ... Martin Sevior -------------- Source: "Frederick J. Sparber" <frederick.spar...@worldnet.att.net> To: <vortex-l@eskimo.com> Subject: Re: Note on electron capture and nuclear remediation Date: Wed, 19 Nov 1997 03:36:42 -0700 -----Original Message----- Source: Horace Heffner <hheff...@corecom.net> To: vortex-l@eskimo.com <vortex-l@eskimo.com> Date: Tuesday, November 18, 1997 11:23 PM Subject: Re: Note on electron capture and nuclear remediation On the bright side you can "orbit" electrons in a simple cylindrical magnetron with an equilibrium radius of 0.1 meters. Only problem is they have to be going at five times the speed of light, which creates a bit of a chore. :-) Regards, Frederick Horace Heffner wrote: It does seem that the E.C. rate should go up measureably though, even if the sample is only shaken accoustically, or placed in a strong gradient. The combination of the two should be even better. So, if that can work, it seems like spinning at 144,000,000,000 rpm should work miracles, if that's not a miracle in itself! Regards, Horace Heffner On the other hand, positron decay of p is not out of the question, but the half life is much longer than we have to discuss the possibility. 8^) When I said "However, I think electron capture from conduction band electrons has been observed," I was clearly not talking about H, as the term "conduction band" then has no meaning. I would also like to make a small joke. A physicist would look at your equations of mass (energy) balance and say the reaction was denied. A mathematician might look at your equations and say the resulting neutrino clearly must have negative mass. 8^) Regards, Horace Heffner -------------- On Sun, 16 Nov 1997 tstol...@aol.com wrote: A question for Martin Sevior or Michael Schaffer (or any other physicist who might be reading Vortex-L): what happens when the proton of hydrogen captures the electron? I've heard that this capture occurs from time to time, since there's a finite, non-zero probability for the event. At 10:14 AM 11/17/97, Martin Sevior wrote: If you mean: e- + p => n + neutrino This reaction does not occur in free space because it is endothermic. It occurs inside Supernovae after all the nuclear fuel has been exhasted and if Gravitational energy is sufficient to drive the reaction. ----------- Source: tess...@oro.net (Ross Tessien) Subject: Re: Electron capture by protons. In "free space" there are energetic electrons and other particles due to the presence of cosmic rays. Thus, even in free space this process will occur from time to time. But it is due to the nucleus , ie the proton, being bombarded by an energetic particle. An energetic electron could blast into the nucleus, blast out a neutrino and form a neutron, at which point the other electron if in the valence would fly away due to a lack of positive charge nearby, or if you had a free proton ion, then you would now have a free neutron. This process can manifest in accelerators, and no doubt occurs all the time in our atmosphere. Though the numbers of reactions are fairly low I would guess, we still are able to detect lots of reactions from cosmic rays via the particles such as muons that are emitted, or via Cerenkov radiation from relativistic particles coming into the atmosphere where c is slower than out in free space. This is how high energy astrophysics gets a substantial amount of information, using ground telescopes to monitor the particles and or the Cerenkov light emissions. -------- Source: hheff...@corecom.net (Horace Heffner) Subject: Re: Electron capture by protons. It appears Tom is talking about electron capture from an orbital electron. Some portion of the electron distribution of psi^2 is inside the nucleus, thus the weak interaction can occur with non-zero probability. Don't know the half life for H but it is a very long time as the range of the weak force is very short, and the hydrogen nucleus very small. The half life of e- + 23V26 --> 22Ti27 + neutrino is about 600 days. The reaction e- + p => n + neutrino has not been observed in "free space", as you put it, because an unbound electron typically can not stay around close to a nucleus long enough for the weak force to do its thing. However, I think electron capture from conduction band electrons has been observed. In seems to me unbound electron capture might be observable in one of those intense electron beams generated by one of those superconducting skin accelerators (don't recall the name of those things that look like metal intestines in peristalsis.) The problem might be signature. The signature is typically provided by shell disruption due to loss of an inner shell electron to the nucleus, and generation of sprectra characteristic of the Z-1 element. Capture of an unbound electron by a non-ionized atom would not necessarily disrupt inner shells. However, it also seems to me highly unlikely that target atoms would remain non-ionized in the presence of such an intense electron beam! Possibly gamma emission when nucleus is left in excited state? Another issue is that positron emission is a competing process. Heavy elements are more likely to capture an orbital electron, while light elements are more likely to decay by positron emission. Might be a *very* long wait for p to decay though. 8^) Regards, Horace Heffner -------- Source: hheff...@corecom.net (Horace Heffner) Subject: Re: Electron capture by protons. Looking at the table of isotopes in the 93-94 edition of the CRC Handbook, the Decay Mode/Energy column, the first electron capture (E.C.) entry is for 7Be, at 0.862 MeV and half life of only 53.28 days. To the far right of the page is a column for gamma ray intensity. The entry for our example 7Be shows 0.4776/10.4, meaning 0.4776 MeV of the 0.862 MeV from the 7Be decay is carried away by gammas in 10.4 percent of the decays. I would assume that in the other 89.6 percent of the decays 100 percent of the enrgy is carried away by the kinetic energy of the nucleus and the neutrino in standard kinetic reaction proportions. Since the neutrino is light, it will carry away the vast portion of the energy when there is no gamma. However, on the bright side, the collapse of the atomic shells resulting from the suddenly missing inner electron is "free" energy in the sense it is stolen from the universe in the form of entropy. Did I get all that right? Happy hunting for the E.C. of choice. Regards, Horace Heffner ---------- On Tue, 18 Nov 1997, Robin van Spaandonk wrote: I'm still trying to find out if the energy associated with electron capture (where this is positive), is carried away by the resultant neutrino, or by gamma radiation. Can anyone tell me? -------- Source: Martin Sevior <msev...@axnd05.cern.ch> Subject: Re: Electron capture by protons. Hi Robin, Like everything to do with Nuclear Physics the answer is "it depends". Sometimes the electron capture process leaves the resultant nucleus in an excited state, in which case the energy of the transition from the excited to ground state is carried by a gamma. Often this is not the case and most of the energy is carried by the neutrino. In all cases there is a cascade of X-rays due to electrons from higher orbitals falling into the hole left when the electron was captured by the nucleus ... Martin Sevior
