Re: [Vo]:Minor progress
Rossi has already exposed it by injecting the high frequencies. Any power meter used to check this would likely be subject to the same inaccuracy. I suggest a simple frquency meter with a lead touched to the dpf. - Original Message - From: Jouni Valkonen jounivalko...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, November 09, 2011 7:18 PM Subject: Re: [Vo]:Minor progress 2011/11/10 Joe Catania zrosumg...@aol.com: requency generator inout? Is there any more info on that? I can tell you one thing- the power company is not going to be too happy with Rossi or whoever runs one of these things when they find out they are meter cheaters! I think too that the falsification of input energy measurements is most plausible way to do the cheat. However this cheat has a hole, because anyone of the guests could just plug a power meter to their iPad and then make a quick check of the calibration of ammeters. These kind of fakes that are based on input electricity, I think, are too easy to expose. –Jouni Ps. it was possible to check for guest also every else variable that was measurable. Including gamma radiation.
Re: [Vo]:Minor progress
requency generator inout? Is there any more info on that? I can tell you one thing- the power company is not going to be too happy with Rossi or whoever runs one of these things when they find out they are meter cheaters! - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Wednesday, November 09, 2011 5:48 PM Subject: Re: [Vo]:Minor progress First let me correct an earlier statement in this thread. In regards to the pipe conduits to the interior box from the front of the outer box I said: There are actually four: 1 water, 1 gas, 2 for frequency generator input. That was meant to say: There are actually four: 1 gas, 1 main power, and 2 for frequency generator input. I think it is especially odd that the two frequency generator conduits, one above the interior box flanges, one below, are 1 1/4 inch pipe, while the conduit for the main power is only 1 pipe. It seems reasonable to speculate as to what might require, and be located inside, the large pipes. On Nov 9, 2011, at 10:35 AM, Jouni Valkonen wrote: 2011/11/9 Horace Heffner hheff...@mtaonline.net: The material I have analyzed fits inside the 30x30x30 cm box. The 50x60x35 cm exterior box to which others refer is irrelevant, except when water levels and temperatures are simulated. I am responding to this post only because words I did not issue have been put in my mouth. If you think that there is a 30×30×30 cm³ black box Black is your wording, not mine, in relation to color. Those dimensions came from Mats Lewan's report which I reference in my paper: http://www.mtaonline.net/~hheffner/Rossi6Oct2011Review.pdf I also determined from the photos that the actual dimension is closer to 30.3 cm. Any reference to a black box I might have made in my writing was not literal, but I don't recall referring to the interior box as black. The color might be called rusty dirty scale deposited on aluminum. (it was not mine impression, but my impression is based on indirect conclusion made that I do not remember anyone saying seen such a large black box inside), If you had read my paper you would have seen a photograph appended of the 30x30x30 cm interior box, with sealed pipe fittings going into it from the front of the larger box. and you think that Rossi is an evil criminal and fraudster, I did not at any time say that. Those are your words, not mine. It is you who repeatedly jumps to the fraud conclusion, not me. Fraud or self delusion are of course possibilities I recognize, as do many others, especially given Rossi's inability numerous times to provide anything other than highly flawed calorimetry data, or refusal to admit the importance of such mundane scientific concepts as controls, etc. The lives of billions of people are affected by Rossi's actions now, regardless the outcome. Why will he never make the tiny incremental effort required to properly demonstrate he produces nuclear heat? If he does not give a damn about the rest of the world, only his marketing strategy, then that indeed does not speak highly of his morality, does it? His bizarre behavior raises logical questions. Has he no faith in himself to produce his claimed results? Has his discovery gone the way of Patterson's beads? Are his results now merely amplified artifacts, or insufficient to be commercially viable? Is he unable to run for multiple days, much less multiple months as claimed? Only Rossi himself is responsible for creating these doubts. What I *would* be happy to do is show the possibility that a logical construction can produce the observed results. Given the 37% extra output heat that I mistakenly built into my spread sheet by biasing the temperature, it does not take an unfeasible error in the Tout reading to accommodate a good match of result by simulation. Given it is not even known for sure the Tout thermocouple was in direct contact with metal, this is not a far reach. However, if I could show even a possible fraud based mechanism exists which simulates the results with the given inputs, that would be sufficient to demonstrate the calorimetry requires improving. It should be sufficient to quell at least some of the ridiculous non-quantitative arm waving true believer arguments made here, but probably won't. You do see the difference between calling Rossi an evil criminal fraudster and showing a logical mechanism exists which reproduces the experiment outputs given only the experiment inputs, don't you? The purpose for the latter is to provide some motivation or justification for a customer demand for appropriate due diligence. The former would serve no purpose. Many people in the blogosphere have said or implied the E-cat is a fraud, so the former would be useless, in addition to being unsubstantiated arm waving. then why do you cannot understand, that it is also trivial to fit internal chemical power source to 30×30×30 cm³ black box? If you had read my paper,
Re: [Vo]:About that Frequency Generator
I've spoken to Lewan about the device producing frequencies. I believe it to be a meter cheater in that it produces high frequency energy that cannot be tracked accurately by the clamp-on ammeter. Notice energy in= energy out in Oct test before dpf is used. After switching on this device all hell beaks loose and the E-cat appears to be producinbg anonmalous energy but this is easily explained by the fact that the device produces more power than the ~100W logged by the meter. - Original Message - From: Robert Leguillon robert.leguil...@hotmail.com To: vortex-l@eskimo.com Sent: Thursday, October 27, 2011 5:54 PM Subject: [Vo]:About that Frequency Generator Has anyone seen a photo? Does anyone know what make/model? Does anyone know the specific purpose it was serving? Does anyone know how it was hooked into the circuit? Was it electrically connected to the heater? Was it electrically connected to the E-Cat at all? Had anyone heard any reference to it before October 6? Was it needed for self-sustaining operation in September? David Roberson dlrober...@aol.com wrote: Here is an analysis that I just completed. It shows that Rossi has achieved what he has been suggesting. LENR is real and will only get better with time. Dave I have been reviewing the data obtained during the September and October tests and can now confirm that there is proof that the ECAT generates a large amount of excess energy. I would assume that the skeptic ones among our group will read this report and realize that the proof has been before us for a long time but is not easy to discern. Start with a graph of the temperature readings at the ECAT output thermocouple referred to as T2 during the October test. You must have a graph that includes all of the temperature-time pairs supplied by Mats Lewan in his Excel file. My analysis is as follows: Mr. Rossi performed a carefully controlled ECAT heating procedure. The pattern of setting the input power to “5”, then “6”, all the way to “9” is intended to slowly allow the internal components to reach ideal operation temperature. The reactor reaches equilibrium somewhere around 13000 seconds into the test. Once this has been achieved, a series of on and off power pulses (“9”) is applied to the core. This series of power applications occur at a frequency that is high enough to be well filtered by the low pass nature of the internal ECAT heat flow mechanism. This is evident by the smooth curve of T2 versus time that shows up from 13000 seconds through about 15500 seconds. It is important to note that the T2 curve is slowly falling throughout this time duration. The average T2 reading is 120.5 C and has a slight negative slope. I realized that the implication was that the ECAT output power would slowly begin to fall along with this curve since that temperature drives the check valve, etc. What can we make of this curve of T2 versus time? It turns out that a lot of information is revealed. I did an analysis of the input power pulse waveform starting at 11400 seconds until 14881 seconds to get the average filtered component of the drive signal and obtained a net power input of 1252 watts. Then I realized that all of this power must be causing the ECAT core module to reach some operational temperature. It then responds to the elevated temperature and the LENR effect within starts to generate extra energy. Next, the energy associated with the input power (1252 joules/second * time) adds to the newly released energy of the core. The two of these energy sources end up as heat which proceeds to add energy to the water contained within the ECAT. The water will now either increases or decrease in temperature, depending upon the heat that is lost from the system. We know of at least three loss paths. The main output leading to the heat exchanger, leakage water or vapor from the case, and heat leaving the case due to radiation or other means. All that we need to prove is that the sum of these loss factors is greater than 1252 watts in order to prove beyond doubt that LENR is functioning within the Rossi device. There is one subtle point to explain. There is a very slight negative slope in T2 versus time during this region. I performed a quick calculation and found that the power lost within the water tank as a result of this slope is ((122-120.7) C x 4.188 joules/(C-grams) x 3 grams)/1860 seconds = 87 joules/seconds or 87 watts. This calculation reveals that a very small increase in the drive power will allow the temperature of the water bath and hence output power to remain constant. This is a very important point to make. The ECAT will continue to put out the same power for as long as this input power (1252 watts) is applied. This may not be the ideal self-sustain mode that we all love, but it is significant. Of course I was not content to leave out the additional knowledge revealed by this region of the T2 temperature
Re: [Vo]:1 MW plant testing is underway.
Physics is natural science. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, October 23, 2011 2:40 PM Subject: Re: [Vo]:1 MW plant testing is underway. That's excellent news. Very open of Rossi. Entirely reasonable. We complain about Rossi's habits, but you have give him credit for allowing a lot of access to this tests, and for giving out a great deal of information. The problem is not that he is unwilling to share data. It is that his tests do not produce good data, and he does not write scientific papers. People have said that Rossi is a liar, or he exaggerates, or he cannot be trusted. As I see it, he has a split personality. When he talks about business or personal matters, I think he gets excited and he blurts out nonsense. I don't take this nonsense seriously. He scapegoats people -- including me. He can be devious, sometimes planting misinformation to cause dissension. I know he does that, because he did it to me several times. However, when it comes to engineering-based technical claims, as far as I know, Rossi is the soul of honestly. He has often made astounding claims that seem utterly impossible. As far as I know, all the ones that have been put to the test turned out to be true. I do not know about that factory heater that ran for a year. Cousin Peter says he cannot believe it. I can't be sure it is real, but I am sure it is unwise to bet against Rossi. I do not think there is a shred of evidence that Rossi has ever tried to use a hidden source of energy, fake instruments, or any other kind of fraud. It would be much harder to do this with his cells and reactors than with any previous cold fusion devices, because the scale of the reaction is so much larger. He is careless with instruments, and sloppy, and this sometimes obscures the results. That is not a deliberate effort to hide results or escape from scrutiny. It is what it appears to be: sloppy. Lots of people are like that. Some geniuses such are Arata are like that. Many programmers write unstructured spaghetti code too. It is not because they are devious or they want to sabotage the project or infuriate their co-workers. It is because they are sloppy. They should be promoted to management where they will cause less harm. Many engineers and inventors have this kind of split personality. Edison is a famous example. He was a sharp dealer as they said in the 19th century. Sharp dealing -- cheating, breaking contracts, and taking unfair advantage -- was widespread and considered normal back then. He put on Dog and Pony show exhibits of his inventions. When investors asked him how much progress he was making, he lied so extravagantly, it would have embarrassed a data processing project manager circa 1972, when computer programming was at the lowest ebb of reliability and projects routinely went off the rails. Edison did all of that, but he would never lie to himself, to his coworkers, or in a serious technical discussion. He did not have it in him to lie. Most engineers and programmers do not. It would be analogous to a farmer who neglects to plant seeds and then expects a crop to grow. Every technician in history has known that you cannot fool Mother Nature. I cannot judge Rossi's assertions about theory or transmutations. Theoreticians tell me they are bunk. I suppose they are, but Rossi is unaware of that. They are not lies. I have also learned to believe everything Rossi says about his operational plans. When he said he was building a 1 MW reactor, I believed him. He says he will try to turn it on. I have no doubt he means it. I just hope he does not blow himself up, or get arrested for operating it without a license. I hope that someone dissuades him but I doubt anyone will. If he changes his mind at the last minute, I would never accuse him of lying. A person who does cutting edge research who does not frequently change his mind, his plans, and his entire approach will fail catastrophically. Flexibility is essential to that job, as it is to a general fighting a battle. As Eisenhower said, no battle plan survives contact with the enemy. You have to respond to things as they are, not as you hoped they would be. I wish Rossi would change course more often, not less often. I think Rossi is careless with instruments because he is old fashioned and he agrees with Fleischmann and me that direct observation is the best science. It is better than proof by instruments and calculation. He does not bother to write down the thermocouple readings, or insert an SD card, because he thinks that the heat continuing for 4 hours is all the proof anyone can ask for. Worrying about the thermocouples when you have a reactor too hot to touch is ridiculous. It is useless nitpicking in the face of definitive, first-principle proof that you can literally feel with your hand. The instruments are the icing on the cake; the
Re: [Vo]:Krivit report on Oct. 6 Rossi test
Strange, I just commented on The EEStory.com thst input energy looks equal to output up to the time Lewan turns on the infamous device that creates frequencies which to my mind is clearly a meter beater. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Thursday, October 13, 2011 2:19 PM Subject: Re: [Vo]:Krivit report on Oct. 6 Rossi test I wrote: it clearly shows that by the end of the warm-up period, I put energy already exceeded input energy. Meant to say: OUTPUT energy already exceeded input . . . The point is, this is balance of heat generation and loss, similar to a financial balance sheet. Not complicated. - Jed
Re: [Vo]:Krivit report on Oct. 6 Rossi test
Yes your imagination is vivid. I'm refering to the clamp on ammeter of course. Nice try. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Thursday, October 13, 2011 2:40 PM Subject: Re: [Vo]:Krivit report on Oct. 6 Rossi test Joe Catania wrote: Strange, I just commented on The EEStory.com thst input energy looks equal to output up to the time Lewan turns on the infamous device that creates frequencies which to my mind is clearly a meter beater. You have a vivid imagination. Perhaps you should inform Termometro and whoever made that analog, non-electronic water meter. - Jed
Re: [Vo]:Krivit report on Oct. 6 Rossi test
I should? I've already done what I should do. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Thursday, October 13, 2011 3:11 PM Subject: Re: [Vo]:Krivit report on Oct. 6 Rossi test Joe Catania wrote: Yes your imagination is vivid. I'm refering to the clamp on ammeter of course. Nice try. Ah, so you should inform Digimaster company that their instrument can be fooled with a simple device. Please let us know their reply. - Jed
Re: [Vo]:Krivit report on Oct. 6 Rossi test
The point is more that the choice of a meter that can't measure high frequency is requisite for this hoax. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Thursday, October 13, 2011 3:56 PM Subject: Re: [Vo]:Krivit report on Oct. 6 Rossi test Joe Catania wrote: I should? I've already done what I should do. Nonsense! You should tell the company. They will be grateful you have discovered this terrible problem with their instrument. They may pay you a large sum of money for helping them find this problem. You should inform all of the other companies that make ammeters. You will become a highly paid industry consultant and they will invite you to give keynote speeches at their trade shows. Think about this. You have discovered a way to fool an instrument that is used throughout the world, often in critical applications. You are the second person to discover an easy way to make this instrument display the wrong numbers. (Rossi was the first.) No doubt thousands of industrial accidents occur everyday because this happens inadvertently. When you tell the world why these instruments do not work, you will be a hero. I had no idea it was so easy to interfere with the operation of such a widely used, critical instrument. Do you also happen know how to make a thermocouple produce the wrong answer? By ESP perhaps? Can you use your superpowers to change the answer on my Casio calculator, while you are at it? I hope that you will use your powers for good and not evil. - Jed
Re: [Vo]:Krivit report on Oct. 6 Rossi test
I suggest you accept my treatment was theoretical. Rossi should comply, not me. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Thursday, October 13, 2011 4:14 PM Subject: Re: [Vo]:Krivit report on Oct. 6 Rossi test Joe Catania zrosumg...@aol.com wrote: The point is more that the choice of a meter that can't measure high frequency is requisite for this hoax. Two different meters, in this case. I suggest you make a box that fools both of these meters to demonstrate that it can be done. - Jed
Re: [Vo]:Krivit report on Oct. 6 Rossi test
Jed if you can't explain your position you are a fraud. Me building a test circuit is not going to vindicate you. Lewan hasn't answered queries about the freq device but most people know that cheap meters cannot follow this well. If current and voltage aren't in phase its no good. If high freqs distrurb meter likewise. I'm saying the coicidence is glaring that excess energy is only produced after this device starts therefore it not measuring power accurately. How that, in your mind, requires me to test it is beyond everyone here. The idea here is not to assume that the power measurements are valid. Proove that! - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Thursday, October 13, 2011 5:02 PM Subject: Re: [Vo]:Krivit report on Oct. 6 Rossi test Joe Catania wrote: I suggest you accept my treatment was theoretical. Rossi should comply, not me. Rossi set up two meters, a Digitmaster DM201 and a Mastech MS2102. You are saying you know a theoretical way to fool both of them, simultaneously, with some sort of external signal generator or electrical waveform. If you are not willing to do an experiment proving this claim of yours, I think you should at least explain your theory here. Otherwise, why should anyone believe that you actually know how to do this? Rossi has already done a credible measurement of input amperage. I would have preferred a wattmeter and something like a battery backup, but using two separate meters does reduce the likelihood of error. I do not know much about these meters but it seems to me that an external signal generator is unlikely to affect both of them the same way simultaneously. It seems to me that you are now making a claim contrary to conventional knowledge, so you should back it up if you want people to take you seriously. The ball is in your court. I was being flippant before but I mean that seriously. A skeptical assertion dismissing evidence does not get a free pass. You have to prove your point just as Rossi must prove his. - Jed
[Vo]:Energy Analyzer for E-Cat
It occurs to me that the means they are using to measure power is prone to error. An energy analyzer would be the best way to do it. If there's any reactance in the circuit they power calculations they use would be inaccurate.
Re: [Vo]:Energy Analyzer for E-Cat
http://www.omega.com/heaters/pdf/HEATER_INTRO_BAND_REF.pdf, as you can see this one uses a coiled wire. If not designed properly this could have high inductance. Also Lewan say he injects high frequency at one point. - Original Message - From: Terry Blanton hohlr...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 12:04 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Yep, it's called power factor. You're really on top of things, Joe! T On Wed, Oct 12, 2011 at 11:59 AM, Joe Catania zrosumg...@aol.com wrote: It occurs to me that the means they are using to measure power is prone to error. An energy analyzer would be the best way to do it. If there's any reactance in the circuit they power calculations they use would be inaccurate.
Re: [Vo]:Energy Analyzer for E-Cat
Nonsense, high frequencies are subject to skin effect. - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 1:31 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Am 12.10.2011 18:39, schrieb Joe Catania: http://www.omega.com/heaters/pdf/HEATER_INTRO_BAND_REF.pdf, as you can see this one uses a coiled wire. If not designed properly this could have high inductance. If you suceed to make a remarkably high inductance without an iron core, then you should patent and market this. You will get rich and famous. Also Lewan say he injects high frequency at one point. A series inductance will shift the current phase and reduce the power. Power maximum is, when inductance is zero. This is even more true with high frequencies. Kind regards, Peter - Original Message - From: Terry Blanton hohlr...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 12:04 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Yep, it's called power factor. You're really on top of things, Joe! T On Wed, Oct 12, 2011 at 11:59 AM, Joe Catania zrosumg...@aol.com wrote: It occurs to me that the means they are using to measure power is prone to error. An energy analyzer would be the best way to do it. If there's any reactance in the circuit they power calculations they use would be inaccurate.
Re: [Vo]:Energy Analyzer for E-Cat
The real point is that line current and voltage may not be in phase to begin with. Heckert is not knowledable and must resort to ad hominems. For instance there are most likely eddy currents induced in the band heaters. - Original Message - From: Jouni Valkonen jounivalko...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 1:58 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat 2011/10/12 Joe Catania zrosumg...@aol.com: It occurs to me that the means they are using to measure power is prone to error. An energy analyzer would be the best way to do it. If there's any reactance in the circuit they power calculations they use would be inaccurate. Indeed, they used very cheap (€40) clamp ammeter (DIGIMASTER DM201) that can only measure AC current in very limited frequency range at 50-60 Hz. But if there are spikes in the feed, it may measure even 50% too low values. However, I think that spikes would show up in voltage meter. It can measure also DC current, but with separate DC settings of course. So could it be plausible to feed DC-current along with AC and clamp ammeter would not notice a thing? Then only conducting wire's capacity could limit how much electric power Rossi is feeding into E-Cat. –Jouni http://www.zetabishop.it/product/8630/Pinza-Amperometrica-Digimaster-DM-201-VCA-ACA.asp
Re: [Vo]:Energy Analyzer for E-Cat
But an analyzer would eliminate doubt. You'd actually be measuring power instead of relying of neglecting something you know nothing about. - Original Message - From: Alan J Fletcher a...@well.com To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 2:07 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat At 10:58 AM 10/12/2011, Jouni Valkonen wrote: 2011/10/12 Joe Catania zrosumg...@aol.com: It can measure also DC current, but with separate DC settings of course. So could it be plausible to feed DC-current along with AC and clamp ammeter would not notice a thing? Then only conducting wire's capacity could limit how much electric power Rossi is feeding into E-Cat. In September Lewan checked the DC periodically, and found it was zero.
Re: [Vo]:Energy Analyzer for E-Cat
Another problem is magnetic and electric field coupling to dipolar matter. This can dissipate energy as well. - Original Message - From: Man on Bridges manonbrid...@aim.com To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 2:42 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Hi, On 12-10-2011 20:08, Joe Catania wrote: But an analyzer would eliminate doubt. You'd actually be measuring power instead of relying of neglecting something you know nothing about. A cheap secondhand CRT oscilloscope up to 10 MHz would show a lot of information as well ;-) Kind regards, MoB
Re: [Vo]:Energy Analyzer for E-Cat
Heckert, why don't you go stand on a corner with a tin cup. Yes skin effect is important at high frequencies especiall in the case of certain pulse shapes. I'm a physicist and I happen to have intimate knowledge of just hgow important skin effect can be. Inductive reactance isn't just proportional to inductance its proportional to frequency as well. No doubt there may be considerable iron nearby the current. Alternating electric and magnetic fields can induce electric polarization and eddie currents which can dissiapte heat. - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 2:54 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Am 12.10.2011 20:00, schrieb Joe Catania: Nonsense, high frequencies are subject to skin effect. So you have studied electrical engineering? I have. Unfortunately I dont know the proper english expressions to explain this, but it is trivial, anyway. For these frequencies that are in question here and with those thick cables you can almost forget the skin effect. - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 1:31 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Am 12.10.2011 18:39, schrieb Joe Catania: http://www.omega.com/heaters/pdf/HEATER_INTRO_BAND_REF.pdf, as you can see this one uses a coiled wire. If not designed properly this could have high inductance. If you suceed to make a remarkably high inductance without an iron core, then you should patent and market this. You will get rich and famous. Also Lewan say he injects high frequency at one point. A series inductance will shift the current phase and reduce the power. Power maximum is, when inductance is zero. This is even more true with high frequencies. Kind regards, Peter - Original Message - From: Terry Blanton hohlr...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 12:04 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Yep, it's called power factor. You're really on top of things, Joe! T On Wed, Oct 12, 2011 at 11:59 AM, Joe Catania zrosumg...@aol.com wrote: It occurs to me that the means they are using to measure power is prone to error. An energy analyzer would be the best way to do it. If there's any reactance in the circuit they power calculations they use would be inaccurate.
Re: EXTERNAL: Re: [Vo]:Energy Analyzer for E-Cat
No you don't understand skin effect. - Original Message - From: Roarty, Francis X francis.x.roa...@lmco.com To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 3:15 PM Subject: RE: EXTERNAL: Re: [Vo]:Energy Analyzer for E-Cat Joe, Peter is correct -XL =6.28fl and real current thru the coil is choked off even though the dc resistance looks like a short. skin effect is only relevant on small diameter wires but in any case would also be choked off by the impeadance just like the DC path. The impedance effectively places itself in series with the circuit limiting any currents even through magnetic couplings - whatever momentary current goes one way is stored in the field and then repaid on the alternate cycle. A Coil would get hot to the touch if it really dropped the power like a resistor but it does not get hot because it is only storing it not dissipating it. Fran -Original Message- From: Joe Catania [mailto:zrosumg...@aol.com] Sent: Wednesday, October 12, 2011 2:01 PM To: vortex-l@eskimo.com Subject: EXTERNAL: Re: [Vo]:Energy Analyzer for E-Cat Nonsense, high frequencies are subject to skin effect. - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 1:31 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Am 12.10.2011 18:39, schrieb Joe Catania: http://www.omega.com/heaters/pdf/HEATER_INTRO_BAND_REF.pdf, as you can see this one uses a coiled wire. If not designed properly this could have high inductance. If you suceed to make a remarkably high inductance without an iron core, then you should patent and market this. You will get rich and famous. Also Lewan say he injects high frequency at one point. A series inductance will shift the current phase and reduce the power. Power maximum is, when inductance is zero. This is even more true with high frequencies. Kind regards, Peter - Original Message - From: Terry Blanton hohlr...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 12:04 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Yep, it's called power factor. You're really on top of things, Joe! T On Wed, Oct 12, 2011 at 11:59 AM, Joe Catania zrosumg...@aol.com wrote: It occurs to me that the means they are using to measure power is prone to error. An energy analyzer would be the best way to do it. If there's any reactance in the circuit they power calculations they use would be inaccurate.
Re: [Vo]:Energy Analyzer for E-Cat
You interst me in the way Rossi may be going about this. It seems you are suggesting Rossi is studying from the book , How to Scam the Masses and Become Rich without Detection. The high-frequency injection certainly would seem to be in the bag of tricks for many scammers. Its well known that pulsed power will blow a fuse which can't be blown by the same DC level. - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 3:57 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Another possibility is to make a small modification to each component: Measure the flow rate a little bit wrong, measure temperatures a little bit wrong, calculate a little bit wrong, introduce so much errors and inaccuracies that a single one -if discovered- would prove nothing, but all together make an energy gain. If he is a real talented chaos experimenter who doesnt doublecheck and who doesnt make plausibility tests, as this seems to be the case, then he might have done just this with a long series of dilletantic experiments and he could really believe in the energy production.
Re: [Vo]:Energy Analyzer for E-Cat
You don't understand skin effect well. Injecting high frequencies obviously may fool the meter. I think it would be safer to heat with DC. - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 3:38 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat The skin effect can be neglected, because it adds a pure ohm resitance to the wire and the resistance is unknown anyway. And inductive resistance means that the power is smaller than U_rms*I_rms because there is phaseshift. Both effects reduce the heating power. So there are two possibilities remaining: Use a large crest factor or a high frequency that the meter cannot detect. I think we can exclude this. This would be too easy to detetect. Fraud would be much easier: The heat exchanger could be manipulated, so that only part of the water was heated. Because the thermal difference was so small, it would be almost impossible to detect. Another possibility is to make a small modification to each component: Measure the flow rate a little bit wrong, measure temperatures a little bit wrong, calculate a little bit wrong, introduce so much errors and inaccuracies that a single one -if discovered- would prove nothing, but all together make an energy gain. Am 12.10.2011 21:15, schrieb Joe Catania: Heckert, why don't you go stand on a corner with a tin cup. Yes skin effect is important at high frequencies especiall in the case of certain pulse shapes. I'm a physicist and I happen to have intimate knowledge of just hgow important skin effect can be. Inductive reactance isn't just proportional to inductance its proportional to frequency as well. No doubt there may be considerable iron nearby the current. Alternating electric and magnetic fields can induce electric polarization and eddie currents which can dissiapte heat. - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 2:54 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Am 12.10.2011 20:00, schrieb Joe Catania: Nonsense, high frequencies are subject to skin effect. So you have studied electrical engineering? I have. Unfortunately I dont know the proper english expressions to explain this, but it is trivial, anyway. For these frequencies that are in question here and with those thick cables you can almost forget the skin effect. - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 1:31 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Am 12.10.2011 18:39, schrieb Joe Catania: http://www.omega.com/heaters/pdf/HEATER_INTRO_BAND_REF.pdf, as you can see this one uses a coiled wire. If not designed properly this could have high inductance. If you suceed to make a remarkably high inductance without an iron core, then you should patent and market this. You will get rich and famous. Also Lewan say he injects high frequency at one point. A series inductance will shift the current phase and reduce the power. Power maximum is, when inductance is zero. This is even more true with high frequencies. Kind regards, Peter - Original Message - From: Terry Blanton hohlr...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, October 12, 2011 12:04 PM Subject: Re: [Vo]:Energy Analyzer for E-Cat Yep, it's called power factor. You're really on top of things, Joe! T On Wed, Oct 12, 2011 at 11:59 AM, Joe Catania zrosumg...@aol.com wrote: It occurs to me that the means they are using to measure power is prone to error. An energy analyzer would be the best way to do it. If there's any reactance in the circuit they power calculations they use would be inaccurate.
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
That appears to be a graph of power noy yemperature. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, October 09, 2011 9:24 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Joe Catania zrosumg...@aol.com wrote: No the band heater is at 900C but that metal block talk was only for illustrative purposes. Newtons LAw is irrelevant. Newton's law governs passive heat loss, which is what this has to be if there is not energy input and the flow rate does change. An insulated metal block that loses heat at a rate of 1W loses heat at the rate of 1W. You mention lack of monotonicity but what's the example (be specific, post link). Right here: http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg The temperature rises several times after the power is turned off. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
If its passive cooling? Excuse me but are we discussing something here? - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, October 09, 2011 9:41 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Excuse me I meant to say that the cooling rate must obey Newton's law if there is NO energy generation and the flow rate does NOT change. In other words, if it passive cooling in unchanging conditions. Lewan's observations and report show that the flow rate and other essential parameters did not change. - Jed
Re: [Vo]:Rossi heat exchanger fitting
Jed I'm not going to bother to comment on your very flawed analysis. It dosen't seem you want us to agree. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, October 09, 2011 10:54 PM Subject: Re: [Vo]:Rossi heat exchanger fitting Alan Fletcher a...@well.com wrote: A ton of water went through the heat exchanger -- but we don't know whether it heated up AT ALL. Oh give me a break Alan! Seriously, get real. There was STEAM going in one side and TAP WATER going in the other. How could it not be heated up AT ALL?!? What the hell do you think a heat exchanger does, anyway? If it does not get heated up AT ALL Rossi needs to get his money back from the heat exchanger company. All we know is that SOME water was boiled, that the internal eCat thermistor measured SOMETHING to be 120C, and that SOME water and/or steam made it to the heat exchanger and was able to affect the output thermocouple. But we don't have ANY idea how much water went through the eCat. You can see the hoses going from the sink to the eCat and the heat exchanger. Lewan measured the flow in both. Besides, it makes no difference how much went through the eCat; there was enough steam to make the inlet 120 deg C. You can quibble about how much boiling water there was, but it had to be enough for Lewan to hear it, and to make the insulated reactor surface. It wasn't 50 ml, that's for sure. It had to be a substantial amount. You know how much cooling power 10 L/min water has. A box of that size cannot produce heat for 4 hours and remain boiling and heating the heat-exchanger water with no input power. You could put the thermocouples anywhere you like in that heat exchanger box, and I guarantee that after an hour they will all register 25 deg C. The loading power could have heated a 90 kg chunk of metal to well over 100C But it didn't. The metal was 80 deg C. And it stayed at 80 deg C. Four hours after the power was cut, it was still at 80 deg C. If it was loaded and then unloaded, the temperature would have to drop! -- and that could have been used to heat a small flow of water to any desired temperature-vs-time pattern -- and would explain why there was the sound of boiling and why the surface of the eCat was hot. For crying out loud, look up the specific heat of metal. Read Heffner's analysis, p. 1, stored heat. Think about what loading or storing heat means. It means heating up the material. When you store, the temperature goes up. When you release the heat, the temperature goes down. When the temperature does not go up or down, there is no storage or release -- by definition. When the temperature is steady over 4 hours ago, no heat has been stored or released during that time. This reminds me of Krivit's latest hypothesis that 33 MJ were stored in the reactor. Before they turned off the power, the reactor and heat exchanger got hot, the heat balanced and then went exothermic so obviously all 33 MJ came out, plus some more. Not stored, right? Then, I suppose, the same 33 MJ did an about face, went back in, and came out again after they turned off the power. Zounds! Heat that appears twice! Call Vienna! -- as Howland Owl put it. I fear that in this test we have a cornucopia of experimental PROBLEMS. Yes there are many problems. I pointed out many of them. However, despite these problems, the first-principle proof is still obvious. You need to stop looking at the problems, and look at the proof instead. Stop inventing ad hoc nonsense about stored heat that does not change the temperature, or heat exchangers that do not exchange heat. Look at the facts, and do not be blinded or distracted by the problems. Those problems cannot change the conclusions this test forces upon the observer. Forget about those thermocouples if you like, and think only about the fact that the water was still boiling and the reactor was still hot 4 hours after the power was turned off. That fact, all by itself, is all the proof you can ask for. - Jed
Re: [Vo]:Rossi heat exchanger fitting
Jed I'm not going to bother to comment on your very flawed analysis. It dosen't seem you want us to agree. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, October 09, 2011 10:54 PM Subject: Re: [Vo]:Rossi heat exchanger fitting Alan Fletcher a...@well.com wrote: A ton of water went through the heat exchanger -- but we don't know whether it heated up AT ALL. Oh give me a break Alan! Seriously, get real. There was STEAM going in one side and TAP WATER going in the other. How could it not be heated up AT ALL?!? What the hell do you think a heat exchanger does, anyway? If it does not get heated up AT ALL Rossi needs to get his money back from the heat exchanger company. All we know is that SOME water was boiled, that the internal eCat thermistor measured SOMETHING to be 120C, and that SOME water and/or steam made it to the heat exchanger and was able to affect the output thermocouple. But we don't have ANY idea how much water went through the eCat. You can see the hoses going from the sink to the eCat and the heat exchanger. Lewan measured the flow in both. Besides, it makes no difference how much went through the eCat; there was enough steam to make the inlet 120 deg C. You can quibble about how much boiling water there was, but it had to be enough for Lewan to hear it, and to make the insulated reactor surface. It wasn't 50 ml, that's for sure. It had to be a substantial amount. You know how much cooling power 10 L/min water has. A box of that size cannot produce heat for 4 hours and remain boiling and heating the heat-exchanger water with no input power. You could put the thermocouples anywhere you like in that heat exchanger box, and I guarantee that after an hour they will all register 25 deg C. The loading power could have heated a 90 kg chunk of metal to well over 100C But it didn't. The metal was 80 deg C. And it stayed at 80 deg C. Four hours after the power was cut, it was still at 80 deg C. If it was loaded and then unloaded, the temperature would have to drop! -- and that could have been used to heat a small flow of water to any desired temperature-vs-time pattern -- and would explain why there was the sound of boiling and why the surface of the eCat was hot. For crying out loud, look up the specific heat of metal. Read Heffner's analysis, p. 1, stored heat. Think about what loading or storing heat means. It means heating up the material. When you store, the temperature goes up. When you release the heat, the temperature goes down. When the temperature does not go up or down, there is no storage or release -- by definition. When the temperature is steady over 4 hours ago, no heat has been stored or released during that time. This reminds me of Krivit's latest hypothesis that 33 MJ were stored in the reactor. Before they turned off the power, the reactor and heat exchanger got hot, the heat balanced and then went exothermic so obviously all 33 MJ came out, plus some more. Not stored, right? Then, I suppose, the same 33 MJ did an about face, went back in, and came out again after they turned off the power. Zounds! Heat that appears twice! Call Vienna! -- as Howland Owl put it. I fear that in this test we have a cornucopia of experimental PROBLEMS. Yes there are many problems. I pointed out many of them. However, despite these problems, the first-principle proof is still obvious. You need to stop looking at the problems, and look at the proof instead. Stop inventing ad hoc nonsense about stored heat that does not change the temperature, or heat exchangers that do not exchange heat. Look at the facts, and do not be blinded or distracted by the problems. Those problems cannot change the conclusions this test forces upon the observer. Forget about those thermocouples if you like, and think only about the fact that the water was still boiling and the reactor was still hot 4 hours after the power was turned off. That fact, all by itself, is all the proof you can ask for. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Newton's Law is irrelevant. Your the type of buffoon who believes that since there's an Ohms LAw every conductor obeys it. The temperature law the e-cat obeys is ostensibly written in the temperature data if we can consider that valid. Whether that confirms its Newton's Law or notr is not relevant to the dubunking of the CF myth. Cf is not being assumed and since it hasn't been shown we are correct in not assuming it. You still aren't able to show me the temperature data you say exists and is increasing. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 10:28 AM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Joe Catania zrosumg...@aol.com wrote: That appears to be a graph of power noy yemperature. It is derived from Lewan's temperature readings. The flow rate was unchanged so correspondence to the temperature is unchanged for the entire dataset. In other words, you could replace the vertical axis power numbers with the corresponding temperatures and it would look exactly the same. If its passive cooling? Excuse me but are we discussing something here? You claim the heat comes from heat storage with no input power. That would mean it is passive cooling, by definition. It has to follow Newton's law of cooling. That is how heat storage and release works. You keep talking about thermal inertia. I suggest you learn what that is, how it works, and what laws of physics govern it. - Jed
Re: [Vo]:Rossi heat exchanger fitting
I already said there was heat storage. We are not contesting me here Jed and that's what is clear. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 10:43 AM Subject: Re: [Vo]:Rossi heat exchanger fitting Joe Catania zrosumg...@aol.com wrote: Jed I'm not going to bother to comment on your very flawed analysis. It dosen't seem you want us to agree. You don't believe that heat storage means the temperature rises? Forget about me. You do not agree with Newton; that's your problem. What the heck do you think heat storage is, anyway? - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Newton's Law is irrelevant. Your the type of buffoon who believes that since there's an Ohms LAw every conductor obeys it. The temperature law the e-cat obeys is ostensibly written in the temperature data if we can consider that valid. Whether that confirms its Newton's Law or notr is not relevant to the dubunking of the CF myth. Cf is not being assumed and since it hasn't been shown we are correct in not assuming it. You still aren't able to show me the temperature data you say exists and is increasing. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 10:28 AM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Joe Catania zrosumg...@aol.com wrote: That appears to be a graph of power noy yemperature. It is derived from Lewan's temperature readings. The flow rate was unchanged so correspondence to the temperature is unchanged for the entire dataset. In other words, you could replace the vertical axis power numbers with the corresponding temperatures and it would look exactly the same. If its passive cooling? Excuse me but are we discussing something here? You claim the heat comes from heat storage with no input power. That would mean it is passive cooling, by definition. It has to follow Newton's law of cooling. That is how heat storage and release works. You keep talking about thermal inertia. I suggest you learn what that is, how it works, and what laws of physics govern it. - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
LOL. That's hypocritical. - Original Message - From: Rich Murray rmfor...@gmail.com To: vortex-l@eskimo.com; Rich Murray rmfor...@gmail.com; Rich Murray rmfor...@comcast.net Sent: Monday, October 10, 2011 12:49 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Jed Rothwell is a serious, intelligent, dedicated, honorable, careful, scientific layman with the highest motives to benefit our world -- he always acknowledges his bias clearly and openly. I think it would be much to his credit to agree that the term pathological skeptic is as unworthy in public discourse as buffoon. within infinite patience, Rich Murray
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
LOL. That's hypocritical. - Original Message - From: Rich Murray rmfor...@gmail.com To: vortex-l@eskimo.com; Rich Murray rmfor...@gmail.com; Rich Murray rmfor...@comcast.net Sent: Monday, October 10, 2011 12:49 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Jed Rothwell is a serious, intelligent, dedicated, honorable, careful, scientific layman with the highest motives to benefit our world -- he always acknowledges his bias clearly and openly. I think it would be much to his credit to agree that the term pathological skeptic is as unworthy in public discourse as buffoon. within infinite patience, Rich Murray
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Funny, you don't seem annoyed. All Jed is capable with regard to this matter is condescension. - Original Message - From: Stephen A. Lawrence sa...@pobox.com To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 12:33 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof On 11-10-10 11:04 AM, Joe Catania wrote: Newton's Law is irrelevant. Your the type of buffoon who ... And you, /Mister/ Catania, are apparently the type of poster who resorts to ad hominems when he's having trouble expressing himself clearly enough to get his point across. Jed's may be a lot of things, possibly even including wrong, but he's no buffoon. And you, /Mister/ Catania, are plonked. I don't need to see this kind of stuff on Vortex. (You are also apparently the type of poster who can't be bothered to proof read his posts for obvious typos before sending them, which also contributes needlessly to the annoyance level of this list.)
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
No that was part of the decor in a restaurant in Taormina. Its nice to know that the only thing that counts here is spelling (and self-affected narcissists). - Original Message - From: Terry Blanton hohlr...@gmail.com To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 1:41 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Quit picking on Catania who does not know the difference between 'your' and 'you're'. He passed away some time ago as is evidenced by this piccy of him surrounded by flowers. RIP JOE! http://www.theeestory.com/posts/199540 T
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
What do my posts matter anyway? Yes please block me. - Original Message - From: OrionWorks - Steven V Johnson svj.orionwo...@gmail.com To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 2:50 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Congratulations, Mr. Catania. Further posts from you will be routed to my block list. I'm sure you could care less. I guess the feeling is mutual. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
Since you know nothing of the e-cat your remarks have been dismissed. Yes it was prooveable in the September e-cat that the effects were purely based on thermal inertia. I suspect the same here. Rothwwell has not been able to substantiate his position which seems to be a blind acceptance of CF before aanyone heard of Rossi. I never made the claims you say I made. Yes there has been conversion and elaborate journalism on this point. You seem to confuse your total ignorance with lack of merit. You will regret that. - Original Message - From: Mark Iverson-ZeroPoint zeropo...@charter.net To: vortex-l@eskimo.com Sent: Monday, October 10, 2011 3:44 PM Subject: RE: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof From one narcissist to another... Seems ol Joe thinks he's converted the lot of us... http://www.theeestory.com/users/1681/posts# 80kgs of metal can easily store over 40MJ. It's not on the level of a discussion. My arguments have been extremely convincing as I think you can tell by the recent conversion of vortex members and Krivit. Joe Catania states, The band heater temp is ~900C. In September test my calculations show that boiling could be produced for many hours. There is certainly a massive amount of metal in the e-cat. Joe: So your reasoning is based on the band heater being 900C, and therefore the majority of the massive amount of metal in the E-Cat is at or near that same temperature. You sincerely think that everything underneath the insulation is anywhere near that temp? The melting point of lead is 327C, so we certainly know that the lead is no more than one-third 900C, or else we'd have a mass of molten lead on the table. In addition, with the irregularity of the shape of the plumbing, at least with the old, tubular design, it is unlikely that there is much physical contact between the lead shielding and the plumbing (water jacket), ergo, poor heat conduction between the plumbing and the lead, ergo, not much heat storage in the lead. Finally, the only thing that could be anywhere near 900C is the (stainless steel) core container that is the transfer medium between the reaction material (Ni-powder-hydrogen-catalyst) and the water outside the core container. Conclusion: Being that the only mass that could possibly be anywhere near 900C is the reactor core container, which might be a few kilograms, would you care to revise your ... not on the level of a discussion heat storage estimate??? -Mark
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
With 40MJ of heat in the system it would be impossible for the temperature to drop suddenly. I heat a block of steel to 900C, then I stop heating it, and drop a gram of water on it. What's the temperature? 900C. Notice there was no precipitous drop. Nor would there be after many grams of water. In fact 40MJ is stored in the metal. This is enough to boil ~20kg of water. Where are you getting 1.8 tons? - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, October 09, 2011 4:59 PM Subject: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Or if it is refutable, let's see someone make a serious effort to refute it. Stop quibbling about details. Get the heart of the matter, and tell us how a box of this size with no input power can boil water for 3 hours and remain at the same high temperature while you cool it with 1.8 tons of water. I wrote to some friends complaining about the test. My conclusion: Despite these problems . . . I think this test produced irrefutable proof of anomalous heat. Here is why I think so -- Look at the graph here: http://a2.sphotos.ak.fbcdn.net/hphotos-ak-ash4/304196_10150844451570375_818270374_20774905_1010742682_n.jpg Nothing happens until 13:22 when the steam begins to flow through the heat exchanger. At 15:13 output is a little higher than input, even though there is a great deal of heat unaccounted for, especially the water from the condensed steam, which they poured down the drain. At 15:50 the power is cut off. If there had been no source of anomalous heat, the power would have fallen off rapidly and monotonically, at the same rate it did after 19:55. It would have approached the zero line by 17:25. Actually, it would have approached zero before that, based on Newton's law of cooling. In other words, it would have been stone cold after 3 hours. During that time, 1.8 tons of water went through the cooling loop. It is inconceivable that an object of this size with no power input could have remained at the same high temperature the whole time. Yet Lewan reports that the surface of the reactor was still hot, and boiling could still be heard inside it. As you see, the temperature did not fall. It went up at 16:26. The cooling water flow rate was unchanged, so only a source of heat could have caused this. You can ignore the thermocouple data, and look only at the fact that it continued to boil for more than 3 hours after the power was turned off, and the reactor surface remained hot. That alone is rock solid proof. It is possible that the placement of the outlet thermocouple was flawed, and it recorded a value midway between the outlet cooling water temperature and the steam in the pipe next to that. I do not think much heat can cross from the steam pipe to the water pipe next to it. Alan Fletcher did a rigorous analysis to demonstrate this. The thermal mass of the cooling water was much larger than the steam, so the average temperature was closer to the water than the steam. However, for sake of argument let us assume the temperature was too high. In that case, we can ignore the actual temperature and look only at the temperature trends. We can look at relative temperatures. Whatever the temperature was, it goes up after the power turns off. It stays up. It stays at a higher level than it was when the power was on! Even if the actual temperature was half this value, it still should have fallen monotonically, as I said. This behavior is simply impossible without some source of heat, at some power level. I think that very little wicking from the hot water pipe occurred, so I expect the peak anomalous power was ~8 kW as shown in this graph. (I also ran this analysis and my complaints past Rossi himself.) - Jed
Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof
No the band heater is at 900C but that metal block talk was only for illustrative purposes. Newtons LAw is irrelevant. An insulated metal block that loses heat at a rate of 1W loses heat at the rate of 1W. You mention lack of monotonicity but what's the example (be specific, post link). - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, October 09, 2011 8:14 PM Subject: Re: [Vo]:Look at the BIG PICTURE and you will see this is irrefutable proof Joe Catania zrosumg...@aol.com wrote: With 40MJ of heat in the system it would be impossible for the temperature to drop suddenly. I heat a block of steel to 900C, then I stop heating it, and drop a gram of water on it. What's the temperature? 900C. Notice there was no precipitous drop. Please see Newton's law of cooling: https://www.math.duke.edu/education/ccp/materials/diffcalc/ozone/ozone1.html The other point you are overlooking is the drop is monotonic, that is Varying in such a way that it either never decreases or never increases. When heat is released from a system the way you describe, the temperature can only drop. It NEVER NEVER RISES. That is a fundamental physical law. Note also that this device was at 80 deg C, not 900 deg C. - Jed
Re: [Vo]:NyTeknik report on October 6th test
Lewan's report states that hydrogen pressure was lowered during shut-down. This is the angle they should have exploited. With constant heating and water flow conditions they should vary the hydrogen pressure and record the results. They should also try an inert gas like helium. - Original Message - From: Jed Rothwell jedrothw...@gmail.com To: vortex-l@eskimo.com Sent: Friday, October 07, 2011 10:59 AM Subject: Re: [Vo]:NyTeknik report on October 6th test peter.heck...@arcor.de wrote: If the heat exchanger has only 60% efficieny, then the energy loss is 5kW * 0.4 = 2kW. Where does the enrgy go? Energy cannot vanish magically, it must go into the ambient. Correct. It radiates into the surroundings, from the reactor and the heat exchanger. Lewan reported the reactor surface was between 60°C and 85°C. I presume he means at different times. I do not know how he measured that. It has a lot of surface area so it is radiating a lot of heat. Someone better physics and I can estimate how much. With a really good calorimeter having a high recovery rate, nearly all the heat ends up captured by the calorimeter. With the flow calorimeter it ends up heating the water. With a Seebeck calorimeter it may radiate out into the room, or if there is a water bath on the outside shell of the chamber to ensure a stable background, it will be captured by the water bath. This reactor is too large for most Seebeck calorimeters. I think even if the heat exchanger at this size (as visible in the video) has no insulation, it cannot lose 2kW. It is well isolated and the loss must be much lower. I believe the heat exchanger plus the reactor itself can radiate 2 kW. They look crude to me. Such things are inefficient. See photo of the two of them (in one box): http://www.nyteknik.se/incoming/article3284962.ece/BINARY/Test+of+E-cat+October+6+%28pdf%29 - Jed
Re: [Vo]:NyTeknik report on October 6th test
I have to disagree that the change in hydrogen pressure wouldn't be almost immediately obvious. IYou should get an immediate rise in delta T across the reactor which would immediately boost heat flow. Helium should confirm a null result- ie no CF and would be used as a control. You should be able to subtract out the helium data to account for thermal inertia and warm up and cool down w/ the heater.--- Original Message - From: Jouni Valkonen jounivalko...@gmail.com To: vortex-l@eskimo.com Sent: Friday, October 07, 2011 12:14 PM Subject: Re: [Vo]:NyTeknik report on October 6th test 2011/10/7 Joe Catania zrosumg...@aol.com: Lewan's report states that hydrogen pressure was lowered during shut-down. This is the angle they should have exploited. With constant heating and water flow conditions they should vary the hydrogen pressure and record the results. They should also try an inert gas like helium. Of course, but unfortunately there was not time to do such thing (doing such correlative analysis would take several days) . And also, reaction speed did not react too much for the reducing the hydrogen pressure. But test excluded all possible hidden power sources (E-Cat was weighted before and after the test). Therefore what would be the point of injecting helium? –Jouni
Re: [Vo]:July 7th E-Cat test report
I wouldn't evn take more output heat as input heat as the sine qua non. In fact there's nothing going on in the e-cat that can proove cold fusion- its not about a cold fusion proof, there just isn't one of those contemplated. If you want CF proof maybe look at the Navy's data. - Original Message - From: Robert Leguillon To: vortex-l@eskimo.com Sent: Thursday, October 06, 2011 1:37 PM Subject: RE: [Vo]:July 7th E-Cat test report I think that you're misunderstanding me. If-And-Only-If the power at the secondary is LESS than the peak power input to the primary, there will be arguments about the heat after death or self-sustaining operation. If the most energy that you put into the E-Cat is 1 kW, and 2 kW is observed at the output, then the H.A.D. operation is totally unnecessary, but may impress some people. However, if you put 1 kW into the input for two hours, seeing a slow build-to-parity at the secondary (where the secondary only achieves 1 kW), then how long the heat takes to decay when power is removed will be a bone of contention. I think H.A.D. could serve as a distraction. What we HAVE TO SEE is more kW at the secondary than is ever applied to the primary. Was that cogent? This was the prediction I'd supplied yesterday - that power gains would be reliant on the no input mode of operation, less than the peak power applied at the primary. And this would leave people arguing over residual, or stored, heat vs. a maintained reaction. I truly hope that they are observing 3kW out, and less than 10 Amps peak power consumption. -- Date: Thu, 6 Oct 2011 12:58:55 -0400 Subject: Re: [Vo]:July 7th E-Cat test report From: jedrothw...@gmail.com To: vortex-l@eskimo.com Robert Leguillon robert.leguil...@hotmail.com wrote: We can only hope and pray that there is more power observed on the secondary than is supplied to the primary during peak energy application. If gains are only observed during heat after death, we will be arguing the results ad infinitum. Why do you say that?!? It is much easier to be sure the heat is real when there is no input power. It is much more definitive, not less. What you say makes no sense to me. Please explain. - Jed
Re: [Vo]:Overall efficiency is not known but it is probably low
Your right after spending millions uselessly Rossi can always promote the e-cat as a very accurate calorimeter ( in fact the one that discovered profitable CF) and thus mark up the sale price even further. - Original Message - From: Jed Rothwell jedrothw...@gmail.com To: vortex-l@eskimo.com Sent: Thursday, October 06, 2011 4:27 PM Subject: [Vo]:Overall efficiency is not known but it is probably low Daniel Rocha wrote: You must not forget the losses due the conversion between the heat exchangers. If it was 70%, that means around 5KW for the core. I pulled 70% out of a hat, by the way. I do not know what the overall efficiency is. I am just guessing, based on large, crude experimental calorimeters I have seen in various labs and at Hydrodynamics, Inc. McKubre's calorimeter is superb, and it recovers something above 95% of the heat, as I recall. Or was it 98%? Anyway, the Rossi's reactor is the opposite of superb. It has a large surface area which must be hot and must be radiating a great deal of heat. Large, uninsulated boxes like this that are not engineered with multiple tubes inside and lots of internal heat transfer surface area recover no more than 80% in my experience. I do not know how efficient the heat exchanger is, but top-notch good industrial ones are about 90% efficient according to on-line sources. I have no idea what this heat exchanger looks like but if it is experimental equipment put together by Rossi or by professors in the last month I'll bet it is well below good industry equipment. So I am guessing maybe 80% again. That would be 64% recovery overall. The right way to do this is to perform a calibration with a joule heater boiling water. That would tell us the recovery rate. Knowing Rossi I'll that they did not do that. Anyway, it can't be anything close to 100%. You can bet the surface of that machine and of the heat exchanger was hot. How hot? I asked several people who attended the demonstration to try to measure that surface temperature but I doubt any of them did it. I don't think they had time to prepare for that. As I said this test was an improvement over previous ones but I expect I will find plenty of ways in which it could have been done better, such as calibrating and using a IR sensor. Having said that, we should not lose sight of the fact that finding out how much heat is lost from the system unaccounted for can only improve the numbers for Rossi. It can only strengthen the claim. I am sure that total output energy exceeded total input by a large measure. With 4 hours of heat after death no other result is possible. You cannot begin to store 4 hours of heat at 3.5 kW in a device this size. That notion is preposterous. If the heat recovery was 98% (which it could not be; that is far too high) this result is definitive. If the recovery was 70% or 40% it is even more definitive. You do not actually need to know what it was. Knowing it would be icing on the cake. In some early cold fusion experiments, there was only excess heat if you took into account of the measured losses from the calorimeter, which are measured by calibrating with a joule heater. In other words, you would only believe there was excess heat if you trusted the calibrations were done right, and the recovery rate was correctly measured. Such results were close to the margin. In Rossi's case, you can ignore the recovery rate. You could pretend it is 100% (which is impossible) and you still get large excess in most tests. This inspires much more confidence than the early marginal tests. Rossi does not trust precision measurements or complicated methods, so he would never ask anyone to trust his recovery rate, and he probably does not even bother to measure it. Still, it would be a good idea to establish the performance of the instrument. - Jed
Re: [Vo]:The faster than light neutrino speed should be determined in a non rotating frame
There would seem to be no other way of explaining a result like: I send a photon from point A to point B and measure the time of flight. I then send a neutrino. The neutrino gets there faster. This should show up the fact that neutrinos are faster than photons unless there's some error. - Original Message - From: Horace Heffner To: vortex-l@eskimo.com Sent: Saturday, October 01, 2011 1:47 PM Subject: Re: [Vo]:The faster than light neutrino speed should be determined in a non rotating frame Hopefully this one is correct. Sorry for the multiple posts on this. I am surprised and happy to see the archives now save and show jpgs. On Sep 30, 2011, at 11:16 AM, David Jonsson wrote: I made a calculation in an inertial system and found that the CERN-OPERA neutrino speed was by some percent due to the rotation of the Earth around its own axis. Do you agree that the calculation should be made in a non rotating system? By the time CERN sends and OPERA receives the Earth rotation makes OPERA to come a bit closer. How many of you agree or disagree with this? Silvertooth, Bryan G. Wallace, GPS and laser gyroscopes also supports this view. It is not suitable to apply the principle of relativity in a non inertial rotating frame. David David Jonsson, Sweden, phone callto:+46703000370 The OPERA experiment neutrino beam is directed from CERN, 46�14'N 6� 3'E, to Gran Sasso LNGS lab, 42�25'N 13�31'E. The geometry of this is shown in Fig.1, in OPERA.jpg, attached. Point C is CERN, the neutrino origin. Point S is San Sasso at the time of neutrino departure. Since San Sasso is east of CERN, the earth rotates away, eastward, from CERN during the time of flight of the neutrino. This makes the distance longer than would be estimated by distance between geodetic coordinates. The neutrino arrives at the new San Sasso location S', which is eastward from S by distance d. Only the neutrinos initially aimed at point S' arrive there. Assume the distance C to S is 730 km stated in the Adam et al. OPERA article. Assume point B to be 730 km from point C on the line from C to S'. The neutrino thus has to travel the additional distance x from B to S' due to the eastward motion of the earth during its time of flight. Let point A be the point due south of CERN and due wet of San Sasso, i.e. at 42�25'N, 6�3' E. The distance C to A s then about 404 km, and A to S 608 km. The angle of the direction of CERN from due wast as seen from San Sasso is thus roughly ATAN(404/608) = 33.6�. The earth's radius if 6371 km. San Sasso is located at latitude 42.42�N. Its radius of rotations is thus cos(42.4)*(6371 km) = 4720 km. Its speed of rotation is thus 2*Pi*(4720 km)/(24 hr) = 343 m/s. The speed of CERN due to earth's rotation is 2*Pi*cos(46.2�)* (6371 km)/(24 hr) = 321 m/s. The 22 m/s speed difference between CERN and San Sasso is not enough to relativistically affect the measurements, especially given the extreme effort put into clock synchronization and geodetic coordinate location. The relative motion however, is enough. A non-rotating linear motion approximation is sufficient to approximate the expected effect. Light travels 730 km in (730 km)/(3x10^8 m/s) = 2.435x10^-3 s. In that time San Sasso moves d = (2.435x10^-3 s) * (343 m/s) = 0.835 m eastward. The distance x added to the travel can thus be approximated as x = cos(33.6�) * d = 0.833 * (0.853 m) = 0.71 m. The travel time of the neutrinos should be increased by (0.71 m)/(3x10^8 m/s) = 2.36x10^-9 s = 2.36 ns. The neutrinos were observed arriving 60.7 ns early. This extra 0.71 m, 2.36 ns, had it not been taken into account, would have made the neutrino arrival time 60.7 ns + 2.4 ns = 63.1 ns early vs speed of light. Failure to account for earth's rotation thus provides approximately a 2.4/60.7 = 4 % error. However, this error is in a direction which makes the anomaly even greater. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ OPERA.jpg
Re: [Vo]:Inexpensive steam/water calorimeter
It might be nice to know the metal mass and temps as well. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: Vortex-L vortex-l@eskimo.com Sent: Tuesday, September 27, 2011 12:41 PM Subject: [Vo]:Inexpensive steam/water calorimeter A simple inexpensive continuously operating steam/water calorimeter can be obtained using a combined barrel and flow calorimetry. A water container, a barrel, or perhaps a trash can which is silicone sealed for leaks, can be used to condense steam via a submerged copper coil, preferably mostly located near the top of the barrel to avoid imposing a steam pressure head on the tested device. This water container can be insulated cheaply using construction foam board and fiberglass. A stirrer can be driven via a shaft through the foam board. A secondary coil can be used for pumped coolant. A fixed flow rate pump can be used to deliver the coolant flow. The coolant flow circuit can be open or closed. A closed secondary coolant temperature can be maintained via either water or air heat exchange or ice heat exchange. The source of the coolant energy is not important if the Tin and Tout are measured close to the water container, and any tubing between the temperature measuring stations and the water container is insulated. Ideally the secondary flow rate would be measured by a digital flow meter, and driven by a variable speed pump. The coolant flow rate can then be adjusted to suit the coolant delta T and the thermal power of the device under test. Alternatively, an accurate fixed flow rate pump can be chosen with a flow rate approximately matching the expected thermal power of the device under test given the expected coolant delta T. A reasonable goal for the water container temperature is the range 50°C to 70°C. Use of a large water container provides some degree of momentary heat pulse energy integration and confidence in the device thermal power measurements. It applies a significant time constant to the thermal data that reduces the frequency temperature data must be taken. It even permits manual temperature reading if a modestly stable condition is established. This is at the cost of being able to see instant response thermal and energy output curves. There is no need to see such fast response curves if the primary goal is to measure total energy in vs total energy out for a long run. The primary circuit water flow can be pumped directly from the water container. Ideally the primary water flow should be measured by digital flow meter. If a low pressure head is presented to the primary circuit flow pump, then a precision fixed flow rate pump can be used. If precision digital flow meters are not used, and reliance is placed on precision flow rate pumps, then at minimum simple (flow integrating) water meters should be monitored periodically to verify assumed pump mean flow rates. Calibration runs on dummy devices should be used to verify the calorimeter over the thermal range expected. A calibration control run should be used with the device under test to determine the water capacity of the device so the volume of water in the barrel is known in order to provide improved intermediate time thermal power measurements. At the conclusion of a run, the circuits should continue to be driven until thermal equilibrium is obtained and essentially all thermal energy is drained form the device under test. A water depth gage for the barrel may be of use, calibrated to depth vs volume, in order to keep track of the amount of water in the device under test. The secondary circuit input and output temperature should be recorded frequently. Alternatively, a direct delta T can be measured frequently using an appropriate dual thermocouple arrangement, thus providing improved data quality and reducing data acquisition required. Flow stirrers should be used, if feasible, in the secondary circuit prior to the thermometer wells. Barrel water temperature should be monitored. Ideally primary circuit water input temperature and room temperature should be monitored as well. A thermal decline curve should be measured for the water container when there is no primary circuit flow, and the water is stirred. The calorimeter constant C(dT) as a function of the difference between room temperature and water contained temperature (dT) should be determined. The curve C(dT) can be fit to a polynomial using regression analysis for convenient use in data analysis. Experience shows this method is not very accurate if the water container is not well insulated. This is due to room drafts, variations in humidity and temperature during the day, etc. Ideally active insulation could be used, whereby an extra envelope surrounds the water container insulation and the temperature there is maintained at the temperature of the water, thereby producing a dT = 0, and no heat loss. This is excessive for this approach, however, the goals of which are cheap, simple, and good enough.
[Vo]:Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011
Congrats! You are doing some good work. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Thursday, September 22, 2011 2:25 AM Subject: Re: [Vo]:Review of Travel report by Hanno Essén and Sven Kullander, 3 April 2011 On Sep 21, 2011, at 7:51 PM, Alan Fletcher wrote: HEAT FLOW THROUGH THE NICKEL CONTAINING STAINLESS STEEL COMPARTMENT If the stainless steel compartment has a surface area of approximately S = 180 cm^2, as approximated above, and 4.39 kW heat flow through it occurred, as specified in the report, then the heat flow was (4390 W)/(180 cm^2) = 24.3 W/cm^2 = 2.4x10^5 W/m^2. The thermal conductivity of stainless steel is 16 W/(m K). The compartment area is 180 cm^2 or 1.8x10^-2 m^2. If the wall thickness is 2 mm = 0.002 m, then the thermal resistance R of the compartment is: R = (0.002 m)/(16 W/(m K)*(1.8x10^-2 m^2) = 1.78 °C/W Producing a heat flow of 4.39 kW, or 4390 W then requires a delta T given as: delta T = (1.78 °C/W) * (4390 W) = 7800 °C The melting point of Ni is 1453°C. Even if the internal temperature of the chamber were 1000°C above water temperature then power out would be at best (1000°C)/(1.78 °C/W) = 561 W. Most of the input water mass flow necessarily must have continued on out the exit port without being converted to steam. That presumes that the heat exchange takes place on the surface of the core. But the heat is (supposedly) produced by thermalization of gamma rays, which could be anywhere nearby. Rossi has said that it is partly in the copper tubing and partly in the lead shielding. The total available area is easily 10 times that of the core, so the delta T could be 780C, not 7800C. This is not likely, because no gammas were detected. As I have shown, if the gamma energies are large, on the order of an MeV, a large portion of the gammas, on the order of 25%, will pass right through 2 cm of lead. The lower the energy of the gammas, the more that make up a kW of gamma flux. Consider the following: EnergyActivity (in gammas per second) for 1 kW -- 1.00 MeV 6.24x10^15 100 keV 6.24x10^16 10.0 keV 6.24x10^17 The absorption for low energy gammas is mostly photoelectic. The photoelectric mass attenuation coefficient (expressed in cm^2/gm) increases with decreasing gamma wavelength. Here are some approximations: Energymu (cm^2/gm) -- 1.00 MeV 0.02 100 keV 1.0 10.0 keV 80 We can approximate the gamma absorption qualities of the subject E- cat as 2.3 cm of lead. Given a source gamma intensity I0, surrounded by 2.3 cm of lead we have an activity: I = I0 * exp(-mu * rho * L) where rho is the mass density, and L is the thickness. For lead rho = 11.34 gm/cm^3. For 1 kW of MeV gammas we have: I = (6.24x10^15 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) * (2.3 cm)) I = 3.7x10^15 s^-1 For 1 kW of 100 keV gammas we have: I = (6.24x10^16 s^-1) * exp(-(1.0 cm^2/gm) * (11.34 gm/cm^3) * (2.3 cm)) I = 2.9x10^5 s^-1 For 1 kW of 10 keV gammas we have: I = (6.24x10^17 s^-1) * exp(-(0.80 cm^2/gm) * (11.34 gm/cm^3) * (2.3 cm)) I = ~0 s^-1 So, we can see that gammas at 100 keV will be readily detectible, but much below that not so. However, it is also true that 0.2 cm of stainless will absorb the majority of the low energy gamma energy, so we are back essentially where we started, all the heat absorbed by the stainless, and even the catalyst itself, in the low energy range. If the 2 mm of stainless is equivalent to 1 mm of lead, for 1 kW of 100 keV gammas we have: I = (6.24x10^16 s^-1) * exp(-(1.0 cm^2/gm) * (11.34 gm/cm^3) * (0.1 cm)) I = 2x10^16 s^-1 and an attenuation factor of (2x10^16 s^-1)/(6.24x10^16 s^-1) = 32%. Down near 10 keV all the gamma energy is captured in the stainless steel or in the nickel itself. To support this hypothesis a p+Ni reaction set including all possibilities for all the Ni isotopes in the catalyst would have to be found that emitted gammas only in the approximately 50 kEV range or below, but well above 10 keV, and yet emitted these at a kW level. This seems very unlikely. If such were found, however, it would be a monumental discovery. And, it would be easily detectible at close range by NaI detectors, easily demonstrated scientifically. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]: About measurement of steam with Galantini probe
Who knows enough about sound velocity in various quality steam? - Original Message - From: Jouni Valkonen To: vortex-l@eskimo.com Sent: Thursday, September 22, 2011 12:06 PM Subject: Re: [Vo]: About measurement of steam with Galantini probe Mattia, you can also measure the steam quality by measuring the speed of sound in steam. This is correlated with amount of liquid water droplets in steam suspension. Therefore you do not need to condense steam in order to find it's quality. In close to room pressure it is really not necessary to condense the steam, but it is enough to measure steam quality and separate hot water and steam with water trap. This gives the mass flow of steam and thus we can calculate the total enthalpy from humidity sensor readings. Usually water boilers are designed thus that there is build in water trap so that only steam escapes. With tube boiler this is however the case due to percolator effect. Of course it would be easier and more reliable to condense the steam by sparging it into the water bucket and measure the change of water temperature. Then we would not need to worry about the amount of overflown water. —Jouni On Sep 22, 2011 6:21 PM, Mattia Rizzi mattia.ri...@gmail.com wrote: It’s the manufacter that say the readings are useless, not me. If you don’t trust the manufacter, then provide a single reference from the literature that say that it’s possibile to measure the entalphy/steam quality/ecc from a RH reading. I challenge you. Nobody do it. ISO standard is to condensate the steam. From: Jouni Valkonen Sent: Thursday, September 22, 2011 4:45 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:About measurement of steam with Galantini probe Peter, in order to measure the enthalpy you need to know the mass flow of steam. This is not known therefore humidity sensor gives only the amount of liquid water in suspension with steam. That was measured 1.2% and thus steam quality was 98.8%. Problem is that critics such as Mattia Rizzi and Krivit has wrong definition for steam quality. Measuring steam quality is irrelevant because it is always 99-98%. Instead what would have been necessary to measure, was the mass flow of steam. This was not measured, therefore steam quality reading is useless. It tells only that 98.8% of steam mass flow was vapor and 1.2% was liquid water droplets in suspension. But indeed this does not tell us how much liquid water was overflown that was not in suspension with water vapor. I wonder how long people will repeat this Krivit's silly misconception! —Jouni On Sep 22, 2011 5:25 PM, peter.heck...@arcor.de wrote: - Original Nachricht Von: Jed Rothwell jedrothw...@gmail.com An: vortex-l@eskimo.com Datum: 22.09.2011 15:53 Betreff: Re: Aw: [Vo]:About measurement of steam with Galantini probe peter.heck...@arcor.de wrote: Now what happens, when an inventor without deep knowledge and experience constructs a steam device, makes it unaccessible and then lets unexperienced scientist measure the steam? Most scientists expect that devices that they use are properly constructed and work as designed because they know nothing else. Some questions for you and other self-appointed experts here: How much deep knowledge and experience do you have? How many steam devices have you constructed? Have you done calorimetry on this scale? What do you know about Galantini's background and his previous work? You are presumptuous. I do repair professional devices and had contact with many professors and doctors in chemical labors using our products. I have experiences with chromatography devices (with the electronic sensors,and computers, not with the chemistry), and with microparticel measurement devices and with continuous flow devices. All these dont only need calibration, fresh calibration is sometimes needed before each measurement. I have no experience with steam measurements, but was reading a lot in the last time and I learned that this are heavily nonlinear problems with many variable known and unknown parameters and it is too easy to make mistakes and too easy to fool others with such measurements.
Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat
The constriction dosen't necessarily matter as flow will tend to spped up when constricted. So you agree that there's no significant extra pressure? - Original Message - From: Alan J Fletcher To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 7:46 PM Subject: Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat At 04:19 PM 9/20/2011, Joe Catania wrote: Have it your way. We can't see inside the tap (or know what type it is), or if it's only partly open -- it is probably more constricted than the outlet. Still there is little pressure necessary. I put up the full table at : http://lenr.qumbu.com/rossi_ecat_sep11_drain_g.php I was using 30 litres ... but the actual water volume was 25L (based on the time to fill the eCat), and it could be even less than that after it's been in operation. A draining time of 7 minutes fits 1 Bar better than 2 Bar.
Re: [Vo]:Debunking Steorn Orbo
Pulses cause significant skin effect because their Fourier components consist of high frequency harmonics. - Original Message - From: Mark Iverson-ZeroPoint To: vortex-l@eskimo.com Sent: Monday, September 19, 2011 11:08 PM Subject: RE: [Vo]:Debunking Steorn Orbo From: Robert Leguillon [mailto:robert.leguil...@hotmail.com] Subject: RE: [Vo]:Debunking Steorn Orbo [deleted] Thus the original question set: Q1) Does this uneven current flow (skin effect) translate to potentially uneven heating - even at equilibrium**? [deleted] R.L. From everything that I've read and experienced, the skin effect doesn't become significant until you are well into the kilohertz frequencies; certainly above Mhz. At the 50 or 60 Hz that is all modern AC power, I highly doubt that ANY skin effect is happening. -Mark
Re: [Vo]:Calulations for 1 MW plant.
One does not have to measure that it is open to the atmosphere since that is a valid datum. It is no assumption. Assuming it is under pressure is worthless. You did not observe pressure. What experience would you be talking about? Its incredible to me that there would be any significant pressure in something open to the atmosphere. That should be your experience. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Monday, September 19, 2011 9:24 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 19, 2011, at 4:35 PM, Joe Catania wrote: The device is open to atmosphere- therefore its at atmospheric pressure. The steam is being created upon water contacting hot metal. That is an assumption, not a measurement. When the valve is opened it looks to me the device is under significant pressure. That is an assumption on my part, but based on observation and experience. It should not be under that much pressure. The other end should be open to the atmosphere via the hose. Steam should be flying out the hole around the thermometer if that much pressure is present. It would obviously be useful to continuously measure the flow and pressure of the supply water (since we know for sure that is variable), and, for safety sake, the pressure just inside the relief valve. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Monday, September 19, 2011 8:29 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 19, 2011, at 2:26 PM, Joe Catania wrote: Why do you think the device is under pressure? See end of: http://www.nyteknik.se/nyheter/energi_miljo/energi/article3264362.ece Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Calulations for 1 MW plant.
I don't know the last time you inverted a gallon jug of water but the water does not come dribbling out. Since its open to the atmosphere it won't dribble. Or if air can infiltrate from the bottom it won't dribble. I'm not saying the overlying water dosen't give it pressure. We also don't know how long it takes to drain. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 10:56 AM Subject: Re: [Vo]:Calulations for 1 MW plant. Joe, could you please explain why the water is ejected at such a high velocity instead of just dribbling out of the tap? On Sep 20, 2011, at 4:55 AM, Joe Catania wrote: One does not have to measure that it is open to the atmosphere since that is a valid datum. It is no assumption. Assuming it is under pressure is worthless. You did not observe pressure. What experience would you be talking about? Its incredible to me that there would be any significant pressure in something open to the atmosphere. That should be your experience. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Monday, September 19, 2011 9:24 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 19, 2011, at 4:35 PM, Joe Catania wrote: The device is open to atmosphere- therefore its at atmospheric pressure. The steam is being created upon water contacting hot metal. That is an assumption, not a measurement. When the valve is opened it looks to me the device is under significant pressure. That is an assumption on my part, but based on observation and experience. It should not be under that much pressure. The other end should be open to the atmosphere via the hose. Steam should be flying out the hole around the thermometer if that much pressure is present. It would obviously be useful to continuously measure the flow and pressure of the supply water (since we know for sure that is variable), and, for safety sake, the pressure just inside the relief valve. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Monday, September 19, 2011 8:29 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 19, 2011, at 2:26 PM, Joe Catania wrote: Why do you think the device is under pressure? See end of: http://www.nyteknik.se/nyheter/energi_miljo/energi/ article3264362.ece Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Calulations for 1 MW plant.
Yes a sealed galon bottle may dribble if a hole is poked but if its vented at the top you should get a steady stream. Or if air enters through the bottom you don't get a dribble! I scan't confirm high velocity flow in the video. Since you can't tell me the rate of flow out the valve we have nothing to discuss. The video runs for about 1 minute 20 seconds before ending and the tank is still emptying. I assume ~20L of water in the tank. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 1:27 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 20, 2011, at 8:41 AM, Joe Catania wrote: I don't know the last time you inverted a gallon jug of water but the water does not come dribbling out. Of course it does. I didn't say dripping. The water flows from a gallon container in an unsteady stream. It doesn't spray out at high velocity as if it were from a pressure washer nozzle. Besides, the opening on the E-cat was much smaller than a typical gallon bottle. If you poke a small hole in a gallon bottle it will dribble or drip. One estimate given for the tank pressure was 2 bar. The water was above 100°C so some of it flashed to steam. It came from the bottom of the tank so was likely entirely water before being ejected. Since its open to the atmosphere it won't dribble. Or if air can infiltrate from the bottom it won't dribble. I'm not saying the overlying water dosen't give it pressure. We also don't know how long it takes to drain. Aha. We have a dribble quibble. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Calulations for 1 MW plant.
They state there is an auxillary heater. - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 3:24 PM Subject: Re: [Vo]:Calulations for 1 MW plant. Am 20.09.2011 20:38, schrieb Horace Heffner: On Sep 20, 2011, at 10:14 AM, Peter Heckert wrote: In all demonstrations, January demo, Essen Kulander demo, 3 Ny Teknik demos, the electrical input energy was not enough to heat the water to 100° Celsius. (I dont know aout the Krivit demo) There was without doubt some considerable boiling in all experiments and so the COP should be larger than 2. This is mass flow calorimetry. There /must/ be more energy than the /measured/ electrical energy. So there is something, lets hope it is not a trick. Peter I don't recall at all that there was not enough power to boil the water in the initial tests. (My memory is not very good though!) Do you mean there wasn't enough power applied to convert all the water flow to steam? Yes. Kullander and Essen have calculated this explicitely and I recalculated it and can confirm. Also I dont think two Physics Professors can do errors here because this is too simple to calculate. Look here: http://www.lenr-canr.org/acrobat/EssenHexperiment.pdf At Page 2 they write: It is worth noting that at this point in time and temperature, 10:36 and 60°C, the 300 W from the heater is barely sufficient to raise the temperature of the flowing water from the inlet temperature of 17.6 °C to the 60 °C recorded at this time. If no additional heat had been generated internally, the temperature would not exceed the 60 °C recorded at 10:36. Instead the temperature increases faster after 10:36, I recalculated this. I did not recalculate the other documents, but reliable persons said this and I made some rule of thumb estimations. I guess one of the problems with making that assertion is not actually knowing the true flow rate at all times. Mattia Rizzi observed pump rates on a video which indicated much less than 2 gm/s. Essen Kullander measured it with a carafe. (See page 1, chapter Calibrations). In the january experiment they measured the weigt of the water bottle. They use a peristaltic pump. I was often in chemical labors in my life. ( I did electronics and computer servicing there) They use peristaltic pumps, (equipped with calibrated hoses) when accurate flow is required. This should be pretty constant and a big variation would be audible. If I recall correctly the Krivit demo was for the most part 1.94 gm/s, input temp 23°C, and 748 W input, which makes for all the flow heated to 100°C plus 83 cc/sec steam generated. All that is hard to know too because apparently Rossi touched the control panel. Manual adjustment is apparently part of the process, as is changing duty factors. This is one reason why a good kWh meter would be of use. Yes but the heater is controlled by a zero crosspoint switch. The heater should be on some seconds and off some seconds. The current that they measured should be the maximum current and it corresponded to the 300W rating of the band heater. A technical problem exists because the thermal mass of the E-cats is so high. Momentary power readings don't mean very much. I think Kullander and Essen where there all the time and they watched carefully what was going on. Of course this cannot prove that there ai no hidden fake energy source and that there are no tricks, but I think in the Kullander and Essen demo we can be sure there was more energy than 300W. 600W would have been required to heat the water flow to 100° and some additional 100 Watts are needed to get reasonable steam and boiling. Only fast sampled power measurements integrated to cumulative energy is meaningful, or first principle energy integrating techniques. Total energy in vs total energy out for a long period is the meaningful number. Yes of course for a scientific publication test this is necessary, but not for a qualitative plausibility test. Best, Peter
Re: [Vo]:Calulations for 1 MW plant.
The point is that a gallon empties very quickly even though not vented at the top. The sound it makes is immaterial and is most like caused by the water hitting the barrel. I don't know why you feel the water is under inordinate pressure. The E-CAt is open to the atmosphere unless Lewan seals the other valve. I doubt this as the water seems to be drainig with venting. Why not ask Lewan how long it took to empty the E-Cat? - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 2:46 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 20, 2011, at 10:36 AM, Joe Catania wrote: Yes a sealed galon bottle may dribble if a hole is poked but if its vented at the top you should get a steady stream. Or if air enters through the bottom you don't get a dribble! I scan't confirm high velocity flow in the video. Since you can't tell me the rate of flow out the valve we have nothing to discuss. The video runs for about 1 minute 20 seconds before ending and the tank is still emptying. I assume ~20L of water in the tank. Sigh. Look at the video! Do you hear a gurgle gurgle gurgle or a high powered woos? The water is obviously under high pressure. The couple atmospheres pressure estimate by others does not seem off. You need a numerical velocity to determine the difference? - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 1:27 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 20, 2011, at 8:41 AM, Joe Catania wrote: I don't know the last time you inverted a gallon jug of water but the water does not come dribbling out. Of course it does. I didn't say dripping. The water flows from a gallon container in an unsteady stream. It doesn't spray out at high velocity as if it were from a pressure washer nozzle. Besides, the opening on the E-cat was much smaller than a typical gallon bottle. If you poke a small hole in a gallon bottle it will dribble or drip. One estimate given for the tank pressure was 2 bar. The water was above 100°C so some of it flashed to steam. It came from the bottom of the tank so was likely entirely water before being ejected. Since its open to the atmosphere it won't dribble. Or if air can infiltrate from the bottom it won't dribble. I'm not saying the overlying water dosen't give it pressure. We also don't know how long it takes to drain. Aha. We have a dribble quibble. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Calulations for 1 MW plant.
Really? - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 2:14 PM Subject: Re: [Vo]:Calulations for 1 MW plant. Am 20.09.2011 19:49, schrieb Horace Heffner: I think my conclusion was good: None of this indicates for sure whether Rossi has anything of value or not. Maybe he does. The continued failure to obtain independent high quality input and output energy measurements prevents the public from knowing. There is one thing that was unfortunately ignored in allmost all public discussions: In all demonstrations, January demo, Essen Kulander demo, 3 Ny Teknik demos, the electrical input energy was not enough to heat the water to 100° Celsius. (I dont know aout the Krivit demo) There was without doubt some considerable boiling in all experiments and so the COP should be larger than 2. This is mass flow calorimetry. There /must/ be more energy than the /measured/ electrical energy. So there is something, lets hope it is not a trick. Peter
Re: [Vo]:Calulations for 1 MW plant.
Still I'm not convinced that those tests you mentioned weren't exactly like the September test. Why shouldn't they be? - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 4:10 PM Subject: Re: [Vo]:Calulations for 1 MW plant. Am 20.09.2011 21:51, schrieb Joe Catania: They state there is an auxillary heater. Yes but they examined all cables and even lifted the devices to see whats below and I think this extra heater was connected to the blue control box where they measured the input current. If not, then they should have reported this.
Re: [Vo]:Calulations for 1 MW plant.
The screaming does not indicate high pressure. It could be a whistle effect as bubbles of steam are forming in the outlet. Why not experiment and see how fast a container drains through an outlet the size of the E-Cat's? - Original Message - From: Terry Blanton hohlr...@gmail.com To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 4:50 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Tue, Sep 20, 2011 at 2:46 PM, Horace Heffner hheff...@mtaonline.net wrote: Sigh. Look at the video! Do you hear a gurgle gurgle gurgle or a high powered woos? The water is obviously under high pressure. The couple atmospheres pressure estimate by others does not seem off. You need a numerical velocity to determine the difference? http://www.mail-archive.com/vortex-l@eskimo.com/msg51256.html http://www.mail-archive.com/vortex-l@eskimo.com/msg51289.html I don't think Joe has bothered to see the video. The steam screams! ;-) T
Re: [Vo]:Calulations for 1 MW plant.
To ay the matter to rest I was not the one to use the word dribble. It was HH. - Original Message - From: Mark Iverson-ZeroPoint zeropo...@charter.net To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 4:41 PM Subject: RE: [Vo]:Calulations for 1 MW plant. Horace: The first thing I thought of when Joe used the word dribble was that he had not seen the video where they opened the water inlet valve on the bottom and a VERY strong stream of liquid water and steam came out! To refer to that as a dribble, is clearly the wrong adjective... forceful expulsion is much closer to an accurate decription. Joe: Perhaps you should go back and watch that video several times, and then look up the word 'dribble' to see if the definition accurately describes what you saw coming out of that valve... if so, then we're looking at wo different videos. -Mark -Original Message- From: Horace Heffner [mailto:hheff...@mtaonline.net] Sent: Tuesday, September 20, 2011 11:46 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 20, 2011, at 10:36 AM, Joe Catania wrote: Yes a sealed galon bottle may dribble if a hole is poked but if its vented at the top you should get a steady stream. Or if air enters through the bottom you don't get a dribble! I scan't confirm high velocity flow in the video. Since you can't tell me the rate of flow out the valve we have nothing to discuss. The video runs for about 1 minute 20 seconds before ending and the tank is still emptying. I assume ~20L of water in the tank. Sigh. Look at the video! Do you hear a gurgle gurgle gurgle or a high powered woos? The water is obviously under high pressure. The couple atmospheres pressure estimate by others does not seem off. You need a numerical velocity to determine the difference? - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 1:27 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 20, 2011, at 8:41 AM, Joe Catania wrote: I don't know the last time you inverted a gallon jug of water but the water does not come dribbling out. Of course it does. I didn't say dripping. The water flows from a gallon container in an unsteady stream. It doesn't spray out at high velocity as if it were from a pressure washer nozzle. Besides, the opening on the E-cat was much smaller than a typical gallon bottle. If you poke a small hole in a gallon bottle it will dribble or drip. One estimate given for the tank pressure was 2 bar. The water was above 100°C so some of it flashed to steam. It came from the bottom of the tank so was likely entirely water before being ejected. Since its open to the atmosphere it won't dribble. Or if air can infiltrate from the bottom it won't dribble. I'm not saying the overlying water dosen't give it pressure. We also don't know how long it takes to drain. Aha. We have a dribble quibble. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/ Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Calulations for 1 MW plant.
That wasn't me. I've never posted to that site. But so what? Is that the best you can do? - Original Message - From: Terry Blanton hohlr...@gmail.com To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 4:54 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Tue, Sep 20, 2011 at 4:50 PM, Terry Blanton hohlr...@gmail.com wrote: On Tue, Sep 20, 2011 at 2:46 PM, Horace Heffner hheff...@mtaonline.net wrote: Sigh. Look at the video! Do you hear a gurgle gurgle gurgle or a high powered woos? The water is obviously under high pressure. The couple atmospheres pressure estimate by others does not seem off. You need a numerical velocity to determine the difference? http://www.mail-archive.com/vortex-l@eskimo.com/msg51256.html http://www.mail-archive.com/vortex-l@eskimo.com/msg51289.html I don't think Joe has bothered to see the video. The steam screams! ;-) I don't see why you bother to waste your time on Catania. Look at his question that no one bothered to answer: http://www.industrycommunity.com/bbs/mfg_1_2805.html Where is the world is there a 5 GW (electric) turbine? Maybe in a UFO! g T
Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat
Clearly your calculations are a bit off. The running time on video is more like 1:20, still greater than drain time for 2 atm, showing there is less than 2atm pressure. But since we don't know for how long the draining continues we dont know how much less. Since the E-Cat is open to atmosphere (by report) we can assume the pressure is 1 atm. Also 1/4 cm seems a bit small for the orifice and drain time would seem to affected by height of water column. - Original Message - From: Alan J Fletcher To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 5:03 PM Subject: Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat At 12:49 PM 9/20/2011, Joe Catania wrote: The point is that a gallon empties very quickly even though not vented at the top. The sound it makes is immaterial and is most like caused by the water hitting the barrel. I don't know why you feel the water is under inordinate pressure. The E-CAt is open to the atmosphere unless Lewan seals the other valve. I doubt this as the water seems to be drainig with venting. Why not ask Lewan how long it took to empty the E-Cat? - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 2:46 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 20, 2011, at 10:36 AM, Joe Catania wrote: Yes a sealed galon bottle may dribble if a hole is poked but if its vented at the top you should get a steady stream. Or if air enters through the bottom you don't get a dribble! I scan't confirm high velocity flow in the video. Since you can't tell me the rate of flow out the valve we have nothing to discuss. The video runs for about 1 minute 20 seconds before ending and the tank is still emptying. I assume ~20L of water in the tank. Sigh. Look at the video! Do you hear a gurgle gurgle gurgle or a high powered woos? The water is obviously under high pressure. The couple atmospheres pressure estimate by others does not seem off. You need a numerical velocity to determine the difference? I just ran the calculations for draining a 30L eCat through a 0.25 cm radius tap. http://lenr.qumbu.com/rossi_ecat_sep11_f.php The drain-time says 2 Bars ! 6. Discharge at the End I can't figure out the dumping of the water at the end, either. Is it 100C water, or is it 118C water? 1 Bar or 2 Bars ? I've never seen 25L of boiling water dumped through a tap, so I don't know what it should look like. It does appear to come out under pressure, and it does seem to flash to steam at the edge of the stream -- both supporting evidence for an internal pressure of 2 Bars. The video ends before the discharge is complete. Time to drain tank The drain is at a depth of 30 cm and 30 liters is to be drained (based on the dimensions of 60 x 50 x 30 cm). The radius of the outlet tap is about 0.25 cm. For atmospheric pressure (1 Bar) the time to drain is 1260.18 secs ( 21.00 min) For a pressure of 2 Bar we can ADD 33 feet of water to the tank height (draining from 33 feet + 30cm to 33 feet + 0 cm). The time to drain is then 108.02 secs ( 1.80 min) Although the video ended before the eCat was completely drained, the time shown on the video (6:44 to 8:05) -- or 1.83 minutes tends indicate 2 bars pressure, not 1 bar. The time to discharge, the fact that the flow did not diminish, and that the water seemed to flash into steam around the edge, all support the pressurized hypothesis. The general argument is the same as for the hose outlet -- 118C water would flash rapidly.
Re: [Vo]:stopping
Take some aspirin and see a doctor. - Original Message - From: OrionWorks - Steven V Johnson svj.orionwo...@gmail.com To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 5:30 PM Subject: Re: [Vo]:stopping Horace, Needless to say... call your doctor or optometrist right away. Could be a number of serious issues. Migraine, retinal detachment, mini-stroke. Don't wait. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat
BTW you should run those time-to-drain numbers again. The outlet looks like its about 2cm in diameter. The sound seems to be mostly water impacting on the side of the pail. - Original Message - From: Alan J Fletcher To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 6:09 PM Subject: Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat At 02:56 PM 9/20/2011, Alan J Fletcher wrote: At 02:33 PM 9/20/2011, Joe Catania wrote: http://lenr.qumbu.com/rossi_ecat_sep11_f.php I seem to have broken my file ... back soon! It's back ... I added a table of draining time vs tap radius, and corrected the video time. I'm still open to revising my conclusion. (!!!)
Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat
A 5-7 min draining time is completely consistent with 1 atm (ie no additional pressure). That represents a flow of ~50ml/s or a velocity of ~15cm/s which is ~ 1/66 of the velocity obtained from dropping for 1 sec in a gravity field. Since mgh=1/2mv^2, h= 1/2 (.15m/s)^2 /10ms^-2 or h=0.1125cm so the water only has to drop a 1/10 cm to gain enough KE to drain the tank at 50ml/s. - Original Message - From: Jouni Valkonen jounivalko...@gmail.com To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 5:50 PM Subject: Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat Alan, excellent work again. Considering Akira's temperature graph, we can take that draining took about 5-7 min. In the beginning pressure was 210 kPa or 122°C. But it is needed to take into consideration, that valve was opened slowly. In the end of video, valve was only half open. http://i.imgur.com/lU42G.png Therefore I think that we have now rather conclusive proof, that indeed, temperature gives us at least approximately the pressure inside E-Cat. It is not anymore just an assumption, but data supports the idea. –Jouni 2011/9/21 Alan J Fletcher a...@well.com: I just ran the calculations for draining a 30L eCat through a 0.25 cm radius tap. http://lenr.qumbu.com/rossi_ecat_sep11_f.php The drain-time says 2 Bars ! 6. Discharge at the End I can't figure out the dumping of the water at the end, either. Is it 100C water, or is it 118C water? 1 Bar or 2 Bars ? I've never seen 25L of boiling water dumped through a tap, so I don't know what it should look like. It does appear to come out under pressure, and it does seem to flash to steam at the edge of the stream -- both supporting evidence for an internal pressure of 2 Bars. The video ends before the discharge is complete. Time to drain tank The drain is at a depth of 30 cm and 30 liters is to be drained (based on the dimensions of 60 x 50 x 30 cm). The radius of the outlet tap is about 0.25 cm. For atmospheric pressure (1 Bar) the time to drain is 1260.18 secs ( 21.00 min) For a pressure of 2 Bar we can ADD 33 feet of water to the tank height (draining from 33 feet + 30cm to 33 feet + 0 cm). The time to drain is then 108.02 secs ( 1.80 min) Although the video ended before the eCat was completely drained, the time shown on the video (6:44 to 8:05) -- or 1.83 minutes tends indicate 2 bars pressure, not 1 bar. The time to discharge, the fact that the flow did not diminish, and that the water seemed to flash into steam around the edge, all support the pressurized hypothesis. The general argument is the same as for the hose outlet -- 118C water would flash rapidly.
Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat
I can't agree w/ a diameter of 1 cm. - Original Message - From: Alan J Fletcher To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 6:49 PM Subject: Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat At 02:33 PM 9/20/2011, Joe Catania wrote: Clearly your calculations are a bit off. The running time on video is more like 1:20, still greater than drain time for 2 atm, showing there is less than 2atm pressure. But since we don't know for how long the draining continues we dont know how much less. Since the E-Cat is open to atmosphere (by report) we can assume the pressure is 1 atm. Also 1/4 cm seems a bit small for the orifice and drain time would seem to affected by height of water column. I corrected the run time. The time to drain goes as 1/orifice_area * sqrt(column_height) 1/4 is the radius -- 1/2cm diameter At 02:50 PM 9/20/2011, Jouni Valkonen wrote: Considering Akira's temperature graph, we can take that draining took about 5-7 min. That's about 23:15 to 23:22 Hmmm since the outlet is still open cool air will be sucked past the temperature probe, cooling it. When it's completely drained this flow will stop, and the thermal mass will cause the air to heat up again. Tank height 20 Radius 0.25Time 1 Bar 25.72 minTime 2 Bar 1.80 min Radius 0.30Time 1 Bar 17.86 minTime 2 Bar 1.25 min Radius 0.35Time 1 Bar 13.12 minTime 2 Bar 0.92 min Radius 0.40Time 1 Bar 10.05 minTime 2 Bar 0.70 min Radius 0.45Time 1 Bar 7.94 minTime 2 Bar 0.56 min Radius 0.50Time 1 Bar 6.43 minTime 2 Bar 0.45 min Tank height 22.5 Radius 0.25Time 1 Bar 27.28 minTime 2 Bar 2.03 min Radius 0.30Time 1 Bar 18.95 minTime 2 Bar 1.41 min Radius 0.35Time 1 Bar 13.92 minTime 2 Bar 1.04 min Radius 0.40Time 1 Bar 10.66 minTime 2 Bar 0.79 min Radius 0.45Time 1 Bar 8.42 minTime 2 Bar 0.63 min Radius 0.50Time 1 Bar 6.82 minTime 2 Bar 0.51 min Tank height 25 Radius 0.25Time 1 Bar 28.76 minTime 2 Bar 2.25 min Radius 0.30Time 1 Bar 19.97 minTime 2 Bar 1.56 min Radius 0.35Time 1 Bar 14.67 minTime 2 Bar 1.15 min Radius 0.40Time 1 Bar 11.23 minTime 2 Bar 0.88 min Radius 0.45Time 1 Bar 8.88 minTime 2 Bar 0.70 min Radius 0.50Time 1 Bar 7.19 minTime 2 Bar 0.56 min Tank height 27.5 Radius 0.25Time 1 Bar 30.16 minTime 2 Bar 2.48 min Radius 0.30Time 1 Bar 20.95 minTime 2 Bar 1.72 min Radius 0.35Time 1 Bar 15.39 minTime 2 Bar 1.26 min Radius 0.40Time 1 Bar 11.78 minTime 2 Bar 0.97 min Radius 0.45Time 1 Bar 9.31 minTime 2 Bar 0.76 min Radius 0.50Time 1 Bar 7.54 minTime 2 Bar 0.62 min Tank height 30 Radius 0.25Time 1 Bar 31.50 minTime 2 Bar 2.70 min Radius 0.30Time 1 Bar 21.88 minTime 2 Bar 1.88 min Radius 0.35Time 1 Bar 16.07 minTime 2 Bar 1.38 min Radius 0.40Time 1 Bar 12.31 minTime 2 Bar 1.05 min Radius 0.45Time 1 Bar 9.72 minTime 2 Bar 0.83 min Radius 0.50Time 1 Bar 7.88 minTime 2 Bar 0.68 min So ... pick a number (or two!) and draw your conclusions.
Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat
But look at the size of the orifice in the video. - Original Message - From: Alan J Fletcher To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 6:54 PM Subject: Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat At 03:36 PM 9/20/2011, Joe Catania wrote: BTW you should run those time-to-drain numbers again. The outlet looks like its about 2cm in diameter. The sound seems to be mostly water impacting on the side of the pail. Tank height 25 Radius 0.20Time 1 Bar 44.94 minTime 2 Bar 3.52 min Radius 0.30Time 1 Bar 19.97 minTime 2 Bar 1.56 min Radius 0.40Time 1 Bar 11.23 minTime 2 Bar 0.88 min Radius 0.50Time 1 Bar 7.19 minTime 2 Bar 0.56 min Radius 0.60Time 1 Bar 4.99 minTime 2 Bar 0.39 min Radius 0.70Time 1 Bar 3.67 minTime 2 Bar 0.29 min Radius 0.80Time 1 Bar 2.81 minTime 2 Bar 0.22 min Radius 0.90Time 1 Bar 2.22 minTime 2 Bar 0.17 min Radius 1.00Time 1 Bar 1.80 minTime 2 Bar 0.14 min 2cm diam is MUCH too quick.
Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat
Have it your way. Still there is little pressure necessary. - Original Message - From: Alan J Fletcher To: vortex-l@eskimo.com Sent: Tuesday, September 20, 2011 7:18 PM Subject: Re: [Vo]:Calculations for 1 MW plant. + Time to Drain the eCat At 04:00 PM 9/20/2011, Joe Catania wrote: But look at the size of the orifice in the video. http://lenr.qumbu.com/steampics/110920_sept_0007.jpg http://lenr.qumbu.com/steampics/110920_sept_0009.jpg 1cm diameter, maximum.
Re: [Vo]:Debunking Steorn Orbo
I'm not going to take it on faith about the AC power being less than DC. I've done these types of calculations before and I can tell you they are not simple. A sawtooth wave can generate some extremely high harmonics which have a large skin effect. I'd need to see the formula used to evaluate them. - Original Message - From: Robert Leguillon To: vortex-l@eskimo.com Sent: Monday, September 19, 2011 2:21 PM Subject: RE: [Vo]:Debunking Steorn Orbo My Two Cents-- I must confess that I'm unfamiliar with the effect of electromagnetism on conductive heating. I thought that I'd throw out a few questions regarding the observations of the 4th paper, hoping to learn: Background for the questions: Alternating current (dependent on the frequency) can produce a more prominent skin effect on conductors than standard direct current. This skin effect can cause the vast bulk of current to flow down only the outer surface of a conductor. Q1) Does this uneven current flow translate to potentially uneven heating - even at equilibrium? Q2) Could the nickel core be cooler in the middle with more heat being concentrated, and subsequently shed, on the surface? Q3) Could the surface of the inductor wires appear hotter, though the entire conductor is dissipating the same amount of total heat? Donating to the World, Two Cents at a Time, R.L. Documents 1-3 were quite interesting - compelling, really. I'm going to have to read up more on Steorn. Document #4 - It's getting hot in here, turn off that Orbo! The fourth report that we were allowed to examine is unique from the others in that it is about a solid state version of Steorn's technology. It is also the most recent of the documents, being written in March, 2011. A solid state Orbo offers the potential of having no moving parts, having no need for bearings (as in permanent manget (PM) or E-Orbo configurations), being simpler to build, and potentially being simpler to test. Other advantages of solid state Orbo include fewer parts to wear out, and perhaps more potential to evolve quickly -- in a similar manner to the way computers evolved during the past twenty years. In this paper the author describes a very simple configuration that involves a coil wrapped around a nickel core (that is both magnetic and conductive) acting as an inductor. The coil and core is placed in a calorimeter composed of a vacuum chamber. Two thermocouples measure the temperature of the coil itself, and the temperature of the air in the room. A metered power supply provides the input power to the coil, and an oscilloscope monitors the current, voltage, and can also calculate total input power by using a math function of the scope. The purpose of the test is to determine if the coil fed with a quantity of AC power, can produce more heat than the same coil fed with the same quantity of DC power. In the paper, the formula needed to calculate the total AC power is presented. The AC input and DC input is configured to be as identical as possible. Actually, the power input during the AC run was .9 (point nine) watts, and in the DC run it was 1 (one) watt. The fact that the input power during the AC run was slightly less than in the DC run actually biases the test against the AC run. This makes the results of the test even more significant. In the first test, 1 watt of DC power is fed into the coil wound around the nickel core. The temperature of the coil increases until it reaches an equilibrium point of 36.1 degrees. This is the point at which the power lost by the coil via heat dissipation matches the electrical input power. Even if the input power stayed on for hours longer, the temperature of the coil would not increase above this temperature. In the second test, .9 watts is fed into the same coil wound around the same exact nickel core. Obviously, this test took place a period of time after the first one, after the temperature of the coil has dropped back to its original value. The result of AC being fed into the coil is that it rises to an equilibrium temperature of 41.1 degrees. This means that in the AC test, the temperature of the coil reached a temperature five degrees higher than in the DC test. The higher equilibrium temperature obtained when the coil was powered with AC, indicates an anomalous gain of energy. The gain of energy is unexplainable, because the input power in both tests were almost identical -- actually slightly less when AC was utilized. As the paper continues, the author indicates that resistive heating cannot be the case for the increased temperature in the AC test run. Here is the conclusion found at the end of the paper. The extra heating effect under the application of an AC signal is not explained simply by the transfer of input power to the coil. Consideration of the energy input to the system does
Re: [Vo]:Debunking Steorn Orbo
Now you are asking me to take it on faith from you. I find you less convincing than Steorn. - Original Message - From: Peter Heckert To: vortex-l@eskimo.com Sent: Monday, September 19, 2011 4:29 PM Subject: Re: [Vo]:Debunking Steorn Orbo Am 19.09.2011 22:22, schrieb Joe Catania: I'm not going to take it on faith about the AC power being less than DC. I've done these types of calculations before and I can tell you they are not simple. It is simple. The simplest way to calculate such problems is to use the law of enery conservation ;-) A sawtooth wave can generate some extremely high harmonics which have a large skin effect. I'd need to see the formula used to evaluate them. ;-)
Re: [Vo]:Debunking Steorn Orbo
Ok, Peter. What I'm saying is I've run into this kind of thing before. There was an electrical engineering professor on TheEEStory.com blog who thought a patent was invalid and falsified because it showed a fuse blowing at a current that (if it were DC) would be insufficient to melt the fuse. I still haven't convinced him that skin effect is the reason it blew. He says that skin effect in the case of this fuse would be negligible but he does not calculate it correctly, One must take into account all the Fourier components in the pulse to get the proper effect. He only traets the fundamental and is thus mislead. But a sawtooth wave has harmonics that stretch theoretically to infinity. Although the amplitudes of these harmonics decrease as their frequency increases there is always the same net contribution to skin effect for each frequency decade. In theory the upper limit of frequency should only be limited by the electron plasma frequency. In other words, if there were no such limitation the series would diverge. This is a known property of the harmonic series (1 + 1/2 + 1/3 + 1/4...) which also diverges and is related tothe sawtooth Fourier components. Where is the paper mentioned? - Original Message - From: Peter Heckert To: vortex-l@eskimo.com Sent: Monday, September 19, 2011 5:21 PM Subject: Re: [Vo]:Debunking Steorn Orbo Am 19.09.2011 22:33, schrieb Joe Catania: Now you are asking me to take it on faith from you. I find you less convincing than Steorn. Let me explain. All known rules about electricity and magnetism are compatible with energy conservation. It is therefore impossible to derive an extra energy mathematically, basing on /known/ electromagnetic effects like skin effect. There must be an energy source. I dont say that the effect is untrue. If it is true then it is not an electromagnetic effect. Possibly the Nickel core contains spurious Hydrogen atoms. Nickel is magnetostrictive. Possibly the AC induces magnetostrictive vibrations in the core or current in microscopic superconductive spots and triggers hydrogen Nickel fusion. The next locical thing to do would be to measure the frequency depency of the effect. Why didnt they do this? Or might they have done? Should I buy the paper? Tell me the price. Best, Peter - Original Message - From: Peter Heckert To: vortex-l@eskimo.com Sent: Monday, September 19, 2011 4:29 PM Subject: Re: [Vo]:Debunking Steorn Orbo Am 19.09.2011 22:22, schrieb Joe Catania: I'm not going to take it on faith about the AC power being less than DC. I've done these types of calculations before and I can tell you they are not simple. It is simple. The simplest way to calculate such problems is to use the law of enery conservation ;-) A sawtooth wave can generate some extremely high harmonics which have a large skin effect. I'd need to see the formula used to evaluate them. ;-)
Re: [Vo]:Calulations for 1 MW plant.
Why do you think the device is under pressure? - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Monday, September 19, 2011 6:11 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 19, 2011, at 11:46 AM, OrionWorks - Steven V Johnson wrote: It's quite odd to notice that on the skeptical side of the fence the subject of CF continues to be perceived as a bogus completely unproven source of energy. Therefore, one would infer from such conclusions that Rossi's 1 MW demonstration couldn't possibly harm a fly. It is not necessarily true that the E-cat can not harm a fly if there is no excess energy produced. This is because purely normal electrical input may be enough to blow the thing up.The 4 metric tons of mostly steel constitute an enormous thermal mass. With a steel heat capacity of 0.49 J/(gm K), the 1 MW E-cat has a possible thermal mass Mt given by: Mt = (0.49 J/(gm K))(4 tons)(1x10^6 gm/ton) = 1.96x10^6 J/K At 200°C, or delta T = 100°C above boiling, this is an energy storage of 196 MJ. This is enough to produce 196 MW seconds of boiling energy if the water being recycled back into the E-cat from a condenser is at 100°C. It is thus critical to know where the heating element is located in the E-cat, and the general geometry of the device, to determine the device safety even if no excess energy is produced. Earlier I estimated the flow rate out the E-cat pipe to be 223 m/s, or 803 km/hr, at 1 MW output with 100°C water recycled. This is over 6 times a reasonable flow rate limit for the pipe size. Each of the new E-cats, if like the one demonstrated briefly, can utilize 2500 W electric input, for a total of 130 kW. If the E-cat is operating at a COP of 6 then it will produce 0.78 MW of thermal output. However, if the thermal mass is heated to a mean temperature of 200°C, the device can periodically produce over a MW of steam without any excess energy input at all ever. This demonstrates why it is important to measure each test run total energy balance vs momentary powers. Instabilities can develop in the water condense cycle flow rate, especially if the condenser capacity can be overrun. If the condenser capacity is overrun an explosion can result due to pressure build up. High pressure steam can drive water within and from the condenser into the E-cat, and then steam as well, creating a momentary feedback loop. If the steam momentarily cannot be condensed at an adequate rate, say due to water slugs in the line, then the input water flow rate is momentarily low and the water entering will end up superheated steam, allowing the thermal mass to overheat. This kind of flow instability then can be the source cause for a periodically over 1 MW feedback loop oscillating condition to form, even without excess energy. This demonstrates the need to control the flow of water into each E-cat independent of the flow rate out of the condenser and dependent on the mean thermal energy stored in the overall device. The new 80 kg E-cat, one 52nd of the 1 MW E-cat, when tested alone, looked like it might have had some unusual transient properties. For example, it is strange the device at the end was under so much pressure, yet steam was not pouring forth from the thermometer well, around the probe. The hose itself should have been able to take much of the pressure off the device. It looked as if possibly some thermostatically controlled orifice closed or the output flow was momentarily blocked for some reason (pure speculation of course.) If true, that a dangerous situation was suddenly perceived by the operators, then this one wild speculation would account for the abrupt lack of will to carry on the experiment through the night, or the next day. The huge thermal mass provided by 80 kg of mostly steel could bring instabilities not only to a 1 MW E-cat made of 52 of them, but internal instabilities to the small E-cats by themselves. There is no way of knowing if this is true without detailed knowledge of the structure of the device. Such knowledge is not required to determine true COP, provided total test run energy balances are accurately determined. Such knowledge is required, however, to make any estimate of the device safety. If a single E-cat catastrophically fails, it will be difficult to enter the container to perform any emergency operation of the remaining devices. Hopefully complete operation can be performed remotely. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Calulations for 1 MW plant.
The device is open to atmosphere- therefore its at atmospheric pressure. The steam is being created upon water contacting hot metal. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Monday, September 19, 2011 8:29 PM Subject: Re: [Vo]:Calulations for 1 MW plant. On Sep 19, 2011, at 2:26 PM, Joe Catania wrote: Why do you think the device is under pressure? See end of: http://www.nyteknik.se/nyheter/energi_miljo/energi/article3264362.ece Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:The September E-Cat
I saw the termination as a resignation that there is no anomalous heat. It showed there is no self-sustaining reaction since the temperature drop is correlated with power off. The write-up that Lewan gives shoes his lack of general physics knowledge and that he is most likely a paid biased spokesperson. A continuation of the demo would have borne out the continuation of temperature drop from the cooling of the thermal mass. - Original Message - From: OrionWorks - Steven Vincent Johnson orionwo...@charter.net To: vortex-l@eskimo.com Sent: Sunday, September 18, 2011 11:25 AM Subject: RE: [Vo]:The September E-Cat From Catania, ... As I've said before I think thermal inertia neatly explains it all. I don't know of anyone who was not disappointed in the abrupt ending of the experiment, after input power had been turned off. Yeah, yeah, I know, they tell us it was late in the evening and they needed their beauty rest. For true skeptics, the abrupt ending is nothing more than further proof that something fishy is going on. I'm suspect many skeptics probably feel vindicated... again. I supposed for true believers the recorded anomalous heat is just more evidence that proof is in the pudd'in... but, oh, what a shame they didn't run it a while longer, perhaps for a couple of hours, but oh well... I guess I'm currently in the camp that feels frustrated by the abrupt ending. Such abruptness tends to make me feel less confident as to the outcome. In the continued vacuum of solid rock-hard evidence such abruptness tends to make me personally want to conjure up unfounded assumptions - to manufacture conjecture based primarily on my emotionally laced suspicions: That the termination was done deliberately, with forethought. I don't know why they terminated it so abruptly. They tell us it was late in the evening, but Hell! Who really knows why. All I know is that basing my conclusions on emotionally based conjecture that neither proves or disproves an extraordinary claim is a fools game. Therefore, I will endeavor to do what I have done in the past: Wait and see. IOW I remain ignorant. Under the current circumstances there is no shame in that. Regards, Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:The September E-Cat
Here are some band heater specs. Notice the max temps, http://www.omega.com/heaters/pdf/HEATER_INTRO_BAND_REF.pdf, As I've said before I think thermal inertia neatly explains it all. Although there is a slight rise in temp after power off its hard to believe that CF knows when we switch the power off and then puts in such a poor showing. Its more likely an anomaly or perhaps due to diffusion time. The amount of energy pumped into the E-Cat before even the first water overflow is quite large as I have said. It would also appear that a band heater can get hot enough to heat the E-Cat metal to the proper temp. The possible amount of steam produced would seem to be less than 3.0 - 1.8= 1.2 g/s (maybe less since overflow and pump inlet are known or checked very well). If I recall correctly it takes about 2250J/g to vaporize. So only 2700W would be necessary to vaporize 1.2g. There may also be other liquid besides overflow entrained in the steam. Too bad no one measured the heat of the overflow. In all there seems to be some heat unaccounted for if you take the overflow and inlet measurements at face value and assume steam is dry. But there is too much inaccuracy in these to seriously conclude. Also thermal inertia would seem to explain everything nicely. Why dosen't someone do a run without hydrogen for comparison? - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Thursday, September 15, 2011 8:42 PM Subject: Re: [Vo]:The September E-Cat On Sep 15, 2011, at 4:29 PM, Jouni Valkonen wrote: [snip] As metal content of the E-Cat is at the same temperature as water content, This is an assumption with no (apparent) foundation. All 80 kg of E- at will not be at the water temperature. If the new E-cat is heated by a band heater, then the outside metal blanket will be *much* hotter than the water. We need to know the structure of the new E-cat. it does not matter where the probe is installed. It matters where the probe is installed. It might not even be in the steam or water. Here is a poser. If the temperature probe is in the steam/water, why is it that when the internal pressure is a couple atmospheres that there is no leakage around the probe. I recall seeing in a video the probe being easily removed from one of the early E-cat demo machines. Even if they do not exactly match, there is still a correlation because heat conduction speed is somewhat constant. We only look for the correlation. Do we actually know what the input flow was, or the water outflow was, after the power was shut off? Yes. Peristaltic pumps are quite predictive. –Jouni So, what then do you predict the flow from the pump would be if a water inlet valve in the machine were closed? It is a good thing to have measurements instead of estimates. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:E-cat news at Nyteknik
From the report:The impression was that the loss of heating power was minor. Consequently the heat produced by the E-cat in self sustained mode should have been clearly larger than the heat from the power that was lost when the electric resistance was switched off. What a crock! A minor loss of heating power is exactly what one expects from thermal inertia. There is no anomalous heat. Also since 1.8 grams were collected as overflow and only ~3 grams flowed in we have 1!.2 grams at most converted to steam. This means about 2700W. That's close enough for me to the 2600W input. - Original Message - From: Finlay MacNab To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 8:49 PM Subject: RE: [Vo]:E-cat news at Nyteknik Excellent observation! If this was a closed system with no FLOWING WATER EXITING THE SYSTEM you would have a point. As it is you have only discredited your argument about thermal inertia. Congratulations! I find your hand waving arguments completely unconvincing. Please describe in detail the geometry of the system you propose could account for the observed changes in temperature taking into account the well known rate of heat exchange between water and metals/other materials and the heat capacities of the various materials. Also, please account for the energy inputs and outputs to the device during its operation. 5 minutes with a text book will convince anyone with half a brain that what you describe is more improbable than cold fusion itself! Please do everyone here a favor and give a rigorous explanation of how thermal inertia can explain the rossi device. Please use equations and data to back up your claims. If you don't want to do this please stop spamming this message board and distracting from more interesting discussion. -- Well, at a setting of 9 you have the same temp rise in 35 minutes as temperature fall in 35 minutes after power-off. - Original Message - From: Mark Iverson-ZeroPoint To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 4:55 PM Subject: RE: [Vo]:E-cat news at Nyteknik JC stated: “(and note that this takes considerable time in the ramp up)” Where he is referring to the long time it takes to ramp up the E-Cat’s internal temperature on startup… Mr. Catania, do you realize that the electrical power into the E-Cat’s resistance heater was NOT started at 100%, it was started at a setting of ‘5’ and RAMPED UP slowly over 40 minutes! Here is the time progression for resistance heater power… Timestamp PLC Setting DeltaTime (minutes) - --- -- 18:59 5 0 19:10 611 19:20 710 19:30 810 19:40 910 We know that the ‘Setting’ is referring to the duty cycle, but we do not know exactly what the relationship is… since 9 is the MAXimum setting, and Lewan states ‘power was at this point constantly switched on’, then a setting of ‘9’ is presumably a 100% duty cycle. (?) Since the PLC’s are programmable, we cannot assume that a setting of ‘5’ is 50% or 60%; it could even be programmed to be 10% duty cycle. So no useful calculations OR conclusions can be made during this ramp-up phase. -Mark From: Joe Catania [mailto:zrosumg...@aol.com] Sent: Wednesday, September 14, 2011 11:58 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:E-cat news at Nyteknik I think it caused a rise. There is no rise. Its your imagination. The temperature at power off is too low and must be discarded. If I bring a piece of metal the size of an E-Cat to some temperature (and note that this takes considerable time in the ramp up) and then I cut the power, the temperature will not instantaneously drop. It will stay at the same temperature and decline slowly. There is much too much mass for what your talking about to happen. I have to laugh at the fact that if you saw the temp drop even a hundredth of a degree at power down you would have declared the thermal inertia regime over and the CF regime to have begun.
Re: [Vo]:The September E-Cat
You're understanding of thermal inertia is incorrect. We don't expect a rapid decline. With Megajoules in storage a 1000W draw will change the temperature but little. Its like your telling me you can slow down a Mack tuck by shooting peas at it. It'll decelerate quickly at first but as it comes to a halt it will be more difficult to slow it dowm. Even a cursory glance at the data will show that enormous energy is being pumped into the E-Cat with very little coming out. In 10 Minutes about 1.5MJ goes in at full power. Nothing comes out until overflow at 20:16. At 20:50 there's 3.7 g overflow at 90C. That's about 1`/3 of what's going in.From 19:00 to 19:40, i.e 40 minutes, the power is increasd from 1/2 to full. I'll count that as 20 min. at full which is 3MJ. From 19:40 to 22:40, 3 hrs @ full gives 27MJ for a total of 30MJ. There would appear to be from 17 to 20L of water stored in the E-Cat. It takes ~5MJ to heat 17L of water from 30C to 100C. So it would appear that there are 25MJ stored elsewhere at this point. That's enough to produce 1000W for over 7 hrs. And there is probably additional heating. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Thursday, September 15, 2011 10:49 AM Subject: Re: [Vo]:The September E-Cat Horace Heffner hheff...@mtaonline.net wrote: More importantly, the claim that all the water was being converted to steam, the repeated, defended, and heralded basis for thinking something practical has been created, the basis for the calorimetry of the public demos, is now shown to be without basis in fact. The hose was taken off. Water pulsed out of the outlet right at the exit of the E-cat in large quantity. It obviously did not condense there. That is true. However, in the Krivit test and other previous tests, the flow rate was lower, so I do not think you can compare them. Also if they had put a probe into this stream of steam and water and withdrawn it, it would have come out wet, whereas in previous tests it was dry. In general I agree that a non-steady state mixture of water and steam is difficult to measure. I wish that Lewan had sparged the steam and water. Before this test, I sent messages to Lewan, Rossi and others urging them to do this, but they did not. They had a perfect opportunity to do this, with that large plastic trashcan. It will easily hold enough water to condense all of the steam. By the way, flow rate was almost exactly 3 g per second. Input power will be enough to vaporize 0.7 g assuming no heat radiated from the device. That is extremely unrealistic. So the fact that about half the water was vaporized does indicate there was excess heat. More to the point, during the 35 min. heat after death event, the temperature did not decline much. This is proof that there was anomalous heat. Stored heat can only produce a temperature that declines rapidly at first and then gradually. After the power went off the temperature did not decline rapidly. Therefore the input power of 2.5 kW was only a fraction of the total power. If the total power was around 5 kW where 2.5 kW was half, the temperature would've fallen a lot faster and sooner. Lewan estimates the water volume of the cell at 22 to 30 L. If there had been no anomalous heat the temperature would have fallen sharply within minutes. You can boil a pot of 22 L of hot water and observe this easily. Turn off the heat, and it stops boiling instantly. It starts to cool a few degrees in minutes. The temperature never rises and never stabilizes, unless you change the insulation (or the flow rate, in this case). In this case the temperature will certainly fall quickly because during the 35 min. 6 kg of cold water was added to the cell. The heat capacity of this water far exceeds the total heat capacity of all the metal in the cell. Now the new E-cat never reaches equilibrium. This is a far more difficult regime in which to do accurate calorimetry, and a far better regime for self deception. That is true, but there is no doubt it was boiling for 35 minutes with no input power. Anyone who ignores this fact is engaged in the worst kind of self-deception imaginable. Further, the E-cat mass has been greatly increased, and the max input power increased. The heat after death from mundane causes will now obviously be much longer. This cannot sustain boiling for more than a few seconds, at this flow rate. Metal cannot store much heat, and this cell was producing excess heat the whole time, so there was no possible storage at all. With 2.5 kW input only, it would have transitioned from boiling about one third of the water to boiling none of it, and that would have taken a few seconds at most. - Jed
Re: [Vo]:E-cat news at Nyteknik
http://www.nyteknik.se/incoming/article3264365.ece/BINARY/Report+E-cat+test+September+7+%28pdf%29, I have to laugh at the hydrogen weight measurement in the Nyteknik Preliminary Report. The report a 2.7 gram drop in weight after filling with hydrogen. But an average air molecule weighs about 28 whereas hydrogen at 60 bar weighs 120 so you should see a gain. - Original Message - From: Jouni Valkonen jounivalko...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 8:00 AM Subject: Re: [Vo]:E-cat news at Nyteknik These test results are indeed difficult to explain. I have one question to those who have some or partial expert knowledge on steam engineering: Does they use superheated steam or steam that is at boiling point of local pressure? My guess is latter of course. However, I cannot explain 130°C temperature if assumed low pressure inside E-Cat, because specific temperature of steam is just too low so that it could produce such a smooth temperature graph. E.g. input power cut off should cause huge bump into graph. Smooth temperature graph should be only plausible, if steam temperature is regulated by the boiling point at local pressure. But for 130°C/170 kPa pressure requirements are quite high, higher than in autoclave, although it is not out of question. Also 5 kg/h water collected from outlet, is consistent that 60-80% of water was evaporated, just like previous e-Cat experiments (excluding March experiment). This would support the idea that steam temperature is regulated by boiling point temperature at local pressure. Could someone calculate the size of orifice for steam exit, to explain 130°C temperature corresponding 170 kPa over pressure? If it is assumed that E-Cat produces steam in ca. 9 kW total power. Using previous E-Cat demonstrations as reference, it should be quite small, just few millimeters. Unlike what Mats Lewan estimated, I think that it may be big enough to enable water to overflow, as pump pumps water with sufficient pressure. Also I have not yet carefully studied the data, but I would guess that 170 kPa over pressure could explain why the water pumping rate was decreased after E-Cat started operating, because pump pumps water only with 300 kPa pressure IIRC. But, this seems more plausible 1MW production plant. I think that later development can boost individual module output power at least few orders of magnitude. It should be possible, if sufficient cooling is arranged, that there is 1 GW power plant fitted to the similar sized container. Anyways, my confidence for E-Cat has increased somewhat due to this new experiment. This really is starting to look commercially viable prototype. This would also decrease the main problem with Rossi that he chose very irrational method for bringing this cat out of the closed. He really seems to be ready to go directly into market without spending lots of public resources for RD. –Jouni
Re: [Vo]:E-cat news at Nyteknik
Good catch. Yes I've commented about how I dtested this method of weighing before. I seem to have forgotten how he did it but I can see it is prone to inaccuracy. He only fills it to 20 bars. He'd have to buy me many dinners to convince me of this. All in all the rest of the report is sloppy or full on inconsistencies. A seemingly bad temperature measurement shows up. He admits to water overflow. He guesses about the 130 degree temperature. The curreny number seems to bounce around from 11A to .11A even when the power is off but most glaringly he attributes what is clearly thermal inertia to CF in so many words! - Original Message - From: Man on Bridges manonbrid...@aim.com To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 9:20 AM Subject: Re: [Vo]:E-cat news at Nyteknik Hi, On 14-9-2011 15:05, Joe Catania wrote: I have to laugh at the hydrogen weight measurement in the Nyteknik Preliminary Report. The report a 2.7 gram drop in weight after filling with hydrogen. But an average air molecule weighs about 28 whereas hydrogen at 60 bar weighs 120 so you should see a gain. It seems you misunderstood the term filling. It means filling the Rossi rector and NOT the Hydrogen bottle. These numbers apply to the Hydrogen bottle only and not the Rossi reactor. So filling in this case means removing or better said using from the bottle of Hydrogen. Kind regards, MoB
Re: [Vo]:E-cat news at Nyteknik
The E-Cat ran for 35 minutes without electrical power? Did anyone tell you that the thermal inertia will run the E-Cat for that long? - Original Message - From: Jouni Valkonen jounivalko...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 10:11 AM Subject: Re: [Vo]:E-cat news at Nyteknik See the E-cat run in self-sustained mode http://www.nyteknik.se/nyheter/energi_miljo/energi/article3264362.ece This video confirms my previous assumption above, that new E-Cat is operating approximately 170 kPa overpressure. Also it confirms that roughly 5 kW excess heat was produced. I have not yet made accurate analysis for calorimetry, but I think that we have now even better data than previously and we can calculate total enthalpy by at least one significant number. This video also disproofs wet steam hypothesis as steam and hot water are clearly separated. There is definitely not Abd's atomization of water, but steam quality is ca. 99-98% as it should be according normal steam physics. –Jouni
Re: [Vo]:E-cat news at Nyteknik
A) You're a fool to tell me that the E-Cat has no thermal inertia. It certainly does. This is unavoidable. B) The data given are certainly consistent withy thermal inertia being the cause. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 10:46 AM Subject: Re: [Vo]:E-cat news at Nyteknik Joe Catania zrosumg...@aol.com wrote: The E-Cat ran for 35 minutes without electrical power? Did anyone tell you that the thermal inertia will run the E-Cat for that long? At 22:35 input electric power was 2.5 kW. All electric power was cut off at this time. The temperature dropped from 131.9°C down to 123.0°C, which is the expected amount. At 22:40, 5 minutes later, the temperature rose to 133.7°C, higher than it was with electric power input. By 23:10 when the run ended, the temperature had fallen to 122.7°C. Stored heat cannot explain this behavior. That would violate the second law of thermodynamics. Since the flow rate remained stable, the temperature cannot rise without some source of energy production within the cell. - Jed
Re: EXTERNAL: Re: [Vo]:E-cat news at Nyteknik
No. Admittedly the temperature drop at powerdown may or may not be valid. In fact if there's any magnetic field associated with the heating coils there could be some EMF from shutting it down. It would seem to be an anomaly if we assume it was measuring anything with thermal mass. Just notice that the next valid reading is at the level it was before power off. There does seem to be some inaccuracy (or at least variation) in the thermometry. For instance the anomalous drop in T1 to 21.4 at 21:10. Aside from a couple of obvious glitches there is nothing thyere that dosen't suggest the temperature decay expected from thermal inertia causes. In fact it is not possible to rule out thermal inertia at all as it must exist. It's as likely that the gravitational field suddenly ceased to exist as thermal inertia was eliminated. In any case even if this was a demo of anomalous heat the explanation certainly wouldn't be CF. There's no way to justify that. In my opinion more study needs to be done on the heating core. - Original Message - From: Roarty, Francis X To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 12:32 PM Subject: RE: EXTERNAL: Re: [Vo]:E-cat news at Nyteknik Any mass has a certain gradient described in temp/time for thermal gain or loss. I think Jed was specifying the period where the temperature rebounded higher than it existed while being heated by input power. That seems anomalous to me made more curious by the initial drop in temp when the input power is initially removed - the extra temp would seem to indicate the reaction has reinitiated without the resistive heating. My posit is that the active heating has opposite effects on the reaction cavities where the dominant heat is being generated by nominal nano scale cavities while there also exist some hot spots of sub nano geometry that are held from runaway by the pulse width modulation - I suspect that these pockets can finally start to run away when the PWM is removed and quickly grow to the point where they start to reignite the larger cavities in place of the PWM. This would also explain Rossi's concern about damage - not only to the pico cavities melting down and losing the ability to operate closed loop but also over stimulating the larger cavities to plastic hot conditions where the stiction forces would alleviate the Casimir geometries.[melting closed or growing perpendicular whiskers] Fran From: Joe Catania [mailto:zrosumg...@aol.com] Sent: Wednesday, September 14, 2011 11:11 AM To: vortex-l@eskimo.com Subject: EXTERNAL: Re: [Vo]:E-cat news at Nyteknik A) You're a fool to tell me that the E-Cat has no thermal inertia. It certainly does. This is unavoidable. B) The data given are certainly consistent withy thermal inertia being the cause. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 10:46 AM Subject: Re: [Vo]:E-cat news at Nyteknik Joe Catania zrosumg...@aol.com wrote: The E-Cat ran for 35 minutes without electrical power? Did anyone tell you that the thermal inertia will run the E-Cat for that long? At 22:35 input electric power was 2.5 kW. All electric power was cut off at this time. The temperature dropped from 131.9°C down to 123.0°C, which is the expected amount. At 22:40, 5 minutes later, the temperature rose to 133.7°C, higher than it was with electric power input. By 23:10 when the run ended, the temperature had fallen to 122.7°C. Stored heat cannot explain this behavior. That would violate the second law of thermodynamics. Since the flow rate remained stable, the temperature cannot rise without some source of energy production within the cell. - Jed
Re: [Vo]:E-cat news at Nyteknik
The data after power off are not consistent with a temperature increase from before power off. In fact there is a steady decline from before power of which is completely consitent with thermal inertia. The thermal inertia is of course more than a two minute effect in this E-Cat as examination of the heat-up data and post power-down data confirm. Also this is inline w/ estimates of the mass of metal in E-Cat. You're confused if you think you see anomalous production after power-off. - Original Message - From: OrionWorks - Steven V Johnson svj.orionwo...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 1:33 PM Subject: Re: [Vo]:E-cat news at Nyteknik Mr. Catania, What I found interesting about latest reply was the fact that you did nothing more than restate your previous comment, basically that the effects of thermal inertia in the recorded measurements have not been accounted for. Meanwhile, Mr. Rothwell replied to your original comment by posting thermal measurements that apparently reveal the interesting fact that thermal inertia had already been taken into account when the temperature initially dropped from 131.9 C down to 123.0 C soon after input power had been cut off. But amazingly, five minutes later, measurements recorded a 10 degree increase. Not only that, this sudden increase was apparently HIGHER than the recorded temperature when the input power was still on - by approximately 2 degrees. This implies that any residual effects pertaining to thermal inertia had already been accounted for long ago. The effects of thermal inertia cannot magically make a device suddenly become HOTTER particularly if previous measurements were revealing the fact that the temperature was already in the process of dropping. It therefore make no sense to imply that the effects of thermal inertia could be responsible for a sudden 10 C increase five minutes after all input power had been cut off - especially when the temperature had been previously recorded to have been dropping. BTW, proclaiming that Mr. Rothwell is a fool is no way to go about winning friends and influencing people to your POV. In fact, I suspect your latest actions have done nothing more than to suggest to most here that Jed has probably done a far better job of analyzing the thermal inertia situation than you. Learn to be civil in the presentation of you POVs or get kicked out of this forum. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:E-cat news at Nyteknik
I think it caused a rise. There is no rise. Its your imagination. The temperature at power off is too low and must be discarded. If I bring a piece of metal the size of an E-Cat to some temperature (and note that this takes considerable time in the ramp up) and then I cut the power, the temperature will not instantaneously drop. It will stay at the same temperature and decline slowly. There is much too much mass for what your talking about to happen. I have to laugh at the fact that if you saw the temp drop even a hundredth of a degree at power down you would have declared the thermal inertia regime over and the CF regime to have begun. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 2:11 PM Subject: Re: [Vo]:E-cat news at Nyteknik OrionWorks - Steven V Johnson svj.orionwo...@gmail.com wrote: Meanwhile, Mr. Rothwell replied to your original comment by posting thermal measurements that apparently reveal the interesting fact that thermal inertia had already been taken into account when the temperature initially dropped from 131.9 C down to 123.0 C soon after input power had been cut off. That data is from: Test of Energy Catalyzer, Bologna, September 7, 2011 Analysis of calorimetry http://www.nyteknik.se/incoming/article3264365.ece/BINARY/Report+E-cat+test+September+7+%28pdf%29 I am glad to see Lewan included a fairly detailed time-stamped data log in this report. We could have used that in previous reports. As Lewan remarks, it is a shame they did not let it run another hour in self-sustaining (heat after death) mode. But it was late at night, after all. I am still working through this report. Someone here suggested that the power supplies might have affected the thermocouples. I don't think so. Thermocouples and interface equipment attached to them are designed to work around machines with power supplies and magnetic fields. If the power supplies produced affected thermocouple performance, the people observing the experiment would have seen that happen immediately when the power went on, and again when it went off. Also this could not explain the temperature rise 10 minutes after the power went off. Catania apparently thinks that thermal inertia can cause a temperature to rise when there is no internal power production and no change in the flow rate (rate of heat loss). This is a violation of the laws of thermodynamics. Thermal inertia can only produce a temperature that falls at some rate. The highest temperature would have to be recorded just before the power was turned off. I believe the temperature could rise because of thermal inertia if you cut the flow rate and if there were a very hot body inside the cell. - Jed
Re: [Vo]:E-cat news at Nyteknik
At the end, when the water input valve is opened, then a mixture out of water and steam comes out with remarkable pressure. Now, how can we have pressure when the steam outlet is still open? This troubled me too and I found it unexplainable until I thought that the valve, valve stem and metal were probably hot from having been previously heated by heater core. If their temperature had'nt dropped below 100C there could be considerable flahing to steam upon exit of water through the valve. - Original Message - From: Peter Heckert To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 2:36 PM Subject: Re: [Vo]:E-cat news at Nyteknik Am 14.09.2011 08:55, schrieb Peter Gluck: a) See the E-cat run in the self sustaining mode http://www.nyteknik.se/nyheter/energi_miljo/energi/article3264362.ece Here my Analysis: At the end, when the water input valve is opened, then a mixture out of water and steam comes out with remarkable pressure. Now, how can we have pressure when the steam outlet is still open? Answer: The steam outlet is not open. Probably there is a pressure reduction valve in the oulet. This opens at 1-2 bar pressure and it closes when the pressure sinks. This means inside the ecat is always 1-2 bar overpressure. Saturated steam temperatures versus pressure tabulated: (This is the over-pressure that is more than air pressure) 1 bar - 120.2° 1.5 bar - 127.4° 2.0 bar - 133.5° 2.5 bar - 138.9° Now this explains why water and steam come out. Water comes out and it has a temperature of 120°. Wenn it flows out it will vaporize partially and produce steam. This also explains the water output flow at the steam hose: The steam inside of the ecat has a pressure between 1 and 2 bar and a temperature between 120 and 133 centigrade. When the steam passes the pressure reduction valve then it will expand to air (over) pressure of 0 bar. To do this, work must be done and the steam will cool down to 100° and partially condensate. This explains the output water flow at the steam outlet. So far my qualitative steam temperature pressure analysis. There is one thing that irritated me. When they show the e-cat in self-sustained mode, then I cannot hear the pump anymore. Did they stop the pump and why? Best, Peter
Re: [Vo]:E-cat news at Nyteknik
They admit themselves that steam quality could be as low as 59%. The pressure in the E-Cat is probably near atmospheric. - Original Message - From: Jouni Valkonen jounivalko...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 2:41 PM Subject: Re: [Vo]:E-cat news at Nyteknik 2011/9/14 Alan J Fletcher a...@well.com: 50% fluid water 2.5% drops 47.5% vapour This must be noted that these estimations are when temperature was ca. 118 °C or 90 kPa overpressure. After that temperature rose to 133°C and overpressure to 170 kPa. Therefore 60-80% of water was evaporated and E-Cat did work exactly as it should work. Actually I am somewhat puzzled that indeed E-Cat is working such a perfect way that Rossi can push output power so close to the maximum of the enthalpy absorption ability of cooling water. This is either sure sign that technology is very commercially mature or it is a hoax. It is no more just a lab prototype, but commercially ready prototype. I was glad to see that he DOES have a simple water trap in the outlet hose, which separates the fluid water. I wonder if there is now enough evidence for the steam quality people to see that even after such high pressure difference hot water and steam are clearly separated. I wonder how history will remember this steam quality chapter, when prominent people (such as Krivit and Ekström) were violently discussing about steam quality without knowing what steam quality actually means. When Rossi opens the outlet the pressure of the water and steam is clearly greater than atmospheric. Indeed, for me it is very consistent pressure difference that of in autoclave although I have never dared to open the valve that fast as they did. I estimated the pressure drop through the mini eCat (March/April) and hose -- it only came out to be (as I recall) about 3% -- assuming a 2cm internal diameter pipe in the reactor and a 1cm diameter hose. (I used an online calculator) Actually the diameter of the orifice where the hose is attached is probably the tightest place. And of course for steam backpressure, the tightest place is what counts most. The diameter of the orifice is considerably less than the inner diameter of the hose. I would estimate it to be 5-10 mm. This should be consistent with ca. 1.0°C / 3.2 kPa overpressure and the steam volume that was produced ca. 2 kW total power. –Jouni
Re: [Vo]:The pump was left running during the self-sustaining event
Pump was stopped at 23:10 - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 3:04 PM Subject: [Vo]:The pump was left running during the self-sustaining event Peter Heckert peter.heck...@arcor.de wrote: There is one thing that irritated me. When they show the e-cat in self-sustained mode, then I cannot hear the pump anymore. Did they stop the pump and why? There is no way they would stop the pump! The temperature would climb and it would blow up. I do not see what you mean. (I don't hear what you mean.) In the video, starting around 5:00 they turn off the power. I hear the pump still running. The pump sound is gone at 6:10 in the video, which is after the test concludes, just before they open the reactor. The log shows that was real-time 23:10. - Jed
Re: [Vo]:E-cat news at Nyteknik
For once? I only been saying that one thing- many times. But you'd better understand that from first principles not from a typo. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 4:35 PM Subject: Re: [Vo]:E-cat news at Nyteknik OrionWorks - Steven V Johnson svj.orionwo...@gmail.com wrote: Meanwhile, Mr. Rothwell replied to your original comment by posting thermal measurements that apparently reveal the interesting fact that thermal inertia had already been taken into account when the temperature initially dropped from 131.9 C down to 123.0 C soon after input power had been cut off. Okay, that's probably a typo, as shown in the video. For once Catania is correct. The temperature did not drop suddenly and then rise. I expect it did drop soon, given the loss of 2.5 kW input at a flow rate of 185 ml/min. See my message Video time synced to real time. I will confirm this with Lewan. - Jed
Re: [Vo]:E-cat news at Nyteknik
Well, at a setting of 9 you have the same temp rise in 35 minutes as temperature fall in 35 minutes after power-off. - Original Message - From: Mark Iverson-ZeroPoint To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 4:55 PM Subject: RE: [Vo]:E-cat news at Nyteknik JC stated: (and note that this takes considerable time in the ramp up) Where he is referring to the long time it takes to ramp up the E-Cat's internal temperature on startup. Mr. Catania, do you realize that the electrical power into the E-Cat's resistance heater was NOT started at 100%, it was started at a setting of '5' and RAMPED UP slowly over 40 minutes! Here is the time progression for resistance heater power. Timestamp PLC Setting DeltaTime (minutes) - --- -- 18:59 5 0 19:10 611 19:20 710 19:30 810 19:40 910 We know that the 'Setting' is referring to the duty cycle, but we do not know exactly what the relationship is. since 9 is the MAXimum setting, and Lewan states 'power was at this point constantly switched on', then a setting of '9' is presumably a 100% duty cycle. (?) Since the PLC's are programmable, we cannot assume that a setting of '5' is 50% or 60%; it could even be programmed to be 10% duty cycle. So no useful calculations OR conclusions can be made during this ramp-up phase. -Mark From: Joe Catania [mailto:zrosumg...@aol.com] Sent: Wednesday, September 14, 2011 11:58 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:E-cat news at Nyteknik I think it caused a rise. There is no rise. Its your imagination. The temperature at power off is too low and must be discarded. If I bring a piece of metal the size of an E-Cat to some temperature (and note that this takes considerable time in the ramp up) and then I cut the power, the temperature will not instantaneously drop. It will stay at the same temperature and decline slowly. There is much too much mass for what your talking about to happen. I have to laugh at the fact that if you saw the temp drop even a hundredth of a degree at power down you would have declared the thermal inertia regime over and the CF regime to have begun.
Re: [Vo]:E-cat news at Nyteknik
Wrong, nothing like that mass is necessary. - Original Message - From: Jouni Valkonen jounivalko...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 4:58 PM Subject: Re: [Vo]:E-cat news at Nyteknik 2011/9/14 Jed Rothwell jedrothw...@gmail.com: OrionWorks - Steven V Johnson svj.orionwo...@gmail.com wrote: Meanwhile, Mr. Rothwell replied to your original comment by posting thermal measurements that apparently reveal the interesting fact that thermal inertia had already been taken into account when the temperature initially dropped from 131.9 C down to 123.0 C soon after input power had been cut off. Okay, that's probably a typo, as shown in the video. For once Catania is correct. The temperature did not drop suddenly and then rise. I expect it did drop soon, given the loss of 2.5 kW input at a flow rate of 185 ml/min. Indeed that temperature graph is suggesting that thermal inertia could explain the behavior. This would work, if there is no inlet water pumped. But as there is pumped about 5 kg of inlet water into E-Cat during the self-sustaining mode, this would require that there is metallic thermal mass something like in order of one ton. Of course as there is lots of water, requirements are not that high, but still thermal inertia cannot explain the behavior of E-Cat not, by two orders of magnitude. –Jouni
Re: [Vo]:Lewan report corrected
Could be significant. LOL. With the glitches and inaccuracies I see in this data I doubt anything that small could be considered significant. I doubt there is even hydriding occuring. Thermal inertis explains it. Definitely I won;t let you ascribe a 0.7C for 5 min glitch to CF. That would be impossible to justify at this point as it would with even a pronounced anomaly. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 5:02 PM Subject: [Vo]:Lewan report corrected A new version of this report has been uploaded: Test of Energy Catalyzer, Bologna, September 7, 2011 Analysis of calorimetry http://www.nyteknik.se/incoming/article3264365.ece/BINARY/Report+E-cat+test+September+7+%28pdf%29 The new version says QUOTE: 22:35 Power to the resistance was cut off. Input AC current was 0.11 A. Over-all AC voltage was 232 volts. DC voltage was zero. AC current through the resistance was 0.11 A. T2=29.0°C, T3=133.0°C. (Typo corrected Sept 14). 22:40 T2=28.9°C, T3=133.7°C. 22:50 T2=28.8°C, T3=131.2°C. END QUOTE There is a slight temperature rise at 22:40. Could be significant. I would like to see second-by-second data after the power cut off. - Jed
Re: [Vo]:E-cat news at Nyteknik
They probably go from 80 to 100% in going from 8 to 9. So its obvious that thermal inertia would take it out about 2hrs. - Original Message - From: Alan J Fletcher a...@well.com To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 5:07 PM Subject: RE: [Vo]:E-cat news at Nyteknik At 01:55 PM 9/14/2011, Mark Iverson-ZeroPoint wrote: We know that the 'Setting' is referring to the duty cycle, but we do not know exactly what the relationship is. since 9 is the MAXimum setting, and Lewan states 'power was at this point constantly switched on', then a setting of '9' is presumably a 100% duty cycle. (?) Since the PLC's are programmable, we cannot assume that a setting of '5' is 50% or 60%; it could even be programmed to be 10% duty cycle. So no useful calculations OR conclusions can be made during this ramp-up phase. Lewan did report that at setting 5 the ON and OFF times were equal. So taking the duty cycle as PLC/9 is about as good as we can guess.
Re: [Vo]:E-cat news at Nyteknik
What was personally communicated to me by JR is, of course, beyond SVJ's ken. You seem to keen to overllok data which shows up the obvious flaw in your CF bias. - Original Message - From: OrionWorks - Steven V Johnson svj.orionwo...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 6:11 PM Subject: Re: [Vo]:E-cat news at Nyteknik From Catania: For once? I only been saying that one thing- many times. But you'd better understand that from first principles not from a typo. From: Jed Rothwell Okay, that's probably a typo, as shown in the video. For once Catania is correct. The temperature did not drop suddenly and then rise. I expect it did drop soon, given the loss of 2.5 kW input at a flow rate of 185 ml/min. See my message Video time synced to real time. I will confirm this with Lewan. It has been a constant observation of mine that when Mr. Rothwell's has suspected a potential mistake or perhaps a typo in published data he has been quick to express his suspicions. Jed often quickly seeks to correct previous assumptions, even if it contradicts previous assessments he may have made. Meanwhile, I noticed that Mr. Catania's response to Mr. Rothwell's retraction appears to hinge on assuming a position of superiority by challenging Jed - such that Jed had better understand the first principals. The implication I derive from Mr. Catania's response is that he does not often seem to consider the possibility that his own crafted assessments might occasionally be prone to similar mistakes. I could say something about that, such as: we are only human. Some more than others. Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:E-cat news at Nyteknik
If you want the response from Sun Tzu study it yourself. If you have nothing to say why refer me to Sun Tzu. Are you saying he does have something to say? - Original Message - From: Terry Blanton hohlr...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 6:22 PM Subject: Re: [Vo]:E-cat news at Nyteknik On Wed, Sep 14, 2011 at 6:11 PM, OrionWorks - Steven V Johnson svj.orionwo...@gmail.com wrote: The implication I derive from Mr. Catania's response is that he does not often seem to consider the possibility that his own crafted assessments might occasionally be prone to similar mistakes. It does seem to imply that there is an inflated ego involved somewhere in his analysis. I suggested he study Sun Tzu and he did not bother to respond. Maybe he is Sun Tzu reincarnated? At least *that* would understandable. T
Re: [Vo]:E-cat news at Nyteknik
When Aristotle explains in general terms what he tries to do in his philosophical works, he says he is looking for first principles (or origins; archai): In every systematic inquiry (methodos) where there are first principles, or causes, or elements, knowledge and science result from acquiring knowledge of these; for we think we know something just in case we acquire knowledge of the primary causes, the primary first principles, all the way to the elements. It is clear, then, that in the science of nature as elsewhere, we should try first to determine questions about the first principles. The naturally proper direction of our road is from things better known and clearer to us, to things that are clearer and better known by nature; for the things known to us are not the same as the things known unconditionally (haplôs). Hence it is necessary for us to progress, following this procedure, from the things that are less clear by nature, but clearer to us, towards things that are clearer and better known by nature. (Phys. 184a10-21) - Original Message - From: Joe Catania zrosumg...@aol.com To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 6:40 PM Subject: Re: [Vo]:E-cat news at Nyteknik If you want the response from Sun Tzu study it yourself. If you have nothing to say why refer me to Sun Tzu. Are you saying he does have something to say? - Original Message - From: Terry Blanton hohlr...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 6:22 PM Subject: Re: [Vo]:E-cat news at Nyteknik On Wed, Sep 14, 2011 at 6:11 PM, OrionWorks - Steven V Johnson svj.orionwo...@gmail.com wrote: The implication I derive from Mr. Catania's response is that he does not often seem to consider the possibility that his own crafted assessments might occasionally be prone to similar mistakes. It does seem to imply that there is an inflated ego involved somewhere in his analysis. I suggested he study Sun Tzu and he did not bother to respond. Maybe he is Sun Tzu reincarnated? At least *that* would understandable. T
Re: [Vo]:E-cat news at Nyteknik
Its a first principle. - Original Message - From: Finlay MacNab To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 8:49 PM Subject: RE: [Vo]:E-cat news at Nyteknik Excellent observation! If this was a closed system with no FLOWING WATER EXITING THE SYSTEM you would have a point. As it is you have only discredited your argument about thermal inertia. Congratulations! I find your hand waving arguments completely unconvincing. Please describe in detail the geometry of the system you propose could account for the observed changes in temperature taking into account the well known rate of heat exchange between water and metals/other materials and the heat capacities of the various materials. Also, please account for the energy inputs and outputs to the device during its operation. 5 minutes with a text book will convince anyone with half a brain that what you describe is more improbable than cold fusion itself! Please do everyone here a favor and give a rigorous explanation of how thermal inertia can explain the rossi device. Please use equations and data to back up your claims. If you don't want to do this please stop spamming this message board and distracting from more interesting discussion. -- Well, at a setting of 9 you have the same temp rise in 35 minutes as temperature fall in 35 minutes after power-off. - Original Message - From: Mark Iverson-ZeroPoint To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 4:55 PM Subject: RE: [Vo]:E-cat news at Nyteknik JC stated: “(and note that this takes considerable time in the ramp up)” Where he is referring to the long time it takes to ramp up the E-Cat’s internal temperature on startup… Mr. Catania, do you realize that the electrical power into the E-Cat’s resistance heater was NOT started at 100%, it was started at a setting of ‘5’ and RAMPED UP slowly over 40 minutes! Here is the time progression for resistance heater power… Timestamp PLC Setting DeltaTime (minutes) - --- -- 18:59 5 0 19:10 611 19:20 710 19:30 810 19:40 910 We know that the ‘Setting’ is referring to the duty cycle, but we do not know exactly what the relationship is… since 9 is the MAXimum setting, and Lewan states ‘power was at this point constantly switched on’, then a setting of ‘9’ is presumably a 100% duty cycle. (?) Since the PLC’s are programmable, we cannot assume that a setting of ‘5’ is 50% or 60%; it could even be programmed to be 10% duty cycle. So no useful calculations OR conclusions can be made during this ramp-up phase. -Mark From: Joe Catania [mailto:zrosumg...@aol.com] Sent: Wednesday, September 14, 2011 11:58 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:E-cat news at Nyteknik I think it caused a rise. There is no rise. Its your imagination. The temperature at power off is too low and must be discarded. If I bring a piece of metal the size of an E-Cat to some temperature (and note that this takes considerable time in the ramp up) and then I cut the power, the temperature will not instantaneously drop. It will stay at the same temperature and decline slowly. There is much too much mass for what your talking about to happen. I have to laugh at the fact that if you saw the temp drop even a hundredth of a degree at power down you would have declared the thermal inertia regime over and the CF regime to have begun.
Re: [Vo]:E-cat news at Nyteknik
You're trying to be too exacting. I'm pointing out facts. Because I'm not giving you a equation of everything dosen't mean thermal inertia has been ruled out. Thus you've made a grave philosophical error. It means its thermal inertia but I haven't given you the equation. Thermal inertia is a first principle. It is accepted without proof. If I add 1 megajoule to a hunk of metal at room temp and its temp goes up to 500C then it seems safe to assume that removing that 1MJ will take the temp back down to room temp. I'll admit that you're saying flow complicates this simple picture but its far from certain that you've established that through proof or equations. For instance in both cases cold water is imput at the same rate and temperature so why should there be a difference? - Original Message - From: Finlay MacNab To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 8:49 PM Subject: RE: [Vo]:E-cat news at Nyteknik Excellent observation! If this was a closed system with no FLOWING WATER EXITING THE SYSTEM you would have a point. As it is you have only discredited your argument about thermal inertia. Congratulations! I find your hand waving arguments completely unconvincing. Please describe in detail the geometry of the system you propose could account for the observed changes in temperature taking into account the well known rate of heat exchange between water and metals/other materials and the heat capacities of the various materials. Also, please account for the energy inputs and outputs to the device during its operation. 5 minutes with a text book will convince anyone with half a brain that what you describe is more improbable than cold fusion itself! Please do everyone here a favor and give a rigorous explanation of how thermal inertia can explain the rossi device. Please use equations and data to back up your claims. If you don't want to do this please stop spamming this message board and distracting from more interesting discussion. -- Well, at a setting of 9 you have the same temp rise in 35 minutes as temperature fall in 35 minutes after power-off. - Original Message - From: Mark Iverson-ZeroPoint To: vortex-l@eskimo.com Sent: Wednesday, September 14, 2011 4:55 PM Subject: RE: [Vo]:E-cat news at Nyteknik JC stated: “(and note that this takes considerable time in the ramp up)” Where he is referring to the long time it takes to ramp up the E-Cat’s internal temperature on startup… Mr. Catania, do you realize that the electrical power into the E-Cat’s resistance heater was NOT started at 100%, it was started at a setting of ‘5’ and RAMPED UP slowly over 40 minutes! Here is the time progression for resistance heater power… Timestamp PLC Setting DeltaTime (minutes) - --- -- 18:59 5 0 19:10 611 19:20 710 19:30 810 19:40 910 We know that the ‘Setting’ is referring to the duty cycle, but we do not know exactly what the relationship is… since 9 is the MAXimum setting, and Lewan states ‘power was at this point constantly switched on’, then a setting of ‘9’ is presumably a 100% duty cycle. (?) Since the PLC’s are programmable, we cannot assume that a setting of ‘5’ is 50% or 60%; it could even be programmed to be 10% duty cycle. So no useful calculations OR conclusions can be made during this ramp-up phase. -Mark From: Joe Catania [mailto:zrosumg...@aol.com] Sent: Wednesday, September 14, 2011 11:58 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:E-cat news at Nyteknik I think it caused a rise. There is no rise. Its your imagination. The temperature at power off is too low and must be discarded. If I bring a piece of metal the size of an E-Cat to some temperature (and note that this takes considerable time in the ramp up) and then I cut the power, the temperature will not instantaneously drop. It will stay at the same temperature and decline slowly. There is much too much mass for what your talking about to happen. I have to laugh at the fact that if you saw the temp drop even a hundredth of a degree at power down you would have declared the thermal inertia regime over and the CF regime to have begun.
Re: [Vo]:The Rossi Radiowave Reactor
Widom-Larsen reaches into a bag of tricks, not plauibility. The theory is uncritiqued and suffers from numerous flaws which render it totally untenable. The other theories mentioned are untenable as well, some by their own admission. As for the others, proud exclamation is not what I want in a theory. I believe you've misunderstood me to mean that I'd entertain a Coulomb barrier lowering theory. In a sense I already have but such a theory has little to do with my view of posssible proton nickel fusion. Its not my view that barrier lowering is necessary. - Original Message - From: Mark Iverson-ZeroPoint zeropo...@charter.net To: vortex-l@eskimo.com Sent: Sunday, September 04, 2011 1:08 AM Subject: RE: [Vo]:The Rossi Radiowave Reactor Joe Catania [mailto:zrosumg...@aol.com] wrote: I'm not sanguine on any of the theories popularized so far like Widom-Larsen, Rossi's theories, Piantelli's, BEC, etc. They've all been discredited. 'Discredited' how??? Please don't say, Because they contradict current theories... Are there any hypotheses that you are sanguine about??? Do you have your own hypothesis??? Don't be shy... -Mark
Re: [Vo]:The Rossi Radiowave Reactor
I'm not sanguine on any of the theories pppularized so far like Widom-Larsen, Rossi's theories, Piantelli's, BEC, etc. They've all been discredited. Ultracold neutrons simply don't work. The effective mass explanation does not wash. Other of W-L papers are simply not believeable. If you read the coverage of Wendt Irion you should see what I mean. Debye-Huckel shielding and other shielding theories are a laugh as well. In short what these theories seem to require is some way of eliminating or lowering the Coulomb barrier. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: vortex-l@eskimo.com Sent: Saturday, September 03, 2011 7:16 PM Subject: Re: [Vo]:The Rossi Radiowave Reactor I commented on possible use of nanopores earlier, and the possible useful relationship to my theory: http://www.mail-archive.com/vortex-l@eskimo.com/msg44662.html http://www.mail-archive.com/vortex-l@eskimo.com/msg44676.html http://www.mail-archive.com/vortex-l@eskimo.com/msg44683.html I should have included the correction posted below too: http://www.mail-archive.com/vortex-l@eskimo.com/msg44845.html Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:September 22 might be Rossi's final deadline
10kW sounds very low, are you in the South? - Original Message - From: Peter Heckert peter.heck...@arcor.de To: vortex-l@eskimo.com Sent: Thursday, September 01, 2011 5:39 PM Subject: Re: [Vo]:September 22 might be Rossi's final deadline Am 01.09.2011 23:12, schrieb Peter Heckert: I want to see if it can replace my 10 kW Gas-boiler. This master craftsman that calculated the gasboiler and radiators for my home would be able to tell it. Why can these top scientists not simply test or measure it and tell it to us? I have 5 radiators here in my home. These are heated with hot water that is made by a 10 kW Gasboiler. In a hard winter this gasboiler runs about 50-80% of time. If you simply take 10 radiators or a cooling sytem of a car, then you could pump water through it and measure the temperature that goes into the radiators and the temperature that comes out. If the water flow is known with 10% accuracy then the energy can be calculated with 10% accuracy. This is so simple to do. Why dont they do it? This is routine for a master craftsman to calculate heating systems that work according to pregiven specifications.
Re: [Vo]:Structure of Rossi device
I have to question that order, esp the heater abutting the lead. Also I'm not entirely convinced there isn't electrical conduction through the water. - Original Message - From: Horace Heffner hheff...@mtaonline.net To: Vortex-L vortex-l@eskimo.com Sent: Thursday, September 01, 2011 12:01 PM Subject: [Vo]:Structure of Rossi device This post is just to check my understanding of the supposed structure of the Rossi device. As I understand it, the device elements, in cross section of the reactor portion, are annular in nature, radially symmetric, except for the band heaters which roughly approximate radial symmetry. The following layers comprise the cross section, from innermost to outermost: 1. Reactor material with hydrogen 2. Steel reactor container 3. Water 4. Copper outer jacket 5. Band heater resistors driven by 220 V 6. A lead shield 7. Insulation An exception to the above is there is a resistance heater located in or around the steel reactor container? There are temperature sensors in the device which allow the controller to individually apply current to the resistance heating elements. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Corrections to heat after death calculations
You should measure the increase in your sparging more accurately for instance in a graduated cylinder. - Original Message - From: Jouni Valkonen jounivalko...@gmail.com To: vortex-l@eskimo.com Sent: Thursday, September 01, 2011 1:07 PM Subject: Re: [Vo]:Corrections to heat after death calculations 2011/9/1 Horace Heffner hheff...@mtaonline.net: Lying is not an important issue with the public tests. What if there was a hidden hydrogen bottle? 200 grams of hydrogen is enough. The issue is whether the calorimetry showed anything at all. Indeed, it showed. I will return this issue tomorrow. The issue is a relative humidity probe does not measure steam quality, or sense whether large amounts of water are overflowing. This silly measurement has nothing to do with Rossi, but all to do with Galantini. The issue is the use of a set-up that is perfect for self delusion and erroneous results, and proves nothing. This is untrue. It is almost trivial to measure enthalpy from steam by sparging. See my recent experiment in other threat. Therefore experimental setup was correct. Such major flaws are not an indication of an appropriate level of science being applied. You should tell this to Levi, Passerini, Bianchini, Galantini, Kullander, Essén, Lewan and various other persons who all failed with calorimetry. Frankly I am stunned when I realized how simple science calorimetry is and how poorly it was conducted during the demonstrations AND in various discussion forums. Lewan had extensive public discussion what to measure before he went to Bologna. He even measured electromagnetic radiation for heat transfer but no one suggested for him to do simple steam sparging calorimetry. This is what surprises me the most. And also Rossi was only passive observer when scientist made all the measurements what they thought to be necessary. Therefore no-one cannot state any objections for Rossi if the level of science was poor. –Jouni
Re: [Vo]:Corrections to heat after death calculations
ahem Mother Nature has authorized me act on Her behalf, as Her agent. I am authorized to forgive these insults. But also to warn you people to Watch Your Step. Next time She may not be so magnanimous. - Jed Jed appears to be pursued by demons. What else would induce a Japlish translator to take up residence in a cold fusion forum. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Wednesday, August 31, 2011 4:25 PM Subject: Re: [Vo]:Corrections to heat after death calculations Joe Catania zrosumg...@aol.com wrote: Not only have I been the subject of ad hominems for a presentaion that is obvious by the very nature of what is being discussed, there have been false allegations and insults to nature ahem Mother Nature has authorized me act on Her behalf, as Her agent. I am authorized to forgive these insults. But also to warn you people to Watch Your Step. Next time She may not be so magnanimous. - Jed
Re: [Vo]:Corrections to heat after death calculations
I begin to see you can be gracious where Mother Nature isn't. - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Wednesday, August 31, 2011 5:57 PM Subject: Re: [Vo]:Corrections to heat after death calculations Joe Catania wrote: ahem Mother Nature has authorized me act on Her behalf, as Her agent. I am authorized to forgive these insults. But also to warn you people to Watch Your Step. Next time She may not be so magnanimous. - Jed Jed appears to be pursued by demons. What else would induce a Japlish translator to take up residence in a cold fusion forum. The answer should be obvious. Mother Nature is Japanese. Why do you think she picked me to be her minister and interpreter, as Bacon put it? What do you think an interpreter does? Hey, I'm not proud of this. There's no profit in being a prophet. It is basically an SM relationship. Just as Bacon said, it features chains and submission, very kinky: For man, as the minister and interpreter of nature does, and understands, as much as he has observed of the order, operation, and mind of nature; and neither knows nor is able to do more. Neither is it possible for any power to loosen or burst the chain of causes, nor is nature to be overcome except by submission. - Jed
Re: [Vo]:Corrections to heat after death calculations
You have no idea what I'm talking about. If I say water flow is not what you need to get on square one with this then its true. Your posts are completely of topic and show a total lack of competence. I never said there was no water flow. I said it is not relevant. One does grasp therma inertia by understanding there is a water flow. Also it is not certain that Levi leaves the flow on. In either case your calculations have prooven to be incorrect. As this has been stated many times you must accept that your are being ignoramus. The heat capacity of water is irrelevant. The heat capacity of water is nowhere near 100 to 1000 times a metal. In fact on a volume basis they are about equal. The time of production of steam has to do with the time it takes the hot metals temperature to decay to 100C. With good insulation and no water flow (but allowing steam flow) its impossible for the temperature of the metal to decay beow 100C. You need to understand that it will produce steam for on the order of 15 minutes not 1 minute. Water flow will not drastically change the situation. Remember this is an order of mag calculation. There are no practical measurements available. Any numbers you suggest are totally biased. Also the calculations you believers suggest do not jibe with length of steam production using those numbers. You are all running away from the truth. There is little point in this since any anomalous heat has already been ascribed to hydride formation. - Original Message - From: Mark Iverson-ZeroPoint To: vortex-l@eskimo.com Sent: Tuesday, August 30, 2011 3:30 AM Subject: RE: [Vo]:Corrections to heat after death calculations Joe: Water flow is most certainly pertinent to any energy calculations concerning the E-Cat. Your statement that we aren't discussing water flow seems to indicate that either we are talking about two completely different calculations or you have no idea what you're talking about. All demonstrations of the E-Cat have had a high quality pump pumping water thru the E-Cat - where do you think the steam comes from? There is some disagreement as to the 'claimed' flowrates which might be in conflict with the apparent flowrate based on the number of 'strokes' per minute, but no one has ever claimed that there is no water flow thru the E-Cat. Given that and the fact that there are few substances that have a higher heat capacity than water, make me seriously question your understanding of the device and/or the physics involved here. Compare the heat capacity of any metal with water and you will see that water can store 100 to 1000 times more heat per mass than any metal. Since the mass of water in the E-Cat and the mass of the metal structure are at least similar, how long the E-Cat could continue to produce steam once the power was turned off is MOST CERTAINLY dependent on the water flow and the temperature of that water. probably much more so than the metal structure. PS: Horace was probably doing these kinds of energy calculations when you were still pissing in your diapers, so I'd suggest that you calm down and stop insisting that all others are wrong and you are right. If you want to gain any credibility on this discussion list then I'd suggest that you stick to facts and figures and calculations to support your points, and stop the personal attacks. -Mark From: Horace Heffner [mailto:hheff...@mtaonline.net] Sent: Monday, August 29, 2011 7:59 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Corrections to heat after death calculations On Aug 29, 2011, at 5:14 PM, Joe Catania wrote: [snip ad hominem and continued mistakes] We aren't discussing water flow. [snip ad hominem and continued mistakes] Of course we are discussing water flow. The device had water pumped into it at a constant rate. If you chose to ignore that then you chose to ignore reality. Looking back, I do see that you simply chose to ignore reality in your discussion with Jed. Joe On Aug 26, 2011, at 5:37 AM, Jed Rothwell wrote: Joe Catania wrote: No, its not out of the question at all. Since we don't know the flow rate of water (whether its flowing or not) and since it isn't particularly relevant I neglect it. The water is always flowing. This is a flow calorimeter. It is completely unrealistic to suppose that you can boil water in device this size, save up heat in metal, and then continue boiling at any observable rate for more than a few seconds after the power goes off. That is out of the question. The temperature of the metal would be far above the melting point. The metal would be incandescent. - Jed Instead of talking imaginary things I suggest a quantitative analysis to see what kinds of numbers make sense. I have taken no position on the reality of input t this point except
Re: [Vo]:Corrections to heat after death calculations
Water flow is irrelevant to what I'm discussing. It is not a certainty that Levi keeps the flow on during his power out. Clearly your grasp of physics is limited. Insulting mother nature won't clinch proof of CF. If the best you have is a dream you may as well join the rational thinkers. - Original Message - From: Horace Heffner To: vortex-l@eskimo.com Sent: Monday, August 29, 2011 10:59 PM Subject: Re: [Vo]:Corrections to heat after death calculations On Aug 29, 2011, at 5:14 PM, Joe Catania wrote: [snip ad hominem and continued mistakes] We aren't discussing water flow. [snip ad hominem and continued mistakes] Of course we are discussing water flow. The device had water pumped into it at a constant rate. If you chose to ignore that then you chose to ignore reality. Looking back, I do see that you simply chose to ignore reality in your discussion with Jed. Joe On Aug 26, 2011, at 5:37 AM, Jed Rothwell wrote: Joe Catania wrote: No, its not out of the question at all. Since we don't know the flow rate of water (whether its flowing or not) and since it isn't particularly relevant I neglect it. The water is always flowing. This is a flow calorimeter. It is completely unrealistic to suppose that you can boil water in device this size, save up heat in metal, and then continue boiling at any observable rate for more than a few seconds after the power goes off. That is out of the question. The temperature of the metal would be far above the melting point. The metal would be incandescent. - Jed Instead of talking imaginary things I suggest a quantitative analysis to see what kinds of numbers make sense. I have taken no position on the reality of input t this point except to say it looks to me that 1 MJ of stored energy seems to be too high to be real. Still, I ran some numbers that support that proposition. Applying logic to a proposition is *not* accepting the proposition as true. The statement: If x then y is not the same as: x is true. It merely provides the opportunity to examine y to see if it is feasibly true. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
