Re: [Vo]:Understandin BLP: Chapter 3
Frank,
At a university (TUe) in the Netherlands some of the experiments described by
R. Mills were successfully reproduced. EUV emission was clearly seen when the
blacklight catalysts where used in combination with hydrogen gas. The grazing
incidence EUV detector could measure emissions in the region between 10 and 30
nm.
Peter
- Original Message -
From: fznidar...@aol.com
To: vortex-l@eskimo.com
Sent: Sunday, January 26, 2014 12:51 AM
Subject: Re: [Vo]:Understandin BLP: Chapter 3
Also the hydrino reaction is much more energetic ( 200x).
So then where are the X-rays?
-Original Message-
From: P.J van Noorden
To: vortex-l
Sent: Sat, Jan 25, 2014 6:48 pm
Subject: Re: [Vo]:Understandin BLP: Chapter 3
Frank,
You say that Mill`s device is a million times hotter than fire.
That is not correct:
The powerdensity of the hydrino reaction is 1 million times higher that that
of gasoline combusion mainly due to the rate of hydrino transition compared to
the rate of gasoiline combustion ( 5000 times higher)
So per second many more reactions can take place then during gasolibe
combustion.
Also the hydrino reaction is much more energetic ( 200x).
The hydrinotransition can cause significant Balmerline broadening ( 50 eV).
The electrodes can hold that kind of temperature, when adequately cooled.
Peter v Noorden
- Original Message -
From: fznidar...@aol.com
To: vortex-l@eskimo.com
Sent: Sunday, January 26, 2014 12:27 AM
Subject: Re: [Vo]:Understandin BLP: Chapter 3
Hydrinos, shmeamos. Jones is on the right track with his energy coherence.
Exaggerated claims do not attract me. Mill's device is a million times hotter
than fire! That would place the temp at 3 billion degrees. That would be
hotter than a Tokamak. I uses very hot hydrodynamics. What does he make the
electrodes out of, neutronium! That's a claim in itself. An what about all
of those hydrinos, where do they go? What about all of the x-rays from the an
inner election reactions. Why is he not dead?
Reminds me of many older inventors who have failed. Henry Moray for
example. Not only did his device produce energy but it could detected the
faintest radio signals. In today's world of satellite communications such a
claim is truly dated.
Chankov, not only did his technology produce energy but he cold make strong
ingredients through a process infinite dilution.
Frank Z
-Original Message-
From: Mike Carrell
To: vortex-l ; cmns
Sent: Sat, Jan 25, 2014 5:53 pm
Subject: [Vo]:Understandin BLP: Chapter 3
This chapter is dedicated to Jones, others are welcome as well.
Please read:
http://www.blacklightpower.com/wp-content/uploads/presentations/TechnicalPresentation.pdf
http://www.blacklightpower.com/wp-content/uploads/pdf/GEN3_Harvard.pdf
In the Technical Presentation, read pp. 2-3, and 23-31 [but anything else
that suits your fancy].
The first two pages of the Technical Presentation is a terse summary of
GUTCP. Page 23 is a listing of methods identifying hydrinos.
The following pages discuss the conditions wherein H can act as a catalyst.
The potential energy of an isolated H atom is 13.6 eV. The energy of the
catalyst must be m(27.2] eV. Such can be supplied by many arrangements,
including 2H.2H > H{1/4] + 2H + 24 eV. Because this is a three-body reaction,
it is seen only where here is a high concentration of H atoms [such as at the
cathode of in electrolytic cell in the apparatus on p.30. [I speculate that H >
[h[1/4] catalysis may be a source of ‘excess heat’ in CF electrolytic cell
experiments. This is my conjecture.]
The cited pages above contain the core of the CIHT chemistry.
The apparatus illustrated on p.30 was built by BLP, but the reported study
and the spectrum of p.31 came from a study sponsored by GEN3 Partners at the
Harvard-Smithsonian Center for Astrophysics. None were employees of LP.
An intense beam of protons is illuminated by an intense bam of electrons,
which combine to form a cloud hydrogen atoms in which many interactions can
occur. The light from these reactions is in the low nanometer range of soft
X-rays. The spectrum is recorded by a vacuuc spectrometer. No continuum
spectrum was produced by helium.
The six ‘Validation’ reports on the website deserve respectful attention.
Although on-site and coached by Mills, the ‘validators’ built he cells
themselves and conducted tests with instruments whose calibrations are
traceable to NIST. The reports and resumes are on the .website. Over the years
experiments by Dr. Conrads in Germany, and Dr. Jonathan Phillips, U. New
Mexico, have done supporting experiments. One of the six ‘validaors’, ENSER
corporation, went on to off-site, independent validation of the CIHT cell
operation. Their new report
Re: [Vo]:Understandin BLP: Chapter 3
Frank,
You say that Mill`s device is a million times hotter than fire.
That is not correct:
The powerdensity of the hydrino reaction is 1 million times higher that that of
gasoline combusion mainly due to the rate of hydrino transition compared to the
rate of gasoiline combustion ( 5000 times higher)
So per second many more reactions can take place then during gasolibe
combustion.
Also the hydrino reaction is much more energetic ( 200x).
The hydrinotransition can cause significant Balmerline broadening ( 50 eV).
The electrodes can hold that kind of temperature, when adequately cooled.
Peter v Noorden
- Original Message -
From: fznidar...@aol.com
To: vortex-l@eskimo.com
Sent: Sunday, January 26, 2014 12:27 AM
Subject: Re: [Vo]:Understandin BLP: Chapter 3
Hydrinos, shmeamos. Jones is on the right track with his energy coherence.
Exaggerated claims do not attract me. Mill's device is a million times hotter
than fire! That would place the temp at 3 billion degrees. That would be
hotter than a Tokamak. I uses very hot hydrodynamics. What does he make the
electrodes out of, neutronium! That's a claim in itself. An what about all
of those hydrinos, where do they go? What about all of the x-rays from the an
inner election reactions. Why is he not dead?
Reminds me of many older inventors who have failed. Henry Moray for example.
Not only did his device produce energy but it could detected the faintest
radio signals. In today's world of satellite communications such a claim is
truly dated.
Chankov, not only did his technology produce energy but he cold make strong
ingredients through a process infinite dilution.
Frank Z
-Original Message-
From: Mike Carrell
To: vortex-l ; cmns
Sent: Sat, Jan 25, 2014 5:53 pm
Subject: [Vo]:Understandin BLP: Chapter 3
This chapter is dedicated to Jones, others are welcome as well.
Please read:
http://www.blacklightpower.com/wp-content/uploads/presentations/TechnicalPresentation.pdf
http://www.blacklightpower.com/wp-content/uploads/pdf/GEN3_Harvard.pdf
In the Technical Presentation, read pp. 2-3, and 23-31 [but anything else
that suits your fancy].
The first two pages of the Technical Presentation is a terse summary of
GUTCP. Page 23 is a listing of methods identifying hydrinos.
The following pages discuss the conditions wherein H can act as a catalyst.
The potential energy of an isolated H atom is 13.6 eV. The energy of the
catalyst must be m(27.2] eV. Such can be supplied by many arrangements,
including 2H.2H > H{1/4] + 2H + 24 eV. Because this is a three-body reaction,
it is seen only where here is a high concentration of H atoms [such as at the
cathode of in electrolytic cell in the apparatus on p.30. [I speculate that H >
[h[1/4] catalysis may be a source of ‘excess heat’ in CF electrolytic cell
experiments. This is my conjecture.]
The cited pages above contain the core of the CIHT chemistry.
The apparatus illustrated on p.30 was built by BLP, but the reported study
and the spectrum of p.31 came from a study sponsored by GEN3 Partners at the
Harvard-Smithsonian Center for Astrophysics. None were employees of LP.
An intense beam of protons is illuminated by an intense bam of electrons,
which combine to form a cloud hydrogen atoms in which many interactions can
occur. The light from these reactions is in the low nanometer range of soft
X-rays. The spectrum is recorded by a vacuuc spectrometer. No continuum
spectrum was produced by helium.
The six ‘Validation’ reports on the website deserve respectful attention.
Although on-site and coached by Mills, the ‘validators’ built he cells
themselves and conducted tests with instruments whose calibrations are
traceable to NIST. The reports and resumes are on the .website. Over the years
experiments by Dr. Conrads in Germany, and Dr. Jonathan Phillips, U. New
Mexico, have done supporting experiments. One of the six ‘validaors’, ENSER
corporation, went on to off-site, independent validation of the CIHT cell
operation. Their new report is on the BLP website.
“Independent verification” is a gold standard. Over the years several groups
have ‘tested’ Mills’ claims. However, they did not *duplicated*the instruments
or protocols, effectively doing *another* non-Mills experiment.
Mike Carrell
Re: [Vo]:BLP's announcement
Hello Jones You mean the experiment in which a very long capillary tube of nickel was pressurised with H2 gas and put in a K2CO3 solution? Peter - Original Message - From: "Jones Beene" To: Sent: Monday, January 20, 2014 7:11 PM Subject: RE: [Vo]:BLP's announcement Mike, I am bit surprised and disappointed that you apparently do not realize that the study in question was indeed gas phase. This was in fact a nickel hydrogen (capillary tube) reactor of Thermacore’s own design, and the study was done for the Air Force at Wright Patterson. This is as close to the Rossi effect as anything seen by others … only it preceded Rossi by over 10 years and it has never been debunked by skeptics. The experiment is stronger than anything even done by Mills IMHO, and there is nothing that comes close from any other third party. The XPS from Lehigh was independent of Mills. _ From: Mike Carrell I don’t know what Jones is attempting to prove by citing a Thermacore electrolytic cell experiment from long ago and neglecting the later years of studies in the gas phase with water bath calorimetery and magnetic resonance spectroscopy of effluent gases which show the presence of hydrinos. Mike Carrell
Re: [Vo]:BLP's announcement
Hello Jeff, Mills only provided the cell which was send to Conrads. Mills was not involved in the experiments which where done in Jüllich by Conrads (and a Phd). Conrads was a very respected plasmaphysicist (Germany). Unfortunateley he died years ago. A collegue of him in the Netherlands continued his work Peter - Original Message - From: Jeff Driscoll To: vortex-l@eskimo.com Sent: Monday, January 20, 2014 6:30 PM Subject: Re: [Vo]:BLP's announcement thank you Peter, Are there any more groups that you know replicated Mills's work - besides Rowan? The link above shows the authors to be H Conrads, R Mills and Th Wrubel, so Mills was involved but it was done outside of BLP laboratories (I assume). here is the abstract from the link you gave: A hydrogen plasma with intense extreme ultraviolet and visible emission was generated from low pressure hydrogen gas (0.1–1 mbar) in contact with a hot tungsten filament only when the filament heated a titanium dissociator coated with K2CO3 above 750�C. The electric field strength from the filament was about 1 V cm−1, two orders of magnitude lower than the starting voltages measured for gas glow discharges. The emission of the H� and H� transitions as well as the L� and L� transitions were recorded and analysed. The plasma seemed to be far from thermal equilibrium, and no conventional mechanism was found to explain the formation of a hydrogen plasma by incandescently heating hydrogen gas in the presence of trace amounts of K2CO3. The temporal behaviour of the plasma was recorded via hydrogen Balmer alpha line emission when all power into the cell was terminated and an excessive afterglow duration (2 s) was observed. The plasma was found to be dependent on the chemistry of atomic hydrogen with potassium since no plasma formed with Na2CO3 replacing K2CO3 and the time constant of the emission following the removal of all of the power to the cell matched that of the cooling of the filament and the resulting shift from atomic to molecular hydrogen. Our results indicate that a novel chemical power source is present and that it forms the energetic hydrogen plasma that is a potential new light source. On Mon, Jan 20, 2014 at 12:15 PM, P.J van Noorden wrote: Hello Jones I have talked to plasmaphysicists and they say that the continuumspectrum ( which was reproduced) proves that there is a until now unknown physical proces going on when hydrogen atoms collide (probably during 3 body reactions). Peter v Noorden - Original Message - From: Jones Beene To: vortex-l@eskimo.com Sent: Monday, January 20, 2014 5:39 PM Subject: RE: [Vo]:BLP's announcement Your spiel is a complete cop out. The Lehigh chart, which I have seen, shows a distinct signature. A so-called “continuum with a cutoff” is NOT a signature. It is a subterfuge. Mills has been frustrated over the years in being unable to show a distinct signature for the first level of redundancy (27.2) and this crap about a “continuum with a cutoff” is his feeble attempt to show what he cannot show otherwise – which is a real signature. He can show line broadening in the visible range - which is somewhat helpful – but you have “drunk to kool-aid” on this “continuum with a cutoff” BS as being anything other than a generalization, meaning nothing. If it were not for the fine study by Thermacore, Mills could probably get away with this kind of intellectual dishonesty. He is looking more and more like a charlatan and this upcoming demo will be an insult. Jones From: Jeff Driscoll As far as I know, Mills's theory does not predict a continuum radiation having a cuttoff at a frequency that corresponds to a 27.2 eV for transitions that start from n = 1 (maybe fractional to fractional transition does, I don't know) see here: http://zhydrogen.com/wp-content/uploads/2011/04/19pn.gif And, Mills theory only has continuum radiation with a cuttoff frequency. There are no photons emitted that have a specific frequency that shows up sharply on a graph. That's why it is hard to detect hydrino photon emission during hydrino creation. I try to explain it all here on pages 52-55: http://zhydrogen.com/wp-content/uploads/2013/04/BLP-presentation.pdf Jeff On Mon, Jan 20, 2014 at 11:13 AM, Jones Beene wrote: From: David Roberson A thought just occurred to me. Is it not possible to ionize a hydrino with high temperatures, gamma radiation, or other energetic processes? This should be able to return the hydrino back into hydrogen again which should be detected. I suppose that if these processes can impact the hydrinos then they should not be considered dark manner by definition.
Re: [Vo]:BLP's announcement
Hello Jones I have talked to plasmaphysicists and they say that the continuumspectrum ( which was reproduced) proves that there is a until now unknown physical proces going on when hydrogen atoms collide (probably during 3 body reactions). Peter v Noorden - Original Message - From: Jones Beene To: vortex-l@eskimo.com Sent: Monday, January 20, 2014 5:39 PM Subject: RE: [Vo]:BLP's announcement Your spiel is a complete cop out. The Lehigh chart, which I have seen, shows a distinct signature. A so-called "continuum with a cutoff" is NOT a signature. It is a subterfuge. Mills has been frustrated over the years in being unable to show a distinct signature for the first level of redundancy (27.2) and this crap about a "continuum with a cutoff" is his feeble attempt to show what he cannot show otherwise - which is a real signature. He can show line broadening in the visible range - which is somewhat helpful - but you have "drunk to kool-aid" on this "continuum with a cutoff" BS as being anything other than a generalization, meaning nothing. If it were not for the fine study by Thermacore, Mills could probably get away with this kind of intellectual dishonesty. He is looking more and more like a charlatan and this upcoming demo will be an insult. Jones From: Jeff Driscoll As far as I know, Mills's theory does not predict a continuum radiation having a cuttoff at a frequency that corresponds to a 27.2 eV for transitions that start from n = 1 (maybe fractional to fractional transition does, I don't know) see here: http://zhydrogen.com/wp-content/uploads/2011/04/19pn.gif And, Mills theory only has continuum radiation with a cuttoff frequency. There are no photons emitted that have a specific frequency that shows up sharply on a graph. That's why it is hard to detect hydrino photon emission during hydrino creation. I try to explain it all here on pages 52-55: http://zhydrogen.com/wp-content/uploads/2013/04/BLP-presentation.pdf Jeff On Mon, Jan 20, 2014 at 11:13 AM, Jones Beene wrote: From: David Roberson A thought just occurred to me. Is it not possible to ionize a hydrino with high temperatures, gamma radiation, or other energetic processes? This should be able to return the hydrino back into hydrogen again which should be detected. I suppose that if these processes can impact the hydrinos then they should not be considered dark manner by definition. Dave, Yes, this procedure you mention is rather obvious - and it has in fact been done; but one reason that you do not hear about this particular finding on a regular basis could be that the results are open to interpretation. I am going to present the interpretation which Mills does not want you to hear. You can make your own judgment on what is really happening. The most convincing paper on hydrinos which is available to view - was not performed by Mills but by Thermacore. Long term excess heat was found as was a time delayed signature. https://www.google.com/url?q=http://www.lenr-canr.org/acrobat/GernertNnascen thyd.pdf&sa=U&ei=e0DdUq3AIsTgyQHUyoGIAg&ved=0CAYQFjAA&client=internal-uds-cs e&usg=AFQjCNG_00ZwiWP5nfDF2NVjs0l9AOKQmQ .and in that paper the nickel capillary tubing, after the very long successful run, gives up the best evidence ever for the existence of the hydrino - since it was tested by ESCA analysis at Lehigh University. There is no doubt the tests were accurate - it is the interpretation that can vary. ESCA is now known as X-ray photoelectron spectroscopy (XPS) and is accomplished by capturing spectra obtained by irradiating a material with a monochromatic beam of relatively soft X-rays. In this case, the results seem to support some of Mills theory but not all of it. The Lehigh University testing in fact finds NO 27.2 eV signature, as Mills theory suggests. However, XPS does find the a 55 eV signal/signature, which is close to Mills' theoretical signature for the hydrino, which is supposed to be 54.4 eV but not exact. However, the XPS device is in fact capable of showing an exact signature, but none is found. Mike Carrel has also mentioned that Mills has lately dropped efforts to find the lower Rydberg signature in favor of the H(1/4). What Mike failed to mention is that the reason for this change in strategy is that BLP HAS NEVER BEEN ABEL TO SHOW THE 27.2 SIGNATURE. and if one is mildly skeptical of Mills, this can be viewed as a disaster since the higher energy signal is itself off target. In fact, it is clear to me that the Mills theory cannot be accurate, given the independent testing, and that there is no signal at the all-important level of 27.2 eV and in fact the higher level signal is itself NOT at the exact Rydberg level but is off by up to 8 percent. The bottom line i
Re: [Vo]: What is Faraday Efficiency?
Hello John Please look at the very important video of the research done by A. de Ninno more then 10 years ago in which she confirmed that the excess heatrelease was related to 4He production. http://www.youtube.com/watch?v=bujrxqwRwc0 Peter van Noorden - Original Message - From: John Franks To: vortex-l@eskimo.com Sent: Sunday, December 15, 2013 3:00 PM Subject: Re: [Vo]: What is Faraday Efficiency? Axil Axil Sat, 14 Dec 2013 18:34:20 -0800 Experimentation with gold nano-particles show LENR+reaction with 100% repeatability. Foks0904 . Sat, 14 Dec 2013 17:35:15 -0800 ...take Energetics in Israel (now at U of M) for example who had reach ~70-80% repeatability in their cells. Delusional, just lies or maybe incompetence. The onus is on you to rule out mistakes and other artifacts before shooting off that it's nuclear reaction in an otherwise chemical setup. So far, no one is listening.
Re: [Vo]:Re:
Hi James Mills atomic model considers the normal ground state as a treshold between photon and photonless transitions between quantumlevels in the atom. So there is no spontaneous transition possible of the electron to subground levels. Only when a catalyst is used which can absorb n x 27.2 eV , the electron orbitsphere destabilizes and contracts and releases additional energy until a new stable situation is reached. Regarding hydrino formation and LENR I suspect that the transition to subground levels is probably the first step in all excess heat phenomenon. If transition progresses to very low fractional quantum levels (according to Mills there are 137), the coulomb barrier is lowered and incidentally normal nuclear reactions are possible with normal particle emissions like in hot fusion. This is seen by many groups. That could also explain transmutations, which could be the result of normal nuclear reactions. These particle emissions and transmutations are not commensurate with the released heat. Now comes the big question. Are there any new nuclear reactions possible which produce for instance 4He without particles or gamma emission? A possible explanation could be that these new nuclear reactions would be the result of 3 body interaction between hydrino`s, which would be most probable when a metallic surface is present. The production of 4He almost commensurate with the heat release remains a big mystery . Why would 4He be produced with a 3 body reaction while a more probable 2 body reaction between deuterium atoms is suppressed? I often wonder if a mass spectrometer could be fooled by a di-deuterino molecule in a very low fractional quantumstate (D 1/37), but i am no expert to decide on that. Peter - Original Message - From: James Bowery To: vortex-l Sent: Sunday, December 01, 2013 10:12 PM Subject: Re: [Vo]:Re: I haven't followed the hydrino theory but the little that I've taken notice of it there seems to be a problem with the idea that entropy favors hydrino production since, if that were the case the ocean would turn into dihydrino oxide, right? Moreover, to add confusion there is the energy emitted in going from hydrogen to hydrino conflated with the energy emitted from LENR. If someone could provide a concise disaggregation of these two points of confusion/conflation, it would help. On Sun, Dec 1, 2013 at 2:28 PM, wrote: In reply to David Roberson's message of Sun, 1 Dec 2013 03:17:14 -0500 (EST): Hi, [snip] > >Have you determined whether or not this is a reversible process? How often does the lower energy hydrino accept energy from a catalyst that has not yet released the same amount of energy in the form of radiation? It is common for energy to be traded in both directions according to thermodynamic laws. > >Dave Entropy works in favour of Hydrino production. Once the energy has been released, it is difficult to get it back again. Furthermore, it is a two stage process. In the first stage a multiple of 27.2 eV is handed to the catalyst. In the second stage more energy is released as the newly formed Hydrino stabilizes. Consequently, an "excited" catalyst doesn't have enough energy to "re-inflate" a stable Hydrino. The simplest case will serve as an example:- Stage I)H + Ar+ => Hy* + Ar++ + e- Stage II) Hy* => Hy + 13.6 eV (UV or kinetic) (I have used Hy* to denote an intermediate state of the Hydrino). Note that in (I) 27.2 eV is transferred to the Ar+, which further ionizes it, but an additional 13.6 eV is lost in (II), so the total energy lost is 40.8 eV. Hence Ar++ recombining with a free electron to produce Ar+ would only generate 27.2 eV, which is 13.6 eV short of the amount required to "re-inflate" the Hydrino back to H. Note the actual ionization energy of Ar+ is 27.6 eV, but I have deliberately left the consequences of that out of the explanation as I am trying to keep it simple. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Re:
Hello Dave Lower energy hydrino can accept energy from another hydrino which is in a higher fractional quantum state. So they switch the quantum state. Collision of atomic H with the catalyst who has first accepted (ionisation and kinetic) energy from atomic H and produced the lower energy hydrino can probably also switch energy. If this occurs depends on the timeframe between the original energytransfer and the second collision. If inbetween energy is lost to other atoms/molecules in the environment the proces is probably not reversible, bcs there is no match for energy transfer. Interesting is that energy transfer can also occur in a 3 body reaction. When 3 atomic H atoms collide ( lower probability) 2 H atoms accept 2x 13.6 eV and the third H atom goes into a fractional quantum state. I think the proces will result in net energy release bcs at moderate temperatures the proces produces more and more H in a fractional quantum state. With increasing temperature the reaction will probably be less effective in producing lower energy hydrino. I think the hydrino producing proces is common in nature but only if atomic hydrogen present and a catalyst. On earth there is virtually no atomic H present, so there are only minute quantities of hydrino componds in the environment. In space atomic hydrogen is abundant. Collision with a catalyst (He+) will probably result in large quantities of H in a fractional quantum state. It would be not surprising that the corona of the sun is the result of this energy transfer. Peter - Original Message - From: David Roberson To: vortex-l@eskimo.com Sent: Sunday, December 01, 2013 9:17 AM Subject: Re: [Vo]:Re: Have you determined whether or not this is a reversible process? How often does the lower energy hydrino accept energy from a catalyst that has not yet released the same amount of energy in the form of radiation? It is common for energy to be traded in both directions according to thermodynamic laws. Dave -Original Message- From: P.J van Noorden To: vortex-l Sent: Sat, Nov 30, 2013 5:46 pm Subject: Re: [Vo]:Re: Collision of atomic hydrogen with a element or compound (called a catalyst) that can accept n times 27.2 eV ( 2 times the ionistaion energy of hydrogen) destabilises the electron, which falls to a lower fractional quantum level, thereby releasing energy. - Original Message - From: Axil Axil To: vortex-l Sent: Saturday, November 30, 2013 6:48 PM Subject: Re: [Vo]:Re: According to Mill theory, what causes the electron/hydrino to enter the fractional charge state? On Sat, Nov 30, 2013 at 12:08 PM, P.J van Noorden wrote: Hello Steven, There have been validation reports about the working of the CIHT cell and in the month of december new validation reports will be relased as well as a press release. I think the main issue is now to prevent the electrodes fro degrading and to increase the surface density Peter - Original Message - From: OrionWorks - Steven Vincent Johnson To: vortex-l@eskimo.com Sent: Saturday, November 30, 2013 5:31 PM Subject: RE: [Vo]:Re: Hi P.J. Thanks for the report. I must confess I have not monitored the BLP website for some time. It would be nice to discover I missed some juicy news. You mention the fact that -a- 10W C1HT cell has been developed (and presumably demonstrated?) within BLP's labs. Do you feel fairly confident that the 10W prototype actually does what BLP claims? If so, there must exist a LOT of prototypes in various evolutionary stages focusing on out how to get the engineering to endure high temperatures without degrading. I’m assuming the C1HT chemistry is highly caustic, particularly at the high temperatures needed. If this is all true I think BLP is more than capable of succeeding. But will they be able to make a commercial success out of their efforts? Depends on how long will it take them to get a product out on the shelves of Wall Mart. One year? Ten more years? Who knows. We all know there are other competitors veiling for the same slice of the pie. Humans are a very clever species. When motivated, such as getting caught in the aphrodisiac of making obscene piles of money, or something as simple as promises of sexual favors, we monkeys can accomplish just about anything. Having been blessed with thumbs does help. Woody Allan was wrong in Sleeper. The thumb should be considered the “second favorite organ”. (Here’s a little weekend down-time humor.) http://www.youtube.com/watch?v=ngizj5FIcjo You appear to be taking the place of Mike Carrell when it comes to reporting on the status of BLP's R&D efforts. Again, thanks for the report, P.J. Hopefully, Mike is listening in. ;-) Regards, Steven Vincent Johnson svjart.
Re: [Vo]:Re:
At this moment the CIHT cell is part of a laboratory setup. More a demonstartion unit. - Original Message - From: James Bowery To: vortex-l Sent: Sunday, December 01, 2013 12:27 AM Subject: Re: [Vo]:Re: Pardon me, I should have asked, "What is the price BLP is charging for a 10W cell in quantity one?" On Sat, Nov 30, 2013 at 6:35 AM, P.J van Noorden wrote: Hello James, According to BLP the material cost of a CIHT cell of 1kW will cost only $100. There are no expensive materials needed. Peter - Original Message - From: James Bowery To: vortex-l Sent: Saturday, November 30, 2013 8:44 AM Subject: Re: [Vo]:Re: How much does a 10W cell cost in quantity one? On Sat, Nov 30, 2013 at 12:36 AM, P.J van Noorden wrote: Hello Steven Blacklightpower has made a 10W CIHT cell which can produce electricity from watervapour. The composition of the electrode is such that the hydrino producing reaction is facilitated and the electronshifts caused by transition of the electrons to sub groundstate levels can be used externally like in a battery. These cells work at high temperature ( few hundred dgrC ) so they must be well insulated. By using a very good insulation the cell should stay hot, bcs the reaction will also produce heat. The focus of BLP lies in the construction of an electrode that can function at such a high temperature for a very long time without degradation and to scaleup the powerdensity of the reaction in order to make the cell more compact. To build a system of 1000W would then be relatively simple by using 100 cells. Peter v Noorden - Original Message - From: "OrionWorks - Steven Vincent Johnson" To: Sent: Saturday, November 30, 2013 12:42 AM Subject: [Vo]: This was recently posted out on the Yahoo group: "Society for Classical Physics", Dr. Mill's Yahoo group: The recent FAQ under the CIHT topic mentions that "The CIHT cell has been scaled to 10 W, and a development projection with the achieved significant increase in surface power density is a 1.5 kW electric module that can be ganged accordingly to serve larger power applications." It also talks about the 1.5kw pre-production prototype expected by the end of 2013. Here we are near the end of 2013 and I have to ask, why aren't we hearing about a public demo of the 10W version? It was actually projected for a couple of years ago and is apparently working at BLP. We keep hearing about the scaling up, but as another poster observed earlier, "nothing to hang your hat on". Whither the 10W demo? -- Lynn I'll be curious to find out what Dr. Mills might choose to say on the matter. I thought the web site was monitored. I was surprised to see it get posted. Regards, Steven Vincent Johnson svjart.OrionWorks.com www.zazzle.com/orionworks tech.groups.yahoo.com/group/newvortex/
Re: [Vo]:Re:
Collision of atomic hydrogen with a element or compound (called a catalyst)
that can accept n times 27.2 eV ( 2 times the ionistaion energy of hydrogen)
destabilises the electron, which falls to a lower fractional quantum level,
thereby releasing energy.
- Original Message -
From: Axil Axil
To: vortex-l
Sent: Saturday, November 30, 2013 6:48 PM
Subject: Re: [Vo]:Re:
According to Mill theory, what causes the electron/hydrino to enter the
fractional charge state?
On Sat, Nov 30, 2013 at 12:08 PM, P.J van Noorden
wrote:
Hello Steven,
There have been validation reports about the working of the CIHT cell and
in the month of december new validation reports will be relased as well as a
press release. I think the main issue is now to prevent the electrodes fro
degrading and to increase the surface density
Peter
- Original Message -
From: OrionWorks - Steven Vincent Johnson
To: vortex-l@eskimo.com
Sent: Saturday, November 30, 2013 5:31 PM
Subject: RE: [Vo]:Re:
Hi P.J.
Thanks for the report. I must confess I have not monitored the BLP
website for some time. It would be nice to discover I missed some juicy news.
You mention the fact that -a- 10W C1HT cell has been developed (and
presumably demonstrated?) within BLP's labs. Do you feel fairly confident that
the 10W prototype actually does what BLP claims? If so, there must exist a LOT
of prototypes in various evolutionary stages focusing on out how to get the
engineering to endure high temperatures without degrading. I’m assuming the
C1HT chemistry is highly caustic, particularly at the high temperatures needed.
If this is all true I think BLP is more than capable of succeeding. But
will they be able to make a commercial success out of their efforts? Depends on
how long will it take them to get a product out on the shelves of Wall Mart.
One year? Ten more years? Who knows. We all know there are other competitors
veiling for the same slice of the pie. Humans are a very clever species. When
motivated, such as getting caught in the aphrodisiac of making obscene piles of
money, or something as simple as promises of sexual favors, we monkeys can
accomplish just about anything. Having been blessed with thumbs does help.
Woody Allan was wrong in Sleeper. The thumb should be considered the “second
favorite organ”. (Here’s a little weekend down-time humor.)
http://www.youtube.com/watch?v=ngizj5FIcjo
You appear to be taking the place of Mike Carrell when it comes to
reporting on the status of BLP's R&D efforts. Again, thanks for the report,
P.J. Hopefully, Mike is listening in. ;-)
Regards,
Steven Vincent Johnson
svjart.OrionWorks.com
www.zazzle.com/orionworks
tech.groups.yahoo.com/group/newvortex/
From: P.J van Noorden [mailto:pjvannoor...@caiway.nl]
Sent: Saturday, November 30, 2013 6:35 AM
To: vortex-l@eskimo.com
Subject: Re: [Vo]:Re:
Hello James,
According to BLP the material cost of a CIHT cell of 1kW will cost only
$100.
There are no expensive materials needed.
Peter
- Original Message -
From: James Bowery
To: vortex-l
Sent: Saturday, November 30, 2013 8:44 AM
Subject: Re: [Vo]:Re:
How much does a 10W cell cost in quantity one?
On Sat, Nov 30, 2013 at 12:36 AM, P.J van Noorden
wrote:
Hello Steven
Blacklightpower has made a 10W CIHT cell which can produce electricity
from watervapour.
The composition of the electrode is such that the hydrino producing
reaction is facilitated and the electronshifts caused by transition of the
electrons to sub groundstate levels can be used externally like in a battery.
These cells work at high temperature ( few hundred dgrC ) so they must be well
insulated. By using a very good insulation the cell should stay hot, bcs the
reaction will also produce heat. The focus of BLP lies in the construction of
an electrode that can function at such a high temperature for a very long time
without degradation and to scaleup the powerdensity of the reaction in order to
make the cell more compact. To build a system of 1000W would then be relatively
simple by using 100 cells.
Peter v Noorden
- Original Message - From: "OrionWorks - Steven Vincent
Johnson"
To:
Sent: Saturday, November 30, 2013 12:42 AM
Subject: [Vo]:
This was recently posted out on the Yahoo group: "Society for Classical
Physics",
Dr. Mill's Yahoo group:
The recent FAQ under the CIHT topic mentions that "The CIHT cell has
been scaled to 10 W, and a development projection with the achieved
significant increase in surface power density is a 1.5 kW electric
modu
Re: [Vo]:Re:
Hello Steven, There have been validation reports about the working of the CIHT cell and in the month of december new validation reports will be relased as well as a press release. I think the main issue is now to prevent the electrodes fro degrading and to increase the surface density Peter - Original Message - From: OrionWorks - Steven Vincent Johnson To: vortex-l@eskimo.com Sent: Saturday, November 30, 2013 5:31 PM Subject: RE: [Vo]:Re: Hi P.J. Thanks for the report. I must confess I have not monitored the BLP website for some time. It would be nice to discover I missed some juicy news. You mention the fact that -a- 10W C1HT cell has been developed (and presumably demonstrated?) within BLP's labs. Do you feel fairly confident that the 10W prototype actually does what BLP claims? If so, there must exist a LOT of prototypes in various evolutionary stages focusing on out how to get the engineering to endure high temperatures without degrading. I'm assuming the C1HT chemistry is highly caustic, particularly at the high temperatures needed. If this is all true I think BLP is more than capable of succeeding. But will they be able to make a commercial success out of their efforts? Depends on how long will it take them to get a product out on the shelves of Wall Mart. One year? Ten more years? Who knows. We all know there are other competitors veiling for the same slice of the pie. Humans are a very clever species. When motivated, such as getting caught in the aphrodisiac of making obscene piles of money, or something as simple as promises of sexual favors, we monkeys can accomplish just about anything. Having been blessed with thumbs does help. Woody Allan was wrong in Sleeper. The thumb should be considered the "second favorite organ". (Here's a little weekend down-time humor.) http://www.youtube.com/watch?v=ngizj5FIcjo You appear to be taking the place of Mike Carrell when it comes to reporting on the status of BLP's R&D efforts. Again, thanks for the report, P.J. Hopefully, Mike is listening in. ;-) Regards, Steven Vincent Johnson svjart.OrionWorks.com www.zazzle.com/orionworks tech.groups.yahoo.com/group/newvortex/ From: P.J van Noorden [mailto:pjvannoor...@caiway.nl] Sent: Saturday, November 30, 2013 6:35 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:Re: Hello James, According to BLP the material cost of a CIHT cell of 1kW will cost only $100. There are no expensive materials needed. Peter - Original Message - From: James Bowery To: vortex-l Sent: Saturday, November 30, 2013 8:44 AM Subject: Re: [Vo]:Re: How much does a 10W cell cost in quantity one? On Sat, Nov 30, 2013 at 12:36 AM, P.J van Noorden wrote: Hello Steven Blacklightpower has made a 10W CIHT cell which can produce electricity from watervapour. The composition of the electrode is such that the hydrino producing reaction is facilitated and the electronshifts caused by transition of the electrons to sub groundstate levels can be used externally like in a battery. These cells work at high temperature ( few hundred dgrC ) so they must be well insulated. By using a very good insulation the cell should stay hot, bcs the reaction will also produce heat. The focus of BLP lies in the construction of an electrode that can function at such a high temperature for a very long time without degradation and to scaleup the powerdensity of the reaction in order to make the cell more compact. To build a system of 1000W would then be relatively simple by using 100 cells. Peter v Noorden - Original Message - From: "OrionWorks - Steven Vincent Johnson" To: Sent: Saturday, November 30, 2013 12:42 AM Subject: [Vo]: This was recently posted out on the Yahoo group: "Society for Classical Physics", Dr. Mill's Yahoo group: The recent FAQ under the CIHT topic mentions that "The CIHT cell has been scaled to 10 W, and a development projection with the achieved significant increase in surface power density is a 1.5 kW electric module that can be ganged accordingly to serve larger power applications." It also talks about the 1.5kw pre-production prototype expected by the end of 2013. Here we are near the end of 2013 and I have to ask, why aren't we hearing about a public demo of the 10W version? It was actually projected for a couple of years ago and is apparently working at BLP. We keep hearing about the scaling up, but as another poster observed earlier, "nothing to hang your hat on". Whither the 10W demo? -- Lynn I'll be curious to find out what Dr. Mills might choose to say on the matter. I thought the web site was
Re: [Vo]:Re:
Hello James, According to BLP the material cost of a CIHT cell of 1kW will cost only $100. There are no expensive materials needed. Peter - Original Message - From: James Bowery To: vortex-l Sent: Saturday, November 30, 2013 8:44 AM Subject: Re: [Vo]:Re: How much does a 10W cell cost in quantity one? On Sat, Nov 30, 2013 at 12:36 AM, P.J van Noorden wrote: Hello Steven Blacklightpower has made a 10W CIHT cell which can produce electricity from watervapour. The composition of the electrode is such that the hydrino producing reaction is facilitated and the electronshifts caused by transition of the electrons to sub groundstate levels can be used externally like in a battery. These cells work at high temperature ( few hundred dgrC ) so they must be well insulated. By using a very good insulation the cell should stay hot, bcs the reaction will also produce heat. The focus of BLP lies in the construction of an electrode that can function at such a high temperature for a very long time without degradation and to scaleup the powerdensity of the reaction in order to make the cell more compact. To build a system of 1000W would then be relatively simple by using 100 cells. Peter v Noorden - Original Message - From: "OrionWorks - Steven Vincent Johnson" To: Sent: Saturday, November 30, 2013 12:42 AM Subject: [Vo]: This was recently posted out on the Yahoo group: "Society for Classical Physics", Dr. Mill's Yahoo group: The recent FAQ under the CIHT topic mentions that "The CIHT cell has been scaled to 10 W, and a development projection with the achieved significant increase in surface power density is a 1.5 kW electric module that can be ganged accordingly to serve larger power applications." It also talks about the 1.5kw pre-production prototype expected by the end of 2013. Here we are near the end of 2013 and I have to ask, why aren't we hearing about a public demo of the 10W version? It was actually projected for a couple of years ago and is apparently working at BLP. We keep hearing about the scaling up, but as another poster observed earlier, "nothing to hang your hat on". Whither the 10W demo? -- Lynn I'll be curious to find out what Dr. Mills might choose to say on the matter. I thought the web site was monitored. I was surprised to see it get posted. Regards, Steven Vincent Johnson svjart.OrionWorks.com www.zazzle.com/orionworks tech.groups.yahoo.com/group/newvortex/
[Vo]:Re:
Hello Steven Blacklightpower has made a 10W CIHT cell which can produce electricity from watervapour. The composition of the electrode is such that the hydrino producing reaction is facilitated and the electronshifts caused by transition of the electrons to sub groundstate levels can be used externally like in a battery. These cells work at high temperature ( few hundred dgrC ) so they must be well insulated. By using a very good insulation the cell should stay hot, bcs the reaction will also produce heat. The focus of BLP lies in the construction of an electrode that can function at such a high temperature for a very long time without degradation and to scaleup the powerdensity of the reaction in order to make the cell more compact. To build a system of 1000W would then be relatively simple by using 100 cells. Peter v Noorden - Original Message - From: "OrionWorks - Steven Vincent Johnson" To: Sent: Saturday, November 30, 2013 12:42 AM Subject: [Vo]: This was recently posted out on the Yahoo group: "Society for Classical Physics", Dr. Mill's Yahoo group: The recent FAQ under the CIHT topic mentions that "The CIHT cell has been scaled to 10 W, and a development projection with the achieved significant increase in surface power density is a 1.5 kW electric module that can be ganged accordingly to serve larger power applications." It also talks about the 1.5kw pre-production prototype expected by the end of 2013. Here we are near the end of 2013 and I have to ask, why aren't we hearing about a public demo of the 10W version? It was actually projected for a couple of years ago and is apparently working at BLP. We keep hearing about the scaling up, but as another poster observed earlier, "nothing to hang your hat on". Whither the 10W demo? -- Lynn I'll be curious to find out what Dr. Mills might choose to say on the matter. I thought the web site was monitored. I was surprised to see it get posted. Regards, Steven Vincent Johnson svjart.OrionWorks.com www.zazzle.com/orionworks tech.groups.yahoo.com/group/newvortex/
Re: [Vo]:Meshuganon, pseudo-scientists and other ridiculing terms...
Meshuganon will probably turn out to be hydrino/deuterino Peter - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Friday, September 13, 2013 7:43 PM Subject: Re: [Vo]:Meshuganon, pseudo-scientists and other ridiculing terms... Alain Sepeda wrote: I remember that some nasty terms was used agains Fleischmann experiments... beside Meshuganon used in some conference, Meshuganon was not an insult at all. Edward Teller used it at at the NSF conference. See: http://lenr-canr.org/acrobat/EPRInsfepriwor.pdf - Jed Geen virus gevonden in dit bericht. Gecontroleerd door AVG - www.avg.com Versie: 2013.0.3408 / Virusdatabase: 3222/6662 - datum van uitgifte: 09/13/13
Re: [Vo]:new solar concentrator
Hello Jed, Yes it is but it is capable to concentrate sunlight on a high efficiency solar cel during the day, while the sun moves. The system is rather flat Peter - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Thursday, April 25, 2013 3:55 PM Subject: Re: [Vo]:new solar concentrator This is an annoying web site. Slow to load and uninformative. What is this thing? A Fresnel lens plus PV cell? - Jed
[Vo]:new solar concentrator
Hello, New solar concentrator without parabolic mirror. System is flat and can produce electricity and heat http://suncycle.nl/# Peter v Noorden
Re: [Vo]:April 2013 Defkalion GT Interview
Hello Axil, Do you really think that QM is the end of the story? How is it possible that an electron can circle around a positive H atom and not radiate. That is only possible when you postulate nonradiation. This is a very unscientific way of explaining. Peter - Original Message - From: Axil Axil To: vortex-l Sent: Sunday, April 07, 2013 12:14 AM Subject: Re: [Vo]:April 2013 Defkalion GT Interview The evidence that Miles provides is spectroscopic. The blue shifts that Mills sites are due to polariton far field frequency emanations from a micro cavity or grain boundary. Hydrinos violate the rules of quantum mechanics. On Sat, Apr 6, 2013 at 5:47 PM, P.J van Noorden wrote: Hello Axil, I just send an article about the theory of R.Mills to vortex. It was an attachment so it is probably rejected. I thought about that only after I send it to vortex. If you are interested I can send it to you. The article is called: The fallacy of Feynman’s and related arguments on the stability of the hydrogen atom according to quantum mechanics. Peter van Noorden the Netherlands
Re: [Vo]:April 2013 Defkalion GT Interview
Hello Axil, I just send an article about the theory of R.Mills to vortex. It was an attachment so it is probably rejected. I thought about that only after I send it to vortex. If you are interested I can send it to you. The article is called: The fallacy of Feynman's and related arguments on the stability of the hydrogen atom according to quantum mechanics. Peter van Noorden the Netherlands
Re: [Vo]:Violante and others are trying the engineering approach
Hello, As far as I know because of the low heat production ( 25-50W for 5 km nickle tubing as an kathode) and the saftely risk of developing this boilersystem (it can explode) thermacore shifted the interest to heatpipe cooling systems for computers. Peter v Noorden - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Thursday, February 21, 2013 9:39 PM Subject: Re: [Vo]:Violante and others are trying the engineering approach Jones Beene wrote: Somewhat amazing that no major lab has taken the initiative to replicate (or debunk), after all these years... Srinivasan thought he replicated this at BARC. He got heat and tritium. Then he spent 6 months at SRI trying to do it again. He finally concluded that his results were caused by recombination. That was a noble effort. He went back to India and tried again, looking for tritium only, with no calorimetry. They saw some tritium this time, but not as much. I recall some other people tried to replicate, without success. The results were not encouraging. I do not understand why Thermacore abandoned this. It is one of many discouraging failures. The failure to follow through. - Jed
Re: [Vo]:Bose Einstein Condensate formed at Room Temperature
Hello Jed, Interesting question. I wonder if this has been tried in the past or recently. Peter - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Tuesday, February 12, 2013 9:02 PM Subject: Re: [Vo]:Bose Einstein Condensate formed at Room Temperature Edmund Storms wrote: I predict that the H-Ni reaction is not the source of energy being claimed by Rossi et al. Any theory that attempts to justify the H-Ni reaction as the source of claimed energy is open to test because, if I'm right, such a theory is useless because it predicts something that does not happen. I predict deuterium is the source of heat based on a mechanism that also explains all other reactions. You are saying the deuterium in ordinary water is reacting in Rossi's cell? What do you predict will happen if it is enriched above the natural ratio? - Jed
Re: [Vo]: MFMP Null Result
test - Original Message - From: David Roberson To: vortex-l@eskimo.com Sent: Thursday, February 07, 2013 4:33 PM Subject: Re: [Vo]: MFMP Null Result I wish I knew how to answer this line of inquiry. If you are suggesting that there should be LENR activity and thus a reading of zero excess power is a false negative, then the program demonstrates that. It is my philosophy to let the results speak for themselves regardless of the outcome. The program does that by fitting the input power variable to the data for the best match. I have no way to change this once it has been told to optimize unless I intentionally lock its value for other purposes. For all of the runs up through the present, the optimized input power has calculated less than the applied power. There have been a few times when the instantaneous power difference has suggested that slightly more is coming out than in, but the longer term average never has. Most times the average excess power has been quite close to the applied input as in the latest run where it was within -.2 watts out of 105.4 watts. I suspect that the noise riding on the data due to external temperature variation, or etc. has enabled the peaks to exceed the input, but there also may be a small component of true excess power. When I make an objective analysis of the program runs so far I come to the conclusion that there is no significant excess power being displayed. Label me a skeptic, but I very much want to see positive results. Dave -Original Message- From: Daniel Rocha To: John Milstone Sent: Thu, Feb 7, 2013 5:53 am Subject: Re: [Vo]: MFMP Null Result Why not doing both? You refer to true positives, that is, a signal actually being measured. So, why not a false negative, that is, something that should be there but it isn't. 2013/2/6 David Roberson If it does not show up, how could it be measured? -- Daniel Rocha - RJ danieldi...@gmail.com
Re: [Vo]:colloquium cold fusion 2013 in eindhoven
Hi Robbie,
I registered late in the afternoon and got a welcome mail.
It will be interesting to attend the colloquium..Are you also living in the
Netherlands like me?
Peter
- Original Message -
From: Teslaalset
To: vortex-l@eskimo.com
Sent: Tuesday, January 15, 2013 9:00 PM
Subject: Re: [Vo]:colloquium cold fusion 2013 in eindhoven
Don't forget to register for this event, Peter.
On Tue, Jan 15, 2013 at 8:40 PM, P.J van Noorden
wrote:
This link is hopefully correct
http://www.tue.nl/en/university/departments/electrical-engineering/department/news-and-press/calendar/colloquium-cold-fusion-2013
- Original Message -
From: Jed Rothwell
To: vortex-l@eskimo.com
Sent: Tuesday, January 15, 2013 8:00 PM
Subject: Re: [Vo]:colloquium cold fusion 2013 in eindhoven
P.J van Noorden wrote:
I just stumbled on this:
http://www.tue.nl/universiteit/nieuws-en-pers/agenda/2013-01-21-cold-fusion-2013/
This link does not work for me.
Jean-Paul is writing a book about cold fusion. I guess he will present
some of material from it.
- Jed
Re: [Vo]:colloquium cold fusion 2013 in eindhoven
This link is hopefully correct http://www.tue.nl/en/university/departments/electrical-engineering/department/news-and-press/calendar/colloquium-cold-fusion-2013 - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Tuesday, January 15, 2013 8:00 PM Subject: Re: [Vo]:colloquium cold fusion 2013 in eindhoven P.J van Noorden wrote: I just stumbled on this: http://www.tue.nl/universiteit/nieuws-en-pers/agenda/2013-01-21-cold-fusion-2013/ This link does not work for me. Jean-Paul is writing a book about cold fusion. I guess he will present some of material from it. - Jed
Re: [Vo]:colloquium cold fusion 2013 in eindhoven
Jean-Paul Biberian will give an update overview of coldfusion research. - Original Message - From: "P.J van Noorden" To: Sent: Tuesday, January 15, 2013 5:19 PM Subject: [Vo]:colloquium cold fusion 2013 in eindhoven Hello Vortexians, I just stumbled on this: http://www.tue.nl/universiteit/nieuws-en-pers/agenda/2013-01-21-cold-fusion-2013/ Coincidentally I am in Eindhoven on that day, so I will attend the colloquium. Peter v Noorden
[Vo]:colloquium cold fusion 2013 in eindhoven
Hello Vortexians, I just stumbled on this: http://www.tue.nl/universiteit/nieuws-en-pers/agenda/2013-01-21-cold-fusion-2013/ Coincidentally I am in Eindhoven on that day, so I will attend the colloquium. Peter v Noorden
Re: [Vo]:OT: The Truth about islam and little girls.
Between protestants and catholics in the Netherlands. It looks a bit as the war between Sunnis and Shiites, but then 350 y earllier. Were I live villages were terrorised and people were beheaded. Peter - Original Message - From: Jojo Jaro To: vortex-l@eskimo.com Sent: Wednesday, January 02, 2013 7:53 PM Subject: Re: [Vo]:OT: The Truth about islam and little girls. Can you elaborate? Which war is this? Which Christian denominations or groups? Jojo - Original Message - From: P.J van Noorden To: vortex-l@eskimo.com Sent: Thursday, January 03, 2013 2:31 AM Subject: Re: [Vo]:OT: The Truth about islam and little girls. Jojo, Here in the Netherlands they(christians) were cutting throats in the 80 y war between 1568 and 1648. Peter - Original Message - From: Jojo Jaro To: vortex-l@eskimo.com Sent: Wednesday, January 02, 2013 2:06 PM Subject: Re: [Vo]:OT: The Truth about islam and little girls. My friend, show me where Christians go after each other's throats like the Sunnis and the Shiites. Last time I checked, Methodists were not cutting off Baptist's heads. Jojo - Original Message - From: Daniel Rocha To: John Milstone Sent: Wednesday, January 02, 2013 7:22 PM Subject: Re: [Vo]:OT: The Truth about islam and little girls. Any heterogeneous group of people fall under this category. And the larger the group is, the better is the example, since it implies a larger variety of people. So, what you are saying about Islam applies much better to Christianity, since it is a bigger group. 2013/1/2 Jojo Jaro I started out thinking that islam is a more or less unified violent religion; now, I know that I was wrong. It is a non-unified violent religion. A rabid mad dog with one head is dangerous, but a rabid mad dog with multiple heads is even more dangerous. -- Daniel Rocha - RJ danieldi...@gmail.com
Re: [Vo]:OT: The Truth about islam and little girls.
Jojo, Here in the Netherlands they(christians) were cutting throats in the 80 y war between 1568 and 1648. Peter - Original Message - From: Jojo Jaro To: vortex-l@eskimo.com Sent: Wednesday, January 02, 2013 2:06 PM Subject: Re: [Vo]:OT: The Truth about islam and little girls. My friend, show me where Christians go after each other's throats like the Sunnis and the Shiites. Last time I checked, Methodists were not cutting off Baptist's heads. Jojo - Original Message - From: Daniel Rocha To: John Milstone Sent: Wednesday, January 02, 2013 7:22 PM Subject: Re: [Vo]:OT: The Truth about islam and little girls. Any heterogeneous group of people fall under this category. And the larger the group is, the better is the example, since it implies a larger variety of people. So, what you are saying about Islam applies much better to Christianity, since it is a bigger group. 2013/1/2 Jojo Jaro I started out thinking that islam is a more or less unified violent religion; now, I know that I was wrong. It is a non-unified violent religion. A rabid mad dog with one head is dangerous, but a rabid mad dog with multiple heads is even more dangerous. -- Daniel Rocha - RJ danieldi...@gmail.com
[Vo]:Moon God, Dozens of wives, and marriageable age
Jojo Jaro, After a year I returned to the vortex list and was unpleasantly surprised to read about the discussion about religion. It would be better to engage in a discussion how the brain generates a believe in god, then to discuss which believe is "real" : Islam or Christianity. The only conclusion one can make from studying the effect of electric stimulation of the parietal lobe of the brain during open brain surgery (to pinpoint the epileptic focus by patients) is that a religeous believe is caused by electrochemical excitation of certain parts of the brain. I have come to the compelling conclusion that there is no reason to think that there is a god, because all thoughts about religion are "brain made". So if there were no human beings on earth there would be no god. Think about that before engaging in an fruitless discussion like we have seen in the past weeks on vortex-l about Moon God etc. So also the God of Jojo is a concoction. For now I wish everybody (religious or not) a happy Christmas. Peter v Noorden - Original Message - From: "Jojo Jaro" To: Sent: Monday, December 24, 2012 11:41 AM Subject: Re: [Vo]:[OT] Moon God, Dozens of wives, and marriageable age Funny thing happened on the way to truth... Lomax first accuses me of lying because I refuse to cite sources. He said I did not have sources. At that time, I wanted people to research this on their own so that they would find their conclusions conclusive. I expressed a concern that whatever source I cite would be considered biased. My friends, that is exactly what happened. Even when I posted a muslim source, the 2 most venerated and respected muslim works recognized by muslims everywhere, Lomax still finds it justified to question it, deny it and spin it away. That is exactly why I wanted people to research this on their own. There seems to be a question as to what constitutes child molestation. Lomax seems to think that if the little girl is sexually mature - this in his definition means a menstrual cycle; a sexual act by a 50 year old with this little girl is not child molestation. My friends, if you are swallowing this argument, you deserve to be deceived by Lomax. For goodness sake, Everyone knows a 9 year old is still immature in every way. Yet Lomax thinks that since she had parental permission, that she enjoyed it, that she had a menstrual cycle, that this does not constitute an abhorrent act. My friends, this is the corruption of islam for all to see. Lomax argues that there is dispute as to her actual age. My friends, even if A'isha was 5 years older at 14, she would still be too young. Only islam thinks this is acceptable. This my friends is the corruption of islam. Lomax argues that since A'isha has had a menstrual cycle, that she is physically big enough for a 50 year old man. What a load of bullcrap. My friends, find me a 9 year old that's physically big enough. Goodness, 9 year olds are tiny, especially middle eastern girls. This my friends is the corruption of islam. Lomax argues that since he found a 5 year old that bore a child, that all little girls would be like that. This of course is argument by generalization. I say a 9 year old is too small. Lomax says its ok. This my friends is the corruption of islam. My friends, this is how Child Molestation is defined in our legal system: "Child molestation is a crime involving a range of indecent or sexual activities between an adult and a child, usually under the age of 14. In psychiatric terms, these acts are sometimes known as pedophilia. It is important, however, to keep in mind that child molestation and child Sexual Abuse refer to specific, legally defined actions. They do not necessarily imply that the perpetrator bears a particular psychological makeup or motive. For example, not all incidents of child molestation are perpetrated by pedophiles; sometimes the perpetrator has other motives for his or her actions and does not manifest an ongoing pattern of sexual attraction to children. Thus, not all child molestation is perpetrated by pedophiles, and not all pedophiles actually commit child molestation. Regardless of the terminology, it is illegal for an adult to touch any portion of a child's body with a "lewd and lascivious" intent. Usually, consent is not a matter of consideration, and is not available as a defense to a charge of child molestation. Even in cases where it can be proven that the minor victim was a willing participant, a sex act or improper touching is still a crime because children cannot legally consent to anything. Criminal penalties are severe for those convicted of child molestation." My friends, in fact, muhammed is guilty of Child Molestation. Sexual maturity, menstrual cycle, parental permission, self permission, size, etc do not negate the fact that an adult performing indecent sexual activities with a child less than 14 is considered
Re: [Vo]:Inspiration
Hello, From Randell Mills I understood that only H can be a catalyst because the atom has to be neutral. He+ is not neutral, so it is difficult / impossible to collaps. Peter van Noorden - Original Message - From: To: Sent: Wednesday, August 15, 2012 7:18 AM Subject: Re: [Vo]:Inspiration In reply to David Roberson's message of Wed, 15 Aug 2012 00:31:31 -0400 (EDT): Hi, [snip] I was wondering about collapsed helium. Does the Mills theory support the concept? My memory on Mills' position is vague. However I fail to see that there is any great functional difference between H and He+ (apart from the double charge on the nucleus), so I think it should be possible. [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
[Vo]:Press release Blacklightpower
See press release and validation reports from Blacklightpower: http://dev.blacklightpower.com/press/052212-2/ http://www.blacklightpower.com/technology/validation-reports/ Peter
Re: EXTERNAL: Re: [Vo]:Mill's and Lu paper define hydrino as fractional Rydberg
Hello Fran, I don`t understand your statement: Now that Mills admits the “hydrino” is actually fractiona Rydberg hydrogen the term hydrino not only becomes redundant but also carries all the baggage of his previously wrong definition that caused so much controversy. I thought Mills has always said that the hydrino = hydrogen in a fractional quantum state. BTW the continuum spectrum in discharges of H2 gas is 100% reproducible and has no known explanation. Peter van Noorden - Original Message - From: Roarty, Francis X To: vortex-l@eskimo.com Sent: Tuesday, November 01, 2011 6:40 PM Subject: RE: EXTERNAL: Re: [Vo]:Mill's and Lu paper define hydrino as fractional Rydberg That is exactly what I was saying… Now that Mills admits the “hydrino” is actually fractiona Rydberg hydrogen the term hydrino not only becomes redundant but also carries all the baggage of his previously wrong definition that caused so much controversy. The term should be eradicated with extreme predjudice. From: Danny Ross Lunsford [mailto:antimatte...@yahoo.com] Sent: Tuesday, November 01, 2011 1:28 PM To: vortex-l@eskimo.com Subject: EXTERNAL: Re: [Vo]:Mill's and Lu paper define hydrino as fractional Rydberg You can forget the hydrino. It does no good to adhere to bad ideas. Angular momentum conservation prevents it. We need to use good physics to get to the bottom of this phenomenon, and ruthlessly eliminate the bad ideas. -- "I write a little. I erase a lot." - Chopin --- On Tue, 11/1/11, Roarty, Francis X wrote: A recent paper “Time-resolved hydrino continuum transitions with cutoffs at 22.8 nm and 10.1 nm” http://www.springerlink.com/content/q8005267210x3568/fulltext.pdf...
Re: [Vo]:European Patent Office observer criticizes Rossi's E-Cat
Hello Jed, If water boils at 99.6 degr C the airpressure must have been 1000 mbar, but according to this site: http://www.worldweatheronline.com/sports/football/weather/Italy/1175869/Bologna-Fc-1909/2638541/info.aspx?day=21 the airpressure on April 28th 2011 was 1011 mbar, so the boilingpoint must have been 99.9 degC. The difference in boilingtemperature can be explained by the accuracy of the thermometer (+/- 0.4 degrC). To conventionally explain the boilingpoint of 100.5 degrC the backpressure in the Ecat must have been 30mbar (for a boilingpoint of 99.6degC) and 20mbar for a boilingpoint of 99.9degC. This compares to resp 30.6 cm and 20.4cm water and this is about the hight of the chimney. The difference in temperature of the steam can ofcourse only be explained if the chimney of the ecat is almost completely filled with water. This is ofcourse the big question. Peter - Original Message - From: "Jed Rothwell" To: Sent: Monday, July 18, 2011 4:08 PM Subject: Re: [Vo]:European Patent Office observer criticizes Rossi's E-Cat P.J van Noorden wrote: It is very important to notice that water boils at 100.5 C when the outside air pressure is 1030 mBar, which can be the case when a high pressure system is covering Italy . . . In the April 28 tests, Lewan reported: "we calibrated the probe by immersing it in a pot with boiling water, and the measured value was then 99.6 degrees centigrade." Later during the test they measured vapor at "about 100.5 degrees centigrade." There is no doubt that was vapor, since it is substantially hotter than the boiling water, plus you can see steam coming out of the pipe. I expect that backpressure is minimal with this system. - Jed
Re: [Vo]:European Patent Office observer criticizes Rossi's E-Cat
Hello Jouni It is very important to notice that water boils at 100.5 C when the outside air pressure is 1030 mBar, which can be the case when a high pressure system is covering Italy ( a normal situation during spring and summer). Look at to calculate the pressure corrected boilingpoint : http://www.csgnetwork.com/prescorh2oboilcalc.html Peter v Noorden - Original Message - From: "Jouni Valkonen" To: Sent: Monday, July 18, 2011 4:12 AM Subject: Re: [Vo]:European Patent Office observer criticizes Rossi's E-Cat 2011/7/18 Abd ul-Rahman Lomax : teapots don't have a fixed water flow input. Rather, water is added when the level declines. This is irrelevant difference. Water flow is there only to ensure that water level does not drop below reactor core, so that core does not expose to air. Water is not there to demonstrate how much E-Cat produces energy, but it's main function is to control reactor temperature and prevent reactor meltdown. This is the very essence of boiling water reactors. See, the purpose water for measurements is irrelevant component, but it is used, because water is very convenient substance as boiling water reactor coolant. That is because the enthalpy of water phase change is so high. This enables to divert much of the heat energy away from reactor core while the temperature of coolant remains constant. This is very crusial, because according to sig. Rossi, his E-Cat is very sensitive for internal temperature of reactor. The problem with this is that dry steam above boiling would require a chamber hotter than boiling, this can't happen unless the chamber substantially empties. It is perfectly possible that pressure rises inside E-Cat so that boiling point is at 100.5°C or 0.8°C higher. But what is impossible without very special setup, is that reactor produces wet steam and such a high pressure simultaneously, that it could cause boiling point to rise. If steam is very wet, then the energy output of the reactor is very low. And it cannot heat up reactor that much that it will cause significant pressure build up. Pressure build up depends on that there is significant amount of dry steam present! But, it is possible that if heating element is very hot, steam temperature can rise somewhat over boiling, because surface tension of water enable the bubble formation. And this gives some time for heating element to heat steam directly in gaseous phase although heating element is under water. Therefore it should not be impossible, that steam temperature finds its equilibrium that is 0.1-2.0 °C higher than actual boiling point. This depends on what is the temperature difference between heating element and boiling water. A teapot with a fixed flow input could overflow, indeed, if that's the only way that water is added, we can predict that, unless there is some complex feedback mechanism either on flow or on heat vs water level, the water will either boil away and the chimney temperature will increase, or water will start to overflow, some portion of the water will flow out. You do not have any evidence for that E-Cat can overflow, therefore this is just empty speculation. If you could even speculate with this possibility seriously, you should know what is the inner volume of the E-Cat. But you do not know even such a rudimentary detail about the E-Cat. This kind of speculation is useless and nonproductive, because first of all, temperature reading would be below 99.7°C, because there cannot be pressure build up without intensive production of dry steam. But as I stated this problem is easy to "fix", that you just introduce a secret heating element near thermometer that feeds false temperature readings. On the other hand, if E-Cat is a hoax, it far more easy to construct such a way that Rossi just hides a internal hydrogen tank. E.g. I have suggested that the stand where E-Cats are mounted could be hollow hand that would be easy way to hide a hydrogen bottle. Therefore Kullander's and Essén's observations about the E-Cat has exactly zero scientific value, because they cannot tell a part, whether they witnessed a catalyzed hydrogen burning or catalyzed cold fusion. –Jouni
Re: [Vo]:Analysis of e-Cat test by E. Storms
Hello, To get an indication if all the water running through the ecat has been fully evaporated it would only be necessary to add a dye (e.g red colour) to the cold water in the tank. If the water in the black hose is completely clear it would be prove that all the water has been evaporated. The water in the black hose would then be distillated water and not overflow. Peter van Noorden - Original Message - From: "OrionWorks - Steven V Johnson" To: Sent: Tuesday, July 05, 2011 8:51 PM Subject: Re: [Vo]:Analysis of e-Cat test by E. Storms I was always taught that, technically speaking, "steam" is an invisible gas. However, most of us quite naturally tend to only notice the clouds of water vapor condensing out from the invisible "steam" as it cools. We tend to incorrectly associate, in the visual sense, those tiny suspended condensed droplets of water as "steam." I continue to make this visualization mistake all the time even today, as do most of us, simply because it's convenient to do so, even though technically speaking I know it's inaccurate. To be honest I think the latest semantics "battle" over the definition of what "steam" really is, is now in danger of turning into silly pointless argument - is the "steam" wet or is it dry. Josh, Correct me if I'm wrong but I gather you believe (or are convinced of the fact) that the videos you viewed proved that tiny suspended condensed water droplets (mist) was observed being expelled directly FROM WITHIN the end of black hose from Rossi's e-cat test. In other words I gather you are arguing from the premise that the "steam" already contained suspended droplets of condense water within the black hose, and through guilt by association, there must have also been condensed water vapor within the chimney of the e-cat prior to the water-gas mixture exiting into the black hose. Is this an accurate assumption on my part? As for me, I was under the impression (an impression that admittedly could be wrong) that those who looked closely at the end of the black hose noticed that the first signs of condensation of tiny suspended water droplets were observed to have formed OUTSIDE of the end of hose... let's say, maybe, about quarter of an inch or so from the tip. Can someone tell me if this is this an accurate assumption on my part or not? Regards, Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: Re: [Vo]:Something more on the steam
Hello, Perhaps this link provides some usefull information about relative humidity RH and temperature above 100 deg C. http://www.macinstruments.com/relatv_c.htm Peter v Noorden - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Tuesday, June 21, 2011 5:24 AM Subject: Re: Re: [Vo]:Something more on the steam Abd ul-Rahman Lomax wrote: The Testo 650 is used for measuring *humidity*, Jed, for, like, food manufacturing and storage, etc. Read that HP literature. The device measures up to 100% humidity, it claims. Wet steam is above 100% humidity. The literature claims that the device measures: CO2, CO, temperature, and relative humidity. Other parameters are calculated from these measurements. http://www.testo.com/online/embedded/Sites/INT/SharedDocuments/ProductBrochures/0563_6501_en_01.pdf It isn't HP; it is testo. The meter also measures "absolute humidity g/m^3" (mass) and enthalpy (kcal/kg), which is what we want to measure. I guess enthalpy is derived from absolute humidity and temperature. Elsewhere you wrote: You misunderstood that, I believe. Look at what the thing actually measures, and look at the humidity measurement operating range. "85% (max), no condensation." This thing doesn't work in the presence of liquid water, as I read it. There would be no point to making a meter like this if it did not work in the presence of liquid water, because there is almost always some liquid water in process steam. It is never purely dry. I think people here should concede that Galantini is expert enough to select the right kind of meter after all, and it is likely he also knows how to use the meter to measure by mass instead of volume. - Jed
[Vo]:How Joshua Cude misrepresents arguments
Dear Joshua, Critics of the so called "cold fusion" field, like you, always ask why there is so less progress in this field in the last 22 years, This interesting video from 2006http://www.youtube.com/watch?v=89dbXdz--eI about the research of cold fusion by Italian researchers (de Ninno e.a) ( article: http://www.lenr-canr.org/acrobat/DeNinnoAexperiment.pdf ) gives an answer to this question. Peter - Original Message - From: Joshua Cude To: vortex-l@eskimo.com Sent: Sunday, June 05, 2011 4:22 PM Subject: Re: [Vo]:How Joshua Cude misrepresents arguments On Fri, Jun 3, 2011 at 4:26 PM, Abd ul-Rahman Lomax wrote: >> "One problem I have with those results. When the current shuts off, the heat dies immediately. It seems implausible that the deuterium would diffuse out of the Pd that quickly. I would expect a more gradual decline. Especially with all the reports of heat after death. That points to artifact to me." > And here Joshua let his assumptions of error lead him into a blatant error, confidently asserted. It turns out that "immediately" is, from the graph, about a hour. You can see the decline, it's not "immediate." And the scale on this chart is one day per division, 24 hours! Right. I guessed wrong. The graph you linked to wasn't labelled. You have to go back to the original to get the scale; I thought the axis was labelled in minutes, and it's actually hours. That weakens the objection, but it doesn't remove it. The complete drop takes about an hour, but it's very steep in the middle, dropping by half the amount in about 12 minutes. That still seems like an unreasonable rate for diffusion, when you consider that a tiny foil in Dardik's experiment maintains its output heat for 4 days. We're told that a very special condition is required in Pd for CF, but now it turns out there are 2 very different special conditions required, one in which the deuterium doesn't diffuse below a critical level in 4 days, and the other in which it diffuses below that level in less than an hour. Fishy! > Joshua just continues to dismiss all this with a wave of the hand. "I'm not convinced." Because if it were true, an obvious demonstration would be easy to design, but in 22 years, there has been no progress. That's why you are trying to convince me with a 16 year-old graph, instead of directing me to demonstrations of isolated devices that are warmer than their surroundings.
Re: [Vo]:RE: Supersizing the BJT
Hello Jed Do you know if there is a independent report about the flow test in February in which the water was not heated to the boilingpoint? Peter - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Sunday, May 08, 2011 4:03 PM Subject: Re: [Vo]:RE: Supersizing the BJT Jones Beene wrote: Everything seen so far – when exposed looks nearly identical, and the only reason to suggest that there ever was a larger model is for Rossi to hide the fact that the results of the January test (Focardi tribute) were wet steam and three times more than actual. Everything, that is, except: 1. Levi et al. looked inside the reactor and saw a 1-liter cell. That is to say, everything seen by looking so far proves it is 1 liter. It is not clear what you mean by "seen." 2. The flow test in February which produced the same results. It is not clear to me what else "everything" consists of. As far as know your imagination is the only source of this information. - Jed
Re: [Vo]:Beene says the error is "at least 1000:1"
Jed, Yes, you are right. The waterflow was higher. During the test on january 14th one can hear a pulsating sound. I suspect that this is caused by a pump which infuses a certain amount of water into the Rossi cell. Do you know if the waterflow is continously or pulsating? Peter - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Tuesday, April 19, 2011 4:49 PM Subject: Re: [Vo]:Beene says the error is "at least 1000:1" P.J van Noorden wrote: I find it difficult to understand that in the Rossi reactor with a flow of 6l/h = about 2 ml/sec and a proposed excess energy of 12 kW no vapour "explosions" are heard. Which experiment was that? You may be confused. On Jan. 14 the flow rate was 18 L/h, power ~12 kW. On March 29, the flow rate was 6.7 L/h but the power was 4.7 kW, not 12 kW. - Jed
Re: [Vo]:Beene says the error is "at least 1000:1"
Hello, I have been looking at the calculations from Jed Rothwell, Robin van Spaandonk and Jones Been. I think the proposed heat transfer of 130 kW would be possible, but only with a pressurised system (as in a nuclear reactor) to prevent steam production. I find it difficult to understand that in the Rossi reactor with a flow of 6l/h = about 2 ml/sec and a proposed excess energy of 12 kW no vapour "explosions" are heard. I.m.o with such a low flow rate the water will reach the boilingpoint before it reaches the end of the nickel catalyst. If you have a hotplate and you put a relatively small amount of water on it, according to me the water will evaporate instantly with a lot of noise. On the video http://www.youtube.com/watch?v=u-Ru1eAymvE the inlet tube is periodically moving and a clicking sound is being heard, but that is probably due to the pump which introduces the water, but I am not sure. In the past I already expressed my surprise that the black tube coming from the Rossi reactor is not moving during the test. Would a steamflow with an energycontent of 12 kW not cause a reactionforce on the hole system? Peter van Noorden - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Tuesday, April 19, 2011 3:28 PM Subject: Re: [Vo]:Beene says the error is "at least 1000:1" I quoted: "Thus there is at least a 1000:1 error in that anecdotal appraisal, which is not a surprise . . ." Furthermore, that appraisal was done by the authors themselves, in NyTeknik. It isn't anecdotal -- it is what they measured and reported. - Jed
[Vo]:Problems with energy calculation Rossi device
Hello Mattia, Ofcourse, you are right. I made a mistake. Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Saturday, April 09, 2011 6:05 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device Absolutely not. It's a flow. Water get in and get out. It's not a closed container of water. With 1 watt you can rise, when dynamic transients go away, an half degree! 1 watt = 1 J/s 1 J/s = 1.8 g/s * 4.2 J/(g*K) * deltaT deltaT = 0.1 kelvin (= 0.1 celsius) If you don't trust me, read the report. It's a simple calculation! From: P.J van Noorden Sent: Saturday, April 09, 2011 5:41 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Problems with energy calculation Rossi device Ofcourse. degr/sec is a typo. But do you agree that if the temperature rise of a waterflow of 6.47 W/h is 30 degr/180sec( 0.15 degr/sec) the thermal power needed is only 1.13 W? Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Saturday, April 09, 2011 5:05 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device >I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. It's not 40 degr/sec. It's 40 degress. 1.8cc = 1.8 g. Flow is 1.8 g/s 1.8 g/s * 4.2 J/(g*K) * 40 K = 300 J/s = 300 watt From: P.J van Noorden Sent: Saturday, April 09, 2011 4:36 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Problems with energy calculation Rossi device Hello Mattia, When using a flow of 6.47 l/hour = 1.8 cc/ sec, and a temperature rise of 30degrC in 180 seconds (from the curve in the article = is 0.15 degr C/sec) the thermal power would be only: 1.8cc x 4.2 J/gr degC x 0.15 = 1.13 W. I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. To calculate the correct temperature increase one must ofcourse take into account the mass of the complete Rossi device and the water inside the rest of the machine. The claim that the Rossi device would produce 4000W would in my opinion lead to a continous evaporation of the incoming water, which is BTW also the claim of Rossi. This is the reason that I stated that in that case the curve would rise very steeply. Because I did a lot of experiments in the past in which I tried to measure the heat release of systems at the bolinigpoint of water, I know that when the system is boiling you must use a very good separation between the boiling vessel and the condensor, otherwise water will flow from one system into the other and the calculation of the excess heat will be wrong. I used in the past colour dyes ( and also radioactive material) to be sure if all water was evaporated. In the Rossi device the direct coupling of the "chimney" with the black tube is not a good way to be sure that no liquid water will flow into the black tube. To remove all doubts Rossi could easily demonstrate the exact amount of excess heat by increasing the waterflow. A flow of 100ml/sec would lead to a continous temperature increase of the water by about 10 degr C if the power is 4 kW. I heard that he did that but I haven`t seen the report. Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Thursday, April 07, 2011 6:11 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device >When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. You are confusing between static and dynamic condition. It's physically impossible to have a instantaneously rise. You are missing intertial thermal mass and dynamic conditions. >The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Check your math. With 300 thermal watt you can rise around 40 degrees celsius with 6.47l/hour. But you will see this 40 degree rise only when the system is stationary, when dynamic is over. From: P.J van Noorden Sent: Thursday, April 07, 2011 11:19 AM To: vortex-l@eskimo.com Subject: [Vo]:Problems with energy calculation Rossi device Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. Th
Re: [Vo]:Problems with energy calculation Rossi device
Ofcourse. degr/sec is a typo. But do you agree that if the temperature rise of a waterflow of 6.47 W/h is 30 degr/180sec( 0.15 degr/sec) the thermal power needed is only 1.13 W? Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Saturday, April 09, 2011 5:05 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device >I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. It's not 40 degr/sec. It's 40 degress. 1.8cc = 1.8 g. Flow is 1.8 g/s 1.8 g/s * 4.2 J/(g*K) * 40 K = 300 J/s = 300 watt From: P.J van Noorden Sent: Saturday, April 09, 2011 4:36 PM To: vortex-l@eskimo.com Subject: Re: [Vo]:Problems with energy calculation Rossi device Hello Mattia, When using a flow of 6.47 l/hour = 1.8 cc/ sec, and a temperature rise of 30degrC in 180 seconds (from the curve in the article = is 0.15 degr C/sec) the thermal power would be only: 1.8cc x 4.2 J/gr degC x 0.15 = 1.13 W. I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. To calculate the correct temperature increase one must ofcourse take into account the mass of the complete Rossi device and the water inside the rest of the machine. The claim that the Rossi device would produce 4000W would in my opinion lead to a continous evaporation of the incoming water, which is BTW also the claim of Rossi. This is the reason that I stated that in that case the curve would rise very steeply. Because I did a lot of experiments in the past in which I tried to measure the heat release of systems at the bolinigpoint of water, I know that when the system is boiling you must use a very good separation between the boiling vessel and the condensor, otherwise water will flow from one system into the other and the calculation of the excess heat will be wrong. I used in the past colour dyes ( and also radioactive material) to be sure if all water was evaporated. In the Rossi device the direct coupling of the "chimney" with the black tube is not a good way to be sure that no liquid water will flow into the black tube. To remove all doubts Rossi could easily demonstrate the exact amount of excess heat by increasing the waterflow. A flow of 100ml/sec would lead to a continous temperature increase of the water by about 10 degr C if the power is 4 kW. I heard that he did that but I haven`t seen the report. Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Thursday, April 07, 2011 6:11 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device >When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. You are confusing between static and dynamic condition. It's physically impossible to have a instantaneously rise. You are missing intertial thermal mass and dynamic conditions. >The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Check your math. With 300 thermal watt you can rise around 40 degrees celsius with 6.47l/hour. But you will see this 40 degree rise only when the system is stationary, when dynamic is over. From: P.J van Noorden Sent: Thursday, April 07, 2011 11:19 AM To: vortex-l@eskimo.com Subject: [Vo]:Problems with energy calculation Rossi device Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Can anybody look at these calculations and figure out what is wrong? Further it would be interesting to know if water can flow through the "chimney" of the reactor directly into the black tube. To figure out what is going on one have to add a substance (dye) to the water and see if the dye can be seen in the " condensed" water. If non vaporised water is carried to the end of the black tube this will have consequences for the calculation of excess heat. Peter van Noorden
Re: [Vo]:Problems with energy calculation Rossi device
Hello Mattia, When using a flow of 6.47 l/hour = 1.8 cc/ sec, and a temperature rise of 30degrC in 180 seconds (from the curve in the article = is 0.15 degr C/sec) the thermal power would be only: 1.8cc x 4.2 J/gr degC x 0.15 = 1.13 W. I agree with you that when 300W is used the temperature rise will be: 300W / 1.8 x 4.2 J/gr degrC = 40 degr/ sec. To calculate the correct temperature increase one must ofcourse take into account the mass of the complete Rossi device and the water inside the rest of the machine. The claim that the Rossi device would produce 4000W would in my opinion lead to a continous evaporation of the incoming water, which is BTW also the claim of Rossi. This is the reason that I stated that in that case the curve would rise very steeply. Because I did a lot of experiments in the past in which I tried to measure the heat release of systems at the bolinigpoint of water, I know that when the system is boiling you must use a very good separation between the boiling vessel and the condensor, otherwise water will flow from one system into the other and the calculation of the excess heat will be wrong. I used in the past colour dyes ( and also radioactive material) to be sure if all water was evaporated. In the Rossi device the direct coupling of the "chimney" with the black tube is not a good way to be sure that no liquid water will flow into the black tube. To remove all doubts Rossi could easily demonstrate the exact amount of excess heat by increasing the waterflow. A flow of 100ml/sec would lead to a continous temperature increase of the water by about 10 degr C if the power is 4 kW. I heard that he did that but I haven`t seen the report. Peter - Original Message - From: Mattia Rizzi To: vortex-l@eskimo.com Sent: Thursday, April 07, 2011 6:11 PM Subject: Re: [Vo]:Problems with energy calculation Rossi device >When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. You are confusing between static and dynamic condition. It's physically impossible to have a instantaneously rise. You are missing intertial thermal mass and dynamic conditions. >The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Check your math. With 300 thermal watt you can rise around 40 degrees celsius with 6.47l/hour. But you will see this 40 degree rise only when the system is stationary, when dynamic is over. From: P.J van Noorden Sent: Thursday, April 07, 2011 11:19 AM To: vortex-l@eskimo.com Subject: [Vo]:Problems with energy calculation Rossi device Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Can anybody look at these calculations and figure out what is wrong? Further it would be interesting to know if water can flow through the "chimney" of the reactor directly into the black tube. To figure out what is going on one have to add a substance (dye) to the water and see if the dye can be seen in the " condensed" water. If non vaporised water is carried to the end of the black tube this will have consequences for the calculation of excess heat. Peter van Noorden
[Vo]:Problems with energy calculation Rossi device
Hello In figure 6 of the article http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 it is stated that the waterflow is 6.47 l/ hour= about 2 ml/sec. This is about the waterflow of a espressomachine. When 4kW is added to such a waterflow the temperature would rise instantaneously to 100 degr C. The curve in figure 6 would have to rise vertically. If only 400 Watts is added the waterflow the temp would rise 50 degr C in temperature continously. The curve in figure 6 rises in about 3 minutes 30 degr C ( steepest part). This would compare to a power of only a few watts. Can anybody look at these calculations and figure out what is wrong? Further it would be interesting to know if water can flow through the "chimney" of the reactor directly into the black tube. To figure out what is going on one have to add a substance (dye) to the water and see if the dye can be seen in the " condensed" water. If non vaporised water is carried to the end of the black tube this will have consequences for the calculation of excess heat. Peter van Noorden
Re: [Vo]:RE: [Vo]:Swedish physicists on the E-cat: "It's a nuclear reaction" / The used powder contains ten percent copper
The energy release of the hydrino producing reaction is 50 MJ/mol hydrogen gas. The prefered reactionproduct seems to be H1/4. See http://www.blacklightpower.com/papers/Eng%20Power050410S.pdf So if 25 kWh is produced (90 MJ) this should correspond to 1.8 moles of H2 gas = 3.6 grams. Peter van Noorden - Original Message - From: "OrionWorks - Steven V Johnson" To: Sent: Wednesday, April 06, 2011 4:30 PM Subject: Re: [Vo]:RE: [Vo]:Swedish physicists on the E-cat: "It's a nuclear reaction" / The used powder contains ten percent copper From Jones: The mundane reason for the appearance of iron an[d] copper is electromigration. Seems like a reasonable conclusion to draw. I must apologize for not being sufficiently clear as to what I was really questioning: What is generating the massive amount of heat? I gather the responsible party still remains an unknown quality - especially considering your concluding remark: Even hydrinos would result in an isotopic imbalance. ...which also seems like a reasonable conclusion to draw. Regarding the hydrino theory, my first impression would be to conclude (with absolutely no math to back this conclusion up with) that not enough hydrogen was consumed (into hydrinos) that would explain the massive amount of heat recorded. I hope someone can clarify whether my uneducated assumption on this point is valid or not. (I suspect it's incorrect.) Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
Re: [Vo]:RE: [Vo]:Swedish physicists on the E-cat: "It's a nuclear reaction" / The used powder contains ten percent copper
On the pictures in the article: http://www.nyteknik.se/incoming/article3144960.ece/BINARY/Download+the+report+by+Kullander+and+Ess%C3%A9n+%28pdf%29 ) it is seen that the copper tubes are corroded from the outside , probably due to the high temperature of the reaction. As Jones says it is very likely that he copper (and steel) migrate from the outside to the nickelpowder which is inside of the reaction chamber. This would explain why a lot of Copper is found. I once noticed that when I was using a copper vessel which was plated with steel on the outside that the vessel got a copper coulor when it was accidentally heated to 300 C. The heat releasing reaction from the Ecat looks like the effect which is seen during the use of Raney nickel ( or TiC WC and other compounds) in combination with Mills catalysts and hydrogen. No transmutations are seen, only upfield shifted NMR peaks of the hydrogencompounds in the reactionproduct which have a very narrow resonance peak. This could indicate that these species have a low probablity to interact with the environment after being formed, which is not at odds with the hydrino concept. Peter van Noorden - Original Message - From: "Jones Beene" To: Sent: Wednesday, April 06, 2011 2:20 PM Subject: [Vo]:RE: [Vo]:Swedish physicists on the E-cat: "It's a nuclear reaction" / The used powder contains ten percent copper -Original Message- From: SHIRAKAWA Akira Thank you for posting this but for the record, the conclusions of Kullander are wrong. Not just wrong but irresponsible and foolish. First he says: "Analyses of the nickel powder used in Rossi's energy catalyzer show that a large amount of copper is formed." The Facts: There is evidence of the presence of copper but that is all. If it were formed by transmutation some of it should be radioactive. In fact there is a mundane explanation for the presence of copper. Sven Kullander considers this to be evidence of a nuclear reaction" "For copper to be formed out of nickel, the nucleus of nickel has to capture a proton. The fact that this possibly occurs in Rossi's reactor is why the concept of cold fusion has been mentioned - it would consist of fusion between nuclei of nickel and hydrogen." The facts: Yes but if this were the case there would be a wide variation in the balance of isotopes. Element and isotopic analysis showed that the isotopic analysis through ICP-MS doesn't show any deviation from the natural isotopic composition of nickel and copper. The mistakes of Kullander are juvenile and silly. There is a mundane explanation for both copper and iron so why invent a reaction that does not exist? Jones
Re: [Vo]:Re: Levi's interpretation of the two Rossi demos does not hold water, decisive critique by Joshua Cude: Rich Murray 2011.02.08
On January 15th 2011 the airpressure in Bologna about 1025 hPa ( High pressure system) . At that airpressure water boils at 101 deg C. So the higher temperature of the steam can be explained by the higher airpressure and not be "superheated "steam. Peter - Original Message - From: "Terry Blanton" To: Sent: Tuesday, February 08, 2011 9:52 PM Subject: Re: [Vo]:Re: Levi's interpretation of the two Rossi demos does not hold water, decisive critique by Joshua Cude: Rich Murray 2011.02.08 On Tue, Feb 8, 2011 at 3:29 PM, Stephen A. Lawrence wrote: In an open boiler with a submerged heating element, the steam won't rise above 100 C, because its temperature is buffered by the liquid water with which it's in contact. In a pipeline, once the water has boiled away to steam, the steam is no longer in intimate contact with liquid water, but it's still in contact with the heating element, and there's nothing to keep it from getting hotter as it moves along the tube. You are making assumptions about the geometry of the device. How do you know there is no liquid water reservoir? T
Re: [Vo]:Your suspicion is not proof there is a problem
Jed In 1990 I have been working with a self deviced electromagnet which was powered by 25 ( and later 50) batteries of 12 V in parallel with a total current of 1000 A continously to generate a very electric strong magnetic field to detect possible neutron emission from an electrolytic experiment with Pd cathode, Pt anode in LiOD . The hollow coil had to be cooled by water which was flowing through it but the amount of heat produced was so high that later I had to cool the coil also from the outside ( in a waterbath). So initially I tried to cool it only from the inside, but the waterflow was restricted because of the small diameter of the coil ( 6mm outside diameter and 4mm inside) and the length of the coil, so the temperature rose to the boilingpoint. The steam escaped very violently from the exit of the coil. The sound was as if explosions took place. Therefore you can understand that I am somewhat critical about the relative quiet experiment I saw on the internet. Peter - Original Message - From: "Jed Rothwell" To: Sent: Tuesday, January 25, 2011 9:04 PM Subject: Re: [Vo]:Your suspicion is not proof there is a problem Terry Blanton wrote: I don't mean to be too harsh. Jed, don't misunderstand. My comments were directed at the same people you directed comments. I didn't mean you any-who. Dr. Levi has too much to lose to participate in a hoax. He stands to profit nothing from a hoax. I agree. If there is fraud, only Rossi is involved, and I think that is extremely unlikely. Levi is no fool. He would catch a fraud. He says he looked for signs of fraud. Quote: "Prudentially I have lifted the control box in search for any other eventually hidden cable and found none. The weight of the control box was of few Kg." . . . "As a prudential check the reactor was lifted to seek any eventually hidden power cord. None was found." (You can see that in the photo.) - Jed
Re: [Vo]:Nagel: Check List for LENR Validation Experiments
The calculations from Jeff Driscoll about the velocity of the ejected steam are very important to understand what is going on. *How is it possible that the black tube is laying quiet on the ground while the steam has a high velocity of about 300 km/hr. The high velocity of the steam would create a reaction force in the tube which would cause the tube to change position. *Further the escape of the steam from the black tube would create a very hard noise, which would attract the attention from people. *Why has nobody noticed the extreme volicity of the steam when the steam quality was measured with a probe ( deafening hissing sound)? *Has anybody paid attention to the situation in the adjacent room during the demonstration? What happened in that room is equally important for the interpretation of the experiment in comparison to the results witnessed in the room with the Rossi experiment. *Also I would expect that the black tube ( which looks like a waterhose) would soften because of the high energycontent of the steam. All in all it looks as if the energy content of the "watervapour" from the Rossi experiment is much lower to the energy content of steam created by for instance a steam engine heated with 12 kW of power. Peter v Noorden - Original Message - From: Jeff Driscoll To: vortex-l@eskimo.com Sent: Wednesday, January 19, 2011 12:56 AM Subject: Re: [Vo]:Nagel: Check List for LENR Validation Experiments The calculations of the steam velocity below translates to a 188 mph jet of steam coming out of a hose having an area of 1 cm^2 (equates to a 1.13 cm inner diameter hose or .44" inner diamter) Double the area of the hose and the velocity will drop by a factor of 2 to 94 mph. The steam should be transparent for many inches beyond the end of the hose if sprayed into the room - did it? How do people describe the velocity and volume of the steam? On Tue, Jan 18, 2011 at 3:35 PM, Jones Beene wrote: Here are some calculations that imply certain water/humidity effects which should have been observed at the demo. This is from an associate LENR researcher - Jeff Morriss, in response to the other issues on steam/vapor raised by Jeff Driscoll and Peter van Noorden, which so far do not have convincing answers. Nagel states that 150 grams of water are boiled every 30 sec, or 5 cc/sec. Taking the density of steam at 100C as .590 Kg/m**3 and ratio-ing it against the density of liquid water as 1000 kG/m**3 yields a volume increase of 1690. So each 5 cc of water is converted into 8450 cc of steam every second. If we estimate the area of the vent hose at ~1 cm**2, then the steam velocity must be 8450 cm/sec of 84.5 m/sec. This is about 1/4 the speed of sound and should produce quite a jet of steam. Did anyone observe this? Also, the steam would condense and quickly produce a saturated atmosphere and condensation on metal surfaces. Again, did anyone observe this? Here is a second sanity check. The specific heat of dry air at 1 atm is 1.14 kJ/Km**3. If we assume a room volume of 300 m**3 (about the size of an average classroom) then it takes (300 m**3)*(1.14 kJ/Km**3) = 342 kJ to raise the temperature of the room by one degree. The energy required to boil 18 liters of water is 4.7E4 kJ. So if no heat escaped the room and we ignored the additional energy change due to an increase in relative humidity then the ambient temperature should have increased by 4.7e4/342 or about 137 degrees C. Even if the air in the room cycled every 6 minutes (and that would require special ventilation) the ambient would still rise by 13.7C, which would be noticeably hot and muggy. Finally, the 4.7E4 kJ/hour is equivalent to 1.31E4J/sec. As a basis of comparison, it would be equivalent to 240V at 54 Amps, which is the capacity of an electric furnace for a large house. You may want to pass my calculations by someone else for checking, but I believe they are correct. Jeff From: Jeff Driscoll Was the "steam" exiting the Rossi device transparent or was it an opaque white? (right at the top where it transitions from the aluminum foil covered "chimney" to the black hose) …If it is transparent then that would mean it is water vapor - and truly 12 kW of steam… But if it was white then that would indicate condensed tiny liquid droplets (or ultrasonic fogging) and fraudulent scamming. Water vapor is virtually invisible…. On a tea kettle, the steam immediately coming out of the kettle is transparent but roughly 1 or 2 inches away the vapor condenses to tiny droplets which become a white fog. On Tue, Jan 18, P.J van Noorden wrote: I wondered why people had no problems with the 8 liters of watervapour which was released into the room during the Rossi experiment. A simple experiment in which I evaporised 8 liters of water in a room of
Re: [Vo]:Re: CMNS: Loss of heavy water as mist instead of as vapor, on open cell CF experiments.
Boilingpoint depends on air pressure and can be 101 C with an airpressure of 770 mmHg. Peter - Original Message - From: Jed Rothwell To: vortex-l@eskimo.com Sent: Saturday, January 22, 2011 7:51 PM Subject: Re: [Vo]:Re: CMNS: Loss of heavy water as mist instead of as vapor, on open cell CF experiments. P.J van Noorden wrote: I can confirm the problems which occur during plasma electrolysis: The excess heat measured by Rossi would be correct if indeed all the water is converted into steam , but in practice water can be taken away with the watervapour . . . This is ruled out. The temperature was measured at 101 deg C and they used a relative humidity meter to confirm that there was no vapor. (To confirm it was "dry steam"). There was some confusion about the meter. I updated my description at http://lenr-canr.org/News.htm to address this: A Delta Ohm monitor [is used] to measure the relative humidity of the steam. This is to confirm that it is “dry steam”; that is, steam only, with no water droplets. This is a model HD37AB1347 IAQ with a high temperature add-on SICRAM module - Jed
[Vo]:Re: CMNS: Loss of heavy water as mist instead of as vapor, on open cell CF experiments.
I can confirm the problems which occur during plasma electrolysis: The excess heat measured by Rossi would be correct if indeed all the water is converted into steam , but in practice water can be taken away with the watervapour in e.g plasma electrolysis experiments. In the past I did many plasma electrolysis experiments and I witnessed ( when I used a small amount of radioactive 201 Tl to investigate the claims of remediating radioactivity by plasma electrolysis and the production of excess heat) that about 20-40% of the water was not evaporated during this test, but was transported in small droplets (mist) to the condensor. This led to an overestimation of the excess power of the system. So the only way to exactly measure the energy output of the Rossi device is to increase the waterflow and to avoid steam formation. One could also measure the recovered heat from the condensed water which would allow an estimate of the claimed energyproduction. Peter van Noorden the Netherlands - Original Message - From: "Michael McKubre" To: Sent: Saturday, January 22, 2011 7:05 PM Subject: Re: CMNS: Loss of heavy water as mist instead of as vapor, on open cell CF experiments. See below On Jan 22, 2011, at 9:08 AM, Abd ul-Rahman Lomax wrote: The first objection is "misting." Misting refers to the loss of heavy water as liquid, from a CF cell, instead of as vapor. Because the model of CF calorimetry as elaborated by Miles and Fleischmann includes a term for water vapor loss, but no term for liquid water loss, if there is misting and the loss of heavy water from an opoen cell is significant, the correction for water vapor will produce an apparent excess heat. In some experiments this might explain all the excess heat, if the amounts are sufficient. Misting would occur most prominently during conditions of high current, even more if there is boiling, it may be speculated. The correction for the chemical energy of D2 and O2 release in thermodynamically open cells is normally and most conveniently made by integrating the product of the current and a thermoneutral voltage*. Water (light or heavy) lost from the cell as vapor or mist results in an uncounted energy loss, and is therefore endothermic or conservative in calorimetric calculation. In fact a possible problem of open cell calorimetry occurs when the amount of water consumed is less than that expected from electrolysis, not more. This criticism seems to be upside down. * The origin of this term and its derivation can be found in Ed's book. What is known about loss of D2O as water? What measures have been taken to prevent misting? What measures have been taken to detect such loss, as distinct from loss of vapor? Over the years we have run cells both thermodynamically closed and open. Open cells running at low currents (most loading studies) have usually relied on manual addition. In bigger, longer experiments we used liquid metering pumps for which the amount of water added was calculated and delivered based on Faraday's Law (relating charge and moles). This law we observe by both methods to be well obeyed and slightly conservative. At "modest temperatures" (30 - 40°C) we measure about 1% more D2O leaving the cells than nFQ** (this percentage increases with temperature and bubbling). This difference we attribute to the release of molecular D2O as vapor. Possibly fine droplets (mist) contribute some, but the egress from the cells is so tortuous that very little of occurs in practical cells. Two points: 1) This is a small term 2) It is thermodynamically conservative ** n = charge/species, F = Faraday constant, Q = integral charge. In plasma electrolysis experiments, misting is a major factor, and some of the work by Naudin or others seems to neglect misting; Kowalski has definitely considered mist, though. I am not referring to plasma electrolysis, but to normal CF work as performed by Pons and Fleischmann and others, with open cells. The situation of plasma experiments (and the Bologna demonstration) are different from FPE studies where the primary agent of water elimination is electrolysis (which is very easily and accurately quantified). Misting would not affect closed cells, nor would it affect work with open cells operated within a Seebeck enclosure. The critic has a separate attempted explanation for McKubre's closed-cell work, I will ask about that in a separate mail. I can't wait. m -- You received this message because you are subscribed to the Google Groups "CMNS" group. To post to this group, send email to c...@googlegroups.com. To unsubscribe from this group, send email to cmns+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/cmns?hl=en.
Re: [Vo]:Nagel: Check List for LENR Validation Experiments
Hello all,
I just understood that the watervapour of the Rossi experiment was released in
the room, next to the location of the experiment, because otherwise it would
have been a sauna ( according to Rossi).
I could also see it on the video ( 37 th second).
http://pesn.com/2011/01/17/9501746_Focardi-Rossi_10_kW_cold_fusion_prepping_for_market/
What I would like to know if they have ever measured the amount of heat
recovered from the heat echanger. This would confirm/ deconfirm if they have
completely evaporated the water.
Peter
- Original Message -
From: Jeff Driscoll
To: vortex-l@eskimo.com
Sent: Wednesday, January 19, 2011 12:56 AM
Subject: Re: [Vo]:Nagel: Check List for LENR Validation Experiments
The calculations of the steam velocity below translates to a 188 mph jet of
steam coming out of a hose having an area of 1 cm^2 (equates to a 1.13 cm
inner diameter hose or .44" inner diamter)
Double the area of the hose and the velocity will drop by a factor of 2 to 94
mph.
The steam should be transparent for many inches beyond the end of the hose if
sprayed into the room - did it? How do people describe the velocity and
volume of the steam?
On Tue, Jan 18, 2011 at 3:35 PM, Jones Beene wrote:
Here are some calculations that imply certain water/humidity effects which
should have been observed at the demo.
This is from an associate LENR researcher - Jeff Morriss, in response to
the other issues on steam/vapor raised by Jeff Driscoll and Peter van Noorden,
which so far do not have convincing answers.
Nagel states that 150 grams of water are boiled every 30 sec, or 5 cc/sec.
Taking the density of steam at 100C as .590 Kg/m**3 and ratio-ing it against
the density of liquid water as 1000 kG/m**3 yields a volume increase of 1690.
So each 5 cc of water is converted into 8450 cc of steam every second. If we
estimate the area of the vent hose at ~1 cm**2, then the steam velocity must be
8450 cm/sec of 84.5 m/sec. This is about 1/4 the speed of sound and should
produce quite a jet of steam. Did anyone observe this? Also, the steam would
condense and quickly produce a saturated atmosphere and condensation on metal
surfaces. Again, did anyone observe this?
Here is a second sanity check. The specific heat of dry air at 1 atm is
1.14 kJ/Km**3. If we assume a room volume of 300 m**3 (about the size of an
average classroom) then it takes
(300 m**3)*(1.14 kJ/Km**3) = 342 kJ
to raise the temperature of the room by one degree. The energy required to
boil 18 liters of water is 4.7E4 kJ. So if no heat escaped the room and we
ignored the additional energy change due to an increase in relative humidity
then the ambient temperature should have increased by 4.7e4/342 or about 137
degrees C. Even if the air in the room cycled every 6 minutes (and that would
require special ventilation) the ambient would still rise by 13.7C, which would
be noticeably hot and muggy.
Finally, the 4.7E4 kJ/hour is equivalent to 1.31E4J/sec. As a basis of
comparison, it would be equivalent to 240V at 54 Amps, which is the capacity of
an electric furnace for a large house.
You may want to pass my calculations by someone else for checking, but I
believe they are correct.
Jeff
From: Jeff Driscoll
Was the "steam" exiting the Rossi device transparent or was it an opaque
white? (right at the top where it transitions from the aluminum foil covered
"chimney" to the black hose) …If it is transparent then that would mean it is
water vapor - and truly 12 kW of steam… But if it was white then that would
indicate condensed tiny liquid droplets (or ultrasonic fogging) and fraudulent
scamming.
Water vapor is virtually invisible…. On a tea kettle, the steam immediately
coming out of the kettle is transparent but roughly 1 or 2 inches away the
vapor condenses to tiny droplets which become a white fog.
On Tue, Jan 18, P.J van Noorden wrote:
I wondered why people had no problems with the 8 liters of watervapour
which was released into the room during the Rossi experiment. A simple
experiment in which I evaporised 8 liters of water in a room of 100 m3 with a
powersource of 9 kW ( 3 heaters of each 3 kW) did produce a very humid
atmosphere ( approaching RH 90%) and the temperature rose to more then 30 degr.
Why wasn`t this detected during the experiment of Rossi? If the aircon was
powerfull enough one would still notice a turbulence of warm and cold airflow
in the room.
Peter
- Original Message -
From: Jeff Driscoll
To: vortex-l@eskimo.com
Sent: Tuesday, January 18, 2011 4:08 AM
Subject: Re: [Vo]:Nagel: Check List for LENR Validation Experiments
That meter that was listed can measure Relative Humidity but it can not
measure the quality of the steam. As you know, relative humidity just means
how saturated the
Re: [Vo]:Nagel: Check List for LENR Validation Experiments
I wondered why people had no problems with the 8 liters of watervapour which
was released into the room during the Rossi experiment. A simple experiment in
which I evaporised 8 liters of water in a room of 100 m3 with a powersource of
9 kW ( 3 heaters of each 3 kW) did produce a very humid atmosphere (
approaching RH 90%) and the temperature rose to more then 30 degr.
Why wasn`t this detected during the experiment of Rossi? If the aircon was
powerfull enough one would still notice a turbulence of warm and cold airflow
in the room.
Peter
- Original Message -
From: Jeff Driscoll
To: vortex-l@eskimo.com
Sent: Tuesday, January 18, 2011 4:08 AM
Subject: Re: [Vo]:Nagel: Check List for LENR Validation Experiments
That meter that was listed can measure Relative Humidity but it can not
measure the quality of the steam. As you know, relative humidity just means
how saturated the air is for for the given temperature - it says absolutely
nothing about the quality (dryness or "wetness") of the steam.
The quality of the steam (a.k.a. dryness on Vortex) gives you the ratio of
the mass of vapor to the total mass of water (liquid and vapor) in a given
sample.
It takes complicated expensive instruments to measure the quality of steam
(one device is called a "throttling calorimeter"). A common or even expensive
Relative Humidity instrument can not do it.
If Rossi used an ultrasonic fogger in boiling water, he could get micron
sized droplets at 100 C. That's close enough to 101 C with errors due to
calibration. They should insulate the black hose and stick it in a barrel of
water. 12 kW of steam that is fed into 50 gallons of water (or some number of
gallons) will raise the temperature at rate that could be easily measurable.
If it can be done, find out exactly what information rules out "wet" steam.
Here is a photo of an ultrasonic fogger using water to produce what looks
like steam, but is in fact micron sized water droplets:
http://www.buzzle.com/articles/ultrasonic-fogger-how-does-it-work.html
Here is a link to a description of a "throttling calorimeter" which is a
device that measures the quality ("wetness") of steam. Basically the
throttling calorimeter involves letting the pressurized steam expand into a
cavity and measuring the temperature of the resulting gas. It only works with
pressurized steam such as 30 psia steam or higher so that it can expand down to
15 psia or atmospheric pressure.
http://www.plantservices.com/articles/2003/378.html?page=full
On Mon, Jan 17, 2011 at 8:38 PM, Jed Rothwell wrote:
Jeff Driscoll wrote:
How can you use an indoor air quality meter (listed in Jed's email) to
measure the dryness of the steam? (you can't)
Apparently you can. The person who did this is reportedly an expert in
steam. I gather this meter measures RH in steam as well as air.
Can it be faked the following way:
Use an ultrasonic fogger operating at 1.6 MHz to create micron size
droplets. Heat the droplets to 90 C and then send it down the black hose.
The temperature of the steam out the outlet is measured with a
thermocouple. It is 101 deg C. So it is definitely steam, or a mixture of steam
and water. The RH meter ensures that is all dry steam.
- Jed
Re: [Vo]:Brief Description of the Calorimetry in the Rossi Experiment at U. Bologna, January 14, 2011
Hello Jed, How do we know that all the water ( 8.8 l) evaporated? Was the Rossi device weighted before and after the test? The diameter of the device is about 10 cm, so there could still be a few liters inside after the experiment. An easy way to measure the heat of this system more accurately would have been to increase the waterflow to e.g 100 ml /sec ( about 20 times higher as the flow that was used). If 12 kW was produced one would have measured a temperature increase of 30 degrees constantly, with a power input of only 700-800W. This would have been a very practical system because normally with 700-800 W you can not have a shower with hot water. You need about 10 kW. If Rossi had demonstrated that he could heat such an amount of water continously for an hour he could have convinced almost anybody. Why didn`t he do that? Peter - Original Message - From: "Jed Rothwell" To: Cc: Sent: Monday, January 17, 2011 3:20 PM Subject: [Vo]:Brief Description of the Calorimetry in the Rossi Experiment at U. Bologna, January 14, 2011 Brief Description of the Calorimetry in the Rossi Experiment at U. Bologna, January 14, 2011 by Jed Rothwell The experiment has been underway at U. Bologna since mid-December 2010. It has been done several times. Several professors with expertise in related subjects such as calorimetry are involved. LIST OF MAIN EQUIPMENT IN EXPERIMENT A hydrogen tank mounted on a weight scale which is accurate to 0.1 g 10 liter tank reservoir, which is refilled as needed during the run Displacement pump Tube from pump to Rossi device (The Rossi device is known as an "ECat") Outlet tube from the Rossi device, which emits hot water or steam Thermocouples in the reservoir, ambient air and the outlet tube An HD37AB1347 IAQ Monitor (Delta Ohm) to measure the relative humidity of the steam. This is to confirm that it is “dry steam”; that is, steam only, with no water droplets. Alternating-current heater used to bring the Rossi device up the working temperature METHOD The reservoir water temperature is measured at 13°C, ambient air at 23°C. The heater is set to about 1000 W to heat up the Rossi device. Hydrogen is admitted to the Rossi device. The displacement pump is turned on, injecting water into the Rossi device at 292 ml/min. The water comes out as warm water at first, then as a mixture of steam and water, and finally after about 30 minutes, as dry steam. This is confirmed with the relative humidity meter. As the device heats up, heater power is reduced to around 400 W. RESULTS The test run on January 14 lasted for 1 hour. After the first 30 minutes the outlet flow became dry steam. The enthalpy during this last 30 minutes can be computed very simply, based on the heat capacity of water (4.2 kJ/kgK) and heat of vaporization of water (2260 kJ/kg): Mass of water 8.8 kg Temperature change 87°C Energy to bring water to 100°C: 87°C*4.2*8.8 kg = 3,216 kJ Energy to vaporize 10 kg of water: 2260*8.8 = 19,888 kJ Total: 23,107 kJ Duration 30 minutes = 1800 seconds Power 12,837 W, minus auxiliary power ~12 kW There were two potential ways in which input power might have been measured incorrectly: heater power, and the hydrogen, which might have burned if air had been present in the cell. The heater power was measured at 400 W. It could not have been much higher that this, because it is plugged into an ordinary wall outlet. Even if a wall socket could supply 12 kW, the heater electric wire would burn. During the test runs the weight of the hydrogen tank did not measurably decrease, so less than 0.1 g of hydrogen was consumed. 0.1 g of hydrogen is 0.1 mole, which makes 0.05 mole of water. The heat of formation of water is 286 kJ/mole, so if the hydrogen had been burned it would have produced less than 14.3 kJ.
Re: [Vo]:Rossi Responds
Hello Peter, On the photo http://translate.google.com/translate?js=n&prev=_t&hl=en&ie=UTF-8&layout=2&eotf=1&sl=it&tl=en&u=http://22passi.blogspot.com/2011/01/bolognia-14111-cronaca-test-fusione_14.html I see a black flexible pipe, which must be the cold water input. The other transparent pipe is ending in a plastic vessel. Is this heated water removed out of the room through a drainpipe? The somewhat younger Peter This heat was removed by condensing the steam- by the cooling water. Peter the Older On Mon, Jan 17, 2011 at 1:45 PM, P.J van Noorden wrote: Hello, What I don`t understand is that with this system producing 15 kW of power the temperature in the room isn`t higher then 23 degrees Celcius. This amount of power corresponds to a group of 150 people or an intense perpendicular solar flux through a large window of 15 m2. It seems that everybody in the room during the Rossi experiments was feeling very comfortable. Normally when such an amount of heat is dumped into a room the aircon will fail. Peter - Original Message - From: "Terry Blanton" To: Sent: Monday, January 17, 2011 1:50 AM Subject: [Vo]:Rossi Responds Three pages of questions and answers at his weblog: http://www.journal-of-nuclear-physics.com/?p=360&cpage=3#comments including: Daniel G. Zavela January 15th, 2011 at 4:28 AM Greetings from California and congratulations on your successful work! Can you simply state what the Watts IN are versus Watts OUT? Can you turn off the input current? Does the reaction become self-sustaining? Andrea Rossi January 15th, 2011 at 5:05 AM Dear Mr Daniel Zavela: Watts in: 400 wh/h Watts out: 15,000 wh/h Yes, we can turn off the input current, but we prefer to maintain a drive and the reasons are very difficult to explain without violating my confidentiality restraints. The reaction becomes self sustaining. Warm Regards, A.R. COP = 37.5 T - Original Message - From: Peter Gluck To: vortex-l@eskimo.com Sent: Monday, January 17, 2011 12:53 PM Subject: Re: [Vo]:Rossi Responds This heat was removed by condensing the steam- by the cooling water. Peter the Older On Mon, Jan 17, 2011 at 1:45 PM, P.J van Noorden wrote: Hello, What I don`t understand is that with this system producing 15 kW of power the temperature in the room isn`t higher then 23 degrees Celcius. This amount of power corresponds to a group of 150 people or an intense perpendicular solar flux through a large window of 15 m2. It seems that everybody in the room during the Rossi experiments was feeling very comfortable. Normally when such an amount of heat is dumped into a room the aircon will fail. Peter - Original Message - From: "Terry Blanton" To: Sent: Monday, January 17, 2011 1:50 AM Subject: [Vo]:Rossi Responds Three pages of questions and answers at his weblog: http://www.journal-of-nuclear-physics.com/?p=360&cpage=3#comments including: Daniel G. Zavela January 15th, 2011 at 4:28 AM Greetings from California and congratulations on your successful work! Can you simply state what the Watts IN are versus Watts OUT? Can you turn off the input current? Does the reaction become self-sustaining? Andrea Rossi January 15th, 2011 at 5:05 AM Dear Mr Daniel Zavela: Watts in: 400 wh/h Watts out: 15,000 wh/h Yes, we can turn off the input current, but we prefer to maintain a drive and the reasons are very difficult to explain without violating my confidentiality restraints. The reaction becomes self sustaining. Warm Regards, A.R. COP = 37.5 T
Re: [Vo]:Rossi Responds
Hello, What I don`t understand is that with this system producing 15 kW of power the temperature in the room isn`t higher then 23 degrees Celcius. This amount of power corresponds to a group of 150 people or an intense perpendicular solar flux through a large window of 15 m2. It seems that everybody in the room during the Rossi experiments was feeling very comfortable. Normally when such an amount of heat is dumped into a room the aircon will fail. Peter - Original Message - From: "Terry Blanton" To: Sent: Monday, January 17, 2011 1:50 AM Subject: [Vo]:Rossi Responds Three pages of questions and answers at his weblog: http://www.journal-of-nuclear-physics.com/?p=360&cpage=3#comments including: Daniel G. Zavela January 15th, 2011 at 4:28 AM Greetings from California and congratulations on your successful work! Can you simply state what the Watts IN are versus Watts OUT? Can you turn off the input current? Does the reaction become self-sustaining? Andrea Rossi January 15th, 2011 at 5:05 AM Dear Mr Daniel Zavela: Watts in: 400 wh/h Watts out: 15,000 wh/h Yes, we can turn off the input current, but we prefer to maintain a drive and the reasons are very difficult to explain without violating my confidentiality restraints. The reaction becomes self sustaining. Warm Regards, A.R. COP = 37.5 T
Re: [Vo]:real heat wrong theory?
Dear Vorticians, It is claimed that the experiment from Rossi released about 10kW of power. Suppose that the experiment produced this amount of power during about 2 hours, then a total amount of 72 MJ is produced. The amount of water that can be heated from 20 to 100 C is about 200 liters.( 20- buckets of water or a large boiler!). It would be easy to see if this amount of hot water has been produced during the reaction. Did the water flow away out of the room or did it stay there? If the water was kept in the room the amount of heat would have make the room a lot hotter and humid .The room temperature was only 23 C, so I wonder what is going on. Peter van Noorden the Netherlands - Original Message - From: To: Sent: Sunday, January 16, 2011 9:04 AM Subject: Re: [Vo]:real heat wrong theory? In reply to Jones Beene's message of Sat, 15 Jan 2011 18:04:47 -0800: Hi, [snip] Robin, We cannot assume that this is directly comparable to a known hot fusion reaction, assuming it is real. Why should we? There is every reason to suspect that LENR is based on previously unknown pathways. I agree. However I am criticizing their theory, not their experimental findings. I simply pointed out that if the theory they propose were the correct one, then one would expect to detect lots of gammas even outside the shielding. However there is a catch. My calculations were based on beta+ decay (as they suggest), and EC may be so enhanced during Hydrino fusion that it completely swamps beta+ decay (it's usually the other way around). That would essentially eliminate most of the annihilation gammas. This could be a truer picture of what's going on. The fusion energy would be emitted as kinetic energy of electrons (& protons?). About 1% of the electrons would create energetic X-rays, and a small percentage of these would be bremsstrahlung X-rays with a top edge equal to the electron energy (about 3.4 MeV). Even so, only about half of all Cu-59 decays go directly to the ground state. Those remaining still emit gammas of varying energies, and at least some of these ought to be detected. The best way to validate the claim is to test a sample of spent fuel for copper isotope ratio. We can probably expect the heavier 65Cu to be completely absent. That would constitute almost indisputable proof. Why wasn't this done? From one document I got the impression that it was done and a ratio tilted toward Cu-63 was detected. Jones [snip] Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
Re: comments on the Cirillo paper
Hello Horace The condenser was made out of glass and had a length of 1.5 meter and was positioned vertically. It was cooled by water which flowed around the glass condenser. Best Regards Peter - Original Message - From: "Horace Heffner" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Friday, December 03, 2004 1:23 AM Subject: Re: comments on the Cirillo paper > At 12:06 PM 12/2/4, Jones Beene wrote: > >Horace, you seem to be saying that the condenser was > >air-cooled instead of water-cooled. Of course this would > >introduce major errors, and it still doesn't address the > >issue of tritium. > > Actually, there is no mention of a condenser in the Cirillio paper. The > standard method of doing boiloff calorimetry is to measure the weight of > water boiled off (that disappears) and then multply by the energy required > to boil that water (which explicitly *is* the method used by Cirillo.) It > appears the plastic cylinder with pyrex lid located above the cell does the > condensing. There is apparently no intent to use the condensation heat > (i.e. mass flow calorimetry on the secondary coil) as a secondary > calorimetric means. Cirillo's method is definitely susceptable to > entrained water droplets. > > I would assume P.J van Noorden (he can clue us in) used an ordinary > laboratory condenser. Such condensers are typically made of glass and used > in either straight through mode or reflux mode. In straight through mode > the steam comes in through one (elevated) end and water comes out the > other. In reflux mode the condenser is usually vertical and steam is > admitted in at the bottom and water comes out the bottom into an attached > flask. Unless you are trying to do dual calorimetry, it doesn't matter how > the condenser is cooled, by gas, by water, or by ice. The heat measurment > is via the mass of water lost in the reactor. > > Boiloff calorimeters are typically calibrated using boil-off runs using > calibration resistors for heat and cool-off runs to determine the > calorimeter constant for ambient losses. P.J van Noorden certianly makes > it clear that such calibration runs may be invalid becuase ultrasound or > other turbulence creates entraind droplets, and tthe calibration resistor > will not cause droplet entrainment like a source of ultrasound does. One > solution to this problem is to include an ultrasound device in at least one > clibration run to test whatever water drop barrier is used. It would not > be possible to calibrate the drop formation rate itself, so some kind of > drop barrier would have to be utilized. > > These principles have ramifications *way* beyond the Cirillo paper. They > are fundamental to all boiloff calorimetry. > > > > > >Only if it had been water cooled could all the heat be > >accounted for, and that is why I assumed it was water cooled > >and that the thallium was turning up in the second circuit. > > > >> This is a very important comment. It means that boiloff > >calorimetry can be very suspect without proper controls. > > > >Yes, proper controls like a second circuit with dual > >calorimetry. > > > You need to account for more than just the enthalpy of condensation. > > > > > >> A radioactive tracer would be good in labs equipped to > >handle them. > > > >Not unless the possibility of tritium can be eliminated, > > > I have done plenty of tritium counting using liquid scintillation counting. > I think it is more difficult to count water borne tritium by other means. > Scintillation couters can reliably and automatically discriminate between > tritium and say carbon 14. There is almost no penetrating power for 20 keV > beta particles, so counting 201 Tl without interference from tritium is > easy. > > Technetium counting and even imaging is readily done using 180 degrees > opposed scintillation couters to track positron annihilation photon pairs. > I had this procedure done to image my heart. I was signifcantly > radioactive for a day. It was a bit scary to turn on my geiger counter and > hear it go wild near me. > > > >or > >unless your tracer has a far more energetic signature than > >tritium. Thallium is just too close IMHO. > > > >After all, your are doing cold fusion. Cold fusion often > >produces tritium. Isn't the cross-connection obvious? BTW > >even though tritium "normally" has a significant spread of > >energy, can we be sure that tritium produced via CF is not > >closer to being mono-energetic? > > > What do you mean significant spread? The peak is fairly confined. > > BTW, my handbook shows 201 Tl decaying b
Re: comments on the Cirillo paper
Hello Jones We analysed the reactionproduct with a multi channel analyser and we where convinced that it was 201 Tl. Peter - Original Message - From: "Jones Beene" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Thursday, December 02, 2004 4:40 PM Subject: Re: comments on the Cirillo paper > - Original Message - > From: "P.J van Noorden" > > > We used 201 Thallium in our nuclear medicine department > > to study the perfusion of the heart.The energy emission of > radioactive > > thallium is about 80 eV > > The amounts of thallium we used was about a few nanograms. > Therefore you can > > inject it in a patient beacuse in this concentration it is > not toxic.The > > amount I used for this experiment is 1% of the amount we > inject into a > > patient. > > Hello Peter, > > Since this tiny amount of thallium works out to only a few > one-hundredths of a nanogram, one must suspect that this > cannot be measured reliably (by mass) on any kind of a > precision scale, so one must further suspect that you > measured it by assuming that any radioactive emission was > due to the thallium... > > ...but, that raises another problem. > > What if the species which you measured "in the second > vessel, where you only would expect distillated water" was > NOT the Thallium? That is, it was not the thallium which had > migrated through the walls of the condenser, but instead was > Tritium, which was the ash of the adjoining CF reaction? > > Tritium of course, easily is transported through most > metals, such as your condenser. I can find no reference on > the web to thallium crossing a metal boundary. Also the 80 > KeV is characteristic of tritium as well as thallium, but > tritium would have a broader spread (did you do spectrometry > ?) > > Although it is somewhat of an affront to Occam, you could > conceivably have witnessed both radioactive remediation (of > the thallium) and at the same time the LENR cold-fusion (ala > Claytor) of the tritium-ash variety, in this cell. But since > the total radioactive reading on your meter of the combined > two sources added up to nearly what you were expecting from > just the thallium, you assumed the simplest underlying > situation? > > Jones > > >
Re: comments on the Cirillo paper
Hello We used 201 Thallium in our nuclear medicine department to study the perfusion of the heart.The energy emission of radioactive thallium is about 80 eV. Now we have a technetium based radiopharmacon which gives a better image quality.( 140eV) The amounts of thallium we used was about a few nanograms. Therefore you can inject it in a patient beacuse in this concentration it is not toxic.The amount I used for this experiment is 1% of the amount we inject into a patient. Peter - Original Message - From: "Jones Beene" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Thursday, December 02, 2004 3:51 PM Subject: Re: comments on the Cirillo paper > - Original Message - > From: "P.J van Noorden" > > > It was very interesting to see that during evaporation a > significant amount > > (25%) of the radioactive Thallium could be found in the > second vessel, > > where you only would expect destillated water. So I > suspect that during > > violent boiling of the electrolyte a significant amount of > small dropplets > > liquid water ( with radioactive Tl ) was transported > through the condensor > > into the second vessel. This could lead to a significant > overestimation of > > the produced heat by about 25 % > > Well, first a caveat - it should be mentioned for the > benefit of any younger readers contemplating CF experiments, > that it takes a knowledgeable researcher to experiment with > thallium (a.k.a. rat poison), which some chemists believe to > be among the most toxic in the periodic table... and that is > the less-radioactive variety. Thallium does occurs in the > environment naturally in trace amounts; and is responsible > for many more deaths than is commonly known because the > human body absorbs thallium very effectively, especially > through the skin, lungs and the digestive tract. Just > touching it can be dangerous. > > but as to the unusual "transport mechanism" (if it did > indeed cross a metal boundary) this anomaly seems to be > similar to what has been witnessed over the years with > Bismuth, which is a similar heavy metal in many ways and > which was the subject of messages last month (below)... it > would be enlightening to understand the dynamics of this > transport mechanism, and whether or not it is somehow > related to gravity, but there appears to be little reliable > information available. > > Nick Reiter wrote: > > > It [bismuth] also was or is one of > > the most promising stars in the odd half integer spin > > nucleon kinemassic gravity claims of Wallace, > > RC Macaulay wrote: > > > Once knew a man that spent his days during WW2 on > > the Manhattan project that remained puzzled by bismuth. > Such an > > oddity that he considered the element unexplainable. > > (which may have been mentioned in the Rhodes book on the > Manhattan project), I remember hearing about some definite > "peculiarities" concerning bismuth during the LMBR and MSR > (liquid metal and salt cooled reactors) days at Oak Ridge in > the > 60s... the problem was "containment" of the molten bismuth. > It seem that you can have a bismuth alloy or eutectic in a > *sealed* > circuit - completely encased in SS tubing... but > miraculously > it will somehow "seep" through metal and appear in the > adjoining circuit - > > Jones > > > > >
comments on the Cirillo paper
A few years ago I did some interesting experiments with a plasma discharge system send to me by Eugene Mallove. This system was equiped with a reaction vessel connected with a long vertical tube . The evaporated gasses could be condensed and where collected in a second vessel.I used a K2CO3 solution as a electrolyte. First I did experiments with carbon electrodes but later I used Tungsten as cathode. I noticed that when a strong glow discharge developed ,a strong EM signal disturbed my neutroncount readings ,which is in accordance with the results in the Cirillo paper. Because I was interested in transmutations, I used radioactive 201 Thallium ( chemically like K) and measured the counts before and after the plasma discharge. ( activity was 1 MBq) First I send a mild current through the electrodes and the 201Thallium was deposited on the cathode. After I raised the voltage and the plasma discharge developed, I saw a drop in gamma emission measured from the outside of the reactionvessel. Later I noticed that the 201Tl was removed from the kathode and could be found in debris on the bottom of the cell.The change in geometry had produced the measured drop in gamma emission! After correction for the change in geometry I could not find a significant change in activity before and after the plasma discharge. It was very interesting to see that during evaporation a significant amount ( 25%) of the radioactive Thallium could be found in the second vessel, where you only would expect destillated water. So I suspect that during violent boiling of the electrolyte a significant amount of small dropplets liquid water ( with radioactive Tl ) was transported through the condensor into the second vessel. This could lead to a significant overestimation of the produced heat by about 25 %. This effect must i.m.o be taken in account before one can conclude that there is excess heat produced. Peter van Noorden nuclear medicine department the Netherlands
Re: WashingtonPost article
For everybody who is interested I have just downloaded the Washington Post article. Greetings Peter van Noorden Warming Up to Cold Fusion Peter Hagelstein is trying to revive hope for a future of clean, inexhaustible, inexpensive energy. Fifteen years after the scientific embarrassment of the century, is this the beginning of something By Sharon Weinberger Sunday, November 21, 2004; Page W22 On a quiet Monday in late August -- a time of year when much of the Washington bureaucracy has gone to the beach -- a panel of scientists gathered at a Doubletree Hotel set between the Congressional Plaza strip mall and a drab concrete office building on Rockville Pike. They sat around a U-shaped table decked with laptops, with three government officials at the front, ready to hear about an idea that, if it worked, could change the world. The panel's charge was simple: to determine whether that idea had even a prayer of a chance at working. The Department of Energy went to great lengths to cloak the meeting from public view. No announcement, no reporters. None of the names of the people attending that day was disclosed. The DOE made sure to inform the panel's members that they were to provide their conclusions individually rather than as a group, which under a loophole in federal law allowed the agency to close the meeting to the public. At 9:30 a.m., six presenters were invited in and instructed to sit in a row of chairs along the wall. The group included a prominent MIT physicist, a Navy researcher and four other scientists from Russia, Italy and the United States. They had waited a long time for this opportunity and, one by one, stood up to speak about a scientific idea they had been pursuing for more than a decade. All the secrecy likely had little to do with national security and more to do with avoiding possible embarrassment to the agency. To some, the meeting would seem no less outrageous than if the DOE honchos had convened for a seance to raise the dead -- and in a way, they had: Fifteen years ago, the DOE held a very similar review of the very same idea. It was front-page news back in 1989. The subject was cold fusion, the claim that nuclear energy could be released at room temperature, using little more than a high school chemistry set. In one of the most infamous episodes of modern science, two chemists at the University of Utah announced at a news conference that they had harnessed the power of the sun in a test tube. It was, if true, the holy grail of energy: pollution-free, cheap and virtually unlimited. If it worked, cold fusion could supply the country's energy needs, with no more smog, no more nuclear waste, no more depending on other countries for oil. For a brief moment, an energy revolution seemed on the horizon. But when many laboratories tried and failed to reproduce the Utah results, scientists began to line up against cold fusion. Less than a year after the announcement, a DOE review found that none of the experiments had demonstrated convincing evidence of cold fusion. Almost as quickly as they had become famous, the scientists involved became the butt of comedians' jokes; they were even lampooned in a Canadian production called "Cold Fusion: The Musical." A Time magazine millennium poll ranked cold fusion among the "worst ideas" of the century. But now, at the Doubletree in Rockville, it seemed all that could change. For the scientists who had risked ostracism to persist in studying cold fusion, the very fact that the Energy Department was reviewing their work this summer seemed like a breakthrough. True, according to two of the presenters who were there, the meeting began with harsh questions. But at 5 p.m., the presenters were ordered to leave the room, and when they returned, the mood had visibly lifted. At the end, the scientists presenting the idea and those reviewing it all shook hands. The reviewers stayed on to discuss the material. The cold fusionists went to a barbecue, feeling celebratory. No one had told them if the presentation had convinced anyone that cold fusion was real. But it was nice, they said, after so many years, just to be treated with respect. "WHERE'S PETER?" It was noon and the sun was shining in California's Bay Area. It was the week before the DOE meeting in Rockville, and at SRI International, a nonprofit research center in Menlo Park, chemist Michael McKubre was gearing up for what he hoped would be cold fusion's big break. He believed that after 15 years, the new DOE review could give him and others a chance to build an energy source that had the potential to revolutionize society. But first he needed to find Peter Hagelstein for a meeting with a reporter. McKubre's secretary poked her head in the office and said she'd ask Jessica, the summer intern. A minute later the secretary was back. No Peter. "Can you call Peter?" he asked. "Tell him to comb his hair and stuff," he added, shaking his head. McKubre checked the time and settled back in h
