Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
I wrote: The following is a bit speculative, but perhaps someone can correct any misstatements I make -- if there is a magnetic field being created by the cables coiling around the tube [1], I believe the field would point along the axis of the tube, creating a theta pinch, even if only momentarily. It now occurs to me that such a field will itself create quite a bit of acceleration of the metal particles in the tube. Eric
[Vo]:high temp condinsate
*http://scitation.aip.org/content/aip/journal/jcp/136/3/10.1063/1.3678015 http://scitation.aip.org/content/aip/journal/jcp/136/3/10.1063/1.3678015* *Quote:* Within the optical cavity, the photons acquire an effective mass as determined by the cut-off frequency of the cavity that can be 6–7 orders of magnitude less than mass of an electron. *Depending upon the density, this allows for a BEC transition temperature that can approach room temperature. *Polaritons are also ultra-light quasiparticles that are known to condense in systems composed of a semiconducting quantum well sandwiched between two reflective mirrors. 2–6 In this case, however, the polaritons act as hard-core Bosons and scattering at high density allows for a rapid thermalization of the gas. Note: the temperature of condensation of polaritons is proportional to the density of the polaritons and so is their effective mass. The Ni/H reactor produces a huge density of coherent polaritons far greater than what a single Nano-cavity can produce. The effective mass of the polariton can drop into the millivolts. Within the Ni/H reactor's reaction, there is a positive feedback mechanism in place that converts nuclear energy into infrared photons and electrons from more vigorous dipole motion. This energy infusion pushes the density of the polaritons to extreme levels causing the condensate to establish at ever higher temperatures. http://www.talk-polywell.org/bb/report.php?f=10p=116454
[Vo]:temperature of the resistor wire.
Some people suspect that the resistor wire can't be Inconel because they are predicted to melt at the reactor's operating temperature. However, since we know the resistor wire casts a shadow in the alumina, the temperature of the wire remains below the operating temperature and therefore can't melt. Harry
Re: [Vo]:temperature of the resistor wire.
how do you know this? How do you know the the wire is not the brightest area? On 15 October 2014 15:06, H Veeder hveeder...@gmail.com wrote: Some people suspect that the resistor wire can't be Inconel because they are predicted to melt at the reactor's operating temperature. However, since we know the resistor wire casts a shadow in the alumina, the temperature of the wire remains below the operating temperature and therefore can't melt. Harry
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
Highly doubtful. Above curie temperture of Nickel so no ferromagnetism, and powder too microscopic hot resistivity too high, and AC frequency, current and number of windings too low for strong magnetic fields or significant eddy currents to form and give push via lenzs law. On 15 October 2014 14:05, Eric Walker eric.wal...@gmail.com wrote: I wrote: The following is a bit speculative, but perhaps someone can correct any misstatements I make -- if there is a magnetic field being created by the cables coiling around the tube [1], I believe the field would point along the axis of the tube, creating a theta pinch, even if only momentarily. It now occurs to me that such a field will itself create quite a bit of acceleration of the metal particles in the tube. Eric
Re: [Vo]:temperature of the resistor wire.
If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. I am 95% confident that is the shadow of the coil. The light areas change in brightness, width, etc. On Wed, Oct 15, 2014 at 3:56 AM, Robert Lynn robert.gulliver.l...@gmail.com wrote: how do you know this? How do you know the the wire is not the brightest area? On 15 October 2014 15:06, H Veeder hveeder...@gmail.com wrote: Some people suspect that the resistor wire can't be Inconel because they are predicted to melt at the reactor's operating temperature. However, since we know the resistor wire casts a shadow in the alumina, the temperature of the wire remains below the operating temperature and therefore can't melt. Harry
Re: [Vo]:temperature of the resistor wire.
I am looking at high zoom at the same photos and finding it easy to draw the opposite conclusion. Confirmation bias on both our parts :) I think it is equivocal at best. On 15 October 2014 19:52, ChemE Stewart cheme...@gmail.com wrote: If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. I am 95% confident that is the shadow of the coil. The light areas change in brightness, width, etc. On Wed, Oct 15, 2014 at 3:56 AM, Robert Lynn robert.gulliver.l...@gmail.com wrote: how do you know this? How do you know the the wire is not the brightest area? On 15 October 2014 15:06, H Veeder hveeder...@gmail.com wrote: Some people suspect that the resistor wire can't be Inconel because they are predicted to melt at the reactor's operating temperature. However, since we know the resistor wire casts a shadow in the alumina, the temperature of the wire remains below the operating temperature and therefore can't melt. Harry
Re: [Vo]:temperature of the resistor wire.
Additionally, look at the darkened photo, the wire exterior to the reactor sourrounded by cooler materials to radiate to are brighter than the bright wires in the reactor. Hard to believe it would be colder inside the reactor surrounded by relatively hotter materials that are harder to radiate to. I think that is pretty strong indication that it is the wires that are the bright areas. On 15 October 2014 20:14, Robert Lynn robert.gulliver.l...@gmail.com wrote: I am looking at high zoom at the same photos and finding it easy to draw the opposite conclusion. Confirmation bias on both our parts :) I think it is equivocal at best. On 15 October 2014 19:52, ChemE Stewart cheme...@gmail.com wrote: If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. I am 95% confident that is the shadow of the coil. The light areas change in brightness, width, etc. On Wed, Oct 15, 2014 at 3:56 AM, Robert Lynn robert.gulliver.l...@gmail.com wrote: how do you know this? How do you know the the wire is not the brightest area? On 15 October 2014 15:06, H Veeder hveeder...@gmail.com wrote: Some people suspect that the resistor wire can't be Inconel because they are predicted to melt at the reactor's operating temperature. However, since we know the resistor wire casts a shadow in the alumina, the temperature of the wire remains below the operating temperature and therefore can't melt. Harry
Re: [Vo]:Testing fuels without a reactor
On Tue, Oct 14, 2014 at 10:14 PM, ChemE Stewart cheme...@gmail.com wrote: Or just ride your bike... http://koin.com/2014/09/05/electric-bike-battery-may-have-caused-bend-house-fire/ I have been studying the 18650, the cell which powers the Tesla. Proper charging of the cell is a far cry from slapping the old trickle charger on the '64 Chev Impala. There are three stages in the charge cycle: http://www.electronicproducts.com/Power_Products/Challenges_in_Li-ion_charging.aspx For your multi-cell laptop battery, there is a 4th stage between the CC and CV stages called cell equalization. Due to variations in the internal resistance of each cell, cells in series will not charge evenly. Your 11.1 V (nom) laptop battery can separately charge each 3.7 V parallel stage. The fully charged laptop will actually read about 4.2 V per cell. It is considered fully discharged at about 2.8 V. Optimum life is achieved by always charging between 25 and 50% discharge. This kid has a series of videos where he builds his homemade electric bike. This is in the middle of the series where he gets into the 18650: https://www.youtube.com/watch?v=Z-VLdE6RxYQ Tesla Motors and Panasonic are building a plant to produce 35,000,000 18650s by 2018 to the tune of $5,000,000,000 capital cost.
RE: [Vo]:temperature of the resistor wire.
From: ChemE Stewart If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. If this is 3-phase 50-cycle, then the photo should be showing the gap of the odd phase at any instant, which gap moves in one direction or the other, which is the marquee-effect of 3-phase (effective directionality). Thus one expects non-uniform width and continuity of the conductors … this is really 3-phase, no?
Re: [Vo]:temperature of the resistor wire.
The photo is an average of radiation from 50 Hz cycles, not instantaneous...temp does not swing that quickly...?... On Wednesday, October 15, 2014, Jones Beene jone...@pacbell.net wrote: *From:* ChemE Stewart If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. If this is 3-phase 50-cycle, then the photo should be showing the gap of the odd phase at any instant, which gap moves in one direction or the other, which is the marquee-effect of 3-phase (effective directionality). Thus one expects non-uniform width and continuity of the conductors … this is really 3-phase, no?
Re: [Vo]:temperature of the resistor wire.
They specifically say in the report the coils are the dark areas. I doubt they're lying about that. On Wed, Oct 15, 2014 at 6:07 AM, Jones Beene jone...@pacbell.net wrote: *From:* ChemE Stewart If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. If this is 3-phase 50-cycle, then the photo should be showing the gap of the odd phase at any instant, which gap moves in one direction or the other, which is the marquee-effect of 3-phase (effective directionality). Thus one expects non-uniform width and continuity of the conductors … this is really 3-phase, no?
Re: [Vo]:temperature of the resistor wire.
Can you attach that photo? I am not sure which one On Wednesday, October 15, 2014, Robert Lynn robert.gulliver.l...@gmail.com wrote: Additionally, look at the darkened photo, the wire exterior to the reactor sourrounded by cooler materials to radiate to are brighter than the bright wires in the reactor. Hard to believe it would be colder inside the reactor surrounded by relatively hotter materials that are harder to radiate to. I think that is pretty strong indication that it is the wires that are the bright areas. On 15 October 2014 20:14, Robert Lynn robert.gulliver.l...@gmail.com javascript:_e(%7B%7D,'cvml','robert.gulliver.l...@gmail.com'); wrote: I am looking at high zoom at the same photos and finding it easy to draw the opposite conclusion. Confirmation bias on both our parts :) I think it is equivocal at best. On 15 October 2014 19:52, ChemE Stewart cheme...@gmail.com javascript:_e(%7B%7D,'cvml','cheme...@gmail.com'); wrote: If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. I am 95% confident that is the shadow of the coil. The light areas change in brightness, width, etc. On Wed, Oct 15, 2014 at 3:56 AM, Robert Lynn robert.gulliver.l...@gmail.com javascript:_e(%7B%7D,'cvml','robert.gulliver.l...@gmail.com'); wrote: how do you know this? How do you know the the wire is not the brightest area? On 15 October 2014 15:06, H Veeder hveeder...@gmail.com javascript:_e(%7B%7D,'cvml','hveeder...@gmail.com'); wrote: Some people suspect that the resistor wire can't be Inconel because they are predicted to melt at the reactor's operating temperature. However, since we know the resistor wire casts a shadow in the alumina, the temperature of the wire remains below the operating temperature and therefore can't melt. Harry
RE: [Vo]:temperature of the resistor wire.
Depends on the camera exposure time… Probably a digital camera. What would the exposure time be? From: ChemE Stewart The photo is an average of radiation from 50 Hz cycles, not instantaneous...temp does not swing that quickly...?... If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. If this is 3-phase 50-cycle, then the photo should be showing the gap of the odd phase at any instant, which gap moves in one direction or the other, which is the marquee-effect of 3-phase (effective directionality). Thus one expects non-uniform width and continuity of the conductors … this is really 3-phase, no?
Re: [Vo]:temperature of the resistor wire.
I think it also depends on high quickly the wire temp oscillates, thermal conductivity, etc On Wednesday, October 15, 2014, Jones Beene jone...@pacbell.net wrote: Depends on the camera exposure time… Probably a digital camera. What would the exposure time be? *From:* ChemE Stewart The photo is an average of radiation from 50 Hz cycles, not instantaneous...temp does not swing that quickly...?... If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. If this is 3-phase 50-cycle, then the photo should be showing the gap of the odd phase at any instant, which gap moves in one direction or the other, which is the marquee-effect of 3-phase (effective directionality). Thus one expects non-uniform width and continuity of the conductors … this is really 3-phase, no?
Re: [Vo]:temperature of the resistor wire.
Not lying, but perhaps again confirmation bias, based on wrong assumptions. How can the inconel wire in Fig 12b be hotter/brighter in the cooler external environment outside the end of the reactor than it is in the hotter internal environment inside the reactor? On 15 October 2014 21:12, Blaze Spinnaker blazespinna...@gmail.com wrote: They specifically say in the report the coils are the dark areas. I doubt they're lying about that. On Wed, Oct 15, 2014 at 6:07 AM, Jones Beene jone...@pacbell.net wrote: *From:* ChemE Stewart If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. If this is 3-phase 50-cycle, then the photo should be showing the gap of the odd phase at any instant, which gap moves in one direction or the other, which is the marquee-effect of 3-phase (effective directionality). Thus one expects non-uniform width and continuity of the conductors … this is really 3-phase, no?
Re: [Vo]:temperature of the resistor wire.
The wire is not in the reactor, it is embedded in the insulating alumina shell On Wednesday, October 15, 2014, Robert Lynn robert.gulliver.l...@gmail.com wrote: Not lying, but perhaps again confirmation bias, based on wrong assumptions. How can the inconel wire in Fig 12b be hotter/brighter in the cooler external environment outside the end of the reactor than it is in the hotter internal environment inside the reactor? On 15 October 2014 21:12, Blaze Spinnaker blazespinna...@gmail.com javascript:_e(%7B%7D,'cvml','blazespinna...@gmail.com'); wrote: They specifically say in the report the coils are the dark areas. I doubt they're lying about that. On Wed, Oct 15, 2014 at 6:07 AM, Jones Beene jone...@pacbell.net javascript:_e(%7B%7D,'cvml','jone...@pacbell.net'); wrote: *From:* ChemE Stewart If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. If this is 3-phase 50-cycle, then the photo should be showing the gap of the odd phase at any instant, which gap moves in one direction or the other, which is the marquee-effect of 3-phase (effective directionality). Thus one expects non-uniform width and continuity of the conductors … this is really 3-phase, no?
[Vo]:The Potential Vortex of Self-consistent Electrodynamics
PIERS Proceedings, Moscow, Russia, August 19–23, 2012 *Self-consistent Electrodynamics https://piers.org/piersproceedings/download.php?file=cGllcnMyMDEyTW9zY293fDFBN18wMTcyLnBkZnwxMjAzMTkwNTU5MTE=* Konstantin Meyl Faculty of Computer and Electrical Engineering, Furtwangen University, Germany Abstract— Even though one usually calculates capacitor losses with a complex epsilon it still offends the principle of a constant speed of light. Maxwell’s term c² = 1/ε·µ suggests a physically inexplicable complex speed. By such an offence against basic principles every physicist is asked to search and to repair the mistake in the textbooks. The contribution clearly explains how vortex losses occur instead of using the postulated and fictive imaginary part of the material constant epsilon. The theory better explains the function of a microwave oven, the welding of PVC foils or how capacitor losses occur. The responsible potential vortices can be derived without postulate them from the established laws of physics. Vortex losses can even be proven experimentally and are clearly shown. *The potential vortex is substituted for the vector potential A, which has controlled electrodynamics as “impurity factors” ever since its introduction. A unified theory of all interactions and physical phenomena is missing without potential vortices.*This theory justifies the efforts and the rationale for rebuilding electrodynamics and in so doing effectively removes contradictions of the vector potential and loss theory. Consequences are discussed such as the discovery of magnetic monopoles by the German Helmholtz Center [1], the extended Poynting vector, and many more effects involved with the new approach of the potential vortex, that is replacing the vector potential in the dielectric.
RE: [Vo]:temperature of the resistor wire.
From: Robert Lynn How can the Inconel wire in Fig 12b be hotter/brighter in the cooler external environment outside the end of the reactor than it is in the hotter internal environment inside the reactor? In FWIW department, here is the grade of Inconel often used for resistor wire Inconel 600. As you can see, it is rated to less than 540 C. http://www.ebay.com/itm/Nickel-600-Inconel-Wire-041-1-04mm-x-10-3m-/320676194894?pt=LH_DefaultDomain_2 http://www.ebay.com/itm/Nickel-600-Inconel-Wire-041-1-04mm-x-10-3m-/320676194894?pt=LH_DefaultDomain_2hash=item4aa9ca6a4e hash=item4aa9ca6a4e As Eric suggests, given the impossibility of Inconel - they must be using something else besides Inconel. I agree. Tungsten comes to mind. This goes along with a growing belief that there is gain here and it could be more than they claim or less … since they did not calibrate - but there is also intentional deception, meaning that this is not a scientific report, but one designed to look that way using cast of PhDs who were essentially asleep at the wheel.
Re: [Vo]:temperature of the resistor wire.
I keep thinking he has built some sort of alumina/ceramic klystron/microwave tube with that reactor. http://www.daenotes.com/electronics/microwave-radar/microwave-tube-devices On Wednesday, October 15, 2014, Jones Beene jone...@pacbell.net wrote: *From:* Robert Lynn How can the Inconel wire in Fig 12b be hotter/brighter in the cooler external environment outside the end of the reactor than it is in the hotter internal environment inside the reactor? In FWIW department, here is the grade of Inconel often used for resistor wire Inconel 600. As you can see, it is rated to less than 540 C. http://www.ebay.com/itm/Nickel-600-Inconel-Wire-041-1-04mm-x-10-3m-/320676194894?pt=LH_DefaultDomain_2hash=item4aa9ca6a4e As Eric suggests, given the impossibility of Inconel - they must be using something else besides Inconel. I agree. Tungsten comes to mind. This goes along with a growing belief that there is gain here and it could be more than they claim or less … since they did not calibrate - but there is also intentional deception, meaning that this is not a scientific report, but one designed to look that way using cast of PhDs who were essentially asleep at the wheel.
Re: [Vo]:temperature of the resistor wire.
ChemE Stewart cheme...@gmail.com wrote: I think it also depends on high quickly the wire temp oscillates, thermal conductivity, etc Macroscopic changes in temperature are very slow compared to the kinds of pulses a camera might capture. Unless you are talking about temperature changes in micro degrees they always take minutes, even when the pulse of heat lasts milliseconds. This is what calorimeters show. - Jed
RE: [Vo]:temperature of the resistor wire.
From: ChemE Stewart I keep thinking he has built some sort of alumina/ceramic klystron/microwave tube with that reactor. Maybe… but is it not fair to say that you are kinda’ obsessed with microwaves :-) BTW in response to James Bowery’s post on Meryl – the SPP is an intense magnetic vortex.
Re: [Vo]:temperature of the resistor wire.
From: ChemE Stewart cheme...@gmail.com Sent: Wednesday, October 15, 2014 6:28:37 AM The wire is not in the reactor, it is embedded in the insulating alumina shell. That's a guess. But it might be true. I drew (and scanned) some diagrams of a possible structure following the March 2013 hotcats -- where there was a ceramic insert inside the outer steel tube, and round which the heater wire was strung, lengthways. But I now think that's not feasible. First, the new tube is too narrow. Secondly, it's not perfectly straight, so a ceramic wire-holder wouldn't slide in. (In the 2013 hotcat the outer cylinder was steel). They might be able to slide in a more flexible holder, eg made of mica, but there's no evidence for that. Setting aside the concerns that the nickel powder would have melted, it seems that the powder was easily poured out, which implies a smooth inner wall. So now we have even more guessing. Figure 2 was probably taken during the dummy run. The heaters are on, but there is no glow through the ceramic, though they are glowing outside the tube. Figure 10 from the dummy run shows slight evidence of spiral banding (orange-ish) from top-left to bottom-right in segments 1-2-3 4-5-6 and 7-8-9 : but this looks to be a coarser spiral than than the distinct bands of fig 12. Since we don't know which end is which, we can't even tell if this is in the same direction as the visible-light banding. Again, I complain that there are no IR pictures during the live run. Since there WAS visible banding they should have taken visible and IR pictures from the same angle and orientation. If the wire IS embedded in the ceramic, there may or may not be a gap around it. And the overall energy pathways could give bright-wires or dark-wires, depending on the relative balance of heat conduction, radiation in any gaps, and transmission through the ceramic. I'll put up my diagrams: they apply to the 2013 test, even if the don't in this one. ps : Overall, I'm sticking at inconclusive.
Re: [Vo]:temperature of the resistor wire.
But the wire cannot be tungsten outside of the reactor where it is exposed to air! Only inconel will survive air exposure at such temperatures for a month, and it maxes out at about 1300-1350°C (even that is pushing it). And that wire (Fig 12b) is visibly brighter than the wire lines in the reactor (or brightest surface areas of reactor), hence hotter. So QED the reactor surface is 1400°C. The thermography is flat out wrong for reasons unknown, and knowing that it is wrong you have to set aside all the conclusions that are based on it! The wire in the reactor in an insulated environment is necessarily hotter than the wire outside the reactor, and while everyone might want to believe that they must therefore be using exotic refractory wires that cannot be the case: There is no way to joining the inconel wire to a refractory metal at a temperature above the melting point of inconel within the insulated and even higher temperature of an oxygen-free sealed environment within the reactor. The only conclusion that makes sense is that the wires in the reactor are at or below the melting temperature of inconel, and in such circumstances the only way that they do not melt and fail is if the reactor surface temperature is at least 2-300°C lower as I have previously shown. As to growing belief in gain, I started out that way, but more I have looked at the thermal physics in play and the inconsistencies it creates the less believable it has become, the pictures and heat transfer physics at play make it a strong possibility that there was no gain. On 15 October 2014 21:40, Jones Beene jone...@pacbell.net wrote: *From:* Robert Lynn How can the Inconel wire in Fig 12b be hotter/brighter in the cooler external environment outside the end of the reactor than it is in the hotter internal environment inside the reactor? In FWIW department, here is the grade of Inconel often used for resistor wire Inconel 600. As you can see, it is rated to less than 540 C. http://www.ebay.com/itm/Nickel-600-Inconel-Wire-041-1-04mm-x-10-3m-/320676194894?pt=LH_DefaultDomain_2hash=item4aa9ca6a4e As Eric suggests, given the impossibility of Inconel - they must be using something else besides Inconel. I agree. Tungsten comes to mind. This goes along with a growing belief that there is gain here and it could be more than they claim or less … since they did not calibrate - but there is also intentional deception, meaning that this is not a scientific report, but one designed to look that way using cast of PhDs who were essentially asleep at the wheel.
Re: [Vo]:temperature of the resistor wire.
You would think after 30 days of sitting around, staring at the reactor , drinking coffee, eating lunch, sleeping and thinking about it, the team of scientists would have discussed all of this and verified. They would have plenty of time to triple check readings, even some type of portable thermocouple, etc. We suffer from a lack of information, possibly on purpose. On Wednesday, October 15, 2014, Robert Lynn robert.gulliver.l...@gmail.com wrote: But the wire cannot be tungsten outside of the reactor where it is exposed to air! Only inconel will survive air exposure at such temperatures for a month, and it maxes out at about 1300-1350°C (even that is pushing it). And that wire (Fig 12b) is visibly brighter than the wire lines in the reactor (or brightest surface areas of reactor), hence hotter. So QED the reactor surface is 1400°C. The thermography is flat out wrong for reasons unknown, and knowing that it is wrong you have to set aside all the conclusions that are based on it! The wire in the reactor in an insulated environment is necessarily hotter than the wire outside the reactor, and while everyone might want to believe that they must therefore be using exotic refractory wires that cannot be the case: There is no way to joining the inconel wire to a refractory metal at a temperature above the melting point of inconel within the insulated and even higher temperature of an oxygen-free sealed environment within the reactor. The only conclusion that makes sense is that the wires in the reactor are at or below the melting temperature of inconel, and in such circumstances the only way that they do not melt and fail is if the reactor surface temperature is at least 2-300°C lower as I have previously shown. As to growing belief in gain, I started out that way, but more I have looked at the thermal physics in play and the inconsistencies it creates the less believable it has become, the pictures and heat transfer physics at play make it a strong possibility that there was no gain. On 15 October 2014 21:40, Jones Beene jone...@pacbell.net javascript:_e(%7B%7D,'cvml','jone...@pacbell.net'); wrote: *From:* Robert Lynn How can the Inconel wire in Fig 12b be hotter/brighter in the cooler external environment outside the end of the reactor than it is in the hotter internal environment inside the reactor? In FWIW department, here is the grade of Inconel often used for resistor wire Inconel 600. As you can see, it is rated to less than 540 C. http://www.ebay.com/itm/Nickel-600-Inconel-Wire-041-1-04mm-x-10-3m-/320676194894?pt=LH_DefaultDomain_2hash=item4aa9ca6a4e As Eric suggests, given the impossibility of Inconel - they must be using something else besides Inconel. I agree. Tungsten comes to mind. This goes along with a growing belief that there is gain here and it could be more than they claim or less … since they did not calibrate - but there is also intentional deception, meaning that this is not a scientific report, but one designed to look that way using cast of PhDs who were essentially asleep at the wheel.
Re: [Vo]:temperature of the resistor wire.
fig 2 : http://lenr.qumbu.com/web_hotcat2_pics/141011_lugano_fig2.jpg No banding, dummy run fig10 : http://lenr.qumbu.com/web_hotcat2_pics/141011_lugano_fig10.png Possible IR banding, dummy run fig 12a : http://lenr.qumbu.com/web_hotcat2_pics/141011_lugano_fig12a.jpg Strong visible banding - Original Message - From: Alan Fletcher a...@well.com To: vortex-l@eskimo.com Sent: Wednesday, October 15, 2014 7:08:57 AM Subject: Re: [Vo]:temperature of the resistor wire. From: ChemE Stewart cheme...@gmail.com Sent: Wednesday, October 15, 2014 6:28:37 AM The wire is not in the reactor, it is embedded in the insulating alumina shell. That's a guess. But it might be true. I drew (and scanned) some diagrams of a possible structure following the March 2013 hotcats -- where there was a ceramic insert inside the outer steel tube, and round which the heater wire was strung, lengthways. But I now think that's not feasible. First, the new tube is too narrow. Secondly, it's not perfectly straight, so a ceramic wire-holder wouldn't slide in. (In the 2013 hotcat the outer cylinder was steel). They might be able to slide in a more flexible holder, eg made of mica, but there's no evidence for that. Setting aside the concerns that the nickel powder would have melted, it seems that the powder was easily poured out, which implies a smooth inner wall. So now we have even more guessing. Figure 2 was probably taken during the dummy run. The heaters are on, but there is no glow through the ceramic, though they are glowing outside the tube. Figure 10 from the dummy run shows slight evidence of spiral banding (orange-ish) from top-left to bottom-right in segments 1-2-3 4-5-6 and 7-8-9 : but this looks to be a coarser spiral than than the distinct bands of fig 12. Since we don't know which end is which, we can't even tell if this is in the same direction as the visible-light banding. Again, I complain that there are no IR pictures during the live run. Since there WAS visible banding they should have taken visible and IR pictures from the same angle and orientation. If the wire IS embedded in the ceramic, there may or may not be a gap around it. And the overall energy pathways could give bright-wires or dark-wires, depending on the relative balance of heat conduction, radiation in any gaps, and transmission through the ceramic. I'll put up my diagrams: they apply to the 2013 test, even if the don't in this one. ps : Overall, I'm sticking at inconclusive.
Re: [Vo]:temperature of the resistor wire.
I've been thinking of tungsten for a while now. Do they make an alloy with tungsten that operates at high temps in an oxygen atmosphere. I ask because, although the tungsten that is embedded in the reactor would be protected from oxygen by the aluminum oxide coating, you have to connect it to power somewhere outside the reactor that would be exposed to air and the wire, if pure tungsten, would decompose rapidly. Also, I think that you continue to use the word deception without proof that Rossi has deceived anyone in this experiment. I realize that all the data goes against current knowledge, but do you think that we know absolutely everything there is to know about reactions at the nuclear level? I think not. I think that there is a reaction that is going on that does not follow our current knowledge and it may be determined that it is not nuclear in the common sense but it is indeed a novel reaction and it needs to be studied and not scoffed at. Robert Dorr On 10/15/2014 6:40 AM, Jones Beene wrote: *From:* Robert Lynn How can the Inconel wire in Fig 12b be hotter/brighter in the cooler external environment outside the end of the reactor than it is in the hotter internal environment inside the reactor? In FWIW department, here is the grade of Inconel often used for resistor wire Inconel 600. As you can see, it is rated to less than 540 C. http://www.ebay.com/itm/Nickel-600-Inconel-Wire-041-1-04mm-x-10-3m-/320676194894?pt=LH_DefaultDomain_2hash=item4aa9ca6a4e http://www.ebay.com/itm/Nickel-600-Inconel-Wire-041-1-04mm-x-10-3m-/320676194894?pt=LH_DefaultDomain_2hash=item4aa9ca6a4e As Eric suggests, given the impossibility of Inconel - they must be using something else besides Inconel. I agree. Tungsten comes to mind. This goes along with a growing belief that there is gain here and it could be more than they claim or less … since they did not calibrate - but there is also intentional deception, meaning that this is not a scientific report, but one designed to look that way using cast of PhDs who were essentially asleep at the wheel. No virus found in this message. Checked by AVG - www.avg.com http://www.avg.com Version: 2014.0.4765 / Virus Database: 4040/8393 - Release Date: 10/15/14
Re: [Vo]:temperature of the resistor wire.
Your conclusion that there is no gain is incorrect. If that were the situation, the behavior that the testers witnessed with increasing temperature could not have happened. I do not know how much gain was actually present due to some of the questions that remain about the true temperature, but I do not doubt that a significant amount is shown. If you question the fact that the COP is greater than 1, then you should make an attempt to explain what the experimenters witnessed as the input power was increased by 100 watts. Gain is the only sensible explanation as far as I can imagine. Dave -Original Message- From: Robert Lynn robert.gulliver.l...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Oct 15, 2014 10:25 am Subject: Re: [Vo]:temperature of the resistor wire. But the wire cannot be tungsten outside of the reactor where it is exposed to air! Only inconel will survive air exposure at such temperatures for a month, and it maxes out at about 1300-1350°C (even that is pushing it). And that wire (Fig 12b) is visibly brighter than the wire lines in the reactor (or brightest surface areas of reactor), hence hotter. So QED the reactor surface is 1400°C. The thermography is flat out wrong for reasons unknown, and knowing that it is wrong you have to set aside all the conclusions that are based on it! The wire in the reactor in an insulated environment is necessarily hotter than the wire outside the reactor, and while everyone might want to believe that they must therefore be using exotic refractory wires that cannot be the case: There is no way to joining the inconel wire to a refractory metal at a temperature above the melting point of inconel within the insulated and even higher temperature of an oxygen-free sealed environment within the reactor. The only conclusion that makes sense is that the wires in the reactor are at or below the melting temperature of inconel, and in such circumstances the only way that they do not melt and fail is if the reactor surface temperature is at least 2-300°C lower as I have previously shown. As to growing belief in gain, I started out that way, but more I have looked at the thermal physics in play and the inconsistencies it creates the less believable it has become, the pictures and heat transfer physics at play make it a strong possibility that there was no gain. On 15 October 2014 21:40, Jones Beene jone...@pacbell.net wrote: From: Robert Lynn How can the Inconelwire in Fig 12b be hotter/brighter in the cooler external environment outsidethe end of the reactor than it is in the hotter internal environment inside thereactor? In FWIWdepartment, here is the grade of Inconel often used for resistor wire Inconel600. As you can see, it is rated to less than 540 C. http://www.ebay.com/itm/Nickel-600-Inconel-Wire-041-1-04mm-x-10-3m-/320676194894?pt=LH_DefaultDomain_2hash=item4aa9ca6a4e As Ericsuggests, given the impossibility of Inconel - they must be using something elsebesides Inconel. I agree. Tungsten comes to mind. Thisgoes along with a growing belief that there is gain here and it could be morethan they claim or less … since they did not calibrate - but there is alsointentional deception, meaning that this is not a scientific report, but one designedto look that way using cast of PhDs who were essentially asleep at the wheel.
Re: [Vo]:temperature of the resistor wire.
12A looks like a damn flame shooting out the end of the thing! Thin/uniform bands spiraling one way and thinker bands going the other. That alumina shell looks wrapped, sort of like a paper mache newspaper wrap, like alumina felt soaked in glue and wrapped. It is not even straight. On Wed, Oct 15, 2014 at 10:46 AM, Alan Fletcher a...@well.com wrote: fig 2 : http://lenr.qumbu.com/web_hotcat2_pics/141011_lugano_fig2.jpg No banding, dummy run fig10 : http://lenr.qumbu.com/web_hotcat2_pics/141011_lugano_fig10.png Possible IR banding, dummy run fig 12a : http://lenr.qumbu.com/web_hotcat2_pics/141011_lugano_fig12a.jpg Strong visible banding -- *From: *Alan Fletcher a...@well.com *To: *vortex-l@eskimo.com *Sent: *Wednesday, October 15, 2014 7:08:57 AM *Subject: *Re: [Vo]:temperature of the resistor wire. *From: *ChemE Stewart cheme...@gmail.com *Sent: *Wednesday, October 15, 2014 6:28:37 AM The wire is not in the reactor, it is embedded in the insulating alumina shell. That's a guess. But it might be true. I drew (and scanned) some diagrams of a possible structure following the March 2013 hotcats -- where there was a ceramic insert inside the outer steel tube, and round which the heater wire was strung, lengthways. But I now think that's not feasible. First, the new tube is too narrow. Secondly, it's not perfectly straight, so a ceramic wire-holder wouldn't slide in. (In the 2013 hotcat the outer cylinder was steel). They might be able to slide in a more flexible holder, eg made of mica, but there's no evidence for that. Setting aside the concerns that the nickel powder would have melted, it seems that the powder was easily poured out, which implies a smooth inner wall. So now we have even more guessing. Figure 2 was probably taken during the dummy run. The heaters are on, but there is no glow through the ceramic, though they are glowing outside the tube. Figure 10 from the dummy run shows slight evidence of spiral banding (orange-ish) from top-left to bottom-right in segments 1-2-3 4-5-6 and 7-8-9 : but this looks to be a coarser spiral than than the distinct bands of fig 12. Since we don't know which end is which, we can't even tell if this is in the same direction as the visible-light banding. Again, I complain that there are no IR pictures during the live run. Since there WAS visible banding they should have taken visible and IR pictures from the same angle and orientation. If the wire IS embedded in the ceramic, there may or may not be a gap around it. And the overall energy pathways could give bright-wires or dark-wires, depending on the relative balance of heat conduction, radiation in any gaps, and transmission through the ceramic. I'll put up my diagrams: they apply to the 2013 test, even if the don't in this one. ps : Overall, I'm sticking at inconclusive.
Re: [Vo]:temperature of the resistor wire.
I wrote: Macroscopic changes in temperature are very slow compared to the kinds of pulses a camera might capture. I mean they all blur together. If there are moving pulses of heat under the alumina, by the time the heat reaches the surface it blurs together and you would not see light and dark areas with an IR camera or by any other technique. They do not exist. The surface is all the same temperature to within micro-degrees. With the right equipment you might see moving waves of tiny temperature differences. - Jed
Re: [Vo]:temperature of the resistor wire.
I agree On Wed, Oct 15, 2014 at 11:03 AM, Jed Rothwell jedrothw...@gmail.com wrote: I wrote: Macroscopic changes in temperature are very slow compared to the kinds of pulses a camera might capture. I mean they all blur together. If there are moving pulses of heat under the alumina, by the time the heat reaches the surface it blurs together and you would not see light and dark areas with an IR camera or by any other technique. They do not exist. The surface is all the same temperature to within micro-degrees. With the right equipment you might see moving waves of tiny temperature differences. - Jed
[Vo]:An galling alternative
One more followup on Bob Ellefson's analysis of the SEM/EDS ToF/SIMS where the unexpected surprise was apparent gallium. (mass 69) I am wondering if the explanation is not exactly nuclear ! but instead, the lithium has been fully absorbed into the K or L shell of the Ni62, giving it high stability and the appearance of mass 69 but not the reality of a nuclear reaction. This is called a ballotechnic reaction, and has been clouded in secrecy over the years. (it does not help credibility to bring up the red mercury connection) http://en.wikipedia.org/wiki/Ballotechnics I guess it would have to be the L2p shell to hold three electrons. Thus the intense beam of SEM electrons would ionize the combined particle but not kick out the lithium from the inner shell. BTW - the intense field of the SPP would probably be capable of forcing the lithium in there. This could be magnetic bonding also. As mentioned before Li7 has a high nuclear magnetic moment and Ni is ferromagnetic. The Curie Temp is a problem with a magnetic explanation - but perhaps once formed, this coupling survives. Where is the energy gain? That is still under consideration, but the reason for saying that this could happen, is that lithium is known to absorb completely into deep shells of a few elements. Nickel is not one of them, so this may not apply. But it is a far more likely explanation than a real fusion of nickel to lithium. Jones Bob, Just so you know, your insight is not being ignored - although you may reasonably suspect it is, since there are few comments. Lots of people are trying to come to grips with this data, as preposterous as it may seem at first glance. But the main problem remains: these are very energetic reactions with extreme Coulomb barriers. Obviously, if Ni-62 is the key to the Rossi effect - which is what his patent would suggest, and it fused with Li7, then there is a putative case for gallium-69. Beyond that, there is no support for anything close to this in the literature. -Original Message- From: Robert Ellefson Given the results of the SIMS analysis from the Lugano report, particularly as detailed in this posting: http://www.mail-archive.com/vortex-l@eskimo.com/msg98596.html
RE: [Vo]:temperature of the resistor wire.
Robert Lynn wrote: But the wire cannot be tungsten outside of the reactor where it is exposed to air! It could be exposed to air in the reactor too. One could imagine the leads into the end-caps are Inconel - and the wire coils are tungsten coated with alumina, or else body is two pieces as Bob Higgins suggests, with the tungsten encased and the two pieces are diffusion bonded and with Inconel leads.
[Vo]:Lockheed says makes breakthrough on fusion energy project - Yahoo News
Lockheed says makes breakthrough on fusion energy project - Yahoo News WHOA ! http://news.yahoo.com/lockheed-says-makes-breakthrough-fusion-energy-project-123840986--finance.html Get this: “Ultra-dense deuterium, an isotope of hydrogen, is found in the earth's oceans…. I am wondering if this is simply a naïve reporter (most likely) or it this truly ultra-dense in the context of f/H?
Re: [Vo]:Lockheed says makes breakthrough on fusion energy project - Yahoo News
Makes you wonder if this is LENR related. 100 megawatts from a reactor that fits on a truck, and no radioactive waste!!! Hmmm Robert Dorr On 10/15/2014 8:19 AM, Jones Beene wrote: Lockheed says makes breakthrough on fusion energy project - Yahoo News WHOA ! http://news.yahoo.com/lockheed-says-makes-breakthrough-fusion-energy-project-123840986--finance.html Get this: “Ultra-dense deuterium, an isotope of hydrogen, is found in the earth's oceans…. I am wondering if this is simply a naïve reporter (most likely) or it this truly ultra-dense in the context of f/H? No virus found in this message. Checked by AVG - www.avg.com http://www.avg.com Version: 2014.0.4765 / Virus Database: 4040/8393 - Release Date: 10/15/14
Re: [Vo]:An galling alternative
Martin Fleiishchmann Memorial Project Facebook page brought up the G69 and said it can be safely ignored as an artifact of the SIMS probe. https://www.facebook.com/MartinFleischmannMemorialProject SIMS stands for Secondary Ion Mass Spectrometry and Gallium 69 is used as the primary ionisation source for the SIMS. This is why it is seen in both the fuel and the ash signatures. Hence it should be considered irrelevant in the reports data. http://en.wikipedia.org/wiki/Secondary_ion_mass_spectrometry - Brad Lowe On Wed, Oct 15, 2014 at 8:05 AM, Jones Beene jone...@pacbell.net wrote: One more followup on Bob Ellefson's analysis of the SEM/EDS ToF/SIMS where the unexpected surprise was apparent gallium. (mass 69) I am wondering if the explanation is not exactly nuclear ! but instead, the lithium has been fully absorbed into the K or L shell of the Ni62, giving it high stability and the appearance of mass 69 but not the reality of a nuclear reaction. This is called a ballotechnic reaction, and has been clouded in secrecy over the years. (it does not help credibility to bring up the red mercury connection) http://en.wikipedia.org/wiki/Ballotechnics I guess it would have to be the L2p shell to hold three electrons. Thus the intense beam of SEM electrons would ionize the combined particle but not kick out the lithium from the inner shell. BTW - the intense field of the SPP would probably be capable of forcing the lithium in there. This could be magnetic bonding also. As mentioned before Li7 has a high nuclear magnetic moment and Ni is ferromagnetic. The Curie Temp is a problem with a magnetic explanation - but perhaps once formed, this coupling survives. Where is the energy gain? That is still under consideration, but the reason for saying that this could happen, is that lithium is known to absorb completely into deep shells of a few elements. Nickel is not one of them, so this may not apply. But it is a far more likely explanation than a real fusion of nickel to lithium. Jones Bob, Just so you know, your insight is not being ignored - although you may reasonably suspect it is, since there are few comments. Lots of people are trying to come to grips with this data, as preposterous as it may seem at first glance. But the main problem remains: these are very energetic reactions with extreme Coulomb barriers. Obviously, if Ni-62 is the key to the Rossi effect - which is what his patent would suggest, and it fused with Li7, then there is a putative case for gallium-69. Beyond that, there is no support for anything close to this in the literature. -Original Message- From: Robert Ellefson Given the results of the SIMS analysis from the Lugano report, particularly as detailed in this posting: http://www.mail-archive.com/vortex-l@eskimo.com/msg98596.html
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
On Wed, Oct 15, 2014 at 1:04 AM, Robert Lynn robert.gulliver.l...@gmail.com wrote: Highly doubtful. Above curie temperture of Nickel so no ferromagnetism, and powder too microscopic hot resistivity too high I assume, then, that magnetic domains and the curie temperature are not relevant. I'm thinking more along the lines of metal vapor and partially ionized nickel and iron atoms and particles, which will carry some amount of electric charge. , and AC frequency, current and number of windings too low for strong magnetic fields or significant eddy currents to form and give push via lenzs law. Again, it would seem, then, that eddy currents in the target would be relevant at the sizes we're talking about. You and others here will know about the AC frequency and number of windings, which are something I can't comment on. Eric
RE: [Vo]:Lockheed says makes breakthrough on fusion energy project - Yahoo News
The video is pretty http://www.youtube.com/watch?feature=player_embedded http://www.youtube.com/watch?feature=player_embeddedv=UlYClniDFkM v=UlYClniDFkM but sparse technical details. Looks a lot like they added magnetic coils to the Farnsworth Fusor. I want to know about the ultra-dense deuterium… From: Robert Dorr [mailto:rod...@comcast.net] Sent: Wednesday, October 15, 2014 8:38 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:Lockheed says makes breakthrough on fusion energy project - Yahoo News Makes you wonder if this is LENR related. 100 megawatts from a reactor that fits on a truck, and no radioactive waste!!! Hmmm Robert Dorr On 10/15/2014 8:19 AM, Jones Beene wrote: Lockheed says makes breakthrough on fusion energy project - Yahoo News WHOA ! http://news.yahoo.com/lockheed-says-makes-breakthrough-fusion-energy-project-123840986--finance.html Get this: “Ultra-dense deuterium, an isotope of hydrogen, is found in the earth's oceans…. I am wondering if this is simply a naïve reporter (most likely) or it this truly ultra-dense in the context of f/H? No virus found in this message. Checked by AVG - www.avg.com Version: 2014.0.4765 / Virus Database: 4040/8393 - Release Date: 10/15/14
Re: [Vo]:temperature of the resistor wire.
I wrote up my analysis of the banding : (Draft -- I'll rename it later). http://lenr.qumbu.com/rossi_hotcat_oct2014_141014a.php Short answer : we don't even know whether the bright bands line up with the wires, or the gaps between them. There are multiple explanations, which depend on the structure used to hold the wires, and on the properties of everything. Insufficient data !
RE: [Vo]:An galling alternative
Thanks Brad AFAIK the SIMS which use gallium use elemental gallium, so there should be a 71Ga signal as well, which is apparently not seen. 40% of elemental gallium is mass-71. Do they have a reference for a device which only uses 69Ga? If so, case closed. -Original Message- From: Brad Lowe [mailto:ecatbuil...@gmail.com] Sent: Wednesday, October 15, 2014 8:46 AM To: vortex-l Subject: Re: [Vo]:An galling alternative Martin Fleiishchmann Memorial Project Facebook page brought up the G69 and said it can be safely ignored as an artifact of the SIMS probe. https://www.facebook.com/MartinFleischmannMemorialProject SIMS stands for Secondary Ion Mass Spectrometry and Gallium 69 is used as the primary ionisation source for the SIMS. This is why it is seen in both the fuel and the ash signatures. Hence it should be considered irrelevant in the reports data. http://en.wikipedia.org/wiki/Secondary_ion_mass_spectrometry - Brad Lowe On Wed, Oct 15, 2014 at 8:05 AM, Jones Beene jone...@pacbell.net wrote: One more followup on Bob Ellefson's analysis of the SEM/EDS ToF/SIMS where the unexpected surprise was apparent gallium. (mass 69) I am wondering if the explanation is not exactly nuclear ! but instead, the lithium has been fully absorbed into the K or L shell of the Ni62, giving it high stability and the appearance of mass 69 but not the reality of a nuclear reaction. This is called a ballotechnic reaction, and has been clouded in secrecy over the years. (it does not help credibility to bring up the red mercury connection) http://en.wikipedia.org/wiki/Ballotechnics I guess it would have to be the L2p shell to hold three electrons. Thus the intense beam of SEM electrons would ionize the combined particle but not kick out the lithium from the inner shell. BTW - the intense field of the SPP would probably be capable of forcing the lithium in there. This could be magnetic bonding also. As mentioned before Li7 has a high nuclear magnetic moment and Ni is ferromagnetic. The Curie Temp is a problem with a magnetic explanation - but perhaps once formed, this coupling survives. Where is the energy gain? That is still under consideration, but the reason for saying that this could happen, is that lithium is known to absorb completely into deep shells of a few elements. Nickel is not one of them, so this may not apply. But it is a far more likely explanation than a real fusion of nickel to lithium. Jones Bob, Just so you know, your insight is not being ignored - although you may reasonably suspect it is, since there are few comments. Lots of people are trying to come to grips with this data, as preposterous as it may seem at first glance. But the main problem remains: these are very energetic reactions with extreme Coulomb barriers. Obviously, if Ni-62 is the key to the Rossi effect - which is what his patent would suggest, and it fused with Li7, then there is a putative case for gallium-69. Beyond that, there is no support for anything close to this in the literature. -Original Message- From: Robert Ellefson Given the results of the SIMS analysis from the Lugano report, particularly as detailed in this posting: http://www.mail-archive.com/vortex-l@eskimo.com/msg98596.html
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
I just suggested the following as the construction of the latest hotCat to MFMP: Here is a section drawing of the end of the new hotCat as I see it ( https://drive.google.com/file/d/0B5Pc25a4cOM2WXBNRjE2bDVVT1U/view?usp=sharing ). I believe IH would have used as many off-the-shelf components as possible. I think the outer 2cm tube is a custom cast tube that used a 3/4 threaded rod as the OD and something like a 5/8 rod as the interior for the cast. For replications, a standard Coorstek tube without the ridges could be used to start - they probably did this themselves before going to the custom cast tube. So, here is the Coorstek catalog of their standard cast and extruded tubes ( http://www.coorstek.com/resource-library/library/8510-1031_tubes_rods.pdf ). The fillers may be high alumina fillers like those from Vitcast: http://www.vitcas.com/insulating-castable ; I believe the large tubes on the end to be thermally insulating supports for the hot central 2 cm tube. Referencing the 4mm bore called out in the report, it means that the core tube for the hotCat is 4mm ID. This will not open to a bigger tube - they are not made that way - it is 4mm uniform all the way through. I believe the central tube is a 4mm ID cast tube with a closed end (probably the rounded one). I will go out on a limb and say that I believe the inside of this tube is probably sealed with an alpha alumina coating ( http://www.azom.com/article.aspx?ArticleID=3560 ) and fired. Then they probably created a Ni-Fe2O3 powder slurry with an organic binder such as PVA and coated the inside of this central tube with the powder. Then they fired this at about 800C with the tube open to burn out the PVA binder and sinter the Ni to the alumina tube inner wall. The heater coils are wrapped around the OD of the center tube, with the wires leads insulated with alumina beads, and the wires are cemented to this tube with something like MTI's high alumina adhesive ( http://www.mtixtl.com/1650c3000fhi-purityaluminaadhesiveforbondingandrepairinghightemperaturefurnace… http://www.mtixtl.com/1650c3000fhi-purityaluminaadhesiveforbondingandrepairinghightemperaturefurnacequart.aspx%C2%A0;). This center tube assembly is probably then fired in an oven to cure the glue which should completely coat the wire. Then I believe this tube assembly is coated with something like the Vitcast 1400 INS-H and pushed into the outer long tube and the assembly is fired again in an oven. Then the ends are sort of cast on by placing the large alumina rings over the tube ends and filling with something like the Vitcast 1200 INS as an insulating support. I think the plug for the center tube is made from a 1 cm piece of Coorstek 0.156 OD tube that has a 0.94 ID which is used to cement in a sheathed thermocouple. You would have to be very careful to pick a thermocouple type that is appropriate for up to 1500C. The thermocouple sheath is cemented into the plug with something like the MTI adhesive. If you go through the thought experiment of constructing a hotCat of your own, it will give you some insight into what the IH team may have done. Here are some of my other thoughts: - I believe the coils are 3-phase to create a moving magnetic field to circulate an interior plasma when it emerges. - I believe the active powder is sintered onto the inside of the central tube - almost like a catalytic converter. - I think what Rossi added in the beginning is primarily hydride - maybe some Ni powder was added for obfuscation. You can't take this for starting material for comparison to an ash (which they didn't get either). It is probably just a consumable + obfuscation. - I think what came out at the end of the experiment (thought to be ash) was some left over loose slag and debris - it is not representative of the active material which is sintered to the inside of the central tube. This debris cannot be used as representative of active powder ash because the real active powder is still inside the central tube. - The heater wires and the termination wires are probably different materials and the actual heater wire coils are fully encapsulated with a high alumina cement. Maybe the ends are terminated in inconel for interconnection. Bob Higgins
Re: [Vo]:Lockheed says makes breakthrough on fusion energy project - Yahoo News
Watching the video makes me a bit suspicious of the no radiation claim. High temperature fusion, magnetically bottled in a small container. Seems to me the container will eventually become radioactive. Robert Dorr On 10/15/2014 9:17 AM, Jones Beene wrote: The video is pretty http://www.youtube.com/watch?feature=player_embeddedv=UlYClniDFkM http://www.youtube.com/watch?feature=player_embeddedv=UlYClniDFkM but sparse technical details. Looks a lot like they added magnetic coils to the Farnsworth Fusor. I want to know about the ultra-dense deuterium… *From:* Robert Dorr [mailto:rod...@comcast.net] *Sent:* Wednesday, October 15, 2014 8:38 AM *To:* vortex-l@eskimo.com *Subject:* Re: [Vo]:Lockheed says makes breakthrough on fusion energy project - Yahoo News Makes you wonder if this is LENR related. 100 megawatts from a reactor that fits on a truck, and no radioactive waste!!! Hmmm Robert Dorr On 10/15/2014 8:19 AM, Jones Beene wrote: Lockheed says makes breakthrough on fusion energy project - Yahoo News WHOA ! http://news.yahoo.com/lockheed-says-makes-breakthrough-fusion-energy-project-123840986--finance.html Get this: “Ultra-dense deuterium, an isotope of hydrogen, is found in the earth's oceans…. I am wondering if this is simply a naïve reporter (most likely) or it this truly ultra-dense in the context of f/H? No virus found in this message. Checked by AVG - www.avg.com http://www.avg.com Version: 2014.0.4765 / Virus Database: 4040/8393 - Release Date: 10/15/14 No virus found in this message. Checked by AVG - www.avg.com http://www.avg.com Version: 2014.0.4765 / Virus Database: 4040/8394 - Release Date: 10/15/14
RE: [Vo]:Lockheed says makes breakthrough on fusion energy project - Yahoo News
If there are neutrons emitted, then everything becomes activated, and this reactor cannot be used on an airplane, as suggested. In the Wiki entry http://en.wikipedia.org/wiki/Aneutronic_fusion …there are two deuterium possibilities. But even those have secondary neutrons. This does not necessarily suggest LENR as they are talking plasma - and the only reason to even think it could be a hybrid - was the mention of “ultra-dense” deuterium… which is probably the product of bad technical writing. Heck, these guys were part of the Glomar explorer team, and everything else secret. Maybe they found ultra-dense deuterium 5 miles down in the Pacific and are finally letting the rest of us in on it :-) From: Robert Dorr Watching the video makes me a bit suspicious of the no radiation claim. High temperature fusion, magnetically bottled in a small container. Seems to me the container will eventually become radioactive. The video is pretty http://www.youtube.com/watch?feature=player_embedded http://www.youtube.com/watch?feature=player_embeddedv=UlYClniDFkM v=UlYClniDFkM but sparse technical details. Looks a lot like they added magnetic coils to the Farnsworth Fusor. I want to know about the ultra-dense deuterium… From: Robert Dorr Makes you wonder if this is LENR related. 100 megawatts from a reactor that fits on a truck, and no radioactive waste!!! Hmmm Robert Dorr
Re: [Vo]:Lockheed says makes breakthrough on fusion energy project - Yahoo News
They state that the device operates at many millions of degrees so it does not make sense to discuss deuterium atoms. Ions are all that exist at that temperature. Dave -Original Message- From: Jones Beene jone...@pacbell.net To: vortex-l vortex-l@eskimo.com Sent: Wed, Oct 15, 2014 12:18 pm Subject: RE: [Vo]:Lockheed says makes breakthrough on fusion energy project - Yahoo News The video is pretty http://www.youtube.com/watch?feature=player_embeddedv=UlYClniDFkM but sparse technicaldetails. Looks a lot like they added magnetic coils to the Farnsworth Fusor. I want to know about theultra-dense deuterium… From: Robert Dorr [mailto:rod...@comcast.net] Sent: Wednesday, October 15, 20148:38 AM To: vortex-l@eskimo.com Subject: Re: [Vo]:Lockheed saysmakes breakthrough on fusion energy project - Yahoo News Makes you wonder if this is LENR related. 100 megawatts from a reactor thatfits on a truck, and no radioactive waste!!! Hmmm Robert Dorr On 10/15/2014 8:19 AM, Jones Beene wrote: Lockheedsays makes breakthrough on fusion energy project - Yahoo News WHOA! http://news.yahoo.com/lockheed-says-makes-breakthrough-fusion-energy-project-123840986--finance.html Getthis: “Ultra-densedeuterium, an isotope of hydrogen, is found in the earth's oceans…. I amwondering if this is simply a naïve reporter (most likely) or it this trulyultra-dense in the context of f/H? No virus found in this message. Checked by AVG - www.avg.com Version: 2014.0.4765 / Virus Database: 4040/8393 - Release Date: 10/15/14
Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible
A calibration curve will bend down. It never bends up. this mean that temperature grow less than the power ? this mean that when you increase the power, and if temperature grows much more that before, something anomalous is happening ? Either excess heat, or some external blanket effect (increase of thermal resistance)... but convection does not diminish with heat? did I undertand well? 2014-10-14 22:09 GMT+02:00 Jed Rothwell jedrothw...@gmail.com: Alain Sepeda alain.sep...@gmail.com wrote: - is there a simple way , with minimal assumption, to be sure that the COP1 Look at the color. If it is dull red, it may be around 750°C which is where you would expect it to be in a straight line extrapolation calibration up to 800 W. If it is white it has to be around 1300°C, which is far higher than the calibration indicates it should be. A calibration curve will bend down. It never bends up. McKubre pointed this out: On page 7 of the report the authors state: “Subsequent calculation proved that increasing the input by roughly 100 watts had caused an increase of about 700 watts in power emitted.” This is interesting. The shape of the output vs. input power curve is observed (or implied) to strongly curve upwards in a manner completely inconsistent with the Stefan-Boltzmann law for radiative heat loss. It is also inconsistent with simple convective heat transfer but several issues need to be addressed before we can claim this as a qualitative or even “semi-quantitative” measure of excess heat production . . . Note that incandescent colors are similar for all materials. - Jed
Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible
You have a good understanding in my opinion. There is no doubt that energy is being generated within the core. Dave -Original Message- From: Alain Sepeda alain.sep...@gmail.com To: Vortex List vortex-l@eskimo.com Sent: Wed, Oct 15, 2014 12:59 pm Subject: Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible A calibration curve will bend down. It never bends up. this mean that temperature grow less than the power ? this mean that when you increase the power, and if temperature grows much more that before, something anomalous is happening ? Either excess heat, or some external blanket effect (increase of thermal resistance)... but convection does not diminish with heat? did I undertand well? 2014-10-14 22:09 GMT+02:00 Jed Rothwell jedrothw...@gmail.com: Alain Sepeda alain.sep...@gmail.com wrote: is there a simple way , with minimal assumption, to be sure that the COP1 Look at the color. If it is dull red, it may be around 750°C which is where you would expect it to be in a straight line extrapolation calibration up to 800 W. If it is white it has to be around 1300°C, which is far higher than the calibration indicates it should be. A calibration curve will bend down. It never bends up. McKubre pointed this out: On page 7 of the report the authors state: “Subsequent calculation proved that increasing the input by roughly 100 watts had caused an increase of about 700 watts in power emitted.” This is interesting. The shape of the output vs. input power curve is observed (or implied) to strongly curve upwards in a manner completely inconsistent with the Stefan-Boltzmann law for radiative heat loss. It is also inconsistent with simple convective heat transfer but several issues need to be addressed before we can claim this as a qualitative or even “semi-quantitative” measure of excess heat production . . . Note that incandescent colors are similar for all materials. - Jed
Re: [Vo]:temperature of the resistor wire.
New version with embedded wires. http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php Here I've also assumed that the wires are a simple single strand, rather than the spiral form used in the earlier tests, and are in good thermal contact with the Alumina.
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
One detail is how does the e-cat work when in after-death mode (SSM). One mystery for me is the powder not to melt, hotter and the around... I feel E-cat are very well engineered... we can be surpised like physicists were by felischmanpons calorimeter... 2014-10-15 18:24 GMT+02:00 Bob Higgins rj.bob.higg...@gmail.com: I just suggested the following as the construction of the latest hotCat to MFMP: Here is a section drawing of the end of the new hotCat as I see it ( https://drive.google.com/file/d/0B5Pc25a4cOM2WXBNRjE2bDVVT1U/view?usp=sharing ). I believe IH would have used as many off-the-shelf components as possible. I think the outer 2cm tube is a custom cast tube that used a 3/4 threaded rod as the OD and something like a 5/8 rod as the interior for the cast. For replications, a standard Coorstek tube without the ridges could be used to start - they probably did this themselves before going to the custom cast tube. So, here is the Coorstek catalog of their standard cast and extruded tubes ( http://www.coorstek.com/resource-library/library/8510-1031_tubes_rods.pdf ). The fillers may be high alumina fillers like those from Vitcast: http://www.vitcas.com/insulating-castable ; I believe the large tubes on the end to be thermally insulating supports for the hot central 2 cm tube. Referencing the 4mm bore called out in the report, it means that the core tube for the hotCat is 4mm ID. This will not open to a bigger tube - they are not made that way - it is 4mm uniform all the way through. I believe the central tube is a 4mm ID cast tube with a closed end (probably the rounded one). I will go out on a limb and say that I believe the inside of this tube is probably sealed with an alpha alumina coating ( http://www.azom.com/article.aspx?ArticleID=3560 ) and fired. Then they probably created a Ni-Fe2O3 powder slurry with an organic binder such as PVA and coated the inside of this central tube with the powder. Then they fired this at about 800C with the tube open to burn out the PVA binder and sinter the Ni to the alumina tube inner wall. The heater coils are wrapped around the OD of the center tube, with the wires leads insulated with alumina beads, and the wires are cemented to this tube with something like MTI's high alumina adhesive ( http://www.mtixtl.com/1650c3000fhi-purityaluminaadhesiveforbondingandrepairinghightemperaturefurnace… http://www.mtixtl.com/1650c3000fhi-purityaluminaadhesiveforbondingandrepairinghightemperaturefurnacequart.aspx%C2%A0;). This center tube assembly is probably then fired in an oven to cure the glue which should completely coat the wire. Then I believe this tube assembly is coated with something like the Vitcast 1400 INS-H and pushed into the outer long tube and the assembly is fired again in an oven. Then the ends are sort of cast on by placing the large alumina rings over the tube ends and filling with something like the Vitcast 1200 INS as an insulating support. I think the plug for the center tube is made from a 1 cm piece of Coorstek 0.156 OD tube that has a 0.94 ID which is used to cement in a sheathed thermocouple. You would have to be very careful to pick a thermocouple type that is appropriate for up to 1500C. The thermocouple sheath is cemented into the plug with something like the MTI adhesive. If you go through the thought experiment of constructing a hotCat of your own, it will give you some insight into what the IH team may have done. Here are some of my other thoughts: - I believe the coils are 3-phase to create a moving magnetic field to circulate an interior plasma when it emerges. - I believe the active powder is sintered onto the inside of the central tube - almost like a catalytic converter. - I think what Rossi added in the beginning is primarily hydride - maybe some Ni powder was added for obfuscation. You can't take this for starting material for comparison to an ash (which they didn't get either). It is probably just a consumable + obfuscation. - I think what came out at the end of the experiment (thought to be ash) was some left over loose slag and debris - it is not representative of the active material which is sintered to the inside of the central tube. This debris cannot be used as representative of active powder ash because the real active powder is still inside the central tube. - The heater wires and the termination wires are probably different materials and the actual heater wire coils are fully encapsulated with a high alumina cement. Maybe the ends are terminated in inconel for interconnection. Bob Higgins
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
I now think the general structure is more like this : http://lenr.qumbu.com/web_hotcat_pics/hotcat_141014_1_010.png See http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php in my other thread for details - Original Message - From: Bob Higgins rj.bob.higg...@gmail.com To: vortex-l@eskimo.com Sent: Wednesday, October 15, 2014 9:24:07 AM Subject: Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo I just suggested the following as the construction of the latest hotCat to MFMP: Here is a section drawing of the end of the new hotCat as I see it ( https://drive.google.com/file/d/0B5Pc25a4cOM2WXBNRjE2bDVVT1U/view?usp=sharing ).
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
Alan, Alumina fires at an extremely hot temperature, 1600-1700C for 96%, and for high purity aluminas, the firing temperature may be over 1800C. Firing must also occur in an oxygen (air) environment or the alumina will lose some of it oxygen during firing. Embedding any metals in this and having them come out as metals is seriously problematic. Most heaters begin with a pre-fired ceramic form and then apply the heater coil and back fill the heater wire volume with a lower temperature glass-ceramic and fire it again at a lower temperature. With the right filler, it can continue to sinter as the heater is operated. The outer hot alumina tube looks like an early prototype. Its finish is poor and it is not fired perfectly straight. There is a problem with having appropriate setters in a kiln for something like this - setters that won't ruin the surface or let it sag at the high temperature at which it is fired. This doesn't appear to have been entirely worked out in what is shown in the Lugano demo. I suspect the outer tube is just that - a tube that was plain on the inside and having 3/4 rod threads cast into the outside. On Wed, Oct 15, 2014 at 11:16 AM, Alan Fletcher a...@well.com wrote: I now think the general structure is more like this : http://lenr.qumbu.com/web_hotcat_pics/hotcat_141014_1_010.png See http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php in my other thread for details -- *From: *Bob Higgins rj.bob.higg...@gmail.com *To: *vortex-l@eskimo.com *Sent: *Wednesday, October 15, 2014 9:24:07 AM *Subject: *Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo I just suggested the following as the construction of the latest hotCat to MFMP: Here is a section drawing of the end of the new hotCat as I see it ( https://drive.google.com/file/d/0B5Pc25a4cOM2WXBNRjE2bDVVT1U/view?usp=sharing ).
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
You Alan have done an amazing job sleuthing out the details of this thing. I suppose you are right, although I cannot tell. If you are right it is a great job and if you are wrong you have a vivid imagination! - Jed
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
Me too, good job. Tube in a tube reminds of the model rockets I used to build. Fin supports between tubes might explain the wider dark band seen as a spiral?. Do you think a lot of the heat might be discharged in a space between tubes and out the ends? Or are the ends completely sealed? On Wednesday, October 15, 2014, Jed Rothwell jedrothw...@gmail.com wrote: You Alan have done an amazing job sleuthing out the details of this thing. I suppose you are right, although I cannot tell. If you are right it is a great job and if you are wrong you have a vivid imagination! - Jed
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
I believe the ends, between the central tube and the outer 2 cm hot tube, are filled in with a refractory cement. In fact, that whole space, and maybe not the ends could be filled in to help improve the thermal conductivity. Most importantly it should keep out the air that could oxidize the heater wire. It would not have to be hydrogen tight or anything - just keep out the air. If you look at the photo, the ends don't look beautifully finished, nor need they be. The central tube is probably a cast tube with one end having a cast in closure. The open end, as the report says, is plugged with an alumina tube having in its center a thermocouple. When the plug is glued into place (available nearly line-to-line fit from Coorstek), it will seal the end of the central tube. On Wed, Oct 15, 2014 at 12:08 PM, ChemE Stewart cheme...@gmail.com wrote: Me too, good job. Tube in a tube reminds of the model rockets I used to build. Fin supports between tubes might explain the wider dark band seen as a spiral?. Do you think a lot of the heat might be discharged in a space between tubes and out the ends? Or are the ends completely sealed? On Wednesday, October 15, 2014, Jed Rothwell jedrothw...@gmail.com wrote: You Alan have done an amazing job sleuthing out the details of this thing. I suppose you are right, although I cannot tell. If you are right it is a great job and if you are wrong you have a vivid imagination! - Jed
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
From: ChemE Stewart cheme...@gmail.com Sent: Wednesday, October 15, 2014 11:08:42 AM Me too, good job. Tube in a tube reminds of the model rockets I used to build. Fin supports between tubes might explain the wider dark band seen as a spiral?. Do you think a lot of the heat might be discharged in a space between tubes and out the ends? Or are the ends completely sealed? The fins are too fine to be a source of the bands. And the ends are completely sealed, so the temperature and heat-loss will taper off at the ends. But your .. and your, and your ... guess on the structure and the banding is as good as mine. Very frustrating.
[Vo]:Today's Confront Journal
Dear Friends, The second issue of this Confront journal: starts with a really bad shock http://egooutpeters.blogspot.ro/2014/10/lenr-confront-journal-october-15-2014.html but ends in a realistic- optimistic note. Peter -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
[Vo]:Gibbsite
This is not Mark Gibbs' site but an aluminum mineral which may be relevant to this discussion. http://en.wikipedia.org/wiki/Gibbsite Gibbsite is Al(OH)3 is one of the minerals found in bauxite. Unlike other hydroxides, it is stable at high temperature. We are told that in the Rossi reactor hydrogen is admitted in the form of LiAlH4. In that case, about 12% of the weight of that compound is hydrogen. We are told elsewhere that Lithium accounts for 1 percent of the total fuel mix. The density of Al is about 2.7 g/cc and lithium is .53 g/cc. or about 5 times less than the Al so that in the total mix, here is what we have to work with, roughly. Lithium - 0.01 grams Aluminum 0.05 grams Hydrogen 0.006 grams We are also told that the fuel powder is put into a cavity filled with air and not evacuated, so it is clear that as soon as the 6 milligrams of hydrogen is released from the carrier, it will oxidize to steam, and then as the temperature rises, and the steam pyrolizes at 1200 C, we will end-up preferentially with a stable hydroxide. That would be Gibbsite, perhaps. Is there a better scenario? At any rate this does not seem to be a hydrogen reactor. Since no radioactive debris is seen in the ash, it may not be a nuclear reactor either, but for certain 6 milligrams of hydrogen is unlikely to provide over a MWhr of heat. If every single atom converted in the Alain's version of the hydrotron reaction we would be left out in the cold by a factor of 10 times too little heat. Jones attachment: winmail.dat
Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible
I agree David. You can verify this by looking at the data for both the caps and the E-Cat body. The caps are not incandescent, so there does not appear to be any transparency issue there. The Delta T/Watt is nearly the same despite an increase in input power of ~100W. You would expect it to be significantly lower. On Wed, Oct 15, 2014 at 12:08 PM, David Roberson dlrober...@aol.com wrote: You have a good understanding in my opinion. There is no doubt that energy is being generated within the core. Dave -Original Message- From: Alain Sepeda alain.sep...@gmail.com To: Vortex List vortex-l@eskimo.com Sent: Wed, Oct 15, 2014 12:59 pm Subject: Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible A calibration curve will bend down. It never bends up. this mean that temperature grow less than the power ? this mean that when you increase the power, and if temperature grows much more that before, something anomalous is happening ? Either excess heat, or some external blanket effect (increase of thermal resistance)... but convection does not diminish with heat? did I undertand well? 2014-10-14 22:09 GMT+02:00 Jed Rothwell jedrothw...@gmail.com: Alain Sepeda alain.sep...@gmail.com wrote: - is there a simple way , with minimal assumption, to be sure that the COP1 Look at the color. If it is dull red, it may be around 750°C which is where you would expect it to be in a straight line extrapolation calibration up to 800 W. If it is white it has to be around 1300°C, which is far higher than the calibration indicates it should be. A calibration curve will bend down. It never bends up. McKubre pointed this out: On page 7 of the report the authors state: “Subsequent calculation proved that increasing the input by roughly 100 watts had caused an increase of about 700 watts in power emitted.” This is interesting. The shape of the output vs. input power curve is observed (or implied) to strongly curve upwards in a manner completely inconsistent with the Stefan-Boltzmann law for radiative heat loss. It is also inconsistent with simple convective heat transfer but several issues need to be addressed before we can claim this as a qualitative or even “semi-quantitative” measure of excess heat production . . . Note that incandescent colors are similar for all materials. - Jed
Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible
Alain Sepeda alain.sep...@gmail.com wrote: A calibration curve will bend down. It never bends up. this mean that temperature grow less than the power ? Right. this mean that when you increase the power, and if temperature grows much more that before, something anomalous is happening ? Exactly. However, there are some open questions about the IR camera and the color. So we cannot be sure the temperature really is getting higher. If the authors can confirm that the temperature shown in the internal thermocouple is also rising, and that the colors are transitioning to ones indicating a higher temperature, then yes, we can be sure there is an anomaly. This is true even though there is only one calibration point. That is inadequate, but it does not preclude certainty. Either excess heat, or some external blanket effect (increase of thermal resistance)... but convection does not diminish with heat? The device is open to air in a reasonably stable environment. I do not think that a change in convection on this scale is possible. - Jed
[Vo]:the true source of energy
If the nickel particles are the ultimate source of 3.5X over-unity heat in the Rossi reactor, it is paradoxical and against common sense that 900 constantly applied watts of heat energy is required to keep the nickel particles active. Furthermore, this COP value is far under what the Hot-cat can do. The real COP is somewhere north of 6. At an external temperature that is hovering at 1400C for days, there is no room for differences in temperature within the guts of the reactor itself. The answer must be that the nickel particles are not the main source of the heat in the reactor. They need lots of heat stimulation to function and they are not getting that heat from over-unity heat production. The isotopic tests confirm that the nickel particles are pure nickel. These particles must melt at 1450C. The conclusion that logic forces us to arrive at must be that there is another place where all that over unity heat is coming from. These particles cannot be producing (900 watts) (3.5) = 3150 watts of output power.
RE: [Vo]:Gibbsite
According to Albert, adding 25 MWhrs (90 gigajoules) of any form of energy to an object increases its mass by 1 milligram, even though no matter has been added... and vice-versa. In the case of the AR glow-tube, where 1.5 MWhr has been reported, the equivalent mass loss would only be about 60 micrograms. This could come from anywhere, although fringe fizzix suggests it should come from the single gram of reactant. But this is new territory. For instance, can SPP or even electrons be actually consumed to provide mass-energy? (say by dark matter or positrons without the gamma signature)? Can SPP act as a gateway to the Dirac field? 1.1*10^21 electrons weigh in 1 microgram. 6.25*10^18 electrons per second = 1 amp 30 days is 2.6*10^7 seconds... ! What's the point? The point is that if there is a real paradigm shift here, then it does not necessarily have to nuclear fusion as the source, not even involve the nucleus. We should be thinking outside the box ... err... make that outside the glow tube. A 'sink' for electrons could be imagined. This is not Mark Gibbs' site but an aluminum mineral which may be relevant to this discussion. http://en.wikipedia.org/wiki/Gibbsite Gibbsite is Al(OH)3 is one of the minerals found in bauxite. Unlike other hydroxides, it is stable at high temperature. We are told that in the Rossi reactor hydrogen is admitted in the form of LiAlH4. In that case, about 12% of the weight of that compound is hydrogen. We are told elsewhere that Lithium accounts for 1 percent of the total fuel mix. The density of Al is about 2.7 g/cc and lithium is .53 g/cc. or about 5 times less than the Al so that in the total mix, here is what we have to work with, roughly. Lithium - 0.01 grams Aluminum 0.05 grams Hydrogen 0.006 grams We are also told that the fuel powder is put into a cavity filled with air and not evacuated, so it is clear that as soon as the 6 milligrams of hydrogen is released from the carrier, it will oxidize to steam, and then as the temperature rises, and the steam pyrolizes at 1200 C, we will end-up preferentially with a stable hydroxide. That would be Gibbsite, perhaps. Is there a better scenario? At any rate this does not seem to be a hydrogen reactor. Since no radioactive debris is seen in the ash, it may not be a nuclear reactor either, but for certain 6 milligrams of hydrogen is unlikely to provide over a MWhr of heat. If every single atom converted in the Alain's version of the hydrotron reaction we would be left out in the cold by a factor of 10 times too little heat. Jones attachment: winmail.dat
Re: [Vo]:Gibbsite
Yeah might be pulling protons out of the Dirac sea to combine with electrons, or something like that On Wednesday, October 15, 2014, Jones Beene jone...@pacbell.net wrote: According to Albert, adding 25 MWhrs (90 gigajoules) of any form of energy to an object increases its mass by 1 milligram, even though no matter has been added... and vice-versa. In the case of the AR glow-tube, where 1.5 MWhr has been reported, the equivalent mass loss would only be about 60 micrograms. This could come from anywhere, although fringe fizzix suggests it should come from the single gram of reactant. But this is new territory. For instance, can SPP or even electrons be actually consumed to provide mass-energy? (say by dark matter or positrons without the gamma signature)? Can SPP act as a gateway to the Dirac field? 1.1*10^21 electrons weigh in 1 microgram. 6.25*10^18 electrons per second = 1 amp 30 days is 2.6*10^7 seconds... ! What's the point? The point is that if there is a real paradigm shift here, then it does not necessarily have to nuclear fusion as the source, not even involve the nucleus. We should be thinking outside the box ... err... make that outside the glow tube. A 'sink' for electrons could be imagined. This is not Mark Gibbs' site but an aluminum mineral which may be relevant to this discussion. http://en.wikipedia.org/wiki/Gibbsite Gibbsite is Al(OH)3 is one of the minerals found in bauxite. Unlike other hydroxides, it is stable at high temperature. We are told that in the Rossi reactor hydrogen is admitted in the form of LiAlH4. In that case, about 12% of the weight of that compound is hydrogen. We are told elsewhere that Lithium accounts for 1 percent of the total fuel mix. The density of Al is about 2.7 g/cc and lithium is .53 g/cc. or about 5 times less than the Al so that in the total mix, here is what we have to work with, roughly. Lithium - 0.01 grams Aluminum 0.05 grams Hydrogen 0.006 grams We are also told that the fuel powder is put into a cavity filled with air and not evacuated, so it is clear that as soon as the 6 milligrams of hydrogen is released from the carrier, it will oxidize to steam, and then as the temperature rises, and the steam pyrolizes at 1200 C, we will end-up preferentially with a stable hydroxide. That would be Gibbsite, perhaps. Is there a better scenario? At any rate this does not seem to be a hydrogen reactor. Since no radioactive debris is seen in the ash, it may not be a nuclear reactor either, but for certain 6 milligrams of hydrogen is unlikely to provide over a MWhr of heat. If every single atom converted in the Alain's version of the hydrotron reaction we would be left out in the cold by a factor of 10 times too little heat. Jones
Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible
Good point Jack. One item of interest is that my simulation shows that one can refer to two different COPs when the core generates heating power that are not so evident with an inactive core. The first is an incremental COP that is calculated by taking the derivative of the output power at a particular input power level. This is the what had the scientists concerned when they calculated the relatively large output power delta of 700 watts with an input delta of 100 watts. The large slope is caused by the non linear nature of both the power generation mechanism as well as the forth order non linear relationship between output radiated power and surface temperature. I consider this effect as demonstrating how the effective thermal resistance of the device is multiplied by the positive feedback that is associated with its operation. The temperature rises or falls much faster than expected by an inactive device when subjected to an incremental input power adjustment. The simulation shows that the maximum slope occurs at the temperature where the internally generated power comes closest to being equal to the power escaping. The device must be designed so that the generated power (as indicated by temperature being a time dependent parameter) never quite reaches the level of the power that is radiated, convected, and conducted away from the core at any time. This should be viewed as a requirement for stable operation under steady input power conditions. Operation outside of this region will result in a latching of output power even when input is removed. Special requirements and restriction to the input waveform can enable one to operate a stable system under careful constraints such as in SSM mode with PWM drive. The second type of COP measurement is our more common way of expressing device performance. This is calculated by taking the total power output and dividing by the total power input. I am assuming steady state conditions with constant input drive power. The number determined by this operation will always be smaller than that of the incremental form of COP mentioned earlier. This is due to the very non linear nature of the relationship between output power and input power when the core power generation is of a significant quantity. This demonstrated behavior solidifies in my opinion the proof that the core is generating a major amount of power during the test. Dave -Original Message- From: Jack Cole jcol...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Oct 15, 2014 3:19 pm Subject: Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible I agree David. You can verify this by looking at the data for both the caps and the E-Cat body. The caps are not incandescent, so there does not appear to be any transparency issue there. The Delta T/Watt is nearly the same despite an increase in input power of ~100W. You would expect it to be significantly lower. On Wed, Oct 15, 2014 at 12:08 PM, David Roberson dlrober...@aol.com wrote: You have a good understanding in my opinion. There is no doubt that energy is being generated within the core. Dave -Original Message- From: Alain Sepeda alain.sep...@gmail.com To: Vortex List vortex-l@eskimo.com Sent: Wed, Oct 15, 2014 12:59 pm Subject: Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible A calibration curve will bend down. It never bends up. this mean that temperature grow less than the power ? this mean that when you increase the power, and if temperature grows much more that before, something anomalous is happening ? Either excess heat, or some external blanket effect (increase of thermal resistance)... but convection does not diminish with heat? did I undertand well? 2014-10-14 22:09 GMT+02:00 Jed Rothwell jedrothw...@gmail.com: Alain Sepeda alain.sep...@gmail.com wrote: is there a simple way , with minimal assumption, to be sure that the COP1 Look at the color. If it is dull red, it may be around 750°C which is where you would expect it to be in a straight line extrapolation calibration up to 800 W. If it is white it has to be around 1300°C, which is far higher than the calibration indicates it should be. A calibration curve will bend down. It never bends up. McKubre pointed this out: On page 7 of the report the authors state: “Subsequent calculation proved that increasing the input by roughly 100 watts had caused an increase of about 700 watts in power emitted.” This is interesting. The shape of the output vs. input power curve is observed (or implied) to strongly curve upwards in a manner completely inconsistent with the Stefan-Boltzmann law for radiative heat loss. It is also inconsistent with simple convective heat transfer but several issues need to be addressed before we can claim this as a qualitative or even “semi-quantitative” measure of excess heat production . . . Note that incandescent
Re: [Vo]:the true source of energy
From: Axil Axil janap...@gmail.com Sent: Wednesday, October 15, 2014 12:47:05 PM The conclusion that logic forces us to arrive at must be that there is another place where all that over unity heat is coming from. These particles cannot be producing (900 watts) (3.5) = 3150 watts of output power. I have always argued that the thermalization may happen away from the particles.
Re: [Vo]:the true source of energy
Perhaps these are the sources. If you look at the literature (I don't remember where, but I think it's a paper o presentation from Kim), it's pointed out that 24MeV particles from within old experiments with Pd look like conic craters with 4um in diameter and similar depth, if my memory is correct. Compare with the nickel powder. If such explosion occurs, a grain of powder will explode. There will be flying hot debris, which will glue to other to other grains. If there are a few thousands of explosion per grain/s. There will be like a dust chaos of debris flying around in the powder. The movement from the dust will heat the air and send most heat out by convection. -- Daniel Rocha - RJ danieldi...@gmail.com
Re: [Vo]:the true source of energy
We may eventually come to the conclusion that the nickel can produce power even in the molten form. That seems to be what is implied. Is there reason to assume that molten nickel can not work? A higher temperature might enhance the process that is not well understood at the moment. I have no opinion about this matter. Dave -Original Message- From: Axil Axil janap...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Oct 15, 2014 3:47 pm Subject: [Vo]:the true source of energy If the nickel particles are the ultimate source of 3.5Xover-unity heat in the Rossi reactor, it is paradoxical and against commonsense that 900 constantly applied watts of heat energy is required to keep the nickel particlesactive. Furthermore, this COP value is far under what the Hot-catcan do. The real COP is somewhere north of 6. At an external temperature that is hovering at 1400C fordays, there is no room for differences in temperature within the guts of thereactor itself. The answer must be that the nickel particles are not themain source of the heat in the reactor. They need lots of heat stimulation tofunction and they are not getting that heat from over-unity heat production. The isotopic tests confirm that the nickel particles arepure nickel. These particles must melt at 1450C. The conclusion that logic forces us to arrive at must bethat there is another place where all that over unity heat is coming from. These particles cannot be producing (900watts) (3.5) = 3150 watts of output power.
Re: [Vo]:the true source of energy
In reply to David Roberson's message of Wed, 15 Oct 2014 16:16:29 -0400: Hi, [snip] We may eventually come to the conclusion that the nickel can produce power even in the molten form. That seems to be what is implied. Is there reason to assume that molten nickel can not work? A higher temperature might enhance the process that is not well understood at the moment. I have no opinion about this matter. Dave If it's a neutron exchange reaction, it shouldn't matter what form the atoms are in. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:the true source of energy
I believe that Ni particles will not work once melted - just intuition, because I don't buy the neutron stripping yet. If we take a leap of faith and say that the central reactor core alumina tube is coated with particles sintered to its inside (like a catalytic converter for example), we don't know anymore whether the active material is Ni. The active material could be a more refractory metal, for example, zirconium, perhaps processed with an additional catalyst as the Ni was. The only clue we have seems to be the very high operating temperatures, suggesting that it is not Ni. Any Ni that Rossi added in the beginning could have been a startup mouse or just simple obfuscation. On Wed, Oct 15, 2014 at 1:47 PM, Axil Axil janap...@gmail.com wrote: If the nickel particles are the ultimate source of 3.5X over-unity heat in the Rossi reactor, it is paradoxical and against common sense that 900 constantly applied watts of heat energy is required to keep the nickel particles active. Furthermore, this COP value is far under what the Hot-cat can do. The real COP is somewhere north of 6. At an external temperature that is hovering at 1400C for days, there is no room for differences in temperature within the guts of the reactor itself. The answer must be that the nickel particles are not the main source of the heat in the reactor. They need lots of heat stimulation to function and they are not getting that heat from over-unity heat production. The isotopic tests confirm that the nickel particles are pure nickel. These particles must melt at 1450C. The conclusion that logic forces us to arrive at must be that there is another place where all that over unity heat is coming from. These particles cannot be producing (900 watts) (3.5) = 3150 watts of output power.
Re: [Vo]:Gibbsite
In reply to Jones Beene's message of Wed, 15 Oct 2014 12:13:20 -0700: Hi, [snip] If 0.006 gm of H has to supply 1.5 MWh, then each atom needs to supply 9.3 MeV. This is not out of the question, if the right reaction is found. However if it only acts as a catalyst for neutron transfer reactions, then nowhere near that amount would be needed. Lithium - 0.01 grams Aluminum 0.05 grams Hydrogen 0.006 grams We are also told that the fuel powder is put into a cavity filled with air and not evacuated, so it is clear that as soon as the 6 milligrams of hydrogen is released from the carrier, it will oxidize to steam, and then as the temperature rises, and the steam pyrolizes at 1200 C, we will end-up preferentially with a stable hydroxide. That would be Gibbsite, perhaps. Is there a better scenario? At any rate this does not seem to be a hydrogen reactor. Since no radioactive debris is seen in the ash, it may not be a nuclear reactor either, but for certain 6 milligrams of hydrogen is unlikely to provide over a MWhr of heat. If every single atom converted in the Alain's version of the hydrotron reaction we would be left out in the cold by a factor of 10 times too little heat. Jones Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:An galling alternative
In reply to Jones Beene's message of Wed, 15 Oct 2014 08:05:31 -0700: Hi, [snip] One more followup on Bob Ellefson's analysis of the SEM/EDS ToF/SIMS where the unexpected surprise was apparent gallium. (mass 69) I am wondering if the explanation is not exactly nuclear ! but instead, the lithium has been fully absorbed into the K or L shell of the Ni62, giving it high stability and the appearance of mass 69 but not the reality of a nuclear reaction. This is called a ballotechnic reaction, and has been clouded in secrecy over the years. (it does not help credibility to bring up the red mercury connection) http://en.wikipedia.org/wiki/Ballotechnics I guess it would have to be the L2p shell to hold three electrons. Thus the intense beam of SEM electrons would ionize the combined particle but not kick out the lithium from the inner shell. BTW - the intense field of the SPP would probably be capable of forcing the lithium in there. This could be magnetic bonding also. As mentioned before Li7 has a high nuclear magnetic moment and Ni is ferromagnetic. The Curie Temp is a problem with a magnetic explanation - but perhaps once formed, this coupling survives. Where is the energy gain? That is still under consideration, but the reason for saying that this could happen, is that lithium is known to absorb completely into deep shells of a few elements. Nickel is not one of them, so this may not apply. But it is a far more likely explanation than a real fusion of nickel to lithium. Jones ...but once there, a neutron transfer reaction would be more likely. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
Re: [Vo]:Gibbsite
On Wed, Oct 15, 2014 at 4:03 PM, Jones Beene jone...@pacbell.net wrote: What's the point? The point is that if there is a real paradigm shift here, then it does not necessarily have to nuclear fusion as the source, not even involve the nucleus. We should be thinking outside the box ... err... make that outside the glow tube. A 'sink' for electrons could be imagined. What are electrons after all? I tend to think of them as a vortex in the sea of negative energy.
[Vo]:Lockheed- Fusion website-Gone????
Greetings vortex-L, Went here after seeing it earlier..now gone. Temporary??? http://www.lockheedmartin.com/us/news/press-releases/2014/october/141015ae_lockheed-martin-pursuing-compact-nucelar-fusion.html Ad astra, Ron Kita, Chiralex
Re: [Vo]:Lockheed- Fusion website-Gone????
Still Listed here: http://www.lockheedmartin.com/us/news/press-releases.html On Thu, Oct 16, 2014 at 10:02 AM, Ron Kita chiralex.k...@gmail.com wrote: Greetings vortex-L, Went here after seeing it earlier..now gone. Temporary??? http://www.lockheedmartin.com/us/news/press-releases/2014/october/141015ae_lockheed-martin-pursuing-compact-nucelar-fusion.html Ad astra, Ron Kita, Chiralex -- Patrick www.tRacePerfect.com The daily puzzle everyone can finish but not everyone can perfect! The quickest puzzle ever!
Re: [Vo]:Lockheed- Fusion website-Gone????
You have a typo in your link. nuCELar ... cheers patrick On Thu, Oct 16, 2014 at 10:17 AM, Patrick Ellul ellulpatr...@gmail.com wrote: Still Listed here: http://www.lockheedmartin.com/us/news/press-releases.html On Thu, Oct 16, 2014 at 10:02 AM, Ron Kita chiralex.k...@gmail.com wrote: Greetings vortex-L, Went here after seeing it earlier..now gone. Temporary??? http://www.lockheedmartin.com/us/news/press-releases/2014/october/141015ae_lockheed-martin-pursuing-compact-nucelar-fusion.html Ad astra, Ron Kita, Chiralex -- Patrick www.tRacePerfect.com The daily puzzle everyone can finish but not everyone can perfect! The quickest puzzle ever! -- Patrick www.tRacePerfect.com The daily puzzle everyone can finish but not everyone can perfect! The quickest puzzle ever!
RE: [Vo]:Testing fuels without a reactor
Ok, after escalating from random-questions-embedded-in-a-long-posting to open-conjecture-with-direct-questions and getting nowhere, I went with the shakily-supported-but-loudly-declared-claim-of-fact route so popular on the internet, and that elicited a couple of responses from kind and knowledgeable researchers (Merci!). Although I did not resolve all of the questions which formed the basis of my conjecture about a potential signal hidden within the noise of the peak-69 artifacts, I am able to infer and presume the following: 1) Although not directly specified, the sputter-cleaning ion source was most likely also isotopically-pure Ga-69. This and many other models of ToF-SIMS analyzers are apparently capable of performing sputter-cleaning and SIMS-scanning with a variety of ion sources, using either the same or different sources, notably including cesium, argon, oxygen, and gold. In this case, given only a single ion source specified, common practice implies that they are both performed with Ga69. 2) The high variability of residual Ga69 in different post-sputtering sample spectra (particularly figure 11b) is apparently within the normal variance range for different sample substances. So, if there is something interesting to be found regarding a mass-69 species, it will need to be discovered using an analytic method that does not employ Ga-69 such as this one does. I hope that other analytic results will be published soon! Unless and until then, I'll just presume this to be artifact. -Bob From: Robert Ellefson Sent: Tuesday, October 14, 2014 5:29 PM Subject: [Vo]:Testing fuels without a reactor Given the results of the SIMS analysis from the Lugano report, particularly as detailed in this posting: http://www.mail-archive.com/vortex-l@eskimo.com/msg98596.html I believe that it is possible to evaluate the nuclear activity of candidate fuel samples simply by sputter-cleaning them as part of a ToF-SIMS analysis. If researchers with access to such analytic equipment were willing to run the experiment, I believe that a successful replication of Rossi's reaction can be observed occurring with before-and-after spectra of the fuel. So, skip the reactors, start evaluating powders in the SEM itself!
Re: [Vo]:temperature of the resistor wire.
the resistor wire expands with respect to the alumina as it heats up, breaking any bonding contact, or lifting the wire of the inner alumina tube in more and more places and leading to less and less conductive contact - prompting the wire to heat up as more as more of the energy it transmits to the reactor must be via radiation and conduction through gas rather than contact-conduction. This is the likely what makes it appear that there is a gain above 1. On 16 October 2014 01:13, Alan Fletcher a...@well.com wrote: New version with embedded wires. http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php Here I've also assumed that the wires are a simple single strand, rather than the spiral form used in the earlier tests, and are in good thermal contact with the Alumina.
Re: [Vo]:temperature of the resistor wire.
Since they are measuring the input energy to the wire that makes no sense On Wednesday, October 15, 2014, Robert Lynn robert.gulliver.l...@gmail.com wrote: the resistor wire expands with respect to the alumina as it heats up, breaking any bonding contact, or lifting the wire of the inner alumina tube in more and more places and leading to less and less conductive contact - prompting the wire to heat up as more as more of the energy it transmits to the reactor must be via radiation and conduction through gas rather than contact-conduction. This is the likely what makes it appear that there is a gain above 1. On 16 October 2014 01:13, Alan Fletcher a...@well.com javascript:_e(%7B%7D,'cvml','a...@well.com'); wrote: New version with embedded wires. http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php Here I've also assumed that the wires are a simple single strand, rather than the spiral form used in the earlier tests, and are in good thermal contact with the Alumina.
Re: [Vo]:temperature of the resistor wire.
Do you have exceptional hearing?
Re: [Vo]:temperature of the resistor wire.
It has nothing to do with measuring of the wire power, except that as the wire heats up, increase the thermal resistance of the heat flow between wire and reactor body (by reducing number of points of physical contact) and of course the wire temperature will go up (given same input power) - I suggest you think a bit longer about it. On 16 October 2014 08:33, ChemE Stewart cheme...@gmail.com wrote: Since they are measuring the input energy to the wire that makes no sense On Wednesday, October 15, 2014, Robert Lynn robert.gulliver.l...@gmail.com wrote: the resistor wire expands with respect to the alumina as it heats up, breaking any bonding contact, or lifting the wire of the inner alumina tube in more and more places and leading to less and less conductive contact - prompting the wire to heat up as more as more of the energy it transmits to the reactor must be via radiation and conduction through gas rather than contact-conduction. This is the likely what makes it appear that there is a gain above 1. On 16 October 2014 01:13, Alan Fletcher a...@well.com wrote: New version with embedded wires. http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php Here I've also assumed that the wires are a simple single strand, rather than the spiral form used in the earlier tests, and are in good thermal contact with the Alumina.
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
These might be pretty similar to Rossi's setup. (Google Superthal smu) Superthal heating modules Prefabricated heating modules consisting of vacuum-formed ceramic fibre with an integral Kanthal Super molybdenum-disilicide (MoSi2) heating element for up to 1750°C (3180°F) element temperature. Geometries Superthal heating modules are available in a variety of geometries and standard sizes. Tailor-made modules can be supplied to optimize the design and function of the particular application. Muffles Cylinders Half cylinders Radiating panels High-power reflectors On October 15, 2014 2:23:03 PM Alan Fletcher a...@well.com wrote: From: ChemE Stewart cheme...@gmail.com Sent: Wednesday, October 15, 2014 11:08:42 AM Me too, good job. Tube in a tube reminds of the model rockets I used to build. Fin supports between tubes might explain the wider dark band seen as a spiral?. Do you think a lot of the heat might be discharged in a space between tubes and out the ends? Or are the ends completely sealed? The fins are too fine to be a source of the bands. And the ends are completely sealed, so the temperature and heat-loss will taper off at the ends. But your .. and your, and your ... guess on the structure and the banding is as good as mine. Very frustrating.
Re: [Vo]:Engineering and materials issues with high temperature hot-cat Lugano demo
such units are not made as 3 phase helically wound assemblies, MoSi2 is non-ductile/brittle and very difficult to make and even worse to bond to, and there is still the unanswered problem of how do you bond inconel wire that can survive only to 1350°C to an insulated heating element that is supposed to be at higher temperature? On 16 October 2014 08:40, John Page johnp...@comcast.net wrote: These might be pretty similar to Rossi's setup. (Google Superthal smu) Superthal heating modules Prefabricated heating modules consisting of vacuum-formed ceramic fibre with an integral Kanthal Super molybdenum-disilicide (MoSi2) heating element for up to 1750°C (3180°F) element temperature. Geometries Superthal heating modules are available in a variety of geometries and standard sizes. Tailor-made modules can be supplied to optimize the design and function of the particular application. Muffles Cylinders Half cylinders Radiating panels High-power reflectors On October 15, 2014 2:23:03 PM Alan Fletcher a...@well.com wrote: *From: *ChemE Stewart cheme...@gmail.com *Sent: *Wednesday, October 15, 2014 11:08:42 AM Me too, good job. Tube in a tube reminds of the model rockets I used to build. Fin supports between tubes might explain the wider dark band seen as a spiral?. Do you think a lot of the heat might be discharged in a space between tubes and out the ends? Or are the ends completely sealed? The fins are too fine to be a source of the bands. And the ends are completely sealed, so the temperature and heat-loss will taper off at the ends. But your .. and your, and your ... guess on the structure and the banding is as good as mine. Very frustrating.
Re: [Vo]:temperature of the resistor wire.
If the wire were the brightest area there would be no excess heat. Harry On Wed, Oct 15, 2014 at 3:56 AM, Robert Lynn robert.gulliver.l...@gmail.com wrote: how do you know this? How do you know the the wire is not the brightest area? On 15 October 2014 15:06, H Veeder hveeder...@gmail.com wrote: Some people suspect that the resistor wire can't be Inconel because they are predicted to melt at the reactor's operating temperature. However, since we know the resistor wire casts a shadow in the alumina, the temperature of the wire remains below the operating temperature and therefore can't melt. Harry
Re: [Vo]:temperature of the resistor wire.
But the reactor heat should drop at the same time. I guess it depends if they are measuring heat loss on all surfaces of the device. The total dissipated heat should not change except for due to the reaction and changes to input power. I have not read the report in detail to see how many measurements they were doing On Wed, Oct 15, 2014 at 8:38 PM, Robert Lynn robert.gulliver.l...@gmail.com wrote: It has nothing to do with measuring of the wire power, except that as the wire heats up, increase the thermal resistance of the heat flow between wire and reactor body (by reducing number of points of physical contact) and of course the wire temperature will go up (given same input power) - I suggest you think a bit longer about it. On 16 October 2014 08:33, ChemE Stewart cheme...@gmail.com wrote: Since they are measuring the input energy to the wire that makes no sense On Wednesday, October 15, 2014, Robert Lynn robert.gulliver.l...@gmail.com wrote: the resistor wire expands with respect to the alumina as it heats up, breaking any bonding contact, or lifting the wire of the inner alumina tube in more and more places and leading to less and less conductive contact - prompting the wire to heat up as more as more of the energy it transmits to the reactor must be via radiation and conduction through gas rather than contact-conduction. This is the likely what makes it appear that there is a gain above 1. On 16 October 2014 01:13, Alan Fletcher a...@well.com wrote: New version with embedded wires. http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php Here I've also assumed that the wires are a simple single strand, rather than the spiral form used in the earlier tests, and are in good thermal contact with the Alumina.
[Vo]:Aviation Week and the Lockheed Fusion Reactor
Greeting Vortex-L, A yawn for the Aviation Week Lockheed Fusion website: http://aviationweek.com/technology/skunk-works-reveals-compact-fusion-reactor-details Ad Astra, Ron Kita, Chiralex Doylestown PA
Re: [Vo]:temperature of the resistor wire.
Does this not indicate that the wire must be producing inductive heating in the powder? On Wed, Oct 15, 2014 at 8:23 PM, Robert Lynn robert.gulliver.l...@gmail.com wrote: the resistor wire expands with respect to the alumina as it heats up, breaking any bonding contact, or lifting the wire of the inner alumina tube in more and more places and leading to less and less conductive contact - prompting the wire to heat up as more as more of the energy it transmits to the reactor must be via radiation and conduction through gas rather than contact-conduction. This is the likely what makes it appear that there is a gain above 1. On 16 October 2014 01:13, Alan Fletcher a...@well.com wrote: New version with embedded wires. http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php Here I've also assumed that the wires are a simple single strand, rather than the spiral form used in the earlier tests, and are in good thermal contact with the Alumina.
Re: [Vo]:temperature of the resistor wire.
I think it is, but irregardless a wire alone cannot create COP1 On Wednesday, October 15, 2014, Axil Axil janap...@gmail.com wrote: Does this not indicate that the wire must be producing inductive heating in the powder? On Wed, Oct 15, 2014 at 8:23 PM, Robert Lynn robert.gulliver.l...@gmail.com javascript:_e(%7B%7D,'cvml','robert.gulliver.l...@gmail.com'); wrote: the resistor wire expands with respect to the alumina as it heats up, breaking any bonding contact, or lifting the wire of the inner alumina tube in more and more places and leading to less and less conductive contact - prompting the wire to heat up as more as more of the energy it transmits to the reactor must be via radiation and conduction through gas rather than contact-conduction. This is the likely what makes it appear that there is a gain above 1. On 16 October 2014 01:13, Alan Fletcher a...@well.com javascript:_e(%7B%7D,'cvml','a...@well.com'); wrote: New version with embedded wires. http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php Here I've also assumed that the wires are a simple single strand, rather than the spiral form used in the earlier tests, and are in good thermal contact with the Alumina.
Re: [Vo]:temperature of the resistor wire.
So how do you imagine it inductively heats the powder given low AC frequency, weak solenoid magnetic field, tiny cross section area powder, and high resistivity of nickel near its melting point? The physics + mathematics to estimate the magnetic field strength and eddy currents induced are high-school /freshman physics level (estimate wire turns, solenoid inductance = applied voltage gives current rate of change, = solenoid magnetic field strength rate of change = eddy currents induced in particles of given diameter - power dissipation, so you could very quickly do some calculation to confirm or disprove your theory, and numbers would at least give foundation to your hope. On 16 October 2014 09:25, Axil Axil janap...@gmail.com wrote: Does this not indicate that the wire must be producing inductive heating in the powder? On Wed, Oct 15, 2014 at 8:23 PM, Robert Lynn robert.gulliver.l...@gmail.com wrote: the resistor wire expands with respect to the alumina as it heats up, breaking any bonding contact, or lifting the wire of the inner alumina tube in more and more places and leading to less and less conductive contact - prompting the wire to heat up as more as more of the energy it transmits to the reactor must be via radiation and conduction through gas rather than contact-conduction. This is the likely what makes it appear that there is a gain above 1. On 16 October 2014 01:13, Alan Fletcher a...@well.com wrote: New version with embedded wires. http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php Here I've also assumed that the wires are a simple single strand, rather than the spiral form used in the earlier tests, and are in good thermal contact with the Alumina.
Re: [Vo]:temperature of the resistor wire.
Why did Rossi say that a DC current applied to the wire would not work? Why does the startup procedure need for a magnetic field to be applied? On Wed, Oct 15, 2014 at 9:56 PM, Robert Lynn robert.gulliver.l...@gmail.com wrote: So how do you imagine it inductively heats the powder given low AC frequency, weak solenoid magnetic field, tiny cross section area powder, and high resistivity of nickel near its melting point? The physics + mathematics to estimate the magnetic field strength and eddy currents induced are high-school /freshman physics level (estimate wire turns, solenoid inductance = applied voltage gives current rate of change, = solenoid magnetic field strength rate of change = eddy currents induced in particles of given diameter - power dissipation, so you could very quickly do some calculation to confirm or disprove your theory, and numbers would at least give foundation to your hope. On 16 October 2014 09:25, Axil Axil janap...@gmail.com wrote: Does this not indicate that the wire must be producing inductive heating in the powder? On Wed, Oct 15, 2014 at 8:23 PM, Robert Lynn robert.gulliver.l...@gmail.com wrote: the resistor wire expands with respect to the alumina as it heats up, breaking any bonding contact, or lifting the wire of the inner alumina tube in more and more places and leading to less and less conductive contact - prompting the wire to heat up as more as more of the energy it transmits to the reactor must be via radiation and conduction through gas rather than contact-conduction. This is the likely what makes it appear that there is a gain above 1. On 16 October 2014 01:13, Alan Fletcher a...@well.com wrote: New version with embedded wires. http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php Here I've also assumed that the wires are a simple single strand, rather than the spiral form used in the earlier tests, and are in good thermal contact with the Alumina.
Re: [Vo]:temperature of the resistor wire.
So how many turns are in the coil? And what are you calculating for a field strength? On Wednesday, October 15, 2014, Robert Lynn robert.gulliver.l...@gmail.com wrote: So how do you imagine it inductively heats the powder given low AC frequency, weak solenoid magnetic field, tiny cross section area powder, and high resistivity of nickel near its melting point? The physics + mathematics to estimate the magnetic field strength and eddy currents induced are high-school /freshman physics level (estimate wire turns, solenoid inductance = applied voltage gives current rate of change, = solenoid magnetic field strength rate of change = eddy currents induced in particles of given diameter - power dissipation, so you could very quickly do some calculation to confirm or disprove your theory, and numbers would at least give foundation to your hope. On 16 October 2014 09:25, Axil Axil janap...@gmail.com javascript:_e(%7B%7D,'cvml','janap...@gmail.com'); wrote: Does this not indicate that the wire must be producing inductive heating in the powder? On Wed, Oct 15, 2014 at 8:23 PM, Robert Lynn robert.gulliver.l...@gmail.com javascript:_e(%7B%7D,'cvml','robert.gulliver.l...@gmail.com'); wrote: the resistor wire expands with respect to the alumina as it heats up, breaking any bonding contact, or lifting the wire of the inner alumina tube in more and more places and leading to less and less conductive contact - prompting the wire to heat up as more as more of the energy it transmits to the reactor must be via radiation and conduction through gas rather than contact-conduction. This is the likely what makes it appear that there is a gain above 1. On 16 October 2014 01:13, Alan Fletcher a...@well.com javascript:_e(%7B%7D,'cvml','a...@well.com'); wrote: New version with embedded wires. http://lenr.qumbu.com/rossi_hotcat_oct2014_141014b.php Here I've also assumed that the wires are a simple single strand, rather than the spiral form used in the earlier tests, and are in good thermal contact with the Alumina.
Re: [Vo]:temperature of the resistor wire.
If the wire inside the reactor was hot enough to glow it should produce a more uniform spiral glow along the entire length of the tube. Harry On Wed, Oct 15, 2014 at 8:19 AM, Robert Lynn robert.gulliver.l...@gmail.com wrote: Additionally, look at the darkened photo, the wire exterior to the reactor sourrounded by cooler materials to radiate to are brighter than the bright wires in the reactor. Hard to believe it would be colder inside the reactor surrounded by relatively hotter materials that are harder to radiate to. I think that is pretty strong indication that it is the wires that are the bright areas. On 15 October 2014 20:14, Robert Lynn robert.gulliver.l...@gmail.com wrote: I am looking at high zoom at the same photos and finding it easy to draw the opposite conclusion. Confirmation bias on both our parts :) I think it is equivocal at best. On 15 October 2014 19:52, ChemE Stewart cheme...@gmail.com wrote: If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. I am 95% confident that is the shadow of the coil. The light areas change in brightness, width, etc. On Wed, Oct 15, 2014 at 3:56 AM, Robert Lynn robert.gulliver.l...@gmail.com wrote: how do you know this? How do you know the the wire is not the brightest area? On 15 October 2014 15:06, H Veeder hveeder...@gmail.com wrote: Some people suspect that the resistor wire can't be Inconel because they are predicted to melt at the reactor's operating temperature. However, since we know the resistor wire casts a shadow in the alumina, the temperature of the wire remains below the operating temperature and therefore can't melt. Harry
[Vo]:Thermography
If the thermography was done reasonably well, then it is proving to be more than just a means of measuring excess heat, it is also a means of probing the LENR phenomena. Harry
Re: [Vo]:temperature of the resistor wire.
Not if it is touching the walls of inner or outer alumina tube in places, intermittent contact due to vagaries of original wire winding around inner tube and subsequent large differential thermal expansion so that the wire is quenched in some places but not in others. Would explain the variation in glow that we see (along with slight translucence of alumina tube), and would change as the wire gets hotter and relaxes pre-existing springiness that might otherwise hold the wire in contact with the inner tube - would lead to wire temperature increasing faster than power input would suggest - ie what we see with supposedly increasing COP. Most likely means of construction is winding wires around an inner tube, or winding them around a different mandrel and then slipping them over the tube. Bonding them to the inner tube is an extra step that (based on inconsistency/variability of surface glow) has likely not been done and for which their would be little initial motive anyway. And massive relative thermal expansion of the wire (~1%) would likely have cracked any ceramic bonding or attempts to rigidly encase the wires or bond them to the inner tube anyway. Differential thermal expansion means that the internal tube/vessel is likely only bonded to the thermocouple end cap, otherwise the external tube would be broken by axial stress due to differential thermal expansion of higher temperature of inner tube compared to external tube. On 16 October 2014 10:58, H Veeder hveeder...@gmail.com wrote: If the wire inside the reactor was hot enough to glow it should produce a more uniform spiral glow along the entire length of the tube. Harry On Wed, Oct 15, 2014 at 8:19 AM, Robert Lynn robert.gulliver.l...@gmail.com wrote: Additionally, look at the darkened photo, the wire exterior to the reactor sourrounded by cooler materials to radiate to are brighter than the bright wires in the reactor. Hard to believe it would be colder inside the reactor surrounded by relatively hotter materials that are harder to radiate to. I think that is pretty strong indication that it is the wires that are the bright areas. On 15 October 2014 20:14, Robert Lynn robert.gulliver.l...@gmail.com wrote: I am looking at high zoom at the same photos and finding it easy to draw the opposite conclusion. Confirmation bias on both our parts :) I think it is equivocal at best. On 15 October 2014 19:52, ChemE Stewart cheme...@gmail.com wrote: If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. I am 95% confident that is the shadow of the coil. The light areas change in brightness, width, etc. On Wed, Oct 15, 2014 at 3:56 AM, Robert Lynn robert.gulliver.l...@gmail.com wrote: how do you know this? How do you know the the wire is not the brightest area? On 15 October 2014 15:06, H Veeder hveeder...@gmail.com wrote: Some people suspect that the resistor wire can't be Inconel because they are predicted to melt at the reactor's operating temperature. However, since we know the resistor wire casts a shadow in the alumina, the temperature of the wire remains below the operating temperature and therefore can't melt. Harry
Re: [Vo]:temperature of the resistor wire.
Thanks for posting your ideas. I hadn't seen that picture of the march 2013 reactor sitting on the scale with heating coils visible. Why don't we just accept that the authors of the 2014 test also know enough about the construction of the reactor to say that the dark bands align with the wires? Harry On Wed, Oct 15, 2014 at 12:21 PM, Alan Fletcher a...@well.com wrote: I wrote up my analysis of the banding : (Draft -- I'll rename it later). http://lenr.qumbu.com/rossi_hotcat_oct2014_141014a.php Short answer : we don't even know whether the bright bands line up with the wires, or the gaps between them. There are multiple explanations, which depend on the structure used to hold the wires, and on the properties of everything. Insufficient data !
Re: [Vo]:temperature of the resistor wire.
This is the likely what makes it appear that there is a gain above 1. So at least we have you now believing that a wire cannot create a gain above 1 and that the wire is not inside the reactor core. I wonder if we can estimate number of coil wraps from the photo(dark bands), we might be able to estimate an inductance On Wednesday, October 15, 2014, Robert Lynn robert.gulliver.l...@gmail.com wrote: Not if it is touching the walls of inner or outer alumina tube in places, intermittent contact due to vagaries of original wire winding around inner tube and subsequent large differential thermal expansion so that the wire is quenched in some places but not in others. Would explain the variation in glow that we see (along with slight translucence of alumina tube), and would change as the wire gets hotter and relaxes pre-existing springiness that might otherwise hold the wire in contact with the inner tube - would lead to wire temperature increasing faster than power input would suggest - ie what we see with supposedly increasing COP. Most likely means of construction is winding wires around an inner tube, or winding them around a different mandrel and then slipping them over the tube. Bonding them to the inner tube is an extra step that (based on inconsistency/variability of surface glow) has likely not been done and for which their would be little initial motive anyway. And massive relative thermal expansion of the wire (~1%) would likely have cracked any ceramic bonding or attempts to rigidly encase the wires or bond them to the inner tube anyway. Differential thermal expansion means that the internal tube/vessel is likely only bonded to the thermocouple end cap, otherwise the external tube would be broken by axial stress due to differential thermal expansion of higher temperature of inner tube compared to external tube. On 16 October 2014 10:58, H Veeder hveeder...@gmail.com javascript:_e(%7B%7D,'cvml','hveeder...@gmail.com'); wrote: If the wire inside the reactor was hot enough to glow it should produce a more uniform spiral glow along the entire length of the tube. Harry On Wed, Oct 15, 2014 at 8:19 AM, Robert Lynn robert.gulliver.l...@gmail.com javascript:_e(%7B%7D,'cvml','robert.gulliver.l...@gmail.com'); wrote: Additionally, look at the darkened photo, the wire exterior to the reactor sourrounded by cooler materials to radiate to are brighter than the bright wires in the reactor. Hard to believe it would be colder inside the reactor surrounded by relatively hotter materials that are harder to radiate to. I think that is pretty strong indication that it is the wires that are the bright areas. On 15 October 2014 20:14, Robert Lynn robert.gulliver.l...@gmail.com javascript:_e(%7B%7D,'cvml','robert.gulliver.l...@gmail.com'); wrote: I am looking at high zoom at the same photos and finding it easy to draw the opposite conclusion. Confirmation bias on both our parts :) I think it is equivocal at best. On 15 October 2014 19:52, ChemE Stewart cheme...@gmail.com javascript:_e(%7B%7D,'cvml','cheme...@gmail.com'); wrote: If you zoom in very closely on the hot reactor photos you can see the the dark lines are of uniform width, continuity and shade. I am 95% confident that is the shadow of the coil. The light areas change in brightness, width, etc. On Wed, Oct 15, 2014 at 3:56 AM, Robert Lynn robert.gulliver.l...@gmail.com javascript:_e(%7B%7D,'cvml','robert.gulliver.l...@gmail.com'); wrote: how do you know this? How do you know the the wire is not the brightest area? On 15 October 2014 15:06, H Veeder hveeder...@gmail.com javascript:_e(%7B%7D,'cvml','hveeder...@gmail.com'); wrote: Some people suspect that the resistor wire can't be Inconel because they are predicted to melt at the reactor's operating temperature. However, since we know the resistor wire casts a shadow in the alumina, the temperature of the wire remains below the operating temperature and therefore can't melt. Harry
Re: [Vo]:temperature of the resistor wire.
Nullis in verba. :) I believe my eyes more than others words. In finding so many potential faults with so little published information (they had a month to investigate!!) I can only say that I am unimpressed by the critical observational skills of the testers. If they had approached this demo with a more critical mindset I might be more inclined to believe them. On 16 October 2014 11:41, H Veeder hveeder...@gmail.com wrote: Thanks for posting your ideas. I hadn't seen that picture of the march 2013 reactor sitting on the scale with heating coils visible. Why don't we just accept that the authors of the 2014 test also know enough about the construction of the reactor to say that the dark bands align with the wires? Harry On Wed, Oct 15, 2014 at 12:21 PM, Alan Fletcher a...@well.com wrote: I wrote up my analysis of the banding : (Draft -- I'll rename it later). http://lenr.qumbu.com/rossi_hotcat_oct2014_141014a.php Short answer : we don't even know whether the bright bands line up with the wires, or the gaps between them. There are multiple explanations, which depend on the structure used to hold the wires, and on the properties of everything. Insufficient data !
Re: [Vo]:temperature of the resistor wire.
Nullis in verba. :) I believe my eyes more than others words. In finding so many potential faults with so little published information (they had a month to investigate!!) I can only say that I am unimpressed by the critical observational skills or reporting of the testers. If they had approached this demo with a more critical mindset I might be more inclined to believe them. There is a mountain to climb to convince the world, and they have not really helped that process. On 16 October 2014 11:41, H Veeder hveeder...@gmail.com wrote: Thanks for posting your ideas. I hadn't seen that picture of the march 2013 reactor sitting on the scale with heating coils visible. Why don't we just accept that the authors of the 2014 test also know enough about the construction of the reactor to say that the dark bands align with the wires? Harry On Wed, Oct 15, 2014 at 12:21 PM, Alan Fletcher a...@well.com wrote: I wrote up my analysis of the banding : (Draft -- I'll rename it later). http://lenr.qumbu.com/rossi_hotcat_oct2014_141014a.php Short answer : we don't even know whether the bright bands line up with the wires, or the gaps between them. There are multiple explanations, which depend on the structure used to hold the wires, and on the properties of everything. Insufficient data !
Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible
You might say I am splitting hairs but what Mckubre has written here is technically incorrect. The Stephan-Boltzman law is relationship between temperature and output power. It is not a relationship between input power and output power so you can't use the law by itself to infer any relationship between input and output power. Additional assumptions/laws are required. Harry On Wed, Oct 15, 2014 at 1:08 PM, David Roberson dlrober...@aol.com wrote: You have a good understanding in my opinion. There is no doubt that energy is being generated within the core. Dave -Original Message- From: Alain Sepeda alain.sep...@gmail.com To: Vortex List vortex-l@eskimo.com Sent: Wed, Oct 15, 2014 12:59 pm Subject: Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible A calibration curve will bend down. It never bends up. this mean that temperature grow less than the power ? this mean that when you increase the power, and if temperature grows much more that before, something anomalous is happening ? Either excess heat, or some external blanket effect (increase of thermal resistance)... but convection does not diminish with heat? did I undertand well? 2014-10-14 22:09 GMT+02:00 Jed Rothwell jedrothw...@gmail.com: Alain Sepeda alain.sep...@gmail.com wrote: - is there a simple way , with minimal assumption, to be sure that the COP1 Look at the color. If it is dull red, it may be around 750°C which is where you would expect it to be in a straight line extrapolation calibration up to 800 W. If it is white it has to be around 1300°C, which is far higher than the calibration indicates it should be. A calibration curve will bend down. It never bends up. McKubre pointed this out: On page 7 of the report the authors state: “Subsequent calculation proved that increasing the input by roughly 100 watts had caused an increase of about 700 watts in power emitted.” This is interesting. The shape of the output vs. input power curve is observed (or implied) to strongly curve upwards in a manner completely inconsistent with the Stefan-Boltzmann law for radiative heat loss. It is also inconsistent with simple convective heat transfer but several issues need to be addressed before we can claim this as a qualitative or even “semi-quantitative” measure of excess heat production . . . Note that incandescent colors are similar for all materials. - Jed
Re: [Vo]:temperature of the resistor wire.
Bob, you appear to be too convinced that the gain is unity and are going to great lengths to obtain that result. The testers are well respected scientists and no one should assume that they are so easily misslead. Besides, there are several measurements that support the fact that the COP is greater than unity which you seem to brush off. I wonder about whether or not the actual temperature is correct as well, but am in no position to prove one way or the other. The most important observation that supports the elevated COP is the slope of output power versus input power that they measure about their chosen operating point. I can think of no way to fake that measurement without a dose of true magic. And then it would be extremely difficult to understand why the measured behavior tends to follow what my simulation predicts. Dave -Original Message- From: Robert Lynn robert.gulliver.l...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Wed, Oct 15, 2014 11:53 pm Subject: Re: [Vo]:temperature of the resistor wire. Nullis in verba. :) I believe my eyes more than others words. In finding so many potential faults with so little published information (they had a month to investigate!!) I can only say that I am unimpressed by the critical observational skills of the testers. If they had approached this demo with a more critical mindset I might be more inclined to believe them. On 16 October 2014 11:41, H Veeder hveeder...@gmail.com wrote: Thanks for posting your ideas. I hadn't seen that picture of the march 2013 reactor sitting on the scale with heating coils visible. Why don't we just accept that the authors of the 2014 test also know enough about the construction of the reactor to say that the dark bands align with the wires? Harry On Wed, Oct 15, 2014 at 12:21 PM, Alan Fletcher a...@well.com wrote: I wrote up my analysis of the banding : (Draft -- I'll rename it later). http://lenr.qumbu.com/rossi_hotcat_oct2014_141014a.php Short answer : we don't even know whether the bright bands line up with the wires, or the gaps between them. There are multiple explanations, which depend on the structure used to hold the wires, and on the properties of everything. Insufficient data !
Re: [Vo]:Gibbsite
On Wed, Oct 15, 2014 at 1:50 PM, mix...@bigpond.com wrote: However if it [hydrogen] only acts as a catalyst for neutron transfer reactions, then nowhere near that amount would be needed. My current theory is that the hydrogen plays no role in this particular instance. Perhaps elsewhere, deep in the IH labs, they are experimenting with lithium hydride with some fraction of deuterium in it, which would play a role. From the standpoint of industrial design, it would be convenient to use the same type of fuel and reactor design in both cases, so, according to this line of thinking, lithium hydride is used in this case even though the hydrogen is not involved. Eric
Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible
To go a bit further, the law applies to black body radiation and not just any emitter. I suspect it is a difficult task to prove that the actual power being radiated is exactly what is expected unless a system is constructed to capture all of the radiation over the entire spectrum. Then that must be reconciled against a known amount of power being supplied to the radiator. It would be a miracle to find that the temperature exactly matched what is expected according to the Stephan-Boltzman equation. The best that we can do is to calibrate our test system with a known amount of power being radiated. That is exactly what the testers did in their dummy run. It would have been better had they calibrated their equipment at the same output power as generated by the device, but that could not be done under the conditions they experienced. I give them a great deal of credit for what they actually were able to accomplish. Dave -Original Message- From: H Veeder hveeder...@gmail.com To: vortex-l vortex-l@eskimo.com Sent: Thu, Oct 16, 2014 12:18 am Subject: Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible You might say I am splitting hairs but what Mckubre has written here is technically incorrect. The Stephan-Boltzman law is relationship between temperature and output power. It is not a relationship between input power and output power so you can't use the law by itself to infer any relationship between input and output power. Additional assumptions/laws are required. Harry On Wed, Oct 15, 2014 at 1:08 PM, David Roberson dlrober...@aol.com wrote: You have a good understanding in my opinion. There is no doubt that energy is being generated within the core. Dave -Original Message- From: Alain Sepeda alain.sep...@gmail.com To: Vortex List vortex-l@eskimo.com Sent: Wed, Oct 15, 2014 12:59 pm Subject: Re: [Vo]:E-cat : Minimum COP assuming worst mistakes possible A calibration curve will bend down. It never bends up. this mean that temperature grow less than the power ? this mean that when you increase the power, and if temperature grows much more that before, something anomalous is happening ? Either excess heat, or some external blanket effect (increase of thermal resistance)... but convection does not diminish with heat? did I undertand well? 2014-10-14 22:09 GMT+02:00 Jed Rothwell jedrothw...@gmail.com: Alain Sepeda alain.sep...@gmail.com wrote: is there a simple way , with minimal assumption, to be sure that the COP1 Look at the color. If it is dull red, it may be around 750°C which is where you would expect it to be in a straight line extrapolation calibration up to 800 W. If it is white it has to be around 1300°C, which is far higher than the calibration indicates it should be. A calibration curve will bend down. It never bends up. McKubre pointed this out: On page 7 of the report the authors state: “Subsequent calculation proved that increasing the input by roughly 100 watts had caused an increase of about 700 watts in power emitted.” This is interesting. The shape of the output vs. input power curve is observed (or implied) to strongly curve upwards in a manner completely inconsistent with the Stefan-Boltzmann law for radiative heat loss. It is also inconsistent with simple convective heat transfer but several issues need to be addressed before we can claim this as a qualitative or even “semi-quantitative” measure of excess heat production . . . Note that incandescent colors are similar for all materials. - Jed
Re: [Vo]:temperature of the resistor wire.
Look at this way the paper is getting peer reviewed in public. Hopefully they will revise the paper to address the criticisms. Harry On Wed, Oct 15, 2014 at 11:58 PM, Robert Lynn robert.gulliver.l...@gmail.com wrote: Nullis in verba. :) I believe my eyes more than others words. In finding so many potential faults with so little published information (they had a month to investigate!!) I can only say that I am unimpressed by the critical observational skills or reporting of the testers. If they had approached this demo with a more critical mindset I might be more inclined to believe them. There is a mountain to climb to convince the world, and they have not really helped that process. On 16 October 2014 11:41, H Veeder hveeder...@gmail.com wrote: Thanks for posting your ideas. I hadn't seen that picture of the march 2013 reactor sitting on the scale with heating coils visible. Why don't we just accept that the authors of the 2014 test also know enough about the construction of the reactor to say that the dark bands align with the wires? Harry On Wed, Oct 15, 2014 at 12:21 PM, Alan Fletcher a...@well.com wrote: I wrote up my analysis of the banding : (Draft -- I'll rename it later). http://lenr.qumbu.com/rossi_hotcat_oct2014_141014a.php Short answer : we don't even know whether the bright bands line up with the wires, or the gaps between them. There are multiple explanations, which depend on the structure used to hold the wires, and on the properties of everything. Insufficient data !
