New Paper By Gullström, Rossi - COP 22,000

Alan - what you don't know is not my loss.

Sorry that your hero is going down by his own hand - that is never fun but after 40 years as a deceiving crook, its way past time for some accountability.

Perhaps L.F. will eventually be able to figure out how to handle jaded moderators in the future.

Planet Rossi - be advised that the countdown has begun.

Regardless of the theory, is the Rossi-Gullström paper describing an experiment similar to what Darden reported in his previously confidential email?

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We are building tiny glass reactors for rapid testing purposes. Our cost per test has dropped from thousands of dollars to about $25. We have been seeing flashes of energy in the fuel when it is heated. Our goal is to be able to see and record the intensity of reactions occurring with different fuel materials. We think this also could help us with patents because it will be harder to deny enablement, or that something happens, which is the basis for the US patent office's anti-LENR policy. If we have a video showing something occurring, that may be persuasive. We like these tests because they are very inexpensive, and they will be important for our patent development if they are in fact easy to see. But broadly, we intend to hand serious, long-term assessment and testing over to large engineering entities that have much greater capability than ours.

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How can anyone rest when the lights are getting turned on?

Said the cockroach to the silverfish

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Another "overwhelming" headline, which surely is nothing then pure fraud.

Of course isn't - the cold fusion in high voltage discharge of hydrogen at the surface of lithium has been published in way more trustworthy sources already (not just Me356). The energy threshold reported in original Minari's experiments was just about 300 eVolts, so that it's feasibility is out of discussion. According to these sources it does run in very narrow temperature range once the bulk molten lithium is used. I presume it's because the atoms near the surface get oriented by surface tension.

At the case of thin layer of molten lithium soaked onto nickel substrate this orientation works even at much higher temperatures.

March 7, 2014

Industrial Heat Update 3/6/14 CONFIDENTIAL

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[..]. We have been seeing flashes of energy in the fuel when it is heated.

March 24, 2016

New LENR Company: Lux Energy — An Industrial Heat Affiliate (Update: Site Now Hidden)

http://www.e-catworld.com/2016…ndustrial-heat-affiliate/

Lux

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From Latin lūx (“light”);

[...]

In the International System of Units, the derived unit of illuminance or illumination; one lumen per square metre. Symbol: lx

Okay regardless how credible this is ... am i the only one seeing parallels to Mr. Suhas reactor that is getting tested by MFMP quite soon?

A tube with 2 "rods" pointing at the fuel from both sides having "a little bit of space" between the rods and the fuel (1,5 cm)? And a voltage (also DC) applied to the rods?

Planet Rossi - you can drop the Luxe gig. It was abandoned after a developer left a back door open exposing the then IH re-brand. Rossi glommed on within hours of the disclosure permanently rendering the effort worthless and useless.

He has a habit of doing that to everything he touches.

Obviously this paper seems to describe what Rossi has been calling his "Quark" device. From the scant description, it sounds like a quasi-gas/metal vapor discharge tube. Per the description, the sealed small diameter tube would have Ni electrodes on each end with a 1.5cm gap. When such a tube reaches its operating discharge temperature, the plasma inside the tube is a far different temperature than the envelope of the tube. Collisions of the ions and electrons with the wall of the tube transfer heat and create a colder layer around the inside of the tube. Think of a neon sign. The hot orange plasma may have an effective plasma temperature of >2000°C but the lead glass tube will soften in the range of 600°C and melt substantially by 800°C. Plasma temperature is best characterized by its spectrum, but it will not necessarily be Boltzmann distribution - it could have strong lines that contain most of the energy which will impart a particular color. In a linear tube the plasma will be hotter in the center and will fall off in temperature with radius as it approaches the envelope boundary. It sounds like from this paper that the tube is not glass since an emissivity of 0.9 is being ascribed to it. Likely it is a thin alumina tube. Note that alumina appears opaque, but it is really translucent, so measuring the spectrum is important. The surface area being used is 1 cm^2 with a length of 1.5cm suggesting that the tube OD is 2.1mm - pretty small. So, at high outer surface temperature, if the spectrum shows a Boltzmann distribution, a blackbody calculation could be used for emitted power. In this case, the spectrum is only measured to 1.1 microns.

From their measurements, they estimate a temperature of the device surface at 2636°K or 2363°C. Alumina would melt at 2072°C, so either the tube is not alumina, or the tube surface temperature is wrong. It is likely that the tube is alumina and the tube surface temperature is wrong. As I mentioned, the tube can be much cooler than the plasma (for example the much cooler glass temperature of a neon sign), particularly if the tube is transparent. Alumina is translucent at visible and near infrared wavelengths - it has a high degree of transparency because its crystallites are randomly oriented sapphire. So it has scattering, but a high degree of transparency. This means that the diameter for the Boltzmann calculation would have to be some kind of average plasma discharge diameter inside the tube because it is not the tube surface that is at that temperature and radiating at the 2363°C Boltzmann spectrum - it is the inner plasma.

So, lets make an guess-timate for the diameter of the plasma. Lets say that the OD of the tube is 2.1mm, the ID of the tube might be 1.1mm. The diameter of the average emitting plasma discharge might be 0.4mm (just to guess). This would make the emitting area (0.4/2.1)x(1cm^2) = 0.19 cm^2 and would reduce the calculated output power to 47 watts. But, this estimate is prone to huge inaccuracy. What is needed is to measure the tube in a calorimeter. If you are going to do it optically, at least make a bolometric measurement of the emitted power/cm^2 and integrate over the sphere.

Once again, we see that Rossi could be fooling himself with poor measurements. That doesn't mean that he does not have excess power. It suggests we don't know how much he has if any due to poor measurements.

@Dewey

"The picture is slowly starting to come into focus.

Sure this can still all be some toxic brew of delusion, greed, ego and fraud. But if so, add another Uppsala physics professor (EDIT: actually still a Doctoral student) to the supposedly duped/part of the conspiracy list. One not warned away by his fellow burned Uppsala colleagues... one unable to competently measure a COP of 22k (ha)... or one throwing his lot in with a conspiratorial crew at a late date with a rather radical theory paper, jeopardizing a promising career that's just getting started."

I should also mention that I am not convinced of adequate input power measurement. When the device is cold, the Li-Al will condense on the inside of the alumina tube. Initially when measured cold, the resistance may appear low as there is a metal-metal electrode connection through the condensed Li-Al and the Ni electrodes. It would appear as a resistor until the alumina tube got to over 700°C when the film would melt changing the resistance. Then at 1342°C the Li would become vapor and at some point the tube would become a Li vapor lamp with the current flowing through the Li vapor. At this point the resistance could be substantially different than the starting resistance. So the resistance is going to change, I believe, from metal-metal resistance to Li vapor nonlinear discharge. Once the Li is in vapor phase and there is a plasma discharge, the tube could actually be lower temperature than the Li vapor temperature.

Quote from Dewey Weaver

Planet Rossi - be advised that the countdown has begun.

I'm not sure we've ever heard such a thing from you. I'm now officially terrified.

Ultrasure - who is shepherding Gullstrom thru at Uppsala? Is it the same folks who either continue drink the Rossi Kool-aid or are complicit in something worse?

Furthermore, who from Uppsala served Gullstrom up for use and abuse to Rossi? They have ruined this kids career before it even got started.

Quote from Dewey Weaver

It was abandoned after a developer left a back door open exposing the then IH re-brand.

IH was re-branding the Quark X?