That's why MFMP and alike groups (even this Forum) are very important to inform the mankind. This is also an emulsion to replicate and and spread the knowledge around the world. If we can have a recipe which can work and easily shared then we will be stronger than APCO.
Arnaud
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Posts by Arnaud
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Fais nous un bon résumé de cette conférence
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Hello Pr Songsheng Jiang,
Would it possible to receive the raw data of the acquisition? In the graph it is very difficult to understand what is the cause and the consequences of all the events that occurred in the experimental setup.
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How about dark energy, does that violate the 2nd law of thermodynamics?
How Dark energy, if it's ever existed, is violating the 2nd law of thermodynamic?
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How is it possible without violating the 2nd law of thermodynamic?
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They are climbed above 1000° ( no written in the paper ) but no more results ..unfortunally
With same electrical currents? -
Thank you David for this paper I didn't know about. Shame the temperature didn't rise above 800°C. This temperature isn't high enough whatever the kind of electrical feed in (square, no square, pulse, no pulse, AC, chopped AC). Parkhomow has shown that a minimum temperature of 1000°C is required.
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150W excess heat inside tube, showing coil shadows. Note crack forming in the tube next to the TC. The tube failed moments later.
Clearly any powerful exothermic reaction needs to be spread evenly over the tube (in time also), or structural failure is…
Hello @Paradigmnoia, Could you explain a little more about this picture? What is the mix? Type of tube? How hid you made your reactor? -
@ogfusionist if there was a constant stream of H2 flowing in to the reactor, then my previous sentence about H2 quantities is no more relevant. This hypothesis can be ruled out.
How long was the reactor with a stream of H2 below 830°C ? Could be that H2 was captured by NiO to form atomic metallic hydrogen as described here http://arxiv.org/ftp/arxiv/papers/1312/1312.6851.pdf The accumulation could be very strong below 830°C (Faster than at room temperatrure as for the KSX25 batteries). Then at 830°C the metastable atomic metallic hydrogen is released back with a lot of energy accumulated when the atomic metallic hydrogen was forming.
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It seems that the quantity of H2 (gas) was not enough to reduce all the NiO present in the reactor: Only a small part get reduced, no change in the color of the NiO.
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@Ecco At 800°C, if H2 is coming close to NiO, it will be reduced whatever there is in the surrounding/support. See http://www.rafaldb.com/papers/…NiO-reduction-in-ETEM.pdf
But if the content of H2 (per atomic number) is lower than the content of NiO (per atomic number), all the H2 could be converted into H2O vapor. And H20 could be the source of hydrogen (H+ + OH-) without killing the the rest of NiO. And as you said there would have also Ni on the surface to increase the reaction rate.
So volume of reactor, H2 pressure and NiO quantities is a must to know before going further. If @ogfusionist could light up our mind to understand his process more clearly ...
In the parkhomov reactor the only oxygen available is coming from the air. Al2O3 and Si02 will not be reduced at those temperatures.
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David, the NiO reactor is easy to replicate if protocol is followed exactly. Not expensive to build or replicate. A ball mill is critical for the particle size of catalyst necessary for a nanoscale reaction.
No sloppiness or it won't work. …
The NiO will be reduced in Ni and H20 in presence of H2. So in your protocol, there is a first step that reduce the NiO particles in to Ni on the fiberfrax support. The Ni formed should be on µm scale with a high surface per volume ratio. The Ni will adsorb a lot of H2 andwill produce a lot of monoatomic H in the surround fiberfrax. The fiberfrax contains Si02 and Al203. One of the 2 elements of fiberfrax should then react with the mono atomic hydrogen to form the reactant. Al203 has been widely used by replicants without success. SiO2 is found in the mulite used by Parkhomov. There is no Li in your protocol. Or could it have a hidden Li somewhere? -
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<span style="color: #FF0000">Response to questions from <a href="http://www.lenr-forum.com/forum/index.php/User/1157-David-Fojt/">@David Fojt</a> and <a href="http://www.lenr-forum.com/forum/index.php/User/873-Arnaud/">@Arnaud</a> </span>
A detail description of the variations of temperature and pressure in the reactor chamber in the first 2 days can be found in Figure 6, and a clean version was provided to Jed Rothwell. Then he made an…
The minimum temperature is 0 Pa which not possible to reach currently. It can't have a negative pressure in absolute. -100kPa doesn't exist! -
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Hello Andrea,
Thank you for the answer. There is still an unanswered question. Where the energy comes from to ionize a nucleus to expulse the naked electron?
Arnaud
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A "simple" way to verify your theory, is to beam protons with the right energy to a target made of the right ionized state ... just need the right ionized state. Then we should see the Hydronion forming. How to detect them?
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Hello Andrea,
I like your theory. But I'm not a theorist and can not judge it in deep.
Nevertheless, there is a point I noticed which is unclear. It is about the Ionisation energies of the ECO (Electron Orbital Core). In the table page 22 of your slide presentation, the ionization state is unclear. The ionization energy is the energy needed or released to move from one state to the next one (+1 or -1). For example, Mg has an ionization energy of 80.17 eV if ionized from Mg +2 to Mg +3. It is a transition.
For the H, to receive 80.17 eV of energy, the Mg must go from +3 state to +2 state. How can you have Mg +3 in a NAE? Which electron is given to the nucleus to lower its ionization state?
The more I write, and the more you theory is becoming foggy. For an e- to move out of the nucleus you need energy, you don't receive energy! So imagine that the coupling exists, the H move close to the nucleus at the right speed (see energy), the e- of the transition ionisation energy is captured. But where comes the energy of the ionization? Asap a second electron will come to fill the gap made by the capture. That releases energy in form of Xray.
So your theory should say that a p/d/t with a sufficient energy comes close the target nucleus. It gives to the nucleus the energy of ionization and take out the gained electron with him. The nucleus will not give the energy, it is the p/d/t.
Do you understand my point?
Arnaud -
It is not said that the outer surface of the reactor (ceramic) is in direct contact with the copper of the inner surface of the cooler. It would probably not otherwise the copper would have melted! It is not the thermal conduction law in play here but the radiation law which take the temperature in Kelvin at the power 4.
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Hello Fabrice,
Can you give us some references of the David-Giles effect?
Merci,
Arnaud