10KW LENR Demonstrator?

[KitemanSA]

Also, I don't understand the resistance to my arguements. M<y communication skills (and several readjustments of my arguements may have fogged the picture some) but my conclusions are merely restatements of published and accepted physics.

The resistance to your arguments is due to the fact that your "conclusions" are misstatements of the published and accepted physics. Or perhaps the better description is due to you appearant confusion between nucleons and nuclei.

[chrismb]

The key to the Rossi reactor is the production of coherent hydrogen. The prolific production of Rydberg hydrogen is what gives the Rossi reaction its great productivity in power.

Kite.. do you see what I mean yet? Definitive, absolutist, yet non-science!

[Giorgio]

The key to the Rossi reactor is the production of coherent hydrogen. The prolific production of Rydberg hydrogen is what gives the Rossi reaction its great productivity in power. I believe that clusters of coherent, entangled and inverted ultra-dense Rydberg hydrogen condensate crystals are formed on the surface of a doped solid carbon compound such as graphite. Such ions attain a long average lifetime due to the high pressure and temperatures maintained within the hydrogen envelope of the reaction vessel.

This long lifetime is sufficient to permit the ions to drift across the hydrogen envelope. Once they reach the nickel oxide nano-powder affixed to the reaction vessel walls, a hybrid hydride reaction occurs with the highly eroded nickel oxide surface layer. The action of quantum lattice vibrations (phonons) synchronizes the nickel atoms local to the lattice defect that confine this Rydberg matter into a coherent state with the Rydberg matter.

Axil, you cannot take pieces of different research papers and stick them together hoping to create something that makes sense.

Really, this is becoming embarrassing to read.

[Axil]

From the Colloquium at MIT today, I hear that the Rossi replication effort has inspired a large number of Ni-H experiments to be undertaken. One of them has produced some level of success.

Any high school student should soon be able to perform a version of Ni-H experiments for less than the cost of an X-Box- since it is an UNPOWERED experiment, without deuterium or platinum. The reaction gives excess heat simply from the nanopowder contact with hydrogen at ambient (like Arata, but better). It is a surface effect and there is no significant absorption and zero radioactivity. The delta-T is low but the heat is apparently continuous and anomalous.

Soon, a student kit will be made available to popularize the subject. Like the Heathkit popularized electronics through education in the 1970’s, once the cold fusion student kit is available; you can get one and be as scientific as you need to be.

[KitemanSA]

The key to the Rossi reactor is the production of coherent hydrogen. The prolific production of Rydberg hydrogen is what gives the Rossi reaction its great productivity in power.

Kite.. do you see what I mean yet? Definitive, absolutist, yet non-science!

Sure. I have ignored Axil since day one. I actually have little hope for him. In you however I do detect the possibility of learning.

Wouldn't it be a shocker if he turned out to be right. Oh my!

[tomclarke]

Indeed. It is pretty difficult to imagine such a mechanism which would not have left other evidence. After all, nuclei, stable or unstable, are well studied and intract in very well defined ways with their environment.

I have read of a mechanism that is known but not often considered called "internal conversion". It is like beta decay but actually employs an atomic electron so there is no change in nuclear charge number. It is also a mechanism (AFAIK) that is used to shed excitation energy on the way to ground state so it is very rapid, rather than potentially taking hundreds of years like some instances of beta decay. For reasons I do not understand, some isotopes can shed a lot of energy this way and some can't shed any.

To me, it is NOT difficult to imagine that a lattice of Ni under unique stimulation like UV lasers (not normally seen in nature) might make atypical use of such an alternate de-excitation process.

How would the lattice (with nuclei very far apart & electron wave functions having zero effect on nuclei) alter the nuclear behaviour of nucleus?

[tomclarke]

From the Colloquium at MIT today, I hear that the Rossi replication effort has inspired a large number of Ni-H experiments to be undertaken. One of them has produced some level of success.

Any high school student should soon be able to perform a version of Ni-H experiments for less than the cost of an X-Box- since it is an UNPOWERED experiment, without deuterium or platinum. The reaction gives excess heat simply from the nanopowder contact with hydrogen at ambient (like Arata, but better). It is a surface effect and there is no significant absorption and zero radioactivity. The delta-T is low but the heat is apparently continuous and anomalous.

Soon, a student kit will be made available to popularize the subject. Like the Heathkit popularized electronics through education in the 1970’s, once the cold fusion student kit is available; you can get one and be as scientific as you need to be.

We'll find out whether you are right, won't we? I would not bet on it.

No strong evidence for.

Strongly "extraordinary" low prior.

[KitemanSA]

How would the lattice (with nuclei very far apart & electron wave functions having zero effect on nuclei) alter the nuclear behaviour of nucleus?

See my new post in "General"... The Kiteman Konjecture.

I stuck it in general cuz it is no where near sufficiently supported to be a hypothesis, let alone a theory!

[Ivy Matt]

Well, it looks like cold fusion is going to be hot this summer. Let's hope this heat produces some light as well. Briefly, what we've got coming down the pike (albeit mostly rumors so far):

• * An announcement of a breakthrough in replication from Brian Ahern at the 2011 LANR/CF Colloquium at MIT this weekend.
  * An important announcement from Defkalion regarding the E-Cat within 10 days. (See comments by "Tizzie".)
  * An announcement from the Piantelli team regarding their results (supposedly 40 kW thermal power/7kW electrical power) in the next few weeks.

There's also the documentary film The Believers as well, although I have not yet seen a release date for it. And the University of Bologna will be examining one of Rossi's devices, although the results of that examination will probably not be made public any time soon.

If Piantelli's team can demonstrate a device that produces 40 kW thermal power/7 kW electrical power, and their demonstration is backed up by a useful theory and/or independent replication, it will add to the credibility of Rossi's claims...and to the squabbling over who did what first.

Any high school student should soon be able to perform a version of Ni-H experiments for less than the cost of an X-Box- since it is an UNPOWERED experiment, without deuterium or platinum. The reaction gives excess heat simply from the nanopowder contact with hydrogen at ambient (like Arata, but better). It is a surface effect and there is no significant absorption and zero radioactivity. The delta-T is low but the heat is apparently continuous and anomalous.

Axil, please cite your source and, when you use a direct quote, put it in quotation marks, or in a quote box.

[D Tibbets]

Dan:

So? The question is not whether the Ni62 + P reaction might release energy. Your example demonstrates this. The question is if there is NET energy release.

There are about two dozen light elements present in the ash from the Rossi process in addition to copper. We cannot know what set of complicated nuclear processes are producing all this stuff. You are fixated on copper because Rossi says so and thinks you should.

I think there is a lot of energy produced during the transmutation of all the other lighter elements that can be found in the ash. Thinking only about Copper is a wild goose chase.

Best regards:
Axil

I don't want to validate the claims, but based on my expertise
such light element fusions within the nickel matrix, makes a lot more sense from an excess energy viewpoint. This is more in line with mainstream LENR reasoning. Nickel serves as the catalyst/ enabler , as opposed to being consumed in a one step reaction. Nickel could be consumed / coveted to copper, as part of a series of lighter element fusions. The net energy output could be positive in this case, but that energy is not coming from the Ni to copper reaction.

Dan Tibbets

[D Tibbets]

So if the approaching particle is a neutron you would allow as how the reaction (n + 62Ni = 63Ni) is substantially exothermic? After all, there would be effectively NO kinetic energy needed to bring the particles together. What about a situation where the particle is an Electron Guided Proton (like whatsisname's mini-atom)? If the EGP has little or no coulomb resistance (no net charge), would the released energy of the reaction ({e,p} + 62Ni = 63Cu + e) be ~8.7 MeV?

I admit the effects with neutrons is more subtile. But look at the picture. The Ni62 is the highest binding nucleus. Other isotopes on nickel with more or fewer neutrons are lower on the graph. Cu62 is lower. Ni63 is lower. Granted if a Ni63 underwent a beta decay, it would be a Cu63. so this localized reaction would release some energy, if it could happen*. But this energy may still be less than the difference between Ni62 and Cu63- you would be making up some of the difference, but not nessisarily matching or exceeding the nuclear binding energy of Ni62.

* If a Ni63 beta decays to a Cu63 the neutron count would decrease by one and the proton count would increase by one. The strong force attraction would be little different. There would be more Coulomb repulsion, so the nucleus would have less binding energy. I'm not sure how to label this reaction, it may be inappropriate to call it either a fusion or fission, as the nucleon count is not changing. I suspect the reaction (beta decay ) would be unlikely to happen it it results in a less stable nucleus. Unless the parent Ni63 had excess excitation to start with, then the beta decay may be a method of shedding this excess isomer energy.. Again, that excitation energy had to be added to the Ni63' to start with.

My guestimate of the energy balance would be something like:
Ni62 + Neutron -> Ni63 +10 units of energy -> Cu63 +electron - 11 units of energy. The initial reactants and final products would represent a mildly endothermic reaction. The intermediate Ni63 (or Ni64 for that matter) would seem to be exothermic, but this is not so based on the Nuclear binding energy graph where Ni62 rules all. The explanation may have to do with the stability of the intermediate, or more likely have to do with some nuclear dynamics concerning the orbitals and quantum effects. There are strict rules to how many neutrons can fit into a nucleus and the resultant stability and morphology of the nucleus. I have spoken of the strong force and electromagnetic repulsive force mostly, but other factors like the weak force, etc cannot be ignored for a more accurate understanding of the nucleus.

Dan Tibbets.

[tomclarke]

I admit the effects with neutrons is more subtile. But look at the picture. The Ni62 is the highest binding nucleus. Other isotopes on nickel with more or fewer neutrons are lower on the graph. Cu62 is lower. Ni63 is lower. Granted if a Ni63 underwent a beta decay, it would be a Cu63. so this localized reaction would release some energy, if it could happen*. But this energy may still be less than the difference between Ni62 and Cu63- you would be making up some of the difference, but not nessisarily matching or exceeding the nuclear binding energy of Ni62.

* If a Ni63 beta decays to a Cu63 the neutron count would decrease by one and the proton count would increase by one. The strong force attraction would be little different. There would be more Coulomb repulsion, so the nucleus would have less binding energy. I'm not sure how to label this reaction, it may be inappropriate to call it either a fusion or fission, as the nucleon count is not changing. I suspect the reaction (beta decay ) would be unlikely to happen it it results in a less stable nucleus. Unless the parent Ni63 had excess excitation to start with, then the beta decay may be a method of shedding this excess isomer energy.. Again, that excitation energy had to be added to the Ni63' to start with.

My guestimate of the energy balance would be something like:
Ni62 + Neutron -> Ni63 +10 units of energy -> Cu63 +electron - 11 units of energy. The initial reactants and final products would represent a mildly endothermic reaction. The intermediate Ni63 (or Ni64 for that matter) would seem to be exothermic, but this is not so based on the Nuclear binding energy graph where Ni62 rules all. The explanation may have to do with the stability of the intermediate, or more likely have to do with some nuclear dynamics concerning the orbitals and quantum effects. There are strict rules to how many neutrons can fit into a nucleus and the resultant stability and morphology of the nucleus. I have spoken of the strong force and electromagnetic repulsive force mostly, but other factors like the weak force, etc cannot be ignored for a more accurate understanding of the nucleus.

Dan Tibbets.

I'm not sure the quote above is me....

Anyway, you can do better than guess.

Take 1u (12C/12) as your relative atomic mass unit.

Measure the RAMs of reactants and products.

In this case:
neutron ~ 1.008u
Cu,Ni ~ 0.999u per nucleon (not much difference)

So clearly we have an excess energy of approx 0.008u

You can look up precise values here:
http://www.sisweb.com/referenc/source/exactmas.htm

Note that these masses include electrons but this is negligible - they balance +/1 1 or so emitted as b+ or b-, and one electron is 0.0005u and so negligible here. Similarly the total electron binding energy (to the nuclei) is negligible.

[KitemanSA]

I admit the effects with neutrons is more subtile. But look at the picture. The Ni62 is the highest binding nucleus.

No it is NOT!!

Dan, until you get this right you will continue to mis-understand.

[KitemanSA]

My guestimate of the energy balance would be something like:
Ni62 + Neutron -> Ni63 +10 units of energy -> Cu63 +electron - 11 units of energy.

And you would be WAY off. If the first step is +10, the second is -0.098, not minus 11. You are off by ~2 orders of magnitude. These are real ratios calculated by known methods. I didn't make them up. I just normalized them to your first step.
[Edit] OOPS! Forgot the electron in my second step! If first is 10, second is -0.00048 ! Now it is 4.5 orders of magnitude. [/Edit]

The initial reactants and final products would represent a mildly endothermic reaction.

Significantly exothermic.

The intermediate Ni63 (or Ni64 for that matter) would seem to be exothermic, but this is not so based on the Nuclear binding energy graph where Ni62 rules all.

Only on a per-nucleon basis.

The explanation may have to do with the stability of the intermediate, or more likely have to do with some nuclear dynamics concerning the orbitals and quantum effects.

The explanation is that you still don't get it.

Dan
Please define nucleon.
Please define nucleus.
Detect the difference?
If not, learn. If so, why can't you apply that knowledge here?

Everyone else... does it strike you that Dan is running a psych experiment on me? Can he REALLY be this obtuse?

[chrismb]

I admit the effects with neutrons is more subtile. But look at the picture. The Ni62 is the highest binding nucleus.

No it is NOT!!

Dan, until you get this right you will continue to mis-understand.

DAN!

See;

viewtopic.php?p=64094#64094