Talk-Polywell.org

Yet to calculate net output energy, you need to also account for energy input into the system (which seems like it ought to be a lot for Ni62), and lost output energy (gamma rays or whatever).

Long, long way between here and there .... at the moment, assuming it does incredibly what they say it does, physically what they have is a stand-alone kettle that runs on nickel powder ... passing the lightbulb test is not as silly as it sounds.

Well, I agree with the fact that it still has not been proven in any way. I still think that it is a hoax. However that does not mean that hot water does not have any value. It does have a lot of value if you can produce it cheaply enough. That was my point. Their machine, whatever it is, still has to proof that it can deliver what they promise.
That demonstration was only 30 minutes or so (if I understand that transcript correctly). You can fake 30 minutes with some sort of trickery or there could even be some chemical reaction that keeps this thing running for 30 minutes. I would like to see this thing running for a day or so... in a neutral lab...
Of course knowing what is inside would help also. The secrecy is suspicious.

Even my cabin in the woods has 200A service at 240 V. That's typically the minimum for a single family home in the US.

For me, it has been plenty.

EMC2 probably had less than that available at the Manassas Park lab, and the experiments ran off a 50 A 240 V range jack. The move to San Diego was an improvement, but they only had 600 A 208 3-phase and it limited the power of the apparatus.

So, how am I going to up my service when I build my lab at the cabin? I gotta build at least a small WB-D machine, right?

The new garage/shop needs a heating system. I'd love to put in one of those LENR foot-warmers. That would save power for the WB machine.

So according to several comments on NBF that make a lot of sense (Goatguy), I am now 99.99% sure that this is quackery. This is bad news for anybody doing serious fusion research because yet another Pons and Fleischman situation will yet again put a bad light on fusion as a whole.

Was this physics journal held in high regard before this? It'd be pretty curious for seasoned professionals to back such vaporware, if that's all it is.

Which physics journal? The Journal of Nuclear Physics? It's been around since February 2010, not much time to gain high regard.

Well, if it works, we'll know what BLP had and were blaming on hydrinos...


IIRC, the touchstone for extraordinary power-producing widgets that may be misdirecting meters and/or have concealed storage, innocent or otherwise, is to have them feed a wall of incandescent lamps...

10 @ '100 watt' per kilowatt should be brought into circuit by twos and threes to avoid inrush issues that might overwhelm a prototype. Bring up the lights and leave them on. Even a concealed uninterruptible power supply would give up the ghost when its amp-hours are exhausted...

cynic:
Wouldn't it be funny if BLP sue the Italian crew *and* US patent office ??
/

The patent is pretty interesting, if not always intelligible in English. The Italian patent can be found here, for those who read Italian.

This part caught my attention:



So, having nothing better to do...

"Method of testing the gate oxide in integrated DMOS power transistors and integrated device comprising a DMOS power transistor"

"Device for stopping a radiant burner automatically in the event of ignition"

"Hall effect device assembly"

"Electrical technique"

"Fusogenic lipsomes and methods for making and using same"

"Optical fibre splicing"

"Current guiding system"

"Beam dancer fusion device"

"Methods for generating catalytic proteins"

Someone with more scientific expertise than I will have to figure out how these patents relate to each other and to the claims in Mr. Rossi's patent. I will note, however, that all of the patents contain the word "fusion" or, in one case, "fuze".

EDIT: Edited to add working links.

The house I own in VA has a 200A panel as well. Plenty, the only issue I have had there is that the well pump was on the garage curcuit. So if the pump cycled while the freezer was running, and someone hit the garage opener, guaranteed trip. I fixed it by splitting the well pump off onto its own breaker. Silly contractors, never think stuff through.

It is 11F out tonight, and I am thinking that the FHW in this house would be better with the Italian rig vice the gas boiler. Hmmm. Have to think about that while skiing tomorrow. I am not sure where to buy all that tin-foil though.

??? Why? What has Ni62 got to do with energy into the system?
I was seeking clarification of the statement by ChrisMB that it would not be possible to produce ANY energy with a nuclear reaction involving 62Ni. I in no way am trying to imply that this particular machine has a viable positive energy production. It seems they say it does, by about 10:1, e.g., 10kW out for every kW in. If this is true, this puts heat pumps to shame and will probably take over the space heating market, a $Billion(?) industry if not the industrial service heat industry.

I see hydrogen, nickel powder, and a short, energetic reaction. I hate to say it, but this looks an awful lot like the Blacklight Power process. Blacklight published detailed instructions for reproducing it not too long ago. Blacklight can show a lot of heat and light, but it's been many months and they haven't been able to produce a system with net energy yield. Blacklight may turn out to have walked the longest garden path since phlogiston. And the folks behind this event may be wayfarers on the same path.

I regret to say that I believe this is all a highly naive argument. Perhaps, even, the one that the researchers have gone along with, without thinking it through.

I'll run this from two different angles:

1) The binding energy of 62Ni and 63Cu are both around 8.7MeV/nucleon. If you throw in a proton into the mix (a single nucleon) and the total extra 'mass-energy' you are adding is 6.15MeV, then why do you think you'd suddenly be able to make a new atom where each of the nucleons has 8.7MeV worth of binding energy? It's a bit like a group of people going to an event paying $8.70 per ticket. An extra comes along and says to the cashier 'I've got an extra $6.15 to get in with this group'. Is the cashier likely to a) let you in, b) turn you away? That group is bound by their contribution of $8.70 each to the party fund. An upstart proton that blags its way in with $6.15 isn't going to be welcome.

2) Let us say, just for the sake of argument, that the proton does manage to muscle its way into the 62Ni, endothermically or by physical processes not yet known [that run against thermodynamics!]. Then we'd have an excited 63Cu. Let us say it has an excess energy of 6.15MeV. ..... What does it do next?

Q1: Is this 63Cu 'hot'?
A: No. It was a reaction that occurred in the inertial frame of the collision, so it isn't going anywhere. It is a 'cold' excited atom that is undetectable by ''thermal' measurements, unless the energy is released from the atom, somehow. So... no heat.

{Q1a: Why does fusion generate heat?
A: Because there are fusion products that take that excess energy away as kinetic energy [which manifests itself as 'heat', in ensembles of such fast, kinetic, products].}

Q2: What would a 63Cu do with this extra 6.15MeV?...
A: Well, I am guessing that the thermodynamic thing to do would be to chuck the proton out again. I do not know for sure, but I think you will find that the energy necessary for an excited 63Cu to shed a proton is 6.15MeV! [I say that, mainly because of the principle of (1) above, but don't have those exact figures to confirm that.] But let us just still hold on there with a glimmer of hope that this 63Cu does something else;
a) it can undergo a strong-force mediated outcome; this means that it will chuck out nucleons. As 62Ni is the bottom of the pile, we know immediately that if it chucks out anything other than that extra proton, then it is going to be an endothermic release of nucleons,
b) it can undergo a weak-force mediated reaction; this means that either a proton gives out a positron and turns into a neutron, becoming 63Ni, or a neutron gives up an electron giving 63Zn. Without even resorting to calculating those energy equations, I am sure those all work out endothermic [again... because 62Ni is already the bottom of the energy barrel].
c) it can undergo an electromagnetic-force mediated reaction; OK, so now we have something that ....er.... could be argued. A 6.15 MeV photon emission would [I presume, not really thought about it too hard] naively gives an energy balance. If so, then it will be very easy to detect 6.15MeV gammas....so...show me the Geiger tube readings while this thing is running...... oh, and also, if this is the reaction, then there will be no heat..... heat will merely demonstrate it is not an EM mediated reaction!



So, in summary; I see no thermodynamic reason for the proton to enter the 62Ni, and that even if it were there, there is no known nuclear mechanism by which heat would be released.

I am looking at the videos of the test of yesterday.

Dott. Celani (the guy measuring the Gamma radiation and is quite skeptical about the test) stated that he requested to measure the spectra of the gamma to understand the energy of the gamma itself and was prevented to do so from Ing. Rossi, under the logic that if he actually measured the energy of the emitted Gamma he would have clearly understood the working principle of the reactor.
Just a couple of minutes before he stated that they are not still understanding the working principle of the machine, and just have some hypothesis that will need a lot of further work.

Quite a contradiction IMHO.

Dott. Celani also stated that the instable increase of the gamma radiation could indicate that no hidden energy source should be present.



More details should be online monday



Link to videos in italian, fot the one willing to give a look at the machine or want to refresh their Italian

http://www.youtube.com/user/efagroup201 ... -Ru1eAymvE

Well, my post did go up, third time around. I guess the moderator thought it might get lost after 60 posts?

So I am now posting;

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Mr Rossi,

I would still like to understand why you think Ni62 could possibly have any exothermic reactions.

Collis Williams has attempted to reply, but it is naive to think that there can be an exothermic reaction from a resultant 6.15MeV excited 63Cu, as its total binding energy is 8.7MeV higher that 62Ni.

The per-nucleon binding energy of 63Cu and 62Ni are around 8.7MeV. So it would be thermodynamically unfavourable for an additional nucleon to be added to 62Ni with only 6.15MeV available.

This would have to be endothermic if you have only 6.15MeV, yet the total binding energy goes up by more than 8MeV.

I have given a more detailed account on; http://www.talk-polywell.org/bb/viewtop ... 4883#54883 , which also covers each possible nuclear reaction (each of which cannot result in heat).

The question is; how can the 6.15MeV mass-energy increase of 62Ni+p->63Cu account for a total binding energy increase of 8.7MeV, and yet also be exothermic?
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