# Home LENR newbie questions

T

#### Tom P

Jan 1, 1970
0
I've been half following some of the discussions on this group, maybe
someone can fill me in some details.

I've gotten the impression that the way these things work is by
accelerating deuterium or tritium ions so that enough collisions occur
to release energy.

So we have for example:
1D2 + 1D2 -->2He3 (0.82 MeV) + 0n1 (2.45 MeV)

This reaction is actually used in a commercial neutron generator, so in
terms of proof of principle, so far so good. There are other reactions,
but let's take this as a working example.

Where I get unstuck is with the numbers. The commercial device in
question consumes around 2-3 kW of power and produces 10^7 neutrons per
second.
That means that as a by-product, it produces 10^7 x 2.45 MeV, or
24500000 MeV of energy per second.
That sounds a lot, but if you convert into watts and joules that comes
out as 0.000003925334385 watts. Hmm.

Ok, well let's suppose that we could up the efficiency by a factor of
several million so that it produces more energy than it consumed - say
10kW. That's 62415064799632350 MeV per second, so you would be
generating around 6x10^24 neutrons per second. These neutrons are also
carrying the energy that you are trying to capture, so you'll need a
suitable moderator to slow them, say water, and something to absorb
them. Whatever absorbs them needs to do so extremely efficiently so that
you don't get too many neutrons leaking out of your basement and making

Maybe there's some other technology, but the bottom line is that it
seems you have a lot of neutrons to look after, and this doesn't really
sound like a sound DIY project to me.
Can someone comment? Are my numbers all wrong?

V

#### Vaughn

Jan 1, 1970
0
... but the bottom line is that it
seems you have a lot of neutrons to look after, and this doesn't really
sound like a sound DIY project to me.
Can someone comment? Are my numbers all wrong?

Your assumption on the physics of the reaction must be wrong. From what
I've read, those things don't generate any significant number of
neutrons. Take it from an old nuke, no government is about to allow you
to have any significant generator of neutrons in your basement.

Vaughn

J

#### j

Jan 1, 1970
0
Your assumption on the physics of the reaction must be wrong. From what
I've read, those things don't generate any significant number of
neutrons.

Yes.

More here:

http://lenr-canr.org/?page_id=263

The key is providing sites in the catalyst for nuclear reactions (not
necessarily fusion) to take place.

Jeff

Take it from an old nuke, no government is about to allow you

T

#### Tom P

Jan 1, 1970
0
Yes.

More here:

http://lenr-canr.org/?page_id=263

The key is providing sites in the catalyst for nuclear reactions (not
necessarily fusion) to take place.

Jeff

Jeff,just what kind of nuclear reactions are you talking about, if not
fusion? Just show me a simple equation like this one:
1D2 + 1D2 -->2He3 (0.82 MeV) + 0n1 (2.45 MeV)

T

#### Tom P

Jan 1, 1970
0
Everybody can comment. ;-)

I'm doing an amateur investigation of one of the "other reactions" -
specifically the Ni/H reaction with which Andrea Rossi (and a few
others) have claimed to produce heat output in excess of heat input, and
with which Rossi has claimed to produce a self-sustaining reaction.

You can find all that is publicly known buried under heaps of ignorant
speculation by doing a Google search on Rossi+ECat+LENR.

There appears to be a growing acceptance in the physics community that
there's /some/ kind of nuclear reaction taking place, but I haven't seen
any movement toward consensus concerning reaction specifics.

I don't have much interest in the physics. I have need of a compact heat
source that can operate for extended periods of time in a third-world
context without a reliable fuel distribution network, and am
willing/able to expend a limited amount of personal resource to discover
whether or not a Ni/H LENR can satisfy my requirement.

Since it's one of your concerns, Rossi and all of the observers have
reported that radiation is not a problem, and that the Ni/H reaction
produces no unstable isotopes. Still, it struck me as prudent to
incorporate Geiger-Müller detector input and program the controller to
do an automatic shutdown if a high count (three times a normal
background peak) is detected in any one-second interval.

Rossi is claiming to produce about 5kW output from a reactor containing
50 grams of fine nickel powder, and it appears that an initial charge of
hydrogen (~50cc) will run his reactor for 4-5 hours. In one of the
interview transcripts I read, he said that the 50 grams of nickel would
power his reactor for about 6 months of operation at that power level.

It all just begs for testing.

Thanks for the tip. I find http://en.wikipedia.org/wiki/Energy_Catalyzer

There is no way you can get any net energy out of a reaction between
hydrogen and nickel, either nuclear or otherwise. If it's chemical just
work out how much energy you need to produce atomic hydrogen.

If it's nuclear, then work out the precise atomic weights and binding
energies.

Naturally occurring nickel (Ni) is composed of five stable isotopes;
58Ni, 60Ni, 61Ni, 62Ni and 64Ni with 58Ni being the most abundant.

There are 29 isotopes of copper. 63Cu and 65Cu are stable, with 63Cu
comprising approximately 69% of naturally occurring copper; they both
have a spin of 3/2.[12] The other isotopes are radioactive, with the
most stable being 67Cu with a half-life of 61.83 hours.[12]

So supposing that the reaction involves stable isotopes, the possible
reactions would seem to be
62Ni28 + H -> 63Cu29 3%
64Ni28 + H -> 65Cu29 0.1%
61Ni28 + D -> 63Cu29 1%

The percentages are the relative natural occurrences of the isotopes.
Deuterium occurs naturally at around 0.015%.

The problem is that all elements around this atomic weight have the
highest binding energies - see
http://en.wikipedia.org/wiki/Nuclear_binding_energy#Nuclear_binding_energy_curve
- this means that it is hard to see how nuclear reactions can produce
any excess energy.

If have yet to find anything published by Rossi that shows how excess
energy can be produced.

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