This is a nonsensical argument. The less hydrogen available for
nuclear reactions the *more* the MeV per reaction that is required to
make the 1 MW output, thus the less effective any shielding would be,
and the *less credible* it is that the MW heat comes from nuclear
reactions.
On Oct 27, 2011, at 11:14 AM, Axil Axil wrote:
There are some ifs and buts associated with this subject. It has
been known for over a hundred years how that hydrogen will defuse
through a hot metal enclosure.
The rate of diffusion is subject to the temperature and pressure of
the hydrogen, together with the exact kind, thickness, and
temperature of the metal. These are all variables in the
calculation of the diffusion rate.
Furthermore, the presence of oxides and/or carbides on the surface
of the metal can reduce the rate of diffusion of hydrogen by up to
5 orders of magnitude.
We don’t know for sure what the accurate values of some of these
variables are and additionally they would vary widely within an
operational range throughout the operational lifetime of the E-Cat.
However, since hydrogen is very slippery and notoriously hard to
contain, a good guess can be made that most of the hydrogen
consumed by the Rossi reactor would be lost through diffusion
through the hot walls of the stainless steel reaction vessel.
Because of all these large uncertainties, calculation of the
nuclear reaction rates as a function of hydrogen consumption
implying a clue to the nuclear processes going on inside the E-Cat
reaction vessel cannot be made in my opinion.
With best regards,
Axil
On Thu, Oct 27, 2011 at 7:48 AM, Horace Heffner
<hheff...@mtaonline.net> wrote:
From:
http://www.rossilivecat.com/
Quote:
- - - - - - - - - - - - - - - - - - - -
Andrea Rossi
October 25th, 2011 at 4:59 PM
Dear Thomas Blakeslee:
Grams/Power for a 180 days charge
Hydrogen: 18000 g
Nickel: 10000 g
Warm Regards,
A.R.
- - - - - - - - - - - - - - - - - - - -
End quote.
At atomic weight of 1.0079 the 18000 gm of H is 1786 mols. At an
atoommic weight of 58.69 the 10,000 gm of Ni is 170.4 mol. This
means 10.48 atoms of H need be provided per 1 atom of Ni.
Assuming the reaction is Ni-H, as claimed, only about 1 in 10 atoms
of H is consumed, thus 170.4 mols of H and a170.4 mols of Ni are
consumed, maximum. This involves the obviously wrong assumption
that all the Ni atoms are transmuted, not a more realistic 3
percent. There is also an outside possibility the H reacts with
daughter products, giving the possibility of 10 subsequent daughter
reactions per primary Ni+H reaction. Three such reactions is an
outside possibility.
One MW for 180 days is 1.556x10^13 J, or 10^7 MJ. That is
(6.241x10^24 eV/MJ)*(1.556x10^13 J)/(170.4 mol * 6.022x10^23 atoms /
mol) = 9.464x10^5 eV/(Ni atom). If there is one reaction per atom
and all Ni is consumed by single reactions than that is 0.9464 MeV
per Ni-H event. The gammas from this would be lethal at short
range, even through 2 cm of lead. If it is assumed that 3% of the
Ni is consumed then that is 0.9464 MeV/0.03 = 31.5 MeV per
reaction. If there are an average of 3 daughter reactions per
primary reactions that is about 10 Mev per reaction.
If 10 MeV gammas are produced then 5 cm of lead shielding will be
of no use in protecting the operators. If near 1 MeV gammas are
produced the lead shielding is inadequate.
One MW of gammas is 6.241x10^24 eV/s, or, for 1 MeV gammas,
6.24x10^18 gammas per second. using:
I = I0 * exp(-mu * rho * L)
where mu for 1 MeV is 0.02 cm^2/gm), and density of lead 11.34 gm/
cm^3, we have for 5 cm of lead:
I = (6.24x10^18 s^-1) * exp(-(0.02 cm^2/gm) * (11.34 gm/cm^3) *(5
cm))
I = 2x10^18 free gammas per second.
About half that, or 10^18 gammas/s would be directed toward the
interior of the container housing the E-cats, and most of the
2x10^18 gammas per second would end up escaping the container.
This is an approximate calculation. Even if it is off by an order
of magnitude, this kind of 1 MeV gamma flux, even 1/32 of it from
one E-cat, would be readily detected by a geiger counter at
significant range.
It does not seem credible the energy from a Ni-H reaction, at least
in the form of one gamma per reaction, provides any explanation for
1 MW of heat, if that thermal power is in fact achieved.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
