You are off on a tangent. My point is that Rossi's claims are in
conflict with the observed results. I will no longer respond for now.
On Oct 27, 2011, at 12:15 PM, Axil Axil wrote:
In the Miley presentation that he has recently released, Miley
shows transmutation to 39 isotopes over possible contamination levels.
The nuclear reactions and transmutation patterns that are going on
inside the Rossi reactor are similar to what Miley documents as
mentioned in Rossi’s original patent.
The presence of a large amount of iron in the Miley results is
interesting and similar iron contamination was found in the Rossi
ash(10%) when they were analyzed by the swedes.
The assumption that the nuclear reactions taking place in the Rossi
reaction are exclusively restricted to copper transmutation is
mistaken in my opinion.
The possibility that the reactions going on are hydrogen only
cannot be ignored with the production of copper as only one of many
reactions going on.
On Thu, Oct 27, 2011 at 3:21 PM, Horace Heffner
<hheff...@mtaonline.net> wrote:
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/
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/