On Thurs Oct 27, 2011 Horace said [snip] 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.[/snip]
Horace,
Assuming the thermal power is in fact achieved, and the reaction is not
Ni-H, what do you feel is the next most credible theory ?
Fran
-----Original Message-----
From: Horace Heffner [mailto:hheff...@mtaonline.net]
Sent: Thursday, October 27, 2011 7:49 AM
To: Vortex-L
Subject: EXTERNAL: [Vo]:Rossi H and Ni consumption
From:
http://www.rossilivecat.com/
Quote:
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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.
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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/
