----- Original Message -----
From: "Robin van Spaandonk" <[EMAIL PROTECTED]>
To: <vortex-l@eskimo.com>
Sent: Friday, October 24, 2008 10:21 PM
Subject: Re: [Vo]:Banking on BLP?
In reply to Mike Carrell's message of Fri, 24 Oct 2008 16:54:12 -0400:
Hi,
[snip]
To:
Robin van Spaandonk
Jones Beene
Ed Storms
Scott Little
[and lurkers]
This has been a very useful discussion. If you have not done so, I
recommend
downloading http://www.blacklightpower.com/papers/WFC102308WebS.pdf and
printing pages 10-14 and 48. Figure 7 on p48 is a scan of NaH using
Differential Scanning Clorimetry. It is most instructive. At 350 C there is
endothermic decompoisition of NaH. Beginning at 640 C is a very strong
exothermic reaction, which I think is conventionally unexpected. The NaH
was
in 760 Torr He.
This is unfortunate given that He+ is also a catalyst.
MC: But He is not a catalyst, it used as a chemically inert heat transfer
medium. When the reaction fires, undoubtedly some He will be ionized and the
H atoms around, may contribute to the energy yield. That is not unfortunate,
it is just a sideshow.
The reactions involved in the test cell are complex, and discussed on pp
10-12, equations 23-35. The next-to-bottom paragraph of p11 is specially
interesting.
NaH apparently qulaifies as a catalyst because heating can intiate a
reaction resulting in H[1/3] which is a hydrino catalyst.
That is secondary. The primary reason it qualifies as a catalyst is that the
sum
of the three components of the dissociation energy into the specified
components
adds to 54.35 eV, which is a close match for 54.4 eV.
Ah, but as a compound those electrons are in place. The riddle here is that
Na in a compound does not appear to manifest the required energy hole. The
molecule may thermally dissociate, with the H taking back it electron. Where
is the energy to ionize the Na as it separates from the H? If Na can act as
a catalyst during the separation with only thermal energy, then the
"resonant raansfer" phenomenon as used/described by Mills apparently has new
aspects. Ignoring this detail, and regarding the H[1/3] rpoduct of the
reaction, then a 'conventional' hydrino catalyst has appeared and can act
with any H around.
It still is not
clear to me where the 54.35 eV for ionizing Na to catalyze H comes from.
Mills has this weird way of writing his equations. Note that the Hydrino
reaction itself on the right hand side of equation 23 actually produces
108.8
eV, half of which goes into the electron hole, and the other half of which
is
just direct free energy.
Any one else would just have written eq. 23 with an excess of 54.45 eV on
the
right hand side, and nothing on the left.
MC: agreed, I have traouble understanding these chemical equations.
He writes it the way he does, in order to indicate that the energy release
occurs in 2 phases, the first resonant energy dump into the "hole" (which in
this case is 54.35 eV), and the second phase release, which is likely in the
form of kinetic energy.
However don't mistake the 54.35 eV on the left as external input to the
reaction. It isn't. (it's just a quantity of -54.35 eV that Mills has
transferred from the right hand side of the equation to the left hand side).
What he should have done was:
NaH -> Na++ + 2 e- -54.35 eV + H[1/3] + 108.8 eV (note that the net on the
right
hand side is 54.45 eV)
This makes it obvious that 54.35 eV is needed to break up the molecule,
while
the shrinkage yields a total of 108.8 eV.
After the Hydrino forming reaction is complete, there is still free Na++ in
the
environment, and when this reacquires its missing electrons and recombines
with
a free H atom, to form a new molecule of NaH, a total of 54.35 eV is
released.
So in total for the two reactions (23 & 24) we get 54.45 (from 23) and
54.35
(from 24) = 108.8, which is precisely the total released during Hydrino
formation.
To make a long story short, when the Hydrino forms, part of the energy
released
is stored in chemical form (Na++ etc.) and part is released directly to the
environment. The part stored in chemical form is then shortly (and
separately)
also released to the environment as per equ. 24.
MC: That helps a bit, Robin, but where does the 54.45 eV come from? The
thermal input from the heater does not seem enough, and there is no
ionization field as in the microwave cell. Yet the DSC plot clearly shows
something happening.
Regards,
Mike Carrell
[snip]
Regards,
Robin van Spaandonk <[EMAIL PROTECTED]>
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