to split a deuteron costs 2.2 MeV. Hot fusion of two deuterons
yields about 4 MeV. At best this would
never yield more than about a factor of 2....and that’s not taking into account
any of the losses. And those losses will
be very significant.
1) Maybe 1% of the electrons will create significant x-rays, of which only a
fraction will have the requisite minimum
energy of 2.2 MeV. => most of the electron energy ends up as heat.
2) Only a fraction of the 2.2 MeV or greater x-rays will split a deuteron
(1%?). The rest just ionize atoms and end up
as heat.
3) Of the split deuterons, only a fraction will produce neutrons with even the
minimal energy required to fuse two
deuterons (5 keV? - but the more the better).
4) Of those neutrons, only a fraction will actually accelerate a deuteron
resulting in a fusion reaction.
5) A fusion reaction will primarily create two energetic particles, both of
which can further accelerate other
deuterons, however only a tiny fraction of them will actually do so. Most will
simply lose energy ionizing surrounding
atoms, and end up as heat.
In all, …. they would be lucky to get even one part in a million of the
electron beam energy out as fusion energy,
if the proposed method were actually an accurate description of what happens in
their reactor.
< https://www.mail-archive.com/vortex-
Jones Beene Thu, 06 Aug 2020 08:23:11 -0700 H LV wrote: ***Remember 10-12 years ago the buzz around x-rays from peeling tape?
https://youtu.be/r63e5y3Z3R8
*** If this way of generating x-rays could be harnessed it would make this lattice confinement fusion more economical.
That is a QM effect which does not scale up. The same could be said for much of LENR. In addition, it would seem that the Lawson criterion of hot plasma fusion would also apply, in a modified (reworded) way to the new and improved semantics for lattice enhanced but no longer "cold" fusion. i.e. when we observe effective temperature and pressure on the femtoscale.
As for input - an external electron beam of hot fusion could be modeled as internal k-shell or l-shell resonant electron. Here is the Wiki site for Lawson.
https://en.wikipedia.org/wiki/Lawson_criterion
IOW - one needs only to reduce the geometry of the active site to its actual minimal dimension to see the similarity to plasma fusion, except for one big distinction.. The lack of gamma radiation remains the main difference between hot and (formerly) cold -- and this is where the lattice itself comes into play.
We have to assume that Hagelstein got that part right, or close - and that the lattice carries away most of the downshifted excess instead of gamma radiation. With that addition, the old "cold fusion" becomes the new QM-lattice-fusion.
It never was cold, was it?
https://www.mail-archive.com/vortex-l@eskimo.com/msg119229.html