Project UFO/Project UFO (Disk 3 of 6).adf/FUSION.THE

Text File | 1980-01-09 | 15KB | 322 lines

Article 4739 of sci.physics:
Path: dasys1!cucard!rocky8!cmcl2!rutgers!ucsd!ames!lll-winken!uunet!mcvax!ukc!etive!bjp
From: [e-mail] (B Pendleton)
Newsgroups: sci.physics
Subject: New theory of room temperature fusion
Message-ID: <[e-mail]>
Date: 31 Mar 89 23:12:49 GMT
Reply-To: [e-mail] (B Pendleton)
Organization: Edinburgh University Computer Services
Lines: 43
Posted: Fri Mar 31 18:12:49 1989

Posted for a friend without access to the net -- *please* do *not* reply
to me.  Responses to [e-mail]


Theory of "Kitchen Table Fusion"
--------------------------------

Most physicists agree that the most puzzling part of the recently announced
room temperature fusion is how two deuterons can overcome the intense coulomb
repulsion and get sufficiently close for the short range attractive nuclear
force to take over and produce a triton plus emitted neutron.  It has been
suggested that there is a catalysis or tunneling process that allows the two
deuterons to get very close.  I shall argue that this is not necessary.  What
may actually happen is the following.  Deuterons are absorbed into the regular
crystal lattice of the Palladium electrode in such a way that the deuterons
are aligned in their internal isospin space.  Now, since isospin is a
continuous symmetry, and this alignment picks out a particular direction in
isospace, then Goldstone's theorem tells us that there must be a massless
Goldstone boson associated with this symmetry breaking.  For the proton and
neutron, this isospin symmetry is not exact -- the u and d quarks are not
exactly massless and the Goldstone boson corresponding to axial symmetry
breaking (namely the isotriplet pion) is not exactly massless -- it has a mass
of 139 Mev/C^2.  However, the deuteron contains equal numbers of u and d
quarks, moreover, it is the *vector* isospin symmetry that is broken in the
Palladium crystal, and the symmetry breaking factor cancels to some extent.
A back-of-the-envelope calculation reveals that the Goldstone of the new
symmetry breaking would have a mass of the order of tens of Kev/C^2 -- or
about the same as the energy in the Coulomb barrier!!  This balance of
energies allows fusion to occur in a controlled fashion.  In other words,
there is little chance of an "accidental H-bomb".

Another way of looking at it is to notice that this new Goldstone gives the
nuclear force a longer range than is usual and so the tunneling or catalysis
mechanism is not necessary -- the two deuterons do not need to get close
together for fusion to occur.  In addition, the energies of the emitted
neutrons will be *small* -- one does not need to "get back" the coulomb energy.

Of course, the theory has to be worked out in more detail, but it does seem
more promising than any other I have seen.

John Butcher
--
Hello!


Article 35 of sci.chem:
Path: dasys1!cucard!rocky8!cmcl2!husc6!ukma!gatech!bbn!rochester!dietz
From: [e-mail] (Paul Dietz)
Newsgroups: sci.physics,sci.chem,sci.research,sci.space
Subject: Re: Reactions described in the Pons seminar summary
Message-ID: <[e-mail]>
Date: 3 Apr 89 19:29:49 GMT
References: <[e-mail]> <[e-mail]> <[e-mail]>
Reply-To: [e-mail] (Paul Dietz)
Organization: U of Rochester, CS Dept, Rochester, NY
Lines: 34
Xref: dasys1 sci.physics:4771 sci.chem:35 sci.research:684 sci.space:8441
Posted: Mon Apr  3 14:29:49 1989

In article <[e-mail]> [e-mail] (William Johnson) writes:

>However, I would like to point out that the most mystifying thing about the
>Fleischmann-Pons experiment -- and many things about it are mystifying -- is
>that *none* of the nuclear physics makes sense.  I say this not implying that
>F&P don't know what they are talking about, but rather that many things about
>the experiment -- notably the enormous dearth of neutrons observed relative to
>the energy allegedly released -- fly in the face of what we *think* we know
>about (d,d) reactions.

Everyone has been assuming that the neutrons are coming from catalyzed
dd reactions.  If, instead, some exotic fusion reaction was occuring
-- say, Li6 + d -- we'd expect some neutrons anyway.  First, a fast
charged fusion product would occasionally break up a deuteron before
stopping.  Second, deuterons would occasionally be scattered and fuse
with another deuteron.

Some proposed experiments:

(1) Measure the ratio of neutron rate/power density as the density
   of d atoms increases.  It should increase if this model is true.

(2) Measure the neutron spectrum -- it should differ considerably from
   that of cold dd fusion.

(3) Try to detect energetic fusion product nuclei by mixing the Pd
   with beryllium and observing the neutron flux.

(4) Try to observe fusion products directly by using a low energy deuterium
   ion beam to saturate a very thin target of Pd.  Turn off the beam
   and observe any charged particles emitted.

    Paul F. Dietz
    [e-mail]


Article 4761 of sci.physics:
Path: dasys1!cucard!rocky8!cmcl2!husc6!bloom-beacon!tut.cis.ohio-state.edu!ucbvax!decwrl!shelby!glacier!jbn
From: [e-mail] (John B. Nagle)
Newsgroups: sci.research,sci.physics
Subject: Energy efficiency of Pons-type cold fusion
Keywords: fusion Pons
Message-ID: <[e-mail]>
Date: 2 Apr 89 18:45:16 GMT
Sender: John B. Nagle <[e-mail]>
Organization: Stanford University
Lines: 23
Xref: dasys1 sci.research:678 sci.physics:4761
Posted: Sun Apr  2 13:45:16 1989


      Does this thing really result in a net gain in energy?  Pons reports
that one must pump energy in for weeks to months before getting energy out.
Further, there seems to be an implication that the energy output doesn't
go on indefinitely once output is achieved.  But Pons's numbers on output
only seem to refer to the situation that exists during the output phase.
He doesn't integrate the total energy input during the startup phase
into his "gain" figures.

      The obvious question to ask is whether this is some kind of energy
storage phenomenon.

      Suppose that there are actually two phenomena taking place here.
One is tunnelling-type fusion as previously reported, which would account
for a modest output of neutrons.  The other is some kind of energy
storage, which may be chemical in nature, as in a battery.  This would
explain the low output of neutrons along with the high energy output.

      From this standpoint, a reasonable question to ask is whether the
energy density observed is totally out of reach for an electrochemical
system.  Is a battery hypothesis totally untenable?

                    John Nagle


Article 4750 of sci.physics:
Path: dasys1!cucard!rocky8!cmcl2!rutgers!cs.utexas.edu!ut-emx!ethan
From: [e-mail] (Ethan Tecumseh Vishniac)
Newsgroups: sci.physics
Subject: Re: Success with cold fusion reported
Summary: making Helium4
Message-ID: <[e-mail]>
Date: 1 Apr 89 22:12:09 GMT
References: <[e-mail]> <[e-mail]>
Organization: The University of Texas at Austin, Austin, Texas
Lines: 40
Posted: Sat Apr  1 17:12:09 1989

As I understand it, from summaries of talks by Pons and Fleischmann
(not from their preprint) they have detected Helium4 as a byproduct
of their reaction.  Their estimation of production rates for
Helium3 and tritium are consistent with each other and much lower
(by about 10^9) than the rates required by the observed (or claimed)
energy production.  It therefore seems impossible to invoke any
reactions that use He3 or tritium as fuel to explain the energy
production.  This leaves 2D into He4, but under normal circumstances
this is down from the production rates for He3 and tritium by one
power of the fine structure constant since it involves emitting
another photon.  So we can either discard this possibility or choose
to believe in an enhancement of the rate due to the emission of
phonons with energies of several MeV.  (Incidentally, this makes
the fusion experiment the shrillest noise on Earth :-)).  I find
the idea of these "super-phonons" unlikely, but I'm hardly an
expert.

The only alternative that comes to mind is Li6 + D into 2He4.  This
has the virtue of avoiding nucleon emission in a reaction whose rate
is not obviously tied to the measured reaction rates for 2D into
something.  (Actually, I assume that this would make Be8 which
would decay into 2He4 on a time scale of 10^(-8?) seconds.)
This means the addition of LiOD to the solution is important for
physical, as opposed to chemical, reasons.  However, is it really
true that the lithium would dissolve into the palladium?  If not,
it is difficult to see how it could be involved.

The results of P and F still don't make much sense to me, but if
they are correct then this would seem like the most likely hypothesis
(which may be a dubious honor).  If this is right then it suggests
that lithium is about to become a little more expensive.  :-)
(sort of).

--
 I'm not afraid of dying     Ethan Vishniac, Dept of Astronomy, Univ. of Texas
 I just don't want to be     {charm,ut-sally,ut-emx,noao}!utastro!ethan
 there when it happens.      (arpanet) [e-mail]
    - Woody Allen            (bitnet) [e-mail]

These must be my opinions.  Who else would bother?


Article 4759 of sci.physics:
Path: dasys1!cucard!rocky8!cmcl2!lanl!hc!ames!amdahl!pyramid!prls!philabs!linus!mbunix!eachus
From: [e-mail] (Robert Eachus)
Newsgroups: sci.physics
Subject: Re: UU Power Requirements
Summary: I don't agree
Keywords: power fusion
Message-ID: <[e-mail]>
Date: 30 Mar 89 00:53:46 GMT
References: <[e-mail]> <[e-mail]>
Sender: [e-mail]
Reply-To: [e-mail] (Robert I. Eachus)
Organization: The MITRE Corporation, Bedford, Mass.
Lines: 32
Posted: Wed Mar 29 19:53:46 1989

In article <[e-mail]> [e-mail] (andrew) writes:

>A check on the claims made here. Pons has reported :
>    Pd wire 0.25" diam, 1" long
>    26 W/cc generated @ 100 degC after "a few minutes"
>    generated energy 4.5 times input energy
(stuff deleted)
>The output would be at 20mV with 4.6W supplied. I doubt whether this
>current could be sustained for more than a minute or two, however.
>In this configuration, a 12V source is dissipating 2900 W internally.
>Maybe the batteries melted the concrete and exploded?! (just kidding).

>Check my figures - I hope they're OK.

     I agree with your 4.6W figure, however as I understand the
apparatus the current is not passing directly through the palladium,
but is being used to electrolyze D2O.  Depending on your asumptions
about drop across the cell you get on the order of 2 Amperes, which is
much more reasonable.  I assumed that batteries were used for the demo
(and the experiment) because for this type of setup a battery,
electrolysis cell, and resistor loop allows you to acurately measure
current and power without worrying about imposed AC.

     The four-inch deep hole in the concrete (assuming that it is
true) would result either from spalling of the concrete or from
hot battery acid removing the water.  Concrete, as such, cannot melt,
since once the cement has given up its water of crystalization all you
have is a mix of dry powders, most of which will break down chemically
before melting at one atmosphere. (CaSO4.2H20 -> 2CaSO4.H2O -> Ca(OH)2
+ CaO + 2S03, etc.)

                    Robert I. Eachus


Article 4780 of sci.physics:
Path: dasys1!cucard!rocky8!cmcl2!lanl!hc!lll-winken!xanth!nic.MR.NET!umn-d-ub!umn-cs!ns!logajan
From: [e-mail] (John Logajan)
Newsgroups: sci.physics
Subject: Fusion: Increasing energy gain.
Message-ID: <[e-mail]>
Date: 3 Apr 89 17:30:39 GMT
Organization: Network Systems Corp. Mpls MN
Lines: 19
Posted: Mon Apr  3 12:30:39 1989


The energy-in/energy-out ratio was, I think, very carefully worded when the
10:1 figure was mentioned.  I think they said the 10:1 ratio EXCLUDED the
losses associated with I*I*R in the heavy-water, and possibly in the
0.8mv drop in the electrode itself.  In fact, it is not clear to me if
they include the several weeks worth of "pumping up" electrolysis loss.

Which brings me to my next point:  Even if a voltage gradient is needed
during fusion (instead of just a pressurized vessel of deutrium gas), there
is no reason to assume that the "pumping up" process couldn't be done
in a pressurized mode.  Maintaining pressure (even if it takes longer) is
much more energy efficient than producing pressure via electrolysis.

So, saturate the electodes in a pressure cooker first, then run them in
fusion mode in the electrolytic bath.

--
- John M. Logajan @ Network Systems; 7600 Boone Ave; Brooklyn Park, MN 55428  -
- ...rutgers!umn-cs!ns!logajan / [e-mail] / [e-mail] -


Article 4781 of sci.physics:
Path: dasys1!cucard!rocky8!cmcl2!lanl!hc!lll-winken!xanth!nic.MR.NET!umn-d-ub!umn-cs!ns!logajan
From: [e-mail] (John Logajan)
Newsgroups: sci.physics
Subject: Notes on fusion
Message-ID: <[e-mail]>
Date: 3 Apr 89 17:29:40 GMT
Organization: Network Systems Corp. Mpls MN
Lines: 24
Posted: Mon Apr  3 12:29:40 1989


I just read about the Mossbaur effect in the book Elementry Modern Physics.

Seems that when a gamma ray is emitted from a free floating nucleus the nucleus
recoils in the opposite direction -- the nucleus gains X kinetic energy and the
gamma ray departs with Y energy.   BUT!!! if the atom of the nucleus is bound
in a lattice, then the recoil energy is smaller and the gamma ray energy is
LARGER.  In other words the fact that the atom was bound in a physical
structure caused the gamma to exit with an increased frequency!!!!!!!

The point is that the lattice chemical forces can affect the results of
nuclear reactions!!!!

Also in the book they say that typical spontaneous alpha fissions of heavy
nuclei are of less energy than the coulomb gradient energy -- meaning that
the alpha particles have tunneled out.  They cite 8Mev as the coulomb force
and 4.5Mev as the alpha particle energy -- in one instance.

They also said that fusions can occur through tunneling, but they did not give
a memorable example.

--
- John M. Logajan @ Network Systems; 7600 Boone Ave; Brooklyn Park, MN 55428  -
- ...rutgers!umn-cs!ns!logajan / [e-mail] / [e-mail] -