United Public Domain Gold 2/United Public Domain Gold 2.iso/utilities/pu729.dms/pu729.adf/Fusionheat.txt

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>>>u_sci/physics 3500 [e-mail](18428)13Nov91 13:08
TITLE: Reproduceable Excess Heat -- From sci.physics.fusion

+From         : [e-mail] (Nathaniel Stitt)
+Organization : Netcom - Online Communication Services  (408 241-9760 guest)
+Keywords     : fusion heat energy atom electron


This is from a thread on sci.physics.fusion.  The following articles
were all posted by Prof. John Farrell, one of the authors of a paper
published in Fusion Technology, V20, Aug. 1991.

The most interesting claim is of a 100% reproducible method of
generating measurable excess heat.  Note that the heat generating
process is NOT claimed to be fusion, but rather a previosly unobserved
physical process.

This repost to sci.physics is by permission of Prof. Farrell.


======= Article posted to sci.physics.fusion on Mon, 4 Nov 1991 21:47:47 GMT

Richard Schroeppel <[e-mail]> asks why the paper by Mills
and Kneizys, Fusion Technology. 20 (1991) 65, has not received more
attention and if there has been any duplication.

1.  The paper was published in Fusion Technology.  Not many
institutions receive this journal--thus, not much attention.

2.   Mills, Kneizys, and Farrell are not calorimetry people. (Although,
when you get this kind of excess heat, how good do you have to be.
Furthermore, we had some excellent calorimetry people come in and
help us.)

3.  Mills and Farrell have tried to publish the theory elsewhere--Phys.
Rev. Lett.,  for example.  Some of the referees have been favorable
(very creative, intriguing, and so on).  But trying to publish a theory
that overturns Schrodinger mechanics is quite difficult (as you might
expect).

4.  Yes, we have had others duplicate the work.  One internationally
famous electrochemist has submitted his results to Nature.  As you
know, Nature has not published **any** positive cold-fusion research.
My guess is that this work will not be published in Nature either.

5.  I am not at liberty to give you the names of all who have had
positive results with our system (0.6 M K2CO3 with a Ni cathode and a
Pt anode).  To my knowledge, six labs have successfully repeated the
work.  I can give you the following :  V. C. Noninski [phone]--his
work has been accepted for publication in Fusion Technology.  James
McBreen, Brookhaven National Labs (Upton, NY).

6.  The work reported in Fusion Technology was essentially a 100 mW
reactor.  We have had a 100 W reactor working for about one month.
30-50 W in, that is, (Vappl - 1.48) i  =  30-50 W.  100-120 W out.  There is
little or no recombination of hydrogen and oxygen.

7.  This week we should have a 1000 W reactor going.

8. We have **never** had a K+/Ni system that did not produce excess
heat (unless we poisoned the electrode).  As far as we can tell this
system is 100% reproducible.

Tips on repeating the experiment:

1.  Use normal water, H2O not D2O, unless you are looking for tritium
or neutrons.  Essentially all of the heat is **not** caused by fusion but
by some other physical process--namely shrinkage of hydrogen atoms
from the n = 1 state to the n = 1/2 state.  (I know this is hard to believe,
particularly for someone like me who has taught quantum chemistry
for 25 years, but life **is** stranger than fiction.)

2.  Ni foil or wire can be used.  The Ni should be clean.  Handle the Ni
carefully with cotton or plastic gloves.  Do **not** clean the Ni with
nitric acid or organic solvents.

3.  About 0.6 M K2CO3 is best.  Lower and higher concentrations work
but not as well.

4.  Use a current density of 1 ma/cm2 with a foil or 2 ma/cm2 with a
wire.  Most researchers are using current densities that are **too
high**.  The object is to form H atoms on the surface of the NiHx.
These H atoms then can undergo a catalytic shrinkage in the presence
of K+ (or other suitable ion).  If a high current density is used the H
atoms are forced off of the surface.  (The Ni does not enter into the
reaction, it simply is a surface on which the H atoms can form.)

5.  It is important to electropolish the Ni cathode before beginning the
calorimetry.  That is, run the electrolysis (preferably in the calorimetry
cell) for about half an hour to an hour with the Ni as the anode and the
Pt as the cathode.


Other possible systems:

Thousands of other systems are possible.  Unfortunately, most of these
are ions or ion combinations that are difficult or impossible to make.
We have tried many of the chemically reasonable ones and the K+
system works best.  Pd2+/ Li+ works, but not as well.  Note Pd2+,  not
Pd metal.  We believe that to the extent that Pd/Li+ works, it is Pd2+
on or near the surface of the Pd that is the active species.  Rb+ works,
but not as well.  Li+, Na+, Cs+ do not work.  Ti2+ does work.  Here
again, to the extent that Ti/D2 gives neutrons, we believe that the
active species is Ti2+.  (Whenever neutrons are given off, enormous
amounts of heat are given off as well.  The heat does not come from
the fusion itself but from shrinkage of the H atoms or D atoms to a size
sufficiently small that fusion can occur.  As far as we can tell, only an
extremely small fraction of the atoms shrink sufficiently for fusion to
occur.)


John Farrell
Franklin & Marshall College



======= Article posted to sci.physics.fusion on 8 Nov 91 15:53:43 GMT

John Moore asks:

>What is the source of energy in this system?

Hydrogen energy levels are given by E(n)  = -13.6 eV/n**2.
H(n = 1) is at -13.6 eV.  Assume, for a moment, the wildly impossible--
that n can not only = 1, 2, 3, ... but that n = 1/2, 1/3, 1/4, ... .
Then, H(n = 1/2) is at 4(-13.6 eV) = -54.4.   Thus, the energy difference
between the two states is 40.8 eV.  That is, a 40.8 eV photon should be
given off in going to the n = 1/2 state.

At -54.4 eV this atom, should it exist, would be *very stable*.  After all,
the ionization energy of He is 24.6 eV.  That is, this atom should be
**extremely stable** and very small-- r =  (1/2) of the Bohr radius.
There would be no cooling when the electrolysis is stopped.  The only
way to get the atom to return to the n = 1 state is for it to absorb a 40.8
eV photon.

One can observe these 40.8 eV (and higher energy) transitions by
putting dental film next to the Ni cathode.  You have to remove the
film from the plastic and from the cardboard wrapping (in a dark
room).  The film must be wrapped in water tight material, but thin (so
that the 40.8 eV photons can penetrate).  (A condom works fine.)  The
whole experiment must be done in a dark room or dark container.
Suitable controls must be used--like Na2CO3 instead of K2CO3.
Electrolyze for a week or so, you will observe dark spots on the film
(K2CO3 solution only) indicating hot spots on the
cathode.  (This is, of course, not conclusive proof of these lower energy
states for hydrogen but we are getting there.)


>What are the values of Vappl and I (in the 100 W reactor)?

Vappl = 4 volts      I =  20 amps  We used a Kepco constant current
power supply, Model ATE6-50M.  This requires about 2000 meters
of 0.127 mm diameter Ni wire (Johnson Matthey).  We used
platinized Ti mesh anodes.

This is about 50 W,  (4-1.48)20  = 50 W.  At 50 W we get about 120 W
out.  (We applied no correction for heat taken away by the escaping
gases, which is considerable, so the actual output is probably greater
than 120 W.

Notes:

1.  When you run 100-200 mW reactors you have to use a well
insulated container with a cell constant of at least 20 C/watt.  We used
small dewar flasks.

2.  When you run a 100 W reactor you need a less well insulated cell--
about 0.5 C/watt.  We use a large nalgene beaker with a fitted nalgene
lid (with holes for letting the gases escape).


John Farrell
Franklin & Marshall College


======= Article posted to sci.physics.fusion on 11 Nov 91 15:39:58 GMT

Chuck Sites asks:

1.  Have you run any AC experiments?

Answer.  We have **pulsed** the voltage and current.  We vary the
voltage in a (almost) square wave.  We get a lot more heat this way, but
this introduces another complication in calculating the excess heat.  At
this point, we would prefer that anyone who wants to duplicate these
results use constant current or constant voltage or, better yet, constant
power.

2.  Are any salts formed on the cathode or anode?

Answer.  The only noticeable salt formation is at the liquid/air interface
as the liquid level lowers.  Also, in the 100W reactor, evolving gases
carry electrolyte out of the cell and we get salt formation on the exit
holes and so on.

3.  Several months back you mentioned KCl as an electrolyte.  Do you
think this is still a good electrolyte to reproduce your H2O experiments.

Answer.  KCl works, but we don't get as much heat as with K2CO3.  In
addition, chloride gas is produced (with KCl) and eventually the
cathode performance is affected.  In general, -2 anions work better than
-1 anions.

4.  How far into and away from the metal/liquid interface do you expect
the formation of H(n = 1/2) to exist.  In other words when does the
transition from this state to the metal band state occur, or the
electrolyte conduction states?

Answer.  We believe that this phenomenon is a surface effect.  Two K+
ions must be very close to a H(n=1) atom.  H (n=1) atoms are formed at
the surface of the cathode and K+ ions are drawn there because the
cathode is negatively charged.  Accordingly, the H(n = 1/2) atoms are
produced at the electrolyte/metal interface.  Unfortunately, we do not
know the fate of these atoms.  They can undergo additional transitions,
n = 1/3, 1/4 etc, but this simply begs the question.  Remember, these
atoms will be extremely stable (that is, inert) and very small (about 1/8
the size of a H(n=1) *atom*.  At the moment, we are searching for
them in the evolved gases.  The hydrogen to oxygen ratio will not be
2:1 if we are correct because some of the hydrogen produced will be
H(n=1/2) and will not form H2 (because H(n=1/2) is so stable).


John Farrell
Franklin & Marshall College



======= Article posted to sci.physics.fusion on 11 Nov 91 15:43:40 GMT

David Taylor asks:

1. On page 72 of the paper published in Fusion Technology, v20, Aug.
1991, Mills states: "The removal of negative Fourier components of
energy m X 27.2 eV, where m is an integer, gives rise to a larger positive
electric field inside the spherical shell, which is a time-harmonic
solution of Laplace's equations in spherical coordinates. In this case, the
radius at which force balance and nonradiation are achieved is a(sub 0) /
(m+1), where m is an integer. In decaying to this radius from the
ground state, a total energy of [(m+1)(m+1) - 1] X 13.6 eV is released. This
process is hereafter referred to a hydrogen emission by catalytic thermal
electronic relaxation (HERTER)."   Mills then gives examples of some
catalysts that can be used (K, Li/Pd, Ti) to cause resonant shrinkage. He
also lists the potentials of the resonating cavities, which vary from 27.28
to 27.54 eV. How close to the 27.2 eV value does one have to be? If not
exactly on, why not?

Answer.  The energy has to be the same as the potential energy of an
electron in tne n = 1 state of the hydrogen atom--at the actual site of the
hydrogen atom that is to undergo the transition.  Published ionization
energies of atoms and ions are for the **gas** phase.  In solution these
ions are hydrated or otherwise chelated and the ionization energy will
be somewhat different than in the gas phase.  In addition, the ions may
have some velocity relative to the hydrogen atom.  Finally, the energy
will be affected by the electric field (and possibly the magnetic field) of
the electrode.  Thus, we allowed some leeway.

2. Doesn't both potassium and hydrogen exist in the required states in
seawater? What about other reaction chains where all the products
exist simultaneously and can combine to produce resonator cavities of
~ 27.2 eV?

Answer.  Most of the hydrogen in seawater is in H2O, HCO3-, and so
on.  Not much hydrogen in seawater exists as **hydrogen atoms**.
Furthermore,  The hydrogen atoms must be in close proximity to
**two** K+ ions, probably at a specific distance.

3. None of the equations that describe the reaction used for this
research (potassium carbonate) mention the platinum anode. Is it
really necessary to use such an expensive piece of wire, or would
copper serve as well? If copper would be bad (for cathode muck buildup
or other reasons), would platinum electroplated over copper work OK?
How about using nickel for both the cathode and the anode and
running high frequency AC?

Answer.  It is not necessary to use Pt.  But you have to use something
that will not go into solution and gum up the cathode.  Recently we
have been using platinized Ti which is much cheaper than Pt.  We
have used Ni/Ni.  It works, but Ni is transferred form the anode to the
cathode.


4. Do the equations for D hold for plain old H? (I assume they do, since
the extra neutron wouldn't do anything to the charge radius or
whatever.)


Answer.  Yes.

5. When deuterium is used, tritium and protons are produced. What
happens when plain old hydrogen is used (as suggested in the recent
post on repeating the experiment)?

Answer.  If you mean --are any nuclear products formed?  We don't
know.  We hope not.

Paul Dietz asks:

In electrolysis in carbonate solution, can't peroxycarbonate ions be
produced?  Would their recomposition account for the extra heat?

Answer.  We titrated the solutions after one month of operation and
found no change in the carbonate concentration.


John Farrell
Franklin & Marshall College



======= Article posted to sci.physics.fusion on Mon, 11 Nov 1991 20:25:59 GMT

Michael Robinson asks:

>So, lets just say, for the sake of amusement, we managed to accumulate
>a liter or so of, say, H(n=1/8), and explosively compressed it.  Would it
>go boom?

No.


Richard Mathews had several question about the angular momenta of
the fractional quantum states.

We haven't specifically looked at this aspect yet.  But I can give a brief
answer as to why it is so difficult to get to the fractional quantum states.
We call the ground state, the n = 1 state, the "no photon state".  The
electron in  H(n =1) is a spherical shell (infinity ttin) at the Bohr
radius, a(sub zero)--kind of like a soap bubble.  (Note the the electron is,
fundamentally, two-dimensional.)  When H(n = 1) absorbs a photon,
the photon is trapped (in the cavity).  The electric field of the trapped
photon reduces the electric field in the cavity.  That is, in the n = 2 state
the electric field caused by the proton is +1, the electric field of the
photon is -1/2, and the **effective** electric field (or nuclear charge) is
+1/2--the atom is twice as big.  That is, the radius is now 2 x a(sub zero).
In the n = 3 state, the effective nuclear charge is +1/3 and the radius is 3
x a(sub zero)--the decrease in the nuclear charge is caused by the electric
field of the trapped photon.  When the effective nuclear charge is zero,
r = infinity and the electron is ionized (the electron is now a *two-
dimensional* plaee wave).

In order to get to the n = 1/2 state, ome must remove electric field from
the cavity such that the effective nuclear charge is +2.  That, is you
have to increase the **effective** nuclear charge from +1 to +2.  This
requires the removal of 27.2 eV of energy and can be accomplished by
removal of negative Fourier components of the electric field of the
proton.

Note that when you absorb a photon, the photon must be of the correct
energy--quantized (quantization comes from the size of the cavity, not
from an intrinsic property of small particles).  The photon is still there,
however, and can be ejected with a return to the "no photon state".
What are the conditions that will allow us to remove 27.2 eV of energy
from the electric field of the proton?  Good question!  All we know
now is that you must have, nearby, an energy hole of 27.2 eV.

Take Ti2+ as an example:

Ti2+  =   Ti3+   +  e-              IE  =  27.49 eV

Is 29.49 eV close enough?  Maybe.  This is a gas phase IE.  Plus, the
atoms have kinetic energy.  What happens to the Ti3+ and the
electron?  We don't know.

The Pd2+/Li+ system is cleaner:

Pd2+   =   Pd3+   + e-                    +32.93
Li+   +   e-    =     Li                   -5.39
__________________________
Pd2+   +  Li+    =      Pd3+   +  Li       27.54 eV

In this case, the electron is taken care of and the Pd3+/ Li would
immediately form Pd2+ and Li+ with -27.54 eV **released**.  Thus, we
could have a catalytic system where Pd2+ and Li+ is an energy hole of
27.54 eV that regenerates itself.

K+ to K and K+ to K2+ is another such energy hole (31.625-4.341 = 27.28
eV).  So far, we have found this to be the best system.

In any case, the simultaneous junction of these species is probably a
rare occurrence in nature.  Nonetheless,  there may be some around.

This may be my last communication for a while.  We have the 1000 W
reactor going.  Sorry , I can't give you **any** more info on it.
Studying the properties of the new reactor, my normal teaching
responsibilities , and family matters are more than enough to keep me
busy.  Between the article in Fusion Technology and what I've given
here, it should be possible to repeat the experiment and observe excess
heat.  A reprcducible experiment that gives excess heat is the most
important factor in this whole affair.  Maybe someone will find a
trivial explanation--we have searched hard for one.  The theory is
another matter.  We know we are battling uphill here.  What we really
need is some physicists who are willing to say, "You are probably
wrong, but it looks interesting enough to really dig into and to try to
make it work."  This isn't likely until the experimental work is
verified.  I don't blame them--these ideas consume an enormous
amount of time.  I, personally,  find explaining the theory over the
network very time consuming and frustrating .

Final thoughts.  I want to thank Dieter Britz, Barry Merriman (sorry
about the recent criticism), and all other regular contributors on this
net.  Even the nonbelievers on this net are truly open-minded on this
subject.  I read the net daily and I am very thankful to those who spend
the time and energy to make contributions.

(Part of my frustration stems from the fact that when I paste
some text from Word onto the VAX letters are (randomly) changed.
Why does this happen?)

Goin' Fishin'

John Farrell
Franklin & Marshall College