fusion heat energy atom electron

 >>>u_sci/physics 3500 nat@netcom.COM(18428)13Nov91 13:08

TITLE: Reproduceable Excess Heat -- From sci.physics.fusion

+From         : nat@netcom.COM (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 <rcs@cs.arizona.edu> 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My guess is that this work will not be published in Nature either.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 (508)879-4457--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