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259) What is this experiment about?
Ludwik Kowalski (9/22/05)
Department of Mathematical Sciences
Montclair State University, Upper Montclair, NJ, 07043
Next week I will be working with Scott Little and Gregory Luce an ETI (Earth Tech International, Inc.) in Austin. The purpose will be to replicate the
experiment described in unit #252. In the past Scott tried to confirm excess heat in similar experiments (performed by Mizuno and then by Naudin) but
all results were negative. He thinks that this attempt to observe excess heat will also be negative. On the other hand, in a message sent two days ago,
he wrote: If we confirm the excess heat effect, Ludwik, I guarantee you we will try EVERYTHING to see what makes it tick....!!!! At the
end he added: Let vision of fame and glory not be the enemy of accurate measurements. I like this kind of attitude. Each of us would be
happy to confirm excess energy but we want to be objective.
To describe our experiment to a person familiar with high school physics I would begin by asking a question. What is the maximum possible amount of water
that can boil out of a beaker in 5 minutes when an electric heater of 400 W is immersed in it? Then we would calculate the electric energy received by
water, E=400*5*60=120000 joules. Dividing this by 2260 the answer would be 53.1 grams. That is because 2260 J/g is the well known latent heat of evaporation.
In reality less water would be evaporated, perhaps only 40 grams, because not all heat is lost via evaporation; some of it is lost via conduction, convection
Suppose that somebody performed the described experiment and reported that the amount of water evaporated was actually 70 grams. What would you think
about such report? I would ask. Then we would discuss two possibilities: a hidden experimental error and a hidden source of thermal energy. The third
alternative, that the law of conservation of energy is not valid, would not even be discussed in our context. Such topic should be discussed at much higher
An experimental error, either on electrical or on thermal side of the procedure, would certainly be the first thing to suspect in an experiment performed
by a student. But in this case the anomaly (more water evaporated than allowed by the law of observation of energy) is reported by top scientists. P.
Clauzon is a retired nuclear physicist who worked on the French breeder reactor for at least two decades, J.F. Fauvarque is a recognized international
authority in the area of electrochemistry. We are trying to replicate experiments they described in a paper to be published. Furthermore, Fauvarque et al.
were not the first to report the anomaly. Similar reports, from a large number of similar experiments, were published on Naudin website. And Naudin was
only confirming the claim made by Japanese researchers, such as T. Mizuno and T. Ohmori. It is not likely that scientists of such caliber, working in
different laboratories, could make large experimental errors.
That is why my inclination is to suspect a hidden source of energy. Is it possible that chemical reactions are responsible for excess energy? That was a
question I sent to Fauvarque. His reply was a categorical no. The only possible source would be oxidation of tungsten. But the maximum possible amount of
chemical energy from tungsten, he wrote, is much smaller than what is needed to evaporate the reported amount of water. I have to take such statements for
granted because I am not a chemist. For the time being the only option I have is to replicate the experiment as carefully as possible. That is exactly what
we will do next week.
So what has to be measured in each test? Electric energy deposited in water, EL, the mass of the evaporated water, MS, and the amount of non-evaporative
thermal losses, TH. Ignoring a hidden source of energy we expect
EL = MS*L + TH . . . . . . . . . (1)
where L=2260 J/g, is the well known latent heat of evaporation. What can be more simple than to verify that the left side of the equation is approximately
equal to the right side? We want to measure excess heat, EH, defined as:
EH = MS*L + TH - EL . . . . . (2)
According to Fauvaque et al., that quantity can exceed 30 kJ in five minutes (100 W). If this is true then the determination of MS, TH and EL should not
be too difficult. Our methods of measuring these quantities are going to be theirs. The MS will be measured from the readings of the digital scale supporting
the entire setup. The precision of measurements is expected to be better than 0.1%. To determine TH we will multiply the rates of non evaporative cooling,
at 100 C, by the test duration, such as 300 seconds. The rates of cooling, in joules per second (watt), will be measured at several temperatures between 90
and 98 C. The rate of cooling at 100 C will then be obtained through the linear extrapolation. The expected uncertainty in the value of TH, approximately 2%,
will be due mostly to fluctuations in room temperature.
The expected uncertainty in the third quantity, EL, will also be close to 2%. Potential sources of error have already been discussed in unit #256. I think
that our method of measuring the average current should be calibrated by using a simple high school calorimeter. That method has been described in unit #255
(in the piece appended on 9/1/05). Scott knows, on the basis of previous experience, that working at voltages exceeding 300 V might introduce difficulties
not encountered at lower voltages. From what Scott wrote I inferred that problems he encountered might be due to the effect of emitted HF on the data
gathering component. If this is true than grounding the anode, which forms a cylindrical screen around the cathode, might be helpful. he has some other
tricks to deal with that issue.
And in the worse possible case we can stop collecting data with the computer. Fauvarque et al., reported 49 kJ of excess, in 5 minutes, at 350 V. That is
41% of EL, on the average. And it corresponds to the production rate of unexplained energy equal to 164 W. Even very crude voltmeters and ammeters could
be used to detect such rates of production. The strategy , however, should be to do as much as possible at 250 V and at 300 V there the rates of production
of EH was reported as 8% and 15%, respectively. Splashing of the electrolyte will probaly also be more intense at high voltages. Than can lead to an
overestimation of MS, and consequently to an underestimation of EH. Condensation of vapor on the wall of the beaker, and return of water dropps to the
electrolyte, can have the opposite effect of EH. I suspect that these will be our most serious difficulties.
My last comment has to do with L. How do we know that L does not becomes samller than 2260 J/g when vapors are formed in the presence of intensive arcing
and sparking? Using an inflated value of L would certainly generate an illusion of excess heat, as one can see from the last equation above. How can one
distinguish a true excess heat from an apparent excess heat (due to unaccounted lowering of L)? By using a method of measurement of EH that does not depend
on L. How can all thermal energy generated in 5 minutes be measured? What comes to my mind is a closed steel container with very strong walls and some
space above the electrolyte. That space would contan the vapor while temperature and pressure are going up rise during a test. Then
EH = (FC + TH) - EL
where FC is the energy removed by the flow calorimeter and TH, as before, is the heat lost by conduction, convection and radiation. TH can be minimized
by surrounding the cell with a thick layer of styrofoam, as Scott did in trying to replicate Mizunos experiments. TH depends only on the cells
geometry; it can thus be determined, at various temperatures, from the cooling curve for the empty container. To avoid excessive temperatures and pressures
the volume available to vapor should be sufficiently large.
P.S. HERE IS A MESSAGE I POSTED TODAY ABOUT L ON THE CMNS LIST.
Naudin, and others, confirmed large excess heat in Mizuno Ohmori type experiments. The most recent confirmation is described
in the paper of Fauvarque et al. That paper is now downloadable from the library at <www.lenr-canr.org> . It shows that excess power increases
with applied voltage (from nearly nothing below 200 V to ~35% at 350 V. The first author is an electrochemist and I asked him about a possibility that
excess heat might come from chemical reactions. He does not think so. The only possible fuel, he wrote, would be tungsten and not enough of it was
consumed to account for the measured excess heat.
So what is the source of excess heat? The name of the device is CFR (where R stands for the reactor). That indicates one possibility. But to confirm
nuclear origin of excess heat one must show that reaction products accumulate (in a high voltage cell) at a reasonable rate. I am thinking about
another interpretation of experimental results. The underlying assumption is that intense electric arcing, the cause of evaporation, has no effect
on the latent heat of evaporation, L. But what evidence do we have for the validity of that assumption? The value of L was actually measured by
Fauvorque et al. and the result came out to be very close to the known value, 2260 J/g. That shows that no significant errors were made by performing
thermal and electrical measurements. But the expected L was found by using an ohmic resistor.
Let me speculate by assuming that L decreases (perhaps by ~35%) when the cell voltage changes from 200 V to 350 V. Not being aware of this, and using
2260 J/g, one would underestimate the expected amount of water to be evaporated. Expecting 40 grams, for example, and observing 50 grams, would then
be interpreted as excess heat. But that would be an apparent excess heat, not real excess heat. Was this possibility discussed at some conferences?
Latent heat of evaporation depends on binding energy of water molecules (surface tension, if you prefer). I am speculating that presence of a strong
electric field might enhance evaporation. What is wrong with this speculation?
Appended on 9/23/05:
I believe that Conservation of Energy tells us that your speculation cannot possibly be correct, Ludwik. Assume for the moment
that we already have a strong electric field in existence (between capacitor plates, for example). Now bring some water into this field. I think we can do
this for free (energy-wise) provided the water is already at the correct potential. We just "slide" the water in between the capacitor plates
crossing all the E field lines orthogonally at the correct potential point. Now assume for a moment that your speculation is correct. With the water in
the strong E field we can evaporate it using less than the usual amount of energy. Now we convey the water vapor back out of the field along the same
path and condense it in a low field region, recovering the usual heat of vaporization. The result: free energy....just from playing games with ordinary
water. My conclusion: Conservation of Energy tells us that the heat of vaporization of water will not be dependent upon the ambient electrical field
My reply was as follows:
I recognize this "reductio ad absurdum" rebuttal. The absurdum is a possibility, at least in principle, of constructing
a perpetual motion engine. This kind of criticism is very powerful and convincing. Let me rephrase the speculation; perhaps this will make it less vulnerable
to the rebuttal.
Suppose there are two ways of turning water into a vapor (not two ways of decomposing it, as mentioned by Dennis Letts). The first is direct, requiring
2260 J/g, and the second is indirect, requiring only 1000 J/g. The products, ejected from the liquid, are not the same in these two processes. The first
process creates water molecules in air and these are able to recombine and generate 2260 J of heat per gram of water. The second process creates something
that can not recombine, unless an additional 1260 J/g is used OUTSIDE THE CELL, to turn them into common H2O molecules. Do not ask me what the intermediate
products are, or through what kind of oxidation are they transformed into H2O outside the cell. This is only an abstract speculation invented to avoid
Scott's rebuttal. Can a perpetum mobile be constructed on the basis of these two processes? I do not think so.
How can these two processes be used to explain the excess heat reported by Fauvarque et al.? Suppose that the second process does not take place unless
the applied voltage is higher than 200 V. Above that threshold the second process becomes more and more probable when the voltage is increased. At 350 V
the second process is responsible for 50% of the lost liquid. Thus one half of the liquid water is lost at the cost of 2260 J/g while another half is
being lost at the cost of 1000 J/g. The effective L is considerably less than 2260 J/g and this is responsible for the apparent excess heat. The only
thing I did was to postulate that the effective L might be voltage-dependent. What right do we have to assume that any process through which liquid water
can be removed from the cell must have the same energy cost?
Yes, I know that all this will remain highly speculative, until the second process is identified and experimentally confirmed. But this observation is
also applicable to other kinds of speculation, including the idea of nuclear reactions, hydrinos, Casimire effect, monopoles, bacteria, etc. etc. The
Mizuno Ohmori effect seems to be real and we must understand it. . . . The whole point of this discussion is to consider a possibility that not all
liquid water is lost at the same energy cost of 2260 J/g. Can water lost via splashing be treated as if it were lost via evaporation? I do not think so.
How to avoid splashing (or how to account for it)? According to Pierre Clauzon (private communication) splashing becomes more and more pronounced as the
wattage goes up at higher voltages. Practical advice on that point would be highly appreciated.
Large bubbles seen in boiling grow from tiny bubbles spontaneously formed in water at any temperature. Boiling occurs when vapor pressure inside tiny
bubbles exceeds the outside pressure (at t=100 C, when the outside P=1 atm). Under such conditions bubbles formed in the liquid do not collapse; they
grow and rise to the surface. This is not the same kind of evaporation as at lower temperatures, when vapors are formed at the surface. It turns out,
however, that the values of L, measured at the boiling temperature (2260 J/g) are nearly the same as the value of L measured below the boiling point.
So much for normal boiling and evaporation.
Is the glowing volume of plasma, surrounding the cathode inside the electrolyte, made of ionized water vapor? If so then one might think that plasma is
an electrically formed vapor made of ions. Why should the energy needed to form one gram of ionized vapor be the same as to form one gram of neutral vapor?
Appended on 9/24/05:
YY wrote: The correct answer to Ludwik's question is that L (change in enthalpy for vaporization) is a STATE function,
hence it can only depend on the initial and final states-it is independent of path. Thus, the change from D2O(liq)
to D2O(vap) at standard conditions will always yield 45.401 kJ/mol or 2266.9 J/g. The cell voltage used cannot
change this value. If this were not true, you could construct a cyclic process that returns the system to the initial state and produces energy,
i.e. a perpetual motion machine in violation of the First Law of Thermodynamics.
And here is my reply: Thanks for your comment, YY. XX made a similar rebuttal last night. That forced me
to modify the speculation. In the subsequent message I introduced two kinds of vaporization and two kinds of products being ejected from the
liquid. Do you think that the subsequent speculation is also in conflict with the first law? In other word. IS MY
SPECULATION, INVENTED TO EXPLAIN EXPERIMENTAL DATA OF THOSE WHO CONFIRMED EXCESS HEAT IN HIGH VOLTAGE ELECTROLYSIS, IN CONFLICT WITH THE
In the normal evaporation (path #1) neutral molecules are ejected from the liquid electrolyte at the cost of 2267 J/g. In the competitive
process (path #2) what is ejected, at the cost of 1000 J/g is not an already made water molecule. It is something that would need an additional
1267 J/g, outside the cell, if it had to become a water molecule. That additional energy would have to be taken from somewhere (for example,
a chemical reaction) outside the cell. In the context of the experiment I want to interpret the only factor that counts is the mass lost
during each 5-min test. We do not care what happens to the lost mass outside the cell. Is this position acceptable? It is an alternative
hypothesis; the other hypothesis is to postulate that an exothermic nuclear process takes place in the cell and that only path #1 is
responsible for the lost liquid.
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