This website contains other cold fusion items.
Click to see the list of links

376) Arata type experiments

Ludwik Kowalski

Montclair State University, New Jersey, USA
September 17, 2009


Arata and Zhang persentation, described in (1), took place in May 2008. A recent paper by Kitamura et al (2) describes attempts to replicate experiments of Arata and Zhang. Actually, their paper is more than a replication. Arata’s experiment did demonstrate simultaneous generation of excess heat and helium. But no quantitative results were produced (how much thermal energy and how much helium). Kitamura’s paper does provide information about the amount of thermal energy but no information about helium. Hopefully amount of helium produced inside the Pd powder will be measured and results will be reported at the upcoming CMNS conference. If helium is the “nuclear ash” from “cold fusion” then the amount of excess energy per atom of created helium should be close to 24 MeV. Quantitative correlation between excess heat and helium has been reported by several investigators, for example, by Mike McKubre et al (3).

Arata’s Experiments
As mentioned in my item #23 (at this website) Arata used the powdered palladium. Initially the powder was placed into a cavity of a palladium cathode. The deuterium atoms, released during electrolysis, migrated into the cavity through the thin wall of that double-structured cathode. This project was summarized in (4). In more recent experiments (1) the powder (mixture of Pd and ZrO2, an alloy of Zr+Ni+Pd, or pure Pd) was placed into a large cell. The cell was first evacuated and then exposed to pressurized gas. Excess heat and helium were produced when the gas was pure deuterim; these byproducts were not produced when the gas was common hydrogen.

Presence of excess heat was inferred from a small difference of temperature (about 0.5 C) between the inside and outside the cell. That difference of temperature remained constant for about 2500 minutes (after the powder was saturated with deuterium). Presence of 4He--most of it in the powder and not in the gas--was demonstrated by using a high-resolution mass spectrometer. The current due to ionized 4He was about 25 pA. This number, unfortunately, cannot be used to calculate the number of helium atoms. To accomplish this one must know the R ratio (the number ions reaching the mass spectrometer’s collector divided by the number of helium atoms produced during electrolysis.) Suppose R=1.0, which is highly unrealistic. In that case the charge collected in 2500 minutes would be 25*10-12*2500*60=3.75*10-6 coulombs, corresponding to
3.75*10-6/1.6*10-19=2.34*1013 produced atoms. The 1.6*10-19 is the net charge of a single helium ion in the spectrometer.

How much excess heat would be produced if 23.8 MeV were released with each helium atom? The answer is 89 joules. The corresponding power would be 0.00059 watts. In reality R was probably much smaller than unity, corresponding to more atoms and to higher power.

Kitamura’s Experiment
This experiment (2) was similar to that of Arata and Zhang. But the heat generated in the cell was measured by using a flow calorimeter. Three kinds of powder were used: pure Pd (0.1 micron grains), Pd-black and a mixed oxide of Zr and Pd. The highest heating power, close to 0.1 watts, during about 800 minutes, was for the mixture of Pd and Zr oxides. It is significant that excess heat was produced when deuterium was used; it was not produced when ordinary hydrogen was used.

How many atoms of helium would be produced during that time if all excess heat was due to D+D nuclear fusion? The answer is 1.3*1015, or about as little as in one cubic centimeter of normal air. Is this detectable? Probably not. But I am only guessing. I agree with authors that “it is crucial to confirm the phenomenon of heat and He generation with fully quantitative reliability.” But is this possible when excess heat is generated for less than 1000 minutes, at the rate of about 0.1 W? The answer depends on the limit of sensitivity of the helium detection method.

Nuclear or not nuclear?
Absence of a nuclear signature, such as “helium ash,” suggests that a chemical fuel of some kind might be responsible for the measured excess heat--thermal energy released in 800 minutes was close to 0.1*800*60=4800 joules. This is about the same as the amount of heat released when 0.1 grams of gasoline is burned. But absence of a chemical fuel in 99.5%-pure gas is not the only argument against chemical origin of excess heat. All possible artifacts, except deuteron-specific, can be ruled out; excess heat was not generated when ordinary hydrogen, rather than deuterium, was introduced into the cell.

An attempt to detect neutrons and gamma rays was made, but as stated in (2), results were negative. What other nuclear reaction, besideds cold fusion, could have produced excess heat without emitting detectable neutrons or gamma rays? I am pleased that the article was published in a very prestigeous mainstream journal.

1) Jed Rothwell and Ed Storms, downloadable from:

2) Akira Kitamura, Takayoshi Nohmi, Yu Sasaki, Akira Taniike, Akito Takahashi,
Reiko Seto, and Yushi Fujita; “Anomalous effects in charging of Pd powders with high density hydrogen isotopes; Physics Letters A 373 (2009) 3109-3112.

3) Michael McKubre, Francis Tanzella, Paolo Tripodi, and Peter Hagelstein; “The Emergence of a Coherent Explanation for Anomalies Observed in D/Pd and H/Pd Systems; Evidence for 4He and 3He Production;” 8th International Conference on Cold Fusion. 2000. Lerici (La Spezia), Italy: Italian Physical Society, Bologna, Italy. (Also see Section 3.1 in )

4) Mike Carrell; “Arata and Zhang’s Cold Fusion: Excess Heat and Helium Production;” Infinite Energy, Issue 18, 1998 (Also see Section 3.1 in (downloadable from ) )

Added on 11/21/2009
Commenting on the above, Ed Storms wrote: “Your comments about the heat having two possible sources is correct, but incomplete.  All heat that is generated by a CF reaction can have 5 sources.

1. Heat from fusion,
2. Heat from other exothermic nuclear reactions,
3. Heat released when the NAE is created, (This creation process must be exothermic.)
4. Chemical heat not related to CF.
5. Error

The measured relationship between energy and amount of He shows that source #1 can be a major source of heat.  However, the other sources can operate when  conditions do not cause fusion.  As we understand the process better, we need to more carefully determine the amount of energy caused by each of these sources.  Such values will go a long way to determine which theories are correct and which are only imagination.”

The term NAE, invented by Ed, stands for Nuclear Active Environment. Yes, to make a new kind of nuclear energy commercially useful the cost of released energy must be lower than the cost of required NAE. But scientific understanding can be made when this condition is not satisfied. Consider the NAE in a tokamak; it is extremely costly. The break-even point, for a sustained tokamak operation, has not bean achieved. But hot fusion scientists continue to believe that a lot can still be learned from very expensive experiments. That is a healthy attitude. Decisions about how much money to spend on research are not made by scientists; each individual scientist wants as much money as possible, for his or her project. Decisions on what fraction of GNP should be spent on research, and how much money should be allocated on different projects, should be make collectively.

Added on 11/23/2009
It is interesting that the CMNS list is sometimes quiet for weeks, and sometimes intensive. It did become intensive in the last last week. The topic was p+p fusion. Due to the energy and momentum conservation, this two-body reaction In stars is extremely slow; it must be proceeded by the slow p+e conversion of p-into-n. But it can be direct in the deuterium lattice; where surrounding atoms act as a large third body. One researcher reminded us of claim (made by Mills in 1991) that measurable excess heat was produced in a cell whose electrolyte was based on ordinary water. The cathode of that cell was nickel, as in Oriani-type cell I am using. That is what caught my attention. In a subsequent message I wrote:

The amount of energy released per D atom would be:

[(2*1.007852-2.0141022)]*931.5 = 1.49 MeV.

I am using the E=m*c2 formula with c2 = 931.5 MeV/amu. Suppose excess heat from the CMNS p+p reaction is generated for 1000 hours at the rate of 0.1 W (or 100 hours at 1 W, or 10 hours at 10 W, etc). Then the total amount of produced energy would be

0.1*1000*60 = 6000 joules= 6000*6.24*1015=3.74*1019 MeV

The corresponding number of D atoms (the cold fusion ash) would be 3.74*1019 / 1.49 = 2.5*1019. Suppose the "ash" is dissolved in 100 cc of ordinary water (in the used electrolyte). That water contains 3.3*1025 molecules of H2O. Concentration of heavy water (mostly DHO2) is about 5 molecules per million molecules of H2O. In other words 100 cc of ordinary water would contain 3.3*1020 atoms of D. This is much
more than 2.5*1019 atoms in the nuclear ash. The task of identifying nuclear ash would thus be very difficult if the excess heat were only 6000 joules. But suppose the excess heat is 600000 J (10 W and 1000 hours, or 100 W and 100 hrs, etc.). In that case detection of nuclear ash would be easier because its relative concentration would be more favorable.

This website contains other cold fusion items.
Click to see the list of links