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Blog

Cold-fusion demonstration: an update

By Jon Cartwright

Several of you have asked when I’m going to give you an update on Yoshiaki Arata’s cold-fusion demonstration that took place at Osaka University, Japan, three weeks ago. I have not yet come across any other first-hand accounts, and the videos, which I believe were taken by people at the university, have still not surfaced.

However, you may have noticed that Jed Rothwell of the LENR library website has put some figures with explanations relating to Arata’s recent work online. I’ve decided to go over them and some others here briefly to give you an idea of how Arata’s cold-fusion experiments work. It’s a bit more technical than usual, so get your thinking hats on.

ColdFusionFig3.jpg

Above is a diagram of his apparatus. It comprises a stainless-steel cell (2) containing a sample, which for the case of the demonstration was palladium dispersed in zirconium oxide (ZrO2–Pd). Arata measures the temperature of the sample (Tin) using a thermocouple mounted through its centre, and the temperature of the cell wall (Ts) using a thermocouple attached on the outside.

Let’s have a look at how these two temperatures, Tin and Ts, change over the course of Arata’s experiments. The first graph below is one of the control experiments (performed in July last year) in which hydrogen, rather than deuterium, is injected into the cell via the controller- (8) operated valve (5):

ColdFusionFig2.jpg

At 50 minutes — after the cell has been baked and cooled to remove gas contamination — Arata begins injecting hydrogen into the cell. This generates heat, which Arata says is due to a chemical reaction, and the temperature of the sample, Tin (green line), rises to 61 °C. After 15 minutes the sample can apparently take no more hydrogen, and the sample temperature begins to drop.

Now let’s look at the next graph below, which is essentially the same experiment but with deuterium gas (performed in October last year):

ColdFusionFig1.jpg

As before, Arata injects the gas after 50 minutes, although it takes a little longer for the sample to become saturated, around 18 minutes. This time the sample temperature Tin (red line) rises to 71 °C.

At a quick glance the temperatures in both graphs, after saturation, appear to peter out as one would expect if heat escapes to the environment. However, in the case of deuterium there is always a significant temperature difference between Tin and Ts, indicating that the sample and cell are not reaching equilibrium. Moreover, after 300 minutes the Tin of the deuterium experiment is about 28 °C (4 °C warmer than ambient), while Tin/Ts of the hydrogen experiment is at about 25 °C (1 °C warmer than ambient).

These results imply there must be a source of heat from inside the cell. Arata claims that, given the large amount of power involved, this must be some form of fusion — what he prefers to call “solid fusion”. This can be described, he says, by the following equation:

D + D = 4He + heat

(According to this equation, there should be no neutrons produced as by-products — thanks to those of you who pointed this out on the last post.)

If any of you are still reading, this graph below is also worth a look:

ColdFusionFig4.jpg

Here, Arata also displays data from deuterium and hydrogen experiments, but starts recording temperatures after the previous graphs finished, at 300 minutes. There are four plots: (A) is a deuterium and ZrO2–Pd experiment, like the one just described; (B) is another deuterium experiment, this time with a different sample; (C) is a control experiment with hydrogen, again similar to the one just described; and (D) is ambient temperature.

You can see here that the hydrogen experiment (C) reaches ambient temperature quite soon, after around 500 minutes. However, both the deuterium experiments remain 1 °C or more than ambient for at least 3000 minutes while still exhibiting the temperature difference between the sample and the cell, Tin and Ts.

Could this apparently lasting power output be used as a energy source? Arata believes it is potentially more important to us than hot or “thermonuclear” fusion and notes that, unlike the latter, it does not emit any pollution at all.

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6 comments

  1. john

    i just dont get it. Why is there so much difficulty in proving whether cold fusion occurs?!?!? If you put in deuterium, and you get out helium, or some other nuclei more massive than helium, then, you’ve got fusion. Its goddamn f’ing that simple, surely.

  2. Kirk Shanahan

    john wrote on July 30, 2008 8:19 AM:
    ” i just dont get it. Why is there so much difficulty in proving whether cold fusion occurs?!?!? If you put in deuterium, and you get out helium, or some other nuclei more massive than helium, then, you’ve got fusion. Its goddamn f’ing that simple, surely.”
    No, it’s not. The amount of apparent excess heat produced, and the amount of nuclear ash detected, are normally at the trace level, and not commensurate with each other when compared to hot fusion expectations. This means a) you have the typical nightmares of trying to do trace level work on an unknown process and b) the prior experience of the field (physics of fusion) is not applicable, and therefore an equivalent amount of data must be produced independently. That latter is made difficult by the fact that the replication obtained so far is only general, i.e. people see effects, but never at the same levels. So, when conventional explanations exist for the observations (the calibration constant shift problem for calorimetry and contamination + concentration or just improper analytical methods for the nuclear ash measurements), it becomes very difficult to promote a new, paradigm-shaking explanation effectively.
    The problem with the cold fusioneers is that they don’t want to do the work required to establish this new physics. Instead they assert loudly and try to gain acceptance of their ideas by simply repeating them many times, instead of responding to criticisms by redesigning experiments to address the noted problems.
    Kirk Shanahan {{My opinions…noone else’s}}

  3. Jim G

    The third graph rather glaringly omits to track pressure. Any time a graph shows pressure increasing, one can expect to see temperatures elevating as well. What are the relative specific heats of the two test gases ?
    Lacking more than one temperature sensor type would also be another omission. The A and B plots on the 3rd graph, also seen to infer differing thermal resistance of Cell-Air. Note that for A Tin-Ts is rather smaller than Ts-Ta , whilst for B Tin-Ts is somewhat similar to Ts-Ta. Those results conflict from a heat-generation model. It does not state the environment condition changes from A to B, but anyway illustrates a large variance.

  4. James W

    Just wondering if there has been any updates on this yet. Is anyone working to repeat this experiment?

  5. Jelle Boersma

    Perhaps the following theoretical idea related to how cold fusion may occur
    is highly naive, but since I could not find any documentation discussing this the
    only way to find out is to bring it up, Thanks for any feedback.
    Could it be that cold-fusion is facilitated by quantum-entanglement of two
    deuterium nuclei with the external degrees of freedom (eg the phonons
    in the metal-lattice in which the D-nuclei are dissolved).
    The idea is that if two D-nuclei are entangled with independent exterior degrees
    of freedom then the density matrix describing the subsystem of two D-nuclei
    will be diagonal and the interaction hamiltonian vanishes even when the two
    nuclei have overlapping wave-functions in space and time .
    In other words, the electrostatic repulsion between two D-nuclei could be temporarily neutralized, along with the weak and the strong interactions.
    For two decohered nuclei at the same location it would take recoherence (through alignment of the exterior degrees of freedom with which the two nuclei are entangled) to restore the interactions. If the two nuclei recohere with sufficient overlap so that the strong attraction exceeds the electrostatic repulsion then fusion may occur.

  6. Zia Tompkins

    PV = nRT my old friend.
    Pressure is directly proportional to Temperature.
    One goes up significantly so must the other.
    Oh and error bars?
    The experiment was repeated over several days, room temperature vary appreciably?
    Individual thermocouple properties?

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