Mizuno reports increased excess heat


  • Jed - you really do not read what I write. You are including the (tiny) gaseous D2, not the larger loaded quantity. Unless the whole loading process was done with valve closed? In that case all is good.

  • Lovely idea. At what concentration might Deuterium burn reliably under those conditions? I can tell you, it would need to be a concentration of +5% by volume.


    Need we go on anymore?

    Alan -- not burn, but catalytically react. Hot stainless steel would act like a catalyst for reacting the D2 with O2 from a leak. (Edit: so will nickel and palladium.)


    However, I think I just read from Jed that the D2 is valved off, i.e. the cylinder is sealed from any D2 getting in once the experiment commences. Assuming I understand him correctly, there is only a few dozen joules of D2 available for chemical reaction with any air that leaks in (i.e. the D2 that is within the volume of the stainless cylinder when it is valved off at the commencement). This few dozen joules over the course of 24 hours would account for under 1 mW of average heat, i.e. effectively zero heat from chemical, thereby ruling it out. I hope I read Jed right as he doesn't want to answer me anonymously as to how Mizuno hooked up the plumbing, but if my guess is correct, this is not an issue.


    Post Edit: I guess from reading more of the give and take that this was suppose to be clear from table 1. Assuming that this is correct, the results are remarkable. I don't think any small percentage error in the calorimeter is going to change the fact that huge excess heat comes from the device. The 300 watt power and the COP of >5 is just too large.


    Anyway, thank you respectfully for continuing the conversation above without me and thank you Jed for translating.


    Jed, as a mild suggestion, when you publish the final version of the paper you should make the above fact (that the cylinder is sealed off from the D2) more clear to the readers to eliminate their confusion, even for those who are "skilled in the art". I also would suggest a small schematic diagram of the piping showing the valves. Thank you for the good work.

  • I think you have this backward. I don't need to explain anything - because I'm making no claims


    You actually do make claims

    You claim that higher COPS mean the reactor IS unstable.


    You have based that on an assumption of linear dependence.


    "Let us take a simple example of F(T) = CT (linear)"


    "I don't need to explain anything"

    You need to explain how you can come to the conclusion of instabilty

    without any knowledge of the temperature dependence

    of the reactions at the atomic level.


    Mizuno's 2017 work indicated a reaction temperature dependence

    which is not linear.


    " The reaction activation energy Ea was calculated on the basis of the linear region

    between 100 and 523◦Cin Fig. 40 to be 0.165 eV/K/atom"


    In addition Mizuno appears to test run the higher COP's for hours.

  • googling you get maybe 120-140MJ / kg energy content of hydrogene. Excess Energy was about 250*24*3600J so thats about 150g hydrogene equivalents.

    That is the figure that essentially need to have been loaded to explain the results, (actually its more like maybe 500g if the run was for three days).

    This sounds quite a lot but i'm pretty ignorant.

  • The enthalpy of combustion as you say can easily be worked out and compared with total output excess energy

    . I'd prefer to be sure it is not an issue


    It is not an issue if you work out the figures. This has been done long ago in September 2017.

    I actually instructed you how to work this out.

    and you replied "I am inclined to agree with you"

    This combustion energy canard is a dead duck.

    Mizuno : Publication of kW/COP2 excess heat results

    Perhaps you should revise that thread.. instead of bringing up the same " issues."

  • Jed - you really do not read what I write. You are including the (tiny) gaseous D2, not the larger loaded quantity.

    I did read what you wrote. That is the only amount loaded. That's the pressure after the cell is filled. The amount that is then absorbed by the Ni is a tiny fraction of that.


    Furthermore, even if the cell were filled to 10 atm it could not sustain this power for 24 hours.


    Furthermore, there is no oxygen in the cell. If air were leaking in, the pressure would rise. It does not rise.

  • Other posters -- I ask how do we _rule_ _out_ air being sucked in at higher temperatures and then chemically recombining with D2 to make D2O plus chemical heat from the enthalpy of reaction.

    This is really highly unlikely, and should only be checked after more likely culprits are checked (e.g., calibration problems due to heat conduction differences or thermocouple movement). In fact, it's close to certain that this is not the explanation given the apparent linear nature of increasing excess heating with input power. Such chemical effects are fairly short lived. Best to not waste time on less likely explanations.

  • However, I think I just read from Jed that the D2 is valved off, i.e. the cylinder is sealed from any D2 getting in once the experiment commences. Assuming I understand him correctly, there is only a few dozen joules of D2 available for chemical reaction with any air that leaks in (i.e. the D2 that is within the volume of the stainless cylinder when it is valved off at the commencement).

    Table 1 shows that the pressure does not change. So no D2 is added, and no air leaks in. Even a tiny air leak would cause a large change. The cell is at 0.01 atm.


    A few dozen joules is correct. As I said, it is ~343 J, whereas the reaction produces 21.6 MJ per day, for several days.

  • [Other posters -- I ask how do we _rule_ _out_ air being sucked in at higher temperatures and then chemically recombining with D2 to make D2O plus chemical heat from the enthalpy of reaction.]


    This is really highly unlikely,

    No, it is not highly unlikely. It is impossible, by many orders of magnitude. Even if the entire cell was filled with D2 gas at pressures 1,000 times higher than this (10 atm), it could not begin to produce this much energy. That would be 2.5 mol H, 1.25 mol water, 357,000 J. It would burn for 1429 s. That's 24 minutes.

  • more likely culprits are checked (e.g., calibration problems due to heat conduction differences or thermocouple movement).

    Given Figs. 2, 3, 4 and 7, how likely are these culprits? As noted in the paper, the thermocouples are calibrated together, in water over a range of temperatures. (As the water cools.) They agree to within a fraction of a degree with a mercury thermometer. Even if they were inaccurate, as long as they agree with one another, that would work. In other words, they are used to measure a temperature difference, so precision is the only thing that matters. In fact, they remain both accurate and precise. They are regularly checked against the mercury thermometer and the Omega handheld thermocouples, throughout the experiment, and they always agree.


    There is not the slightest chance they are wrong by 10 deg C, given this error checking. If the excess heat produced only a fraction of a degree difference, that would be in the noise, given the unstable ambient room temperatures shown in Fig. 7. The difference is 10 deg C (Fig. 5), which is far above any error or noise. No lab grade thermometer or thermocouple intended for this range of temperatures (0 to 100 deg C) made in the last 150 years would produce an error that large. It would break altogether. The mercury would leak out; the thermocouple would register random numbers. An error of this size is out of the question. Since they are regularly checked against 3 other thermometers it is 3 times out of the question.

  • Cooperation, not division, is the way.


    We're getting a lot of good nuggets on this thread. Most of which I agree with. Unfortunately, it is not those in the CF space who have sought to divide. To the contrary, the gatekeepers of the broader scientific community have eschewed cooperation and not only divided, but obliterated those who dared to differ.

  • In addition Mizuno appears to test run the higher COP's for hours.

    I don't know how many days at the highest COP, but Table 1 shows over 100 days at high COPs with R19. It adds up to roughly 500 MJ. That is about 1.5 million times more than the total energy you can get by burning the D2 in the cell. Which you cannot burn, because there is no oxygen in the cell.

  • JedRothwell

    Quote

    "At room temperature and atmospheric pressure (standard ambient temperature and pressure), palladium can absorb up to 900 times its own volume of hydrogen."

    - various sources


    So the amount of palladium metal in the reactor is very small, at least in the most current design? And forgive my ignorance but does nickel absorb or otherwise bind hydrogen?

    (probably beating a cadaverous equine)

  • ["At room temperature and atmospheric pressure (standard ambient temperature and pressure), palladium can absorb up to 900 times its own volume of hydrogen."]


    So the amount of palladium metal in the reactor is very small, at least in the most current design?


    ~50 mg, as noted on p. 11. (Perhaps that should be 150 mg. Will check.) In these conditions, at this pressure, the highest loading is 0.7 according to Mizuno's electrochemistry textbooks. That's 4.7 * 10^-4 mol of Pd. Holding 0.000329 moles of D. If there were any oxygen, that would form 0.000165 moles of water, which would produce 47 J of heat.


    Maybe we should double check that by assuming the volume is 900 times at STP. 1 g of Pd is 0.08 ml. 50 mg is . . . 4.17 * 10^-6 L. Multiply by 900 equals 0.003753 L. That's 1.5e-4 moles. If there were any oxygen, which there isn't, that's 43 J. Close enough for government work!


    Cheat sheet:


    https://www.aqua-calc.com/calculate/weight-to-volume


    Ideal gas law:


    https://www.cactus2000.de/uk/unit/massigg.shtml



    And forgive my ignorance but does nickel adsorb or otherwise bind hydrogen?

    See p. 11 and Ref. [2]. According to the textbooks, normal bulk Ni absorbs roughly 2,800 times less H than Pd does. However, this Ni seems to be absorbing a lot more than that. It is unclear how much is being absorbed and how much is crammed in between the Ni and Pd.