MIZUNO REPLICATION AND MATERIALS ONLY

  • So, if anyone is still paying attention, the 120 W experiment above, the peak Measured power was about 183 W, whereas the loss-corrected power is about 350 W.

    So, in other words, even at 180 W the calorimeter recovery was just less than 50% of the total heat output.

    That is nearly the same as the recent 500 W plots posted by the gracious Dr. Mizuno a few days ago.


    And that is way less than the more recent tests which range from about 80% to nearly 99% heat recovery, so I wonder, when did the reflective insulation get installed?

    Otherwise I have a hard time understanding why the heat recovery is so low.


    Measured power graph from the Spreadsheet.

    I love the power up arc. It looks like it would keep going up, and up...

    .

  • 古いデーターは正確では無い。測定系も日々改良している。その度に校正試験を行う。補正係数も日々正確になる。新しいデーターを見てほしい。総合的な報告書は特許申請後行う。

    Old data is not accurate. The measurement system is improving daily. A calibration test is performed each time. The correction coefficient is also update every day. Look at the new data. Comprehensive report will be made after patent application.

  • So then I had to make this

    There is no need to make anything . I used the inlet/outlet temperature data from Figures 20/26 years ago.

    The ks.degree data can be calculated by area of the pixels.

    The calibration factor can be calculated as 1.27using the original specs for that expt.

    For 100 W input

    The uncorrected output of the active reactor is 131 W

    The corrected output is 166.4 using 1.27 from the inactive calibration.

    This 1.27 may be an under estimate because of averaging and other effects

    For the 120W case, to get to 200W...the calibration factor would probably be bigger than 1.27.

    https://www.google.com/url?cli…Vaw3gqR-Dsn_dd53UIllFYlZ7



  • The amount of escape from the heat box is not corrected.

    The amount of escape from radiation and convection

    is not corrected in Figure 20 or Figure 26..


    the corrected power will depend on the calibration factor..

    which depends on assumptions


    The corrected power for the active reactor

    depends on the actual calibration factor

    which is not stated in the 2017 paper

    ... this is a rough way of calculating it.

    by using ( Temp-time) areas in yellow


    the spreadsheet method used by MizunoSan is more accurate

    but it needs the numerical data


    the uncorrected heat can be calculated roughly by using


    delta H(J) = delta T (K) x flowrate ( m3/ s) x time(s) x density(kg/m3) x Cp(J/kg.K)

    =deltaTemp x time (ks,deg)  x flowrate x density x Cp


    The power P can be calculated by dividing by TIME 85 ks/82 ks .... from the graph..

    there is an averaging error..(time/TIME ) which underestimates both uncorrected powers.

    there is also the use of average air densities.


    in addition there is an error due to the fact that the active reactor

    is hotter than the inactive reactor , despite equal 100W inputs,

    which increases the heat escape.,, so the 1.27 factor is too small..


    However the uncorrected power for the active reactor will

    be much more than the input electrical power of 100W.

    In addition the uncorrected output of the active reactor

    is much more than the inactive reactor.

    Comparing the ( TEMP-time)areas makes this obvious..

    590 versus 335. is equivalent to 73% more

    The Fig 29 .. derived by spreadsheet anlysis

    gives a corrected output for the 100W

    case of approximately 74% more (74W) for the active reactor

    corrected power.

  • Perhaps a better question is, since the time that the calorimeter recovery has been improved from about 50% to about 80% (or better), have any experiments had the Measured power exceed the input power?

    The calorimeter recovery rate has not changed. It was never 50%. It varies with input power. Measured power exceeds input power in Fig. 5, here:


    https://www.lenr-canr.org/acrobat/MizunoTincreasede.pdf


    But this question makes no sense, and has no scientific point. Suppose, for the sake of argument, there were no examples of this. Suppose that even with excess, output recovered in the airflow was a little lower than input. However, the calibrations showed that at least 20% of the heat is not recovered in the air flow. What would you say about that (pretend) situation? Would you claim that the heat magically vanishes from control cells and non-activated cells, perhaps taking a wormhole shortcut to Mars, and then the heat decides to hop into the air flow with activated cells? Why would it do that? What property of the activated cell would radically change the performance of the calorimeter? Why would heat suddenly stop radiating from the walls? Any answer you come up with to these questions would be along the lines of the nonsense "theory" Shanahan proposes, which actually explain nothing, while somehow invoking sentient heat, magic palladium, and what my mother called "the innate perversity of inanimate objects."


    To put it another way, if the calibrations do not reveal the actual performance of the calorimeter, what would reveal them? How can you know what the recovery rate is, except by measuring it repeatedly with a variety of control reactors?


    I get that if the recovery rate cannot be measured accurately, and the apparent excess is close to the margin, you would not have confidence in it. But that is not the situation.

  • Thanks for that pointer to figure 5, so at least sometimes the measured heat power does exceed the input heat power. That is good.



    As for the 50 % heat recovery, both a 120 W test (images I posted above) and a 500 W test (Images recently posted here by Dr. Mizuno) do show about 50 % recovery at 350 W output and 800 W output respectively. I realize that they are both older tests, and that the recovery has been improved somehow since then. I am curious as to how the recovery was improved.

    .

    13465-%E3%82%B9%E3%83%A9%E3%82%A4%E3%83%892-jpg
    .

    13558-120-w-measured-jpg
    .

    13562-pasted-from-clipboard-png

  • As for the 50 % heat recovery, both a 120 W test (images I posted above) and a 500 W test (Images recently posted here by Dr. Mizuno) do show about 50 % recovery at 350 W output and 800 W output respectively. I realize that they are both older tests, and that the recovery has been improved somehow since then. I am curious as to how the recovery was improved.

    The lowest recovery rate I have seen is 70% at 370 W. See Fig. 2:


    https://www.lenr-canr.org/acrobat/MizunoTincreasede.pdf


    At much higher power levels it will be 50%. Hypothetically, at very high power it would approach zero but the instrument would fail long before that.


    I don't think the instrument has changed or improved but I have no info on it. Maybe it is a new one?



    There was a reference to me claiming it makes no difference whether the calibration reactor is "a peanut or an elephant." A humorous way of saying little or big. I would amend that to: all peanuts and elephants we have tested so far have made no measurable difference. If the test reactor was much smaller or much larger than the ones we have tested, the size might have an impact. I can only comment on tests that have actually been done, producing data that I have seen. I cannot speculate about extreme tests that might produce results outside of the ones I know about. In this case, I know about calibrations up to 370 W. I am not familiar with higher powered ones.


    All calorimeters are limited to a given range of power, and a range of temperatures. They are designed to work within these ranges. When you operate them outside these ranges, they might work, but maybe not. The results may be non-linear and you won't know what to make of them. If you have not tested the calorimeter for a given power level above or below the levels you calibrated for, you cannot predict what it will do. I can't, anyway. See p. 8 of my recent paper:


    https://www.lenr-canr.org/acrobat/RothwellJreviewofth.pdf

  • The lowest recovery rate I have seen is 70% at 370 W. See Fig. 2:


    https://www.lenr-canr.org/acrobat/MizunoTincreasede.pdf


    At much higher power levels it will be 50%. Hypothetically, at very high power it would approach zero but the instrument would fail long before that.


    I don't think the instrument has changed or improved but I have no info on it. Maybe it is a new one?


    Please explain the images I posted in the quote you replied to.

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