Mizuno : Publication of kW/COP2 excess heat results

  • Zeus45,

    Although why 2000/3000?

    that's the problem with average power.. is the input defined to be over 20000, 30000.??


    I was wondering if you could check the xs energy for Figure 26 for an input of 8.25 MJ


    Using average temp =28.1, Hc =1006J/kg/deg, density =1.17 kg/m3, flowrate = (3.87 m/s , area 4.4/1000 m2?)


    I got 11.8 MJ output.. xs = 3.55


    I do not think humidity and temp variations make a significant difference.

  • bocijn ,


    My way of looking at it is 4.96MJ (or 61 units) of energy go in and 121 units of energy come out... So... 4.96*121/61 = 9.84MJ out


    9.48MJ/30000s = 316W average power out


    is the input defined to be over 20000, 30000.??


    Maybe you could say 248*20000/30000 = 161W effective average power in


    But It doesn't make much sense to me personally to compare the average powers if the time span is different... I'm not quite sure what it tells us that's more useful than looking at either the total energy or the instantaneous difference between POut and PIn.


    I was wondering if you could check the xs energy for Figure 26 for an input of 8.25 MJ


    Using average temp =28.1, Hc =1006J/kg/deg, density =1.17 kg/m3, flowrate = (3.87 m/s , area 4.4/1000 m2?)


    Where does the 8.25MJ and average temp = 28.1 come from?


    I do not think humidity and temp variations make a significant difference.


    I agree.

  • The 28.1 is a red herring, as to work out the average power output using Hc =1006J/kg/deg etc. you need to know average output temperature *minus* average input temperature.


    But a faster way is just to say this would be proportional to the top yellow area, and compare that to the size to the control. (590/335)


    So, as you say: 1.76x as much energy Out vs In


    But because the time span is the same, I guess you could also say average COP is also = 1.76 over the whole experiment.


    I was wondering if you could check the xs energy for Figure 26 for an input of 8.25 MJ


    I got 11.8 MJ output.. xs = 3.55


    8.25MJ x 1.76 = 14.5MJ output... xs = 6.25MJ ?

  • Zeus 45". I'm not quite sure what it tells us that's more useful than looking at either the total energy or the instantaneous difference between POut and PIn"


    Energy (MJ )ratios are more useful for me.


    Currently the price here for me in NSW is 4.5 cents per Natural Gas Megajoule and 9.4 cents per Electricity Megajoule.


    Mizuno knows that he has to get the COP way up over 300% for this thing to be attractive.

    at 200% it is not even breakeven.


    For the 248W case the electricity cost of 4.96 Megajoules in is 46.6 cents

    The output heat energy of 9.8 MJ could be supplied by gas at 44.1 cents .

  • Thanks.

    The control/inactive input energy calculates as 7.15 MJ = (100Wx71.5ks)..it was run in for a slightly shorter time than for the active reactor.

    The use of the average temp of 28.1 does account for some of the variation with density of the temperatures in and out

    ( there is not much Hc variation )

    But the calculation is done much more accurately in the spreadsheet calculations, over 3485 time points, with the use of actual in/out temps in the enthalpy calcs

    rather than average temperatures.

    This is just a roughasguts way to look at how much Xs there is

  • Alan Smith... "ultrasound/sonic triggering to reduce the input energy cost"

    Maybe Suhas , Mizuno and Egely can get together


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  • Just to add an extra check. Enthalpy of combustion of H2 is 280kJ/mol or 0.14MJ/g


    So if (let us say) atmospheric O2 could get in to the reactor we have available chemical energy at approximately this level - I'm ignoring the fact that we have H here as Hydride, which would maybe increase the enthalpy a bit?


    So for this output you'd need about 17g of H2 stored (in this case 34g of D2). At a loading of 1 that (whether H2 or D2) turns into roughly 900g of Ni - so we cannot possibly store anything like enough H or D in the 20g Ni mesh.


    However the reactor is 20kg stainless steel. The average loading, given 17g of H (or 34g D) is about 5%.


    This does not look impossible to me - but I'm no expert on the diffusion of hydrogen into stainless steel so cannot say.


    The conditions here look vaguely plausible for this mechanism. High pressure for the loading. near vacuum for the unloading. You would need a slow O2 leak and some way to take the H2O out of the system (I guess a vacuum pump would do that).


    There are a lot of unknowns to check here which could knock this down, but worth checking.


    EDIT - not high pressure for loading, if 500Pa!

  • but worth checking.

    probably checked already by Mizuno last year.


    There are 3.4 mg of D2 in the reactor.

    Enthalpy of combustion 0.07 Mj/g of D2

    If Xs energy due to combustion of D2


    3.4 mg x 0.07 MJ/g = 0.000238 MJ.


    Xs energies observed are much larger


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  • Mizuno report

    "

    A test with this reactor lasts almost 30 days. Typical excess heat during the time is estimated as 300 W. Total energy is thus ~2.6×108 J. The amount of D2 used was 20 cm3 STP. Assuming that the reaction is D+D fusion, and assuming that all gases react, the amount of gas required to generate this much energy is approximately 12 cm3 STP. Although this is a very rough calculation, this value coincides with the amount of gas consumed. "


  • How is this 3.4mg of D2 computed? During the prep phase it does not seem clear to me how much D2 was added? In the report it refers to this quantity as being used during the test, but does not say whether that is just the heat production part, though this is implicit because it is comparing this usage with possible fusion enthalpy. In which case it leaves open how much was used in the prep part.

  • 1 mole of H2 is 2.0158 grams.



    Density = mass/volume.



    In g/L it would be (2.0158)/(22.4) =0.089 g/L 0.089 mg/ml


    D2 =twice = 0.178 mg/ml


    20 ml = ~3.5 mg at STP give or take a few nanograms


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  • bocjin - I did the same calculations up above, you will note (though a different way round).


    The issue at hand here is whether the amt of D2 during the prep phase was limited to 20ml.

  • You could take into account the pressure.


    ballpark


    760 pascals or so.. compared to atmospheric 100.000

    then factor in the volume of the reactor chamber say 2.5 litres


    760/100,000 x 2500 mls = 20???


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  • Kirk.

    There's another one. Found them in my first reading a while back

    Preprint is being corrected.


    You should see the booboos I made in my submission to a journal.

    Its very hard to proofread your own stuff in another language.

    Not sure what David Nagel was doing.

    Now I really need to get some kip.

  • This is ruled out. The pressure is monitored and recorded in the spreadsheet. If enough O2 got in to cause significant heat, the pressure would rise.


    The real world is a bitch, and you learn not to make assumptions.


    In this case we have:


    2D2 [+ O2] -> 2D2O


    so adding oxygen to make water actually keeps pressure the same, in theory.


    Beyond that, since for significant heat we need D out from metal, the issue is what happens to the D2O? If it stays as a gas you are correct, but in a complex environment like this I would not want to bet on what it does chemically.

  • correct me please if i am wrong here but you have 3 mol of gas on the left side of your equation and 2 mol of gas/liquid on the other side. From my understanding the pressure should drop. If you have a Knallgas reaction occur of course thats another thing...


    You are wrong. The suggestion is ingress of O2 from outside, so one of the mol on the LHS does not count. That is why I put it in brackets...