MIZUNO REPLICATION AND MATERIALS ONLY

    Quote from Paradigmnoia

    The support is that the power curve looks just like a resistance heater.


    In physics many phenomena share common math what leads to same shape curves.


    As experiments show: classic hot dry LENR has no ignition point. We see LENR in mixed/loaded powders at room temperature. If the full energy down-scaling path over all steps is in place then, then temperature defines the level of phonon coupling and finally the rate of energy transfer.


    Thus the argument above only shows a lack of real LENR Lab experience.

    • Official Post
    Quote from Wyttenbach

    As experiments show: classic hot dry LENR has no ignition point. We see LENR in mixed/loaded powders at room temperature. If the full energy down-scaling path over all steps is in place then, then temperature defines the level of phonon coupling and finally the rate of energy tran


    That is one reason why I find 'power off' cooling curves so interesting - there a LENR system can deviate unmistakably from any 'normal' system.

    Quote from Wyttenbach

    As experiments show: classic hot dry LENR has no ignition point. We see LENR in mixed/loaded powders at room temperature. If the full energy down-scaling path over all steps is in place then, then temperature defines the level of phonon coupling and finally the rate of energy transfer.


    If you believe that the rate of energy release is temperature-dependent then I don't understand why there shouldn't be an ignition point. "Ignition point" here is not the temperature at which the LENR reaction would begin but instead the point at which energy release cannot be compensated by cooling. Once this point is reached, the fuel should begin to heat up, which increases the reaction rate, which in turn increases the temperature even more, and thus evokes more heating, and so on ... a positive feedback cycle.


    Entry into this positive feedback cycle should produce an upward inflection (trigger point) in either the temperature or output power traces and this is what people are saying they don't see in the Hokkaido University data. I don't see it either although it could come very near the beginnings of the trace and the time resolution isn't good there. Once the external input power is turned off, the feedback cycle should still be working to some extent and this should produce a plateau which is the sort of "power off" phenomenon that Alan mentioned. The existence of a trigger point and the power-off behaviour are two expressions of the same thing.


    Ignoring the thermal conduction loss down the gas fill pipe to the exterior. However this should be small relative to the radiation loss from the reactor to the walls which should be the same at the same temperature assuming the reactor external emissivity is the same on both the test run and the control run.


    There is also the slight but small amount of H2 that could diffuse thru the hot stainless steel reactor wall and recombine with the O2 in the exterior air, creating combustion heat. However, the unit is (under the Mizuno protocol) valved off so that the small amount of H2 has a known amount of potential combustion energy which would be quickly exhausted, thereby not effecting the test.


    Because the calibrated vs measured power difference is large, I think we can rule out the H2 gas or gas thermal conductivity difference in this experiment. The only thing that could provide this level of experimental error is if the emissivity of the calibration unit was much different (thus getting more heat out thru the walls in the calibration run). Someone else can put a slide-rule to this, but my gut feel is that the Hokkaido experiment has sufficient delta-T in the output air to more than compensate for such unreported emissivity differences. And if they were running the SAME reactor body, then there is no difference.

    Quote from Bruce__H

    If you believe that the rate of energy release is temperature-dependent then I don't understand why there shouldn't be an ignition point. "Ignition point" here is not the temperature at which the LENR reaction would begin but instead the point at which energy release cannot be compensated by cooling. Once this point is reached, the fuel should begin to heat up, which increases the reaction rate, which in turn increases the temperature even more, and thus evokes more heating, and so on ... a positive feedback cycle.


    There are many historically documented examples of unbalanced LENR reactions that did lead to a run away situation. But this was not the misunderstanding of Paradigmnoia . The Mizuno style reaction will only work well up to about 450C with no runway potential due to the used fuel material. But do we really know all details..?

    Quote from anonymous

    There is also the slight but small amount of H2 that could diffuse thru the hot stainless steel reactor wall and recombine with the O2 in the exterior air, creating combustion heat.

    No H2 will escape, because the pressure inside the cell is far lower than atmospheric pressure. Gas will only go into the reactor.


    Even if the reactor were at 1 atm or somewhat higher, you could not possibly measure the heat from the H2 escaping through the walls of a stainless steel reactor. Even a microcalorimeter could not detect this. It would probably be less than 1 J per year. The Swagelok connections are so good, the pressure remains the same for weeks, to within a few Pascals. Most of the H2 that leaves the reactor would go out through the pipe. A sealed and welded stainless steel reactor with walls this thick would hold a few atmospheres of H2 for decades without a measurable leak in or out, just as a heavy duty incandescent light will.


    Quote from anonymous

    Because the calibrated vs measured power difference is large, I think we can rule out the H2 gas or gas thermal conductivity difference in this experiment.

    We can ignore gas thermal conductivity because it cannot affect the calorimetry. If the heat were measured as a function of the temperature inside the reactor, gas conductivity would affect it. But conductivity cannot affect a measurement based on the inlet and outlet temperatures of cooling air. The heat is measured after it leaves the reactor. Low gas conductivity in the reactor will make the reactor core hotter, and it will delay the heat from leaving at first, but once the reactor surface reaches the terminal temperature, conductivity makes no difference.

    Quote from Wyttenbach

    There are many historically documented examples of unbalanced LENR reactions that did lead to a run away situation. But this was not the misunderstanding of Paradigmnoia . The Mizuno style reaction will only work well up to about 450C with no runway potential due to the used fuel material. But do we really know all details..?


    I don't understand what you are trying to say. Do you think that there should be no inflection/ignition point in the temperature or power traces? Why not? According to your picture of a temperature-sensitive reaction rate there should be one.


    I also don't know exactly what you mean by "runaway". I think it means when the positive feedback cycle I mentioned previously begins to dominate the time course of the system. It is what causes the inflection point. It doesn't need to mean that the system melts down or explodes or anything like that. It can just appear as the era of transition up to a new equilibrium when heat generation is once again balanced by cooling.

    Quote from JedRothwell

    No H2 will escape, because the pressure inside the cell is far lower than atmospheric pressure. Gas will only go into the reactor.


    Even if the reactor were at 1 atm or somewhat higher...


    We can ignore gas thermal conductivity because it cannot affect the calorimetry. If the heat were measured as a function of the temperature inside the reactor, gas conductivity would affect it. But conductivity cannot affect a measurement based on the inlet and outlet temperatures of cooling air.


    1) "No H2 will Escape". H2 is not "escaping" like a high pressure leak, it would be diffusing from an area of higher H2 density thru the steel to an area of lower h2 density in the surrounding air. But with a sealed volume at 0.1 Torr (my recollection of the reactor pressure when it is valved off, i.e. 1e-4 ATM), we are talking about next to nothing in fuel available. But because the valve is shut with a total volume of say 2 liters and a energy density of 140 MJ/kg and mass density of 0.09 g/L, total energy is around 1e-4*2*.09*0.001*140 MJ = 2.5 joules. The fact that it is valved off means even if the H2 slowly diffused over 1e4 seconds, then its 2.4e-4 watts, i.e. 1/4 mW, i.e. nothing.


    2) "We can ignore gas thermal conductivity because..." I agree we can almost certainly ignore it, because it hardly would effect the calibration. The only thing that could effect the calorimetry is if there is another path for heat that has less thermal resistance in the active run vs the control. In this case, the path could be the loading tube if it still has H2 in it which is an excellent conductor. However, the loading tube inner diameter is only around 5 mm judging by the photos, so there is not much of a path for heat conduction thru the H2. I can't quickly compute that but I think it is next to nothing. Thus, unless someone can prove otherwise analytically, I think an identical rig loaded with 0.1 Torr of H2 instead of air or a vacuum has essentially the same calibration.

    • Official Post
    Quote from anonymous

    1) "No H2 will Escape". H2 is not "escaping" like a high pressure leak, it would be diffusing from an area of higher H2 density thru the steel to an area of lower h2 density in the surrounding air. But with a sealed volume at 0.1 Torr (my recollection of the reactor pressure when it is valved off, i.e. 1e-4 ATM), we are talking about next to nothing in fuel available. But because the valve is shut with a total volume of say 2 liters and a energy density of 140 MJ/kg and mass density of 0.09 g/L, total energy is around 1e-4*2*.09*0.001*140 MJ = 2.5 joules. The fact that it is valved off means even if the H2 slowly diffused over 1e4 seconds, then its 2.4e-4 watts, i.e. 1/4 mW, i.e. nothing.


    2) "We can ignore gat thermal conductivity because..." I agree we can almost certainly ignore it, because it hardly would effect the calibration. The only thing that could effect the calorimetry is if there is another path for heat that has less thermal resistance in the active run vs the control. In this case, the path could be the loading tube if it still has H2 in it which is an excellent conductor. However, the loading tube inner diameter is only around 5 mm judging by the photos, so there is not much of a path for heat conduction thru the H2. I can't quickly compute that but I think it is next to nothing. Thus, unless someone can prove otherwise analytically, I think an identical rig loaded with 0.1 Torr of H2 instead of air or a vacuum has essentially the same calibration.

    As I said, nit picking. There is no evident way in which such a large difference in temperature with the same electric energy input can be explained by anything conventional. You could explain a fraction of a Watt difference this way, but not a tens of Watts difference.

    Quote from Bruce__H

    I don't understand what you are trying to say. Do you think that there should be no inflection/ignition point in the temperature or power traces? Why not? According to your picture of a temperature-sensitive reaction rate there should be one.


    Don't mix your picture with my picture, that is also the physics picture behind LENR. The maximum amount of energy that can drain out of D*-D*, D*-A, H*-H*-A is given by the coupling magnetic moment.


    May be some time next year you will learn more.


    Here we go.


    There might be thus-and-such a problem.


    Specifically, a difference between active and cal runs that results in 30% less heat output from the calorimeter on cal run than on active.


    All that is needed for this is for the reactor, or possibly some elements inside the reactor with thermal bridge to outside, to be much hotter during cal conditions cf active. As Jed points out an equilibrium is reached in which the heat out is still the heat in (roughly). However this can be with different temperature of the reactor and therefore different heat loss. The reactor temperature depends on the cooling of the air, the internal heater temperature depends on configuration and internal gasses (relevant if this thermally bridges outside the reactor.


    How would I check? Increase by factor of 2 the insulation everywhere and see whether apparent heat excess reduces.


    The way these results scale makes them look like some such issue to me, but it is easy enough to check and provide convincing support if they are real.


    THH

    • Official Post

    again an hypothetical scenario that in no way can be applied to the configuration reported.

    As I said, nit picking. There is no evident way in which such a large difference in temperature with the same electric energy input can be explained by anything conventional.


    The issue is what is the heat loss. I agree differences between cal and active cannot be larger than the total cal losses: but we do not have much info about what these are at the moment.


    One caveat, if the in/out temperatures as measured are a not fair average of the air stream that would be another error that could be reactor temperature dependent (altering turbulence etc). Do we have info about that?

    Quote from RobertBryant

    "Why not?"

    Why?


    Show your mathematical model that causes an inflection. and predicts the size of f ′′′ (t)..


    I published it in about 10 large posts on the old Atom Ecology thread in August/September of 2018. You won't find it now because the entire thread was deleted by the site administrators because of something to do with Russ George. If you think it is important then I can ask to have the particular poss resurrected. Shane tried last summer but got the wrong posts.

    Quote from THHuxleynew

    As I said, nit picking. There is no evident way in which such a large difference in temperature with the same electric energy input can be explained by anything conventional.


    The issue is what is the heat loss. I agree differences between cal and active cannot be larger than the total cal losses: but we do not have much info about what these are at the moment.


    One caveat, if the in/out temperatures as measured are a not fair average of the air stream that would be another error that could be reactor temperature dependent (altering turbulence etc). Do we have info about that?

    I am working on the last part, interrupted by New Years silliness, at present.


    Some things I am testing is the changing the internal volume (reactor/calibration size), changing the location of items (can direct passage of inlet air to the outlet dilute the heated air, or is this balanced by even hotter air that has longer residence in the calorimeter), and how restricting inlet air affects temperature (by creating a weak vacuum, and thereby increasing the temperature due to same amount of heat supplied to less air mass).

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