Research Team in Japan Reports Excess Heat - (Nissan Motors among otheres)

  • Are you suggesting that the HEX error is up to 0.26 +20%= 0.31W??? or =+20% of 5W = +1W??.


    I am calculating the fractional standard deviation of the measured output power per the simplfied equation given in the prior post. To 'suggest' an actual error I need reliable error estimates on all the relevant experimental parameters, which are not available. But that is neither of which you suggest. In fact in my last post I explicitly told you what I was guessing were the noise bands based on a standard error calc. I suggest you re-read my last post.


    You wrote B-O-E . What values of ɳ , ΔT , C and ρ are you plugging into your B-O-E?


    Doesn't matter. When you use the simplified fractional standard deviation approach, all you need to discuss is the fractional (easily converted to percentage) error of each experimental variable, which were not all given, so I am unable to give a definitive error estimate. I can only approximate from reasonable assumptions, which you may disagree with. But I won't argue about it because the paper's authors were supposed to give them, not make us guess.



    For the 80W input energy values Iwamura et al have written error estimates.


    oil flowrate error = 0:012(ml/min);

    oil temperature change error = 0:261(K);

    input energy rate error = 0:031(W);

    excess heat energy rate error (HEX) = 0:260 (W):



    How did they determine these? Since they do NOT seem to know how to do standard error analysis, I have to suspect their given estimates are likewise determined in a non-standard fashion, and thus are effectively meaningless, since I have no idea how to apply them to calculate the standard errors. Usually, for some strange reason, when people determine errors in a non-standard way, they are underestimated. Go figure.


    What are your estimates

    oil flowrate error = ? (ml/min);

    oil temperature change error =? (K);

    input energy rate error = ? W);


    I did not estimate these numerically. I applied the usual 'sensitivity analysis' approach of examining the impact if the standard deviations were 1, 5 and 10% of the nominal or maximal variable values. That is all that is needed to estimate the fractional standard deviation of the output power. Then I took the 134W reported output power and calculated my estimated noise bands. My conclusion was they are working in the noise, always a bad thing to do if you want to draw definitive conclusions.


    The excess power is another calculation, both for the value and for the standard deviation, but they are simple equations that you can figure out easily. Historically, input energy is usually reasonably well measured, as are temperatures (and thereby differences), but flowrate can be a problem in some cases.



    excess heat energy rate error (HEX) = ? (W):



    Of course that error will be the square root of the sum of the variance in the input and output powers. (There, I told you one of the equations from the answer just above.)


    input energy rate error = ? W);


    Pin = I * V or Pin = I^2*R . Use the POE equation to answer your question. You will probably have to guess at the errors since we are not informed how the supposed 'delta' is related to the 'sigma', and error calculations use 'sigma' (technically 's', the estimate of sigma).

  • Kirkshanahan

    "I can only approximate from reasonable assumptions,which you may disagree with"


    Iwamura et al estimate

    excess heat(HEX) energy rate error (HEX) = 0:260 (W): for the 80W input, 5W excess .

    The error in the HEX energy rate is crucial.

    You said something like 20% error in variables. Does this mean that you estimate the error could be up to 20% of 5 W = 1W?


    For HEX energy rate value a range 5W+/-1W is much less robust than one of 5W+/-0.26W


    Plug in any ɳ , ΔT , C and ρ etc values, reasonable assumptions you think realistic.


    What error in the HEX do you estimate for the 5W HEX case ?


    1W, 5W,10W, 100W?

  • bocijn


    Having trouble following things aren’t you? I already answered you above...


    OK, last time…


    As show earlier, our best estimate of the relative standard deviation (which is the standard deviation of a measured quantity divided by the nominal value of that quantity) (RSD) of the output power of a flow calorimeter, Pout (= k*F*r*Cp*(Tout-Tin), where k = 1/eta=1/n, and Tout-Tin=dT) is given by the Propagation of Uncertainty equation, i.e.


    ( s_Pout / Pout )^2 = (s_k/k)^2 + (s_F/F)^2 + (s_r/r)^2 + (s_Cp/Cp)^2 + (s_dT/dT)^2 [1]


    where (s_dT/dT)^2 = (s_Tout/Tout)^2 + (s_Tin/Tin)^2 [2]


    I’m doing a two-step process here where I first calculate the RSD of dT, and then use that in the calculation of the RSD of Pout. (Actually here I'm not ever really worrying about the dT error per se.)


    Technically, I need to do the same thing for r and Cp, since they are given as functions of temperature, and I showed that they varied significantly over the temperature range specified. Using an average temperature to calculate them as Takahashi, et al do makes this and makes it difficult to estimate the real variation in those quantities. I would also note a) I suggested they stop using Tavg and put the equations in directly and b) the equations themselves are fits to data and they have an error too, which should be folded in to the computation as well. But these are fine points.


    The big problem is that Takahashi, et al, state the error as being (‘d’ is delta)

    d(Hex) ~= d(F) * r*Cp*dT/n + d(dT) * F*r*Cp/n + d(W) (W=Pin, Pex = Hex = Pout-Pin) (I’ve left off the absolute value symbols.)

    So, by their logic, d(Pin) + d(Pout) = d(Pex) and thus

    d(Pout) ~= d(F) * r*Cp*dT/n + d(dT) * F*r*Cp/n

    Presumably they want us to believe their delta is the standard deviation. Dividing by the calorimetric expression for Pout we then can supposedly calculate the RSD by their logic.


    d(Pout)/Pout = d(F)/F + d(dT)/dT [3]


    So now compare equations [1] and [3] and you will see that their method ignores 3 terms, and incorrectly adds the standard deviations instead of the variances. If they were correct in only including two terms, their approach will overestimate the error. To explain, assume for the moment the RSD of the two variables is the same. Their method then gives the RSD of Pout as twice the RSD of the variables. The correct method squares them, add them, and then takes the square root, which means in this case multiplying one of them by the square root of 2 (~1.4) instead of 2 as they use.


    Getting back to eq. [1], the reason one converts to RSD is so that one can talk of fractional or percentage errors (as %RSD). One does that because there are common values for %RSD that reflect the ease of conducting an analysis. A 10% RSD technique is easy to do. 5% requires a bit more work to get and maintain, and 1% techniques require considerable work to achieve and lots of work to maintain. Getting better that 1% RSD is very, very hard and usually is limited to certain classes of measurements such as voltages and currents. In any case, if one hasn’t explicitly evaluated the RSDs, use of 1, 5 and 10% RSDs is typical to evaluate the level of effort that would be required to ensure you are working ‘above the noise’.


    By eqn. [1], if only one term was important to the Pout’s RSD, then whatever %error it had would be the %error in Pout. Since I have no information on the variation of n, I did the standard B-o-E thing and calculated %RSD_Pout for a 1%, 5% and 10% RSD in n, while ignoring the other terms. Likewise for B-o-E purposes, I temporarily assume the error in Pin is unimportant, thus the RSD of Pout = the RSD of Pex.


    In one case Pout was 134W for example. 1, 5, and 10% of that is 1.34, 6.7 , and 13.4W (we have now converted from RSD to SD as these Watt numbers are the actual estimated standard deviations). To decide if you are working out of the noise you must decide how wide the error band is. The error band width is usually calculated as some integer multiple of the SD (SD is sometimes referred to as ‘sigma’ which is usually technically incorrect, statistically speaking). In routine work, ‘3’ is the usual choice, but as I noted before, for higher importance work, that is sometimes increased to 5. That gives us:

    3-sigma band = +/- 4.0, 20.1, 40.2W for 1, 5, and 10%RSD

    5-sigma band = +/- 6.7, 33.5, 67.0W for 1, 5, and 10%RSD


    The numbers above are excess heats that would be considered ‘in the noise’ for the 6 cases we have considered here.


    Reported excess heats are on the order of 10W or less. That would imply they would need to be working with 1%RSD techniques. That is very difficult to do. As such my tendency is to believe they are actually ‘in the noise’. Without more replication, they cannot expect people to accept their results blindly.


    You should observe at this point that your repeated requests for ‘what value of x did I use’ are not relevant. I used ‘standard’ values for %RSD to evaluate the work, primarily because no actual statistical info was given and it seemed they were ignoring potentially very important parameters in their attempt to guess at their error. And as I noted before, their guesses are grossly too small.


    You should also note that when you are attempting to compute the standard error of a calculated quantity as we do above, you shouldn’t be minimizing things. You do that when you design your experiment and choose you instruments. If you instead play games with the POE calc, you just set yourself up for exactly what we see here, namely ‘convincing’ yourself that you are out of the noise band when you’re not. If anything, you should slightly increase the numbers in the POE calc just to increase the confidence your results aren’t in the noise.

  • OK, last time…


    As show earlier, our best estimate of the relative standard deviation (which is the standard deviation of a measured quantity divided by the nominal value of that quantity) (RSD)

    One last time: Shanahan's theories cannot explain:


    Why the calibrations produce a zero balance.


    Why palladium and heavy water work in platinum and light water do not. These differences cannot affect the calorimetry.


    Why helium is correlated with the anomalous excess heat.


    How a cell could remain probably hot for days, producing ~100 W, with no input power. Shanahan's explanation for this is a gross violation of common sense and the Second Law of Thermodynamics. He claims that the hot object is not a heater and an object that remains hot for days is not being heated. If he believes this he is a crackpot, and if he does not believe it he is trying to deceive stupid people in the audience.

  • Here we are, talking about powder samples in a gas cell and how to compute random error and up pops Jed talking about F&P systems that only have the use of flow calorimetry in common. Does Jed hate me or what? ;)



    One last time: Shanahan's theories cannot explain:


    Why the calibrations produce a zero balance.


    Why palladium and heavy water work in platinum and light water do not. These differences cannot affect the calorimetry.


    Already done many, many times, but for the newbies...


    'Calibrations' are always assumed to be Pout = Pin. Your comment is actually nonsensical, i.e it doesn't make sense.


    And I think you meant "Why palladium and heavy water work *AND* platinum and light water do not". First off, platinum does, just look at Ed Storms excellent work. Light water also works but with Ni electrodes and as I recall a different electrolyte. Second, anyone who thinks about the proposed chemical mechanism for the Fleischmann and Pons Effect will understand why this is so. And it is the getting of the Effect that affects the calorimetry, not the constituents of the cell. The constituents affect the getting of the Effect.


    Why helium is correlated with the anomalous excess heat.


    Technically, it is known as 'small sample statistics'. IOW, it is an accident of the data set. This is easy to recognize when you realize there is little evidence there has ever been any real excess heat in F&P-type experiments (and probably elsewhere as well).


    How a cell could remain probably hot for days, producing ~100 W, with no input power. Shanahan's explanation for this is a gross violation of common sense and the Second Law of Thermodynamics. He claims that the hot object is not a heater and an object that remains hot four days is not being heated. If he believes this he is a crackpot, and if he does not believe it he is trying to deceive stupid people in the audience.


    Ah, the Mizuno bucket anecdote again...fourth or fifth time you've repeated the same old lies isn't it?


    Keep beating that dead horse Jed...


    There are no liars here, Kirk, only people with different beliefs to you. Please remember that. Alan.

  • Kirkshanahan

    "I already answered you above"

    No you didn't

    You wrote. "The excess power is another calculation, both for the value and for the standard deviation, but they are simple equations that you can figure out easily"


    What error in the excess power (HEX )do you estimate for the 5W HEX case ?

    1W, 5W,10W, 100W?


    btw "CCS" = "clear communication society"? "Capsanthin/capsorubin synthase"?

  • Ah, the Mizuno bucket anecdote again...fourth or fifth time you've repeated the same old lies isn't it?

    I am curious. What do you mean?


    1. Are you saying the anecdote is a lie? That it never happened to Mizuno?


    OR


    2. Are you saying that you never claimed a hot object is not a heater, and an object that remains hot for days is not being heated?


    #2 is a violation of the Second Law. If you deny saying that, I can look up your original quotes. Or, perhaps, you could explain what you meant. OR, you could take it back and agree this is not true.

  • Giving JedRothwella chance: what do you mean by that? And how is it evidence for the hypothesis?

    It is unclear to me who this message is addressed to, or what it means.


    Anyway, I meant that Shanahan's theory cannot explain why calibrations work. That is, why calorimeters show a balance of input and output energy where there is no excess (or endothermic deficit). If his theories were correct, you would expect to see Pt-H produce excess heat as often as Pd-D, and both would produce an endothermic deficit as often as anomalous heat.


    Shanahan claims his theories cover this, but I disagree, as do the authors he thinks are mistaken. See:


    http://lenr-canr.org/acrobat/MarwanJanewlookat.pdf

  • KIrkshanahan

    "CCS problem" From your first post on this thread.

    Perhaps you expect CCS to be common knowledge.. but it isn't. not on Google at least

    Eureka!!http://lenr-canr.org/acrobat/MarwanJanewlookat.pdf It does not mean "Clear Communication Society" after all

    CCS = Calibration Constant Shift.

    Did you employ your CCS hypothesis in your intermmetalic tritium research in the last few decades? such as


    Tritium aging effects in LaNi4.25Al0.75 https://doi.org/10.1016/S0925-8388(03)00139-7

  • Wyttenbach " one source of Helium is Soba'

    Cold soba is nice with the temp. 35C like today..but helium is unlikely unless its a cold soba birthday party.

    Tritium has been associated with LENR by number of researchers

    Benyo et al 's 2017 Investigation of Deuterium Loaded Materials Subject to X-Ray Exposure

    concludes "tritium by an unexpected nuclear effect".. https://arxiv.org/abs/1704.01183


    4070-ddduntitled-png


    Claytor, T. N., Tuggle, D. G., Taylor, S. F.; Evolution of Tritium from Deuterided Palladium Subject to High Electrical Currents, Frontiers Science Series No. 4, Proceedings of the Third International Conference on Cold Fusion., October 21-25 Nagoya Japan., Ed. H. Ikegami, Universal Academy Press Tokyo Japan., 1993, p217.


    Tadahiko Mizuno, Tadashi Akimoto, Kazuhisa Azumi and Norio Sato, “Tritium evolution during cathodic polarization of palladium electrode in D2O solution”, The Electrochemical Society of Japan Vol.59, No.9, (1991) 798-799. in Japanese.


    Radhakrishnan, T.P., et al., Tritium Generation during Electrolysis Experiment, in BARC Studies in Cold Fusion,

    P.K. Iyengar and M. Srinivasan, Editors. 1989, Atomic Energy Commission: Bombay. p. A 6.


    N. J. C. Packham, K. L. Wolf, J. C. Wass, R. C. Kainthla and J.O’M. Bockris Production of tritium from D2 O  J. Electroanal. Chem., 270 (1989) 451-458, Elsevier Sequoia S.A., Lausanne.






  • There are no liars here, Kirk, only people with different beliefs to you. Please remember that. Alan.


    Oh really? Well, a) I don't think it's possible to be human and not have told a lie at one point or another (Honey, does this dress make me look fat?) , and what is one who tells a lie called? _______ (fill in the blank). So, we are all in that boat Alan.


    But more to the point, b) what do you call a person who continually writes:



    Shanahan's explanation for this is a gross violation of common sense and the Second Law of Thermodynamics. He claims that the hot object is not a heater and an object that remains hot for days is not being heated. If he believes this he is a crackpot, and if he does not believe it he is trying to deceive stupid people in the audience.


    when he knows this is false? A falsifier? Delusional? What?


    Jed can't support this contention without grossly distorting what I've written multiple times Alan. What do you call a person who deliberately does that then?

  • Anyway, I meant that Shanahan's theory cannot explain why calibrations work. That is, why calorimeters show a balance of input and output energy where there is no excess (or endothermic deficit). If his theories were correct, you would expect to see Pt-H produce excess heat as often as Pd-D, and both would produce an endothermic deficit as often as anomalous heat.


    Also repeating something false after having been corrected multiple times.

  • Quote
    Quote

    Quote
    Shanahan's explanation for this is a gross violation of common sense and the Second Law of Thermodynamics. He claims that the hot object is not a heater and an object that remains hot for days is not being heated. If he believes this he is a crackpot, and if he does not believe it he is trying to deceive stupid people in the audience.


    when he knows this is false? A falsifier? Delusional? What?


    Jed can't support this contention without grossly distorting what I've written multiple times Alan. What do you call a person who deliberately does that then?



    I think you might have a point here Kirk... It was the first law that you horrifically cocked-up previously.

  • The usual Shanahan obfuscation, the fact is that in your 'calculations' of the Mizuno energy balance, you made energy appear out of thin air. (Edit: By choosing an outrageously high water temperature for your evap calcs.)


    In first law terms, that always counts as major cock-up.

    ...And I'm unsurprised you "don't understand" my rendition of the problem... The only question is, whether it's your ego, or just a poor grasp of thermo, that's the main cause?

  • when he knows this is false? A falsifier? Delusional? What?

    This was with regard to Shanahan's repeated claims that a hot object is not heater. For example, here:


    Anecdote about Mizuno's bucket of water


    QUOTE FROM THAT MESSAGE:


    Quote

    In my mind "a large, heavy stainless steel cell in the bucket. It was hot. Too hot to touch. The thermocouple showed it was over 100 deg C inside." is not a 'heater'. It is a hot object.

    What is the difference between a heater and a hot object? He should explain why a hot object is not a heater. If he did not mean this, he should explain what he did mean.


    If there is some difference, why does this difference invalidate Mizuno's claim that the device produced heat beyond the limits of chemistry?


    This statement introduces only confusion, without clarifying anything. Indeed, it does not seem to mean anything, as far as I can tell.


    Shanahan has a habit akin to throwing rocks from a mile away. They never hit the target. He is so far away, no one can tell what target he is aiming for, or why he throws the rocks in the first place. This statement a good example. He repeatedly claims that a hot object is not a heater. No one knows what that mysterious, unnamed difference could be (what the target is), or why it would invalidate Mizuno's claim (why he is throwing). To top it off, he then says he never threw those rocks (he never said it)!


    Another example is his claim that Mizuno's description of the event is "a lie." This could mean that Mizuno is lying; or that I am lying about what Mizuno said; or even, in the weird context of the statement, that Shanahan himself is lying. He never explains what he means. We have to guess.


    I think you might have a point here Kirk... It was the first law that you horrifically cocked-up previously.


    The Second Law as well. Heat must pass from a hotter object to a cooler object. Therefore a hot object is a heater. It heats the surroundings. It does this until it cools down. In case of Mizuno's cell, it did this for several days. Given the thermal mass of the cell, that means something inside it was producing heat. Shanahan denied this, as well -- another violation of the Second Law.


    I suppose this example would violate this First Law if we interpret him to mean that heat is not always heat (not work); i.e., there is some kind of heat that raises the temperature but does not pass from one object to another.

  • The usual Shanahan obfuscation, the fact is that in your 'calculations' of the Mizuno energy balance, you made energy appear out of thin air.

    Ah. I see what you mean by a First Law violation.


    As I said, the Second Law is violated by his claim that the object is hot but it is "not a heater." I suppose that means the hot object is not what makes the water heat up and evaporate. Or it could mean that in general a hot object is not a heater -- it does not heat the surroundings. Or perhaps it means a heater is defined as something that is being heated internally. Of course Mizuno's cell was being heater internally. If it had not been, it would have cooled down quickly. Shanahan also denies that.


    I honestly cannot tell what he means, assuming he means anything. I suspect he may be saying these things only to confuse the issue and mislead people. I can't tell, but as you say, this is obfuscation -- deliberate or accidental.

  • To illustrate my point regarding Jed Rothwell's deliberate distortions, I note he just posted a diatribe that contains this comment:


    What is the difference between a heater and a hot object? He should explain why a hot object is not a heater. If he did not mean this, he should explain what he did mean.


    As he usually does, he failed to fully quote what I wrote. I wrote, IMMEDIATELY FOLLOWING what Jed quoted from my previous post:



    A 'heater' has a power source that adds energy to the system from an external (or perhaps internal, like a kerosene space heater) source. A hot object only has the energy it contains at the nominal 'start point', no additional. So for the too hot to touch, large, heavy stainless steel cell in the bucket to be a heater it would need either a) power inputs, like wires from a power supply, or b) an internal heat source, such as kerosene, a battery, or maybe even a LENR reactor.


    'Nuff said.

  • As he usually does, he failed to fully quote what I wrote. I wrote, IMMEDIATELY FOLLOWING what Jed quoted from my previous post:

    Quote immediately following:

    Quote

    A 'heater' has a power source that adds energy to the system from an external (or perhaps internal, like a kerosene space heater) source.


    Ah, so Shanahan says this is not a heater because it is not heated from an external source or from kerosene. Yet it remains hot for days at the same temperature with no input power. The temperature even went up at times. This is proof that it is being heated internally. There is no kerosene, but there has to be something else producing that heat. How could it remain hot or even heat up more if it is not being heating internally? That would violate the Second Law, as I said. When heat passes from a hot object to a cooler object, the hot object must cool down. Yet it does not cool down.


    Presumably, Shanahan agrees that a cell with plutonium-238 inside it would also be a heater. The source of heat does not have be a chemical reaction; a nuclear reaction also makes it a heater. The cold fusion cell has no chemical fuel, and no chemical changes occur in it. The heat far exceeds the limits of chemistry. Therefore, it has to be nuclear reaction, and a nuclear heater. It produces only helium, so that means it is nuclear fusion.


    Shahanan denies all this, but his denials make no sense.


    Here, for a moment, he seems to agree, but then in other messages he goes back to disagreeing, or he says someone is lying:


    Quote

    So for the too hot to touch, large, heavy stainless steel cell in the bucket to be a heater it would need either a) power inputs, like wires from a power supply, or b) an internal heat source, such as kerosene, a battery, or maybe even a LENR reactor.

    As Mizuno wrote (and I translated):


    There was no power input; the wires were disconnected from the power supply. a) is ruled out.


    So there has to be b) an internal heat source. There was no kerosene or other chemical fuel. There were no chemical changes. So that leaves only a LENR reactor. Since LENR produces helium, it must be fusion.


    As I said, Shanahan says this is a lie but I have no idea who he is accusing of lying. Many researchers have seen cells produce heat with no input. Maybe Shanahan thinks they are all lying?

  • [Responding to Jed: "As I said, Shanahan says this is a lie but I have no idea who he is accusing of lying. Many researchers have seen cells produce heat with no input. Maybe Shanahan thinks they are all lying?"]


    For those of you following this little kerfluffle, you might want to try to find where I said what Jed says I did...

    Right here, message #65:

    Ah, the Mizuno bucket anecdote again...fourth or fifth time you've repeated the same old lies isn't it?

    As I said, it is unclear to me whether you mean Mizuno lied or I lied.

  • Barring that you may have been successful in your attempt to create confusion and misdirect, prior to that activity, it was clear who I was referring to.

    It is not clear to me! Who do you mean? WHAT do you mean? Was Mizuno lying when he claimed the cell remained hot for days with no wires attached? Or am I lying about it?


    Oh Tell Us Please!


    I am joking. Of course you will not say who is lying.


    You have a bad habit of making accusations here and then pretending you did not make them. If you are going to accuse people of lying, you should at least say which people you mean. Mizuno? Me? Fleischmann, McKubre and everyone one else who has observed heat after death? You will not say. It is a disconnected accusation, all the more insidious for that reason. It resembles Sen. McCarthy's accusations that someone, somewhere on a secret list betrayed the nation in some dreadful unnamed way. No one can pin you down or dispute your accusation, because you refuse to say what it is you are accusing us of lying about, or even who you are accusing.