Posts by Zeus46

    It's ridiculous that Maryyugo gets upset and denies people's assertions that he doesn't read the papers he discusses, when the above posts are just the latest example in a long line of evidence that he doesn't.


    To repeat it once more: Mizuno wrote that a close to a bucketful (10L) of water evaporated in the first 24hrs. Mary, if you had bothered to read it, you would know that.


    Jed is absolutely correct in what he says about KShanahan's argument, however, he can't provide a link to a quote, but/because it is IMPLICIT to Kirk's "calculations".


    I am trying to compile some information to share on this matter too.


    You're a little late to the party Roger: The Playground

    </sarcasm>


    Note to people unfamiliar with the subject: we have had that since 1992.


    ...</sarcasm>


    Note to people unfamiliar with the subject: There are world-class experts, and also self-proclaimed world-class experts*... One needs to be careful when distinguishing between the two.


    Eh, Mary?


    *Sorry, but relevant links are (correctly) considered to be 'doxxing'.

    Rossi is pretty well unique in this area in being so concerned with his day-to-day interactive communication with fans, WE, JoNP, etc


    That fact should condition strongly any objective evaluation of his work, and I'm sure does for most observers.



    I wonder which of you has the higher daily posting rate?

    Spare me the "You've never done a literature search" shtick, please. It's getting tiresome, and the suggestion that you have done so many yourself, that a great deal of time is saved by shortening it to an acronym, gives a further sense of the importance to which you ascribe yourself...


    ...And for the fourth time, it's 'Zeus'. Like the god, you know?


    I also suppose that, after doing the necessary number of literature searches to arrive at your conclusions about The Motives of Most Writers, you never pondered the notion that another common thread running through the texts, was that it was you yourself that read them all? Each was filtered through your personal schemata, and as such, any (behavioral) conclusions you reach will be highly biased, by definition.


    But, because I reckon you are bright enough to understand such problems - It seems more reasonable to assume that this apparent knowledge must have come from reading their minds instead... No?



    PS. Whenever I see a 'maryyugo likes this' comment, I get this mental image of Mutley snickering, enthralled by his own Dick Dastardly.

    The preliminary results indicate that between 200° and 650° C., and pressures up to 0.2 mm. Hg, there is no measurable adsorption of hydrogen, but an appreciable absorption (homogeneous solution). The results are in good agreement with Sieverts' measurements with much thicker nickel wire at higher pressures and temperatures1. The amount of absorbed hydrogen at a constant temperature is, within the experimental error, proportional to the square root of the pressure, and increases at constant pressure with increasing temperature, obeying the simple equation:


    0.2 mmHg = 0.0039 PSI

    So I stretch my thinking again under the principle that I am seeking to see if a ‘mundane’ explanation can encompass the observations and assume 10% RH as a practical minimum.


    Pluck! This isn't the Mojave desert on a summer's day...



    "April 25. Mizuno and Akimoto note that temperature is elevated. It has produced 1.2 H 107 joules since April 22, in heat-after-death. The cell is removed from the underground lab and transferred to Mizuno’s lab. Cell temperature is >100 deg C.

    April 26. Cell temperature has not declined. Cell transferred to a 15-liter bucket, where it is partially submerged in water. "


    So, the cell was at 100C when placed in the bucket initially (per Mizuno/JedR). Assuming it cools to 16C and does not stay at 100C as asserted by Mizuno/Rothwell,


    What you assert only applies if you mix up the two sentences of the April 26 entry, effectively reading them backwards. So I don't believe that Mizuno says this, and I'm pretty sure that Jed has previously called you a crackpot for making inferring this same error.



    I would need to insert an exponential cooling curve into the swimming pool equation. And I would also need to account for any heating due to H2 combustion, but that is probably small. This affects the vapor pressure of the water in air. Instead I assumed a simple linear decrease and used a rounded average of 60C.


    Just so that I can understand... Are you saying that after submersing the reactor, all 15L of water heats up to approx. 100C, and then exponentially cools to 16C. But to avoid the maths we can take a linear average of about 60C?


    Because that's just crackers...The water will never get anywhere near 100C - or even 60C.


    For example, putting a 40Kg, 150C lump of solid steel (Cp = 465J/kg.C) into 15L of 16C water (Cp = 4184J/kg.C), could only ever raise the water temperature to 46C


    This can be calculated using T = ( m1.C1.T1 + m2.C2.T2 ) / ( m1.C1 + m2.C2 ).... Or to give it another name, The First Law Of Thermodynamics.


    Perhaps Jed could shed some light on Mizuno's actual reactor design, but I can almost guarantee that it doesn't contain 40Kg of steel. And as a side-bet, I'll also predict that any reasonable estimates as to the unsubmerged reactor's maximum embodied heat energy - will lead to a negligible rise in water temperature (i.e. < 4C 8C - thanks Bocijn) once submerged.


    How about if we use 75C instead of 60.


    For real?



    ...So now slightly bigger area which increases evap rate. All else being the same, more should have evap’d....

    ...So, now the rate is 1/3rd. More cooling? Less ventilation? Higher relative humidity?...

    ...So today I went a little further and looked at this last datum, which means 7.5L evap’ed in 5 days, i.e. 1.5 L/day – almost down to 10% of original.

    ....Do you all get it that now the data seems unreasonable? Rapid evaporation from a cool cell...


    Your reasoning as to why this must be strange, is based on the implied assumption that any heat-after-death must evolve in a constant unchanging manner, before blinking out like a lightbulb one day. (I think) the data you present could also be explained by a tailing off of HAD - from a maximum to nothing.



    So what do we do now? Go out and buy CF stocks? No, we expect replication, as MY has already expressed. Let’s see.


    ...maryyugo likes this.


    "[You] two should write a heat transfer paper together. Nobel prizes all round for sure."


    :)


    AND SOME MORE:



    Likewise, one point I made back there was the lack of knowledge of ventilation. What Jed fails to mention is that the bucket was put in a hood in an abandoned lab. Even without the holes in the windows, what do you think the air flow over the bucket would be in a hood, even with no blower fan to suck air in?


    With no blower fan, it would seem obvious that any air flow would be minimal. However, that's not very interesting. Fortunately, several branches of the US Government like to stick their nose into the business of freedom-loving fume hood manufacturers... They offer up all kinds of standards about how just how hard a fume hood should suck, in order to protect the user's health:


    http://ateam.lbl.gov/hightech/…nts/su00/Fox/FHSafety.htm


    Basically, they agree that if you want a top-spec fume hood, say for working with carcinogens, it should have a face velocity of 150 feet per minute. Or 1.7 miles per hour, for the sake of my spreadsheet: mizuno bucket.zip


    Wind Speed = 1.7mph

    Bucket Diameter = 0.3m

    Record Hokkaido Temperature in March = 19C

    Cats = 0


    This equates to an EPA mandated evaporation rate of 0.54 litres per day.


    But let's suppose the fume hood has an interior geometry that doubles the air velocity inside the chamber, as compared to the front opening.


    Wind Speed = 3.4mph

    Bucket Diameter = 0.3m

    Record Hokkaido Temperature in March = 19C

    Cats = 0


    = 0.93 litres per day.


    In other words, there's still at least 28L of evaporated water to account for* over the whole 10 days. According to the EPA...


    But wait a minute! There's still another variable that can also be wildly exaggerated:


    So I've been piddling around with the equation bocijn found, using 10% relative humidity and 1 m/s air movement over the bucket for the first datum that JedR supplied in his into he referenced (10L evap'd in 1 day, but starting at 100C). The 100C starting point means we have to assume some sort of cooling curve, if we assume no CF heat. I assumed an average temp of 60C.


    Ummm... What?! A 100C starting point. Are you sure?


    That sounds to me like another figure plucked from God-knows-where, for the purposes of, sorry to say it, 'stretching reality to your pathologically-skeptical heart's content'.


    In real reality, for every kilogram of hot (100C) steel you dump into a 15L bucket of 15C water, it's equilibrium temperature increases by 0.68C


    Or to put it simply, Mizuno would struggle to lift a reactor capable of raising the starting water temperature much above 35C.


    So:


    Wind Speed = 3.4mph

    Bucket Diameter = 0.3m

    Water temperature = 35C

    Cats = 0


    Which gives an evaporation rate of 2.2 litres per day, ignoring any cool down effects over time (and Mizuno's sore back).  Which is why I think your reasoning here is well faulty:


    I realized I could play around with parameters and come up with conditions that could produce the desired 10 L evaporation in 24 hours, as well as other sets giving other results. HOWEVER, at *that* point, it becomes a waste of time, because THE CONCLUSION is that you will actually need the REAL DATA for the REAL CONDITIONS during the event, not averages, possibilities or guesses.


    AS SUCH, it would seem to me that a more proper CONCLUSION would be:


    Using UNREASONABLE EXTREME assumptions prevents you making a REASONABLY ACCURATE estimate of evaporation rates.


    Which when I think about it... Is just your CONCLUSION reworded, really.



    * Incidentally, if the latent heat of vapourisation is 2246J/kg, evaporating 28L water should require an average "Heat After Death" power output of 73W over the 10 days.

    As I mentioned, the last set of data I looked at got the 10L evaporation done in 1.75 days. At *that* point, I realized I could play around with parameters and come up with conditions that could produce the desired 10 L evaporation in 24 hours, as well as other sets giving other results.


    Yes, as me and Bocijn suggest, it's possible to get weird answers if you use lunatic inputs.


    And until you show the numbers your 'playings around' are based on, the assumption has to be that you are firmly inside that camp.



    The idea is vary the controlling parameters to see what impact that variation has on the derived conclusions. Zeus doesn't see to get this at all,


    You're not controlling parameters, or providing a sensitivity analysis either. You are plucking impossible numbers out of thin air, and hoping no-one pulls you up on them.



    THE CONCLUSION is that you will actually need the REAL DATA for the REAL CONDITIONS during the event, not averages, possibilities or guesses.


    see here: The Playground

    Following the successful investigation into yet another performance of the Unbounded Error Gambit, a similar, but potentially more egregious, move has come to light...


    The self-proclaimed world-class expert in theoretical calorimetry, master statistician, long-term LENR sidethorn, and suspected Hot Fusion gravy-trainer Kirkshanahan, is also alleged to have played a lesser-known variation of the Unbounded Error Gambit...


    So, as part of an ongoing series, in which one delves into the proclivities of some of the more vociferous/blessed forum characters, it's surely about time for... ,,,



    KIRK WATCH!



    The problem: Mizuno once put a hot experimental reactor in a bucket of water and left it there for 10 days. Lots of water (38 litres) evaporated. Some say this is evidence the reactor must have kept on producing heat... This is bad, for some reason...


    The solution: Try to prove the water could have evaporated naturally by finding a suitable evaporation equation, maxing out the variables beyond all common sense, then adding on a huge error margin... And hey presto! The Hyper-Bounded Error Attack is almost complete! (Beware...It still has to be snuck into the general consciousness however - Often requiring a few subtle appeals to your own authority).



    Regarding an anecdotal account of the event noted here: http://lenr-canr.org/acrobat/MizunoTnucleartra.pdf, Kirk had this to say:


    My examination of evaporation rate equations put out by DOE for swimming pools led me to believe that it [could just have evaporated normally] if the ventilation and humidity characteristics were correct.



    WHAT! 38L of water can just evaporate normally in 10 days!? That sounds little weird to me....


    But let's see if we can somehow arrive at a similar calculated result:


    First we need an equation... You gotta appreciate all that Government research money, so let's use a modified version of the EPA's very own pool evaporation equation, found here:


    https://dengarden.com/swimming…on-Rate-for-Swimming-Pool


    ...Of course, it also helps that this same equation overestimates actual, measured, swimming pool evaporation rates by approximately 30%. But let's just ignore that for now...


    Second, despite the record daytime temperature in May in Hokkaido being 19C, we'll assume the heating has been left on in this 'abandoned building' and it's a balmy 22C... At all times, day and night.


    Third, Kirk says the building has a lot of broken windows, so it's definitely hot and windy inside. Let's say a 5mph wind speed... Leafs and small twigs are moving about on the floor, and the unfortunate janitor constantly feels the wind on his face. This indoor Light Breeze howls through the building all day and all night long... For 10 days straight.


    Fourth, did you know that for various cultural reasons, Japan has very oddly shaped buckets? Really wide and flat? So much so that a 20L bucket has a diameter of 40cm, and hence has almost double the surface area of a standard-issue US Government bucket.


    Lastly, and stick with this one, please... Kirk says there's a thirsty cat prowling around this formerly proud research building (and everybody knows that a cat, when eating dry food, drinks 250ml of water per day. NB: cats are notoriously territorial - so I'm sticking with just the one cat).


    In the above circumstances* the EPA predicts 2.9 litres/day will be lost to evaporation/cats... But Mizuno claims an average evaporation rate of 3.8 litres/day.


    So we're nearly there! All that's left to do is add a 31% error margin on top (multiplying all the other erroneous margins) And we've done it! Hurrah! Hot Fusion is saved!


    (For another funding round at least).


    ----------------------------------------------------------



    Fellow students of the hyper-bounded error attack can tweak parameters stretch reality to their pathologically-skeptical heart's content, on the spreadsheet here... mizuno bucket.zip. (Inputs are in red text).


    Or check out Bocijn's scoop/take on the matter here.




    * Under slightly more believable circumstances (i.e. An all-day-all-night temperature of 17C, a 1mph indoor wind speed, and a normal bucket), the EPA predicts an evaporation rate approximatly 90% lower. But then again, the average outdoor temperature in Hokkaido in May is only 9C, and aren't the windows broken?....


    ...And before anyone moans that I don't consider the relative humidity... It averages 79% in Hokkaido in May... i.e. Similar too, or higher than, almost all of the US - where that pool evaporation equation is calibrated for. ;)

    As you see here, he claims that a DoE database relating to swimming pools shows that a bucket of water left at room temperature overnight might all evaporate. It just happens. It doesn't mean anything.



    I think he also claims a flock of birds/herd of cats might have drank some of the water. No joke.


    Edit: Rossi vs. Darden aftermath discussions.



    Edit2: Nice one bocijn , you beat me to it... I have a half written post which says much the same thing, but still needs a bit of editing - I might (have to) post it in The Playground if/when I finish it.

    WHERE IS THE HEAT CAPACITY TERM???


    My calcs (using degK) from Mizuno's data give Cp's of approximately 1.007, and I note you need to convert from kg to g, which adds a multiplier of 1000 in there, so the 1006 in Mizuno's equation *could be* that factor times a constant Cp of 1.006. But what are the 2.119 and 1.1429?


    1006 is the heat capacity, as you say kg to g.


    2.119 and 1.1429 don't make any sense to me, but I haven't looked at the spreadsheet. The only reason I can think of for those to be there is as some kind of conversion factors, maybe to include some non-metric data inputs?


    Have I done this right?


    Yep, that's the right way to work out POut.


    Is this subtracted graph you've made the exact negative of Mizuno's excess power curves, Kirk?


    Seems a little odd, to say the least.

    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 ?

    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.