The Playground

Quote from THHuxleynew

IH fought Rossi when they would have been better off (in terms of money spent) giving him a sum less than the lawyer fees. That takes balls, and helps their reputation.

Balls?!? LOL!! Rossi have denounced them and they have been forced to defend themselves (or at least to try ...). Rossi expected $ 89 million, why would he have to be happy for much less? Maybe if they had immediatly give him the IP they could have avoided the controversy .... who knows .... but probably IH also considered the IP very important, even though now they deny it.

"Defect generation in Pd layers by "smart" films with high H-affinity"

Nature - Scientific Reports 7 (2017)

https://www.nature.com/articles/s41598-017-09900-z

Writing skills model for cold fusion amateurs - excellent lab research report on tritium-titanium systems.

"Titanium tritide radioisotope heat source development: palladium-coated titanium hydriding kinetics and tritium loading tests"

http://prod.sandia.gov/techlib…ntrol.cgi/2012/120232.pdf

("SAES Getter" cited in the report is the US branch of an Italian company - https://www.saesgetters.com/it…/getters/hydrogen-getters )

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-train rider 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:

Quote from kirkshanahan

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 a bit crazy 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 April in Hokkaido being 6C, 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 April is only 6C, 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 - ie. Where that pool evaporation equation is calibrated for.

AND SOME MORE:

Quote from kirkshanahan

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:

Quote from kirkshanahan

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, the water 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:

Quote from kirkshanahan

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.

Quote from Zeus46

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'.

[emphasis added]

Quote from Zeus46

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

Really?

From http://lenr-canr.org/acrobat/MizunoTnucleartra.pdf pages 2-3

"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. "

{emphasis added}

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, 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.

From my CRC Handbook, the vapor pressure of water at 60C is 149.38 Torr or 11.92 kPa.

I stretch my thinking, and wonder if the relative humidity ever varies in Japan, and guess "YES". 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. 0% would be too low.

So the swimming pool equation, which transforms into ( 1 – relhumid/100) * Vp * (.089+.0782*v), becomes [0.9 * 11.92 * (.089+.0782*v)]/2272.

So what is v (air velocity)? We have no idea. So, you start at zero and go up until it becomes unreasonable. Let’s look at the 2 m/s velocity though. That gives the evaporation rate per unit area of [.9 * 11.92 *(.089 + .1564)] = 1.159e-3 kg/sm^2, which means that for a 0.049 m^2 evap area, that 4905.6 g of water would evaporate in 1 day (under all the above assumptions). Half-way there. So, maybe not a linear representation. How about if we use 75C instead of 60. The Vp = 38.55 kPa then, that’s a multiplier in our formula of 38.55/11.92 = 3.233, giving 15,863 grams evap’d instead. Wow- 50% over now…so clearly what the temp was and its time profile is very important to these calculations.

What about the velocity? Unreasonable? 2 m/s ~ 4.5 mph, a brisk walk. Would this be unreasonable in an abandoned building? I don’t know. Think about the air flow around your face when you are walking fast. Do you notice it? Maybe. You’d probably notice a 5 mph breeze of course. Would Mizuno have noticed that in the lab? When he was worried about his cell exploding? Would there be a constant flow or would it gust a little depending on outside conditions? When the ventilation drops in the building I work in it can make the doors very difficult to open and I briefly get a ‘stiff’ breeze when I do. The hood in the room has a connection to the rooftop via the ventilation system. In an operating lab, flow would be controlled by a blower. But what would control it in an abandoned lab? Personally I can’t reject 2 m/s. Maybe even a bit faster.

So you see the temp profile is really the key factor here, once you make some *reasonable* assumptions about humidity and ventilation. Again the point is that to *actually* understand what is happening in the anecdotal story, you need more information, which we will likely never get.

Continuing to quote Mizuno/JedR:

“April 27. Most of the water in the bucket, ~10 liters, has evaporated.

The cell is transferred to a larger, 20 liter bucket. It is fully submerged in 15 liters of water. “

So now slightly bigger area which increases evap rate. All else being the same, more should have evap’d. Less evap’ing means T probably a little lower.

“April 30. Most of the water has evaporated; ~10 liters.

More water is added to the bucket, bringing the total to 15 liters again. “

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

“May 1. 5 liters of water are added to the bucket.”

Rate at ½ now.

“May 2. 5 more liters are added to the bucket.”

Still at ½.

“May 7. The cell is finally cool. 7.5 liters of water remain in the bucket.”

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.

Let’s see now what nominal conditions might be needed to do that. Assume 16C, since Mizuno/JedR says it was ‘cool’. 1.5L/day means 15000 g/day = 15 kg/day = 1.74e-4 kg/s

Bocijn originally calculated 1.12e-3g/s and 2.3 10^-5 kg/s/m^2 for 16C and 70% RH. For the 0.049 m^2 area, we need 3.54e-3 kg/s/m^2.

So we have [.9 * 1.8 * (.089 + .0782*v)]/2272 = 3.54e-3.

That gives a velocity of 62.3 m/s. Wow! That’s probably better than a rocket engine exhaust (not really)! So yes, now we are unreasonable. What does increasing the temp do? Not too much. If the T was 20C, the Vp is only ~1.3X, cuts the velocity to about 48 m/s. Still unreasonable. Increase the velocity (windy weekend?), probably not too much help. Gosh, maybe the cell was still at 100C! No…Mizuno said it was cool….hmmm…. (Do you all get it that now the data seems unreasonable? Rapid evaporation from a cool cell...)

So what do we do now? Go out and buy CF stocks? No, we expect replication, as MY has already expressed. Let’s see. This incident happened in 1991, so Mizuno should have a cold fusion water heater by now right? Or be drinking cold-fusion-power brewed tea right?

I don’t know what happened in this anecdotal story. That’s the problem with anecdotes. As I showed above, 10L in 1 day during a cool down from 100C in a ventilated hood is possibly not unreasonable. 1.5 L/day at 16C is apparently. The anecdote seems that it should have driven Mizuno to replicate. But just like Storms and his Pt-based CF, the most controlled example of ‘excess heat’ I have seen, he went off and stared working on something else. As MY says, no one seems to be able to produce an incontrovertible demonstration experiment. In the end, this is what leads me to believe there is no LENR, but lots of hidden errors like the Mizuno ‘2.42’ factor from nowhere. No more discussion about anecdotes…

Quote from kirkshanahan

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...

Quote from kirkshanahan

"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.

Quote from kirkshanahan

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 for the 7.45kg, Bocijn) once submerged.

Quote from kirkshanahan

How about if we use 75C instead of 60.

For real?

Quote from kirkshanahan

...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.

Quote from kirkshanahan

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".

Quote from Zeus46

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

There appears to be no thread for website layout.

I was wondering if someone could come up with an alternative name for "Native language forums"

English is one of my 'native' tongues but it is not there.

I thought of ".. NESB...but may be others have a better idea

Zeus46 wrote: "but I can almost guarantee that it doesn't contain 40Kg of steel"

I emailed Mizuno today. then I phoned him. He rang back after awhile.

He is busy working on several reactors or more reactors

I was happy to hear him speak for the first time. His English is good.

We exchanged pleasantries and commiserated each other on being bimbo gakusha.

I asked him what was the name of the newest reactor..Hikobosu or Hi o Kobosu..not quite sure...some pun involved.

Means "spill from the sun".

Mizuno did not know why I asked him but he answered that the metal reactor in the spring 1991 excursion weighed 7.45 kg"

Including the solution and the electrode was a bit more.

I guess this would help a fit 46 year old to lift it, plus maybe 15 litres water from the cold tap.

Quote from bocijn

I asked him what was the name of the newest reactor..Hikobosu or Hi o Kobosu..not quite sure...some pun involved.

Maybe Hikoboshi?

https://en.wikipedia.org/wiki/Hikoboshi

https://en.wikipedia.org/wiki/Tanabata

Can.Yes. Well found. Hikoboshi. That was it. There is a part pun on the Tadahiko name. Also he mentioned something about Altair. An anime character.

There are limited number of consonants and vowels so double meanings and associations are more frequent than in English.

I was thinking. That he would name them after his children.. but there you go

Not anime.. that's about Turkey which isn't so happy http://www.japan-suite.com/blo…f-two-star-crossed-lovers

It's about tanabata.. star... lovers- so wonderful.. I didn't know.. so soosooteki desu

itsuka sono hikoboshi o mite mitai desu..

Quote from Dewey Weaver

Shane - no confusion. Off-white paint. We have the brand and product number.

We also have the same batch of Durapot 810 and a leftover virgin reactor from that production run. Planet Rossi may hear more about that later.

No comment on Levi having the material sample tested and coming back with 99% alumina except....say no more say no more.

Rossi vs. Darden developments [CASE CLOSED]

Dewey,

Can you disclose the paint brand and product number?

I might want to test it out.

Was it Pyro Paint 634 AL?

https://www.aremco.com/news-it…-temp-refractory-coating/

https://www.aremco.com/high-temp-refractory-coatings/

After my first Durapot 810 test, it became clear that Durapot 810, notwithstanding the thermal conductive descriptor on the promotional material, is a poor conductor of heat. Maybe it's better than your average alumina casting material, but compared to zirconium oxide, etc., it is poor at 2.16 W/mK.

Even 5 mm of it is enough to greatly slow down the heat from a wire cast into it. The widening temperature gap between the outside and the inside as the overall temperature rises is evidence of this.

Using this calculator, and inserting the appropriate values, the temperature in the region of the Lugano device heater coil can be estimated. (The temperatures for Lugano can be taken at face value for the present.). Heat calculator

Area of main Lugano body (not Caps): 0.0144 m²

Thickness: fins 1.6 mm, tube 1 cm, overall diameter 2.32 mm so : 4 to 5 mm (0.4 to 0.5 cm)

Thermal conductivity: 2.16 W/mK Metric Durapot specifications

T hot: we are solving for this. Click on the T_hot part of the equation, (in blue at top) when the remaining numbers are filled in to get the answer.

T cold : this is the external temperature. 1400 or 1410 C will do here.

Q/t: might be some conjecture here. 2890 W reported for the device, Caps included. This Calculator will enable one to sort out the Cap contribution. Turns out it is fairly low. Use 4 cm (0. 04 m) diameter, 8 cm (0.08 m) long [ 2 x 4 cm long Cap], 0.7 emissivity, 600 C for Caps and 0.2 m long, 0.023 m diameter, 0.4 emissivity and 1400 or 1410 C temperature. Anyways, about 300 W of the reported 2890 W output of the device (we are ignoring the Rods, Joule heat cables; just looking at the main ribbed tube part) belongs to the Caps. I used 2500 W here for the main tube (to be conservative). Try 910 W (Lugano input power) instead of 2500.

One version is pictured below, with 4 mm depth (wall thickness). This would be close to the depth of the heater coil wires in the Lugano device from surface.

The melting point of Kanthal A1 is 1500 C.

(I suspect that the thermal conductivity of Durapot 810 drops significantly from room T to 1350 C. Possibly to half.)

WTF is with the power factor reported by a Kill A Watt for a purely resistive load fed by a SSVR set at less than full conduction/voltage?

Does a PCE meter get tripped up by a triac?

Has anyone tried it?

Maybe the MFMP can try measuring a coil powered by a triac with theirs and report the results.

Edit: part 1 worked out. The meter uses the V before the SSVR to calculate the PF.

Not sure if the W or VA is correct. Neither seems to be correct...

I can see that doing these tests on the cheap side is not going to work, which is what I was working on, so anyone can do these hot alumina tests.

I just bought a new true RMS multimeter with duty cycle functions etc. to sort this out.

...

The Kill A Watt shows correct W and A, and cross checks with the steamengine.org calculator. (Minor differences based on the included minor load of the SSVR circuit and indicator lights). Volts reading on my other DVM was terrible. 50.4 V in my last test should have been around 76 V.

Luckily my workhorse "Vulcan" coil has the same wire and coil resistance as the square I cast, so a quick test was easy. (The Vulcan has windings on the outside, and only gets to 800 C maxed out at 115 V, so the effect of insulating with Durapot is quite a contrast).

Lesson: Ignore VA and PF on the Kill A Watt when using a purely resistive load and partial sine AC.

(I don't think the PCE is so easily fooled).

I love that you're still following up on the Lugano test. I won't call this OCD.

Eric Walker ,

Oh, it may be OCD...

I'm working up to a design with the same apparent peak COP as Lugano, but fuelled with 1 g of Kool-Aid powder.

In the meantime, the KISS principle is strangely difficult in practice.

Who is Paragdimonia, any relation to Paradigmnoia?