Posts by Paradigmnoia

Quote from Tom Paulsen

The formulas for calculation of joule heat in the Lugano test report (page 14) are wrong and should be corrected as follows:

WC1 = sqrt(3)*(R1*(I1^2)) = 1,732*(0,004375*(19.7^2)) = 2,94 [W] (9)

WC2 = sqrt(3)*2*(R2*(I2^2)) = 1,732*2*(0,002811*(9.85^2)) = 0,94 [W] (10)

W tot dummy = WC1 + WC2 = 2,94 + 0,94 = 3,88 [W] (11)

General usage of these (correct) formulas eliminates the extremely nonlinear differences in joule heating values between the dummy reactor and the active E-Cat.

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The C1 cables are not in the delta. They have full line voltage and current. There are three of them.
The C2 cables are inside the delta. There are 3 pairs of them. They have delta phase current. This is neither 9.85 A nor 9.85 X sqrt(3)
The watt losses from the C1 cables must be subtracted from the total W before calculating the delta.
The phase current for the delta is the line current divided by sqrt(3)

@randombit0
I also did an experiment with a quartz tube and a Kanthal heater coil. You may be interested in the results, although not very well constrained overall.

The quartz tube was heated to a set temperature (516°C), and radiant power calculated by subtracting total input power W from the calculated convective loss (by standard formula). ε was determined by comparing to a thermocouple and correcting an IR probe ε until T matched.
Power by ε was calculated by the standard formula, and compared to the power calculated by subtracting convection from power in.

Then the quartz tube was painted barbeque black, and retested at the same input power.

Temperature increased almost 10%, but this merely increased the convective losses, so that total radiated power remained the same.

Within the admittedly large possible experimental errors, defined (by me) as the difference between a ε of 1.0 for total power output plus convective losses and the necessary ε to get the power to balance (0.96 vs 1.0), the radiated losses unaccounted for by transparency, due to the spectral sensitivity of the IR probe (8 to 14 microns), was about 10% . This is the 10% that was converted to convective losses but a higher temperature.

Therefore, although flimsy evidence (but could be repeated for better confidence), in open air, a selective emitter is a more effective radiator of electrical power converted to IR than a blackbody, since the blackbody ends up with a higher temperature and therefore higher convective losses.
This is, of course, below the temperature where radiant emission exceeds convection by a much larger percentage. At, say, 1000°C, this might not hold up as well.

In other words, perhaps 10% of the heat energy could be "lost" (remain unmeasured/calculated) in the dummy. But as the alumina begins to become incandescent, the transparency decreases. In the Manara et al plot, the maximum temperature is about 777°C, which is just at the tipping point of radiance overtaking convection as the primary heat mover, based on my experiments. Another trace at 1273K would have been very helpful.

@randombit0
The transparency issue means that any power radiated past the alumina does not heat the alumina, therefore does not raise the temperature of the alumina, and therefore does not get calculated into power output equations based on only on temperature as calculated by the Optris. Certainly some power can "sneak by" the Optris in the transparent range, because it cannot see it. I have argued this point many times in the past. But the power calculations based were based on Optris-measured temperature, so the transmitted power was not, and could not be, included in power output calculations made in the report. The amount of possible extra power that was not measured or calculated due to transparency is an open question.

An Optris-specific spectral view emissivity value of about 0.8 to 0.85 would suffice to lower the apparent Optris temperature view so that the 0.95 Dot looked hotter to the Optris. 235°C is very close to the lower effective range of the modified high temperature version of the Optris, as well as only about 10 X the ambient temperature, which accentuates the apparent temperature contrast in the figure 7 image. That is my best explanation, although possibly not the correct one.

The Joule heat in the active is low in the report about 6 to 7 W, once corrected. Not a deal-breaker, but a nuisance for accurate R values.
To fix this you must reverse the original math, as it appears in the report, to get the correct starting current, then re-calculate with the correct delta equations.

@Tom Paulsen
Thank you for reviewing my plots. I sent this to several people for review, and received no answer.

In regards to your comment, the phase voltage and line voltage are the same in a delta. Only the phase current is reduced.

Edit: my favorite on-line delta-wye calculator seems to have disappeared (supposedly improved, but not). I will double-check my math.
Edit2: I stand by by voltage figures.

What Optris says:

@Thomas Clarke
I do get a slight drop in R from the dummy to the active runs, about 0.05 Ω.
Considering the potential measurement inaccuracies, that is probably insignificant.
I also did not consider any inductance in my analyses.
The resistance calculated does not increase relative to temperature in the way any type of calibrated resistance wire I could find normally does.

See what you think of this:

Maybe the DW on this shift didn't read the notes from the one on the previous shift....

@Thomas Clarke ,
The response of the microbolometer array is affected by leveling corrections internal to the Optris.
I know that some work was done on it, but how the unit levels and how Bob or others have leveled the response compares is an open question.
I am just eyeballing the curve, and mentally weighting an average response.
Between 0.8 and 0.9 is something I can agree on, without buying an Optris.

Maybe we can suggest that the MFMP use the Optris on the Glowstick cover tubes at 235°C and 450°C and fiddle with the emissivity setting at the end of the present experiment for maybe half an hour.

Edit: Image attached
Edit2: Notice that in the Manara et al plot, the big alumina spectral ε peak actually broadens with increasing temperature, especially in the far long wave side of the Optris view window, so that the Optris normal ε increases (a bit) with temperature for alumina.

@Thomas Clarke ,
From the plot I posted above, one could argue that a normal emissivity as low as 0.8 is feasible in the Optris range.
0.8 normal emissivity actually fits the dummy fairly well for power, in my cherry picking exercise, when the hemispherical emissivity is also fixed up in the low temperature range.
Data for total hemispherical emissivity of alumina was not very consistent at around 500K, however.

So, lets look at:
(I superimposed the magenta lines on the Manara et al 2008 image)

Quote from randombit0

No my dear, is not nonsense if you use the right curve. Of course if you use another function or curve you will converge on that but THAT only means that you are using the wrong emissivity plot.

BINGO!

Or should I say GIGO!

Sorry, definite a typo.

Try this: WO2015127263
Look for Pamphlet 131, near the bottom. It came out here much before the US version appeared.

(I'll fix the number above)

Try this link : https://patentscope.wipo.int/search/en/detail.jsf?docId=WO2015127263&recNum=9&office=&queryString=FP%3A(Andrea+Rossi)&prevFilter=&sortOption=Pub+Date+Desc&maxRec=220
Under the Documents tab, look for the 157 page Prior Document.

This is all available now on the US side, but this particular document has been available to the public since August 27, 2015.
I mainly refer to this document since it was the first to display the Lugano device blow-up schematic, with photos, now replaced by drawings in the most recent iteration. (with M.C. Escher-like coils...LOL)

Quote from randombit0

Are you ok ? This is nonsense !

I'm very well, thank you. And you?

But this is nonsense (image below). You can prove it to yourself in about 10 minutes, using the red line, then the Lugano line (separately) and an IR camera or suitable facsimile.

And, for the iteration method, as seen in tables 2a and 2b, Lugano report, I have made this as a helpful guide to test the recursive emissivity method, which I assert only self-satifies it's own graph for any given radiance detected by the Optris.
This can be used as a standard reference for anyone who cares to discuss and test this further. The red line has been traced from a scaled monthly rainfall plot, from Ascona Switzerland (chosen for its nice ups and downs). The line was smoothed from the 12 monthly bars in the plot and taken to the edges.

What I recommend to test the validity of the recursive emissivity method is substituting the values on the red line for the Lugano line. This requires a spectral radiance model, or an IR camera (or possibly a spot pointer) using the 7.5 to 13 micron IR range.

Further to the discussion of dots, I submit the following. Maybe it's nothing... but maybe another emissivity dot?

.

Randombit0 said:

Quote

Kanthal like alloys have a very special resistance vs temperature behavior. Knowing just the diameter but NOT the material can hardly help you to calculate the resistance.

Well, the application I refer to, WO2015127263, has a link to PCT/US2015/016897, and on page 10 this this document says "For example, a 15 gauge wire with resistance of 2.650 ohms/ft is one example of suitable wire." {0076}

So, find me a wire, any wire, in 15 Ga that has a resistance of (or even close to) 2.650 ohms/ ft.

15 Ga wire is mentioned twice in the same section, with a possible partial mention of Kanthal, "In particular embodiments, the wires 16 are 2 guage 15 KA resistance wires, and prior to wrapping the wires 16 around the reactor chamber"

You may quickly find that in order to get 2.650 ohms/ft, that something in the range of 28 Ga is required.

I have some wonderful delta vs wye plots. But I became bored of that ages ago, one I was satisfied that at least the electrical end of things was, or at least could be, consistent with normal reality. I could have fixed up my horrible spreadsheet for all the Power, Joule heating and such, when I found a minor error while making the aforementioned plots, that changes the final derived resistance values by a tenth of an ohm or so... But it was far more challenging getting the required resistances into an actual coil. Luckily IH spilled the beans with a patent application, so that the trade secret windings were revealed. The resistance figure they mention in the application is nonsense, in my opinion. But revealing the wire size was good enough to work it out. Putting just the right spin, I mean twist on the wires makes a world of difference.

@randombit0
Figure 7 shows them doing the work, but they did not show their work, as in adjusting the literature-based emissivity plot to reflect the empirical data.

“We therefore took the same emissivity trend found in the literature as reference; but, by applying emissivity reference dots along the rods, we were able to adapt that curve to this specific type of alumina, by directly measuring local emissivity in places close to the reference dots (Figure 7).”
(emphasis mine)

"The temperature at the rectangle next to the circle (237.5 °C) is obtained by setting an emissivity value for alumina found in the literature [3.]"

In fact, the total spectral radiance in the 7.5 to 13 micron IR range at an ε 0.71 at 237.5 °C (ε from the Al ε vs T, plot 1), is not equivalent to the total spectral radiance at an ε of 0.95 at 235 °C in the 7.5 to 13 micron band.
About 287°C at 0.71 is much closer. But that isn't right either, because the values in the Plot 1 chart are for total hemispherical emissivity of alumina, not Optris spectral detection bandwidth emissivity for alumina.
So something weird is going on in Figure 7.

(Maybe Thomas can try this out and see if I'm out to lunch on this, since he has the code already. Anyone else can give it a go, too.)

Now, if they did the iteration thing (in the style of table 2a, & 2b) they might have ended up at around 272°C with an ε of 0.77, in which case the total spectral radiance would match, but the 272°C and 235°C do not match.

The photo (figure 7) does seem to indicate a lower ε than 0.95 is appropriate. But the values do not seem to add up with other assertions.