@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.
QuoteThe calculation of the joule heating values is wrong in the report. They used factor 3, but sqrt(3) would have been correct. So I think that it doesn't make sense to evaluate other configurations, based on these false values.
Yes. I remember some ambiguity in what they said because they got the Joule heating currents wrong by a factor of sqrt(3) (they thought the phase currents were half of the line currents forgetting that you are adding two different phasors). That error must cancel, because it is the same for the two setups. It makes the Joule heating power wrong but that has very little effect on anything.
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.
@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 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.
Quote from paradigmnoiaBut 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.
As far as the Lugano Report main conclusions go this is fine details, but I'd make some comments:
(1) The alumina - for the frequencies it is translucent, can dissipate either more or less power than expected from its spectral emissivity, depending on what is the temperature and emissivity and effective surface area of the object inside that is radiating. But this uncertainty relates only to part of the spectrum - where it is not translucent its spectral emissivity will correctly determine radiated power.
For the Optris measurement, 7-13u, the alumina is mostly opaque so we do not have this problem. But the temperature read will be that of the alumina surface not any possible inner radiating object.
(2) The Figure 7 evidence tells us (if it is an IR view) that the band emissivity of the alumina is lower than 0.95. But whether 0.9 or 0.8 is not resolved since we have no calibration for the picture colour vs temperature.
QuoteThe 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.
Agreed Tom. But the interesting feature, the X3 difference in ratio between Joule and heater powers, remains.
@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.
Display MoreThe 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.
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)
Not quite, it means that it has low total emissivity. However, at longer wavelengths it remains opaque, and therefore has high spectral emissivity.
So you agree that it has low "hemispherical total emissivity" = integral of (spectral). Good.
Now how can you understand that for the Alumina pipes the Authors obtain exactly the literature value ?
Note that the lower emissivity respect to the dot is evident.
Most materials have similar total and band emissivity
No Mr. Clarke they are two different quantities one being the integral of the other !
One can be represented by only one value the other by a function !
Paradigmoia please read the report look at the photo ! The camera looking at the rods was not the Hi-temp one (that has anyway a low temp range) but a standard Optris Pi-160 in range able to read even the lower temperatures of the whole long pipe. The difference of luminosity of the image is striking !
And the Authors in that case applied correctly the method for measuring the emissivity ( only one number ! not a function ! ) described in every manual.
QuoteGeneral usage of these (correct) formulas eliminates the extremely nonlinear differences in joule heating values between the dummy reactor and the active E-Cat.
No, I can't see that?
The exact equation for effective total wire resistance does not matter. It stays the same between the two setups, and so the ratio of Joule power / heater power should stay the same and equal the ratio of effective resistance to heater resistance.
A X3 difference in this ratio is not posible, so something must give. Variable resistance heater, or change in setup.
Agreed Tom. But the interesting feature, the X3 difference in ratio between Joule and heater powers, remains.
If you do a re-calculation (using the correct formulas) of the joule heating values for the active E-Cat, then there is no more factor X3 difference.
@Tom Paulsen
The corrected Joule heat for the dummy is 7.276 W.
This value was determined by several persons about 18 months ago, in different but converging ways.
Quote from Thomas ClarkeSo: perhaps you now could do me the courtesy of stating whether of not your analysis addresses this issue of the band and total emissivity being very different in alumina at temperatures above 600C?
It would be very helpful for our communication is you could answer this, since it colors all the rest of the analysis?
QuoteNo Mr. Clarke they are two different quantities one being the integral of the other !One can be represented by only one value the other by a function !
I'm afraid you are wrong. I think you are thinking of spectral emissivity.
to keep things simple consider just one temperature. Then:
Spectral emissivity is a function of frequency.
Band emissivity is a scalar: this function multiplied by the bolometer (Optris sensor) sensitivity and integrated. It is an average of the spectral emissivity weighted by the bolometer response curve B(v) which is zero outside of 7-13um.
Band emissivity is:
eb = integral (e(v)*B(v)*P(v) dv / integral B(v)*P(v) dv
(I wish this site let us do latex formulae, as in Mathjax )
eb = band emissivity
e(v) = spectral emissivity (v is optical frequency)
B(v) = Optris bolometer spectral response (v is optical frequency)
P(v) is the Planck function for the given temperature - this weights the average higher where the Planck radiation is higher
Total emissivity is:
etot = integral e(v)*P(v)dv / integral P(v) dv
In this case the weighting comes from the whole Planck curve without the optris sensitivity
QuoteIf you do a re-calculation (using the correct formulas) of the joule heating values for the active E-Cat, then there is no more factor X3 difference.
I still can't see this. I argue as follows. The Joule heating power is not the issue. But the Joule heating and the heater heating are related (in a fixed ratio) by the ratio of the two resistances. There is a constant factor which I agree was incorrectly calculated in the report and changes this ratio because the currents are different. But the constant factor is the same for the two cases dummy and active. Therefore the ratio of Joule heating / heater power for the two cases must be the same.
Pjoule / PHeater = K(Rwire/Rheater)
K is fixed but depends on topology (Wye or Delta).
@randombit0
How do you know which camera was used for the picture?
The difference between an ε of 0.95 and 0.85 is almost 11%.
The difference from background (blue) to maximum brightness (bright yellow) is only about 11 times. So one tenth less bright around the dot seems about right.
WC1:
The single phase current is 19.7 A. The system current is 19.7 A * sqrt(3), because the current of 19.7 A runs not all times through all cables. What we want to calculate is the joule heating, not only for a single line/phase, but for the whole system. This is done by calculating the joule heating for a single line and then multiplying this value by sqrt(3) for the whole system.
WC2:
If you simply split the cabels, you don't get more or less total current. In this case the phase current is divided by two, because both lines have the same resistance. So we can calculate the heating for a single wire with half the current (9.85A). But there are two wires, so we multiply by two and to get the value for the complete system we multiply by sqrt(3).
@Tom Paulsen
The 9.85 A thing is totally wrong. This is where the essence of the Joule heat error occurs.
The C2 cables are conducting to some degree in all phases, in one direction or another, at all times. More than half of the connected C1 cable current directly connected to a C2 pair is being conducted through each of those C2 cables in a delta configuration.
Two connected C2 cables are in series for various parts of the phase cycles, and they are in are parallel probably almost never.
<a href="https://www.lenr-forum.com/forum/index.php/User/1569-Tom-Paulsen/">@Tom Paulsen</a>
The 9.85 A thing is totally wrong. This is where the essence of the Joule heat error occurs
.
The C2 cables are conducting to some degree in all phases, in one direction or another, at all times. <i>More</i> than half of the connected C1 cable…
OK, this is really neither here nor there, and not the source of the X3 error, it affects the value of K in my last post.
The C2 currents are sinusoids separated by 120 degrees. The C1 phase is half way between these. To work out the currents you can use Kirchoff's Law but need to use phasors (complex numbers):
C1 = C2a + C2b
a 1A 60 degree phase shift current is represented by phasor: sqrt(1/3) +/- j*sqrt(2/3)
Suppose C2a and C2b have magnitude 1 so are represented by the above phasors:
C2a+C2b = (sqrt(1/3)+j*sqrt(2/3)) + (sqrt(1/3) - j*sqrt(2/3)) = 2*sqrt(1/3) = C1
Thus C1 is 2*sqrt(1/3) times larger than C2a or C2b
In the DC case (used by the authors) C1 is 2 times larger than C2a or C2b
@Thomas Clarke
Your right, but there is no heater coil current data for the active E-Cat. We don't know the currents exactly. All we know from the report is that they are in total in a range of 40 to 50 A. If you re-calculate the joule heatings (16 values for the active E-Cat) and use these values to estimate the currents, you'll get a nearly linear ratio.
To give an example:
Joule heating corrected for the dummy reactor (486W), given line current 19.7 A, is 3,88 W.
Joule heating for the active E-Cat (815,86W), assuming a line current of 25.5 A, is about 6.51 W.
815,86 / 486 = 1,678
6,51 / 3,88 = 1,677
Joule/Heater
3,88 / (486-3,88) = 3,88 / 482,12 = 0,008
6,51 / (815,86-6,51) = 6,51 / 809,35 = 0,008
