Over night we recorded some interesting data that may focus our testing today.
We did 18 small steps to span the range of temperatures that Celani's cell was working in durring the ICCF-17 demo.
The recorded T_mica and T_well look like this.
Nothing particularly stands out here to me. These corresponded to power steps that look like this:
The first two steps are bigger. The next steps were increases in voltage, so as they increased, the power increased more, and shows bigger steps here.
The pressure was a little bit choppy, but it was also on the edge of the noise limit for the pressure sensor, I think. Still, the pressure was definitely dropping after the last step. This seems to be the same kind of drop we saw in the 24 hour run.
The impedance is what I found interesting, though. There was a definite range in which the impedance would decrease with time.
I wanted to zoom into that area. When it is scaled up, it show up very clearly.
After the 3rd through the 7th step, impedance declined. After the res of them, the impedance was flat or inclined after the power step.
When I plot it vs T_mica, I get this funny looking graph. The little curvy bits happen between 202 C and 220 C. With a peak at 208 or so.
When I look at the excess power calculation, we see the at these points correspond to a smaller dip when the power steps up.
That is a tenuous link, but it intrigues me.
I am most intrigued by the wire's impedance dropping over time. I think that there might be something there to learn. I am interested in seeing how low the impedance will go when I go back to 208C. The fact that this is right near the operating conditions of the demo I saw with my own eyes has something to do with this. The fact that the actual operating temperature of the wire may be significantly hotter than the relatively massive mica might mean that there is an actual sweet spot in that temperature range.
Let's watch together.
Comments
I have no idea as to how to allocate the radiation with the other possible heat escapes at this time. We need to find some method that allows us to actually measure the radiation.
This is one of those times when having too many good possibilities prevents us from finding the true function.
I agree, it is time to let this puppy rest for a while and maybe later something will come to us that solves the problem. Where is a good stroke of lightning when you need it?
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f(x) = a*x**2+ b*x +c
a = 0.00190794 +/- 4.171e-05 (2.186%)
b = -0.817727 +/- 0.03134 (3.833%)
c = 75.0295 +/- 5.78 (7.704%)
rms of residuals: 0.25192
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Conclusions?
I redo everything using the 2nd column:
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f(x)= a*5.67E-8*(x)** 4 +c
a = 0.0492754 +/- 0.0007454 (1.513%)
c = -19.8548 +/- 1.19 (5.995%)
RMS = 1.33571
---------
f(x)= a*5.67E-8*(x)** 4 +c + b*x
a = 0.0394318 +/- 0.001125 (2.854%)
b = 0.125228 +/- 0.01412 (11.28%)
c = -54.3717 +/- 3.904 (7.18%)
RMS = 0.328594
------------------
"free to move absolute zero" (parameter b)
f(x)= a*5.67E-8*(x-b) **4 +c
a = 0.0245443 +/- 0.00175 (7.13%)
b = -100.98 +/- 11.53 (11.42%)
c = -34.6504 +/- 1.727 (4.985%)
RMS = 0.26854
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Is "Calculated power" a T_ambient corrected version of P_in?
I used the third column.
However you're right, with these I get:
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f(x)= a*5.67E-8*(x)** 4 +c
a = 0.0492969 +/- 0.0007695 (1.561%)
c = -19.8906 +/- 1.229 (6.178%)
RMS = 1.37898
---------
f(x)= a*5.67E-8*(x)** 4 +c + b*x
a = 0.0392439 +/- 0.001385 (3.529%)
b = 0.127891 +/- 0.01738 (13.59%)
c = -55.1415 +/- 4.805 (8.714%)
RMS = 0.404462
------------------
"free to move absolute zero" (parameter b)
f(x)= a*5.67E-8*(x-b) **4 +c
a = 0.0240849 +/- 0.002048 (8.502%)
b = -104.123 +/- 13.83 (13.29%)
c = -35.1599 +/- 2.074 (5.9%)
RMS = 0.318427
Absolute 0 is way off (not zero).
--------
Temperature K Power_In(Watts) Calculated Power
295.6985 0 0
331.4818 13.6581 13.59494
376.827 37.9368 37.82081
401.499 54.3546 54.28809
427.494 73.7296 74.14389
439.918 84.4947 84.54173
452.173 95.622 95.37347
P(T)=.001902*T*T -.813*T +74.106
The forth order fit was not very good when these temperatures were applied. I will see how my quadratic works with yours.
Can you post me your data set, so that I can check too?
Yes I let "b" free to test if goes to zero by itself (or - 273.15 if in Celsius).
I get b = 19.576 +/- 3.88 which is incompatible with 0.
If I set b=0 I get a higher rms, 0.4 W instead of 0.2.
If I exclude the data points at lower temperatures (keeping only 6 of, it seems that the computed b is compatible with 0. I think this is due to the fact that convection/cond uction contribution is affecting the fit at lower temperatures.
I was looking over the posts and do not see the values that you are using for your constants. What does your fit want the a,b,c, etc. components to be? Sorry if you posted them and I missed finding them.
Could you post your final equation so that I can evaluate it as compared to the one I have derived?
Thanks!
You need to eliminate the 'b' term of your equation if it is to represent the Stefan-Boltzman n function. The Temperature input is absolute in the real world and can not have an offset.
If the offset is required, then your equation has a problem. Give your curve fitting routine another try without that offset and lets evaluate the match.
If it still exhibits the performance you demonstrate, then you may be on to something. Of course a 4th order curve fit should be tighter than a second order one since the 4th order contains within it the second order terms.
My proceedure should be entirely valid as shown. Taking the derivative of the assumed Stefan-Boltzman n ideal equation eliminates the constant term due to ambient radiation. I am confident that you agree with this.
Next, taking the ratios of the derivatives at different temperatures also eliminates any unknown coefficients of the 4 th order term. Do you agree?
Oh sorry... I missed that you already told me your rms of residuals, about 1W.
Suggestion: in your interpolation program, explicitly put the coefficient sigma (5.67E-8) in front of x^4 to help the program, if you don't for example, gnuplot fails to find the fit curve.
I.e. something like f(x)=a*(5.67E-8 )*(x-b)^4+c.
Thank you
Then I took the ratio of each adjacent derivative to eliminate the unknown forth order coefficients.
Next, the ratio of temperatures at which each pair of derivatives were obtained was cubed. I was expecting to see these pairs of numbers match, but was disappointed by the large error.
A chart of the ratios of adjacent derivatives versus temperature ratios was plotted and found to be a sloped straight line instead of the expected cubic. This implied that the function I was seeking was quadratic. For this reason, I performed a second order curve fit and obtained good correlation.
That is how I derived my function. It is a little complicated and the wording suffers from late night writing.
P.S. The worst case error is .4 watts at one temperature, others less than .25 watts. Used 7 measurements for fit.
Ah-hah! That explains everything then :D! Looks like the equations have come together nice, I wonder if we can apply them with this data...
0.04 m^2 :)
I forgot to divide by the area of the cylinder surface (which is around 0.4 m^2, can you check?)
So the emissivity is about 1, comparable to a blackbody!
The formulas are missing the radiation area, of course.
Sorry
No excess heat conclusion is possible yet because wire temperature is far below to the operate temperature 350C. Assume the heat release will double every 10F or 5.56C and will be about 18w at 350C. We would expect power to double (350-260)/5.56 = 16 times. Consequently the Excess power at 260C might be lower than 18*2^-16 = 0.0003w. The measurement error is about +/- 1w. We can remind ourselves to be patient.
I love your calculation work. Since we're measuring T_GlassOut, the calculations should be based on its emissivity and values, I would argue. What power is being lost by the wire itself and not captured in the T_GlassOut... is hard to say. A different glass type will definitely illuminate more of what's going on there. I think assuming an emissivity of 1 makes everything the most conservative for your calculations, so we could do that?
Also, the cell has never been able to run at the full "activation" temperature one would expect, quite yet as far as I know. It's gotten very close, but watching for a transition from negative P_xs to positive is definitely a fingerprint sign we should be on the look out for. I agree with you fully.
This is an indication that there isn't any excess heat, or if there is, there is at any temperature (unlikely).
No?
i m curious how that works.
how many steps up to which power in per wire ?
greetz
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