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 continued to experiment with the curve fitting routines and have come to a conclusion. Either of the two series will do a reasonable job of fitting the data I posted. A combination of the two fits even better with the following equation especially well matched.
P(T)=.015*5.67E-8*(T)**4 + (T-320.0883178) *(T-68.56979702 )*.0011832 .
The quadratic portion translates to:.0011832*(T)**2-.45986028*T+25.969336.
The fact that an entire family of combinations of the forth order term plus the quadratic terms work together allows us to allocate any portion of the escaping power to either path and still obtain an excellent match to the data.
It is now necessary for us to find some way to pin down the proportions in order to have a good total model of the process.
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?
Your parameters are better than mine (gnuplot's)
with P(T)=.001902*T* T -.813*T +74.106
your RMS = 0.206826
-------------
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
-------------
Conclusions?
I redo everything using the 2nd column:
---------------------
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
---------------------------
Is "Calculated power" a T_ambient corrected version of P_in?
I used the third column.
However you're right, with these I get:
---------
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?
Ahah! I wrote 276.15 instead of 273.15 before, oops! :)
I did not choose timestamps by myself, I used the values provided in RunHe2_USA.xls found in docs.google.com/.../...
Page: "calibration points" in that xls file.
Here are the data points (11)
T_GlassOut T_Glassout_Kelvin P_In
24.4696190476 297.6196190476 0.058247619
25.3697857143 298.5197857143 0.3230761905
36.0126952381 309.1626952381 3.6153666667
53.9884380952 327.1384380952 10.457647619
76.9723761905 350.1223761905 20.7743095238
102.3640904762 375.5140904762 34.5922238095
128.4058190476 401.5558190476 51.7301857143
154.2049666667 427.3549666667 72.4241714286
179.096052381 452.246052381 96.2592095238
191.432347619 464.582347619 108.5820809524
202.5709 475.7209 121.7431095238
I would like for us to continue pursuing the heat loss mechanisms until we are confident that the data makes sense.
I have found an excellent curve fit that is quadratic in form which predicts the Power_In at a given T_GlassOut. An additional function of forth order is also available that so far does not match my source data.
I am expecting to see some radiation from the test device that should be of the forth order. It is not clear as to how large the radiation term should be and I would like to see all of the various processes incorporated into one total function.
You and I must be using a different data set for the function generation. When I tried your values with b=0 on my data, the match was quite far off. The final error was a factor of 10 times worse than my simple quadratic results.
I then let my curve fitting program work on your values and it improved a bit, but still not close to the simple second order case. Could you let me know the time stamps from which you derived your data for the fit? My time stamps were posted earlier.
I updated my equation slightly and it now performs better than before.
One point, if your intent is to prove that Stefan-Boltzman n is responsible for the heat loss then you have no choice but to eliminate the 'b' value. I suggest that you add other terms for convection.
Setting b=0 I get
data set:
RunHe2_USA (in Kelvin, 11 data points)
function:
f(x)= a*5.67E-8*(x)** 4 +c
rms of residuals: 0.421663
(before, setting "b" free, it was 0.211224)
Final set of parameters
a = 0.0495287 +/- 0.0001444 (0.2916%)
c = -21.6839 +/- 0.238 (1.098%)
correlation matrix of the fit parameters:
a c
a 1.000
c -0.845 1.000
data set:
RunHe2_USA (in Kelvin, 11 data points)
(note: by mistake I added 276.16 instead of 276.15 to get Kelvins)
function:
f(x)= a*(5.67E-8)*(x- b)**4 +c
(note: ** means "power" in gnuplot)
Final set of parameters
a = 0.0428829 +/- 0.001199 (2.795%)
b = -19.576 +/- 3.88 (19.82%)
c = -24.6493 +/- 0.6029 (2.446%)
correlation matrix of the fit parameters:
a b c
a 1.000
b 0.999 1.000
c 0.969 0.979 1.000
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.
Of course you can interpolate with a second order polynomial ax^2 + bx + c.
If you take the Taylor series you can approximate
x^4 around "a" is x^4 =~ a^4 + 4*a^3*(x-a) + 6*a^2*(x - a)^2 + 4a*(x-a)^3 + .... no wonder a second order polynomial fits good too. How is the rms of your residual btw? Mine is about 0.2/ W.
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.
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