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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.






0 #46 David Roberson 2012-11-20 17:09
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.
0 #45 David Roberson 2012-11-20 02:11
@123Star This is maddening. I have used your forth order curve along with a modified quadratic and can get good fits with several different combinations.

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?
0 #44 123star 2012-11-20 01:45
Last note (I go to sleep)
Your parameters are better than mine (gnuplot's)
with P(T)=.001902*T* T -.813*T +74.106
your RMS = 0.206826
0 #43 123star 2012-11-20 01:35
Quadratic interpolation with gnuplot (the parameters are a bit different from yours)
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
0 #42 123star 2012-11-20 01:29
Ohh I get it, "Calculated power" are just the interpolated values (it fitted TOO nicely with your function, :D )!
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
0 #41 123star 2012-11-20 01:19
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).
0 #40 David Roberson 2012-11-20 00:37
Ok Star, I obtained mine from the viewer. The times can be found in one of my posts below but here they are:

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.
0 #39 123star 2012-11-19 23:39
Can you post me your data set, so that I can check too?
0 #38 123star 2012-11-19 23:37
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 https://docs.google.com/document/pub?id=1e3t4J-x208AIlt1dwQ2Wo2MVgjgnocAWH63ivL5R0oM
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
+1 #37 David Roberson 2012-11-19 23:09
To convert Centigrade to Kelvin one needs to add 273.15. It is so easy to get this value mixed up that I wanted to have it posted for us to see. ;-)

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.

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