Rossi Lugano/early demo's revisited. (technical)


A question for you,
Have you any idea how much heat is leaking away through the wires which connect to the heating element.
For Lugano we have some values, although I think the values should be somewhat lower (Something to address in a future post)
For the MFMP tests I have no idea, but maybe it can not be discarded and should be known in order to calculate the power balance.
Maybe there is some literature about it ?
And at what distance from the device the power dissipation in the wire becomes so low that it can be discarded ?
If there is no data I will have to resort again to the simulator to get a ballpark figure

LDM ,
The heat escaping through conduction by the cables connected to the heater wires should be mostly captured in the Rods. The rods are long enough to go to "infinity" more or less. They were 0.5 m long. Something like 17W was lost due to the big cables just from Joule heat. And then we have the 4 or so cm of heater wire entering the rods. The Rods were a royal pain to work with, the last time I messed with them.

Memorylane time!
a) My thermal analysis of the Lugano system (June 2013)http://lenr.qumbu.com/rossi_hotcat_may2003_spice_130605.php
b) My Spice analysis of the heat exchanger (Dec 2011)http://lenr.qumbu.com/rossi_ecat_oct11_spice.php
I know it's possible, but I never did figure out how to use arbitrary formulae in Spice (not in the version i was using). 
The "late" heat pulse with bulging decay curve fits fairly mundanely with a nested thermal conductive cylinder model that has a moderately insulating material between them, causing thermal lag. Pi helps this along. Lack of thermal steady state between energy pulses also contributes a fair bit.
I don't have good control on the power onoff times relative to the temperature recording (left image), but the T went down a bit after turning power on, and continued going up for a bit after turning power off. There was only about 5 mm between the heater coil and the external thermocouple, and it is a solid mass of Durapot, so not a perfect comparison by any means.
Note that the normal decay curve shape is barely expressed in the left image. I only did the one experiment, but I bet I could fine tune it to match the hot cat version.


Memorylane time!
a) My thermal analysis of the Lugano system (June 2013)http://lenr.qumbu.com/rossi_hotcat_may2003_spice_130605.php
b) My Spice analysis of the heat exchanger (Dec 2011)http://lenr.qumbu.com/rossi_ecat_oct11_spice.php
I know it's possible, but I never did figure out how to use arbitrary formulae in Spice (not in the version i was using).Thanks Allan for posting your Spice analyses.
It's a lot of information to digest, but will try to free some time for it and see what I can lean from it.
Concerning using arbitrary formulae in Spice, it is my understanding that in Spice, being based on nodal analysis, this is not possible. But that is what I learned when Spice was still running on IBM mainframes and at the university I studied we developed our own nodal analysis program for which I wrote the transient analysis part.
Since then some time has passed, so maybe they have developed some methods to circumvent this problem and might it be possible these days.
There is also another method of electronic circuit analysis, called the state variable method, where formula are possible, but that method was not fit for analyzing large networks, and I don't think that method is used anymore .


Concerning using arbitrary formulae in Spice, it is my understanding that in Spice, being based on nodal analysis, this is not possible.
I used LTSpice (free .. with builtin schematic editor), which DOES allow userdefined functions. (eg see https://www.ece.uic.edu/~vahe/…troduction_to_LTspice.pdf page 16).
Quote.FUNC  User Defined Function
Syntax: .func <name>([args]) {<expression>}
Example: .func Pythag(x,y) {sqrt(x*x+y*y)}
I've seen examples with at least one IF in the expression, but can't find it in the documentation.
Edit2 : .func GDS(Vgs,Vds) {IF Vgs<=vth, 0.1F, IF(CTRL(Vds,Vgs)>=0,((W*KP/L)*(VgsvthVds)),(((W*KP)/(2*L))*(Vgsvth)*(vgsvth)*lambda)}I think it's IF condition,true_expression,false_expression ... where IF's can be nested.

THH,
Can you now agree with me, based on my earlier post, that
A. The ratio of the area with fin to the ratio of the bare tube is not sqrt(2) for fins with a 90 degree angle between them
B. The ratio of the area with fins compared to the bare tube is for the Lugano case is larger then 2
Regards, LDM

LDM,
I've been a bit busy with other stuff, still am, and don't want to post properly till I have it sorted out fully.
But, yes, I agree with you. In fact if the reactor is 10mm and the fins are 2.3mm that is an increase (for max diameter at peak) of 46%  I was halving this! to get av of 11.5. So with av of 23%, and the counterintuitive but true fact that the extra area varies with square of fin diameter, we have:
sqrt(2)*1.23^2 as a decent approximation to area, so I'll take 2X + a little.
Why I'm keeping quite still is that this does not correspond to your ray tracing which should therefore give an average Fbg of < 0.5. We are only out by 20% or so, but till I've resolved that I'm reluctant to agree anything for sure...
There are then the overall effects of this on the Lugano calcs, which require some care, and as I say I just don't have the (large) amount of time I need to do this properly yet. I will certainly do it within a shortish time (< 1 month).
THH


LDM ,
There can be no reflection of consequence in regards to heat radiating from one fin to another. It is all effectively absorbed. The alumina simply is not reflective to the bands of radiation that it emits. This would be a violation of Kirchhoff's law.
Edit: this is all
Some thoughts, I might be wrong, if so you can correct me.
Kirchhoffs law is valid under thermal equilibrium, being that the object and it's environment have the same temperature. That also means that all parts of the object must have the same temperature in order to be in thermal equilibrium. In circumstances that the device (as for example a dogbone) has a temperature gradient (Which it has), Kirchhoffs law at least deviates as research has shown.
Also for the dogbone the device is (due to the heater present) at a different temperature then the environment and that makes Kirchoffs law not applicable anymore because there is no thermal equilibrium (The dogbone and the environment have different temperatures)
For fin to fin, where both fins have about the same temperature we might assume thermal equilibrium between the fins and in that case we could possibly use Kirchhoffs law between the fins.
(I don't know if that is formally correct)
Kirchoffs law states that for an object, having temperature T facing an environment of the same temperature T at thermal equilibrium , that then e = a (e the emissivity, a the absorptivity).
But there is also the relationship that r + a + t = 1 (with r the reflectivity, t being the transmissivity).
For Alumina beyond a few mm thickness we can ignore the transmissivity and the equation becomes r + a = 1.
Now if as you state there is no reflectivity in the bands of radiation (r = 0), then the absorptivity in the bands of radiation becomes 1.
Following Kirchhoffs law that e = a then the emissivity in the bands of radiation must also be 1 and we know that this is not true for Alumina.
LDM


LDM ,
To keep the correct version short, the following is my summarized position:
Each time a ray impinges on an opposite fin (or any other surface), the radiant power is absorbed directly proportional to that surface emissivity (which is the absorptivity). The remainder is reflected or transmitted. In the case of alumina, the transmissivity is very limited, to less than 1 mm generally, and so can be ignored in the generalized case. That means the power directly proportional to the reciprocal of the surface emissivity is the reflected portion. If that remaining powerray strikes another surface, it then imparts power again proportional to the absorptivity/emissivity and reflects the again the reciprocal portion, continuously dissipating the original power level until it is completely absorbed or goes to infinity and disappears from the system.
Total emissivity at the correct temperature is correct for the power absorption/reflection calculations.
This is the same as always, and results in no emissivity greater than one, or less than zero paradoxes.

To keep the correct version short, the following is my summarized position:
Each time a ray impinges on an opposite fin (or any other surface), the radiant power is absorbed directly proportional to that surface emissivity (which is the absorptivity).
A part of the fins sees a background at the background temperature and another part sees the opposing fin at another temperature.
That means that the system as a whole is not in thermal equilibrium and as such Kirchhoffs law does not apply (At least that is what I think). In that case emissivity does not have to be equal to absorptivity.
The remainder is reflected or transmitted. In the case of alumina, the transmissivity is very limited, to less than 1 mm generally, and so can be ignored in the generalized case.
Totally agree that the transmissivity can be ignored.
That means the power directly proportional to the reciprocal of the surface emissivity is the reflected portion. If that remaining powerray strikes another surface, it then imparts power again proportional to the absorptivity/emissivity and reflects the again the reciprocal portion, continuously dissipating the original power level until it is completely absorbed or goes to infinity and disappears from the system.
Reciprocal of x equals 1/x
I think your intention was to refer to 1  x
Not all energy is absorbed by the other fin.
Of the energy from a fin directed to the other fin a part will be reflected back to the originating fin, another part will be reflected to the background. That part reflected to the background, dependent on the view factor, will not be absorbed by the other fin anymore.
Thus not all radiation towards the other fin will be absorbed.
What is absorbed an what not is determined by the infinite reflection method.
Total emissivity at the correct temperature is correct for the power absorption/reflection calculations.
Correct,
When the system is not at thermal equilibrium, the power balance needs to be correct.
This is the same as always, and results in no emissivity greater than one, or less than zero paradoxes.
Agreed

I've been a bit busy with other stuff, still am, and don't want to post properly till I have it sorted out fully.
Same for me, figuring out these issues takes time
But, yes, I agree with you. In fact if the reactor is 10mm and the fins are 2.3mm that is an increase (for max diameter at peak) of 46%  I was halving this! to get av of 11.5. So with av of 23%, and the counterintuitive but true fact that the extra area varies with square of fin diameter, we have:
sqrt(2)*1.23^2 as a decent approximation to area, so I'll take 2X + a little.
It is indeed counterintuitive, also for me when I calculated the areas for the first time.
Why I'm keeping quite still is that this does not correspond to your ray tracing which should therefore give an average Fbg of < 0.5. We are only out by 20% or so, but till I've resolved that I'm reluctant to agree anything for sure...
For the 69 fins case, fin height 2.3 mm, fin base 2.9 mm the angle between the fins of 64.46 degree.
For this angle we calculate the view factor for two plates of the same width with a common edge.
The value is 1 sin(64.46 /2) = .467 (view factor fin  fin)
That view factor to background then becomes 1  .467 = .533
However since the fins are bend, the view to the background must be larger.
The conclusion then is that the view factor to the background will always be > .533
As a result a value smaller then .5 can never be obtained.
I don't know what the prerequisites are for the view factor formula given in the paper you referred to. Maybe for example that it is only valid for flat surfaces ?
There are then the overall effects of this on the Lugano calcs, which require some care, and as I say I just don't have the (large) amount of time I need to do this properly yet. I will certainly do it within a shortish time (< 1 month).
Take it easy
LDM


OK, so I still want to look at this properly, and that must wait later.
But:
I expect the calculation with ridges to be comparable with the calculation from a surface area which is the ridges outline multiplied by pi, with emissivity adjusted as per TC paper (but never more than 1.
Which means in this case that the correction for the ridges is roughly (maybe exactly) 1.23 X more power on top of Tc correction.
And that may be compatible with your surface area * Fbg or not. It does now look in right ballpark. if compatible all is as it should be, but need still to flow from this consequences re Lugano results.
THH

Finally found some time again to comment on your last post.
OK, so I still want to look at this properly, and that must wait later.
But:
I expect the calculation with ridges to be comparable with the calculation from a surface area which is the ridges outline multiplied by pi,
Can you explain why you expect this ?
I showed you how to calculate the correct surface area and what the value is.
Thus there is no reason "to expect" another surface area
with emissivity adjusted as per TC paper (but never more than 1.)
I already agreed (now for the second time) that the corrected emissivity value will not be more then one.
However I can not agree with the value of the emissivity adjustment value in TC 's paper, since that value was derived from the view factor for two plates of infinite length and of the same width with a common edge. Calculated with an angle of 90 degree between the plates instead of the angle between the fins and for another geometry.
So we need to recalculate the emissivity correction in TC's paper for the correct view factor of the ridges taking into account the geometry of the ridges. (see note at the end)
Which means in this case that the correction for the ridges is roughly (maybe exactly) 1.23 X more power on top of Tc correction.
I disagree.
The correct view factor gives an other value for the emissivity correction and also another value for the surface area x Fbg . And the total correction for emissivity and surface area will then be larger then 1.23
And that may be compatible with your surface area * Fbg or not. It does now look in right ballpark. if compatible all is as it should be, but need still to flow from this consequences re Lugano results.
IT IS NOT "my surface area * Fbg ".
The factor surface area * Fbg is part of the formula derived from theory. Existing theory, not my theory.
For me the "right ballpark" should be the correct calculated value.That means :
A. The formula right
B. The fin Area right
C. The view factor right
If these three are determined in the right way it becomes only a matter of calculating.
LDM
Note :
Ray tracing as a check the view factor for two plates of infinite length and of the same width with a common edge, with 90 degree between them, I got instead of .293 a value of .307 (5% off)
For the view factor of the ridges , due to the fact that the rays only hit a limited area I expect the error to be less. Nevertheless the ray tracing seems to be on the somewhat conservative side.
My suspicion is that it has to do with a conversion between Polar and Cartesian coordinates and will try to find the reason for this deviation and see if I can correct it.

View factor value correction
In my post of Dec 16th 2017 on this forum thread I stated that my simulation of the view factor between the fins of the Lugano dogbone was somewhat off.
I have tried to correct the error, but came to the conclusion that, while spending a lot of my time on it, the mathematics involved was too complex for a quick fix.
Then last week I found a program called View3D which was specially written for the calculation of view factors. This program was developped at the National Institute of Standards and Technology
(NIST) in the USA. The program is freely available and is one of the fastest and most accurate programs for calulating view factors. As such the code of View3D is also contained in many professional heat transfer calulation programs. The disadvantage of the program is however that the areas to be simulated need to be polygons, eg flat surfaces. This means that the curved surfaces of the Lugano dogbone fins need to be approximated by a number of polygons. With View3D I calculated the view factor between the fins when the fins where made up of 12, 24, 48 and 96 polygons. The results are shown below :
Polygons View factor
12 0.419109
24 0.426402
48 0.428044
96 0.428163
In order to make sure that the calculations are correct I tested View3d for some known view factor configurations and the simulated results where in agreement with theory.
As can be seen the view factor between the fins on the dogbone stabilizes at a value of about .428
This is about 18 % higher then the value of .363 which I got with my own ray tracing and as such has a larger value then I had expected. (Never be too old to discover new insights)
The result also means that the view factor of a fin to the background has a value of 1  . 428 = .572
The found view factor can be used to adjust thermal calculations where Lugano type fins are involved.

Well done. Basically I agree with this. Overall the Lugano guys neglected the VF correction, but more importantly took the reactor diameter as its minimum which underestimates its surface area.
TC included VF  in an approx form  but did not notice the reactor diameter issue.
The reactor diameter issue is more serious than you might think from a 2D approx analysis because the surface is curved.
So overall we have total estimates of power output up by a bit on both dummy and (a bit more) active reactors.

Lugano finned area radiated power correction
Having found the correct view factor between the fins, we can for the Lugano dummy run calculate what the effect is on the radiated thermal energy of the finned area of the dogbone.
The radiated energy for the finned area of the dogbone is proportional to :
Af x e x Fbg x { 1/(1  (1Fbg)(1e))}
For the finned area we have found earlier that :
Af = 0.0263 m^2 (Area of the finned tube)
Fbg = .572 (View factor of the fins to the background)For the emissivity of the dummy run we take the value of .69
The value of Af x e x Fbg x { 1/(1  (1Fbg)(1e))} then becomes 0.011968
The lugano testers did not correct for the view factor and assumed that the radiated energy was proportional to :
At x eThe values are
At = .0125 (Area of the bare tube the Lugano testers used in their calculation)e = .69 (at 450 degree C)
The value of At x e then becomes 0.008625
The ratio of both is 0.011968/0.008625 = 1.387594
Or for the dummy test the radiated power of the finned area of the dogbone is about 38 % higher then what the Lugano testers calculated.
Note however that the radiated power of the finned area of the dogbone is only a part of the total power budget and as a result the percentage power increase for the total device is substantial lower.
