Posts by LDM

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 - (1-Fbg)(1-e))}

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 - (1-Fbg)(1-e))} 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 e

The 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.

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.

Quote from Paradigmnoia

634-ZO !! .... >95% zirconium oxide? Dewey, I do recall you mentioning Pyro Paint, but just not the specific version.

Not 634-AL ? (Alumina based)

You can see perhaps why I like clarification.

(The spectral emissivity of zirconia is pretty close to that of alumina, and within the Optris camera spectral sensitivity band, the two oxides are nearly the same. They sure would look different by any chemical analysis, though.)

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What can be the advantage of coating the dogbone with zirconium oxide paint ?

It beats me ! But I am open to any suggestion.

Could it be that the construction of the dogone started with a tube where the heating coils where wound on ?

And that this then was coated with the zirconium oxide paint to settle the heating coils and preventing possible shorts due to movement of the heating element due to temperature changes.

And that as a final step the rest of the dogbone was casted on this with an alumina casting ?

Remember that the Lugano testers took a sample of the outside of the dogbone and analyzed it and reported that it was more then 99 % alumina.

So where was the zirconium oxide ?

The above possible construction also explains the possible extension of the heater coil under the end-caps (And thereby the flatter temperature profile compared to the MFMP dogbone)

Maybe Dewey has some more details to share ?

THHuxleynew

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.

THHuxleynew

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 counter-intuitive, 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

Paradigmnoia

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 power-ray 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

Quote from Paradigmnoia

Regarding the oscilloscope used for the demo, is not the reference ground of the probe grounded to Earth ground? Can it just be attached to the circuit without consequence?

Should the probe connections be swapped so cavalierly?

See PDF page 133 (document page 113) of the user manual, plus other notes (PDF page 13, doc page vii)

Quote from Paradigmnoia

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

THHuxleynew

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

Quote from Alan Fletcher

Memory-lane 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).

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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 .

Paradigmnoia

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

Paradigmnoia

The reflected radiation is equal to being directly radiated, I would think. I don't see how it is additive.

Direct radiation from a surface to the background is proportional to the emissivity.

Reflected radiation is proportional to (1 - emissivity)

But the reflected radiation is reflected from an opaque diffuse surface and thus is reflected in all directions. As a result a part of the reflected radiation is directed to the originating area and an other part is reflected to the background. That last part directed to the background aids in getting rid of the thermal energy and thus is additive.

At least that is my reasoning and also the underlying theory for the formula of the emissivity correction factor in TC's paper. His statement : "This correction to emissivity is well understood and deterministic". And with well understood he probably meant well understood in theory (the infinite reflections method).

On a flat surface, the radiation is hemispherical from a diffuse emitter. When ribs are added, the total area of hemispherical emittance does not increase per steradian when the view distance is larger in magnitude than the rib dimensions.

Note that for a point on the rib the hemisphere is perpendicular to the surface of the rib, not perpendicular to the tube if it was in the case the tube had no fins. (At least that is what I think)

And the underlying area of the fin is larger then the area of the tube.

The degree or roughness/diffuseness is slightly increased, which improves emissivity slightly. If the surface was perfectly diffuse, the ribs would make no difference to the roughness, and could make it less.

The degree of roughness and porosity is in my opinion largely determined by how the alumina object is made. I guess that you with all your experiments have discovered this and can tell us more about it. From literature I have seen that porosity is also largely dependent on the curing temperature of the alumina potting compound and as a result can give quite different values in parts of the emissivity spectrum. It seems that in the Optris band the effect of different porosities is quite small, so it seems that the porosity does not have much influence for the Optris temperature measurement.

This differences in emissivity spectrum's can possibly explain a limited part of the difference between the Lugano test and the MFMP tests. From the photographs it looks like there are indeed differences in surface structure.

The tips of the ribs are heated more, due to absorbing some of the radiated heat (when there is a gradient from rib tip to valley) whereas in a flat emitting surface, self absorption of radiation is not possible. This does not improve emissivity much (if at all) because it interferes with the gradient that is required to remove heat from the rib by conduction, increasing the valley temperature, but does increase the radiant power of the tips and valleys slightly, because it is hotter.

Makes sense

THHuxleynew

In fact the Lugano authors take the smallest radius for area (10mm) when a better approximation to the area would be half way between smallest to largest radius. That gives +11.5% on the area, a factor of 1.115. The fin area, measured properly, will be roughly sqrt(2) times this (lower bound) area or 1.55X. So I still can't get your >2X.

Since answering to all your remarks will make this post too long and also because I still have not much time available , I am first addressing the point why the ratio between the area's I calculated is more then 2. (other point will be addressed in later posts, but there could be again days between posts)

To show this we are going to calculate the area of the fins.

We do that by subtracting the areas (without bottom) of two cones.

The figure showing both cones is given below

Instead of doing this for the Lugano dimension I have shown in the figure the dimensions for the case there is a 90 degree angle between the fins.

For one fin of 2.3 mm height and having a 90 degree angle with respect to the other fin, the area of one side of the fin (halve fin) can be derived by subtracting the area of the two cones .

The area of a cone without the base is pi() x Radius of base x length of side

Large cone area : pi() x 12.3 x 17.392 = 672 mmxmm

Small cone area : pi() x 10.0 x 14.142 = 444 mmxmm

Area of halve fin : 672 - 444 = 228 mmxmm

The area under the (halve) 90 degree fin is 2 x pi() x 10 x 2.3 = 144 mmxmm

Ratio in areas : 228/144 = 1.58

As you can see this in not sqrt(2) = 1.41

The difference comes from that the sqrt(2) ratio for the 90 degree case is only be valid for the 2 dimensional case or if the two dimensial figure is linear extruded in the direction perpendicular to the figure.

(As was the case in two plates with common edge which TC used for calculating an approximate view factor). For the case where the fins have a cirular form this does not apply. (But there might be exceptions, I suspect for example for a tube radius zero)

Above I showed you how you can calculate the area of a fin by subtracting the area of two cones. From the Lugano pictures I counted 69 fins (somewhat less then the authors assumed).

This results in a base length of the fin of 2.9 mm. (1.45 mm for a halve fin)

For these dimensions you will now be able to calculate the halve fin area, multiply it by 2 to get the area of a whole fin and then multiply it by 69 to obtain the area of all fins.

The calculated area will then be .0263 M^2 and since the area of the bare tube is .0125 M^2

the ratio is 2.10

This explains where the factor > 2 is coming from.

Quote from THHuxleynew

Appreciating your comments and the discussion

I have at least for two days not the time to digest the information you provided and to comment on it.

But I will be back.

Regards,

LDM

Quote from Paradigmnoia

LDM ,

Basically, I was using for a while an online calculator which seemed to work quite well.

I was basically musing that a bare cylinder with increased area to incorporate the fin view area and a minor emissivity adjustment to incorporate that effect of the fins could be used as a simplified model. Such a model could be compared to the advanced model to see how well it works.

If you besides the view factor also incorporate the emissivity factor for reflected radiation, then I calculated a total factor 1.51.

That would bring the diameter of the equivalent bare tube to about 30 mm.

This factor 1.51 is less then the ratio of the areas of a bare tube compared to a finned tube, which is 0.0263/0.0125 = 2.1

If we assume that the fin efficiency for the convected energy stays around 1, then using a bare tube of 30 mm will dissipate less convected heat then the finned tube.

However since at higher temperatures the convected heat is less then the radiated heat, the error induced by this will probably be not be large.

Quote from Paradigmnoia

So, according to LDM, the equivalent diffuse surface cylinder without fins (at the same length of 0.2 m) would be 2.67 cm in diameter. Or is it better to make it longer at the 2 cm diameter?

.0167/.0125 = 1.34

For the fin height of 2.3 mm the "effective surface area " increases by a factor 1.34 compared to a bare tube.

Thus the new effective surface area becomes 1,34 * 0.0125 = 0.0167 square meter

For a bare tube of 20 cm length you need a tube diameter of 26.57 mm to get the same area of 0.0167 square meter

However since you don't have fins anymore, the added advantage of the reflections disappears.

Increasing the height of the fins will increase the fin area, but will also reduce the view factor of the fins to the background.

And indeed removing the fins and making the cylinder longer will be more effective for the radiated heat.

If you want, you can specify a new fin height to me and I will do a new Monte Carlo Ray tracing to determine the new view factor.

With that new view factor we can then exactly calculate the new equivalent area and also the new emissivity correction

But that will not be today anymore.

Quote from THHuxleynew

Neither TC nor the Lugano authors used Fbg and therefore they were both also correct to use A not Afin.

That they both used A and not Afin * Fbg is not a prove that they where correct.

It can also mean they both where wrong and in my opinion they are.

So:

(1) Please confirm you agree that TC used A instead of Afin*Fbg

If you mean by A the area of the central part of the dogbone without fins,

Then TC used A/A = 1. Since it is 1 it showed not up in his calculation.

(2) shall we calculate what is Fbg*Afin/A? I'm only saying it is 1 on general principles. Perhaps it is not 1.

If you had read my earlier posts you would have seen that Fbg*Afin/A = 1.34

The values where :

Area of the cenral part of the dogbone wihout fins .0125 M^2

Area of the central part of the dogbone with fins .0263 M^2

View factor Fbg (By Monte Carlo Ray tracing) .637

I repeat here a definition of view factor which can be found in the following document :

http://webserver.dmt.upm.es/~i…tion%20View%20factors.pdf

The view factor F12 is the fraction of energy exiting an isothermal, opaque, and diffuse surface 1 (by

emission or reflection), that directly impinges on surface 2 (to be absorbed, reflected, or transmitted).

View factors depend only on geometry.

Thus the view factor depends on geometry but your formula wanting the view factor to be Fbg = A/Afin depends only on how large the surface areas are but does not take into account the geometry aspects.

As an example, if we take the ratio of the surface areas we get per your definition Fbg = 0.0125/.0263 = .475. That is quite different from the view factor that Monte Carlo ray tracing gives, which method does take into account the geometry aspects. Thus your assumption can not be correct.

PS - the ambiguity I referred to is real. Some people call Fbg VF, and some call Fff VF. However TC got it right (as does everyone, it is just you need to use things the right way round). In this case your error is not this ambiguity - but what I have outlined above.

You have not proved that I made an error and that TC was correct. Instead I think I have proved that your assumptions are wrong.

If you want to prove that I am wrong you have to prove that the contents of the document about the view factors and reflections is wrong.

Quote from Paradigmnoia

You can buy them !

See http://uk.farnell.com/multicom…-film-0r-0-25w/dp/9339027

Quote from THHuxleynew

So (and no idea why this good topic is in playground) I think your work here is sound (with one slight issue see below) and it is great to check these things.

I'm not so sure however about the "larger area" argument. In fact, I am sure about it and you are wrong. Here is an extreme example that shows this. Suppose the ridges were like fins, with a 10-1 height/width ratio. The surface area would be roughly 10X greater. Of course, the VF correction would be much larger as well. The problem is how the VF correction works. The way TC used it (check his equation) the VF correction makes the effective emissivity higher, but never more than 1. So it actually has no affect on a radiation from a surface which already has emissivity 1. That corresponds to our experience. A rough surface has a higher emissivity - measuring surface area along the average outline not the microscopic serrations that make it rough.

Now, what I'm not certain is which VF you are using. Maybe there is some definition of VF where the radiation for emissivity 1 decreases with higher (or lower, depending how it is defined) VF in which case we could have radiation proportional to microscopic surface area and for all surfaces (even when e=1 initially) the VF compensates. But, I think TC's definition is easier to use and AFAIK it is consistent with everyone else's - though there is some ambiguity about whether you count it as VF or 1-VF, or something like that.

Perhaps you could test the steps of your argument against the counterexample above and see which assumption fails?

Regards, THH

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The total thermal radiation from the finned area is as derived in my earlier post

(A_fin x F_bg) x {e x{ 1/(1 - (1-F_bg)(1-e))}

The second part of this formula is e^ = {e x{ 1/(1 - (1-F_bg)(1-e))} and since 1-F_bg = F_ff
This yields e^ = e x { 1/(1 - F_ff(1-e))} where F_ff is de view factor between fins.
This is as stated earlier the same formula as Tc used. I agree that e^ can only have values between 0 and 1.
Now according to the write down in the Lugano report the team used A_tube * e
The ratio between both thermal radiation calculations is

{(A_fin x F_bg) / A_tube} x {e x{ 1/(1 - F_ff(1-e))}/e

TC used the ratio for his calculation and applied the second term which is { 1/(1 - F_ff(1-e))} in his ratio calculation but he did not apply the first term {(A_fin x F_bg) / A_tube} which is 1.34 but instead assumed, in my opinion wrongfully, that the effective area was A_tube which can be seen from his comment in the program.

Concerning the view factor, this is the fraction by which (thermal) radiation emitting from an opaque surface is received by another surface. If you have one opposing surface, then the rest of the radiation must be directed to the background, and in that case F_ff + F_bg = 1, otherwise stated the amount of radiation arriving at the other fin and the radiation directed to the background has together be the total radiation. For some geometries view factors can be calculated and can be found in the literature, if they can not be calculated then you try to find a case which you think is about the same (As TC did with using the formula for inclined plates of equal width with a common edge). A more accurate method in case there is no formula (or sometimes measured graphs or graphs derived by numerical integration) is using Monte Carlo ray tracing. You take random points at the surface and take from that point a vector with a random direction and calculate if the vector will hit the other surface or not. Doing this for enough cases will give you an amount of vectors which hit the other fin and the ratio of this number to the total number of vectors gives you the view factor to the other fin. That is a known method for calculating view factors and was the method I used in order to get a more accurate number and as you can see that number is somewhat different from the number TC calculated by using an approximation.

Now since F_bg is the fraction of radiation of a surface which is directed to the background and A_fin being the area of the fin surface, then A_fin x F_bg is the radiation from that surface which is going to the background. And note that when you want to derive the formula for the emissivity correction you need to include that radiation to the background, being e x A_fin x F_bg, otherwise you will not end up with the correct formula for the emissivity correction.

I think if you go over the calculation how the formula's were derived and with the explanation what a view factor is that you will be able to understand and see that there is "no ambiguity about whether you count it as VF or 1-VF , or something like that".

If not feel free to post another question.

Best regards, LDM

In my post "View factor and the influence on thermal radiation of finned areas" of 19 November in this forum thread I explained which equations apply to the calculation of radiated heat from finned areas as on the Lugano dogbone. I did a quick scan of the Lugano report and TC's paper to find out if and how these equations where used in their respective thermal calculations.

If we look at the Lugano report, then for the dummy run the testers did not take into account the effective fin area nor did they take into account the effect of reflected radiation between the fins in their calculations for radiated heat. On one hand this under estimates the amount of power calculated from the radiated heat . On the other hand when doing the calculation with effective fin area and reflection between fins, the calculated total power becomes about 550 Watt, more then the 479 watt applied and not possible for a run without working fuel.

An other document to which these type of calculations apply is the report of Thomas Clarke in which he comments on the report of Levi et al. In that report he compensated for the effects of reflected radiation on emissivity, he did however not compensate for the effective fin area. This can be seen when analyzing his Python code. As a consequence there is an error of 34 % with respect to the amount of radiated heat from the finned area of the dogbone. I did not investigate what the consequences of this omission are for the conclusions of that report.

My conclusion is that using thermal camera's for measuring temperatures should in general not be a problem if conducted properly, but that using the measured temperatures for calculating radiated and convected heat can be the more complex part of the exercise.