Posts by LDM

    If the tips of the fins were only 20 C cooler, then nobody would use them.


    That is a misconception.


    Newton's law states that :


    -----Q = h x A x (Ts - Tamb)


    h being the heat transfer coefficient, A the surface area , Ts the surface temperature and Tamb the ambient temperature.

    Now consider two surfaces of the same amount of area, the first surface at the bottom of the fin, the second at the top.

    If the temperature at the top of the fin is lower then at the bottom of the fin, Qtop will be lower then Qbottom.

    If the temperature at the top of the fin is equal to the temperature at the bottom of the fin then Qtop will be equal to Qbottom.

    In the last case where the temperatures at the bottom and the top are equal Qtop + Qbottom will be larger then when the temperature at the tip is lower then that at the bottom.

    Otherwise stated, the convected heat transfer of the fin becomes better if the temperature drop between bottom of the fin and top of the fin is smaller.

    That is one of the reasons why fins in heat exchangers are made of materials with good thermal conductivity such as aluminum and copper. The good thermal conductivity keeps the temperature difference between top and bottom of the fin as small as possible. The other reason being that heat transfer from the base into the fin is large.


    More important factors for a good conductive heat transfer are a high heat transfer coefficient and a high surface area.


    Radiant heat transfer is the dominant heat transfer mode, over around 750 C. Below that rough temperature, convection does the bulk, and fins are the primary convection improver.


    Agreed that at higher temperatures radiation is dominant for heat transfer.

    At lower temperatures it depends on the way how the heat is exchanged. With larger fins (fin height 3 cm) there is good conductive heat transfer and less good radiated heat transfer due to the fins largely blocking the radiation.

    For the Lugano fins, which have only a small height and where radiation is less blocked and due to the small surface area convective heat transfer is limited. As a result even at the temperatures of about 450 degree C, the radiated heat transfer is still dominant. (See Lugano dummy run data)

    The hot valley vs cooler tip of the ribs is easily replicated. At about 800 C, about 50-80 C difference is typical. Testing the tip temperature is tricky, though. The valleys are much easier to get a thermocouple tightly affixed. A glob of cement to hold the thermocouple skews things significantly when ribs are involved.


    I wonder if the thermocouples are partly the cause of the difference in measured temperature. The connecting wires to the thermocouple tip are good thermal conductors, and are thus conducting thermal energy away from the tip, thereby lowering the tip temperature.

    Since at the bottom of the fin there is more thermal mass then at the top of the fin the temperature drop due to this thermal conductance through the thermocouple leads will be less at the bottom of the fin then at the top. This will probably cause an (unknown) temperature difference in the thermocouple measurements.

    Note also that at the MFMP dogbone retest for a temperature of about 970 degree C, the thermocouple showed around 840 degree C, a 130 degree C difference ! In my opinion also caused by the draining away of heat through the thermocouple wires.


    Convection here, however, is the main rib temperature gradient maker.


    In general in heat exchangers where ribs are involved, convection will indeed be the major source of heat transfer and thus the cause of the temperature gradient on the fin. Fins in heat exchangers have normally a larger ratio between height and base width of the fin then the fins on the dogbone. Due to this radiation is largely blocked and thus convection will indeed become the major source for heat transfer. (Also because in heat exchangers there is often a forced air flow)

    In Lugano the hight/base width ratio is small and thus radiation can play a much larger role then for the larger ribs used in heat exchangers.


    The valleys convect poorly (probably very rarified air there, in the extreme valley V location, due to the high heat and small volume) the tips are increasingly in denser, cooler air towards the tips.


    This sounds logical, however I have not found this confirmed in documents on fin heat transfer.


    The ribs also conduct heat semi-poorly, being alumina, a poor heat conductor in general, so that the highly convecting outer rib area is in competition with conducted heat replenishment. As well as radiated heat transfer to the fins.


    While alumina is not a very good heat conductor, it is also not a very bad one either.

    I have simulated a single Lugano style rib in the finite element thermal simulation program I use, and for a fin base temperature of 750 degree C the difference between top and bottom of the fin is only 20 degree C, indeed indicating that the thermal conductance is not too bad.

    However that simulation does not take into account the possible difference in air flow between bottom and top of the fin you suggest.

    Maybe I will be able to adjust that similation for difference in conducted heat transfer from bottom to top and run it again.

    Paradigmnoia


    Good to have you back in the discussion again !


    So..., in other words, since the ribbed device is actually so effective at radiating power, the effective emissivity when calculating the radiated power based on area, when assuming it is a diffuse cylinder, must be very high (?)


    In essence we have two effects here

    First of all is that you should use in the calculation of radiated energy the real surface area,

    That means the area of the tube with fins, not that of the tube without fins.

    However thermal radiation of a fin is partly blocked by the opposing fin. That makes the area of the fin less effictive and thus you need to calculate with an effective area of Af * Fbg in which Af is the area of the fin and Fbg the view factor to the background.

    The second effect effect is that due to the reflection between the fins the effective emissivity changes (Infinite reflection method). The effective emissivity becomes

    e x { 1/(1 - (1-Fbg)(1-e))}. (see also TC's paper)

    The total effect is Af x e x Fbg x { 1/(1 - (1-Fbg)(1-e))} , which is the formula shown in

    an earlier post.

    This makes the finned area more effective in radiating thermal power then without fins. However less effective then the total area of the fins without view factor correction would suggest and not very high, but surely more effective then without fins.


    Or the device must be cooler than they said it was, since it is so efficiently radiant it could not get so hot

    I have currently no idea if the device was cooler then what the testers reported. However having found the correct view factor is a first step in recalculating the results and see where that ends. After having solved this and other issues it gives me the opportunity to restart my thermal simulations of the dogbone and possibly solve the question between the different results of Lugano, MFMP and the thermal simulations

    However there is also possibly another issue with the calculation of the convected power of the stacked rods , which I want to work on before doing new simulations.


    Another issue I currently am investigating is why, if I remember well, there was such a large temperature difference reported between top and bottom of a fin. I do not know where and by whom that was reported, maybe you know and can point me to the information ?


    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.




    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


    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



    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.


    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.

    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.


    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.



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


    (Afin x Fbg) x {e x{ 1/(1 - (1-Fbg)(1-e))}


    The second part of this formula is e^ = {e x{ 1/(1 - (1-Fbg)(1-e))} and since 1-Fbg = Fff
    This yields e^ = e x { 1/(1 - Fff(1-e))} where Fff 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 Atube * e
    The ratio between both thermal radiation calculations is

    {(Afin x Fbg) / Atube} x {e x{ 1/(1 - Fff(1-e))}/e


    TC used the ratio for his calculation and applied the second term which is { 1/(1 - Fff(1-e))} in his ratio calculation but he did not apply the first term {(Afin x Fbg) / Atube} which is 1.34 but instead assumed, in my opinion wrongfully, that the effective area was Atube 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 Fff + Fbg = 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 Fbg is the fraction of radiation of a surface which is directed to the background and Afin being the area of the fin surface, then Afin x Fbg 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 Afin x Fbg, 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.

    LDM ,

    The work on the view factor for the fins looks great.


    How is the thermal gradient of the fins dealt with?


    The thermal gradient of the fins was not dealt with as you can see from the calculations.

    However i believe that if you take average temperatures over an area, as is done with the Optris, that the results will be about what the formula's give.

    Since I discovered recently that I can put linear piecewise thermal constants in the simulation program, I can possibly see what the simulator gives from bottom of the fin to the top.

    But I have currently not much time to spent on the simulator, so something for the future maybe.

    From my past simulations with fixed thermal constants I did not see a large difference between bottom and top of the fins.


    PS : I redid the transient with other thermal constants in the program, more matched to the higher temperatures. It did only give a minor increase of about 2 seconds in the time constant when using an exponential function.

    I also digitized the plot from the Lugano report and did a Fourier analysis. This yielded not any additional information except that the phase diagram which increased linearly had two small flat area's which I have yet not an explanation for.


    View factor and the influence on thermal radiation of finned areas


    I have worked out the formula's having to be applied to calculate radiated heat from opposed surfaces (eg fins as on the dogbone). These formula's are already known, but it might be interesting to know how they can be derived.
    We know that heat radiated by a heated surfaces has to be radiated to the background in order for that surface to get rid of it.
    The amount of radiated heat of a fin directly radiated to the background is proportional to


    e * Af * Fbg


    e is the emissivity and Af is the area of the fin surface and Fbg the view factor, giving the amount of surface which directly sees the background in case there are opposing surfaces such as an opposing fin which can partly block radiation. If we apply this to the Lugano ECAT we have the following figures :

    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

    This gives that for a dogbone with fins we have to calculate with an equivalent area of

    .637 * .0263 = .0167 M^2, this opposed to the .0125 M^2 without fins

    Thus with fins we are able to dissipate a factor .0167/.0125 = 1.34 more thermal radiation then without fins, or a 34% increase.

    In addition to this direct increase in apparent surface area we also have to take into account the effect of the reflected radiation between the fins.
    The radiation directed towards the other fin is reflected, and this reflected energy by the other fin is with the factor Fbg directed to the background, thereby aiding in getting rid of the thermal radiation, and partly radiated back by the factor (1-Fbg) to the originating fin. Then we get again reflection and this process continues to infinity, albeit with every next reflection involving less radiated energy. For the first reflection we have :

    Af x e (1-Fbg) arrives at next fin

    at the next fin Af x e (1-Fbg)(1-e) is reflected

    of which Af x e (1-Fbg)(1-e)Fbg is to back ground

    and Af x e (1-Fbg)(1-e)(1-Fbg) is directed back to the other fin

    Of the reflected back energy Af x e (1-Fbg)(1-e)(1-Fbg)(1-e) is reflected


    of which Af x e (1-Fbg)(1-e)(1-Fbg)(1-e)Fbg to the background

    and Af x e (1-Fbg)(1-e)(1-Fbg)(1-e)(1-Fbg) to the other fin


    Summing all reflection to the background :


    Af x e (1-Fbg)(1-e)Fbg
    Af x e (1-Fbg)(1-e)(1-Fbg)(1-e)Fbg
    .
    .
    etc


    We can express the n th reflection as : Af x e x Fbg x ((1-Fbg)(1-e))^n


    So in our case the sum of all reflected radiation to the background is :


    Af x e x Fbg x Sum ((1-Fbg)(1-e))^n with n = 1 to infinity


    If we combine direct radiation to the background with reflected radiation to the background this becomes :

    Af x e x Fbg + Af x e x Fbg x Sum ((1-Fbg)(1-e))^n with n = 1 to infinity

    which is equal to

    Af x e x Fbg x Sum ((1-Fbg)(1-e))^n with n = 0 to infinity  

    Which represents all radiation directed to the background
    Since the sum of a^n with n = 0 to infinity is 1/(1-a) ( a < 1 )


    Then Sum ((1-Fbg)(1-e))^n - 1} with n = 0 to infinity equals 1/(1-(1-Fbg)(1-e))

    Thus total energy radiated to the background is

    Af x e x Fbg x { 1/(1 - (1-Fbg)(1-e))}

    We can check the formula with the following two boundary conditions :
    For e = 0 the total radiated energy will be zero
    For e = 1 the total radiated energy will be proportional to Af x e x Fbg which is correct because there is no reflection.
    We can rewrite the above formula as

    Aeq x e^


    With Aeq = Af x Fbg being the equivalent surface area and e^ = e x { 1/(1 - (1-Fbg)(1-e))} being the effective emissivity


    Since both the view factor of radiated heat to the background Fbg and that directed towards the other fin ( Fff ) has to be equal to the total radiated heat, it follows that :


    Fff + Fbg = 1

    And thus since since 1 - Fbg = Fff we rewrite the effective emissivity as

    e^ = e x { 1/(1 - Fff(1-e))} with Fff = 1 - .637 = .363 for the dogbone


    This is the formula which also can be found in TC's paper since he used in his paper the view factor between the fins and not the view factor to the background.
    The increase in emissivity van be expressed as


    e^/e = 1/(1 - Fff(1-e))}

    Taking the Lugano dummy run as an example with e = .690 it follows that the increase in apparent emissivity is a factor 1.127.
    If we apply both the factor of 1.34 due to the larger apparent fin area and the factor 1.127 due to the increase in apparent emissivity the finned area of a dogbone will be able to dissipate a factor 1.34 x 1.127 = 1.51 more thermal radiation then if it did not have fins. (a 51 % procent increase).

    (3) However, if you do argue their deficiency in this way you must similarly discount the apparently positive Ferrara test results.


    Best wishes, THH


    I don't find this much help.

    (1) TC found that the COP disparity between the two high temp tests disappears when you correct the emissivity using the data from the paper and realistic IB emissivity. That is an independent check that the data from the paper is correct. It would be unlikely to happen by chance, and was got with no fudge factors, whereas this speculation on what error could rescue things is exactly trawling through fudge factors till you find one that vaguely fits.


    THH, I am really at this moment not interested if whether the COP was larger then one or just one in the Lugano test. For the moment I leave that to other people to decide.


    (2) If the method in the paper for collecting data is not as described then even the experimental data from that test is useless. It is always possible to discount a negative test by supposing (on no evidence) that very carefully collected data was just wrong, or not a described. And except for the validation from (1) above I'd say the Lugano authors are so deficient in theoretical analysis that you cannot be sure there are no other errors.

    I totally agree with you about that ! That's why I think that I cannot trust what was written in the report and have to be very critical about the contents. Analyzing the data gives many possible inconsistencies.


    (2) If the method in the paper for collecting data is not as described then even the experimental data from that test is useless. It is always possible to discount a negative test by supposing (on no evidence) that very carefully collected data was just wrong, or not a described. And except for the validation from (1) above I'd say the Lugano authors are so deficient in theoretical analysis that you cannot be sure there are no other errors.


    What' s bothering me most at the moment is that I see differences in my thermal simulations on the dogbone (Cop = 1) ,the results of the MFMP tests and the Lugano dummy test. However I have recently come up with some idea's which might explain those differences. To test those idea's I have to do some new FEM simulations, but that does take a lot of my spare time, so it will be a while before they are completed.
    Why I do this ? Because if we can explain those differences we all will have gained additional knowledge about the issues involved. What to do with that knowledge may everybody decide for them self.

    I myself am very interested in the technical aspects as a hobby.


    (3) However, if you do argue their deficiency in this way you must similarly discount the apparently positive Ferrara test results.


    As I said. Currently I am not interested in COP values greater then one. So Ferrara is for me not an option, also because making a model of the HOTCAT is much more difficult then the dogbone. That I sometimes give alternative explanations is to make people aware that there are sometimes other ways how things can be explained because many have fixed their opinions without considering alternatives. Does that make the ECAT work or not? Certainly not, because we don't in my opinion have currently the information to make the decision and as I stated above, I agree with you that we can not be sure about errors (But I would make a bet that there are !) .


    Best wishes, THH


    Best wishes to you also !






    Gotcha... the NASA-USGS radiance calculator can do that, since we don't have the original data. We could also use the Optris software with a suitable hot object. (Crunching the numbers manually is feasible also.)

    Using your new graph? I think that is what you mean. OK...

     

    Yes, that is what I meant

    Using the Lugano Plot 1 ? I can visualize how that could work.


    Indeed


    I'm not sure how that is better than doing it the other way, but it does not seem wrong.


    I am not saying it is a better method, I am saying that it could be the way the Lugano team did it because it largely fits the description in the report.


    Seems to me like it adds more uncertainty because the alumina total emissivity plot (Plot 1) is used several times to extract information for the conversion, and that plot is built from fairly coarse data.


    You could indeed use that plot for extracting information. It can also be done in other ways. for example extracting the relationship between in-band and broadband black body temperatures using the Optris and an Alumina sample and test it at several temperatures afterwards in a lab since you don't need to have the information available during the tests itself.

    Also I think that the team, coming from an academic background, could have had access to the original data from which the plot was made through their contacts.

    Can we choose a convenient surface temperature and run it through both processes and see where they end up?


    I have no preferences. I think that you with your knowledge of the subject will choose a good temperature if you want to test both processes.


    In addition to this, I think the transient analysis step times where too short since the alumina physical properties where set for low temperatures and thermal conductivity and heat capacity are different at high temperatures. So I am going to adapt them and will run a new test tomorrow. (Takes the computer about 3 hours in number crunching).

    The problem with the recursive adjustments in the report is that it is messing with the in band emissivity for the camera by inserting a derivative of the broadband emissivity. If the in band emissivity was used in a plot for the recursive method it might have worked properly.


    If you read my post correctly, the in band emissivity used on the Optris during the measurements is not important to arrive at the correct in-band black body temperature of the Optris.

    You arrive at the in-band black body temperature of the Optris by setting the in-band emissivity on the Optris to one after the measurements.

    Then you do a change from in-band to broad-band black body temperature using the data from the graph..

    After that you do the recursive adjustments using the plot.


    Thus in the method posted above we do not mess with in-band emissivities on the Optris during the measurements.

    The recursive method (please review the Lugano report for details) is a feedback method that reiterates the temperature to get a new emissivity to get a new temperature several times in succession, using Plot 1. Try it with the Optris software if you have it. It shifts the final temperature reported from an original temperature reading based on the slope of the Plot 1 graph, (which is for total emissivity, not the Optris IR spectral sensitivity range).


    I have studied the recursive method described in the Lugano report and agree that if we follow the letter of the text it will indeed give inflated temperatures.
    But that's "IF"
    There is another method to measure temperatures with the Optris which does not need calibration of emissivities or even using the correct emissivity but which will still lead to correct calculated powers. That method also closely follows (in part) what was written down in the Lugano report. Let me explain it in the following steps :

    1. Set after completing the test the emissivity of the Optris back to 1

    Setting back the emissivity on the Optris back to 1 gives you the equivalent in band black body temperature. Now if you had used a wrong emissivity when measuring the temperature, setting back the emissivity to 1 would give you the correct in band black body temperature, where it not for the term (1 – ε)·TambN. But for higher temperatures that term becomes small with respect to the term ε·TobjN. For example discarding this term for a temperature of 800 degree C gives a maximum error in the in band black body temperature of .1 % and at 1000 degree C the error is less then .05%.
    So setting back the emissivity back to 1 will, even when you used the wrong emissivity, still give you the in band black body temperature with great accuracy.

    2. Relationship between in band black body temperature and broad band black body temperature.

    It is possible from data which can be found in literature to determine for Alumina the relationship between the in band black body temperature and the broad band black body temperature. I did those calculations. The result can be found in the graph below. (Note : I did not know a detail of the Optris spec, as such the graph might be slightly of)




    3. Calculate the broad band black body temperature

    Having found the relationship between in band black body temperature and broad band black body temperature we can use the in band black body temperature found under 1 to calculate the broad band black body temperature.
    Having found the broad band black body temperature you will be able to calculate the power.

    4. Determine the temperature and emissivity

    If besides the power, you also want to know the temperature and emsissivity you can using the found black body temperature to iterate through a temperature versus emissivity graph to find the temperature and emissivity. Note that even is this graph is not without errors, the found combination of temperature and emissivity will always represent the correct power.



    Unlucky for all of us who want to find out the truth we do not now if the Lugano team followed the text in the report and got inflated temperatures, followed the method which I presented above or did even use another method.