Revisiting the power calculation in the Lugano report

  • Dewey Weaver ,

    Can you confirm whether the Lugano device was painted with Aremco Pyro-Paint 634-AL?

    (or identity for me the correct paint)


    I would like to test its emissivity when painted onto my Durapot slab.


    Is the test report done on the emissivity performed for IH ever going to made public?


    Conditions of dropping the lawsuits perhaps prevent these disclosures. Can you comment on that?

  • Running around 300 C, the spectral emissivity between 8 and 14 microns (IR thermometer sensitivity band) of Durapot 810 seems to be about 0.9 . It will be interesting to see if this changes much as it heats up.

    At least the new device isn't as jumpy when I turn up the input, and the temperature is much more stable when I move around (moving the air around) due to the increased mass.

  • LDM ,

    Using your model, do you have the ability to work out the time required to reach equilibrium temperatures relating to various data points in the Lugano report, based on nominal input power?

    Such as reaching 450 C from 20 C, 1260 C to 1400 C, etc. ?

    It is obvious that LT did those tests :-)


    The finite element thermal simulation program I am using has the capability to do transient analyses. Never played with it and I don't know what it can step.

    However the program. as other thermal simulations programs has shortcomings.

    You can only set a fixed emissivity and thermal transfer coefficient to a surface. So what I did was dividing the dogbone in several sections, run the program, read out sections temperatures and based on those temperatures adjust the emissivity, the reflectivity and the thermal transfer coefficient. Then run the program again. I continued those iterations until the temperature differences of the sections between two runs was not more then one degree C. That is also one of the reasons why those simulations took so much of my time (I did it for the same heater powers as the MFMP)

    Now if you do a transient, you will not be able to adjust the emissivity, thermal transfer coefficient and reflectivity in between and you will end up at the wrong temperature. But for fixed values we could adjust the starting heater power and final heater power such that the begin temperature and end temperature are about the ones you would like, do a transient ( if it indeed can be done) and measure the time to settle at the new value.

    I don't know why you want to know those settling times, but as outlined in my posts I find currently no time to do new tests. And if I have time again I prefer to first simulate the Lugano dummy run with an extended heater and compare the results of such a test with the data in the Lugano report. It will show us if the reported Lugano temperatures for the dummy run where correct or not so that we can conclude if wrong emissivities where used on the Optris for the dummy test or not.


    That said. I have worked in the past for a company (both in Europe and the USA) that designed and manufactured diffusion ovens for the semiconductor industry. Being part of the design team I have seen the FEM thermal simulations which our research lab did on new designs. While the simulated temperatures where in general not exact, they where close to the temperatures measured in practice. As such we could base our design decisions on it. Since the difference between the MFMP test and my FEM simulation are large I must conclude, based on my experience that the MFMP thermal dogbone test has been wrong. Thus in my opinion we can not use that report as a prove that in Lugano wrong emissivities on the Optris where used until the MFMP is redoing their test as they promised and their new results are known. And I realize that the results of a new MFMP test can show that my simulations where flawed instead of the earlier MFMP test, because we all can make errors. But there is nothing wrong with that, as long as they get corrected and we learn from it.

  • LDM ,

    The heating rate for the object, combined with the thermal mass, if properly characterized, should clearly show that the range of input power that is most appropriate to reach one steady state temperature to another. If the steady state temperatures on are grossly exaggerated, the heating rate will probably be very different from that which was reported.


    Purely for example, imagine a dummy unit that takes a very long time to reach steady state, but the input power and temperatures are reported as though they were at steady state. This could result in grossly underestimating the possible output power possible when steady state is actually achieved.

    In case of Lugano, the "700 W" jump in output power with the 100 W increase of input power will still require a certain period of time to heat an object of a certain mass, etc. to the new temperature. If the "real" increase was only 100 W output matching the 100 W input increase, the difference in the heating time period will be very obvious, and only one of the two versions will closely reflect a real situation with a real object.

  • LDM ,

    The heating rate for the object, combined with the thermal mass, if properly characterized, should clearly show that the range of input power that is most appropriate to reach one steady state temperature to another. If the steady state temperatures on are grossly exaggerated, the heating rate will probably be very different from that which was reported.


    Purely for example, imagine a dummy unit that takes a very long time to reach steady state, but the input power and temperatures are reported as though they were at steady state. This could result in grossly underestimating the possible output power possible when steady state is actually achieved.

    In case of Lugano, the "700 W" jump in output power with the 100 W increase of input power will still require a certain period of time to heat an object of a certain mass, etc. to the new temperature. If the "real" increase was only 100 W output matching the 100 W input increase, the difference in the heating time period will be very obvious, and only one of the two versions will closely reflect a real situation with a real object.


    I understand these issues since I had to deal with them when developing control algorithms and PID constants for the diffusion ovens we worked on.

    However I was wondering where you would compare these time constants with. So I looked again at the Lugano report and must admit that I never looked in detail to Plot 5 which gives the step response when the heater power was increased from about 800 Watt to about 915 Watt. Also I took a brief look at the options of the thermal simulation program and think that if I postpone an activity planned for the weekend, I might be able to do a transient analysis, stepping the power from 0 to 800 Watt and then after stabilizing to 915 Watt.

    The emissivity and heat transfer settings will be those of for the latest simulated MFMP test (894.94 Watt). So the reported temperatures will not be correct, but at least you will see the time constants
    So my question is if this will be useful for you ? If so I will give it a try, but can not guarantee that I will succeed.in doing the transient analysis since I have not yet experience with that option of the program.

  • LDM ,

    I am merely curious (always). By all means, don't spend tons of your time if you are not interested in it yourself, or if it is very onerous. I was hoping that it was relatively easy, since you have a functioning model. Years have gone by already, so there is no rush...


    Below are some images of last night's curiosity test...

    I wonder if using the mean temperature (plot 3, Indication of, on RH side of Image 2) for the whole E-Cat affects the shape of the temperature curves?

  • LDM ,

    I am merely curious (always). By all means, don't spend tons of your time if you are not interested in it yourself, or if it is very onerous. I was hoping that it was relatively easy, since you have a functioning model. Years have gone by already, so there is no rush...


    Thanks for your concern about spending my time.

    From a quick look at the program I came however to the conclusion that this would not take much time if I used the settings of the last test. Maybe the program does a wile to do all the steps, but I don't have to attend. And I am curious also, because looking at plot 5 of the Lugano report I think there are two major time constants in that plot. One shorter term an one long term.

    If I had time I would digitize that plot and do a Fourier analyses to see what that gives but that takes too much time at the moment (anybody else who would like do this ? ) . But since from the outcome of the simulation we maybe can conclude if there was a secondary, non resistive heating effect it would be nice to know. So I am going forward anyway.

    If I succeed I will post the results.

  • @Paradigmnoia 

    I had a discussion with Bob Greenyer on ECW about the FEM simulation.
    He disagrees with the results and I then asked him about the promise to redo the MFMP dogbone thermal test. It turned out that this test was already done and he provided the links to the available data. See :

    http://e-catworld.com/2017/11/…ition/#comment-3600322673

    I was not aware that this test had already be done and his answer was :


    " I was meant to widely publicise them but, you might understand, got a little sidetracked."

    Taking a quick look at the results it turns out that the results of the MFMP measurements are about the same as for their first test. I still do not understand while there is such a large difference between my FEM analysis and also my radiated and convected energy calculations which did not match the heater power settings . However the data the MFMP now supplied is much more then for their earlier test and there are Optris ravi files available (the largest almost 600 MB). I am curious if their temperature profiles where more symmetric this time such that I have more confidence in doing recalculations.

    After a few initial trials to find the optimum settings for presenting the results the transient analysis is now running. Hope to post the results in a few hours


  • LDM ,

    Thanks for doing that. Still on coffee #1, so will be in a position to evaluate your work in a short while...

    I have downloaded all of the MFMP Ravi files a long time ago. (I think) I will check that link you provided and see if there are any I don't have there.


    Alan G provided a link here to the recent data when I asked about the first Dogbone test .ravi data (which was lost when the Optris was damaged).

  • LDM ,

    Your plot estimates about 20 seconds to go from around 720 C to 760 C (?)

    The Lugano report shows about 400 seconds to go from 1295 to 1400 C.

    (The Lugano time period seems rather long).

    Clarke estimated the Lugano temperatures to be 705 C rising to 775 C, rather than 1295 C to 1400 C. I thought those two re-calculated temperatures of Clarke should be slightly higher, but only by around 30 C, probably not enough to have a significant effect on the time period.


    Here is my estimate for one of the three Lugano device coil wires. (It can be fine tuned). The heat flux per coil can be worked out/adjusted, per input wattage, near the bottom (where the little coloured flame is).

    The long time period for the Lugano coil seems consistent with the weak temperature output of the coil at low wattages relative to its wire size (less than 1000 W, here, each coil providing 1/3 of the total input). Inputting 35 mW/mm2 into the adjustable field makes for 297 W per coil in this example. (In comparison, my Durapot slabs are at the overheat range at around 120 mW/mm2, using a much smaller diameter 24 AWG wire. I should figure out what 35 mW/mm2 works out to for input W for the present slab and see what that heating rate looks like).


    15 AWG Lugano Coil - one of three



  • Your plot estimates about 20 seconds to go from around 720 C to 760 C (?)


    I get 22 seconds from the exported data


    The Lugano report shows about 400 seconds to go from 1295 to 1400 C.

    (The Lugano time period seems rather long).

    Clarke estimated the Lugano temperatures to be 705 C rising to 775 C, rather than 1295 C. I thought those two temperatures to be slightly higher, but only by around 30 C, probably not enough to have a significant effect on the time period.


    The exported simulation data goes from 720 C to 763 C, So the starting temperature from the simulation is higher, but the end temperature is lower. This makes the estimated delta from TC 70 degree C and the delta from the simulation 43 degree C.
    But TC used for his emissivity correction the view factor of two endless plates under 90 degree. That is not the situation on the dogbone, where the angle is less end the view to the sides is less restricted because the fins are bending away due to their circular form. The smaller angle increases the view factor between the fins, the other decreases the view factor between the fins and the last one wins as the Monte Carlo ray tracing showed. As a result we can expect somewhat lower temperature then TC calculated.


    Here is my estimate for one of the three Lugano device coil wires. (It can be fine tuned). The heat flux per coil can be worked out/adjusted, per input wattage, near the bottom (where the little coloured flame is).

    The long time period for the Lugano coil seems consistent with the weak temperature output of the coil at low wattages relative to its wire size (less than 1000 W, here, each coil providing 1/3 of the total input). Inputting 35 mW/mm2 into the adjustable field makes for 297 W per coil in this example. (In comparison, my Durapot slabs are at the overheat range at around 120 mW/mm2, using a much smaller diameter 24 AWG wire. I should figure out what 35 mW/mm2 works out to for input W for the present slab and see what that heating rate looks like).


    I still think that possibly two major time constants are involved in the Lugano plot.
    I succeeded in digitizing that plot and will subtract the response of the simulation from that plot. Maybe the difference can tell us more. If that is the case then I will report back.


  • LDM , for what is worth, I just checked my slab results from the first test of the 6x7 cm unit.

    At the input of 207 W, the 35 mW/mm2 is obtained. It happens that I had a 208 W input step, at 61.02 V true RMS, with a peak internal T of 722.5 C, and external T of 631.9 C at steady state. So I can easily dial up 61 V and see how long it takes a small slab to heat up to steady state. Not really comparable to Lugano, but possibly informative, grossly.


    I re-calculated the temperatures for Lugano using radiant power matching to arrive at the equivalent temperatures required to reach the same radiant power at different emissivities per steradian, which bypasses some complex calculations. The Optris software agrees very closely with the results of that method.


    Where a complication is introduced into the Lugano figures is the use of the recursive emissivity method applied to the camera emissivity using the Plot 1 values, so that the original emissivity is not directly calculable by any method (if at all) and therefore the original temperature that was detected at some "original" emissivity setting, is in doubt. So we can calculate alternate temperatures based on alternate emissivities at some constant radiant power level, perhaps even perfectly, but the original temperature-emissivity measurement has been obfuscated by the manipulation through the recursive method, and therefore the actual measured temperature-radiant power level detected by the Optris is uncertain. Perhaps this is why you get lower temperatures from the model.


    It should also be remembered that the temperatures reported are composites of those of the respective measurement areas for the main tube and and the caps. This introduces yet more uncertainty. Interestingly, this means that some locations would be hotter than the reported 1410 C maximum, in turn meaning that the coil temperatures would be even hotter than some already very high temperature estimates, if the reported values were taken at face value. (Regardless of whether there was reaction heating or simple Joule heating, or some combination of the two).

  • Where a complication is introduced into the Lugano figures is the use of the recursive emissivity method applied to the camera emissivity using the Plot 1 values, so that the original emissivity is not directly calculable by any method (if at all) and therefore the original temperature that was detected at some "original" emissivity setting, is in doubt. So we can calculate alternate temperatures based on alternate emissivities at some constant radiant power level, perhaps even perfectly, but the original temperature-emissivity measurement has been obfuscated by the manipulation through the recursive method, and therefore the actual measured temperature-radiant power level detected by the Optris is uncertain. Perhaps this is why you get lower temperatures from the model.


    It is my understanding that the broadband temperature/emissivity curve in the Lugano report (are we talking about that curve , Plot 1 ) was from literature, not from using the recursive method.

    For temperatures higher then a few hundred degrees that curve matches closely emissivity data supplied by NASA. In my model I am not using any other data from Lugano, except that curve, which was also used by the MFMP for their dogbone test. And to match my reporting with that of the MFMP I used the same curve.
    All other material constants where from litterature.

    For a static situation the power from the heating element must be dissipated to the outside, and that is determined by emissivity on the outside, thermal exchange coefficient, temperatures and area. Those are determining the temperature at the surface and that is what the FEM simulation program is using, not any Lugano data.

    The FEM simulation program can also calculate back the radiated and convected power by selecting the surfaces and if I do that, that power is in agreement with the power assigned to the heater.

    So I see not how Lugano can in any way have influence on the FEM thermal simulations. But maybe I did not understand the meaning of your statement above

    It should also be remembered that the temperatures reported are composites of those of the respective measurement areas for the main tube and and the caps. This introduces yet more uncertainty. Interestingly, this means that some locations would be hotter than the reported 1410 C maximum, in turn meaning that the coil temperatures would be even hotter than some already very high temperature estimates, if the reported values were taken at face value. (Regardless of whether there was reaction heating or simple Joule heating, or some combination of the two).


    I have stated earlier on ECW that based on calculations I did on the data of the Lugano report, that the reported temperatures in the second column of table 7 of the Lugano report might have been reported in error in that those temperatures are reported in Kelvin instead of Celcius. In that case the Kantal wire will withstand the temperatures and calulated radiated and convected power are in agreement with those reported.

    After having completed further analyzing the transient plot I intend to calculate the radiated and convected power from an x axis temperature curve from ravi data supplied by the MFMP. It will tell us if that power matches their reported heater power.


    What I also wonder about is if different types of casting materials where used for the MFMP and the Lugano device with different broad band emissivity curves. The MFMP said they used the same, but I don´t know how they knew what was used for the Lugano device. If you see the differences in gray-white levels between the two I wonder if it was the same material.

    I tried to find a broadband emissivity curve for Durapot 810 to compare it with standard alumina, bit could not find it. I understand that you are using Durapot so wonder if you have such a curve from the supplier.

  • LDM ,

    I have only one thermal conductivity number for Durapot 810, and have no idea what temperature is is valid for. It is much lower than is typically reported for extruded high density alumina, which has a large decrease from room temperature to ~ 1000 C, typically about 20 W/mK dropping to 10 W/mK in that range. The MFMP used a 99% alumina based casting material, the brand is noted in the construction page. Durapot 810 has less alumina, and unknown other components. Based on dark stains made when heating, I think some manganese is in there (forming pyrolusite?) . I will try some dilute HCl and see if there is a carbonate component. Likely some polymer flow modifier (burned off eventually, based on smell during the first strong heating), some defloculant, and maybe some phosphate. If I send some out for analysis, it will not be soon.


    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 don't think K and degrees C were conflated.


    Edit: I mis-read the beginning of your post. I have only just done the emissivity test of Durapot 810, and it is only applicable to the 8 to 14 micron band. It was only one test, with one pyrometer, but over 10 temperature data points. It very slowly dropped from 0.9 to 0.87, in that band, from 300 C increasing to the maximum external temperature tested of 990 C. (0.88 at 500 C). My intuition (and assisted by many examinations of various other refractory materials) is that the broad band emissivity is similar to alumina, but perhaps slightly higher in the shortwave, so that a single total emissivity figure would be only slightly higher than that for alumina. But that is a guess, really. If the Lugano reactor was painted with alumina paint, like Aremco Pyro-Paint 634-AL then the alumina total emissivity should be about right.

  • 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.



  • 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.


    (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.


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


    Best wishes, THH

  • (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 !





  • 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).

  • 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.

  • LDM ,

    Agreed that the Optris average T matching should do for the most part.


    The bottom to top of fin T difference is typically 50 to 100 C , temperature dependent. The difference will have more influence at lower temperatures (roughly below 600 C) where convection has a higher portion of heat transfer.

  • 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.


  • 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



  • 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

  • Good catch, Eric.


    Just pondering the view factor emissivity bit, and thinking that as far as the alumina is concerned, for the most part, it is sort of like blackbody to itself. In other words, the alumina absorbs and emits the IR in the same bands, and therefore it should act as though it is a perfect absorber of its own emission, and if it is already as hot or hotter, it will re-emit that absorbed radiation instantly, so that it looks like reflection. So in this way, in an ideal case, all rays in the hemisphere equal to the origin position and pointing toward the valley of a rib from a given point in the opposite visible rib are effectively reflected, while those that point away from the valley can be absorbed and add heat to the rib, or miss a rib altogether and continue to infinity.



  • LDM. I have a habit of sanity checking first and then working through maths afterwards, with the sanity check in mind so I can make sure things make sense.


    In this case, since you have not taken my point, I'll do both.


    Sanity check:

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


    If e = 1 this reduces to Afin X Fbg. This checks as long as A = Afin * Fbg, which it probably does. The point is that if you correct for Afin instaed of A you also need to correct for Fbg, which cancels out the increase from Afin.


    Otherwise the sanity check fails because my example (v high Afin) would have some weird emissivity.


    Your point is IMHO ignoring the fact that neither TC nor the Lugano authors used Fbg and therefore they were both also correct to use A not Afin.


    TC did not use Fbg nor Afin. He, like the Lugano authors, used A. Which I contend is correct but will explicitly prove in a moment if needed (we need to show that for this geometry Fbg = A/Afin).


    Otherwise we still have the 1/(1-(1-Fbg)(1-e)) correction. This is quite small, and vanishes for e=1. It is however significant for lower e. This is what TC noted (and I believe included). It is not very large but still worth including.


    So:

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

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


    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.


  • 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.