Posts by LDM



    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.


    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.



    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.


    @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 ,

    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.

    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 ,

    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.

    Since P is proportional to T4, a simple average biases the calculated power to a higher value.


    If you project your calculations for alumina, then for temperatures above 400 C, the decrease in emissivity compensates somewhat for the bias for high temperatures.

    Thus also taking emissivity into account brings the temperatures much closer.


    I was discussing the Lugano report, but referred to the "indication of ...." paper to show that Rossi did use two phase operation before. Thus it is possible in Lugano two phase operation could also have been used during the actual runs. That two phase operation was not used at the Lugano dummy run, but instead three phase operation is clear,.

    If two phase operation was used during the actual runs it explains that Rossi was right about the divisions of the currents for that case and also explains in large part that the calculations of the heater coils resistances showed a very large decrease from dummy run to the actual runs.

    Concerning your remark about the spreadsheets of Andrea S, I know them, have them and analyzed them (even found an error in it which he corrected). For my own analysis I prefer using an electronic circuit simulation program which can for all type of wave forms can calculate the correct amount of dissipated energy/power.

    Unfortunately, the problems I listed are almost all demonstrably wrong, not pathosceptic quibbles.


    Rossi nearly blew a gasket over the lack of using the proper delta configuration math, posting a JoNP diatribe on how much idiots we were for pointing that out, and how correct the formula in the report was. But he is still wrong. (The overall impact is relatively insignificant in the report, but using the wrong formula in the report helps obfuscate the math for figuring what was going on electrically). He did not understand that current flows through the C2 cables from other phases even if it does not "enter" or "exit" the respective connected C1 cable, which means the current from the C1 cable cannot be simply divided in two for the attached C1 cable to get a correct answer for the current in the connected C2 cables.


    In the Levi report you refer to, you can see from figure 3 in the Appendix, that the ECAT-HT was operated in two phase mode.
    My hypothesis is that at the Lugano test the ECAT during the dummy run was operated in three phase mode, while at the actual runs it was operated in two phase mode.
    That also explains that Rossi was right about the divisions of the currents (If he was referring to the actual runs). It also explains why people , assuming three phase operation at the actual run, but with real operation in two phase mode calculated the large negative resistance drop of the heater wires.


    And yes, the professors where confused about this too seeing some of the assumptions they made in the report.




    @IH Fanboy LDM


    It is at this very point that we traverse from hard evidence, photographs, plans, and measurements to things that are unseen and perhaps unseeable. Although I understand the attraction of arguing over these phantoms, maybe it is more suitable to do so elsewhere. I suggest that it is more in the spirit of this particular thread to talk only of things that are tangible and can be measured.


    What you call "Hard evidence" has on this forum and elsewhere been interpreted in many different ways. So That evidence seems not to be so hard as you suggest but is only hard for those people who have fixed their opinion. That I brought up those quotes is that maybe, indeed through evidence, we can confirm or reject what is stated.

    As such I believe that it is suitable to do it in this thread, because we are trying to solve things related to the Prominents in this thread.



    The question remains whether a recirculator pump provides such inlet pressure. Dewey recently alluded to a round of emails between Rossi and Bass near the start of the test regarding the recirculator pump. Dewey suggested that the recirculator pump belongs in the same category as the heat exchanger. However, given that there are any emails at all discussing such a recirculator pump at the beginning (at least according to Dewey) suggests to me that it was a serious consideration, and probably put in place.


    From the interview of Mats Lewan with Rossi :


    Rossi explained a couple of things with regard to the heat exchanger.

    1. A ‘circulator’ (positioned in connection to the heat exchanger) was used to stabilize the flow of steam and water through the whole system. Rossi wouldn’t comment on further data of the circulator since he said he was preparing a patent for this device.
    2. This circulator had nothing to do with a pump of the model ‘Grundfos’ that was brought up by the defense’s expert witness Rick Smith who suggested that the Grundfos pump was used to make hot water flow through the system and that no steam was produced. The real use for the Grundfos pump was instead to push the water through a by-pass with a filter about once a week to make it cleaner.

    So at the heat exchanger there was a pump for circulating the flow through the system

    In a comment from "engineer48" to Bruce on ECW he stated that Rossi had told him :


    Engineer48 Bruce__H • a year ago

    Hi Bruce,

    Rossi told me there are 2 parallel condensate circulating system.
    One which delivers 85% of the volume to ALL reactors and then for each reactor, separate topping up pumps that provide the other 15% and slightly vary their flow to maintain the desired water level in each reactor


    If those topping pumps where the Prominents, it means that they where on an average only providing 15% of the total flow. Well within the normal specs of the Prominents.


    Besides that Rossi also stated (As a part of Point 15 in his comments on the Smith report)


    There was not only the pump system to push the water but also the recirculator (I guess he meant the circulator at the heat exchanger)








    So it sounds like the head of pressure to the pump inlets was due to gravity.


    Since both the internal tank and the outside tank are much lower then the higher rows of Prominents, just gravity alone can not feed the Prominents.

    If we had just gravity, those higher Prominents would have had a negative inlet pressure

    The Prominents can maybe overcome this negative pressure by the sucking action as they increase their internal volume during the inlet stroke but I find this an unlikely scenario.


    @IH Fanboy


    People keep bringing up this idea that the internal reservoir for the E-Cat plant is sealed and has a head of pressure on it. But in this case I don't understand how the external reservoir sitting outside the E-Cat plant, feeds the internal reservoir by gravity as described by Penon and Barry West.


    In case the internal tanks have a pressure on it, then that pressure can feed the Prominents with a positive inlet pressure while at the same time due to the hight difference between the external tank and the external tank can feed the internal one. The water pressure is possibly provided by the additional pump on the JM side which AR describes in his notes about the Smith report.

    I thought about asking that .... but decided to limit my questions to cut-and-dried answers.

    If I had contact with a pump DESIGNER then I'd ask that question.


    If the tech answering your question does not know the answer he will probably contact another person within the organization who knows the answer.

    That might be a pump designer.

    Question : Can you please clarify exactly what the manual means by "back pressure" ... is it JUST the pressure at the discharge end (eg a 5 foot head of water = 0.15 bar), or does it depend in some way on the pressure at the suction port?


    Can you also ask them under which exact circumstances the pump can achieve several times the stated feed rate.




    That is correct. The total lift height of the water column has to be paid for- no free rides.


    Has to be paid for in electric energy to operate the pump.


    However I disagree about adding the negative column seize at the inlet with the column height at the outlet (But correct me if I am wrong)

    Why ? Because when the pump is pulling the water in, the outlet valve is closed and the water column at the inlet will not see the column height at the outlet.

    And when pushing the water out of the pump, the inlet valve will be closed and the water column at the outlet will not see the column at the inlet.

    But one of the things that confuses me is the picture (2017 Catalog) of a suggested set-up, which is basically what I have


    Alan,


    Since you are the owner of a Prominent pump you should have the right to contact the technical support and discuss the "confusions" you have with them


    Technical Support Tier 1
     
    Please call 412-787-2484 or email [email protected] and you will be directed to the appropriate technical representative.


    I am totally agreeing with you that on the picture you are referring to the external ports for controlling the stroke rate are not connected.

    Good point that you looked at the connectors on the picture, should have done that myself.

    However I am not totally convinced since on the picture you refer to you can see that besides the black power cables in the cable ducts before the Prominents, there are also grey cables which seems to have no purpose. I wonder if those cables can be connected to the Prominents.

    I am afraid that we can't determine that and Rossi won't tell us.



    "When metering at atmospheric pressure the pump can achieve several times the stated feed rate" ...

    0.2 bar is pretty close to atmospheric.


    The question is under which circumstances it can be achieved. They are not providing additional details, but maybe the following applies.


    As I understand the maximum standard stroke rate programmed into the Prominent is 200.
    In my opinion this means that for the specified back-pressure of 2 bar, there is enough drive force in the solenoid to maintain the maximum specified stroke rate of 200.
    If the back pressure is much lower and less force is needed, higher stroke rates then 200 can possibly be achieved. How can you achieve that higher stroke rate ?
    Possibly by controlling the stroke rate yourself by an external signal pulse signal.
    The Prominent manual states :

    10.5.3 "Contact" operating mode settings

    Alongside the setting menus, which are described in more detail in
    Chapter 10.6 ‘Programmable function settings ("Settings" menu )’
    on page 47, in ‘Contact’ operating mode in the ‘Settings’ menu, the
    ‘Contact’ menu is also available.


    ‘Contact’ operating mode allows you to trigger individual strokes or a
    stroke series.


    You can trigger the strokes via a pulse sent via the "External control" terminal.


    The purpose of this operating mode is to convert the incoming pulses with
    a reduction (fractions) or small step-up into strokes.



    Further on the documentation states :



    The number of strokes per pulse depends on the factor which you input.
    By use of the factor you can multiply incoming pulses by a factor between
    1.01 and 99.99 or reduce them by a factor of 0.01 to 0.99.


    Number of strokes executed = factor x number of
    incoming pulses



    Could this be the solution to the question how the higher flow rate for the Doral plant was achieved? It was stated that flow rates of 72 Liters/Hour at low back pressure could be achieved. Was that maybe with external controlled stroke rates higher then the rate of 200 in the specification ?