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

  • 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.

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


    .0167/.0125 = 1.34


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

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

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

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


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

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


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

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


    But that will not be today anymore.

  • LDM ,

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

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

  • 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.



    Both A/Afin and Fbg depend on geometry. So let us look at this a bit more closely. I'll not do a first principles calculations here where I can refer explicitly to one done by someone else (which is not behind a paywall). Wen and Mudawar have such a calculation, summarising the work of Abarov on the emissivity of gray-diffuse rough surfaces.


    pp4284 of their paper (snip above) calculates the equivalent surface emissivity for a V-shaped "roughness feature". This part of the calculation assumes feature dimensions are large enough for this to be optical.


    In the above equations alpha is the surface absorbity which for a gray body is equal to the emissivity (I imagine you assume that for the alumina).


    These authors note that for such a V-shaped feature (regardless of exact geometry) Fr,r = 1 - As/Ar. This equation then leads to their result (9) which you can see makes my point, because of the factor Ar/As relating er to es.


    Let us now return to your write-up. You say that Fbg(Afin/A) = 1.34. That seems unlikely to me, and it is the only place I can see that we disagree (except that you also do not accept the general point that emissivity from a diffuse surface can never be higher than 1 - your calculation requires this).


    You say:


    • 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


    First for the ray-traced VF. TC assumed 90 degree exact V-shaped corrugations (as did the Lugano authors). While this is obviously not exactly correct it allows analytical calculation of Fbg as it does not differ much from your VF calculation above and gives Fbg = sqrt(2)/2=0.707. We could revisit the discrepancy if you like, since I'm not certain it should be smaller than ideal (round edges etc) but I'm not myself concerned with this at the moment.


    Where I disagree is how you obtain Afin/A from your statement above. Afin/A is clearly a geometric problem relating to the way the fins increase the area. For the (idealised, flat) problem considered by TC (and also the Lugano authors, though they did not do the VF bit themselves) we obviously have Afin/A = sqrt(2) and the product is 1.


    Thus Fbg*(Afin/A) in the case of ideal 90 degree ridges on a flat surface is indeed 1.


    For the Lugano reactor I can't see Afin/A being that different.


    So this area difference of more than X2 is where we disagree. That is clearly incorrect for any proper extrapolation of the flat surface figures. It comes from comparing apples with pears. Maybe it comes from taking the fin bottom is the "without fin" diameter, so that the fins make the average diameter larger? That would skew the figures, but I'd not expect it to do so that much.


    The reactor radius is given is 10mm, the ridge height as 2.3mm. So comparing these we get a 20% discrepancy between measuring (cylindrical) area from top or bottom of the fins.


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


    I agree with you, TC followed the authors here and did not correct the reactor body area. He should have done so, because the +11% is significant. So I'll allow you a +11% error, not from the VF calculation (all correct) but from the fact that TC took the reactor measurements from the Lugano authors and they underestimated the reactor body diameter by 11% as here. That corresponds to a similar underestimate of calculated radiant power.


    Bottom line: thanks for this reexamination. I agree (I think) that TC should have added a correction for the reactor diameter since the Lugano authors got this wrong. It was careless of TC not to do this. That amounts to +11% on the power budget for the reactor body, a bit less (8%?) overall. Significant, but still at lot smaller than the intrinsic errors. Note also this correction does not change TC's most compelling (in terms of validation) result, that the claimed difference in COP between the two HT tests vanishes when things are recalculated.

  • LDM ,

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

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


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

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

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

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

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

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


    On a flat surface, the radiation is hemispherical from a diffuse emitter. When ribs are added, the total area of hemispherical emittance does not increase per steradian when the view distance is larger in magnitude than the rib dimensions. The degree or roughness/diffusiveness is slightly increased, which improves emissivity slightly. If the surface was perfectly diffuse, the ribs would make no difference to the roughness, and could make it less.


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


    The benefit of the ribs is primarily given to improvement of convection rates, where the extra surface area will have a relatively large effect. (But is still greatly inferior to radiation above around 650 C).

  • THHuxleynew


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

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


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

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

    The figure showing both cones is given below






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

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

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

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

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

    Area of halve fin : 672 - 444 = 228 mmxmm


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

    Ratio in areas : 228/144 = 1.58

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


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

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


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

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

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

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

    the ratio is 2.10


    This explains where the factor > 2 is coming from.


  • Paradigmnoia



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


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

    Reflected radiation is proportional to (1 - emissivity)

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

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


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


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

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


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


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


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


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


    Makes sense