Revisiting the power calculation in the Lugano report

  • 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

  • Paradigmnoia


    A question for you,


    Have you any idea how much heat is leaking away through the wires which connect to the heating element.

    For Lugano we have some values, although I think the values should be somewhat lower (Something to address in a future post)

    For the MFMP tests I have no idea, but maybe it can not be discarded and should be known in order to calculate the power balance.

    Maybe there is some literature about it ?

    And at what distance from the device the power dissipation in the wire becomes so low that it can be discarded ?



    If there is no data I will have to resort again to the simulator to get a ballpark figure

  • LDM ,

    The heat escaping through conduction by the cables connected to the heater wires should be mostly captured in the Rods. The rods are long enough to go to "infinity" more or less. They were 0.5 m long. Something like 17W was lost due to the big cables just from Joule heat. And then we have the 4 or so cm of heater wire entering the rods. The Rods were a royal pain to work with, the last time I messed with them.

  • Alan Fletcher ,

    The "late" heat pulse with bulging decay curve fits fairly mundanely with a nested thermal conductive cylinder model that has a moderately insulating material between them, causing thermal lag. Pi helps this along. Lack of thermal steady state between energy pulses also contributes a fair bit.


    I don't have good control on the power on-off times relative to the temperature recording (left image), but the T went down a bit after turning power on, and continued going up for a bit after turning power off. There was only about 5 mm between the heater coil and the external thermocouple, and it is a solid mass of Durapot, so not a perfect comparison by any means.

    Note that the normal decay curve shape is barely expressed in the left image. I only did the one experiment, but I bet I could fine tune it to match the hot cat version.


    3418-pulse-test-1-zoom-to-33-section-jpg


  • Thanks Allan for posting your Spice analyses.


    It's a lot of information to digest, but will try to free some time for it and see what I can lean from it.


    Concerning using arbitrary formulae in Spice, it is my understanding that in Spice, being based on nodal analysis, this is not possible. But that is what I learned when Spice was still running on IBM mainframes and at the university I studied we developed our own nodal analysis program for which I wrote the transient analysis part.

    Since then some time has passed, so maybe they have developed some methods to circumvent this problem and might it be possible these days.

    There is also another method of electronic circuit analysis, called the state variable method, where formula are possible, but that method was not fit for analyzing large networks, and I don't think that method is used anymore .


  • LDM ,

    There can be no reflection of consequence in regards to heat radiating from one fin to another. It is all effectively absorbed. The alumina simply is not reflective to the bands of radiation that it emits. This would be a violation of Kirchhoff's law.


    Edit: this is all

  • Concerning using arbitrary formulae in Spice, it is my understanding that in Spice, being based on nodal analysis, this is not possible.

    I used LTSpice (free .. with built-in schematic editor), which DOES allow user-defined functions. (eg see https://www.ece.uic.edu/~vahe/…troduction_to_LTspice.pdf page 16).

    Quote

    .FUNC -- User Defined Function

    Syntax: .func <name>([args]) {<expression>}

    Example: .func Pythag(x,y) {sqrt(x*x+y*y)}

    I've seen examples with at least one IF in the expression, but can't find it in the documentation.

    Edit2 : .func GDS(Vgs,Vds) {IF Vgs<=vth, 0.1F, IF(CTRL(Vds,Vgs)>=0,((W*KP/L)*(Vgs-vth-Vds)),(((W*KP)/(2*L))*(Vgs-vth)*(vgs-vth)*lambda)}


    I think it's IF condition,true_expression,false_expression ... where IF's can be nested.

  • THHuxleynew


    THH,


    Can you now agree with me, based on my earlier post, that


    A. The ratio of the area with fin to the ratio of the bare tube is not sqrt(2) for fins with a 90 degree angle between them

    B. The ratio of the area with fins compared to the bare tube is for the Lugano case is larger then 2


    Regards, LDM

  • LDM ,


    I've been a bit busy with other stuff, still am, and don't want to post properly till I have it sorted out fully.


    But, yes, I agree with you. In fact if the reactor is 10mm and the fins are 2.3mm that is an increase (for max diameter at peak) of 46% - I was halving this! to get av of 11.5. So with av of 23%, and the counterintuitive but true fact that the extra area varies with square of fin diameter, we have:

    sqrt(2)*1.23^2 as a decent approximation to area, so I'll take 2X + a little.


    Why I'm keeping quite still is that this does not correspond to your ray tracing which should therefore give an average Fbg of < 0.5. We are only out by 20% or so, but till I've resolved that I'm reluctant to agree anything for sure...


    There are then the overall effects of this on the Lugano calcs, which require some care, and as I say I just don't have the (large) amount of time I need to do this properly yet. I will certainly do it within a shortish time (< 1 month).


    THH

  • LDM ,

    There can be no reflection of consequence in regards to heat radiating from one fin to another. It is all effectively absorbed. The alumina simply is not reflective to the bands of radiation that it emits. This would be a violation of Kirchhoff's law.


    Edit: this is all


    Some thoughts, I might be wrong, if so you can correct me.


    Kirchhoffs law is valid under thermal equilibrium, being that the object and it's environment have the same temperature. That also means that all parts of the object must have the same temperature in order to be in thermal equilibrium. In circumstances that the device (as for example a dogbone) has a temperature gradient (Which it has), Kirchhoffs law at least deviates as research has shown.


    Also for the dogbone the device is (due to the heater present) at a different temperature then the environment and that makes Kirchoffs law not applicable anymore because there is no thermal equilibrium (The dogbone and the environment have different temperatures)


    For fin to fin, where both fins have about the same temperature we might assume thermal equilibrium between the fins and in that case we could possibly use Kirchhoffs law between the fins.

    (I don't know if that is formally correct)

    Kirchoffs law states that for an object, having temperature T facing an environment of the same temperature T at thermal equilibrium , that then e = a (e the emissivity, a the absorptivity).

    But there is also the relationship that r + a + t = 1 (with r the reflectivity, t being the transmissivity).

    For Alumina beyond a few mm thickness we can ignore the transmissivity and the equation becomes r + a = 1.

    Now if as you state there is no reflectivity in the bands of radiation (r = 0), then the absorptivity in the bands of radiation becomes 1.

    Following Kirchhoffs law that e = a then the emissivity in the bands of radiation must also be 1 and we know that this is not true for Alumina.


    LDM


  • LDM ,

    To keep the correct version short, the following is my summarized position:


    Each time a ray impinges on an opposite fin (or any other surface), the radiant power is absorbed directly proportional to that surface emissivity (which is the absorptivity). The remainder is reflected or transmitted. In the case of alumina, the transmissivity is very limited, to less than 1 mm generally, and so can be ignored in the generalized case. That means the power directly proportional to the reciprocal of the surface emissivity is the reflected portion. If that remaining power-ray strikes another surface, it then imparts power again proportional to the absorptivity/emissivity and reflects the again the reciprocal portion, continuously dissipating the original power level until it is completely absorbed or goes to infinity and disappears from the system.

    Total emissivity at the correct temperature is correct for the power absorption/reflection calculations.

    This is the same as always, and results in no emissivity greater than one, or less than zero paradoxes.

  • Paradigmnoia


    To keep the correct version short, the following is my summarized position:

    Each time a ray impinges on an opposite fin (or any other surface), the radiant power is absorbed directly proportional to that surface emissivity (which is the absorptivity).


    A part of the fins sees a background at the background temperature and another part sees the opposing fin at another temperature.

    That means that the system as a whole is not in thermal equilibrium and as such Kirchhoffs law does not apply (At least that is what I think). In that case emissivity does not have to be equal to absorptivity.

    The remainder is reflected or transmitted. In the case of alumina, the transmissivity is very limited, to less than 1 mm generally, and so can be ignored in the generalized case.


    Totally agree that the transmissivity can be ignored.


    That means the power directly proportional to the reciprocal of the surface emissivity is the reflected portion. If that remaining power-ray strikes another surface, it then imparts power again proportional to the absorptivity/emissivity and reflects the again the reciprocal portion, continuously dissipating the original power level until it is completely absorbed or goes to infinity and disappears from the system.

    Reciprocal of x equals 1/x

    I think your intention was to refer to 1 - x

    Not all energy is absorbed by the other fin.

    Of the energy from a fin directed to the other fin a part will be reflected back to the originating fin, another part will be reflected to the background. That part reflected to the background, dependent on the view factor, will not be absorbed by the other fin anymore.

    Thus not all radiation towards the other fin will be absorbed.

    What is absorbed an what not is determined by the infinite reflection method.


    Total emissivity at the correct temperature is correct for the power absorption/reflection calculations.

    Correct,


    When the system is not at thermal equilibrium, the power balance needs to be correct.

    This is the same as always, and results in no emissivity greater than one, or less than zero paradoxes.

    Agreed


  • THHuxleynew


    I've been a bit busy with other stuff, still am, and don't want to post properly till I have it sorted out fully.

    Same for me, figuring out these issues takes time


    But, yes, I agree with you. In fact if the reactor is 10mm and the fins are 2.3mm that is an increase (for max diameter at peak) of 46% - I was halving this! to get av of 11.5. So with av of 23%, and the counterintuitive but true fact that the extra area varies with square of fin diameter, we have:

    sqrt(2)*1.23^2 as a decent approximation to area, so I'll take 2X + a little.

    It is indeed counter-intuitive, also for me when I calculated the areas for the first time.


    Why I'm keeping quite still is that this does not correspond to your ray tracing which should therefore give an average Fbg of < 0.5. We are only out by 20% or so, but till I've resolved that I'm reluctant to agree anything for sure...

    For the 69 fins case, fin height 2.3 mm, fin base 2.9 mm the angle between the fins of 64.46 degree.

    For this angle we calculate the view factor for two plates of the same width with a common edge.

    The value is 1 -sin(64.46 /2) = .467 (view factor fin - fin)

    That view factor to background then becomes 1 - .467 = .533

    However since the fins are bend, the view to the background must be larger.

    The conclusion then is that the view factor to the background will always be > .533

    As a result a value smaller then .5 can never be obtained.


    I don't know what the prerequisites are for the view factor formula given in the paper you referred to. Maybe for example that it is only valid for flat surfaces ?


    There are then the overall effects of this on the Lugano calcs, which require some care, and as I say I just don't have the (large) amount of time I need to do this properly yet. I will certainly do it within a shortish time (< 1 month).

    Take it easy


    LDM


  • OK, so I still want to look at this properly, and that must wait later.


    But:


    I expect the calculation with ridges to be comparable with the calculation from a surface area which is the ridges outline multiplied by pi, with emissivity adjusted as per TC paper (but never more than 1.


    Which means in this case that the correction for the ridges is roughly (maybe exactly) 1.23 X more power on top of Tc correction.


    And that may be compatible with your surface area * Fbg or not. It does now look in right ballpark. if compatible all is as it should be, but need still to flow from this consequences re Lugano results.


    THH