# Revisiting the power calculation in the Lugano report

• Interesting. Has this been applied to correct the dummy also?

• Interesting. Has this been applied to correct the dummy also?

Running his original code gives a COP value of about .75 for the dummy run.

After applying the corrections the COP of the dummy run is about .85

But TC had some possible valid arguments that apply to the dummy run.

These arguments can be found as comments in the Python code.

I personally did not investigate these comments.

• LDM,

Thank you for the work you put into this. Your adjusted Lugano COP puts it in line with what Higgins and MFMP found. Small, but still something. BEC reports about the same COP 1.3, or so.

I wonder if the Swedes got about the same in their attempt to replicate their own Lugano attempt.

• LDM,

Thank you for the work you put into this. Your adjusted Lugano COP puts it in line with what Higgins and MFMP found. Small, but still something. BEC reports about the same COP 1.3, or so.

I wonder if the Swedes got about the same in their attempt to replicate their own Lugano attempt.

Thank's

I used the average of both the values with no bias and high bias to get to about the same values as Thomas Clarke reported.

However if you are not using the high bias (Which brings the values down), but only the normal bias, meaning the standard emissivity curve used in Lugano, then the COP value increases to 1.3.

Indeed in line with what Bob Higgins and the MFMP found.

• The MFMP also found, indirectly, that the unfuelled thermal validation ecat mock-up produced a false COP of about 3.8 if one uses the total emissivity from the alumina plot instead of the integrated in band spectral emissivity of alumina specific to the IR camera.

• Lugano rods convection heat transfer coefficients

The convective heat transfer coefficients used in calculating the convective powers for the rods can be calculated.

We do this by using Newtons law. Newtons law is given by the following formula :

Q = hA(Ts-Ta)

Q being the power

h the convective heat transfer coefficient

A the surface area

Ts the surface temperature

Ta the ambient temperature

h can then be calculated as

h = Q/(A(Ts-Ta))

Since the convective data for the rods in the Lugano report give the values for the power (Q), the value of the area of each section (A) being 4.71E-3 m2, the section temperatures Ts and the ambient temperature (21 degree C), we can for each section calculate the convective heat transfer coefficient used for each section.

Tthe results of these calculations can be found for both the upper and lower rods in the following tables

Upper rod h values

Area----Tu (C)-----Conv (W)------h

-1-------151.52-------5.84-------9.50

-2-------125.13-------4.45-------9.07

-3---------90.85-------2.81-------8.54

-4---------68.17-------1.72-------7.74

-5---------58.26-------1.28-------7.29

-6---------54.12-------1.11-------7.12

-7---------46.33-------0.80-------6.71

-8---------40.02-------0.56-------6.25

-9---------35.34-------0.40-------5.92

10--------31.82-------0.28-------5.49

Lower rod h values

Area----Td (C)-----Conv (W)------h

-1-------147.98-------5.65-------9.45

-2-------118.89-------4.13-------8.96

-3---------87.71-------2.66-------8.47

-4---------68.15-------1.72-------7.75

-5---------58.21-------1.28-------7.30

-6---------52.82-------1.06-------7.07

-7---------45.06-------0.75-------6.62

-8---------38.89-------0.52-------6.17

-9---------34.30-------0.36-------5.75

10--------31.09-------0.26-------5.47

Since I wanted to know if I could correctly calculate the convective heat transfer coefficients myself, I did the calulation of the convective heat transfer coefficients of the rods myself.

The result was that they differed from the values calculated back from the report.

To understand how I did the calculations we recalculate the convective heat transfer coefficients for area 3 of the lower rods.

For the calculation we use the data from the following table which gives k (thermal conductivity), alpha (thermal diffusivity) and v (kinematic viscosity) as function of the film temperature.

(These values are from the same table in the book Fundamentals of heat and mass transfer as the Lugano testers used)

--film temp-----k--------alpha-------v

-------------------E-3--------E-6--------E-6

------100-------9.34--------2.5--------2.00

------150------13.80-------5.8--------4.43

------200------18.10------10.3-------7.59

------250------22.30------15.9------11.44

------300------26.30------22.5------15.89

------350------30.00------29.9------20.92

------400------33.80------38.3------26.41

------450------37.30------47.2------32.39

------500------40.70------56.7------38.79

For a temperature of 300K the tabulated values are in agreement with the values mentioned at the bottom of page 11 of the Lugano report.

The reported ambient temperature during the dummy run was 21 degree C (284.15K) and the surface temperature of the rod was 87.81 degree C (360.86K)

Thus the film temperature is (284.15 + 360.86) / 2 = 327.51 K

The closest tabulated value is 350K and we now take this value, one value lower in the table and one value higher.

For the thermal conductiveity k this gives :

--------T----------k

------300------26.30

------350------30.00

------400------33.80

We now do a 3 point curve fit with the Newton method to solve the k value for the temperature of 327.51.

This gives a k value of 0.0283. In the same way the values for v ( 1.978E-05 ) and alpha ( 2.645E-05 ) can be obtained.

Using for the Fluid thermal expansion coefficient B the value of 1/Tfilm as specified in the Lugano report and doing the same calculations with the same formulas as used in the report, we obtain a value of the convective heat transfer coefficient h of 8.12.

The value we obtained as being used by the test team was 8.47. The difference is -4.1 %.

The outcome of the calculations for all zones of both the upper and lower rods are presented in the following tables.

Upper rod differences

Area-----h---------my h-----Error %

-1-------9.50-------9.57--------0.7

-2-------9.07-------9.06--------0.1

-3-------8.54-------8.21--------3.9

-4-------7.74-------7.47--------3.5

-5-------7.29-------7.05--------3.3

-6-------7.12-------6.85--------3.7

-7-------6.71-------6.41--------4.4

-8-------6.25-------5.96--------4.7

-9-------5.92-------5.56--------6.1

10------5.49-------5.18--------5.7

Lower rod differences

Area------h-------my h-------Error %

-1-------9.45-------9.51--------0.7

-2-------8.96-------8.92--------0.4

-3-------8.47-------8.12--------4.1

-4-------7.75-------7.47--------3.6

-5-------7.30-------7.05--------3.5

-6-------7.07-------6.78--------4.1

-7-------6.62-------6.32--------4.5

-8-------6.17-------5.87--------4.9

-9-------5.75-------5.46--------5.0

10------5.47-------5.09--------7.0

As can be seen especially for the lower temperatures the differences become quite large.

A possible explenation might be that the Lugano testers in the report used in the calculation of the convective heat transfer most of the values in two digit accuracy while i used three digits.

With two digits accuracy the error margin become then quite large and can then possibly add up to the errors in the ranges calculated above.

Suggestions for other reasons why the values differ are welcome.

• Might be that Newtons law of cooling is too over-simplified to give an accurate result, especially over a wide variation in temperatures. It assumes that each unit area transfers the same amount of heat (And also that h isn't affected by T).

Instead of using h = Q/(A(Ts-Ta)), a more accurate method is to calculate an average h (taking the shape of the object into account) - done by first working out the Prandtl, Raleigh & Nusselt numbers for the system.

Can't paste the formulae for those easily, but this should give you what's needed:

• LDM ,

Although the error % seems large for the coolest end of the rods, the final W difference, in terms of the total power budget, should still be rather minor. I believe that I had a similar result when I did it, and it did seem to be rounding that made the difference.

The calculations for working out the Rod segment temperatures and Rod temperature gradient, and therefore power, during the active period is much more complex. There're is very little in the report to go on, since just the final total power is reported for each run.

You should be close enough, however, with your model that it should be possible to test if the 2/3 factor was in fact applied, or not, to the active period Rod power in the report, since a total 33% overestimate error should be obvious. (It is my opinion that the 2/3 adjustment was not made to the active period Rod power in the report, and therefore the Rod power reported is too high for all active runs. (Easily fixed, since they lump those results into two periods anyways, an indication of how concerned the Professors were about the Rods contribution)).

Don't forget the minor Joule heating in the rods added in the report from the cables, and consider carefully the contribution of the heater coil extensions entering the Rods for 4 to 5 cm, which affect calculations for both the temperature of the initial Rod segments (Cap end) and the power budget of the Main Tube since these wire extensions, six in all, reduce the amount of input Joule heat available to the Main Tube and Caps, possibly explaining a portion of the previously calculated excess (COP 1.2 or similar).

Once, here in the Forum, I calculated exactly the parameters of the twisted 15 ga Kanthal resistance wire for the three parallel coils, from which the 6 X 4 cm (estimated) could be subtracted so that the end lead power % of the entire heater windings could be correctly attributed. Off the top of my head, each of the 3 coils have about 1.5 m of wire, which is twisted to make a pre-coiled length of slightly less than half of that. Probably better for me to look it up again. I believe that the Caps and wire extensions combined contain about 30% of the total calibrated resistance heater wire, and therefore the wire extensions could net about 15% of the Total input power.

• Since no one representing TC has raised any objections to the upwardly revised COP, what does it mean that we now have 3 Lugano analysis (Higgins/MFMP/TC) indicating a small, yet significant output>input?

• The dummy is messed up in Lugano too. Once that gets squared away, there is no basis, within uncertainties and real errors, to claim any COP other than what Nature normally provides.

Consider this little crumb: the camera (spectral in band) emissivity is being argued about endlessly, and yet who has tackled the question of the real Total emissivity from which the output power is finally calculated? No one. Why? Because even if we had perfect thermocouple corroborated IR camera temperatures, no one has any real basis to say what the Total emissivity of the device with ribs, uncertain coatings, uncertain roughness, etc., really is. Oh, sure we can grab a textbook value for that. That should be real close to an object we have ourselves only seen in a few photos (one of which could be the one that broke, not the actual working one.)

How much off does the Total emissivity have to be to eat up the 'excess' 7 to 20% ? Not much.

Higgins' report, as good as it is, has an error which moves his estimate closer to TC's when corrected.

The MFMP version does not deal with the thermal distribution very well, and crashes like a Jato-powered sedan into a cliff wall when dealing with the Dummy.

And TC's version can have that 0.07 (?) excess wiped out with the most minor adjustment to the total emissivity.

LDM's version is more complex, but still has ad-hoc parts put in, like total emissiviy, as is necessary, because acurate information simply does not exist.

All of these agree that COP 2+ did not likely happen. (Let alone COP > 1.3 even). The little bit that that might look like excess can be wiped away in a moment in the noise of the uncertainties. More certainty is not going to be forthcoming. Unless maybe IH leaks out their report on tests with the last Reactor.

• Newtons law of cooling is likely too over-simplified to give an accurate result, especially over a wide variation in temperatures. It assumes that each unit area transfers the same amount of heat (And also that h isn't affected by T).

The thermal exchange coefficient h is affected by T, but in an indirect way.

First of all the Rayleigh number is calculated with :

The volumetric thermal expansion coefficient

The kinemathic viscosity

The thermal diffusity

All three are dependent on the temperature.

Then using the Raleigh number to calculate h, we also multiply by k, the thermal conductivity of air, also temperature dependent. Thus the calculated value of h is largely depent on temperature.

But I agree that we work with an average temperature and that h over the area is constant.

That is indeed a simplified approach, which however gives for situations which have been researched many times in literature, such as horizental tubes, adequate results.

An other approach would be to use CFD (Computional Fluid Dynamics) software to calculate the convective heat transfer of a part of the rod for a given temperature. Additional we can then also calculate h.

I have CFD software, but have still to learn how to use it beyond the basics. But I have seen in literature several quite complex simulations with CFD software which give errors of 1% to 2 % compared to the real measured data. This because CFD is not using approximations in calculating the convective heat transfer coefficient, but is instead solving the differential equations governing the heat transfer. So maybe this is the way to go.

Instead of using h = Q/(A(Ts-Ta)), the more accurate method is to calculate an average h (taking the shape of the object into account) - done by first working out the Prandtl, Raleigh & Nusselt numbers for the system.

Using h = Q/(A(Ts-Ta)) is using the forumula as used by the Lugano team in a reverse way using their data. As such that should give the exact value of h they used.

Can't paste the formula's for those easily, but this should give you what's needed:

I had the link myself, but must concede that I did not look at the contents lately.

The formula showed in the link is somewhat different from the one the Lugano team and I used.

If time allows it migh be an idea to use the formula in the link and see how much the data differs.

But there are many formula's for calculating the convective heat transfer coefficient of a horizental tube (or round heated wires).

The formula used by the Lugano testers is normally used for much larger Rayleigh numbers then those valid for the Lugano case.

As such it might be worth the effort to find a better (more accurate) formula for the lower Rayleigh numbers.

• I understand now why the manufacturer of the IR cam advise to calibrate, and not to trust any emissivity number...

We cannot replicate the active rod, but maybe can Dewey Weaver obtain a dummy from the stock? or Maybe is the dogbone of MFMP not far enough from the shape (maybe it can be painted with the ZrO2 paint cited by Dewey Weaver ) to give interesting approximation of emmissivity over temperature?

• I understand now why the manufacturer of the IR cam advise to calibrate, and not to trust any emissivity number...

Alain,

While radiated heat transfer is part of the calculated power and emissivity can then be important, the above posts are about convective heat transfer.

As far as calculating radiated heat transfer, then indeed if you measure the temperature with the Optris, then for a correct temperature measurement you need to use the correct in band emissivity.

Hoevever for the in band emissivity often a fixed value of .95 is quoted, but in reality the in band emissivity is also somewhat dependent on temperature.

In an earlier post I showed that you can also calculate the power using the Optris without calibration. see

Revisiting the power calculation in the Lugano report

We cannot replicate the active rod, but maybe can Dewey Weaver obtain a dummy from the stock? or Maybe is the dogbone of MFMP not far enough from the shape (maybe it can be painted with the ZrO2 paint cited by Dewey Weaver ) to give interesting approximation of emmissivity over temperature?

The ZrO2 paint from Aremco is a special paint which is as Armeco says specially fitted for fixing heater coils. So they could have fixed the heater coil with it before casting the ECAT with Alumina.

That would fit both Dewey's story and also fit the analysis of the Lugano report that the outher casting was almost pure Alumina.

• I understand now why the manufacturer of the IR cam advise to calibrate, and not to trust any emissivity number...

We cannot replicate the active rod, but maybe can Dewey Weaver obtain a dummy from the stock? or Maybe is the dogbone of MFMP not far enough from the shape (maybe it can be painted with the ZrO2 paint cited by Dewey Weaver ) to give interesting approximation of emmissivity over temperature?

I did IR camera emissivity vs temperature tests for Durapot 810. I don't have the notes handy, but about 0.8 was typical, with a very small drop (0.02) at the lower end of temperatures.

The Zr paint is a bit of a surprise, and I haven't tried it yet. Presumably the Al version should be tried also.

But does not help the total emissivity used for the calculation of the output power. There are companies that can test materials and characterize their complete IR spectral response at various temperatures. Not to cheaply, but not terribly expensively either. But what to send them?

• LDM ,

IH built the Lugano reactors, three of them, and were surprised by the paint, so your theory doesn't work.

• IH built the Lugano reactors, three of them, and were surprised by the paint, so your theory doesn't work.

The argument goes also otherwise. The outside of the ECAT was by analyses almost 100% Alumina.

So in the same way we can argue that the theory that it was painted with Zr paint doesn't work.

So the question is who is right and who is wrong.

On that question I don't know the answer.

Right, so grossly similar to alumina (and many other ceramic non-conductors).

However, note the range of emissivity values at each temperature step.

• Although the error % seems large for the coolest end of the rods, the final W difference, in terms of the total power budget, should still be rather minor. I believe that I had a similar result when I did it, and it did seem to be rounding that made the difference.

I agree, except for the dummy run where convected and radiated power are of the same order.

The calculations for working out the Rod segment temperatures and Rod temperature gradient, and therefore power, during the active period is much more complex. There're is very little in the report to go on, since just the final total power is reported for each run.

Agreed. And that also means that I have currently no plans to do any calculations for estimating the COP of the active runs.

You should be close enough, however, with your model that it should be possible to test if the 2/3 factor was in fact applied, or not, to the active period Rod power in the report, since a total 33% overestimate error should be obvious. (It is my opinion that the 2/3 adjustment was not made to the active period Rod power in the report, and therefore the Rod power reported is too high for all active runs. (Easily fixed, since they lump those results into two periods anyways, an indication of how concerned the Professors were about the Rods contribution)).

First of all can you explain what you mean by my model ?

Any ideas how to test with my model if the 2/3 factor was applied ?

If the Lugano testers made the error as per your opinion then indeed the COP value would be less.

But in a previous posts we showed that they made an other error, being that they did not calculate with the fin area combined with the view factor to the background and also did not correct the emissivity for the reflection between the fins. And that total correction has about the same value as when the 2/3 factor was not applied, but with the opposite polarity.

So we end up with about the same power budget

Don't forget the minor Joule heating in the rods added in the report from the cables,

Thanks for the warning . I was already aware of it

and consider carefully the contribution of the heater coil extensions entering the Rods for 4 to 5 cm, which affect calculations for both the temperature of the initial Rod segments (Cap end) and the power budget of the Main Tube since these wire extensions, six in all, reduce the amount of input Joule heat available to the Main Tube and Caps, possibly explaining a portion of the previously calculated excess (COP 1.2 or similar)

Good point !

Once, here in the Forum, I calculated exactly the parameters of the twisted 15 ga Kanthal resistance wire for the three parallel coils, from which the 6 X 4 cm (estimated) could be subtracted so that the end lead power % of the entire heater windings could be correctly attributed. Off the top of my head, each of the 3 coils have about 1.5 m of wire, which is twisted to make a pre-coiled length of slightly less than half of that. Probably better for me to look it up again. I believe that the Caps and wire extensions combined contain about 30% of the total calibrated resistance heater wire, and therefore the wire extensions could net about 15% of the Total input power.

I am still wondering about the heater wire seize of AWG15 they supposedly used.

AWG15 wire has a diameter of 1.45 mm. But they seem to have been braided with two wires.

Twisting two wires will double the diameter, thus 2.9 mm

For about 70 windings total, without inter winding spacing, the length of the heater coil would have been 70 x 2.9 mm = 203 mm. But the photographs of the heated Ecat gives me the impression that a reasonable spacing was used between the windings.

Proper engineering practice for heater coils uses strech factors between 2.5 to 4 for the inter spacing between the coil windings to prevent shortages due to the coil windings moving when heated.

Using a stretch factor of 2.5 (the minumum) gives a minimum length of the coil of 2.5 x 203 = 507 mm. That is much longer then the total Ecat length of 280 mm.

Or is my calculation wrong ?

• LDM ,

Excuse my being brief with some comments.

The stretching of the wires is potentially a problem, but it seems OK in practice, for the most part.

I had 3 coils, each with 10 wraps, with something like 7 mm between each successive wrap (or 21 mm between each wrap on the same coil winding). That gives a 200 mm long coil section, plus leads. The twisted wire seems to "absorb" some stretch, and is flexible. It winds into a 10 mm inside diameter coil surprisingly cooperatively for its combined wire diameter.

Your stretch factor I think, in general, is more important to very long coils (especially of small diameter), like used in a kiln. Those windings can expand and sag, especially with age, although shorting out would be unusual due to the protective oxide coating on the wire. The expansion of Kanthal due to heat itself is (engineerered) low, in general, but it can stretch considerably in open air coils due to gravity when the wire is orange hot and fairly soft. Large numbers of wraps per coil length does add up the in the overall expansion length, so that bunching is more likely when the wire expands. The heater wire expansion is noted both in the coil diameter increase and the coil length increase.

The Durapot cement seems strong enough to keep coil expansion in check, although the coil is likely under some tension at high temperatures. It likely would be a problem to double the length of the coil section of a Lugano type device, as coil expansion might then be too much for the cement.

• Not I think inconsistent

The Lugano report says that the alumina sample was taken from one of the rods, not the reactor body.

THH

• Not I think inconsistent

The Lugano report says that the alumina sample was taken from one of the rods, not the reactor body.

THH

The following quote is from appendix 2 of the report

In order to determine the nature of the material covering the reactor, a sample from one of the ridges was analyzed.

As far as I know the rods had no ridges, but the reactor body had

• However, the reactor (including ridges) was cast from Durapot 810, which is not made of 100% alumina, nor is it 99.9 % alumina with no detectable other elements in concentrations to make up the 0.1% , although it is alumina based.

• New

The following quote is from appendix 2 of the report

In order to determine the nature of the material covering the reactor, a sample from one of the ridges was analyzed.

As far as I know the rods had no ridges, but the reactor body had

I must have misremembered!

I think I did so because it is still inconclusive. The "fragments" analysed would not be all paint, and might in fact contain no paint at all. So, given the reactor was painted, we don't have much info, unless as P says above we are sure that the ridges were not case from something that bakes to 99% alumina.

Should this analysis be incorrect it is another example of the Rossi effect. Not only do small ceramic tubes miraculously deliver a X3 power gain when used as radiant heaters, but also lab analyses deliver (completely) incorrect results! For example the 62Ni.

Axil will have the answer, it is probably all about muonic BE condensates (and yes, since muons are leptons I know that does not make much sense, but then it is par for the course. It would be muons bound in pairs and then forming the BEC).