Posts by LDM

Quote from Bruce__H

Thanks for the re-explanation.

This still leaves open, however, the issue of how you would use the Wien relation for Rossi's SK data. How would you do that?

I wouldn't

As outlined by others and looking at the spectrum, it is not dense for all frequencies.

And I realy have no idea how much energy is involved in the thin part and the thick part.

As such I have no idea if we can apply the Stevan-Boltzmann black body equation to the SK and how well it would fit the situation.

Also we need the correct plasma temperature and I don't know if that was measured correctly.

Only if Rossi would have explained why he could use the black body Stefan_Bolzmann law during his sales pitch we might have known more.

But he is not very good in explaining the technical things he does, wether they work or not.

Quote from Bruce__H

I am having trouble following your reasoning. It seems to me to both contradict what you said earlier and even (in the final sentence) be self-contradictory. Could you restate it please?

OK, Using the theory for plasma radiation , when calculating the power of a thick plasma, this leads to the formula for the thick plasma power of J = σ T⁴

This is the same formula as for the power of a black body, but was not calculated assuming that the plasma was a black body, but is derived in another way

Since the formula of the thick plasma power is the same as that of a black body, one can incorrectly assume that the thick plasma is a black body.

But it isn't, only the formula for calculating the power is the same.

And since the thick plasma doesn't have the properties of a black body. you can't match both temperature curves.

Quote from Bruce__H

What blackbody spectrum to these data? What temperature?

P.S. Your diagram is schematic. Here is Bob Greenyer's superposition of 17 data grabs (supplied by can of this forum and Pekka Janhunnen: see http://disq.us/p/1zgv87d). What blackbody spectrum would you attach to this data set?

The answer is simple.

A thick plasma does not have a black body spectrum.

From that the formula for calculating the energy of a thick plasma is the same as that for a black body you can not conlude that it has the same physical properties as that of a black body and thus that you can match a black body spectrum to that of a plasma.

Plasma radiation

There is a difference in opinion if you can treat the plasma as a black body or not.

So I want to address this subject in this post.

Basically a plasma can be considered to be a thin plasma or a thick plasma.

In a thin plasma radiation from the inside of a plasma can (almost) without obstruction radiate out of the plasma.

For a thick plasma almost all radiation in the inside is blocked (By the particles in the plasma) to flow to the outside.

From a plasma spectrum you can see which part of the frequency range must be considered "thin: and which must be considered "thick".

See the following figure :

In the part of the spectrum where you see the high discrete lines the plasma is thin for those frequencies, where the spectrum is more continuous the plasma can be considered as being thick.

If we now look at the spectrum during the SK demonstration, we can see that the plasma can be considered thick in the range of 350 - 425 nm. Outside this range we see on the left side the high single peaks where the plasma is thin for the radiation.

The temperature of a plasma is due to the movement (speed) of the particles in the plasma.

The speed distribution can be considered to have a Maxwellian distribution.

Based on this a representatative equivalent temperature can be measured.

(The plasma temperature)

In case almost all radiated energy is coming from the thick part of the plasma and if one assumes the radiation from the thick plasma to be uniform from the surface of the plasma then integrating the radiation over all frequencies and all directions of the hemisphere yields the following formula for the total radiation from the thick plasma :

J = σ T4

This is equal to the Stefan-Boltzmann law of a black body.

The conclusion is then that in order to calculate the energy flow from a thick plasma you can consider that plasma as being a black body.

See for the theory the book "Plasma Physics and Engineering" by Alexander Fridman and Lawrence A.Kennedy from which i gathered the above information.

Quote from zorud

DO you have any other information, other than Rossi says, that Leonardo has a robotic factory, many employees in R&D, marketing, manufacturing, a service, installation and sales team ready to roll? Anything that proves that Rossi has a customer and will be installing in the coming weeks more plants to sell heat? I know future will tell, but what we can tell from this point in time, is, that Rossi is still a one man show, doing terrible presentations from his condo. This we know for sure, thanks to Frank Ackland, who was surprised to hear where to travel for this important show...

I don't think that Leonardo has what you stated above.

In my opinion his partner is assisting in the product development and also doing the manufacturing, service and installation.

And if they as stated are a large international company they would have enough resources.

That solves also the problem of having the needed licences to do those installations.

In that case Rossi is not a one man show.

Lugano active run period 16 recalculation if temperatures where inflated.

The attached spreadsheet contains a recalculation of the Lugano active run period 16 if the temperatures where inflated due to using broadband instead of in-band emissivities on the Optris thermal camera.

(Calculation when temperatures where not inflated was published in post #522 )

The recalculation assumes that a factor of 2/3 was already included in the reported rod powers.

However since it was not expliciy stated that the Lugano testers applied this factor 2/3, the sub page for the rod powers in the spreadsheet also includes the calculation for the case that the factor 2/3 was not applied.

Note also that since the calculations are based on average temperatures, the results must be interpreted as approximate values.

The recalculation with the established lower temperatures results in a total convective and radiated power of 1329 Watt assuming the factor of 2/3 was applied to the reported rod powers.

If the factor of 2/3 was not applied to the rod powers the total calculated power would have been 1258 Watt

The total applied electrical power for this run was 865 Watt.

Rod temperatures Lugano active run period 16 if factor 2/3 was not applied

When the factor 2/3 was not applied to the rod power, we have the following data

Convective power--------87.94 Watt

Radiated power-----------88.47 Watt

Total power---------------176.41 Watt

Iterating with both the convective heat tranfer coefficient h and the emissivity on the total power of 176.41 watt for one set of three rods gives the following result

--------------------------------------------Temp----convective power----radiated power----total power

-----------------------------------------------C---------------Watt-------------------Watt-------------------Watt

h and emissivity iteration-------103.18----------104.84-----------------71.57-----------------176.41

Using the temperature of 103.18 degree C with its associated broad band emissivity of 0.684 the recalculated average rod temperature in case broadband instead of in-band emissivities where used on the Optris becomes 85.72 degree C ( n = 4.171).

When making the spreadsheet for Lugano dummy run period 16 in case the temperatures where inflated also a rod power calculation for the temperature of 85.72 degree C will be included.

Paradigmnoia

Are you also in some way going to simulate the interaction with the rods.

This since the interaction with the rods is limiting the heat transfer of the outer side of the end caps.

Quote from Paradigmnoia

The 2/3 Rod adjustment factor is not explicitly shown to be applied to the Active Runs in the report.

The convective and radiant powers are shown in a short table after a brief discussion, and must be considered to be each to have been adjusted already by the 2/3 factor before being listed if the 2/3 was indeed applied. However, the report does not indicate if this was the case or not.

For the Dummy run, the convective and radiant power were first totalled, then the 2/3 factor was applied.

It is indeed correct that this is not explicitly shown in the report and your further remarks are to the point.

So i propose based on my previous post to base the calculation of Lugano run period 6 for the case the temperatures where inflated on the presumption that the factor 2/3 was applied.

But in addition to that I will also calculate the average rod temperatures for the case the factor 2/3 was not appllied and based on that temperature also the rod powers.

It will tell us if this makes a large difference for the calculation the active run period 6 if the temperature wold have been inflated.

Rod temperatures of Lugano active run period 16 if broadband insted of in-band emissivities on the Optris where used

In order to be able to calculate the lower rod power of the Lugano active run period 16 in the case that the measured temperatues where inflated, we need to know the rod temperatures.

However only the total convective and the total radiated power of the rods are given.

To get an approximate temperatures we have three possibilties to determine average temperatures.

- Iterate between h and the temperature till the reported convective power is reached

- Iterate between the emissivity and the temperature untill the reported radiated power is reached

- Iterate with both changing h and emssivity temperatures untill the total power is reached

For Lugano active run 16 the following data was given for the power of one set of three rods

Convective power--------87.94 Watt

Radiated power-----------88.47 Watt

However the Lugano testers applied afterwards a factor 2/3.

Before continuing we need to undo this by multiplying both the convective and radiated heat by 3/2

The corrected powers become

Convective power--------131.91 Watt

Radiated power-----------132.71 Watt

Total power-----------------264.62 Watt

With the above corrected data all three types of iterations where done with the following results

--------------------------------------------Temp----convective power----radiated power----total power

-----------------------------------------------C---------------Watt-------------------Watt-------------------Watt

h iteration----------------------------120.28----------131.90------------------95.08----------------226.98

emissivity iteration----------------145.33----------173.04----------------132.70----------------305.74

h and emissivity iteration-------132.55----------151.87----------------112.75----------------264.62

As expected none of the iterations get both the convective and radiated power about equal to the reported powers.

This is due to the fact that the convective power is a linear function of the temperature, while the radiated power depends on the 4th power of the temperature (in Kelvin).

For a recalculation it is proposed to use the combined h and emissivity iteration since the power is equal to the calculated power and the temperature calculated is about the average of both the h iteration and the emissivity iteration.

It must however be understood that the found temperature is an approximation.

Using the temperature of 132.55 degree C with its associated broad band emissivity of 0.696 the recalculated average rod temperature in case broadband instead of in-band emissivities where used on the Optris becomes 111.17 degree C ( n = 3.938).

The found average temperature of 111.17 degree C for the rods can be used to calculate the approximate power of the rods in case the temperatures of the Lugano active run period would have been inflated by using broadband emissivities on the Optris.

Quote from Paradigmnoia

The heat distribution in the Lugano Ribs area is remarkably flat. This is because the Caps are heated, as well as the Rod ends, which limits thermal losses outwards from the Ribs area.

In my opinion there is another effect in play

The ribs closest to the caps and the inner side of the heated caps are viewing each other for a significant part.

Since both are at elevated temperature, the heat exchange of the involved areas becomes much smaller.

This smaller heat exchange must be compensated by a higher temperature in order to get the power out.

This will result in that the temperature of the ribs close to the inner side of the end caps will be raised.

(for the ribs farther away from the end caps the effect is also present, however to a less extend)

This might also an explanation for the flatter profile.

That the MFMP test had a more parabolic profile is also due to a much larger heater wire length under the ribs compared to the heater wire length under the caps then was the case for the Lugano ECAT.

Quote from Paradigmnoia

LDM ,

What I was hoping to assess was the temperature of the Caps relative to the Ribs with power applied within the Caps.

Unfortunately there is no data for the Caps temperature until the single phase version was run.

The Fat Coil version may have had more input to the Caps, but I haven't found any data for that version at all (yet).

Is there any Optris data for the Fat Coil Glowiness test?

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What I learned from interpreting some simuations is that the following might be the case

1. To get to the reported cap temperatures, the heater coil might have been running for some length also under the end caps

2. You can not from radiation/convection get much power into the rods, thus almost all power to the rods needs to be delivered by the heater wire. 4 cm into the rods is in my opinion not enough.

3. If most of the power to the rods is delivered by the heater wire, then the rods cause a kind of thermal barier near the end cap. This barier limits to a great extend the loss of power at the side of the end cap, thereby reasing the temperature in the end cap

The question now is which heater configurations are fullfilling these requirements and have also the required total resistance.

If it is limited to a few cases, then you can for each case calculate the power under the ribs.

Then test your model with the lowest and highest power setting required for getting the same power under the ribs.

It will give you the maximum and minimum center temperature of the ribs.

You can then compare those with those in the Lugano report and those of the MFMP test.

Maybe you can draw some conclusions from it.

Quote from Paradigmnoia

I have been going over the construction of the MFMP Dogbone replica of the Lugano device.

An important aspect is the lead passing through the Cap. The '3 phase' version (3 windings, but wired in series outside of the caps due to single phase power supply) failed before the arrival of the Optris, so only the single phase version was imaged by the Optris IR camera.

Any reason why the Kantal wire failed ?

In general Kantal (if produced by Kantal) is very reliable.

Maybe Alan S has some suggestions ?

Anyway, note that the Caps are essentially entirely heated by conduction from the Ribs area, since a copper sleeve was placed over top of each of the leads, crimped to the supply winding lead at the Ribs-Cap interface, and the supply cable clamped to the lead while inside that copper sleeve.

(This was to prevent the lead from burning off, as was experienced with one of the leads in the three winding version)

Since, as you also noted in one of your posts, the conduction, even for a reasable conductive material as Alumina, is quite limited sideways, the center temperature of the ribbed area is almost not dependent what is happening in the caps and to the caps rod interface.

I see that also in my FEM simulations.

So if you apply to the ribbed area the power density you want to test with, then the center rib area temperature can in my opinion still be used for comparative measurements.

Quote from Paradigmnoia

Please could you make two more spreadsheets in the same form factor as the Dummy one, (I realize that this quite onerous), for both the recalculated lower temperature Run 16 and your COP 5 version of Run 16 ? Then it will easier to see what the simulation is doing. The Ribs can be treated as a single section for now. I am more interested in chasing why ~ 24% of the electrical input goes into the Rods and Caps, but ~ 50% goes out, when ~76 % of the input power goes into the thin Ribs section in the Dummy cases.

I completed as per your request the spreadsheet for the COP 5 version.

See attachment.

I still need to make a start on the one for the recalculated lower temperatures.

(Working now on how to get the temperatures of the rods, most likely a separate post)

Length of heater wire in the rods

A few weeks ago I did some analysis using data from the MFMP thermal dogbone test based upon the adapted emissivities which I found to limit the errors between thermal and applied powers.

I assumed two different length of heater wire in the rods, 4 cm and 6 cm.

These leads to two different heating distributions and thus also to two different power densities under the ribs.

Based on the found power densities I recalculated from the MFMP data the temperatures of zones 5 through 9 if the emsissivities would have been those of alumina.

For 4 cm of heating wire in the rods I calculated a temperature of 510.69 degree C.

For 6 cm of heating wire in the rods the temperature found was 440.78 degree C.

Since the temperature of 440.78 for the 6 cm heater wire in the rods is the closest to the reported 449.9 degree C in the Lugano report, it is an indication (as far as my assumptions about the MFMP test hold) that indeed the length of the heating wire in the rods is probably more then the 4 cm assumed.

Note also that the if we assume inflated temperatures due to using broadband emissivities on the Optris, that the average temperaure for zones 5 through 9 in Lugano would have been 376.7 degree C.

This value is largely outside the temperature range of 441 through 511 found by recalculating the MFMP data for different heater wire lengths in the rods.

Quote from Paradigmnoia

More annoying is that Coorstek wants $145 for one 500 mm long 30 mm OD 99.8% alumina tube.

What about using as a first approach a 30 mm stainles steel tube coated with a Durapot layer to get the emissivity in line ?

Much cheaper.

Suggestion for the rods:

Take a 3 cm diameter, 50 cm long ceramic pipe (or from another material which can stand the temperature).

Stick the halve of a heating element (coil, power resister) in at one side.

The part sticking out simulating the heated environment of the end cap.

Put the power on.

I think that after a short time it will be acting like a horizental flue pipe in that there will be hot air current flowing in the rod from the hot heater side to the other end which is cold.

It would possibly explain why the rods are hotter near the cold side then would be expected from conduction only.

Paradigmnoia

Your modified e in your Dummy spreadsheet due to IRM and View Factor seems OK to me in general.

When viewing the output power distribution Rods%/Caps%/Ribs% the Rods seem (to me) to have a larger than expected proportion of the output for the Dummy cases, both yours and Lugano, relative to the electrical input for the respective segment. (I’ll post a little summary below in a moment)

I already was wondering about that too and think that the length of the heater coil wire inside the rods could have been more then the 4 cm you suggested.

For the dummy run simulation it would bring the temperatures of the ribbed area possibly a little bit down and maybe even more in agreement with the reported data but would boost significantly the power in the rods.

A possible arangement would be a coil of 9 windings (same number of turns as shown in the IH patent application drawing) with 24 cm length (so somewhat extending under the end caps), 2 cm wire in the end caps and then 6 cm in the rods.

As far as the output power distribution is concerned : For the two sets of rods, based on the data in the Lugano report I calculated about 118 watt while the Lugano report calculated about 130 watts.

The dummy run recalc gave (see page 1 of the spreadsheet)

Rods 118.38 Watt

Caps --87.33 Watt

Ribs --281.77 Watt

The total power is then calculated as 487.48 Watt and the distribution becomes 24.3/17.9/57.8

That is different from the 28.89/21.05/50.06 you are giving.

EDIT : I now see that you compare the inflated case.

Maybe you should add that to the text so that it is clear what data you are presenting

What range of W/mK are you using for the Rods and Caps?

I am assigning the powers to the heater sections in the model as per your suggestion for the power division.

A section power is then evenly distributed over the length of the heater wire in the section.

So I am not using W/mk values.

Please could you make two more spreadsheets in the same form factor as the Dummy one, (I realize that this quite onerous), for both the recalculated lower temperature Run 16 and your COP 5 version of Run 16 ?

I could do that, but you can see that the time between posting calculations is often 2 or more weeks due to the time I have available.

So yes, I can do it and am prepared to do it , but don't expect a quick response. (Also the comming two days and the weekend I am not able to spend time on it)

If you want to continue faster you could take for example the dummy run spread sheet and modify it yourself. (they are not protected and you are in my opinion capable enough to do it)

Let me know what you prefer.

Then it will easier to see what the simulation is doing. The Ribs can be treated as a single section for now.

They must be treated as a single section since the Lugano report gives only accumulated data for the active runs, not the detailed data per section.

And for the rods no average temperatures are given.

So we have to find a solution for that if we want to include the rods in the calculations

I am more interested in chasing why ~ 24% of the electrical input goes into the Rods and Caps, but ~ 50% goes out, when ~76 % of the input power goes into the thin Ribs section in the Dummy cases.

See the comment above about the wire distribution.

Paradigmnoia

Actually, does the View Factor adjustment take into account the normal-viewed area of the tips of the ribs versus the V-cavity area?

The view factor is the value which indeed takes into acount the change in view from the valey to the top.

The V-grooves/Ribs are not ideal triangles with tips vanishing to zero on the outer surface.

The view factor given is for ideal triangle shaped fins

The normal-viewed area is more like a series of rectangles for the valleys, separated by thin rectangles which are the rib tips. So the valleys are perhaps only 90% of the normal view area, while cooler tips are perhaps the remaining 10% (for example).

As stated, the view factor was based on triangle shaped fins.

If the shape differs somewhat from an ideal triangle, i think that on the average the value will still be about the same. For example if at the bottem the shape is more rectangular it will cause the view factor between the fins increase in the valey, but then the shape at the top must bend more inwards, decrasing the view between the fins there. So this wll somewhat average out.

Paradigmnoia

Perhaps the apparent low power, after reducing the T from near 1400 C to near 800 C, is due to not correcting the emissivity for the ribs in the output power calculation?

It is not clear for me what you are meaning by the above text.

The 800 degree C is the real temperature if we assume that incorrect emissivities are used.

The power for my simulation was derived from that the total coil wire power was 906.31 - 41.25 = 865.06 Watt,

But using your proposed wire configuration of 4 cm in the rods, 4 cm under each end cap and under the ribs 10 windings, then the total power under the end caps and under the rips (the power of the ECAT) becomes 735 Watt.

With that power assigned to the FEM model i get a surface temperature of just over 600 degree C in the simulation

So the power of 735 Watt is not from a simulation, but based on the applied power and taking into account the distribution of that power.

If this does not answer your question maybe you can explain your reasoning/question a little bit more ?

IE: the already-high in-band emissivity of the alumina (~ 0.95) gets a small bump to near 0.97 , since there is not so much improvement to be made in the emissivity in that band by self-reflection/absorption (the rib valleys are a weak cavity).

Correct

In the Optris IR spectral sensitively range, the reactor is almost a black body already. However, the total emissivity used for radiant power calculations could have greater overall improvement to the emissivity made by the ribs, so that maybe a total hemispheric emissivity of 0.65 (for a flat surface) could become 0.75 when ribbed, boosting the output power at the lower re-calculated temperatures.

The boost of the emissivity is defined by the view factor, which is fixed for the ECAT and the emissivity, which is variable and dependent on the temperature.

The new emissivity is e /(1- Fv*(1-e)) with Fv = .42812

With the formula applied the emssivity of .650 becomes 0.765

However with only heater power applied, the increased emissivity must be compensated by a lower surface temperature in order to keep the total power the same.