Good to have you back in the discussion again !
So..., in other words, since the ribbed device is actually so effective at radiating power, the effective emissivity when calculating the radiated power based on area, when assuming it is a diffuse cylinder, must be very high (?)
In essence we have two effects here
First of all is that you should use in the calculation of radiated energy the real surface area,
That means the area of the tube with fins, not that of the tube without fins.
However thermal radiation of a fin is partly blocked by the opposing fin. That makes the area of the fin less effictive and thus you need to calculate with an effective area of Af * Fbg in which Af is the area of the fin and Fbg the view factor to the background.
The second effect effect is that due to the reflection between the fins the effective emissivity changes (Infinite reflection method). The effective emissivity becomes
e x { 1/(1 - (1-Fbg)(1-e))}. (see also TC's paper)
The total effect is Af x e x Fbg x { 1/(1 - (1-Fbg)(1-e))} , which is the formula shown in
an earlier post.
This makes the finned area more effective in radiating thermal power then without fins. However less effective then the total area of the fins without view factor correction would suggest and not very high, but surely more effective then without fins.
Or the device must be cooler than they said it was, since it is so efficiently radiant it could not get so hot
I have currently no idea if the device was cooler then what the testers reported. However having found the correct view factor is a first step in recalculating the results and see where that ends. After having solved this and other issues it gives me the opportunity to restart my thermal simulations of the dogbone and possibly solve the question between the different results of Lugano, MFMP and the thermal simulations
However there is also possibly another issue with the calculation of the convected power of the stacked rods , which I want to work on before doing new simulations.
Another issue I currently am investigating is why, if I remember well, there was such a large temperature difference reported between top and bottom of a fin. I do not know where and by whom that was reported, maybe you know and can point me to the information ?
LDM
The hot valley vs cooler tip of the ribs is easily replicated. At about 800 C, about 50-80 C difference is typical. Testing the tip temperature is tricky, though. The valleys are much easier to get a thermocouple tightly affixed. A glob of cement to hold the thermocouple skews things significantly when ribs are involved.
I wonder if the thermocouples are partly the cause of the difference in measured temperature. The connecting wires to the thermocouple tip are good thermal conductors, and are thus conducting thermal energy away from the tip, thereby lowering the tip temperature.
Since at the bottom of the fin there is more thermal mass then at the top of the fin the temperature drop due to this thermal conductance through the thermocouple leads will be less at the bottom of the fin then at the top. This will probably cause an (unknown) temperature difference in the thermocouple measurements.
Note also that at the MFMP dogbone retest for a temperature of about 970 degree C, the thermocouple showed around 840 degree C, a 130 degree C difference ! In my opinion also caused by the draining away of heat through the thermocouple wires.
Convection here, however, is the main rib temperature gradient maker.
In general in heat exchangers where ribs are involved, convection will indeed be the major source of heat transfer and thus the cause of the temperature gradient on the fin. Fins in heat exchangers have normally a larger ratio between height and base width of the fin then the fins on the dogbone. Due to this radiation is largely blocked and thus convection will indeed become the major source for heat transfer. (Also because in heat exchangers there is often a forced air flow)
In Lugano the hight/base width ratio is small and thus radiation can play a much larger role then for the larger ribs used in heat exchangers.
The valleys convect poorly (probably very rarified air there, in the extreme valley V location, due to the high heat and small volume) the tips are increasingly in denser, cooler air towards the tips.
This sounds logical, however I have not found this confirmed in documents on fin heat transfer.
The ribs also conduct heat semi-poorly, being alumina, a poor heat conductor in general, so that the highly convecting outer rib area is in competition with conducted heat replenishment. As well as radiated heat transfer to the fins.
While alumina is not a very good heat conductor, it is also not a very bad one either.
I have simulated a single Lugano style rib in the finite element thermal simulation program I use, and for a fin base temperature of 750 degree C the difference between top and bottom of the fin is only 20 degree C, indeed indicating that the thermal conductance is not too bad.
However that simulation does not take into account the possible difference in air flow between bottom and top of the fin you suggest.
Maybe I will be able to adjust that similation for difference in conducted heat transfer from bottom to top and run it again.
If the tips of the fins were only 20 C cooler, then nobody would use them.
That is a misconception.
Newton's law states that :
-----Q = h x A x (Ts - Tamb)
h being the heat transfer coefficient, A the surface area , Ts the surface temperature and Tamb the ambient temperature.
Now consider two surfaces of the same amount of area, the first surface at the bottom of the fin, the second at the top.
If the temperature at the top of the fin is lower then at the bottom of the fin, Qtop will be lower then Qbottom.
If the temperature at the top of the fin is equal to the temperature at the bottom of the fin then Qtop will be equal to Qbottom.
In the last case where the temperatures at the bottom and the top are equal Qtop + Qbottom will be larger then when the temperature at the tip is lower then that at the bottom.
Otherwise stated, the convected heat transfer of the fin becomes better if the temperature drop between bottom of the fin and top of the fin is smaller.
That is one of the reasons why fins in heat exchangers are made of materials with good thermal conductivity such as aluminum and copper. The good thermal conductivity keeps the temperature difference between top and bottom of the fin as small as possible. The other reason being that heat transfer from the base into the fin is large.
More important factors for a good conductive heat transfer are a high heat transfer coefficient and a high surface area.
Radiant heat transfer is the dominant heat transfer mode, over around 750 C. Below that rough temperature, convection does the bulk, and fins are the primary convection improver.
Agreed that at higher temperatures radiation is dominant for heat transfer.
At lower temperatures it depends on the way how the heat is exchanged. With larger fins (fin height 3 cm) there is good conductive heat transfer and less good radiated heat transfer due to the fins largely blocking the radiation.
For the Lugano fins, which have only a small height and where radiation is less blocked and due to the small surface area convective heat transfer is limited. As a result even at the temperatures of about 450 degree C, the radiated heat transfer is still dominant. (See Lugano dummy run data)
I see your point. If the fin tips are too cool compared to the base, then that could indicate that the ribs are probably too tall to be maximally efficient.
It is indeed so that if you increase the fin dimensions that the temperature drop between bottom and top becomes larger and thus the convected heat near the top becomes less. However this is (partly) offset by the increase in surface area.
In practice it is about finding the optimum dimensions, meaning where the convected heat per surface area devided by the material/production cost per surface area is optimal.
On the other hand, if there is no base to tip rib temperature gradient, then heat will not flow into the ribs and convection will not be improved by ribs. (The case of being in a vacuum would be like this). It would merely be a surface area increase radiation improvement.
Totally agree with you
The (heat) current has to flow, however to get a high heat flow at a low temperature drop you use materials with a high termal conductivity.
Stacked tubes convective heat transfer correction factor
During the Lugano test at each side of the reactor, three tubes containing the wires to the heater coils where stacked in a triangular geometry.
The stacking reduced both the radiated heat transfer and and the conducted heat transfer of those tubes.
In order to correct for this reduction the test team applied a factor 2/3 to both the radiated and convected heat transfer.
For the radiation this factor seems to be right since a geometrical analysis shows the view of each tube is for 1/3 blocked by the other tubes (For closely stacked tubes).
Since all tubes have about the same temperature this means that for 1/3 of the surface of a tube radiation can not be exchanged.
The conclusion is then that indeed as far as radition is concerned the area of the tubes are only for about 2/3 effective.
For the convected heat of the tubes the testers applied the same correction factor of 2/3. The question is now if that is correct.
Befor we can answer this question we need to know how the convective heat transfer correction factor of the stacked tubes can be calculated.
A method often presented in texbooks on heat transfer is to calculate the equivalent tube diameter for the stacked tubes.
The method recalculates the diameter of the tubes based on the so called cross sectional area and this new diameter then represents the change in conductive heat transfer due to the stacking of the tubes.
The method is presented below for for the convective heat of the three stacked tubes.
Figure 1
Stacked tubes cross sectional area
Above the cross sectional flow area used in the calculation of the correction factor for convective heat transfer of three stacked tubes is shown.
To obtain the cross sectional flow area we are drawing lines between the centerpoints of the tubes. This gives the light green/dark green equilateral triangle as shown in the above figure.
From this area we have to subtract the parts of the triangle covered by the circles (light green sections).
This leads to the following calulation:
1. Area of triangle
The area of an equilateral triangle is L2 x Sqrt(3)/4, L being the distance between the center of the tubes For 30 mm tubes spaced at a distance of 5 mm L is 30 + 5 = 35 mm
This makes the area of the triangle 530.441 mmxmm
2. Three equal parts of a circle
The area of each circle is ( pi x D2 ) / 4 = pi x 302 / 4 = 706.858 mmxmm
Each part of a circle spans 60 degree, thus for all three circles we have 3 x 60 = 180 degree, or a halve circle.
The area of this halve circle is 706.858 / 2 = 353.429 mmxmm
We obtain the cross sectional area by subtracting the three equal parts of the circle from the area of the triangle.
Thus the cross sectional area becomes :
530.441 - 353.429 = 177.012 mmxmm
Equivalent diameter
==================
The equivalent diameter is defined as the diameter of a single tube which has approximate the same effective area as that of a stacked tube. The formula presented in literature for the equivalent diameter is :
De = ( 4 x Cross sectional flow area) / Perimeter of all tubes to the cross sectional area
The perimiter of a single tube is 2 x pi x radius = 2 x pi x 15 = 94.2478 mm'
The perimeter of the three tubes which fall inside the triangle is halve of the perimeter of a single tube is 94.2478 /2 = 47.124 mm
We can now calculate the equivalent diameter as 4 x 177.012 / 47.124 = 15.025 mm
Or the correction factor for the tube diameter is 15.025/30 = .501
The above correction of .501 was calculated for a distance between the tubes of 5 mm.
For different distances between the 30 mm tubes the factors can be found in the following table :
Distance(mm) De(mm) Factor
3---------------------10.03------.334
4---------------------12.49------.416
5---------------------15.03------.501
6---------------------17.63------.589
7---------------------20.32------.677
8---------------------23.07------.769
As can be seen the correction factor is quite dependent on the distances between the tubes.
Thus finding the correct factor for the Lugano stacked tubes means that we need to find the correct distances between the stacked tubes.
Fin efficiency
For the triangular shaped fins used on the Lugano ECAT the test team applied in their conductive heat transfer calculation a correction.
This correction was based on a plot showing the fin efficiency of a circular fin having triangular profile.
The reasoning behind this correction is that the temperature of the fin drops towards the top of the fin.
Thus the fin for it's surface is less efficient then if the whole fin has a temperature equal to the base of the fin.
The Lugano testers also noted this, they state :
Since the driving potential for convection is expressed by the difference in temperatures between a body and its exchange fluid, it is obvious that the maximum thermal flow for a fin would be had if its entire surface were at the same temperature as its base. However, as each fin is characterized by a finite resistance to thermal conduction, there will always be a thermal gradient along it, and the condition given above is a mere idealization.
Indeed fin efficiency can be defined as :
Actual heat transfer from fin / Heat transfer if whole fin was at base temperature
The formula shows that you have to correct your convective heat transfer with the fin efficiency if you calculated the heat transfer from the temperature of the base of the fin.
However the Lugano team did not use in their calculations the base temperatures of the fins.
Instead they used in their calculations an average temperature of the fin area based on their Optris measurement.
Doing that they already compensated for the fact that the temperature of the fin is not at it's base temperature but instead has a temperature gradient.
Thus they should not have corrected the convective heat transfer with the fin efficiency.
Regarding the stacked tubes, there may, or maybe not, be an error in the Lugano report regarding the treatment of the two sets of Rods.
For the dummy, they used the 2/3 method, (imperfect as it is).
Indeed imperfect, especially for the convection
For the active version, there is no indication that the 2/3 adjustment was applied.
If I can derive what the distances between the tubes where, I can then recalculate the Lugano dummy run and learn what is important for new FEM simulations.
Maybe then it will then also be possible to calculate if the correction was also applied to the active run
I believe that there is no significant gap between Rods in the bundle of three. They are cemented with blobs of ceramic to hold them together as a bundle.
I wonder if they where cemented.
If you look to figure 2 in the Lugano report, you will see that the ECAT itself is resting on what I believe to be high temperature thermal pads.
The picture is not of a high quality but what I think to see that it are actually two pads.
If you compare the hight with the hight of the end cap, the hight is about 6 mm.
That would then make two pads of 3 mm, or about 1/8" thickness which seems to be a standard for thermal pads. ( Both 1/8" and 1/4")
If that is the case then they could also have used thermal pads between the tubes.
But I agree with you that it looks more like they where cemented.
Cementing them with a gap would be unnecessarily complicated for little to no gain, and would be much more fragile, IMO. There should be a good photo of the cables exiting the end of the Rods available, (possibly the high definition photo by Hoistad).
If you look at your second zoomed picture, you can see that there are indeed gaps between the rods.
Can you give me a link to the photo by Hoistad you are referring to ? (Or any other higher quality photographs)
In the meantime I am trying to derive some projective information from the pictures in the report which will make it possible with 3D CAD to calculate placement information of the tubes.
Will at least give it a try.
Thanks !
If I can derive what the distances between the tubes where, I can then recalculate the Lugano dummy run and learn what is important for new FEM simulations.
Maybe then it will then also be possible to calculate if the correction was also applied to the active run
Just to clarify a bit about this:
The calculated power of the Rods for the active portion of the demo is displayed in the Lugano report (Table 5, page 20).
They would have had to have applied the 2/3 factor to each displayed value before inserting them into this table.
It is not applied later, as shown in the Rods column, Table 7, page 22, where the total power from all segments is calculated.
Just to clarify a bit about this:
The calculated power of the Rods for the active portion of the demo is displayed in the Lugano report (Table 5, page 20).
They would have had to have applied the 2/3 factor to each displayed value before inserting them into this table.
The correction factors are not dependent on temperature and as such you may apply them to the totals.
factor * A + factor *B + ...... + factor * Z = factor* (A + B + .... + Z)
However since the factors for convective heat and radiated heat are most likely different, you have to apply them seperately, not to the total of convective and radiated heat.
It is not applied later, as shown in the Rods column, Table 7, page 22, where the total power from all segments is calculated.
Depends, maybe they applied the correction before inserting the data in the table.
That is again something we probably never will know.
However lets first concentrate on the dummy run.
My priority is to get the dummy run calculations right so that we know more precise which part of the energy was dissipated by the rods and which part by the ECAT itself.
I hope that that then can be the starting point to find out if the heater extended under the end caps by comparing the thermal profile of the dummy run with that of simulations.
With the heater layout confirmed we can then continue with other analysis.
Thomas Clarke's Band emissivity curve
In his report Thomas Clarke calculates in his Python program the in band emissivity curve for Alumina.
While his Python program presents this curve in a separate window, the curve was not included in his document.
For those interested I post it here.
