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Author: Subject: Homebuilt 1300°C tube furnace
Magpie
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[*] posted on 9-3-2008 at 10:07


Tim, I hear what you are saying. I'm not comfortable with the situation either. It does bother me, however, that the heat sink is designed for the TO-220 component style and has the 2.6C/w rating that indicates that it will do the job. Yes, the circuit board and the case are plastic.

If the sink gets too hot then would an externally mounted fan, blowing through the holes drilled in the enclosure possibly suffice?

If I do have to mount the quadrac on a bigger heat sink outside of the box it too will have to be "enclosed" for safety as the terminals are at 230VAC.
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[*] posted on 9-3-2008 at 14:48


It would, but then you have the hassle of a fan, which must run at all times, is noisy (compared to no noise whatsoever), attracts dust, may need another power supply, and probably other things.

Even a small fan moves a lot more air than convection does; one of those dinky microprocessor fans would suffice. Poke plenty of holes (or an opening covered with a fan guard), and use a loose filter type deal if you'd like to keep it somewhat cleaner at the expense of some airflow. And, of course, what goes in must come out... you need two sets of holes!

If you mount it outside, can't you mount it with insulating hardware, so the thyristor faces the chassis and no voltage is exposed? Some sheetmetal could block the ends of the slot the thyristor mounts in, so even a very determined pinky finger couldn't reach the leads.

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[*] posted on 9-3-2008 at 16:41


I ran temperature tests this afternoon with the sink mounted in the enclosure. Conditions were about 20C ambient and the power draw was 1800w (7.5a). With the lid off a steady-state temperature of 46C was reached after 25 minutes. I then put the lid on. After another 9 minutes the temperature had risen to 60C and was still rising. At this time I terminated the test.

I'm going to have to exercise one of the options we have been discussing.
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[*] posted on 10-3-2008 at 14:06


The quadrac/heat sink was remounted about 1" (2.5cm) outside the enclosure. When operating at 8.7a (2000w) the sink temperature quickly stabilized at 36C. Ambient was 27C and rising:D. I will have to make sure I don't get my "pinky" in the wrong place, however. It will help that nobody else will be operating this puppy but me.
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[*] posted on 10-3-2008 at 15:40


Just put some duct tape over it. :D



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[*] posted on 19-6-2008 at 18:19


In consultation with garage chemist I have completed the tube furnace construction that I put aside last fall after completing the power control thyristor circuit. This furnace could almost be given serial no. 2 of the garage chemist design shown upthread.

Differences are as follows:

1. I used 8" (200mm) stove pipe vs GC's 4.8" (120mm) stovepipe.
2. I used an alumina tube vs GC's mullite tube.
3. I used ITC-100-HT to coat the Kanthal wire vs GC's Al2O3/MgO mix.
4. I used 1" (25mm) Kaowool insulation vs GC's 25mm insulation that he suspects contains some ZrO.

My design is for 2kW, same as GC. I used 18 ga Kanthal A-1 wire (1.02mm), GC used 1mm Kanthal A-1.

Edit: My tube is 40mmx34mmx500mm, GC's is 40mmx32mmx500mm.

(GC please make corrections where required.)

Below is a picture of the furnace in operation. It reached 1252C and rising before I shut it down. Maximum amps I put to the furnace was 9. At the end, temperature was rising at about 1C/min with Kaowool plugs in both ends. I am quite pleased with the performance and saw no reason to push my luck by taking the furnace to higher temperatures.

It should be noted that even though I had an extra 40mm of insulation the stovepipe was too hot to touch at the higher temperatures.

I will post 2 more pictures and comment about my use of ITC-100-HT.

[Edited on 19-6-2008 by Magpie]

[Edited on 20-6-2008 by Magpie]

furnace at 1252C.jpg - 59kB
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[*] posted on 19-6-2008 at 18:33


Here's a picture of the tube geing coated with ITC-100-HT. I had never worked with this before. The directions on the bottle said "Add water and mix thoroughly." That's all. So I added just the minimum amount of water to make it workable and brushed it on with a 1" paintbrush. It is sticky and was easy to apply, not dripping off at all. I just kept applying it until I had a thickness of about 5mm. I let it dry for about 3 days at room temperature.

putting on ITC-100-HT.jpg - 84kB
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[*] posted on 19-6-2008 at 18:38


This last picture shows the ITC-100-HT after drying. As you can see it cracked all over. But it stayed on and never spalled off. Dryed pieces of waste coating showed that the product has good strength.

cracked ITC-100-HT.jpg - 58kB
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[*] posted on 19-6-2008 at 21:05


Beautiful! Great work!;)

I am very pleased that your project became such a success. So my design has proved to be workable even with somewhat different components.

Using the ITC-100 paste was a very good choice, I see. My Al2O3/MgO mix became very powdery and flaked off in some spots. If I wanted to build another tube furnace, I would definately not use this again (I'd probably use calcium aluminate cement, even though this would be good to 1250°C only).
ITC-100 does not seem to be easily available here.

A question: what kind of heatup rates did you get with your furnace on full power? I get over 100°C/min at low temperatures, and at 1000°C still 60°C/min.
You said you got 1°C/min at over 1200°C- I think mine still heats up much faster at those temperatures!
Does your alumina tube + ITC-100 have such a large heat capacity, or are there other reasons, like heat loss?

My mullite tube has quite a low thermal conductivity. I remember alumina ceramic having a much higher conducitvity. Does this lead to increased heat loss at the ends?
The free ends of the ceramic tube on my furnace stay below 600°C (no visible glow) even at 1250°C inside the furnace.
Is that different with your alumina tube?




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[*] posted on 19-6-2008 at 21:31


Thank you gc and yes this shows that your design is robust. But I'm sure that the selection of materials is still critical because of the high temperatures.

Yes, the ITC-100 was nice to work with. Time will tell how it proves out in the long run. Perhaps someone with more experience can tell us how to prevent it from cracking. How thick was your Al2O3/MgO layer?

The relatively slow heat up rate is something that I noticed also as I remember you saying how fast your furnace heats up. I have to think that you are right about the tube/coating heat conductivities being higher. My Kaowool conductivity may also be higher. I will see if I can find some conductivity data for these materials.


Quote:

The free ends of the ceramic tube on my furnace stay below 600°C (no visible glow) even at 1250°C inside the furnace.


This was true for me too.

[Edited on 19-6-2008 by Magpie]
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[*] posted on 20-6-2008 at 09:23


I would've used much more kaowool than you did Magpie. I think 2" as a minimum.

I have a forced-air propane furnace that I've lined for 1760C/3200F and I used 3" of 2400F kaowool and 1" of the 2700F material. That is topped off with 3/4" inch of 3200F and an ITC wash. It is a large furnace, perfectly accommodating a #16 SiC crucible. Insulated as it is, if I hold that furnace at about 1600C for an hour or so, the outside of the furnace becomes hot enough to melt the rubber off of my shoes! Around 750C for aluminum casting, the shell is not warm at all.




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[*] posted on 20-6-2008 at 10:51


Quote:

I would've used much more kaowool than you did Magpie. I think 2" as a minimum.


Fleaker, I don't understand this comment. The alumina tube w/coating is 1"radius and the stovepipe has a 4" radius. My Kaowool thickness is therefore 3".
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[*] posted on 20-6-2008 at 11:23


GC I found the data for the 99.5% alumina. First of all I have an ID of 34mm vs your 32mm. Therefore your wall thickness is 4mm whereas mine is 3mm. This means your tube has nearly 33% more mass (assuming equal densities) than mine.

The alumina has a sp. gr. of 3.8.

Its thermal conductivity is 50 BTU-in/(ft2-hr-F) at 800C, which, if I have done the conversion correctly, is 7.2W/(m-K).

I'm still looking for the conductivity for the 8 lb/ft2 Kaowool.
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[*] posted on 20-6-2008 at 13:09


Here's a nice data sheet that compares mullite to alumina:

http://www.ortechceramics.com/material_compare.htm

It shows the mullite thermal conductivity at 3.5 W/(m-K) vs a value of 30 W/(m-K) for alumina. But this is at 20C. At 800C the k for the alumina drops to 7.2 W/(m-K). I don't know what happens to the k for mullite at 800C. But it is likely still no more than half of that for the alumina.

2nd Edit: This website gives k = 3.3W/(m-K) for mullite at 800C:

http://www.techceramics.com/pdf/mulliteMV20.pdf

[Edited on 20-6-2008 by Magpie]

[Edited on 20-6-2008 by Magpie]
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[*] posted on 20-6-2008 at 15:51


Quote:
Originally posted by Magpie
I'm still looking for the conductivity for the 8 lb/ft2 Kaowool.


Of course, you would have to integrate over the radius, because the thermal gradient cross-section has cylindrical geometry and is therefore expands with radius, while at the same time the temperature falls and the kaowool becomes a much better insulator. If you want real numbers, that is. As I recall, in the limit it comes out logarithmic, so you don't gain much by using a lot of insulation around something narrow.

120V * 8A is just shy of 1kW. Evidently, this furnace has an efficiency of roughly 1.3C/W.

Tim




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[*] posted on 20-6-2008 at 16:07


Tim I'm not trying to calculate a thermal resistance. I'm just trying to compare conductivities of my materials to those used by garage chemist in an effort to explain my relatively slow heatup rates.

2nd Edit: The maximum voltage I can put to my furnace using my thyristor control circuit is 230VAC. At 230 volts and 9 amps the power would be 9*230 = 2070W.

[Edited on 20-6-2008 by Magpie]

[Edited on 21-6-2008 by Magpie]
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[*] posted on 20-6-2008 at 21:13


GC I have measured the electrical resistance of my furnace at 18 ohms at room temperature. It should be 26.5 ohms. So instead of delivering 2070W at 9 amps I was actually delivering something like 1458W. This is no doubt caused by internal shorts (coils touching). I suspect this is the main culprit for my slow heatup rate.

It was difficult to keep the wire coils tight and evenly spaced. I tried to correct uneven spacing but the coils wouldn't stay in place. I should have given this more attention, throwing on more nooses or even rewinding.

Could you measure your furnace resistance for comparison?
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[*] posted on 21-6-2008 at 00:30


Not sure if this is of interest, but the resistance of my 2070W Lindberg is 6 ohms. BTW it did not come with the controller that connects to the 4 contacts at the base...if anyone happens to have an extra one...Not that I really need it, because the Watlow 988 controller and Din-a-mite SSR are outstanding little items. Works great on my little electric bunsen even. Glas-Col's too, using their built-in thermocouple, after switching the controller from K to J. Highly recommended.

Quote:
Originally posted by Magpie
I have measured the electrical resistance of my furnace at 18 ohms at room temperature. It should be 26.5 ohms. So instead of delivering 2070W at 9 amps I was actually delivering something like 1458W.


Now, how did you figure your wattage? Please humor someone a little slower than many here with electricity; why isn't the observed potential and current combined with Ohm's thingy relevant to wattage calculation in this circumstance?

[Edited on 21-6-2008 by S.C. Wack]
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[*] posted on 21-6-2008 at 07:55


Quote:

Please humor someone a little slower than many here with electricity; why isn't the observed potential and current combined with Ohm's thingy relevant to wattage calculation in this circumstance?


Yes, Ohm's law is relevant here. E = IR and P=IE, where E is voltage, I is current, R is resistance, and P is power.

By substitution then P = (I^2)R. So for a power of 2000W at 9 amps, the resistance should be:

R = P/(I^2) = 2000/81 = 26.7 ohms

With R at 18 ohms and I at 9 amps, P = (I^2)R = (81)18 = 1458W

For your Lindberg you say you measured a resistance of 6 ohms. Assuming your voltage is 120 volts, then the power calculation would be

I = E/R and P = EI = E (E/R) = (E^2)/R = (120^2)/6 = 2400W. I don't know why the discrepancy with your rating of 2070W, but the calculated result is in the ballpark.
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[*] posted on 21-6-2008 at 09:18


Power dissipated by a resistive load is defined as V*I. If you measured 240V and 9A, you were dissipating 2160W. Evidently, the resistance at that moment was 240/9 = 26.7 ohms. Anything else contradicts either the voltage or current you in fact measured; 9A through 18 ohms develops V = I*R = 162V, far less than the 240V you know you have; likewise, P = V^2/R = 240^2 / 18 = 3200W is a current of 13.3A, which is not the 9A you measured.

The cold resistance will, of course, be much lower than the hot resistance. A factor of two difference shouldn't be at all surprising; over the same range, the resistivity of pure iron increases something like tenfold.

If you measure the current draw at the instant of turn on, I think you will find nearly 13.3A. It's a much safer assumption that line voltage is constant rather than current draw!

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[*] posted on 21-6-2008 at 09:33


I'm using a thyristor circuit to supply power to the furnace so don't really know the effective voltage applied to the furnace. I did not have it set at full voltage.

I used a clamp-on ammeter to measure the 9 amps, which I'm assuming is a fairly reliable value.

The resistance of Kanthal A-1 increases 4% from 100C to 1200C as can be seen here:

http://www.kanthal.com/

[Edited on 21-6-2008 by Magpie]
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[*] posted on 21-6-2008 at 10:04


The clamp-on amperemeter only gives a sensible reading if you don't use a thyristor circuit.
The thyristor circuit "chops up" the sine wave of the current and makes an accurate measurement impossible for the amperemeter, which needs a sine wave to function properly (that's what I think, although I don't have a clamp amperemeter. What I've said is certainly true for a normal electronic amperemeter.).
You would have to use an analogous, electromechanic amperemeter in order to obtain the true value of the current.

With your ca. 18 ohms (as you've correctly pointed out, the resistance of Kanthal wire changes only little with temperature), the power of your furnace will indeed be over 3000W, which is currently reduced by the thyristor circuit.

Try running the furnace without the thyristor circuit (not over 1000°C!)- I am fairly sure that you will get a temperature rise of over 100°C/min then.

You didn't use preoxidised Kanthal wire, did you? Preoxidised wire avoids short-circuited windings due to the insulating properties of the Al2O3 oxide layer on the wire.
The wire you used looks bright, so it was probably not preoxidised- this seems to be the reason for the too low resistance.
I did not use preoxidised wire either, but I did not have any touching windings.
The resistance of my furnace was the calculated 26,5 ohms.
Now, after repeated use, it has increased to 27,5 ohms (measured yesterday) due to oxidation of the wire, so the power has dropped to ca. 1926W (not that this would be a big problem, but it demonstrates that the power of an electric furnace that uses resistance wire will slowly decrease with continued use, and heating spirals are always a wear part).




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[*] posted on 21-6-2008 at 10:27


GC thanks for measuring your resistance.

No, I did not pre-oxidize my wire. Never had heard or thought of such a technique.

I know the voltage sine wave is being chopped up but I still suspect that I'm measuring reasonable values for effective current at the higher amperage levels.

I just now plugged in the furnace, turning up the power until I measured 9 amps. Then I measured the voltage at the furnace terminals. It was 140 volts. This would yield a power consumption of 140*9 = 1260W. Again, I know this is just a rough value.

The reason I have been limiting my current to 9 amps is concern over burning out my Kanthal wire.

Edit: Using the two crude measurements of 9 amps and 140 volts R = E/I = 140/9 = 15.6 ohms is calculated. This is not too far off from my measured 18 ohms, which should be solid.


[Edited on 21-6-2008 by Magpie]
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[*] posted on 21-6-2008 at 18:05


@Magpie,

Sorry, for some reason I read
"4. I used 1" (25mm) Kaowool insulation vs GC's 25mm insulation that he suspects contains some ZrO."
and actually thought you'd be fool enough to use an inch of insulation. Reading it over again, I wonder what exactly I was thinking when I read that!

Still though, it seems odd that your furnace is becoming too hot to touch on the outside and that it takes so long to heat up?




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[*] posted on 21-6-2008 at 18:44


Fleaker I was pretty sure you had just misread something, but I wanted to clear it up. ;)

Now that I have realized my lower power input, it is not surprising to me that my furnace heats up much slower.

My shell being too hot to touch did surprise me, but when you think about it the power has to be dissipated somewhere. Garage chemist said his shell might char wood at 1300C. I don't know if mine was that hot but it was getting there. It seems that the commercial tube furnaces all have a secondary shell of perforated metal, probably for the same reason.

An interesting experiment would be to compare the power required to keep our respective furnaces at a selected temperature, say 1000C or 1200C. Getting believeable, or at least comparable, E or I measurements would be tricky, however, since we are both chopping the sine wave with thyristors.
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