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Donovan Brooke group: registered: GMT - 05/09/2007	Subject: Combination Top load annealer / fuser
	This new clam shell kiln will fill a void in EDC's studio. It will wide bowls to be annealled as well as allow for some serious kiln worked glass... such as thick glass steps, sinks or what have you. The interior dimensions are: 42 X 40 X 14 inches. The pic below is the starting of the shell.
	DATE: GMT - 05/16/2007 | 02:43 PM         |     QUOTE
Donovan Brooke group: registered: GMT - 05/09/2007	Subject: Re: Combination Top load annealer / fuser
	photo of the hinge...
	DATE: GMT - 05/16/2007 | 02:44 PM         |     QUOTE
Donovan Brooke group: registered: GMT - 05/09/2007	Subject: Re: Combination Top load annealer / fuser
	Counter weight system is installed and sheet metal floor is being put in.
	DATE: GMT - 05/18/2007 | 05:07 AM         |     QUOTE
Donovan Brooke group: registered: GMT - 05/09/2007	Subject: Re: Combination Top load annealer / fuser
	The shell is pretty much finished, minus the front door handle.
	DATE: GMT - 05/18/2007 | 05:08 AM         |     QUOTE
Donovan Brooke group: registered: GMT - 05/09/2007	Subject: Re: Combination Top load annealer / fuser
	Hello, I thought I'd share some logic about how to come up with the power configuration for a kiln. The interior width / depth (inside the refractory) will be: 41" X 37".. or about 3.41' X 3.08'. Element config. will be 3 phase... probably 2 in the lid, 1 around the side. I'll go over how I figure out how to impliment the heating aspects. Please take this info as information from a do-it-yourselfer and know that I'm not an electrician, so there could be mistakes. Cubic Feet of heating space: With an interior height of 15" and the widths above, it appears that I have a volume of 22,755 cu. inches or 13.16 cu.ft. to heat. cu. inches: wXhXL Cu inches to Cu. feet: 1 cubic inch = 0.000578703704 cubic feet Also, I will take off a couple of cu. ft. to take in account that I will have a couple of 2" thick vermiculite shelves or other shelving in the kiln... so rather, the final cu.ft. would be around 12. So, lets say that I'm shooting for 1,500 watts per cu. foot. This kiln would then be a 18Kw kiln. (1,500 X 12) Now, to complicate this, I am going to run 3phase to this oven! So, how do I select the right elements to go inside this oven to reach the desired 18kw you ask?? Since I currently have a spool of 11ga. Kanthal A-1 in house. I am going to take a look at that first. Note that this is probably over-kill for this application... even if I can get the numbers to work. So here we go: The first thing we need to find out is, how many ohms of wire do we need to hit that 18Kw mark. The formula for watts in a 3phase delta is: P (watts) = 1.73 * I * E (voltage) So we know two of the variables... 18,000 = 1.73 * I * 250 (my shop has 250 v. legs) reducing.. 18,000 = 432.50 * I 41.61(amps) = I Alright, so, now to ohms... which I'll use the formula to figure out current. I(line) = I(one phase) * 1.73 or: I(line) = (V\R(one phase)) * 1.73 Plugging in the variables: 41.61 = (250\R) * 1.73 10.39(ohms) = R So, we now have the required ohms for each wire... 10.39. The idea now is that we need to find the right wire to *fit* into the kiln we are making nicely. In this process, we need to keep a few things in mind. Coils of wire seem to work best at a stretch of 3 to 1... meaning, you could fit two widths of wire in between the stretch in the coil. To go from here, one needs to have some reference of the wire specs on hand. I will end this post to go and download kanthals specs on their 11ga. A-1 wire. I'll then continue after I have some reference. Donovan
	DATE: GMT - 05/24/2007 | 09:53 AM         |     QUOTE
Donovan Brooke group: registered: GMT - 05/09/2007	Subject: Re: Combination Top load annealler / fuser
	Alright, ready for round two. For Kanthal A-1, I went to their site and found the handbook with all the specs that I need. http://tinyurl.com/2g7fsu Then click on the "Appliance Heating Alloys" of your country or preference. We are trying to figure out how many feet of this 11 ga. wire we would need to fit into the kiln to hit the numbers we have come up with for power requirements. ** one note ** I am under the belief that Kanthal is selling 12ga. wire when you ask for 11ga. So dbl check and adjust accordingly! Kanthal states that their 12ga A-1 wire has a resistance of .0773 ohms per foot. Ohms per feet can vary slightly with Kanthal depending on temp.. but not enough for me to worry about. From this, we can calculate how much wire will be in each element. 10.39 = .0773 * x 134.41 feet = x The reason why we need to figure out feet is to get an idea if we will be able to fit the element in the kiln nicely. We wouldn't want to start winding 10ohms of element wire if only to have to trash it because it won't fit in the kiln! So, 134.41 feet seems like a lot... but remember, we are going to coil this wire. Somewhere there is a formula to figure out coiled length.. but for me, it's easier to wind an inch and estimate from there. I get about 17" of wire to a coiled inch. Using a basic ratio.. total feet converted to inches: 134.41 * 12 = 1,612.92 17x=1,612.92 x = 1,612.92 divided by 17 x = 94.87" 94.87" dividedby 12 = 7.905 feet of coiled wire (unstreatched). I can actually, visually double check this as I currently have coils that are almost 5 ohms in this wire... which I'm off about a foot. :-) However, what is off by a foot can be made up in the stretch... which is what we come to. THE STRETCH. For average's sake, lets say that this 10.39 Kanthal element comes to 7.2 ft. Ideally, we would like to have a 3 to 1 stretch.. meaning two diameters of this element wire would fit between the stretch. 7.2 * 3 = 21.6 feet. Will we be able to fit 21.6 feet, 3 times, inside this kiln? The first element we will assess is the bottom element. (remember, the plan is to put two elements in the lid, and one around the inner circumference of the base wall. reminder: 41" X 37" I will note that final dim.s are a bit larger because of the element grooves.. however, the idea that the pig tales going out the back will have a gap, sort of cancels this out. So, the inner circumference is 156".. which is about 13 feet. Obviously this is not going to work with a 3 to 1 stretch. It is more towards a 2 to 1 stretch. There are options here. We can continue to try to fit the 11 ga. elements in, or we can go to another type of element.. more specifically, a smaller ga. element. One solution would be to run two 7 ohm coils around the circumference. A 7 ohm coil would have a 3 to 1 stretch of closer to the 13 feet we need. This would mean our total power would go down slightly to about 1113 watts per cu. ft. But actually, that is about right as 1500 watts per cu. ft. is overly strong. So to be clear on my thoughts. An element solution for this kiln would be 3 sets of 2 - 7ohm elements. 1 set around the base circumference, and two sets (4 elements) winding around in the top lid. This is a very valid solution and probably what I will do.. since I have the element in stock. I hope this helps with the cauldron of questions that is often a part of element design. Donovan
	DATE: GMT - 05/25/2007 | 02:03 PM         |     QUOTE
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