Tom Yocky on thu 19 jan 06
Hello all, I am tinkering with making myself a small test size electric =
kiln using leftover/surplus materials. I have a good amount of fiber =
blanket left over from a home made raku kiln, and a partial spool of =
kiln element wire from when I last made replacement elements for my =
standard size kiln. =20
My question is that I'm trying to figure out how much element wire do I =
need to heat the small space in my test kiln. The test kiln will be 12" =
diameter, 12" tall (aprox .75 cubic feet inside area) =20
I forget the brand name of the wire I have on hand, but it is 14 guage, =
.210 ohm/ft (according to the label)
I have Fredrick Olsens "The Kiln Book" 3rd ed. which has a section that =
covers calculating wire elements but it's difficult for me to understand =
what he's talking about.
I'm hoping somebody out there with the right technical knowledge can =
walk me though determing the length, ohms, and guage of kiln element =
wire I would need to heat that small space to cone 6=20
Thanks
Tom Yocky
Arnold Howard on fri 20 jan 06
Tom, you might be better off buying a used test kiln. You can find them at
garage or estate sales or in the newspaper classifieds. You can obtain them
for a song. I've seen used kilns here at the factory that people dropped off
for repair and never returned to pick them up. They sell for the cost of the
repair. One time I rescued a used kiln that someone was hauling to the
trash. We gave it to an employee.
On the other hand, your intention might be to learn about kilns by making
your own. When I was 18, I rebuilt several VW engines because I was curious
to know how they worked.
Your kiln size is probably too large for 120 volt unless you want to fire
only raku. So you should first determine voltage. Once you know size and
voltage, you might want to buy elements for that type of kiln.
Sincerely,
Arnold Howard
Paragon Industries, L.P., Mesquite, Texas USA
ahoward@paragonweb.com / www.paragonweb.com
----- Original Message -----
From: "Tom Yocky"
My question is that I'm trying to figure out how much element wire do I need
to heat the small space in my test kiln. The test kiln will be 12"
diameter, 12" tall (aprox .75 cubic feet inside area)
I'm hoping somebody out there with the right technical knowledge can walk me
though determing the length, ohms, and guage of kiln element wire I would
need to heat that small space to cone 6
Bob Masta on sat 21 jan 06
Regarding element calculations, note that the current
drawn will depend upon the total resistance. Let's
say you are planning on using standard 110/120 V
and your breaker and circuit can reliably deliver 15A
Then the total resistance must be 120 / 15 = 8 ohms.
Now if your element wire is 0.210 ohm/foot, you will
need 8 / 0.120 = 66.67 feet of wire. That assumes
the 0.210 value is *hot* resistance. It will be lower
when the element is cold, so the initial inrush current
will be higher than 15A. Might need a slow-blow circuit
for this, or a circuit rated for the extra amps.
That was the easy part. Now you have to determine
how much insulation is required to get your kiln up
to the desired temperature. In theory, with enough
insulation any amount of power will work, but reality
usually rears its ugly head here.
Total power P = V * A = 120 * 15 = 1800 watts.
To find the temperature this will reach, you need to
know the total inside surface area, the thickness of
the insulation, and the thermal conductivity K of the
insulation AT TEMPERATURE. (That last is very
important, since conductivity always increases with
temperature, and a lot of specs are for relatively
low temperatures.) I prefer to work in the metric
system, which works out nicely with Watts,
so that's how I'll show this.
The formula for Celsius degrees is
C = Watts * Thickness / (Area * K)
where Thickness is in meters
Area is in meters squared
thermal conductivity K is Watt/m*C
You will often see K given in BTU*in/hr*ft^2*F,
which can be converted to the above by dividing by 6.9.
Let's say you are using Kaowool at 8 pounds per
square foot. According to the info I have, it's
conductivity at 1800F (highest listed) is 1.69 BTU(etc),
which is 0.24 Watt/m*C. (Less-dense Kaowool has
higher conductivity. For 3 pounds/ft^2 it would be
about twice this.)
The area of a 12 inch diameter x 12 inch long cylindrical
chamber is pi * r^2 for top and bottom, or 113 in^2 each.
The wall area is pi * D * l = 452 in^2, so the total
area is 113 + 113 + 452 = 678 in^2. There are 39.37
in/m, or 1550 in^2/m^2, so we have 678/1550 = 0.437 m^2
Now you can either try plugging in various thicknesses
to see what temperature you can hit, or if you know the
temperature you want you can solve for thickness.
Let's say you want cone 6, about 1200C.
Thickness = C * Area * K / Watts = 1200 * 0.437 * 0.24 / 1800
= 0.07 meter = 2.75 inches of 8 pound Kaowool, assuming
the K value for 1800F holds up at 1200C = 2200F. Thicker is
better!
You may note that nothing in the above gives any indication
of how LONG it will take to reach temperature, and in fact
these calculated values are the minimum to just reach it
after a looong time. So again, more insulation is better.
More power would be better, as long as your wiring and
breaker can handle it. (It would take less element wire,
as well, if you are going to do this all in one element.
Or you can put 2 in parallel, with each having twice the
total resistance required.)
Making the kiln smaller would reduce the area, thus increasing
the speed of firing and/or maximum temperature.
Heavier loads of pots will take more time to heat, but will
still reach the same temperature.
Hope this helps!
Bob Masta
potsATdaqartaDOTcom
| |
|
