Hey

last summer I've made concrete furnace with a friend, powered by coal, to melt aluminium. It turned out pretty cool. We did some hand castings and stuff.

This time I wanted to step it up a bit and make it powered by electricity.

The main reason is electric furnace will also be able to heat proffesional casts (like Kerr Cast material) according to their profile with temperature control, which enables making good-looking castings.

All the other reasons, like coal-handling, weight, preparation time also prove electric furnace advantage.

I've got some questions regarding the "design", though.

This time, I've decided to make a truly engineering approach towards this build

I'm assuming a temperature of 1000*C as this should easily melt alloys I have, plus it gives some heat capacity margin when pouring to a form.

Assuming I want to heat

**just** 0.6L of aluminium (I assumed 0.45L of alu and 0.15L for it's crucible, very roughly).

`903 J/(kg*K) * 2702 kg/m^3 * 0.6L ~= 1500 J/K`

Which gives, for 1000*C:

1.5kJ/K * 1000*C ~= 2MJ

(the little fail here is, I took specific heat at room temperature, but should be OK...)

Now, providing 2kW just to aluminium:

2MJ / 2kW ~= 17 min

Seems doable.

So now I get to thermal insulation, which seems to be the most important part of the furnace.

I've used special insulating bricks before not to damage the lawn. I've looked at them again in a shop and they specify heat coefficient.

Basically there are 3 bricks available in my area (10cm, 7.5cm and 5cm thickness). They have 1.4W/(m^2*K), 1.9W/(m^2*K) and 2.8W/(m^2*K) coefficients respectively.

If I divide their coefficients by respective thickness I get 0.14W/(m*K) for each of them, which seems to be heat conductivity of the material.

Anyway, the coefficient is provided for room temperature. Somewhere, I've found it can change by 50% at 1000*C.

So now I calculate losses into the walls (assuming they are 25x25cm as well as height):

`(1.4W/(m^2*K) + 60%) * (1000*C) * (25cm)^2 * 6 ~= 1070W`

So I need a heating element of >3kW.

For example, at 14Amps 230VAC, I get 3.2kW.

Unfortunately I'm limited to 16A per phase and 25A in total. It's a house not a workshop

Now I need a heating wire,

`R = (230V)^2/(3.2kW) = 16.5 Ohm`

Or I could go with two wires at

**5kW total**:

`R = (230V)^2/(2.5kW) = 21.2 Ohm`

I've already bought 20 meters of

**Kanthal A1 type wire** (operating temperature 1400*C, melting point 1500*C), 1mm diameter.

It has 1.85 Ohm/m, so 9 meters should be enough.

Or will it be?

I can't find any info about the max current, or power per meter for that wire even though the datasheet is pretty awesome.

What I did found though, is a "Maximum recommended surface loads" (W/cm^2) chart in Kanthal industrial type datasheet (

http://www.kanthal.com/Global/Downloads/Furnace%20products%20and%20heating%20systems/Heating%20elements/Metallic%20heating%20elements/S-KA041-B-ENG_2011-09.pdf).

From the other Kanthal home appliances datasheet (

http://www.kanthal.com/Global/Downloads/Materials%20in%20wire%20and%20strip%20form/Resistance%20heating%20wire%20and%20strip/S-KA026-B-ENG-2012-01.pdf):

I can read the surface area of my 1mm wire: 31.4cm^2/m.

9 meters of that wire will have 282.6cm^2, which at 3.2kW gives 11 W/cm^2.

Seems too much according to the datasheet, but I don't think this is the way to calculate it (take the surface of the wire itself). Or is it?

Anyway, let me know what you guys think,

is there any righteousness in this approach and my very rough calculations?

Thanks for reading "TL;DR" post