# How thick do I need to make the insulation in my homemade furnace?

(This is probably a simple question, but I've been staring at it so long I'm not sure I know how to do it any more. I have an Excel spreadsheet with like 3 different ways to calculate this, and they all give different results, so I know I've screwed up somewhere.)

I would like to build a do-it-yourself 1700-degree-Celsius (though will settle for 3000F) electric muffle furnace because I like electricity and playing with fire. I am trying to get an idea of the cost, which means I need to know how thick the insulation need to be.

Unfortunately, materials rated for those kinds of temperatures tend to both be expensive and have relatively high thermal conductivity values, so I plan to use enough high-temperature insulation to reduce the chamber temperature to about 1300C/2400F, surrounded by cheaper, lower-conductivity insulation to bring the exterior temperature down to the target. I want the inside of the furnace as large as I can affordably make it, so instead of a rule of thumb I'd like some real equations I can play with the parameters of.

• The furnace will run on 120V AC, with a target power dissipation of 1200W.
• For convection calculations, I am assuming a 10 W/m^2K heat transfer coefficient as this will be free-standing. I also assume room temperature is 21C and I'd like to target a maximum of 43C on the exterior surface.
• My current calculations use a 12" x 12" x 8" furnace chamber, though I'll probably settle for whatever it turns out I can afford.

I am considering the following refractory materials as insulation:

• Kast-O-Lite 30 LI Insulating Castable: Max 3000F/1650C, max ~0.75 W/mK
• Alumina Fiberboard: Max 3100F/1700C, max ~0.45 W/mK
• NC-28 Insulating Fire Brick: Max 2800F/1540C, max ~0.44 W/mK
• NC-26 Insulating Fire Brick: Max 2600F/1430C, max ~0.32 W/mK
• NC-23 Insulating Fire Brick: Max 2300F/1260C, max ~0.27 W/mK
• 8# Ceramic Fiber Wool: Max 2000F/1100C, max ~0.3 W/mK

(I'm thinking about using Satanite, ITC-100HT, and ITC-296A as coating materials, but I can't find data about how much that would improve the conductivity ratings.)

I'm also pretty new to this, so any advice at all is appreciated. (Hell, maybe I'm overthinking this, the ratings mean nothing, and I should just put "a bunch" of insulation in an see what happens.) I plan to use MoSi2 elements and a nitrogen atmosphere, and will be sintering various powdered metals, including steel. I'll be slapping a PID controller on it for controlled ramp-soak phases.

• Exterior temperature is nowhere near as important as thermal conductivity. Compare what happens if you touch a metal surface at 150 Celsius to touching a Space Shuttle tile at 2000 Celsius. Aug 21, 2017 at 14:53
• You may be right here. Presuming 1200W at steady state, a 1 ft^3 cube interior, and 6 inches of insulation, heat flux at the exterior surface is ~540W/m^2. NIST's fire dynamics page says that a sunny day is 1000W/m^2, so this may be insignificant. Does this pass the sniff test, and would you be willing to put your hand on something with that kind of heat flux? Aug 21, 2017 at 15:14
• Well, again, I'll put my hand on a log in bright sun, but not the hood of a car. It's all in the conductivity Aug 22, 2017 at 12:32
• Put your hand on a black car in bright sun compared to a white car... Aug 24, 2017 at 8:40

First off, the design problem is not trivial and it looks like you have done your homework. On high temperature, small vessels like this you need to consider the 3 dimensional transfer of heat (the wall area is increasing as you move out). Or you can be conservative and much more easily calculate it with all the insulation acting at the outside dimensions.

So lets say you want a 6"x6"x6" firebox with 6 inches (0.152m) of insulation surrounding. Your external surface area of the 18"x18"x18" box would be 18*18*6walls = 5184in^2 or 3.4m^2

At a 1700-20 =1680C differential and 0.4 W/mK Thermal Conductivity Equation

1680K * (3.4m^2 / 0.152m) * 0.4 W/mK = 15kW worst case (Notice how units cancel)

To check ourselves, this commercially available 6"x6"x6" unit uses 4kW. So our numbers were probably a bit too conservative. Calculating it at the 3/4 thickness (14x14x14) gives us 0.76m^2 and 3.4kW. This is likely pretty close. The manufacturer will make sure their amount of heating is some amount greater than the heat loss at maximum temperature so the max temperature can be reached in a reasonable amount of time.

Radiation losses will also become significant at 1700C. A stainless or aluminum reflective shield should be used for your outer layer.

Stacking different rated materials is possible, by running a duplicate calculation (furnace in a furnace). Make sure to have lots of safety factor on the maximum material temperature, as those conductivity ratings are not always very close. They will often provide you "at least" the value they are publishing. More insulation is better for most customers, but if you are counting on that external oven to only get to a certain temperature; too high of thermal resistance will result in that material over heating and failing.

Again, a good test of what is possible in a DIY type environment is to look at what is commercially available. Look at the watts vs insulation vs firebox size of what vendors are selling. Then remember they do this for a living so your diy design (while much cheaper for you) will at best require 50% more power, 50% more insulation, or achieve say 200F less temperature.

Sounds like a fun and challenging project! Targeting a 1300C furnance might be a better starting point if you have not built any high temperature devices before. Good luck!