# Calculating Insulation Thickness

1. Jan 6, 2013

### Voltux

I'm trying to calculate insulation thickness for a foundry, and can't figure it out after spending a few hours searching.

Refractory: Kast-O-Lite

Material Required: 1.44g/cm^3
Thermal Conductivity: 0.65 W/m*C

Maximum inside temperature is 1648*C

I figured there would be a minimum of 10,440 BTU/3057 Watts Required to melt 4.16L of Aluminum

Inside Dimensions "Empty Space" would be 16cm Diameter x 23cm Tall
Outside Dimensions: 31cm Diameter x 44.5cm Tall

The closest thing I found to this would be Insulation thickness calculation for a pipe - EngCyclopedia

However, I cannot figure out how to determine r2

Q=2*pi*k*N*((T1-T2)/ln(r2/r1)

T1=50*C
T2=1648*C
r1=0.1524m
r2=??
k=0.65 (W/m*C)
N=0.2032m

Q=2*pi*0.65*0.2032

Q=0.829883115*(-5198)/(ln(r2/0.1524)

Can anyone help me out, or point me in the right direction?

They say "Rule of Thumb" is 4inches/10cm of refractory. If I could just figure out what the outside temperature of that would be I think it would be helpful.

Thank you very much!

2. Jan 6, 2013

### Simon Bridge

That's a cylindrical chamber?
The total area of the walls would be $A=\pi hd^2/4$
You need to relate the heat flow through a thickness L of insulating material to the rate that you are supplying heat.

For walls thickness L and thermal conductivity $k$ .. the thermal resistance is $$R=\frac{L}{k A}=\frac{4L}{\pi k hd^2}$$... I think the equation you want is: $$T_{inside}-T_{outside}=\dot{Q}R$$...

The outside temperature would be whatever the working equilibrium temperature of the room should be - take a guess.
If it is just sitting in air, the room will probably get quite hot. With no other information, I'd guess 300K for the outside temperature - which would be uncomfortably warm, and hope there's some sort of ventilation. If you have experience of the kinds of workplaces that use these devices you may be able to make a better guess than that.

3. Jan 6, 2013

### haruspex

r1 is the internal radius, 16cm/2=0.08m, r2 is the external radius, 31cm/2=0.155m. Where did you get 0.1524 from? For the length of cylinder, use the internal height, 0.23m. Where did you get 0.2032m from?
What about the top and bottom of the cylinder?