- #1
Voltux
- 30
- 3
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!
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!