Toke,
I did a quick BOTE on this. Assuming uniform, infinite steel (actually iron) plate, uniformly coated with 100µ of thermite, under adiabatic conditions...
Numbers from:
http://physics911.net/thermite
Calculations on the Possible Use of Thermite to Melt Sections of the WTC Core Columns
by D. P. Grimmer
Iron
Density = 7874 kg/m3
Specific heat = 25.1J/mol K = 449 J/kg K = 0.449 kJ/kg K
Thermite
Thus the physical density of our densely-packed Fe2O3/Al thermite mixture is 213.66 g/61.5 cm3 = 3.474 g/cm3 = 3.474 x 10+6 g/m3
and our energy density (per volume) is given by 849 kJ/61.5 cm3 = 13.805 kJ/cm3 = 1.3805 x 10+7 kJ/m3.
1 square meter x 100µ = 1x10e-4 m^3 of thermite.
heat content of 1 m^2 x 100µ thickness of thermite => (10e-4 m^3) (1.4e7 kJ/m^3) = 1.4e3 kJ.
one square meter of steel, @ 5" thick = (.127 m^3) = (.127 m^3) (7873 kg/m^3) = 1000 kg steel.
Raising this mass of steel (per degree K) takes (1000 kg)*(0.449 kJ/kg °K) = 449 kJ/°K
@ 5" steel thickness, 100µ of thermite will give you a rise of: 1.4e3kJ/449 kJ/°K = 3.1°K = 3.1°C once equilibrium has been reached.
1" thick piece of steel rises 15°C.
1/2" thick rises 30°C.
Someone check the math. I was in the middle of a boring meeting...
tom
Without checkking the details I still belive that the neccecary amount of thermite to heat a steel collum to weakening is thicker than a coat of paint. By two or tre orders of magnitude.
I did a quick BOTE on this. Assuming uniform, infinite steel (actually iron) plate, uniformly coated with 100µ of thermite, under adiabatic conditions...
Numbers from:
http://physics911.net/thermite
Calculations on the Possible Use of Thermite to Melt Sections of the WTC Core Columns
by D. P. Grimmer
Iron
Density = 7874 kg/m3
Specific heat = 25.1J/mol K = 449 J/kg K = 0.449 kJ/kg K
Thermite
Thus the physical density of our densely-packed Fe2O3/Al thermite mixture is 213.66 g/61.5 cm3 = 3.474 g/cm3 = 3.474 x 10+6 g/m3
and our energy density (per volume) is given by 849 kJ/61.5 cm3 = 13.805 kJ/cm3 = 1.3805 x 10+7 kJ/m3.
1 square meter x 100µ = 1x10e-4 m^3 of thermite.
heat content of 1 m^2 x 100µ thickness of thermite => (10e-4 m^3) (1.4e7 kJ/m^3) = 1.4e3 kJ.
one square meter of steel, @ 5" thick = (.127 m^3) = (.127 m^3) (7873 kg/m^3) = 1000 kg steel.
Raising this mass of steel (per degree K) takes (1000 kg)*(0.449 kJ/kg °K) = 449 kJ/°K
@ 5" steel thickness, 100µ of thermite will give you a rise of: 1.4e3kJ/449 kJ/°K = 3.1°K = 3.1°C once equilibrium has been reached.
1" thick piece of steel rises 15°C.
1/2" thick rises 30°C.
Someone check the math. I was in the middle of a boring meeting...
tom