1. A sheet of BCC iron 1.3 mm thick was exposed to a carburizing atmosphere on one side and a
decarburizing atmosphere on the other side at 725
A sheet of BCC iron 1.3 mm thick was exposed to a carburizing atmosphere on one side and a
decarburizing atmosphere on the other side at 725 degree C. After having reached steady state,
the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces
were determined to be 0.013 and 0.0067 wt%. Calculate the diffusion coefficient if the diffusion
flux is 3.9 times 10 - 8 kg/m2·s, given that the densities of carbon and iron are 2.25 and 7.87
g/cm3, respectively?
Solution
Let us first convert the carbon concentrations from weight percent to kilograms carbon per meter
cubed using Equation For 0.013 wt% C
= 0.013 * 10^3 / (0.013/2.25 + (100-0.013)/7.87)
= 1.0227 kg C/m^3
similarly for 0.0067 wt%
= 0.0067*10^3 / (0.0067/2.25 + (100-0.0067)/7.87)
= 0.5272 kg C/m^3
D = -3.9*10^-8 *(-10^-3 / (1.0227-0.5272)) = 7.87 * 10^-11 m^2 / s