Question

A sheet of BCC iron 4.6-mm thick was exposed to a carburizing atmosphere on one side and a decarburizing atmosphere on the other side at 725°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 0.010 and 0.0063 wt%. Calculate the diffusion coefficient if the diffusion flux is 3.7 × 10-8 kg/m2-s, given that the densities of carbon and iron are 2.25 and 7.87 g/cm3, respectively.

Answer 1

`Diffusion\ coefficient = 5.848* 10^(-10) m^2/s`

Step-by-step explanation:

convert the carbon concentration from weight percent to kilogram carbon per meter cubed for 0.010%

`C_A = (C_c)/((C_c)/(\rho_c) + (C_b)/(\rho_b)) * 10^3`

where,
`C_c` is carbon concentration 0.010

`C_b`is remaining BCC concentration ( 100 - 0.010 = 99.99)

where
`\rho_c and \rho_b` is density of carbon and bcc respectively

`C_A = (0.010)/((0.010)/(2.25) + (99.99)/(7.87)) * 10^3`

`C_A =0.786`

convert the carbon concentration from weight percent to kilogram carbon per meter cubed for 0.0063%

`C_B = (C_c)/((C_c)/(\rho_c) + (C_b)/(\rho_b)) * 10^3`

where,
`C_c` is carbon concentration 0.010

`C_b`is remaining BCC concentration ( 100 - 0.0063 = 99.993)

where
`\rho_c and \rho_b` is density of carbon and bcc respectively

`C_B = (0.0063)/((0.0063)/(2.25) + (99.99)/(7.87)) * 10^3`

`C_B =0.495`

Determine Diffusion coefficient

`D = -J [(X_A -X_B)/(C_A - C_B)]`

`= -3.7* 10^(-8) [(-4.6* 10^(-3))/(0.786-0.495)]`

`Diffusion\ coefficient = 5.848* 10^(-10) m^2/s`