Question

A sheet of BCC iron 1.5-mm thick was exposed to a carburizing atmosphere on one side and an carburizing 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 0.012 and 0.0069 wt%. Calculate the diffusion coefficient if the diffusion flux is 3.6 × 10-8 kg/m2-s, given that the densities of carbon and iron are 2.25 and 7.87 g/cm3, respectively.

Answer 1

Answer:

Diffusion coefficient D = 2.6613 × 10⁻¹¹ m²/s

Explanation:

From the given information:

Carbon concentration $C_c$ = 0.012

Iron concentration $C_{Fe}$ = 100 - 0.012 = 99.988

Density of carbon = 2.25 g/cm³

Density of iron = 7.87 g/cm³

We can convert the carbon concentrations from weight percentage to kilograms carbon per meter cubed by using the formula.

$C^*_c = \dfrac{C_c}{\dfrac{C_c}{\rho_c}+\dfrac{C_{Fe}}{\rho_{Fe}}}\times 10^3$

$C^*_c = \dfrac{0.012}{\dfrac{0.012}{2.25 \ g/cm^3}+\dfrac{99.988}{7.87 \ g/cm^3}}\times 10^3$

$C^*_c = \dfrac{0.012}{0.00533 \ g/cm^3 + 12.71 \ g/cm^3} \times 10^3$

$C^*_c = 0.9437 \ kgC/m^3$

However; for 0.0069 wt% C

$C^*_c = \dfrac{C_c}{\dfrac{C_c}{\rho_c}+\dfrac{C_{Fe}}{\rho_{Fe}}}\times 10^3$

$C^*_c = \dfrac{0.0069}{\dfrac{0.0069}{2.25 \ g/cm^3}+\dfrac{99.9931}{7.87 \ g/cm^3}}\times 10^3$

$C^*_c =0.5429 \ kgC/m^3$

Thus; the diffusion coefficient can be computed by using the formula:

$D = - J \begin {bmatrix} \dfrac{x_A-x_B}{C_A-C_B}\end {bmatrix}$

$D = - (3.6 \times 10^{-8} \ kg/m^2-s) \begin {bmatrix} \dfrac{-15^{-3} \ m }{0.9437 \ kgC/m^3 - 0.5429 \ kgC/m^3}\end {bmatrix}$

D = 2.6613 × 10⁻¹¹ m²/s