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.

Question

contestada

Respuesta :

rejkjavik rejkjavik · Answer 1 · 2019-10-13T19:14:46+02:00

Answer:

[tex]Diffusion\ coefficient = 5.848\times 10^{-10} m^2/s[/tex]

Explanation:

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

[tex]C_A = \frac{C_c}{\frac{C_c}{\rho_c} + \frac{C_b}{\rho_b}} \times 10^3[/tex]

where, [tex]C_c[/tex] is carbon concentration 0.010

[tex] C_b [/tex]is remaining BCC concentration ( 100 - 0.010 = 99.99)

where [tex]\rho_c and \rho_b[/tex] is density of carbon and bcc respectively

[tex]C_A = \frac{0.010}{\frac{0.010}{2.25} + \frac{99.99}{7.87}} \times 10^3[/tex]

[tex]C_A =0.786[/tex]

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

[tex]C_B = \frac{C_c}{\frac{C_c}{\rho_c} + \frac{C_b}{\rho_b}} \times 10^3[/tex]

where,[tex] C_c[/tex] is carbon concentration 0.010

[tex] C_b [/tex]is remaining BCC concentration ( 100 - 0.0063 = 99.993)

where [tex]\rho_c and \rho_b[/tex] is density of carbon and bcc respectively

[tex]C_B = \frac{0.0063}{\frac{0.0063}{2.25} + \frac{99.99}{7.87}} \times 10^3[/tex]

[tex]C_B =0.495[/tex]

Determine Diffusion coefficient

[tex]D = -J [\frac{X_A -X_B}{C_A - C_B}][/tex]

[tex]= -3.7\times 10^{-8} [\frac{-4.6\times 10^{-3}}{0.786-0.495}][/tex]

[tex]Diffusion\ coefficient = 5.848\times 10^{-10} m^2/s[/tex]