A sheet of BCC iron 4.4-mm thick was exposed to a carburizing atmosphere on one

Question

A sheet of BCC iron 4.4-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 0.011 and 0.0065 wt%. Calculate the diffusion coefficient if the diffusion flux is 3.4 × 10-8 kg/m2-s, given that the densities of carbon and iron are 2.25 and 7.87 g/cm3, respectively.

Hint: Use Equation 4.9a—that is, for this problem

to convert concentrations from weight percent carbon to kilograms of carbon per cubic meter or iron.

Explanation / Answer

For steady state diffusion we will use Fick’s first law,

J = -D dC/dx

where,

J = diffusion flux = 3.4 x 10^-8 kg/m^2.s

D = diffusion coefficient

dC = concentration

dx = position = -4.4 x 10^-3 m

First, convert the concentration in wt% to the concentration in kg/m3

For 0.011 wt% C

C'(C) = [0.011/(0.011/2.25 g/cm^3) + (99.988/7.87 g/cm^3)] x 10^3

         = [0.011/(4.89 x 10^-3 + 12.70] x 10^3

         = 0.87 kg. C/m^3

For 0.0065 wt% C

C''(C) = [0.0065/(0.0065/2.25 g/cm^3) + (99.9925/7.87 g/cm^3)] x 10^3

          = 0.0065/(2.89 x 10^-3 + 12.71] x 10^3

          = 0.51 kg.C/m^3

Now, using a the above equation,

D = -J[(x(A) - x(B))/(C(A) - C(B))]

    = -(3.4 x 10^-8)[(-4.4 x 10^-3)/(0.87-0.51)

    = 4.16 x 10^-10 m^2/s

So, the diffusion coefficient as calculated would be 4.16 x 10^-10 m^2/s

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