Question

A sheet of BCC iron 2.0-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 0.012 and 0.0074 wt%. Calculate the diffusion coefficient if the diffusion flux is 2.2 × 10-8 kg/m2-s, given that the densities of carbon and iron are 2.25 and 7.87 g/cm3, respectively.

Answer 1

The diffusion coefficient is 1.22 * 10 ^-10 m²/s

Step-by-step explanation:

Step 1: Data given

A sheet of BCC iron 2.0-mm thick

The diffusion flux is 2.2 * 10^-8 kg/m²*s

Density of Carbon = 2.25 g/cm³

Density of Iron = 7.87 g/cm³

Step 2: Define the formula for the carbon concentrations

C"c = Cc / ((Cc/rc)+ (Cfe/rfe))

with C"c = The concentration carbon

with Cc = The % concentration

with rc = The density of Carbon

with Cfe = The % concentration iron

with rfe = the density of Iron

Step 3: For 0.012 wt% C

C"c = 0.012 / ((0.012/2.25)+(99.988/7.87)) *10³

C"c = 0.944 kg C/ m³

Step 4: For 0.0074 wt% C

C"c = 0.0074 / ((0.0074/2.25)+(99.9926/7.87)) * 10³

C"c = 0.5823 kg C/m³

Step 5: Calculate the diffusion coefficient

D = - (diffusion flux)* ((- thickness)/(0.012 wt% C - 0.0074 wt% C)

D = -(2.2 *10^-8 ) *(( -2*10^-3)/(0.944 - .5823)

D = 1.22 * 10 ^-10 m²/s

The diffusion coefficient is 1.22 * 10 ^-10 m²/s