Question

No. 5 (20 points) (a) Determine the time necessary to achieve a carbon concentration of 0.30 wt.% at a position of 4 mm into an iron-carbon alloy that initially contains 0.10 wt. % C. The surface concentration is to be maintained at 0.90 wt.% C, and the diffusion coefficient of C is 7.0 x 10-11 m2/sec. (10 points) (b) The diffusion coefficient of Ni in a stainless steel (fcc) is 1 x 1022 m2/sec at 500°C and 1 x 10-15 m2/sec at 1000°C Calculate the energy of activation for diffusion in Ni in this alloy over the temperature range of 500°-1000°C. (10 points)

Answer 1

Ans a

From the Ficks second law of diffusion for unsteady state

(Cx - Co) / (Cs - Co) = 1 - erf [x/2$\sqrt$Dt]

Initial concentration Co = 0.10

Surface concentration Cs = 0.90

position x = 4 mm x 1m/1000mm

= 0.004 m

Concentration at x

Cx = 0.30

(0.30 - 0.10) / (0.90 - 0.10) = 1 - erf [x/2$\sqrt$Dt]

0.25 = 1 - erf [x/2$\sqrt$Dt]

erf [x/2$\sqrt$Dt] = 0.75

z erf z

0.80 0.7421

z 0.75

0.85 0.7707

(z - 0.80) / (0.85-0.80) = (0.75 - 0.7421)/(0.7705 - 0.7421)

(z - 0.80) / (0.05) = (0.013811)

z = 0.8138

[x/2$\sqrt$Dt] = z = 0.8138

Diffusion coefficient D = 7.0*10^-11 m2/s

[0.004/2$\sqrt$( 7.0*10^-11 m2/s) * t] = 0.8138

[$\sqrt$( 7.0*10^-11 m2/s) * t] = 0.002457

( 7.0*10^-11 m2/s) * t] = 6.04 x 10^-6

t = 8.63 x 10^4 s x 1hr/3600s

= 23.97 hr