Question

During the non-steady state diffusion of nitrogen in a semi-infinite bar of steel, it is found that it takes 5 hours to reach a concentration of 1.38 wt% nitrogen at a depth of 1.0 mm. How much time would it take to reach the same concentration of nitrogen at a depth of 2.0 mm assuming the same surface and initial concentrations. Co = 1.38 wt%, Given T = 800 °C, D = 2.22 x 10-10 m?s-1, Cor = Owt%, Cs = 5wt%

Answer 1

This is a non stedy state difusion related problem. This can be solved by using Fick's second law.

20 hr would take to reach the same concentration of nitrogen at a depth of 2.0 mm.

The detail solution is shown below.

Here T=800c, D = 2.22 x 1010 m. s-1. Co= 1.38 wt% , Ca=0 wts (s=5 wt%. if x= distance or depth 4 = time . For non- steady state diffusion, we have the formular, Co-Co = - eof (m ) (s- Co Here at the particular temp groc, D is constant; and all others parameters are constant. in x = constant Here t= 5 hours for = 1 mm we have to calculate the time for no 2mm

in ✓ 5 ha 1 mm = -2mm ✓tz Vez - 2.5 hour t = 20 hour time = 20 hour for a= 2mm