Question

Nitrogen from a gaseous phase is to be diffused into pure iron at 700°C. If the surface concentration is maintained at 0.26 wt% N, what will be the concentration (in weight percent) 2.8 mm from the surface after 6.3 h? The diffusion coefficient for nitrogen in iron at 700°C is 2.2 × 10-10 m2/s.

Answer 1

Answer:

$Cx = .376 \times 0.26 = 0.0978 wt% N$

Explanation:

Given details

concentration of surface is Cs 0.26 wt%

distance from surface is x 2.8 mm

time t = 3.6 h

original concentration is Co 0

we know that

$\frac{Cx - Co}{Cs - Co} = 1 - erf(\frac{x}{2\sqrt{Dt}}$

$\frac{Cx - 0}{0.26 - Co} = 1 - erf(\frac{2\times 10^{-3}}{2\sqrt{2.2\times 10^{-11} 5 hr (3600 /1 hr)}}$

$\frac{Cx}{0.26} = 1 - erf(0.626)$

FROM TABLE OF ERROR FUNCTION WE HAVE

Z           erf(z)

0.60        0.6039

0.65         0.6420

so by interpolation technique we have

for z = 0.626 we have value of erf(0.626) =   0.623

$\frac{Cx}{0.26} = 1 - 0.623$

$Cx = .376 \times 0.26 = 0.0978 wt% N$