When Nitrogen from a gaseous phase is to be diffused into pure iron, the Do and

Question

When Nitrogen from a gaseous phase is to be diffused into pure iron, the Do and Qa are 5.0 x 10 77,000 , respectively. It is assumed that the initial concentration of Nis0wt% N. During the diffusion process, the surface concentration is maintained at 0.2 wt% N. If the required the concentration of N at 2 mm from the surface is 0.08734 wt% N at the temperature 700 oC, what is the time for this diffusion process? (20%) and Qd are 5.0 × 10-7 (m)and Table 5.1 Tabulation of Error Function Values 0.5633 0.6039 0.6420 06778 0.7112 0.7421 0.7707 0.7970 13 1.4 1.5 1.6 0.9340 0.9523 0.9661 0.0% 0.9838 0.9891 09928 0.9953 0.9081 0.9993 0.998 0999 0025 005 0.10 0.15 0.0282 00564 0.1125 0.1680 0.2227 0.2763 0.65 0.70 18 1.9 20 0.35 0.40 0.45 0.50 03794 0.4284 04755 0.5205 0.90 0.95 1.0 1.I 12 08427 26

Explanation / Answer

Surface concentration Cs = 0.20 wt%

Initial concentration Co = 0

Concentration at 2mm = Cx = 0.08734 wt%

Temperature T = 700 + 273 = 973 K

Diffusion coefficient D = D0 exp (-Qd/RT)

= 5*10^-7 exp (-77000/8.314*973)

= 5*10^-7 * 7.351*10^-5

= 3.675*10^-11 m2/s

From Ficks second law of diffusion

(Cx - Co) / (Cs - Co) = 1 - erf [ x/2(Dt)0.5]

1 - [(0.08734) / (0.20)] = erf [ x/2(Dt)0.5]

0.5633 = erf [ x/2(Dt)0.5]

0.5633 = erf [ z]

From the given table, z = 0.55

x/[2(Dt)0.5] = 0.55

x = [2(Dt)0.5] * 0.55

2*10^-3 = 2 * (3.675*10^-11)0.5 * t0.5 * 0.55

299.922 = t0.5

t = 89953.33 s x 1h/3600s

= 24.98 h

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