An iron-carbon alloy initially containing 0.224 wt% C is exposed to an oxygen-ri

Question

An iron-carbon alloy initially containing 0.224 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1090°C. Under these circumstances the carbon diffuses from the alloy and reacts at the surface with the oxygen in the atmosphere; that is, the carbon concentration at the surface position is maintained essentially at 0.0 wt% C. At what position will the carbon concentration be 0.168 wt% after a 5 h treatment? The value of D at 1090°C is 2.2 x 10-11 m2/s. z erf(z) z erf(z) z erf(z) 0.00 0.0000 0.55 0.5633 1.3 0.9340 0.025 0.0282 0.60 0.6039 1.4 0.9523 0.05 0.0564 0.65 0.6420 1.5 0.9661 0.10 0.1125 0.70 0.6778 1.6 0.9763 0.15 0.1680 0.75 0.7112 1.7 0.9838 0.20 0.2227 0.80 0.7421 1.8 0.9891 0.25 0.2763 0.85 0.7707 1.9 0.9928 0.30 0.3286 0.90 0.7970 2.0 0.9953 0.35 0.3794 0.95 0.8209 2.2 0.9981 0.40 0.4284 1.0 0.8427 2.4 0.9993 0.45 0.4755 1. 0.8802 2.6 0.9998 0.50 0.5205 1.2 0.9103 2.8 0.9999

Explanation / Answer

we know that

(Cx-CO)/(CS-CO)= 1-erf(x/2Sqrt(Dt) , CS= surface % concentration, Cx = % concentration at point x, Co= initial %concentration

given Cx= 0.168, CS= 0 wt%, Co= 0.224

(Cx-CO)/(Cs-CO)= (0.168-0.224)/(0-0.224)= 0.25= 1-erf(x/2*Sqrt(Dt)

erf{x/(2*sqrt(Dt)= 0.75, from the data tables, at , erf(Z)= 0.7421, Z=0.8 and erf(Z)= 0.7707, Z=0.85

(Z-0.8)/(0.85-0.8)= (0.75-0.7421)/ (0.7707-0.7421), Z=0.8138

hence Z= x/(2*sqrt(Dt)

2*sqrt(Dt)*Z= 2*Sqrt(2.2*10-11*5*60*60) =0.00125 m

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