An iron-carbon alloy initially containing 0.308 wt% C is exposed to an oxygen-ri

Question

An iron-carbon alloy initially containing 0.308 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1120°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 0.231 wt% after a 5 h treatment? The value of D at 1120°C is 1.4 × 10-10 m2/s. erf(z) erfz) zerfz) 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 .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 2.0.9993 0.45 0.4755 1.1 0.8802 2.6 0.9998 0.50 0.5205 1.2 0.9103 2.8 0.9999 00165 the tolerance is +/-2%

Explanation / Answer

we know that

(Cx-Co)/(Cs-Co)= 1-erf(x/2*sqrt(Dt)), D= diffusivity

where Cx= concentration at x= 0.23%, Cs= 0, Co=0.308 wt%

hence (Cx-Co)/(Cs-Co)= (0.23-0.308)/(0-0.308)= 0.253= 1-erf(x/2*Sqrt(Dt)

erf(x/(2*Sqrt(Dt)= 0.747

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

hence erf(Z)=0.747

when erf(Z)=0.7421, Z=0.8 and erf(Z)=0.7707, Z= 0.85

(Z-0.8)/(0.85-0.8) = (0.747-0.7421)/(0.7707-0.7421)

Z=0.8085

Z= x/(2*Sqrt(Dt), t =time in seconds =5hr= 5*60min/hr*60sec/min

x= Z*2*sqrt(Dt)= 0.8085*2*sqrt(1.4*10-10*5*60*60) =0.0025 m

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