Problem 5.14 An iron-carbon alloy initially containing 0.242 wt% C is exposed to

Question

Problem 5.14 An iron-carbon alloy initially containing 0.242 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1140°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 0.182 wt% after a 5 h treatment? The value of D at 1140°C is 2.3 x 10-11 m2/s 2 3 5 2 5 2 z erf(z) z erf(z) Z erf(z) 0.00 0.0000 0.55 0.56331.3 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 0.74211.8 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 1.1 0.8802 2.6 0.9998 0.50 0.5205 1.2 0.9103 2.8 0.9999

Explanation / Answer

we know that

(Cx-C0)/(Cs-Co)= 1-erf(x/2*sqrt(Dt)

given Cx=0.182, Co=0.242, Co=0

(0.182-0.242)/(0-0.242)= 0.2479= 1-erf(x/2*sqrt(Dt)

ert(x/2*sqrt(Dt)=0.752006, let X= x/2*sqrt(Dt)

from the data given, the value of Z is calculated baed on extrapolation of data.

erf(Z)= 0.7421 at Z=0.8

and erf(Z)=0.7707 at Z=0.85

erf(Z)=0.752006, at Z=Z

(0.85-Z)/(0.85-0.80)=(0.7707-0.752006)/(0.7707-0.7421)= 0.6536

0.85-Z =0.6536*0.05

Z=0.8173

hence x/(2*sqrt(Dt)=0.8173, t= 5 hr= 5*60min= 5*60*60 sec

2*sqrt(Dt)*0.8173=x

x= 2*sqrt(2.3*10-11*5*60*60)*0.8173=0.00105 m =1.05 mm

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