Question

Question: An iron-carbon alloy initially containing 0.308 wt% C is exposed to an oxygen-rich and virtually ...

An iron-carbon alloy initially containing 0.308 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 0.231 wt% after a 5 h treatment? The value of D at 1090°C is 4.8 × 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		1.1		0.8802		2.6		0.9998
0.50		0.5205		1.2		0.9103		2.8		0.9999

Answer 1

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

Cx= wt% at distance x=0.231 wt%, Co= initial weight % = 0.308 wt%, , Cs= wt% at the surface=0

D= 4.8*10^-11 m2/s t= 5 hr= 5*60*60 seconds =18000 seconds, x need to be determined

(0.231-0.308)/(0-0.308)= 1-erf(x/2*Sqrt(Dt)

erf(x/2*Sqrt(Dt)= 0.75

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

given Z= 0.8 erf (Z)=0.7421

Z= ?, erf(Z)= 0.75

Z= 0.85, erf(Z)= 0.7707

hence (Z-0.8)/ (0.85-0.8)= (0.75-0.7421)/(0.7707-0.7421)

Z= 0.8138

hence x/2*sqrt(Dt)= 0.8138

x= 0.8138* 2*Sqrt(4.8*10^-11*18000)= 0.0015 m