Question

Determine the carburizing time (in s) necessary to achieve a carbon concentration of 0.44 wt% at a position 1.6 mm into an iron-carbon alloy that initially contains 0.031 wt% C. The surface concentration is to be maintained at 1.2 wt% C, and the treatment is to be conducted at 1180°C. Assume that D₀ = 5.1 x 10^-5 m²/s and Q_d =154 kJ/mol. You will find the table below useful.

z	erf(z)	z	erf(z)	z	erf(z)
0	0	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

At unsteady state

Ficks second law of diffusion

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

At x = 1.6 mm = 1.6*10^-3 m , Cx = 0.44

Initial concentration Co = 0.031

surface concentration Cs = 1.2

(Cx - Co)/(Cs - Co) = (0.44 - 0.031)/(1.2 - 0.031) = 0.34987

1 - erf (x/2 ) = 0.34987

erf (x/2 ) = 0.6501

From linear interpolation

(z - 0.65)/(0.70 - 0.65) = (0.6501 - 0.6420)/(0.6778 - 0.6420)

z = 0.6613

(x/2 ) = 0.6613

Now calculate the diffusion coefficient at 1180 C (1453 K)

D = D0 exp (-Qd/RT)

= 5.1 x 10^-5 m2/s exp (-154000 J/mol / 8.314 J/mol·K x 1453K)

= 1.483 x 10^-10 m2/s

[1.6*10^-3/2(1.483 x 10^-10 t)] = 0.6613

?(1.483 x 10^-10 t) = 0.0012097

1.483 x 10^-10 t = 1.463 x 10^-6

t = 9868.28 s x 1h/3600s = 2.74 hr