An iron-carbon alloy initially containing 0.275 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1110°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 0.206 wt% after a 5 h treatment? The value of D at 1110°C is 5.6 × 10-10 m2/s.

5.12 mm

Explanation:

Let's convert the time taken from hours to seconds

t = 5 h

t =  

t = 18000 s

The relation expressing the concentration, position and time together is given by the formula:

where:

= concentration at depth  = 0.206  wt%

= initial concentration = 0.275 wt%

= concentration at the surface position = 0.0 wt%

D = diffusion coefficient =

t = time = 18000 s

Replacing the value into the previous formula; we have:

0.2509 = 1 - erf (157.85 x)

erf (157.85 x) = 1 - 0.2509

erf (157.85 x) = 0.7491

So, Let's assume the value of z to be 157.485x ; we have:

z = 157.485 x       ------------------ Equation (1)

We obtain the value of Z corresponding to erf (Z) = 0.7491 from the Table 5.1 , 'Table of  Error Function Values'

- 0.001895 = (0.75 - z ) 0.0334

- 0.001895 = 0.02505 - 0.0334 z

0.0334 z = 0.02505 + 0.001895

0.0334 z = 0.026945

z =

z = 0.806737

Substituting 0.806737 for z in equation (1)

0.806737 = 157.485 x

x =

x = 0.00512 m     to mm; we have

x = 5.12 mm

Thus, the position at which the carbon concentration is 0.206 wt% after a 5 h treatment = 5.12 mm

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not too sure thinking all accelerations are pointed down