Question

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.

Answer 1

5.12 mm

Step-by-step explanation:

Let's convert the time taken from hours to seconds

t = 5 h

t =
`5h * (3600s)/(1h)`

t = 18000 s

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

`(C_x-C_0)/(C_s-C_0) = 1 - erf ((x)/(2√(Dt) ) )`

where:

`C_x` = concentration at depth = 0.206 wt%

`C_0` = initial concentration = 0.275 wt%

`C_s` = concentration at the surface position = 0.0 wt%

D = diffusion coefficient =
`5.6*10 ^(-10)m^2/s`

t = time = 18000 s

Replacing the value into the previous formula; we have:

`(0.206-0.275)/(0.0-0.275) = 1 - erf (\frac{x}{2\sqrt{5.6*10^(-10)*18000}})`

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'

`(0.75-x)/(0.75-0.70) = (0.7112-0.7491)/(0.7112-0.6778)`

`(0.75-x)/(0.05) = (-0.0379)/(0.0334)`

- 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 =
`(0.026945)/(0.0334)`

z = 0.806737

Substituting 0.806737 for z in equation (1)

0.806737 = 157.485 x

x =
`(0.806737)/(157.485)`

x = 0.00512 m to mm; we have

`x = 0.00512 m * (1000mm)/(1m)`

x = 5.12 mm

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