Question

An iron-carbon alloy initially containing 0.258 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1120°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 0.194 wt% after a 6 h treatment? The value of D at 1120°C is 6.9 × 10-11 m2/s.

Answer 1

Answer:

1.95 mm

Explanation:

Let's calculate the carbon conc. 0.194 wt% after a 6 h treatment using the expression:

$\frac{C_x - C_0}{C_s - C_0} = 1 - erf (\frac{x}{2 \sqrt{Dt}})$

$\frac{0.194 - 0.258}{0 - 0.258} = 1 - erf (\frac{x}{2 \sqrt{Dt}})$

$0.248 = 1 - erf (\frac{x}{2 \sqrt{Dt}})$

$erf(\frac{x}{2 \sqrt{Dt}} = 0.752$

From the erf z table, at erf(z) = 0.752

z = 0.80

Thus,

$\frac{x}{2 \sqrt{Dt}} = 0.80$

Where d = $6.9*10^-^1^1$

Time, t = 6 hrs = (6 * 60mins * 60secs)

Substituting values:

$\frac{x}{2 \sqrt{6.9*10^-^1^1 * (6*60mins*60sec)}} = 0.80$

$\frac{x}{2 \sqrt{1.49*10^-^6}}) = 0.80$

$\frac{x}{2.44*10^-^3} = 0.80$

Solve for x:

$x = 0.80 * 2.44*10^-^3$

$x = 1.95*10^-^3$

x = 1.95 mm

The position will the carbon concentration be 0.194 wt% after a 6 h treatment is at 1.95 mm