A sheet of BCC iron 1.5-mm thick was exposed to a carburizing atmosphere on one side and an carburizing atmosphere on the other side at 725°C. After having reached steady state, the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces were determined to be 0.012 and 0.0069 wt%. Calculate the diffusion coefficient if the diffusion flux is 3.6 × 10-8 kg/m2-s, given that the densities of carbon and iron are 2.25 and 7.87 g/cm3, respectively.

Diffusion coefficient D = 2.6613 × 10⁻¹¹ m²/s

Explanation:

From the given information:

Carbon concentration = 0.012

Iron concentration = 100 - 0.012 = 99.988

Density of carbon = 2.25 g/cm³

Density of iron = 7.87 g/cm³

We can convert the carbon concentrations from weight percentage to kilograms carbon per meter cubed by using the formula.

However; for 0.0069 wt% C

Thus; the diffusion coefficient can be computed by using the formula:

D = 2.6613 × 10⁻¹¹ m²/s

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s