Question

A thin sheet of iron 0.5 mm thick has a nitrogen concentration on one side of 0.5 kg/m^3 and a nitrogen concentration on the other side of 3.5kg/m^3 . the diffusion coefficient is 3.44x10^-15 m^2/s. What is the flux of nitrogen through the iron?

Answer 1

Answer:

2.064× 10⁻¹¹ kg/m²s

Explanation:

The expression for the Fick's first law is shown below as:

$J=-D\times \frac {dC}{dx}$

Where, J is the diffusion flux.

D is the diffusion coefficient = $3.44\times 10^{-15}\ m^2/s$

dC is the concentration change ( 3.5 - 0.5 kg/m³ = 3 kg/m³)

dx is the thickness = 0.5 mm = 0.0005 (As , 1 mm = 0.001 m )

Thus,

$J=-3.44\times 10^{-15}\times \frac{3}{0.0005}\ kg/m^2s$

Magnitude of the flux = 2.064× 10⁻¹¹ kg/m²s