Explanation:
It is known that relation between diffusion flux (
) and the diffusion coefficient of
(
) for the linear concentration profile at the steady state is as follows.
![J_{N_{2}} = -D_{N_{2}} \times \frac{C_{2} - C_{1}}{x_{2} - x_{1}}](https://tex.z-dn.net/?f=J_%7BN_%7B2%7D%7D%20%3D%20-D_%7BN_%7B2%7D%7D%20%5Ctimes%20%5Cfrac%7BC_%7B2%7D%20-%20C_%7B1%7D%7D%7Bx_%7B2%7D%20-%20x_%7B1%7D%7D)
The negative sign shows that along the positive x-direction there occurs drop in concentration.
where,
and
are the concentrations of
at position 1 and 2 on steel sheet.
and
are the distances corresponding to distances on the sheet.
Now, putting the given values into the above formula as follows.
![1 \times 10^{-7} kg/m^{2}s = -1.85 \times 10^{-10} m^{2}/s \times \frac{0.5 kg/m^{3} - 2 kg/m^{3}}{x_{2} - 0}](https://tex.z-dn.net/?f=1%20%5Ctimes%2010%5E%7B-7%7D%20kg%2Fm%5E%7B2%7Ds%20%3D%20-1.85%20%5Ctimes%2010%5E%7B-10%7D%20m%5E%7B2%7D%2Fs%20%5Ctimes%20%5Cfrac%7B0.5%20kg%2Fm%5E%7B3%7D%20-%202%20kg%2Fm%5E%7B3%7D%7D%7Bx_%7B2%7D%20-%200%7D)
![x_{2} - 0 = 0.00275 m](https://tex.z-dn.net/?f=x_%7B2%7D%20-%200%20%3D%200.00275%20m)
= 2.75 mm
Thus, we can conclude that 2.75 mm far into the sheet from given high-pressure side will the concentration be 0.5
.