Respuesta :
Explanation:
It is known that relation between diffusion flux ([tex]J_{N_{2}}[/tex]) and the diffusion coefficient of [tex]N_{2}[/tex] ([tex]D_{N_{2}}[/tex]) for the linear concentration profile at the steady state is as follows.
[tex]J_{N_{2}} = -D_{N_{2}} \times \frac{C_{2} - C_{1}}{x_{2} - x_{1}}[/tex]
The negative sign shows that along the positive x-direction there occurs drop in concentration.
where, [tex]C_{1}[/tex] and [tex]C_{2}[/tex] are the concentrations of [tex]N_{2}[/tex] at position 1 and 2 on steel sheet.
[tex]x_{1}[/tex] and [tex]x_{2}[/tex] are the distances corresponding to distances on the sheet.
Now, putting the given values into the above formula as follows.
[tex]J_{N_{2}} = -D_{N_{2}} \times \frac{C_{2} - C_{1}}{x_{2} - x_{1}}[/tex]
[tex]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}[/tex]
[tex]x_{2} - 0 = 0.00275 m[/tex]
[tex]x_{2}[/tex] = 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 [tex]kg/m^{3}[/tex].
