Question

A sheet of steel 2.5-mm thick has nitrogen atmospheres on both sides at 900°C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 1.85 × 10–10 m2/s, and the diffusion flux is found to be 1.0 × 10–7 kg/m2.s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 2 kg/m3. How far into the sheet from this high-pressure side will the concentration be 0.5 kg/m3

Answer 1

Explanation:

It is known that relation between diffusion flux ($J_{N_{2}}$) and the diffusion coefficient of $N_{2}$ ($D_{N_{2}}$) for the linear concentration profile at the steady state is as follows.

       $J_{N_{2}} = -D_{N_{2}} \times \frac{C_{2} - C_{1}}{x_{2} - x_{1}}$

The negative sign shows that along the positive x-direction there occurs drop in concentration.

where,   $C_{1}$ and $C_{2}$ are the concentrations of $N_{2}$ at position 1 and 2 on steel sheet.

              $x_{1}$ and $x_{2}$ are the distances corresponding to distances on the sheet.

Now, putting the given values into the above formula as follows.

     $J_{N_{2}} = -D_{N_{2}} \times \frac{C_{2} - C_{1}}{x_{2} - x_{1}}$  

     $1 \times 10^{-7} kg/m^{2}s = -1.85 \times 10^{-10} m^{2}/s \times \frac{0.5 kg/m^{3} - 2 kg/m^{3}}{x_{2} - 0}$

                $x_{2} - 0 = 0.00275 m$

                       $x_{2}$ = 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 $kg/m^{3}$.