sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200°C and ispermitted to achieve a steady-state diffusion condition. The diffusion coefficient fornitrogen in steel at this temperature is 6 x 10-11 m2/s, and the diffusion flux is found tobe 1.2 x 10-7 kg/m2/s. Also, it is known that the concentration of nitrogen in the steelat the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressureside will the concentration be 2.0 kg/m3? Assume a linear concentration profile.
Question
sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200°C and ispermitted to achieve a steady-state diffusion condition. The diffusion coefficient fornitrogen in steel at this temperature is 6 x 10-11 m2/s, and the diffusion flux is found tobe 1.2 x 10-7 kg/m2/s. Also, it is known that the concentration of nitrogen in the steelat the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressureside will the concentration be 2.0 kg/m3? Assume a linear concentration profile.
Solution
To solve this problem, we can use Fick's first law of diffusion, which states that the diffusion flux is proportional to the concentration gradient. The law can be written as:
J = -D * (dC/dx)
where: J is the diffusion flux, D is the diffusion coefficient, dC is the change in concentration, and dx is the change in position.
We are given that J = 1.2 x 10^-7 kg/m^2/s, D = 6 x 10^-11 m^2/s, and the concentration changes from 4 kg/m^3 to 2 kg/m^3. We need to find dx, the distance from the high-pressure side where the concentration is 2 kg/m^3.
Rearranging the equation for dx gives:
dx = -D * (dC/J)
Substituting the given values gives:
dx = -(6 x 10^-11 m^2/s) * ((2 kg/m^3 - 4 kg/m^3) / (1.2 x 10^-7 kg/m^2/s)) = -(6 x 10^-11 m^2/s) * (-2 kg/m^3 / 1.2 x 10^-7 kg/m^2/s) = 1 m
So, the concentration will be 2.0 kg/m^3 at a distance of 1 meter from the high-pressure side.
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