A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6  10-11 m2/s, and the diffusion flux is found to be 1.2  10-7 kg/m2 -s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3? Assume a linear concentration profile.

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01-05-2020
Physics

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A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6  10-11 m2/s, and the diffusion flux is found to be 1.2  10-7 kg/m2 -s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3? Assume a linear concentration profile.

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ogbe2k3 ogbe2k3 · Answer 1 · 2020-05-04T14:57:02+02:00

Answer:

The answer is 0.001 m

Explanation:

Solution

Recall that,

A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at = 1200°C

The diffusion coefficient for nitrogen in steel at this temperature is =6 * 10-11 m2/s

The diffusion flux is = 1.2 *10^-7 kg/m2s

Concentration of nitrogen in the steel at the high-pressure surface is= 4 kg/m3.

The high-pressure side will the concentration is estimated to be = 2.0 kg/m3

Now,

The flux = -D dC/dx

1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx

∫ˣ₀ dx = -5x^10-4 ∫²₄ dC

so,

x = (2-4) kg/m3 (-5x10-4 m4/kg)

where x = .001 m

Therefore x = 0.001 m

andromache andromache · Answer 2 · 2022-02-24T14:13:26+01:00

The distance from this high-pressure side will the concentration is 0.001 m

Calculation of the distance:

Since the diffusion coefficient for nitrogen in steel at this temperature is 6 x 10-11 m2/s, the diffusion flux is found to be 1.2x 10-7 kg/m2 -s.

We know that

The flux =[tex]-D\ dC\div dx[/tex]

So,

1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx

dC/dx = -2000

Now

-2000 = 4 - 2/0-x_B

x_B = 2/2000

x_B = 1*10^-3m

x_B = 0.001m

Hence, The distance from this high-pressure side will the concentration is 0.001 m

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