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A sheet of steel 1.4 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 7.0 × 10-11 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 4.9 kg/m3. How far into the sheet from this high-pressure side will the concentration be 3.6 kg/m3? Assume a linear concentration profile.

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Answer:

9.1 × 10^-4m

Explanation:

From Fick's first law of diffusion:

J = -D dc/dx

Where J ,= diffusion flux

D = diffusion coefficient and dc/dx = concerntration gradient

Since we can assume a linear concentration profile, we rewrite the equation above as:

J = - D ◇c/◇x = -D (Cs - Cx)/(0 - x)

Where Cs = surface concerntration

C = concerntration of depth

X = target variable

J = diffusion flux = 1.0×10^-7kg/m^2/s

D ,= diffusion coefficient = 7.0×10^-12m^2/s

Substituting into the equation

1.0 ×10^-7 = (-7.0×10^-11)(4.9 -3.6) / ( 0 - X)

1.0 × 10^-7 = (-9.1 ×10^-11)/(0 - X)

Cross multiply

(1.0 ×10^-7) ×(0-X) = (-9.1 ×10^-11)

X = (-9.1 × 10^-11)/(-1.0 × 10^-7)

X = 9.1 × 10^-4m

The distance that should be far into the sheet should be 9.1 × 10^-4m

Fick's first law of diffusion:

Here we applied the following equation

J = -D dc/dx

Since

J ,= diffusion flux

D = diffusion coefficient

and dc/dx = concerntration gradient

Now the following equation should be

J  = -D (Cs - Cx)/(0 - x)

Here Cs = surface concerntration

C = concerntration of depth

X = target variable

Now

J = diffusion flux = 1.0×10^-7kg/m^2/s

D ,= diffusion coefficient = 7.0×10^-12m^2/s

So, the distance should be

1.0 ×10^-7 = (-7.0×10^-11)(4.9 -3.6) / ( 0 - X)

1.0 × 10^-7 = (-9.1 ×10^-11)/(0 - X)

Now we have to cross multiply

(1.0 ×10^-7) ×(0-X) = (-9.1 ×10^-11)

X = (-9.1 × 10^-11)/(-1.0 × 10^-7)

X = 9.1 × 10^-4m

Therefore, The distance that should be far into the sheet should be 9.1 × 10^-4m

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