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.

Question

Leonel

Physics

27 August, 08:39

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.

+4

Answers (1)

Know the Answer?

Abigayle · Answer 1

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