Question

2- A sheet of steel 3.0-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 à 10â10 m2/s, and the diffusion flux is found to be 1.0 à 10â7 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? Assume a linear concentration profile. (40 pts.)

Answer 1

Answer:

2.78x10⁻³ m

Explanation:

Nitrogen diffusion in the steel can be expressed using the following equation:

$J = - D*\frac{\Delta C}{\Delta x} = - D*\frac{C_{1} - C_{2}}{x_{1} - x_{2}}$  

Where:                        

J: is the diffusion flux = 1.0x10⁻⁷ kgm⁻²s⁻¹    

D: is the diffusion coefficient = 1.85x10⁻¹⁰ m²/s  

C₁: is the concentration of nitrogen in the steel at the surface = 2 kg/m³

C₂: is the concentration of nitrogen at the distance x₂ = 0.5 kg/m³

x₁: is the distance at the surface = 0

x₂: is the distance to find    

Hence, we have:

$J = - D*\frac{C_{1} - C_{2}}{- x_{2}}$    (1)

Solving equation (1) for x₂:

$x_{2} = D \frac{C_{1} - C_{2}}{J} = 1.85 \cdot 10^{-10} m^{2}/s \frac{2 kg/m^{3} - 0.5 kg/m^{3}}{1.0 \cdot 10^{-7} kgm^{-2}s^{-1}} = 2.78 \cdot 10^{-3} m$                                        

Therefore, the nitrogen will enter 2.78x10⁻³ m into the sheet.

I hope it helps you!