Question

A sheet of steel 3.8 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 5.1 × 10^-11 m²/s, and the diffusion flux is found to be 4.8 × 10^-7 kg/m²-s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4.7 kg/m³. How far into the sheet from this high-pressure side will the concentration be 1.8 kg/m³? Assume a linear concentration profile.

Answer 1

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Answer 2

To find the distance into the sheet from the high-pressure side where the concentration of nitrogen is 1.8 kg/m³, we can use Fick's second law of diffusion, which is given by:

$�=-�{\displaystyle \frac{��}{��}}$

where:

• $�$ is the diffusion flux (4.8 × 10⁻⁷ kg/m²-s),

• $�$ is the diffusion coefficient (5.1 × 10⁻¹¹ m²/s),

• $�$ is the concentration of nitrogen at a distance $�$ from the high-pressure surface, and

• $�$ is the distance into the sheet from the high-pressure side (what we need to find).

Since we are assuming a linear concentration profile, we can integrate the equation to solve for $�$:

$�=-�{\displaystyle \frac{��}{��}}$

${\int }_{{�}_1}^{{�}_2}��=-�{\int }_0^{�}{\displaystyle \frac{��}{�}}$

${�}_2-{�}_1={\displaystyle \frac{�}{�}}�$

Now, we can plug in the given values:

${�}_1=4.7\text{kg/m³}$${�}_2=1.8\text{kg/m³}$$�=5.1\times 10^{-11}\text{m²/s}$$�=4.8\times 10^{-7}\text{kg/m²-s}$

${�}_2-{�}_1={\displaystyle \frac{�}{�}}�$

$1.8-4.7={\displaystyle \frac{5.1\times 10^{-11}}{4.8\times 10^{-7}}}�$

Solving for $�$:

$-2.9={\displaystyle \frac{5.1\times 10^{-11}}{4.8\times 10^{-7}}}�$

$�={\displaystyle \frac{-2.9\times 4.8\times 10^{-7}}{5.1\times 10^{-11}}}$

$�\approx -2.71\text{m}$

Since the distance cannot be negative, it means the concentration of nitrogen at 1.8 kg/m³ is approximately 2.71 meters into the sheet from the high-pressure side.

Answer 3

To find the distance into the sheet from the high-pressure side where the concentration of nitrogen is 1.8 kg/m³, we can use Fick's second law of diffusion for steady-state diffusion in a one-dimensional system. The equation is as follows:

J = -D * (dc/dx)

where: J = diffusion flux (given as 4.8 × 10⁻⁷ kg/m²-s) D = diffusion coefficient (given as 5.1 × 10⁻¹¹ m²/s) dc/dx = concentration gradient

Since we assume a linear concentration profile, the concentration gradient dc/dx is constant. Let's call this constant gradient as 'm.'

So, J = -D * m

We are given the concentration at the high-pressure surface as 4.7 kg/m³ and the desired concentration as 1.8 kg/m³. Let's call the distance into the sheet as 'x' (in meters) from the high-pressure side where the concentration is 1.8 kg/m³.

Now, we can set up the following equations:

At x = 0 (high-pressure side): Concentration (c) = 4.7 kg/m³

At x = distance into the sheet (where the concentration is 1.8 kg/m³): Concentration (c) = 1.8 kg/m³

Using the concentration gradient 'm,' the change in concentration (dc) is: dc = 1.8 kg/m³ - 4.7 kg/m³ = -2.9 kg/m³

The change in distance (dx) is: dx = x - 0 = x

Now, we can calculate the concentration gradient 'm' using the given diffusion flux 'J': J = -D * m 4.8 × 10⁻⁷ kg/m²-s = -5.1 × 10⁻¹¹ m²/s * m

Solving for 'm': m = (4.8 × 10⁻⁷ kg/m²-s) / (-5.1 × 10⁻¹¹ m²/s) m = -9.41176 × 10³ kg/m⁴-s

Now, we can find the distance 'x' using the concentration change (dc) and the concentration gradient 'm': dc/dx = m (dx)/dx = -2.9 kg/m³ / x

dx = -2.9 kg/m³ / m dx = -2.9 kg/m³ / (-9.41176 × 10³ kg/m⁴-s)

dx ≈ 3.079 × 10⁻⁴ m

So, the distance 'x' into the sheet from the high-pressure side, where the concentration is 1.8 kg/m³, is approximately 3.079 × 10⁻⁴ meters (or 0.3079 mm).