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Chapter 5: Problem 7

A sheet of steel \(1.5 \mathrm{~mm}\) thick has nitrogen atmospheres on both sides at \(1200^{\circ} \mathrm{C}\) and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is \(6 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\), and the diffusion flux is found to be \(1.2 \times\) \(10^{-7} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\). Also, it is known that the concentration of nitrogen in the steel at the highpressure surface is \(4 \mathrm{~kg} / \mathrm{m}^{3} .\) How far into the sheet from this high-pressure side will the concentration be \(2.0 \mathrm{~kg} / \mathrm{m}^{3}\) ? Assume a linear concentration profile.

Short Answer

Expert verified
Answer: The concentration of nitrogen will be 2.0 kg/m³ at a distance of 1.0 mm from the high-pressure side of the steel sheet.

Step by step solution

01

Write down the given information.

We are given the following information: - Thickness of the steel sheet, \(t = 1.5\) mm - Temperature, \(T = 1200^{\circ} \mathrm{C}\) - Diffusion coefficient for nitrogen in steel, \(D = 6 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\) - Diffusion flux, \(J = 1.2 \times 10^{-7} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) - Concentration at the high-pressure surface, \(C_{1} = 4 \mathrm{~kg} / \mathrm{m}^{3}\) - Required concentration, \(C_{2} = 2.0 \mathrm{~kg} / \mathrm{m}^{3}\)
02

Convert thickness to meters.

Convert the thickness of the steel sheet from millimeters to meters: \(t = 1.5 \mathrm{~mm} \times \frac{1 \mathrm{~m}}{1000 \mathrm{~mm}} = 1.5 \times 10^{-3} \mathrm{~m}\)
03

Apply Fick's first law of diffusion.

Fick's first law relates the diffusion flux to the concentration gradient and the diffusion coefficient: \(J = -D \frac{dC}{dx}\) Since we have a linear concentration profile and the boundaries(Concentrations), we can use the following relation for the concentration gradient: \(\frac{dC}{dx} = \frac{C_1 - C_2}{x}\) Combining the above, we get: \(J = -D \frac{C_1 - C_2}{x}\)
04

Solve for the distance, \(x\).

Now we will solve the above equation for \(x\): \(x = -D \frac{C_1 - C_2}{J}\) Plugging in the given values, we get: \(x = - (6 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}) \frac{( 4 \mathrm{~kg} / \mathrm{m}^{3} ) - ( 2.0 \mathrm{~kg} / \mathrm{m}^{3})}{(1.2 \times 10^{-7} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s})}\) Simplifying, we obtain the distance: \(x = 1 \times 10^{-3} \mathrm{~m}\) So, the concentration will be \(2.0 \mathrm{~kg} / \mathrm{m}^{3}\) at a distance of 1.0 mm from the high-pressure side of the steel sheet.

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Most popular questions from this chapter

Nitrogen from a gaseous phase is to be dif? fused into pure iron at \(700^{\circ} \mathrm{C}\). If the surface concentration is maintained at \(0.1 \mathrm{wt} \% \mathrm{~N}\) what will be the concentration \(1 \mathrm{~mm}\) from the surface after \(10 \mathrm{~h}\) ? The diffusion coefficient for nitrogen in iron at \(700^{\circ} \mathrm{C}\) is \(2.5 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\).

Briefly explain the difference between selfdiffusion and interdiffusion.

The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of \(0.20 \mathrm{~m}^{2}\) at \(500^{\circ} \mathrm{C}\). Assume a diffusion coefficient of \(1.0 \times 10^{-8} \mathrm{~m}^{2} / \mathrm{s}\), that the concentrations at the high- and low-pressure sides of the plate are \(2.4\) and \(0.6 \mathrm{~kg}\) of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

An FCC iron-carbon alloy initially containing \(0.35 \mathrm{wt} \% \mathrm{C}\) is exposed to an oxygen-rich and virtually carbon-free atmosphere at \(1400 \mathrm{~K}\) (1127 \(\left.^{\circ} \mathrm{C}\right)\). Under these circumstances the carbon diffuses from the alloy and reacts at the surface, with the oxygen in the atmosphere; that is, the carbon concentration at the surface position is maintained essentially at \(0 \mathrm{wt} \%\) C. (This process of carbon depletion is termed decarburization.) At what position will the carbon concentration be \(0.15 \mathrm{wt} \%\) after a 10 -h treatment? The value of \(D\) at \(1400 \mathrm{~K}\) is \(6.9 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\).

Phosphorus atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat treatments; the background concentration of \(\mathrm{P}\) in this silicon material is known to be \(5 \times 10^{19}\) atoms \(/ \mathrm{m}^{3}\). The predeposition treatment is to be conducted at \(950^{\circ} \mathrm{C}\). for 45 minutes; the surface concentration of \(P\) is to be maintained at a constant level of \(1.5 \times 10^{26}\) atoms \(/ \mathrm{m}^{3}\). Drive-in diffusion will be carried out at \(1200^{\circ} \mathrm{C}\) for a period of \(2.5 \mathrm{~h}\). For the diffusion of \(\mathrm{P}\) in \(\mathrm{Si}\), values of \(Q_{d}\) and \(D_{0}\) are \(3.40 \mathrm{eV} /\) atom and \(1.1 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\), respectively. (a) Calculate the value of \(Q_{0}\). (b) Determine the value of \(x_{j}\) for the drive-in diffusion treatment. (c) Also for the drive-in treatment, compute the position \(x\) at which the concentration of \(P\) atoms is \(10^{24} \mathrm{~m}^{-3}\).
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