The outer surface of a steel gear is to be hardened by increasing its carbon content. The carbon is to be supplied from an external carbon-rich atmosphere, which is maintained at an elevated temperature. A diffusion heat treatment at \(850^{\circ} \mathrm{C}(1123 \mathrm{~K})\) for \(10 \mathrm{~min}\) increases the carbon concentration to \(0.90 \mathrm{wt} \%\) at a position \(1.0 \mathrm{~mm}\) below the surface. Estimate the diffusion time required at \(650^{\circ} \mathrm{C}(923 \mathrm{~K})\) to achieve this same concentration also at a) 1.0-mm position. Assume that the surface carbon content is the same for both heat treatments, which is maintained constant. Use the diffusion data in Table \(5.2\) for C diffusion in \(\alpha\)-Fe.

Short Answer

Expert verified
Based on the step by step solution provided, here's a short answer to the problem: To find the required diffusion time at a higher temperature for the same carbon concentration at a specific position below the surface of a steel gear, follow these steps: 1. Calculate the diffusion coefficient (D) at the given temperature 600°C (873K) for carbon diffusion in alpha-iron using the Arrhenius equation. 2. Estimate the semi-infinite solid solution for carbon concentration at a position of 0.5mm below the surface using Fick's second law and the calculated value of D(873K). 3. Calculate the diffusion coefficient (D) at the higher temperature 900°C (1173K) for carbon diffusion in alpha-iron. 4. Estimate the required diffusion time at the higher temperature to achieve the same carbon concentration at a 0.5mm position using Fick's second law and the calculated value of D(1173K).

Step by step solution

01

Find the diffusion coefficient at the given temperature

Calculate the diffusion coefficient (D) at the given temperature 600°C (873K) for carbon diffusion in alpha-iron using the Arrhenius equation: $$ D = D_0\exp\left(\frac{-Q}{RT}\right) $$ where \(D_0\) is the pre-exponential factor, \(Q\) is the activation energy for diffusion, \(R\) is the gas constant (8.314 J/molK), and \(T\) is the temperature in Kelvin. From Table 5.2 for carbon diffusion in alpha-iron, \(D_0\) = \(2.3 \times 10^{-5}\) m²/s and \(Q\) = 1.47 x 10^5 J/mol. At 873K, $$ D(873\,\mathrm{K}) = (2.3 \times 10^{-5}\mathrm{ m^2/s})\exp\left(\frac{-1.47\times 10^5\, \mathrm{J/mol}}{(8.314\,\mathrm{J/molK})(873\,\mathrm{K})}\right) $$ Calculate the value of \(D(873\,\mathrm{K})\).
02

Estimate the semi-infinite solid solution for carbon concentration

Use Fick's second law for the semi-infinite solid solution to estimate the required carbon concentration at a position of 0.5mm below the surface: $$ C(x,t) = C_s - (C_s - C_0)\text{erf}\left(\frac{x}{(2Dt)^{1/2}}\right) $$ where \(C(x,t)\) is the carbon concentration at a position \(x\) and time \(t\), \(C_s\) is the surface carbon concentration, \(C_0\) is the initial carbon concentration, and erf is the error function. We are given that the heat treatment at 600°C (873K) for 100 minutes increases the carbon concentration to 0.75 wt.% at a 0.5mm (5 x 10^-4m) position. Solve for \(C_s\) using the calculated value of \(D(873\,\mathrm{K})\).
03

Find the diffusion coefficient at the higher temperature

Calculate the diffusion coefficient (D) at the higher temperature 900°C (1173K) for carbon diffusion in alpha-iron: $$ D(1173\,\mathrm{K}) = (2.3 \times 10^{-5}\mathrm{ m^2/s})\exp\left(\frac{-1.47\times 10^5\, \mathrm{J/mol}}{(8.314\,\mathrm{J/molK})(1173\,\mathrm{K})}\right) $$ Calculate the value of \(D(1173\,\mathrm{K})\).
04

Estimate the required diffusion time at the higher temperature

Use Fick's second law for the semi-infinite solid solution to estimate the required diffusion time at the higher temperature to achieve the same carbon concentration at a 0.5mm position: $$ \frac{x}{(2Dt)^{1/2}} = \text{erf}^{-1}\left(1 - \frac{C_0 - C(x,t)}{C_s - C_0}\right) $$ We know the value of \(C_s\), \(C(x,t)\) (0.75 wt.%), \(x\) = 0.5 mm (5 x 10^-4 m), and the diffusion coefficient \(D(1173\,\mathrm{K})\). Solve for \(t\) (time).

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Most popular questions from this chapter

The activation energy for the diffusion of carbon in chromium is \(111,000 \mathrm{~J} / \mathrm{mol}\). Calculate the diffusion coefficient at \(1100 \mathrm{~K}\left(827^{\circ} \mathrm{C}\right)\), given that \(D\) at \(1400 \mathrm{~K}\left(1127^{\circ} \mathrm{C}\right)\) is \(6.25 \times\) \(10^{-11} \mathrm{~m}^{2} / \mathrm{s}\)

Aluminum atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat treatments; the background concentration of \(\mathrm{Al}\) in this silicon material is known to be \(3 \times 10^{19}\) atoms/m \(^{3}\). The drive-in diffusion treatment is to be carried out at \(1050^{\circ} \mathrm{C}\) for a period of \(4.0 \mathrm{~h}\), which gives a junction depth \(x_{j}\) of \(3.0 \mu \mathrm{m}\). Compute the predeposition diffusion time at \(950^{\circ} \mathrm{C}\) if the surface concentration is maintained at a constant level of \(2 \times 10^{25}\) atoms \(/ \mathrm{m}^{3}\). For the diffusion of \(\mathrm{Al}\) in \(\mathrm{Si}\), values of \(Q_{d}\) and \(D_{0}\) are \(3.41\). \(\mathrm{eV} /\) atom and \(1.38 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\), respectively.

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.

(a) Compare interstitial and vacancy atomic mechanisms for diffusion. (b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion.

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.

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