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Chapter 5: Problem 15

For a steel alloy it has been determined that a carburizing heat treatment of 10 -h duration will raise the carbon concentration to \(0.45 \mathrm{wt} \%\) at a point \(2.5 \mathrm{~mm}\) from the surface. Estimate the time necessary to achieve the same concentration at a 5.0-mm position for an identical steel and at the same carburizing temperature.

Short Answer

Expert verified
Use Fick's Second Law of Diffusion in your answer. Answer: It would take 40 hours for the carbon concentration to reach the same level at a 5 mm position in steel.

Step by step solution

01

Determine the formula for Fick's Second Law

For a diffusion process, Fick's Second Law is given as follows: \[C = \frac{C_s - C_0}{2} \times \text{erf}\left(\frac{x}{2\sqrt{Dt}}\right) \] Where: - \(C\) is the carbon concentration after a certain time - \(C_s\) is the carbon concentration at the surface - \(C_0\) is the initial carbon concentration - \(x\) is the distance from the surface where the concentration is measured - \(t\) is the time - \(D\) is the diffusion coefficient - erf is the error function, defined as erf(z) = \(\frac{2}{\sqrt{\pi}}\int_{0}^{z}e^{-t^2} dt\)
02

Rearrange for time

We need to find the time necessary, so we need to rearrange the equation for \(t\). To do this, we first need to isolate the error function on one side. \[ \text{erf} \left(\frac{x}{2\sqrt{Dt}}\right) = \frac{2(C - C_0)}{C_s - C_0}\] Now, we will find the inverse error function, which we denote as \(\text{erf}^{-1}(z)\). Then, we can isolate \(t\) on one side: \[t = \frac{x^2}{4D \left[\text{erf}^{-1}\left(\frac{2(C - C_0)}{C_s - C_0}\right)\right]^2}\]
03

Determine values for given information

Now we can use the given information to find the time for carbon concentration at a 5 mm position. We are given: - \(C = 0.45 \mathrm{wt}\%\) - \(x_1 = 2.5 \mathrm{mm}\) - \(x_2 = 5.0 \mathrm{mm}\) - \(t_1 = 10 \mathrm{h}\) From this information, we need to determine the relationship between the times when the carbon concentration reaches the same value at different positions. \(\frac{t_2}{t_1} = \frac{x_2^2}{x_1^2}\) Now we can solve for \(t_2\): \[t_2 = t_1 \frac{x_2^2}{x_1^2}\]
04

Calculate required time \(t_2\)

Now we can plug in the given values and solve for \(t_2\): \[t_2 = 10 \mathrm{h} \times \frac{(5.0 \mathrm{mm})^2}{(2.5 \mathrm{mm})^2}\] \[t_2 = 10 \mathrm{h} \times \frac{25 \mathrm{mm^2}}{6.25 \mathrm{mm^2}}\] \[t_2 = 10 \mathrm{h} \times 4\] \[t_2 = 40 \mathrm{h}\] Therefore, the time necessary to achieve the same carbon concentration at the 5.0-mm position is 40 hours.

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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 \(2.5 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\).

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 \(6.9 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\).

An FCC iron-carbon alloy initially containing \(0.20 \mathrm{wt} \% \mathrm{C}\) is carburized at an elevated temperature and in an atmosphere wherein the surface carbon concentration is maintained at \(1.0 \mathrm{wt} \%\). If after \(49.5 \mathrm{~h}\) the concentration of carbon is \(0.35 \mathrm{wt} \%\) at a position \(4.0 \mathrm{~mm}\) below the surface, determine the temperature at which the treatment was carried out.

A sheet of BCC iron \(1 \mathrm{~mm}\) thick was exposed to a carburizing gas atmosphere on one side and a decarburizing atmosphere on the other side at \(725^{\circ} \mathrm{C}\). After reaching steady state, the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces of the sheet were determined to be \(0.012\) and \(0.0075 \mathrm{wt} \%\). Compute the diffusion coefficient if the diffusion flux is \(1.4 \times 10^{-8} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\). Hint: Use Equation \(4.9\) to convert the concentrations from weight percent to kilograms of carbon per cubic meter of iron.

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 \(6.25 \times\) \(10^{-11} \mathrm{~m}^{2} / \mathrm{s}\)
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