Question 6 of 8: Diffusion (20 marks)

Question VI — Diffusion (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Non-steady-state (Fickian) carburizing: bulk composition $C_0=0.2\%\text{C}$, surface composition (held fixed) $C_s=1.0\%\text{C}$, depth $x=0.3\ \text{mm}=0.03\ \text{cm}$, time $t_1=10\ \text{h}=36{,}000\ \text{s}$, $D=0.298\times10^{-6}\ \text{cm}^2/\text{s}$ (constant, single temperature — matches published $D$ for C in $\gamma$-Fe near 1000°C, so no anomaly flag is needed here).

Find. (i) $C(x{=}0.3\ \text{mm},\,t{=}10\ \text{h})$; (ii) the time $t_2$ for $C(0.3\ \text{mm})$ to reach 0.6%C.

Approach. Apply the standard semi-infinite-solid solution to Fick's second law, $\dfrac{C_s-C_x}{C_s-C_0}=\text{erf}\!\left(\dfrac{x}{2\sqrt{Dt}}\right)$ (appendix), reading/interpolating the appendix erf table.

1. Part (i) — compute the erf argument. $$\sqrt{Dt_1} = \sqrt{(0.298\times10^{-6})(36{,}000)} = \sqrt{0.010728} = 0.10358\ \text{cm}$$ $$z_1 = \frac{x}{2\sqrt{Dt_1}} = \frac{0.03}{2(0.10358)} = 0.1448$$
2. Read/interpolate erf$(z_1)$. From the appendix table, $\text{erf}(0.10)=0.1125$ and $\text{erf}(0.15)=0.1680$; interpolating at $z_1=0.1448$ (fraction 0.897 of the way from 0.10 to 0.15): $$\text{erf}(0.1448) \approx 0.1125+0.897(0.1680-0.1125) = 0.1623$$
3. Solve for $C_x$. $$\frac{1.0-C_x}{1.0-0.2} = 0.1623 \ \Rightarrow\ 1.0-C_x = 0.1623(0.8) = 0.1298$$ $$C_x = \boxed{0.870\ \%\text{C}}$$
4. Part (ii) — set up for $C_x=0.6\%$ at the same depth. $$\frac{C_s-C_x}{C_s-C_0} = \frac{1.0-0.6}{1.0-0.2} = 0.500 = \text{erf}(z_2)$$ From the appendix table, $\text{erf}(0.45)=0.4755$ and $\text{erf}(0.50)=0.5205$; interpolating for $\text{erf}(z_2)=0.500$ (fraction 0.544): $$z_2 \approx 0.45+0.544(0.05) = 0.477$$
5. Solve for $t_2$. Re-arranging $z_2=x/(2\sqrt{Dt_2})$: $$\sqrt{Dt_2} = \frac{x}{2z_2} = \frac{0.03}{2(0.477)} = 0.03145\ \text{cm} \ \Rightarrow\ Dt_2 = 9.89\times10^{-4}\ \text{cm}^2$$ $$t_2 = \frac{9.89\times10^{-4}}{0.298\times10^{-6}} = 3319\ \text{s} = \boxed{0.922\ \text{h}\ (\approx 55.3\ \text{min})}$$

Part (ii)'s answer being less than the 10 h of part (i) is the expected direction: 0.6%C is a smaller carbon pickup than the 0.87%C already reached after 10 h at that depth, so the shallower target is met sooner.