Question 4 of 8: Diffusion

Question IV: 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.

Part 1(a). Diffusion flux $J$ is the mass (or number) of atoms crossing a unit cross-sectional area per unit time, $J=\dfrac{M}{At}$ (units $\text{kg}/(\text{m}^2\cdot \text{s})$). Under steady-state conditions it is related to the concentration gradient by Fick's first law, $$J=-D\,\frac{dC}{dx},$$ where $D$ is the diffusion coefficient; the negative sign shows flux runs opposite to (i.e. down) the gradient.

Part 1(b). Diffusion proceeds from regions of high concentration to regions of low concentration — net atomic motion acts to eliminate the concentration gradient and drive the system toward a spatially uniform (minimum free-energy) composition, exactly analogous to heat flowing from hot to cold. The negative sign in Fick's first law encodes this directly: $dC/dx$ is positive in the direction of increasing concentration, so $J=-D\,dC/dx$ must point the opposite way.

Part 1(c). Diffusion rate increases with temperature, because $D$ follows an Arrhenius relation, $$D=D_0\exp\!\left(-\frac{Q_d}{RT}\right),$$ where $Q_d$ is the activation energy for the diffusive jump and $D_0$ a temperature-independent pre-exponential. Raising $T$ increases the fraction of atoms with enough thermal energy to surmount the energy barrier $Q_d$ for a jump into a neighbouring vacant site (or interstitial void), so $D$ — and hence the diffusion rate — rises exponentially with $T$.

Part 2 — Given. $C_s=1.00$ wt%, $C_0=0.20$ wt%, target $C_x= 0.60$ wt% at $x=0.75$ mm; error-function table above.

Find. The carburizing time at 900°C and at 1050°C.

Approach. Use the semi-infinite-solid constant-surface-concentration solution to Fick's second law to get a single target value of $Dt$, then divide by $D(T)$ (Arrhenius) at each temperature.

1. Target error-function argument. $$\frac{C_s-C_x}{C_s-C_0} =\frac{1.00-0.60}{1.00-0.20}=0.500=\text{erf}\!\left(\frac{x}{2\sqrt{Dt}}\right).$$ Interpolating the table between $Z=0.45\to0.4755$ and $Z=0.50\to0.5205$ for $\text{erf}(Z)=0.500$ gives $Z=0.4772$.
2. Solve for Dt. $$\frac{x}{2\sqrt{Dt}}=0.4772\ \Rightarrow\ \sqrt{Dt}=\frac{0.75\times10^{-3}}{2(0.4772)}=7.858\times10^{-4}\ \text{m},\qquad Dt=\boxed{6.175\times10^{-7}\ \text{m}^2}.$$ This target is the same at both temperatures — only $D$ changes.
3. Diffusion coefficient at each temperature (Arrhenius). $$D(900^{\circ}\text{C})=D_0e^{-Q_d/RT}=2.3\times10^{-5}\exp\!\left(\frac{-148\,000} {8.31\times1173.15}\right)=\boxed{5.87\times10^{-12}\ \text{m}^2/\text{s}},$$ $$D(1050^{\circ}\text{C})=2.3\times10^{-5}\exp\!\left(\frac{-148\,000}{8.31\times1323.15}\right) =\boxed{3.28\times10^{-11}\ \text{m}^2/\text{s}}.$$
4. Time at each temperature. $$t=\frac{Dt}{D}:\qquad t_{900}=\frac{6.175\times10^{-7}}{5.87\times10^{-12}}=1.052\times10^{5}\ \text{s} =\boxed{29.2\ \text{h}},\qquad t_{1050}=\frac{6.175\times10^{-7}}{3.28\times10^{-11}}=1.882\times10^{4}\ \text{s} =\boxed{5.23\ \text{h}}.$$ Raising the treatment temperature by 150 °C cuts the required time by better than a factor of five, the practical payoff of the Arrhenius $D(T)$ dependence.