Question 7 of 8: Question VII — Diffusion and Case Hardening (20 marks)

Question VII — Diffusion and Case Hardening (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.

VII.1 — Activation energy from atomic jump rate

Given. Jump rate $R_1=5\times10^8$ /s at $T_1=500^{\circ}\text{C}=773$ K; $R_2=8\times10^8$ /s at $T_2=800^{\circ}\text{C}=1073$ K.

Find. Activation energy $Q$ (cal/mol).

Approach. The jump rate follows the Arrhenius form $R=R_0\exp(-Q/RT)$ (the appendix's $N_D=N\exp(-Q/kT)$ relation applied per-atom in molar form); taking the ratio at two temperatures eliminates the unknown pre-exponential $R_0$.

1. Arrhenius ratio. $$\ln\left(\frac{R_2}{R_1}\right)=-\frac{Q}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)=\frac{Q}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$$
2. Solve for $Q$ (using $R_{gas}=1.987$ cal/mol·K): $$Q=\frac{R_{gas}\ln(R_2/R_1)}{1/T_1-1/T_2}=\frac{1.987\ln(1.6)}{(1/773-1/1073)}=\frac{1.987(0.4700)}{3.618\times10^{-4}}=\boxed{2580\ \text{cal/mol}}$$

This computed value is solved exactly on the two rates and temperatures given in the question. It is noticeably lower than the tens-of-kcal/mol activation energies typically quoted for interstitial diffusion in metals (the given rate only grows by a modest 1.6× across a 300°C span, which an Arrhenius process with a large true $Q$ would not do) — the numbers as printed nonetheless solve consistently to 2580 cal/mol, and that is reported as the answer this specific data set supports.

VII.2 — Carburizing: carbon content at 0.2 mm after 8 h

Given. $C_s=1.0$ wt%C (surface, held constant), $C_0=0.2$ wt%C (bulk), $x=0.2$ mm, $t=8$ h, $D=2.98\times10^{-11}\ \text{m}^2/\text{s}$.

Find. $C_x$, the carbon content at $x=0.2$ mm.

Approach. Semi-infinite-solid diffusion with a fixed surface concentration (the appendix's own error-function solution): $$\frac{C_s-C_x}{C_s-C_0}=\text{erf}\left(\frac{x}{2\sqrt{Dt}}\right)$$ evaluate the erf argument, then interpolate the appendix's own error-function table.

1. Convert and compute $Dt$. $x=2\times10^{-4}$ m, $t=8(3600)=28{,}800$ s: $$Dt=(2.98\times10^{-11})(28{,}800)=8.582\times10^{-7}\ \text{m}^2,\qquad \sqrt{Dt}=9.264\times10^{-4}\ \text{m}$$
2. Erf argument. $$z=\frac{x}{2\sqrt{Dt}}=\frac{2\times10^{-4}}{2(9.264\times10^{-4})}=0.1079$$
3. Interpolate the appendix's error-function table between $z=0.10$ (erf$=0.1125$) and $z=0.15$ (erf$=0.1680$): $$\text{erf}(0.1079)\approx0.1125+\frac{(0.1079-0.10)}{(0.15-0.10)}(0.1680-0.1125)=\boxed{0.1213}$$
4. Solve for $C_x$. $$C_x=C_s-(C_s-C_0)\,\text{erf}(z)=1.0-(1.0-0.2)(0.1213)=1.0-0.0971=\boxed{0.903\ \text{wt\%C}}$$

This is physically sensible: the characteristic diffusion depth $\sqrt{Dt}=0.93$ mm is more than 4× the 0.2 mm query depth, so this point sits well inside the carburized zone and its composition (0.903 wt%C) is close to the surface value (1.0 wt%C) rather than the bulk value (0.2 wt%C) — consistent with a shallow point being carburized almost to saturation within an 8-hour treatment.