21-Mat-A4 Deformation Behaviour and Properties of Materials · May 2013
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Exams, May 2013 — 10-Met-A4, Structure of Materials. Three hours, closed book, one approved calculator (Casio or Sharp). Eight questions of 20 marks each; the rubric asks for any five, and only the first five in the answer book are marked. All eight are solved here, because this set is a study resource rather than an exam script. All necessary constants and equations are provided in the exam's own appendix; standard SI values (Planck's constant, electron mass, Avogadro's number) are used below and are noted where that happens.
The printed exam header reads 10-Met-A4, Structure of Materials. Two of the eight questions (V and VI) are genuinely deformation/mechanical-properties questions, but the paper as a whole is a broad introductory materials-science survey — bonding, crystallography, defects, diffusion, dislocations, XRD and phase diagrams — and is answered as such below.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
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.
(a) Diffusion flux $J$ is the mass (or number) of atoms crossing a unit cross-sectional area per unit time, $J = \frac{1}{A}\frac{dM}{dt}$, with SI units $\text{kg}/(\text{m}^2\cdot\text{s})$ or $\text{atoms}/(\text{m}^2\cdot\text{s})$. Fick's first law states that the flux is proportional to, and driven by, the local concentration gradient: $$J = -D\frac{dC}{dx}$$ where $D$ is the diffusion coefficient and $dC/dx$ is the concentration gradient along the diffusion direction. The negative sign is essential: it makes $J$ point down the gradient.
(b) Diffusion proceeds from regions of high concentration to regions of low concentration (down the gradient), never the reverse under a purely diffusive (non-driven) process. This follows directly from the negative sign in Fick's first law and, more fundamentally, from the second law of thermodynamics: random atomic jumps are equally likely in every direction, but because there are simply more atoms available to jump away from a high-concentration region than into it, the net flux is always down-gradient, which increases the system's configurational entropy (mixing) until the concentration is uniform.
(c) The 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 diffusion mechanism (e.g. vacancy migration) and $R$ is the gas constant. Higher $T$ exponentially increases the fraction of atoms with enough thermal energy to jump over the local energy barrier $Q_d$, so $D$ — and hence the flux for a given gradient — rises sharply with temperature.
Given.
| Quantity | Value |
|---|---|
| Surface carbon content, $C_s$ | 1.00 wt% |
| Initial (bulk) carbon content, $C_0$ | 0.20 wt% |
| Target carbon content, $C_x$, at depth $x$ | 0.60 wt% |
| Depth, $x$ | 0.75 mm $=7.5\times10^{-4}$ m |
| $D_0$ (carbon in γ-Fe) | $5\times10^{-5}\ \text{m}^2/\text{s}$ |
| Activation energy, $Q_d$ | 284 kJ/mol |
Find. The carburizing time at 900°C and at 1050°C.
Approach. Use the constant-surface-concentration solution of Fick's second law to find the dimensionless argument $Z$ from the given concentrations, look up $Z$ via the supplied error-function table, compute $D$ at each temperature from the Arrhenius relation, then solve $Z=x/(2\sqrt{Dt})$ for $t$.
Check. These times follow directly and consistently from the exam's own stated $D_0$ and $Q_d$, but they are far larger than a real carburizing treatment (hours, not decades). The supplied $Q_d=284\ \text{kJ/mol}$ matches published values for iron self-diffusion in γ-Fe, not for interstitial carbon diffusion in γ-Fe (textbook values for C in γ-Fe are typically $Q_d\approx148\ \text{kJ/mol}$, $D_0\approx2.3\times10^{-5}\ \text{m}^2/\text{s}$, which would give hour-scale times). The calculation above is carried through exactly as instructed on the exam's own given data; the magnitude is flagged here rather than silently smoothed over, per the "state your assumptions" instruction on the paper's cover page.
| Quantity | Value |
|---|---|
| $Z$ | 0.4772 |
| $D$ at 900 °C | $1.13\times10^{-17}$ m²/s |
| $D$ at 1050 °C | $3.07\times10^{-16}$ m²/s |
| Time at 900 °C | $5.46\times10^{10}$ s ($\approx$1732 yr) |
| Time at 1050 °C | $2.01\times10^{9}$ s ($\approx$63.8 yr) |
