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21-Mat-A4 Deformation Behaviour and Properties of Materials · May 2014

Question 4 of 8: Diffusion (20 marks)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2014 — 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 equations, constants and the error-function table are provided in the exam's own appendix and are used directly below.

Note on the exam title. The printed exam header reads 10-Met-A4, Structure of Materials. Only two of the eight questions (VI and VIII) are substantially deformation/mechanical-properties content; the paper as a whole is a broad introductory materials-science survey — bonding, crystallography, polymers, diffusion, XRD, phase diagrams, dislocations — and is answered as such below.

Check — figure-read values. Question VI.3's stress-strain curve and Question VII's Cu–Ag solvus/liquidus positions are read from the printed figures rather than given numerically. Graphically-read values carry a few percent uncertainty that closed-form calculations do not — this is flagged again at the point of use.

Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:

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.

IV.1 — Steady-state diffusion flux

Given. $C_1=1.2\ \text{kg/m}^3$ at $x_1=5$ mm, $C_2=0.8\ \text{kg/m}^3$ at $x_2=10$ mm beneath the carburizing surface; $D=3\times10^{-11}\ \text{m}^2/\text{s}$.

Find. Steady-state diffusion flux $J$.

Approach. Fick's first law, $J=-D\,dC/dx$, with the concentration gradient taken directly from the two given points (linear in steady state).

  1. Concentration gradient. $\dfrac{\Delta C}{\Delta x} = \dfrac{0.8-1.2}{(10-5)\times10^{-3}} = -80\ \text{kg/m}^4$ (concentration falls with depth, as expected moving away from the carburizing face).
  2. Flux. $$J = -D\frac{\Delta C}{\Delta x} = -(3\times10^{-11})(-80) = \boxed{2.40\times10^{-9}\ \text{kg/(m}^2\text{s)}}$$

IV.2 — Non-steady-state carburizing time

Given. Surface concentration $C_s=1.20$ wt%, bulk (initial) $C_0=0.25$ wt%, target $C_x=0.80$ wt% at $x=0.5$ mm; $D=1.6\times10^{-11}\ \text{m}^2/\text{s}$; error-function table from the exam appendix.

Find. Time $t$ to reach $C_x$ at that depth.

Approach. Fick's second-law solution for a constant-surface-concentration semi-infinite solid, $\dfrac{C_s-C_x}{C_s-C_0}=\text{erf}\!\left(\dfrac{x}{2\sqrt{Dt}}\right)$; find $z$ from the table by interpolation, then solve for $t$.

  1. Target erf value. $$\frac{C_s-C_x}{C_s-C_0} = \frac{1.20-0.80}{1.20-0.25} = \frac{0.40}{0.95} = 0.4211$$
  2. Interpolate $z$ from the appendix table. Table brackets: $\text{erf}(0.35)=0.3794$, $\text{erf}(0.40)=0.4284$. $$z = 0.35 + (0.4211-0.3794)\frac{0.40-0.35}{0.4284-0.3794} = \boxed{0.3925}$$
  3. Solve for time. $z = x/(2\sqrt{Dt}) \Rightarrow t = \dfrac{x^2}{4Dz^2}$: $$t = \frac{(0.5\times10^{-3})^2}{4(1.6\times10^{-11})(0.3925)^2} = \boxed{2.54\times10^{4}\ \text{s} \approx 7.04\ \text{h}}$$
Question IV — final results
QuantityValue
Steady-state flux, $J$$2.40\times10^{-9}$ kg/(m$^2$s)
Interpolated $z$$0.3925$
Carburizing time, $t$$2.54\times10^4$ s $\approx 7.04$ h