Question 5 of 8

Question 5 (15 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 (a) — phase field labeling. The Mg–Pb system is two simple eutectic sub-systems joined at a congruently-melting intermetallic compound (essentially stoichiometric Mg₂Pb, whose 81 wt% Pb composition matches $M_{Pb}/(2M_{Mg}+M_{Pb}) = 207.2/255.8 = 81.0\%$). Reading left to right across the diagram:

[Figure not reproduced: Mg–Pb binary phase diagram redrawn from the source's key-point readings, with all phase fields labeled and the 75 wt% Pb alloy's cooling path (dashed) and 400°C tie line shown in red. See the official exam paper.]

1. Part (a) — the eight fields. $L$ (all-liquid, above both liquidus branches); $\alpha$ (Mg-rich terminal solid solution, upper left); $\alpha+\text{Mg}_2\text{Pb}$ (below the 465°C eutectic on the Mg-rich side); $L+\text{Mg}_2\text{Pb}$ (two mirror-image wedges flanking the compound, between each liquidus branch and the vertical compound line); $\text{Mg}_2\text{Pb}$ (the essentially vertical single-phase field at 81 wt% Pb); $\text{Mg}_2\text{Pb}+\beta$ (below the 250°C eutectic on the Pb-rich side); $\beta$ (Pb-rich terminal solid solution, upper right, above its solvus).

Part (b) Given.

Find. (i) first solid composition; (ii) last-liquid composition; (iii) microstructure and phases at 400°C; (iv) mass fraction of each phase at 400°C.

Approach. Since $C_0=75\%$ falls between the eutectic-1 liquid (41%) and the compound (81%), track the liquidus branch that rises from (41%, 465°C) to (81%, 550°C), apply the lever rule at each stage, and use the eutectic reaction to find the last liquid.

1. Part (b)(i) — first solid. At 600°C the alloy is single-phase liquid at 75 wt% Pb. Cooling, the composition point crosses the $L$/($L+\text{Mg}_2\text{Pb}$) liquidus (the branch running from the 465°C eutectic up to the 550°C compound peak) before reaching any other boundary, since 75% lies between 41% and 81%. Because Mg₂Pb is a line (stoichiometric) compound, every particle of primary solid that nucleates has the compound's fixed composition regardless of the melt's own composition: $$\boxed{\text{first solid} = \text{Mg}_2\text{Pb},\ 81\ \text{wt\% Pb}}$$
2. Part (b)(ii) — last liquid. As Mg₂Pb (81% Pb) precipitates from a liquid that started at 75% Pb, the remaining liquid must become progressively leaner in Pb to conserve the overall mass balance ($C_L f_L + 81\%\,f_S = 75\%$, with $f_S$ growing from 0). The liquid composition therefore slides down the liquidus toward the eutectic point as temperature falls, and the last liquid to exist — immediately before it is fully consumed by the eutectic reaction $L\to\alpha+\text{Mg}_2\text{Pb}$ — is the eutectic liquid itself: $$\boxed{\text{last liquid} = 41\ \text{wt\% Pb, at } 465^\circ\text{C}}$$
3. Part (b)(iii) — microstructure at 400°C. 400°C is below the 465°C eutectic, so solidification is complete. The microstructure consists of primary (proeutectic) Mg₂Pb grains, formed above 465°C, embedded in a fine eutectic mixture of $\alpha$ and Mg₂Pb formed when the last liquid (41% Pb) crystallized at the eutectic. Only two phases are present overall: $\alpha$ and Mg₂Pb — the eutectic $\alpha$ and eutectic Mg₂Pb are the same phases as the pre-eutectic Mg₂Pb, just in a different (lamellar/fine) morphology, so the two-phase count does not increase.
4. Part (b)(iv) — mass fractions via the lever rule. At 400°C, the tie line runs from the $\alpha$ solvus to the fixed Mg₂Pb composition (81%). The diagram's own numeric readings give the $\alpha$ solvus composition only at the eutectic temperature (19 wt% Pb, at 465°C); no lower-temperature solvus value is given in the source, so that 19% boundary is used here as the best available approximation (see check note below). Lever rule with $C_0=75\%$, $C_\alpha=19\%$, $C_{\text{Mg}_2\text{Pb}}=81\%$: $$f_\alpha = \frac{C_{\text{Mg}_2\text{Pb}}-C_0}{C_{\text{Mg}_2\text{Pb}}-C_\alpha} = \frac{81-75}{81-19} = \frac{6}{62}$$ $$\boxed{f_\alpha \approx 9.7\%,\qquad f_{\text{Mg}_2\text{Pb}} = 1-f_\alpha \approx 90.3\%}$$