Question 8 of 8: Question VIII — Phase Diagram (20 marks)

Question VIII — Phase Diagram (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.

Given. The printed Cu–Ag eutectic diagram: eutectic point $T_E=779^{\circ}\text{C}$ at $C_E=71.9\ \text{wt\%Ag}$; $\alpha$-phase eutectic solubility $C_{\alpha E}=8.0\ \text{wt\%Ag}$ (point B); $\beta$-phase eutectic solubility $C_{\beta E}=91.2\ \text{wt\%Ag}$ (point G); pure-Cu and pure-Ag melting points read off the diagram at 1085°C and 962°C respectively (points A, F).

VIII.1 — Phases and compositions at 800°C, 60 wt%Ag alloy

Approach. Locate $C_0=60$ wt%Ag on the 800°C isotherm; since $800^{\circ}\text{C}$ lies just above the eutectic ($779^{\circ}\text{C}$), read the liquidus and $\alpha$-solidus boundary compositions at that temperature directly off the diagram (both curves are nearly straight over this narrow 21°C interval above the eutectic).

1. Read the phase-boundary compositions at 800°C. The $\alpha$/($\alpha$+L) boundary (solidus) reads $C_\alpha\approx8.3$ wt%Ag (very close to its eutectic value of 8.0, since the solidus is steep just above $T_E$); the ($\alpha$+L)/L boundary (liquidus) reads $C_L\approx66.5$ wt%Ag (interpolated from the printed liquidus between A at $(0,1085^{\circ}\text{C})$ and E at $(71.9,\,779^{\circ}\text{C})$).
2. Locate $C_0=60$ relative to the boundaries. Since $C_\alpha(8.3) < C_0(60) < C_L(66.5)$, the 800°C tie-line through $C_0=60$ falls inside the two-phase $\boxed{\alpha+L}$ region.
3. Report compositions. $$\begin{aligned} C_\alpha &\approx \boxed{8.3\ \text{wt\%Ag}\ (91.7\ \text{wt\%Cu})} \\ C_L &\approx \boxed{66.5\ \text{wt\%Ag}\ (33.5\ \text{wt\%Cu})} \end{aligned}$$

VIII.2 — Maximum solid solubility at 700°C

Approach. Read the two solvus curves (the $\alpha/(\alpha+\beta)$ boundary running from B down toward C, and the $\beta/(\alpha+\beta)$ boundary running from G down toward H) at $700^{\circ}\text{C}$, below the eutectic, where solid solubility is normal (not retrograde) and falls monotonically as temperature drops from $T_E$.

1. (b) Ag in Cu — the $\alpha$ solvus. At $779^{\circ}\text{C}$ the $\alpha$ boundary sits at 8.0 wt%Ag (point B); reading the solvus curve down to $700^{\circ}\text{C}$ gives $$C_\alpha(700^{\circ}\text{C}) \approx \boxed{5.2\ \text{wt\%Ag}}$$
2. (a) Cu in Ag — the $\beta$ solvus. At $779^{\circ}\text{C}$ the $\beta$ boundary sits at 91.2 wt%Ag (8.8 wt%Cu, point G); reading the solvus curve down to $700^{\circ}\text{C}$ gives a boundary composition of $\approx94.1$ wt%Ag, i.e. $$100-94.1 = \boxed{5.9\ \text{wt\%Cu}}$$

Both solubility limits fall as $T$ drops from the eutectic (8.0→5.2 wt%Ag for the $\alpha$ phase; 8.8→5.9 wt%Cu for the $\beta$ phase) — the normal behaviour of a simple eutectic system's terminal solid solutions, consistent with the solvus curves both bowing back toward their respective pure-metal corners (C near pure Cu, H near pure Ag) at low temperature.

VIII.3 — Eutectic reaction

A eutectic reaction is an invariant (zero-degree-of-freedom, fixed-temperature) three-phase reaction in which, on cooling, a liquid of one fixed composition transforms isothermally and simultaneously into two distinct solid phases of different, fixed compositions: $L \rightleftharpoons \alpha+\beta$. It is invariant because, by the Gibbs phase rule, three phases coexisting in a binary system pin both temperature and all compositions at a single point (the eutectic point) — there is no freedom to vary $T$ while three phases remain in equilibrium.

For the Cu–Ag system, using the eutectic point and the two flanking solid solubility limits read off the diagram: $$\boxed{L\,(71.9\ \text{wt\%Ag}) \xrightarrow{\ 779^{\circ}\text{C, cooling}\ } \alpha\,(8.0\ \text{wt\%Ag}) + \beta\,(91.2\ \text{wt\%Ag})}$$

VIII.4 — Lever rule: 55 wt%Ag–45 wt%Cu alloy at 800°C

Given. $C_0=55$ wt%Ag; from VIII.1, $C_\alpha(800^{\circ}\text{C})\approx8.3$ wt%Ag, $C_L(800^{\circ}\text{C})\approx66.5$ wt%Ag (same tie-line as VIII.1, different $C_0$).

Find. $W_\alpha$, $W_L$.

Approach. Since $C_\alpha(8.3)<C_0(55)<C_L(66.5)$, the alloy is also in the $\alpha+L$ field at 800°C; apply the lever rule across the same tie-line.

1. Lever-rule fractions. $$W_L = \frac{C_0-C_\alpha}{C_L-C_\alpha} = \frac{55-8.3}{66.5-8.3} = \frac{46.7}{58.2} = \boxed{0.802\ (80.2\%)}$$ $$W_\alpha = \frac{C_L-C_0}{C_L-C_\alpha} = \frac{66.5-55}{58.2} = \frac{11.5}{58.2} = \boxed{0.198\ (19.8\%)}$$
2. Check. $W_L+W_\alpha = 0.802+0.198=1.000$. ✓

Because $C_0=55$ sits closer to the liquidus ($C_L=66.5$) than to the solidus ($C_\alpha=8.3$) on the composition axis, the lever rule correctly assigns the larger mass fraction to $\alpha$'s "far" phase, liquid ($W_L=80.2\%$) — the lever always weights each phase by the opposite arm's length.