Question 6 of 7: Phase and constituent fractions in an Al–8 wt% Cu alloy

Question 6: Phase and constituent fractions in an Al–8 wt% Cu alloy (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. An alloy of overall composition $C_{0} = 8$ wt% Cu, and the aluminium-rich end of the Al–Cu phase diagram reproduced on page 4. The invariant reaction printed on that diagram is in fact a eutectic at 548 °C, in which liquid decomposes into the aluminium-rich solid solution $\alpha$ and the intermetallic $\theta$ (CuAl₂); the question calls it the eutectoid temperature, and the arithmetic requested is identical. Compositions read from the printed boundaries:

Find. (a) the mass fractions of primary (proeutectic) $\alpha$ and primary $\theta$ immediately above 548 °C; (b) the total mass fractions of $\alpha$ and $\theta$ immediately below it; (c) the mass fractions of the eutectic $\alpha$ and eutectic $\theta$ alone; and (d) the mass fractions of $\alpha$ and $\theta$ at 350 °C.

Approach. Apply the lever rule on the tie line at each temperature. The alloy is hypoeutectic (8 wt% Cu lies between $C_{\alpha} = 5.65$ and $C_{E} = 33.2$), so on cooling it first deposits primary $\alpha$ and the last liquid to solidify has the eutectic composition. Immediately above the invariant the phases are $\alpha$ and liquid; immediately below they are $\alpha$ and $\theta$; and the eutectic constituent is simply whatever the liquid turned into, so the eutectic fractions are the differences between the totals and the primary amounts.

1. Part (a) — the tie line just above 548 °C. The two phases are $\alpha$ at 5.65 wt% Cu and liquid at the eutectic composition 33.2 wt% Cu. The lever rule puts the alloy composition at the fulcrum, so$$W_{\alpha'} = \frac{C_{E} - C_{0}}{C_{E} - C_{\alpha}} = \frac{33.2 - 8.0}{33.2 - 5.65} = \frac{25.2}{27.55} = \boxed{0.915}$$and the liquid that remains is $W_{L} = (8.0-5.65)/27.55 = 0.085$, which checks because the two must sum to unity.
2. Part (a) — primary $\theta$. Primary (proeutectic) $\theta$ can only form in a hypereutectic alloy, one whose composition lies to the copper-rich side of 33.2 wt% Cu, because only then does the alloy enter the $\theta + L$ field on cooling. At 8 wt% Cu the alloy enters the $\alpha + L$ field instead, so$$W_{\theta'} = \boxed{0}$$There is no primary $\theta$ in this alloy at any temperature. Saying so explicitly, with the reason, is what the part is testing.
3. Part (b) — the tie line just below 548 °C. The liquid has now transformed and the two phases present are $\alpha$ at 5.65 wt% Cu and $\theta$ at 52.5 wt% Cu. The tie line is much longer, so$$W_{\alpha} = \frac{C_{\theta} - C_{0}}{C_{\theta} - C_{\alpha}} = \frac{52.5 - 8.0}{52.5 - 5.65} = \frac{44.5}{46.85} = \boxed{0.950}$$$$W_{\theta} = \frac{C_{0} - C_{\alpha}}{C_{\theta} - C_{\alpha}} = \frac{2.35}{46.85} = \boxed{0.050}$$These are total phase fractions: they count every atom of $\alpha$ and of $\theta$ in the specimen, no matter which constituent it belongs to.
4. Part (c) — separate the eutectic constituent from the primary. The primary $\alpha$ formed above the invariant does not change as the liquid transforms, so the eutectic $\alpha$ is whatever total $\alpha$ is left over:$$W_{\alpha,\text{eut}} = W_{\alpha} - W_{\alpha'} = 0.9498 - 0.9147 = \boxed{0.035}$$All of the $\theta$ came out of the eutectic liquid, because there was no primary $\theta$, so$$W_{\theta,\text{eut}} = W_{\theta} = \boxed{0.050}$$The check is that the whole eutectic constituent must equal the liquid present just before the reaction: $0.0351 + 0.0502 = 0.0853 = W_{L}$, exactly as required.
5. Part (d) — the tie line at 350 °C. Below the invariant the $\alpha$ solvus falls steeply, and at 350 °C the printed curve gives $C_{\alpha} \approx 1.0$ wt% Cu while $\theta$ remains essentially stoichiometric at 52.5 wt% Cu. The tie line is longer still, so more $\theta$ must be present:$$W_{\alpha} = \frac{52.5 - 8.0}{52.5 - 1.0} = \frac{44.5}{51.5} = \boxed{0.864}$$and the balance of the specimen is the intermetallic,$$W_{\theta} = \frac{8.0 - 1.0}{51.5} = \frac{7.0}{51.5} = \boxed{0.136}$$The extra $\theta$ — the total rose from 0.050 to 0.136 — is the copper rejected from supersaturated $\alpha$ as the solvus falls. That rejection is the whole basis of precipitation hardening in the 2xxx aerospace alloys: solution treat above the solvus, quench to trap the copper, then age so that the $\theta$ appears as fine coherent $\theta''$ / $\theta'$ particles rather than the coarse equilibrium phase.