Be–Si Eutectic System — Diagram and Solidification Paths: 04-BS-11, December 2016, Q3

Question 3: Be–Si Eutectic System — Diagram and Solidification Paths (20 marks)

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Given. $T_{m,Si}=1414^\circ$C, $T_{m,Be}=1252^\circ$C, complete liquid solubility, zero solid solubility (simple eutectic, pure-Si and pure-Be solid phases only); eutectic point $x_E=39$ wt% Be, $T_E=1090^\circ$C.

Find. The phase diagram with labelled fields; the solidification sequence and % eutectic for (a) 90% Be, (b) 30% Be.

Approach

Because the two elements are completely insoluble in the solid state, this is the simplest possible binary eutectic: only three solid-state phases ever appear — pure Si, pure Be, and the eutectic mixture of the two. The liquidus falls from each pure-element melting point down to the eutectic point; the amount of eutectic constituent in any alloy is found from the lever rule applied at the eutectic isotherm, using the eutectic liquid composition and the relevant pure-element composition as the tie-line endpoints.

Fig. Q3 — Be–Si thermal equilibrium diagram. $L$ = liquid; $L+\text{Si}$ and $L+\text{Be}$ = two-phase liquid + primary-solid fields; below 1090 °C the entire diagram is the two-phase solid field Si + Be (eutectic mixture plus whichever primary phase formed above the eutectic isotherm). Dashed lines mark the two alloy compositions.

Fields. Above both liquidus branches: all-liquid, $L$. Between the Si-liquidus (0–39% Be) and the 1090 °C isotherm: $L+\text{Si}$ (primary/proeutectic Si plus liquid). Between the Be-liquidus (39–100% Be) and the isotherm: $L+\text{Be}$ (primary Be plus liquid). Below 1090 °C, for every composition strictly between 0 and 100% Be: Si + Be (two solid phases, since there is no solid solubility at all).
(a) 90% Be alloy — solidification path. Cooling from the liquid, the alloy first crosses the Be-liquidus (Be-rich side, since 90%>39%): primary (proeutectic) Be crystals begin to nucleate and grow, and the remaining liquid’s composition slides down the liquidus away from pure Be and toward the eutectic point (39% Be) as more Be solidifies out. At 1090 °C the remaining liquid has reached exactly the eutectic composition (39% Be) and transforms isothermally into the fine, alternating Si+Be eutectic constituent. The final room-temperature microstructure is: primary Be grains (large, first-formed) embedded in a eutectic (Si+Be) matrix.
(a) % eutectic by the lever rule. At the eutectic isotherm, the tie line runs from pure Be ($x=100$) to the eutectic liquid ($x=39$); the overall alloy composition ($x=90$) divides it: $$\%\text{eutectic}=\frac{100-90}{100-39}\times100=\frac{10}{61}\times100$$ $$\boxed{\%\text{eutectic}\approx16.4\%}\qquad(\text{primary Be}\approx83.6\%)$$
(b) 30% Be alloy — solidification path. This composition is on the Si-rich (hypoeutectic) side. Cooling crosses the Si-liquidus first: primary (proeutectic) Si crystals form, and the remaining liquid slides down the liquidus toward the eutectic point (39% Be) as Si is removed from it — note this means the liquid becomes richer in Be as solidification proceeds, the opposite trend from case (a). At 1090 °C the remaining liquid (now at 39% Be) transforms to the Si+Be eutectic. Final microstructure: primary Si grains in a eutectic (Si+Be) matrix.
(b) % eutectic by the lever rule. Tie line from pure Si ($x=0$) to the eutectic liquid ($x=39$); overall composition $x=30$: $$\%\text{eutectic}=\frac{30-0}{39-0}\times100=\frac{30}{39}\times100$$ $$\boxed{\%\text{eutectic}\approx76.9\%}\qquad(\text{primary Si}\approx23.1\%)$$

Quantity	Result
(a) 90% Be: primary Be	83.6%
(a) 90% Be: eutectic	16.4%
(b) 30% Be: primary Si	23.1%
(b) 30% Be: eutectic	76.9%

Concept & Theory

Topic. Simple (zero solid-solubility) binary eutectic phase diagrams; proeutectic phase identification; the lever rule.

Key relations. Lever rule at the eutectic isotherm: $\%\text{eutectic}=\dfrac{|x_{alloy}-x_{pure}|}{|x_E-x_{pure}|}\times100$, where $x_{pure}$ is whichever pure-element composition (0 or 100) is on the same side of the eutectic as the alloy.

Why this works. With zero solid solubility, the tie line at any temperature just above $T_E$ always runs from the pure primary phase to the liquidus — so the lever rule directly gives the fraction of liquid present the instant before the eutectic reaction, and that liquid is exactly what transforms into the eutectic constituent.

Common pitfall. Using the wrong tie-line endpoint (e.g. pure Si when the alloy is on the Be-rich side) or forgetting that primary-phase fraction and eutectic fraction must sum to 100% — a quick check on every lever-rule answer.

Source. Callister & Rethwisch, Materials Science and Engineering, 9th ed., Ch. 9 (binary eutectic phase diagrams, the lever rule, microstructural development).

Question 3 of 7: Be–Si Eutectic System — Diagram and Solidification Paths

Question 3: Be–Si Eutectic System — Diagram and Solidification Paths (20 marks)

Approach