Question 4 of 7: Pb-Sn Eutectic Phase Diagram — Lever-Rule Analysis of a 30% Sn and an 80% Sn Alloy
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — May 2016. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Candidates attempt any
five of the seven questions for a complete paper, all questions of equal value. All seven
questions are solved below for completeness.
Given. Pb-Sn eutectic diagram: eutectic point $183^\circ$C at
$61.9\,$wt%Sn; $\alpha$ (Pb-rich) terminal solubility $19\,$wt%Sn and $\beta$ (Sn-rich) terminal
solubility $97.5\,$wt%Sn, both at the eutectic temperature; pure-metal melting points
$T_{Pb}=327^\circ$C, $T_{Sn}=232^\circ$C (diagram end-points); two alloys, 30wt%Sn (part a) and
80wt%Sn (part b), both slow-cooled from $350^\circ$C.
Find. For each alloy: (i) first-solid composition, (ii) freezing range,
phase amounts/compositions at (iii) $184^\circ$C, (iv) $182^\circ$C, (v) $25^\circ$C.
[Figure not reproduced: Fig. Q4 — Pb-Sn eutectic diagram redrawn from the source (eutectic $183^\circ$C/61.9%Sn, terminal solubilities 19% and 97.5%Sn), with the two alloy compositions of parts (a) and (b) marked. See the official exam paper.]
Approach
Slow cooling means every step is treated as equilibrium. The liquidus locates where the first
solid appears (approximated as straight lines between the diagram’s labelled points, since
both hypoeutectic and hypereutectic liquidus branches are drawn essentially straight on the
source figure); the two-phase tie lines at each requested temperature are read from the
liquidus/solidus (above the eutectic isotherm) or solidus/solvus (below it) boundaries, and phase
fractions follow from the lever rule. Room-temperature ($25^\circ$C) solvus compositions are read
from the sharply-curving low-temperature boundaries of the diagram ($\approx2$wt%Sn for
$\alpha$, $\approx99$wt%Sn for $\beta$).
(a)(i) First solid, 30% Sn. $30\%<61.9\%$ (hypoeutectic side), so cooling
first crosses the hypoeutectic liquidus. Interpolating that branch (from
$(0\%,327^\circ$C$)$ to $(61.9\%,183^\circ$C$)$) at 30%Sn:
$$T_{liq}=327-\frac{327-183}{61.9}(30)\approx\boxed{257^\circ\text{C}}$$
Because the alloy is left of the eutectic composition, the first solid to form is
$\alpha$ (Pb-rich solid solution), at a composition well below 19%Sn (read from
the steeply-falling $\alpha$ solidus near $257^\circ$C, close to pure Pb).
(a)(ii) Freezing range, 30% Sn. Solidification is complete only when the
last liquid is consumed by the eutectic reaction at $183^\circ$C, so
$$\text{freezing range}=257-183=\boxed{74^\circ\text{C}}$$
(a)(iii) Phases at $184^\circ$C (just above eutectic). The tie line spans
the $\alpha$ solidus ($\approx19\%$Sn, essentially flat just above the eutectic isotherm) and the
liquidus ($\approx61.5\%$Sn at $184^\circ$C). By the lever rule with $C_0=30\%$:
$$\%\alpha=\frac{61.5-30}{61.5-19}\times100\approx\boxed{74.1\%},\qquad
\%L=\frac{30-19}{61.5-19}\times100\approx\boxed{25.9\%}$$
(a)(iv) Phases at $182^\circ$C (just below eutectic). The alloy is now fully
solid: primary $\alpha$ (19%Sn) plus eutectic structure, which by the lever rule between the
eutectic-line boundaries (19% and 97.5%Sn) partitions into overall $\alpha$ and $\beta$:
$$\%\alpha=\frac{97.5-30}{97.5-19}\times100\approx\boxed{86.0\%},\qquad
\%\beta=\frac{30-19}{97.5-19}\times100\approx\boxed{14.0\%}$$
(a)(v) Phases at $25^\circ$C. Using the room-temperature solvus
compositions ($\alpha\approx2\%$Sn, $\beta\approx99\%$Sn):
$$\%\alpha=\frac{99-30}{99-2}\times100\approx\boxed{71.1\%},\qquad
\%\beta=\frac{30-2}{99-2}\times100\approx\boxed{28.9\%}$$
(b)(i)–(ii) First solid and freezing range, 80% Sn. $80\%>61.9\%$
(hypereutectic side), so the first solid is $\beta$ (Sn-rich solid solution).
Interpolating the hypereutectic liquidus (from $(61.9\%,183^\circ$C$)$ to $(100\%,232^\circ$C$)$)
at 80%Sn:
$$T_{liq}=183+\frac{232-183}{100-61.9}(80-61.9)\approx\boxed{206^\circ\text{C}}$$
reading the near-vertical $\beta$ solidus at this temperature gives a first-solid composition
$\approx99\%$Sn (close to pure Sn). Freezing completes at the eutectic isotherm, so
$$\text{freezing range}=206-183=\boxed{23^\circ\text{C}}$$
(b)(iii) Phases at $184^\circ$C. Tie line spans the liquidus
($\approx62.7\%$Sn) and the $\beta$ solidus ($\approx97.5\%$Sn) at $184^\circ$C. With
$C_0=80\%$:
$$\%\beta=\frac{80-62.7}{97.5-62.7}\times100\approx\boxed{49.7\%},\qquad
\%L=\frac{97.5-80}{97.5-62.7}\times100\approx\boxed{50.3\%}$$
(b)(iv) Phases at $182^\circ$C. Same eutectic-line boundaries as part
(a)(iv) (19% and 97.5%Sn), but now with $C_0=80\%$:
$$\%\alpha=\frac{97.5-80}{97.5-19}\times100\approx\boxed{22.3\%},\qquad
\%\beta=\frac{80-19}{97.5-19}\times100\approx\boxed{77.7\%}$$
(b)(v) Phases at $25^\circ$C. Same room-temperature solvus compositions as
(a)(v), with $C_0=80\%$:
$$\%\alpha=\frac{99-80}{99-2}\times100\approx\boxed{19.6\%},\qquad
\%\beta=\frac{80-2}{99-2}\times100\approx\boxed{80.4\%}$$
Quantity
30% Sn (a)
80% Sn (b)
First solid
α (Pb-rich)
β (Sn-rich)
Liquidus temp.
≈257°C
≈206°C
Freezing range
≈74°C
≈23°C
184°C phases
α≈74.1%, L≈25.9%
β≈49.7%, L≈50.3%
182°C phases
α≈86.0%, β≈14.0%
α≈22.3%, β≈77.7%
25°C phases
α≈71.1%, β≈28.9%
α≈19.6%, β≈80.4%
Check
The room-temperature ($25^\circ$C)
solvus compositions ($\approx2\%$Sn for $\alpha$, $\approx99\%$Sn for $\beta$) are read from the
diagram’s sharply-curving low-temperature boundaries, which the source labels only at the
eutectic isotherm (19%/97.5%) — the $25^\circ$C values are a graphical extrapolation
consistent with the printed curve shape and with the well-documented literature Pb-Sn
room-temperature solubilities, not additional numeric data given on the page.
