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21-Mat-A4 Deformation Behaviour and Properties of Materials · December 2016

Question 7 of 7: Question VII — Phase Diagram (20 marks)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2016 — 10-Met-A4, Structure of Materials. Three hours, closed book, one approved calculator (Casio or Sharp). Seven questions of 20 marks each (Roman numerals I–VII); the rubric asks for any five, with only the first five in the answer book marked. All seven are solved here, because this set is a study resource rather than an exam script. All necessary equations, constants and an error-function table are provided in the exam's own appendix (reproduced where used below).

Note on the exam title. The printed exam header reads 10-Met-A4, Structure of Materials. Only Question VI (dislocation theory) is genuinely deformation/mechanical-properties content; the paper as a whole is a broad introductory materials-science survey — electron structure, bonding, crystal structure, crystallographic directions/planes, solid solubility, XRD, microscopy and phase diagrams — and is answered as such below.

Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:

Question VII — 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.

Check — how the tie-line/solvus values below were obtained. The diagram's own printed labels give the eutectic point exactly (E: 71.9 wt% Ag at 779°C; B: 8.0 wt% Ag; G: 91.2 wt% Ag, all at 779°C) but do not print numeric values at 800°C or 700°C. Typical diagram-reading uncertainty is about ±1 wt%.

VII.1 — Phases at 40 wt% Cu–60 wt% Ag, 800°C

Given. Overall composition $C_0=60$ wt% Ag (= 40 wt% Cu); $T=800^{\circ}\text{C}$, i.e. $21^{\circ}\text{C}$ above the eutectic isotherm (779°C).

Find. Phase(s) present and their compositions.

Approach. Locate $(C_0,T)$ on the diagram: at 800°C the two-phase $\alpha+L$ field is bounded on the left by the solidus (descending from A) and on the right by the liquidus (descending from A toward E); reading those two boundaries at 800°C gives the tie-line.

  1. Read the 800°C tie-line. At 800°C the solidus (α boundary) sits at $C_{\alpha}\approx8.1$ wt% Ag — only slightly left of its 779°C value (8.0 wt%, point B), because the solidus is very steep just above the eutectic. The liquidus (boundary with the liquid) sits at $C_L\approx65.9$ wt% Ag.
  2. Locate $C_0=60$ wt% Ag on this tie-line. $$C_\alpha\,(8.1\%) \;\lt\; C_0\,(60\%) \;\lt\; C_L\,(65.9\%)$$ $C_0$ falls strictly between the two boundaries, so the alloy sits inside the $\boxed{\alpha+L}$ two-phase field.
Question VII.1 — phases at 800°C
PhaseComposition
α (solid solution)≈8.1 wt% Ag–91.9 wt% Cu
L (liquid)≈65.9 wt% Ag–34.1 wt% Cu

VII.2 — Maximum solubility at 700°C

Given. $T=700^{\circ}\text{C}$, i.e. $79^{\circ}\text{C}$ below the eutectic isotherm.

Find. (a) Max solubility of Cu in Ag ($\beta$-solvus); (b) max solubility of Ag in Cu ($\alpha$-solvus).

Approach. Both solvus lines exhibit retrograde solubility — each drops off (moves toward its own pure-metal axis) as $T$ falls below the eutectic temperature, from their maxima at B (8.0 wt% Ag) and G (91.2 wt% Ag, i.e. 8.8 wt% Cu) at 779°C. Read each solvus curve at 700°C.

  1. α-solvus (left branch): solubility of Ag in Cu. At 700°C the solvus reads $\approx5.3$ wt% Ag — markedly less than the 8.0 wt% at the eutectic temperature, the expected retrograde-solubility drop. $$\boxed{\text{max solubility of Ag in Cu at }700^\circ\text{C}\approx5.3\ \text{wt\% Ag}}$$
  2. β-solvus (right branch): solubility of Cu in Ag. At 700°C the solvus reads $\approx94.0$ wt% Ag, i.e. $100-94.0=6.0$ wt% Cu — again below the 8.8 wt% Cu at the eutectic temperature. $$\boxed{\text{max solubility of Cu in Ag at }700^\circ\text{C}\approx6.0\ \text{wt\% Cu}}$$

VII.3 — Eutectic reaction

Definition. A eutectic reaction is an invariant (fixed-temperature, fixed-composition) three-phase reaction in which, on cooling, a liquid of one specific composition transforms isothermally and simultaneously into two distinct solid phases: $L\to\alpha+\beta$. It is "invariant" because, per the Gibbs phase rule at fixed pressure ($F=C-P+1$ with $C=2$ components), three phases coexisting in a binary system pins both temperature and all three compositions at a single point — the eutectic point.

Cu-Ag eutectic reaction. Reading directly off the diagram's labelled eutectic point (E, at 779°C and 71.9 wt% Ag), with the two solid phases at their own eutectic-temperature solvus limits (B and G): $$\boxed{L\,(71.9\ \text{wt\%\ Ag})\ \xrightarrow{\ 779^{\circ}\text{C}\ }\ \alpha\,(8.0\ \text{wt\%\ Ag})\ +\ \beta\,(91.2\ \text{wt\%\ Ag})}$$

VII.4 — Lever rule at 55 wt% Ag–45 wt% Cu, 800°C

Given. $C_0=55$ wt% Ag; $T=800^{\circ}\text{C}$; tie-line endpoints from VII.1, $C_\alpha=8.1$ wt% Ag, $C_L=65.9$ wt% Ag.

Find. Mass fractions $f_\alpha$, $f_L$.

Approach. Lever rule: each phase's mass fraction is the length of the opposite tie-line segment divided by the total tie-line length.

  1. Mass fraction of α. $$f_\alpha=\frac{C_L-C_0}{C_L-C_\alpha}=\frac{65.9-55}{65.9-8.1}=\frac{10.9}{57.8}=\boxed{0.189\ (18.9\%)}$$
  2. Mass fraction of liquid. $$f_L=\frac{C_0-C_\alpha}{C_L-C_\alpha}=\frac{55-8.1}{57.8}=\frac{46.9}{57.8}=\boxed{0.811\ (81.1\%)}$$ (check: $f_\alpha+f_L=0.189+0.811=1.000$.)
Question VII — summary
PartResult
VII.1: phases at 60 wt%Ag, 800°Cα (≈8.1 wt%Ag) + L (≈65.9 wt%Ag)
VII.2a: max Cu in Ag at 700°C≈6.0 wt% Cu
VII.2b: max Ag in Cu at 700°C≈5.3 wt% Ag
VII.3: eutectic reaction$L(71.9\%)\to\alpha(8.0\%)+\beta(91.2\%)$ at 779°C
VII.4: mass fractions at 55 wt%Ag, 800°C$f_\alpha=18.9\%$, $f_L=81.1\%$
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