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17-Phys-B7 Structure of Materials · December 2018

Question 7 of 7: Question VII: Phase Diagram

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

Notes on this paper

Paper format. 17-Phys-B7 Structure of Materials, National Examination December 2018 — a closed-book examination (Casio or Sharp approved calculators only; all necessary equations, constants and diagrams supplied in the paper's own appendix). Candidates attempt any five of the seven questions, each worth 20 marks; every question is nonetheless answered in full below so the paper remains a complete study resource.

Reference texts. W. D. Callister Jr. & D. G. Rethwisch, Materials Science and Engineering: An Introduction, 10th ed. (atomic bonding, crystal structure and packing, point defects, diffusion, dislocations and slip, mechanical properties, phase diagrams and the lever rule, precipitation hardening, X-ray diffraction).

Check — two points on the printed paper. (1) Question I.2 prints the ion as "Cl- (Z = 16)"; Z = 16 is sulfur, not chlorine (Cl is Z = 17) — a printed typo. The electronic structure below uses the correct Z = 17 for chlorine. (2) Question III.1(d) prints the hexagonal plane as $(2\bar{2}10)$, i.e. Miller–Bravais indices $h=2,\,k=-2,\,i=1$; the third index of a valid Miller–Bravais symbol is never independent — it is fixed by $i=-(h+k)$, here $i=-(2-2)=0$, not 1. Reducing the self-consistent index $(2\bar200)$ by its common factor of 2 gives $(1\bar100)$, a standard prism-type plane. The drawing below uses the symmetry-equivalent, non-degenerate face $(10\bar10)$ of the same $\{1\bar100\}$ family (chosen because it renders visibly in the cell projection used here) and states this substitution explicitly.

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 — figure-label correction. Point y cannot lie in the "α+γ" field for a 1.5 wt% C alloy (1.5 wt% C is above the 0.76 wt% eutectoid composition, so this is a HYPEREUTECTOID steel and its cooling path never re-enters the α+γ field at all). The printed diagram shows point y sits between the 727°C eutectoid line and the γ/(γ+Fe₃C) solvus ("Acm" line) that runs from (2.14 wt%C, 1147°C) down to (0.76 wt%C, 727°C) — i.e. y is in the two-phase γ+Fe₃C field, consistent with question 4 asking for "pearlite and PROEUTECTOID CEMENTITE" (not proeutectoid ferrite, which would only apply to a hypoeutectoid alloy below 0.76 wt% C). This field is used throughout below.

Part 1 — invariant points and reactions. The diagram has THREE invariant (three-phase, zero-degree-of-freedom) horizontal lines:

Invariant points on the Fe–Fe₃C diagram
ReactionTemperatureComposition
Peritectic (δ+L→γ)1493°Cδ and L meet near 0.1–0.5 wt% C (not separately labelled on this print)
Eutectic1147°C4.30 wt% C
Eutectoid727°C0.76 wt% C

The question specifically asks for the eutectic and eutectoid REACTIONS:

Eutectic (at 1147°C, 4.30 wt% C): $$L\,(4.30\ \text{wt\%C}) \xrightarrow{1147^\circ\text{C}} \gamma\,(2.14\ \text{wt\%C}) + \text{Fe}_3\text{C}\,(6.70\ \text{wt\%C})$$

Eutectoid (at 727°C, 0.76 wt% C): $$\gamma\,(0.76\ \text{wt\%C}) \xrightarrow{727^\circ\text{C}} \alpha\,(0.022\ \text{wt\%C}) + \text{Fe}_3\text{C}\,(6.70\ \text{wt\%C})$$

(the eutectoid product, alternating lamellae of α and Fe₃C, is given the special name "pearlite").

Part 2 — phase(s) at point x. At 1.5 wt% C and 1100°C, point x lies inside the single-phase γ (austenite) field (bounded below by the 912°C/727°C region and above/right by the γ/(γ+L) and γ/(γ+Fe₃C) boundaries, which at 1.5 wt% C sit well above 1100°C and well below 1100°C respectively). Because only ONE phase is present, its composition must equal the overall alloy composition:

Point x
Phase presentComposition
γ (austenite) — single phase1.5 wt% C (= overall alloy composition)

Part 3 — microstructure evolution on cooling. Given. 1.5 wt% C alloy (hypereutectoid, since 1.5 > 0.76 wt% C), cooled along the dashed line from 1100°C (point x) through point y to point z below 727°C. Find. A schematic microstructure at each point. Approach. Track which phase field each point falls in and, for point z, apply the standard hypereutectoid-steel cooling sequence (proeutectoid Fe₃C nucleates at the prior-austenite grain boundaries first, then the remaining austenite — whose composition has drifted down the Acm line to the eutectoid 0.76 wt% C — transforms entirely to pearlite at 727°C).

  1. At point x (1100°C). Single-phase γ: uniform, equiaxed austenite grains, no second phase anywhere.
  2. At point y (between the Acm line and 727°C). Two-phase γ+Fe₃C: proeutectoid cementite has begun precipitating as a thin, continuous film decorating the PRIOR γ grain boundaries, with untransformed γ still filling the grain interiors.
  3. At point z (below 727°C). Two-phase α+Fe₃C, but with a distinctive hypereutectoid microstructure: the grain-boundary proeutectoid-cementite network laid down above 727°C is retained, and the grain INTERIORS — which were still γ at 0.76 wt% C the instant the alloy crossed 727°C — have all transformed to pearlite (fine alternating lamellae of α and Fe₃C).
γ γ γ γ γ γ γ γ
At point x: single-phase γ (austenite), uniform grains.
γ γ γ γ γ γ γ γ proeutectoid Fe₃C film
At point y: γ grains outlined by a thin proeutectoid-Fe₃C grain-boundary film.
grains: pearlite (α+Fe₃C, magnified) boundaries: proeutectoid Fe₃C
At point z: pearlite (lamellar α+Fe₃C, magnified inset) filling the former γ grains, still outlined by the retained proeutectoid-Fe₃C network.

Given. Hypereutectoid alloy, overall composition $C_0=1.5$ wt% C; eutectoid composition $C_{\text{eutectoid}}=0.76$ wt% C; cementite composition $C_{\text{Fe}_3\text{C}}=6.70$ wt% C.

Find. Mass fraction of pearlite and of proeutectoid cementite at point z.

Approach. Apply the lever rule across the tie line just ABOVE 727°C (the proeutectoid-cementite amount is fixed at that instant and does not change further as the remaining γ converts to pearlite on crossing 727°C); the tie line spans from the eutectoid composition (the composition of the γ that is about to become pearlite) to the cementite composition.

  1. Proeutectoid cementite fraction. By the lever rule, using the segment of the tie line on the γ-rich (pearlite) side: $$W_{\text{Fe}_3\text{C}^\prime} = \frac{C_0-C_{\text{eutectoid}}}{C_{\text{Fe}_3\text{C}}-C_{\text{eutectoid}}} = \frac{1.5-0.76}{6.70-0.76} = \frac{0.74}{5.94} = \boxed{0.1246\ (12.5\%)}.$$
  2. Pearlite fraction. The remainder of the microstructure is pearlite: $$W_{\text{pearlite}} = 1-W_{\text{Fe}_3\text{C}^\prime} = 1-0.1246 = \boxed{0.8754\ (87.5\%)}.$$
Final results — Question VII
QuantityValue
Eutectic point1147°C, 4.30 wt% C
Eutectoid point727°C, 0.76 wt% C
Phase(s) at point xSingle-phase γ, 1.5 wt% C
Proeutectoid cementite fraction at z0.1246 (12.5%)
Pearlite fraction at z0.8754 (87.5%)
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