Question 1 of 8: The Fe–C Phase Diagram and Microstructural Design

Question 1: The Fe–C Phase Diagram and Microstructural Design (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.

1.1 — (a) Phases and compositions at 1200°C for Fe-3.0 wt%C

Given. An Fe-3.0 wt%C liquid is slowly (equilibrium) cooled from 1600°C to 1200°C. The attached diagram is a Callister-style partial Fe-Fe₃C diagram; the γ+L two-phase field is bounded on the eutectic side by the invariant point at 4.30 wt%C, 1148°C.

Find. The phase(s) present at 1200°C and their compositions.

[Figure not reproduced. See the official exam paper.]

Approach. Locate C₀=3.0 on the 1200°C isotherm, identify the field it falls in from the diagram, then apply the lever rule to the tie line bounding that field.

1. Locate the field. At 1200°C the two boundaries of the γ+L field are the γ-solidus (left) and the liquidus (right). From the digitized figure these read C_γ≈1.85 wt%C and C_L≈3.86 wt%C at 1200°C. Since $1.85 \lt 3.0 \lt 3.86$, the alloy at 1200°C sits inside the γ + L two-phase field.
2. Apply the lever rule. With the tie line endpoints C_γ=1.85 and C_L=3.86: $$W_L=\dfrac{C_0-C_\gamma}{C_L-C_\gamma}=\dfrac{3.0-1.85}{3.86-1.85}=\boxed{0.57}\qquad W_\gamma=1-W_L=\boxed{0.43}$$

1.2 — (b) Full transformation sequence, Fe-1.5 wt%C, 1600→400°C

Fe-1.5 wt%C lies between the eutectoid (0.77) and the maximum γ solubility (2.11), so it is a hypereutectoid steel. Cooled slowly (equilibrium) from 1600°C, it passes through five distinct microstructural regimes before reaching room temperature:

1. 1600→≈1435°C — single-phase liquid, L. The alloy is entirely liquid; no solid phase is present. As the liquidus is approached, compositional undercooling is negligible under slow cooling and solidification begins by heterogeneous nucleation of γ dendrites at the liquidus temperature for 1.5 wt%C (the digitized liquidus crosses 1.5 wt%C at ≈1435°C).
2. ≈1435→≈1265°C (liquidus to γ-solidus at 1.5 wt%C) — γ + L. Primary γ (austenite) dendrites nucleate and grow, rejecting carbon into the shrinking liquid pool (the liquid composition rises along the liquidus toward 4.30 wt%C while the solid composition rises along the γ-solidus toward 2.11 wt%C, per the lever rule at each temperature). The two-phase mixture consists of coring dendrites of γ surrounded by carbon-enriched liquid.
3. ≈1265→≈995°C — single-phase γ, austenite. Once the alloy crosses the γ-solidus (interpolated at 1.5 wt%C), solidification is complete and the microstructure is 100% γ grains (FCC), homogenizing by solid-state diffusion as cooling continues; the microstructure is a normal polygonal austenite grain structure with no second phase.
4. ≈995→727°C — γ + Fe₃C (proeutectoid cementite). Below ≈995°C, where the 1.5 wt%C vertical crosses the A_cm (γ/γ+Fe₃C) solvus, the alloy enters the γ+Fe₃C two-phase field (since 1.5>0.77, the alloy is on the hypereutectoid side and the FIRST solid to form on further cooling is proeutectoid cementite, not proeutectoid ferrite). Fe₃C nucleates preferentially at γ grain boundaries (highest-energy, easiest-diffusion sites) and grows as a thin, continuous network outlining the prior austenite grains, while the remaining γ loses carbon and its composition slides down the γ-solvus toward 0.77 wt%C as 727°C is approached.
5. At 727°C — eutectoid reaction. The remaining austenite, now at exactly the eutectoid composition (0.77 wt%C), transforms isothermally: $\gamma(0.77\%\text{C})\rightarrow\alpha(0.022\%\text{C})+\text{Fe}_3\text{C}(6.70\%\text{C})$, nucleating cooperatively as alternating lamellae of ferrite and cementite — pearlite.
6. 727→400°C — α + Fe₃C, final microstructure. Below 727°C no further phase change occurs (only a very small further drop in the α solubility limit, from 0.022 toward ≈0.005 wt%C, which is usually neglected). The room-temperature microstructure is pearlite colonies (lamellar α+Fe₃C) set inside a continuous proeutectoid cementite network that decorates the prior austenite grain boundaries — the classic, brittle hypereutectoid microstructure (Question 1(c) below quantifies why the cementite network content matters for toughness).

1.3 — (c) Critical hypereutectoid composition for <10 wt% total cementite

Given. At room temperature the two-phase field for any Fe-C alloy above 0.022 wt%C is α+Fe₃C, bounded by C_α≈0.022 wt%C and C_Fe₃C=6.70 wt%C (fixed stoichiometric compound). The total cementite (proeutectoid + eutectoid, i.e. all of the Fe₃C in the microstructure) obeys the ordinary lever rule across this whole field.

Find. The composition C₀ at which total cementite = 10 wt%, and whether it is genuinely attainable on the hypereutectoid side.

Approach. Apply the room-temperature lever rule for total Fe₃C as a function of C₀, set it equal to 0.10, and solve for C₀; then compare the result against the eutectoid composition to check the hypereutectoid constraint is even satisfiable.

1. Total-cementite lever rule. $$W_{\text{Fe}_3\text{C}}(C_0)=\dfrac{C_0-C_\alpha}{C_{\text{Fe}_3\text{C}}-C_\alpha}=\dfrac{C_0-0.022}{6.70-0.022}$$
2. Solve for the critical composition. Setting $W_{\text{Fe}_3\text{C}}=0.10$: $$C_0=C_\alpha+0.10\,(C_{\text{Fe}_3\text{C}}-C_\alpha)=0.022+0.10(6.678)=\boxed{0.690\ \text{wt\%C}}$$
3. Check against the eutectoid. Evaluate the same lever rule AT the eutectoid composition itself (pure pearlite, no proeutectoid phase at all): $$W_{\text{Fe}_3\text{C}}(0.77)=\dfrac{0.77-0.022}{6.678}=\boxed{0.112\ (11.2\%)}$$ Since $0.690 \lt 0.77$, the critical composition found in Step 2 is below the eutectoid — it is a hypoeutectoid, not a hypereutectoid, composition.
4. Alternative reading: limit the proeutectoid cementite network. In practice the embrittling constituent of a hypereutectoid steel is the continuous grain-boundary network of proeutectoid Fe₃C. If the 10% is read as applying to that constituent, the lever rule is taken just above 727°C between the eutectoid and cementite: $$W_{\text{Fe}_3\text{C,pro}}=\dfrac{C_0-0.77}{6.70-0.77}=0.10\ \Longrightarrow\ C_0=0.77+0.10(5.93)=\boxed{1.36\ \text{wt\%C}}$$ so hypereutectoid steels with 0.77 < C₀ < 1.36 wt%C would qualify. The question's own wording (“total of the cementite phase”) supports the first reading, so that remains the primary answer; state the interpretation adopted, as Note 1 of the paper invites.