---
title: "Cooling of 3% C cast iron"
subject: "Material Science"
chapter: 6
key: "semester_3/material_science:hypoeutectic-cast-iron-3-carbon"
source: https://www.bctengineer.com/q/semester_3/material_science/hypoeutectic-cast-iron-3-carbon
site: BCT Engineering
---

# Cooling of 3% C cast iron

Material Science · Chapter 6: Iron – Iron Carbide diagram and Heat Treatment of Steels

- Asked in: year not given, 5 marks
- Video: [Material Science, The Iron Carbon Phase Diagram, Part 1](https://www.youtube.com/watch?v=HdlCehVdNc4) — Welt der Werkstoffe
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/material_science/hypoeutectic-cast-iron-3-carbon
- Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-6

## Question

Describe the solidification and cooling of a 3 wt% C hypoeutectic white cast iron from the liquid state. Calculate the fractions of primary austenite and eutectic (ledeburite) just above 1147 °C, and the total cementite and austenite just below 1147 °C. Take austenite 2.14% C, eutectic liquid 4.3% C, cementite 6.67% C.

## Short answer

Primary austenite dendrites form first, then remaining liquid solidifies as ledeburite at 1147°C. Just above 1147°C: 60.2% primary austenite, 39.8% liquid (ledeburite). Just below: 81.0% austenite, 19.0% cementite.

## Answer

#### Cooling path
1. Above the liquidus, the alloy is liquid.
2. Between the liquidus and 1147 °C, **primary austenite** dendrites separate; the liquid becomes richer in carbon, reaching 4.3% C at 1147 °C.
3. At 1147 °C the remaining liquid undergoes the eutectic reaction $L \rightarrow \gamma + \text{Fe}_3\text{C}$, giving **ledeburite**.
4. Below 1147 °C, the austenite loses carbon as secondary cementite and at 727 °C reaches 0.76% C and transforms to pearlite.
5. At room temperature: pearlite (dendrites) + cementite + transformed ledeburite. The structure is hard and brittle white cast iron.

#### Fractions just above 1147 °C
Primary austenite (tie line 2.14 to 4.3):

$$
W_{\gamma'} = \frac{4.3 - 3.0}{4.3 - 2.14} = \frac{1.3}{2.16} = 0.602
$$

Liquid, which becomes ledeburite:

$$
W_{L} = \frac{3.0 - 2.14}{4.3 - 2.14} = \frac{0.86}{2.16} = 0.398
$$

#### Phases just below 1147 °C
Total cementite (tie line 2.14 to 6.67):

$$
W_{Fe_3C} = \frac{3.0 - 2.14}{6.67 - 2.14} = \frac{0.86}{4.53} = 0.190
$$

$$
W_{\gamma} = 1 - 0.190 = 0.810
$$

| Stage | Result |
|---|---|
| Just above 1147 °C | 60.2% primary $\gamma$, 39.8% liquid |
| Just below 1147 °C | 81.0% austenite, 19.0% cementite |

**Answer: primary austenite 60.2% and ledeburite 39.8%; just below 1147 °C austenite 81.0% and cementite 19.0%.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In 3% C hypoeutectic cast iron cooling from liquid, what separates first between the liquidus and 1147 °C?
   A. Primary cementite
   B. Primary austenite dendrites
   C. Ferrite
   D. Pearlite
   Answer: B. Austenite dendrites form first and the liquid is enriched in carbon until it reaches 4.3% C. (Key point: Primary solid)

2. What happens to the remaining liquid at 1147 °C?
   A. It solidifies as ferrite
   B. It forms martensite
   C. It undergoes the eutectic reaction to give ledeburite
   D. It undergoes the peritectic reaction
   Answer: C. The liquid has reached the eutectic composition of 4.3% C and transforms into austenite + cementite, ledeburite. (Key point: Fate of remaining liquid)

3. What is the fraction of primary austenite just above 1147 °C?
   A. (3.0 - 2.14)/(4.3 - 2.14) = 0.398
   B. (4.3 - 3.0)/(4.3 - 2.14) = 0.602
   C. (3.0 - 2.14)/(6.67 - 2.14) = 0.190
   D. (6.67 - 3.0)/(6.67 - 2.14) = 0.810
   Answer: B. The tie line is 2.14 to 4.3; austenite is the arm toward the liquid composition: 1.3/2.16 = 0.602. (Key point: Lever rule just above 1147 °C)

4. Just below 1147 °C, which tie line gives cementite at 19.0%?
   A. 2.14 to 4.3
   B. 0.76 to 6.67
   C. 2.14 to 6.67
   D. 0.022 to 6.67
   Answer: C. Below the eutectic the phases are austenite (2.14%) and cementite (6.67%): (3.0 - 2.14)/(6.67 - 2.14) = 0.190. (Key point: Phases just below 1147 °C)

## Related in Material Science

- [Slow cooling of 0.4% C steel](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Aslow-cooling-04-carbon-steel) — Material Science, chapter 6
- [Slow cooling of 1.2% C steel](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Ahypereutectoid-steel-12-carbon) — Material Science, chapter 6
- [Cast structure and solidification defects](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Acast-structure-and-solidification-defects) — Material Science, chapter 5
- [Types of cast iron](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Atypes-of-cast-iron) — Material Science, chapter 7
- [Solidification: nucleation and grain growth](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Asolidification-nucleation-and-grain-growth) — Material Science, chapter 5

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