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Qwestrum Engineering360 · Mechanical Engineering · Iron-Carbon Diagram

Cooling Curves of Steel

Cooling curves plot temperature versus time; arrests (plateaus/inflections) mark phase transformations. Faster cooling shifts transformations to lower temperatures, and section size affects cooling rate (Chvorinov analogy), per physical-metallurgy texts.
Exam tip: keep SI units consistent end-to-end, write the governing relation symbolically before substituting, and sanity-check magnitude and sign.

Key formulas & points

Skim these first — then read the full notes below.

  • Thermal arrest at invariant temperatures on cooling curve
  • Proeutectoid phase forms before eutectoid reaction
  • Faster cooling: finer pearlite, then bainite, then martensite

Topic details

Introduction

Cooling curves link the thermal history of steel to its microstructure and are used to locate transformation temperatures. This entry treats how cooling rate and section size govern the phases that form.

Scope in B.Tech and GATE syllabus

A pure metal shows a horizontal arrest at its freezing point; an alloy shows a sloping two-phase region between liquidus and solidus. Solid-state transformations (like the eutectoid) also produce arrests because latent heat is released.

Why this topic matters in practice

Cooling rate — set by quench medium and section size — determines whether equilibrium products (pearlite) or non-equilibrium products (bainite, martensite) form, foreshadowing the TTT/CCT diagrams. Interpreting arrests and relating cooling rate to microstructure are the exam skills.

Key relations & formulas

tsolidification=f(coolingrate,sectionsize)t_{solidification} = f(cooling rate, section size)
(Chvorinov analogy)

Formulas (Indian textbook notation)

  • UndercoolingΔTbelowA1beforepearlitenucleationUndercooling \Delta T below A_{1} before pearlite nucleation
Fractiontransformedf(t)=1−exp(−ktn)Fraction transformed f(t) = 1 - exp(-kt^n)
(Avrami equation)

Formulas (Indian textbook notation)

  • Hardness ∝ \frac{1}{\sqrt}{pearlite interlamellar spacing}

Notation and sign conventions

Relation 1 —
tsolidification=ft_{solidification} = f
tsolidification=f(coolingrate,sectionsize)t_{solidification} = f(cooling rate, section size)
(Chvorinov analogy)
Write this relation with symbols exactly as in Metallurgical Thermodynamics — Dekkar before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
UndercoolingΔTbelowA1beforepearlitenucleationUndercooling \Delta T below A_{1} before pearlite nucleation

Formulas (Indian textbook notation)

  • UndercoolingΔTbelowA1beforepearlitenucleationUndercooling \Delta T below A_{1} before pearlite nucleation
Write this relation with symbols exactly as in Metallurgical Thermodynamics — Dekkar before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
FractiontransformedfFraction transformed f
Fractiontransformedf(t)=1−exp(−ktn)Fraction transformed f(t) = 1 - exp(-kt^n)
(Avrami equation)
Write this relation with symbols exactly as in Metallurgical Thermodynamics — Dekkar before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 4 —
Hardness∝1/√Hardness ∝ 1/√

Formulas (Indian textbook notation)

  • Hardness ∝ \frac{1}{\sqrt}{pearlite interlamellar spacing}
Write this relation with symbols exactly as in Metallurgical Thermodynamics — Dekkar before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

On a temperature-time cooling curve, a thermal arrest (plateau) or change of slope signals a phase change, because the latent heat evolved slows the temperature drop. Pure iron freezes at a single temperature (flat arrest); carbon steels freeze over a range with liquidus and solidus inflections.

Governing relations in practice

Below solidification, further arrests occur at the allotropic and eutectoid transformations (A₃, A₁), letting the critical temperatures be measured experimentally.

Design and analysis considerations

Cooling rate strongly affects the product: slow cooling follows the equilibrium diagram (coarse pearlite), while rapid cooling suppresses diffusion and produces fine pearlite, bainite, or martensite. This is the practical bridge from the equilibrium diagram to real (non-equilibrium) heat treatment.

Advanced theory and extensions

Section size matters: a thick section cools more slowly at its core (large volume-to-surface ratio, a Chvorinov-type effect), so the core and surface can have different microstructures — the mass effect that hardenability addresses. Reading cooling curves and inferring microstructure are the core competencies.

Assumptions and validity limits

State assumptions explicitly before using any relation for cooling curves of steel — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Iron–Carbon Diagram viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Iron–Carbon Diagram papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to cooling curves of steel.
4. Use equation 1:
tsolidification=ft_{solidification} = f
.
5. Use equation 2:
UndercoolingΔTbelowA1beforepearlitenucleationUndercooling \Delta T below A_{1} before pearlite nucleation
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Cooling Curves of Steel appears in heat treatment shop decisions. In Indian mechanical curricula this topic is tested because it connects theory to phases and transformations in steels.
GATE and semester exams often combine cooling curves of steel with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use cooling curves of steel?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Treating an alloy's freezing as a single-temperature arrest like a pure metal
• Ignoring latent-heat arrests from solid-state transformations
• Assuming uniform cooling rate through a thick section (core lags surface)
• Forgetting that faster cooling lowers the transformation temperature

Quick revision checklist

Before attempting cooling curves of steel problems, confirm you can:
1. Thermal arrest at invariant temperatures on cooling curve
2. Proeutectoid phase forms before eutectoid reaction
3. Faster cooling: finer pearlite, then bainite, then martensite
Revise the solved examples in Metallurgical Thermodynamics — Dekkar and one previous-year GATE or university paper for this unit.

Worked examples

Try the problem first — open the solution when you are ready to check.

Interpret a cooling arrest

Problem

A 0.4 % C steel shows an arrest during cooling at 727 °C. What transformation is occurring?

Solution

At 727 °C the remaining austenite transforms to pearlite (the eutectoid reaction), releasing latent heat that causes the arrest.

Conceptual check — Cooling Curves of Steel

Problem

In a Iron–Carbon Diagram semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of cooling curves of steel." What should a complete answer include?

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

  1. 1
    What is Cooling Curves of Steel, and why does it appear in B.Tech / GATE syllabi?

    Model answer

    Cooling curves plot temperature versus time; arrests (plateaus/inflections) mark phase transformations. Faster cooling shifts transformations to lower temperatures, and section size affects cooling rate (Chvorinov analogy), per physical-metallurgy texts.
  2. 2
    State the relation t_solidification = f and name each symbol.

    Model answer

    The governing relation is tsolidification=ft_{solidification} = f. Write every symbol with SI units before substituting numbers.
  3. 3
    State the relation Undercooling ΔT below A₁ before pearlite nucleation and name each symbol.

    Model answer

    The governing relation is UndercoolingΔTbelowA1beforepearlitenucleationUndercooling \Delta T below A_{1} before pearlite nucleation. Write every symbol with SI units before substituting numbers.
  4. 4
    State the relation Fraction transformed f and name each symbol.

    Model answer

    The governing relation is FractiontransformedfFraction transformed f. Write every symbol with SI units before substituting numbers.
  5. 5
    State the relation Hardness ∝ 1/√ and name each symbol.

    Model answer

    The governing relation is Hardness∝1/√Hardness ∝ 1/√. Write every symbol with SI units before substituting numbers.
  6. 6
    Explain: Thermal arrest at invariant temperatures on cooling curve

    Model answer

    Thermal arrest at invariant temperatures on cooling curve — state the assumption range and one exam trap linked to this point.
  7. 7
    Explain: Proeutectoid phase forms before eutectoid reaction

    Model answer

    Proeutectoid phase forms before eutectoid reaction — state the assumption range and one exam trap linked to this point.
  8. 8
    Explain: Faster cooling: finer pearlite, then bainite, then martensite

    Model answer

    Faster cooling: finer pearlite, then bainite, then martensite — state the assumption range and one exam trap linked to this point.
  9. 9
    How would you correct this error in a viva: Treating an alloy's freezing as a single-temperature arrest like a pure metal?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  10. 10
    How would you correct this error in a viva: Ignoring latent-heat arrests from solid-state transformations?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  11. 11
    How would you correct this error in a viva: Assuming uniform cooling rate through a thick section (core lags surface)?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  12. 12
    How would you correct this error in a viva: Forgetting that faster cooling lowers the transformation temperature?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.

Exams & GATE

  • 1
    Sketch cooling curve for 0.4% C steel — arrests at A₃ and A₁.
  • 2
    Avoid: Treating an alloy's freezing as a single-temperature arrest like a pure metal
  • 3
    Avoid: Ignoring latent-heat arrests from solid-state transformations
  • 4
    Avoid: Assuming uniform cooling rate through a thick section (core lags surface)

📖 Standard books (India)

  • Metallurgical Thermodynamics — Dekkar

    Read: Syllabus unit

    Iron-carbon, heat treatment, and alloys

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