Phase Diagrams

Chapter 11: Phase Diagrams

Issues to Address

  • When combining two elements, the main questions are:
    • What is the resulting equilibrium state?
    • If we specify:
      • The composition (e.g., wt% Cu - wt% Ni)
      • The temperature (T)
    • Then:
      • How many phases form?
      • What is the composition of each phase?
      • What is the amount of each phase?

Solubility Limit

  • Definition: Maximum concentration for which only a single phase solution exists.
  • Example: Sugar in water at 20°C.
    • Answer: 65 wt% sugar.
    • At 20°C, if C < 65 wt% sugar: syrup (single phase)
    • At 20°C, if C > 65 wt% sugar: syrup + sugar (two phases)
    • Phase Diagram:
      • Illustrates the solubility limit of sugar in water as a function of temperature.
      • L (liquid solution, i.e., syrup) exists below the solubility limit.
      • L (liquid) + S (solid sugar) exists above the solubility limit.

Solutions vs. Mixtures

  • Solution: Solid, liquid, or gas, single phase.
  • Mixture: More than one phase.

Components and Phases

  • Components: The elements or compounds present in the alloy (e.g., Al and Cu).
  • Phases: The physically and chemically distinct material regions that form (e.g., α and β).

Altering Temperature (T) and Composition (C)

  • Altering T can change the number of phases: path A to B.
  • Altering C can change the number of phases: path B to D.
  • Example (Water-Sugar System):
    • A (20°C, C = 70 wt% sugar): 2 phases
    • B (100°C, C = 70 wt% sugar): 1 phase
    • D (100°C, C = 90 wt% sugar): 2 phases

One-Component (or Unary) Phase Diagrams

  • Example: Pressure-temperature phase diagram for H₂O.
  • Shows the conditions (pressure and temperature) at which different phases (solid, liquid, vapor) are stable.
  • Intersection of the dashed horizontal line at 1 atm pressure with the solid-liquid phase boundary (point 2) corresponds to the melting point at this pressure (T = 0°C).
  • Similarly, point 3, the intersection with the liquid-vapor boundary, represents the boiling point (T = 100°C).

Simple System (e.g., Ni-Cu solution)

  • Ni and Cu are totally soluble in one another for all proportions.
  • Hume-Rothery Rules: Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii, suggesting high mutual solubility.
    • Ni: FCC, Electronegativity = 1.9, r (nm) = 0.1246
    • Cu: FCC, Electronegativity = 1.8, r (nm) = 0.1278

Phase Diagrams

  • Definition: Shows the relationships among temperature T, composition C, and phases present in a particular alloy system at equilibrium.
  • For this course:
    • Binary systems: just 2 components.
    • Independent variables: T and C (P = 1 atm is almost always used).
  • Example: Phase Diagram for Cu-Ni system.
    • 2 phases: L (liquid), α (FCC solid solution).
    • 3 different phase fields: L, L + α, α.

Cu-Ni Phase Diagram

  • The Cu-Ni system is:
    • Binary: i.e., 2 components: Cu and Ni.
    • Isomorphous: i.e., complete solubility of one component in another; a phase field extends from 0 to 100 wt% Ni.

Rules for Phase Diagrams

  • Rule 1: If we know T and C_0, then we know which phase(s) is (are) present.
    • Examples:
      • A(1100°C, 60 wt% Ni): 1 phase: α
      • B(1250°C, 35 wt% Ni): 2 phases: L + α
  • Rule 2: If we know T and C_0, then we can determine the composition of each phase in the 2-phase region.
    • Examples: Consider C_0 = 35 wt% Ni
      • At TA = 1320°C: Only Liquid (L) present, CL = C_0 = 35 wt% Ni
      • At TB = 1250°C: Both α and L present, CL = C{liquidus} = 32 wt% Ni, C\alpha = C_{solidus} = 43 wt% Ni
      • At TD = 1190°C: Only Solid (α) present, C\alpha = C_0 = 35 wt% Ni
  • Rule 3: If we know T and C_0, then can determine the weight fraction of each phase.
    • Examples:
      • At TA: Only Liquid (L) present, WL = 1.00, W_\alpha = 0
      • At TD: Only Solid (α) present, WL = 0, W_\alpha = 1.00
      • At T_B: Both α and L present
        • W_L = S / (R + S) = (43 - 35) / (43 - 32) = 0.73
        • W_\alpha = R / (R + S) = (35 - 32) / (43 - 32) = 0.27

Tie Line (Isotherm)

  • Connects the phases in equilibrium with each other.
  • Used to determine the fraction of each phase using the lever rule.
  • Think of the tie line as a lever (teeter-totter).
  • ML * S = M\alpha * R
  • WL = (C\alpha - C0) / (C\alpha - C_L)
  • W\alpha = (C0 - CL) / (C\alpha - C_L)

Microstructural Changes During Cooling

  • Consider microstructural changes that accompany the cooling of a C_0 = 35 wt% Ni alloy
  • Examples:
    • A: L: 35wt%Ni
    • B: L: 32 wt% Ni, α: 43 wt% Ni
    • C: L: 24 wt% Ni, α: 46 wt% Ni
    • D: α: 36 wt% Ni

Cored vs. Equilibrium Structures

  • Slow rate of cooling: Equilibrium structure.
  • Fast rate of cooling: Cored structure.
    • First α to solidify: 46 wt% Ni.
    • Last α to solidify: < 35 wt% Ni.
    • C_\alpha changes as we solidify.
  • Cu-Ni case:
    • First α to solidify has C_\alpha = 46 wt% Ni.
    • Last α to solidify has C_\alpha = 35 wt% Ni.
    • Uniform C_\alpha: 35 wt% Ni.

Effect of Solid Solution Strengthening

  • Tensile Strength (TS) increases with Ni content.
  • Ductility (%EL) decreases with Ni content.

Types of Phase Diagrams Showing Partial Solid Solubility

  • Eutectic Diagrams: Liquid transforms to two solid phases.
  • Eutectoid Diagrams: One solid phase transforms to two other solid phases.
  • Peritectic Diagrams: Liquid and one solid phase transform to a second solid phase.
  • Peritectoid Diagrams: Two solid phases transform to one solid.

Eutectic System (Cu-Ag)

  • 3 single phase regions (L, α, β).
  • Limited solubility:
    • α: mostly Cu
    • β: mostly Ag
  • TE: No liquid below TE: Composition at temperature T_E.
  • CE: Composition at temperature TE.
  • Eutectic reaction: L(CE) \rightarrow \alpha(C{\alpha E}) + \beta(C_{\beta E})
    • Eutectic - liquid transforms to two solid phases.
    • Consists of 2 components and has a special composition with a min. melting T. (C_E)
    • L(71.9 \text{ wt\% Ag}) \rightarrow \alpha(8.0 \text{ wt\% Ag}) + \beta (91.2 \text{ wt\% Ag})

Pb-Sn System

  • For a 40 wt% Sn-60 wt% Pb alloy at 150°C:
    • Phases present: α + β
    • Phase compositions: C\alpha = 11 wt% Sn, C\beta = 99 wt% Sn
    • Relative amount of each phase:
      • W\alpha = (C\beta - C0) / (C\beta - C_\alpha) = (99 - 40) / (99 - 11) = 0.67
      • W\beta = (C0 - C\alpha) / (C\beta - C_\alpha) = (40 - 11) / (99 - 11) = 0.33
  • For a 40 wt% Sn-60 wt% Pb alloy at 220°C:
    • Phases present: α + L
    • Phase compositions: C\alpha = 17 wt% Sn, CL = 46 wt% Sn
    • Relative amount of each phase:
      • W\alpha = (CL - C0) / (CL - C_\alpha) = (46 - 40) / (46 - 17) = 0.21
      • WL = (C0 - C\alpha) / (CL - C_\alpha) = (40 - 17) / (46 - 17) = 0.79

Microstructure of Pb-Sn Alloys

  • For alloys for which C_0 < 2 wt% Sn:
    • Result: at room temperature - polycrystalline with grains of α phase having composition C_0.
  • For alloys for which 2 wt% Sn < C_0 < 18.3 wt% Sn:
    • Result: at temperatures in α + β range - polycrystalline with α grains and small β-phase particles.
  • For alloy of composition C0 = CE:
    • Result: Eutectic microstructure (lamellar structure) - alternating layers (lamellae) of α and β phases.

Lamellar Eutectic Structure

  • Alternating layers of α and β phases.
  • Eutectic growth direction

Alloys with Compositions Between Eutectic Point and Max Solubility

  • For alloys for which 18.3 wt% Sn < C_0 < 61.9 wt% Sn:
    • Result: α phase particles and a eutectic microconstituent.
    • Just above TE: W\alpha = R / (R + S) = 0.50, WL = (1 - W\alpha) = 0.50, C\alpha = 18.3 wt% Sn, CL = 61.9 wt% Sn
    • Just below TE: C\alpha = 18.3 wt% Sn, C\beta = 97.8 wt% Sn, W\alpha = S / (R + S) = 0.73, W_\beta = 0.27

Eutectic Micro-Constituent

  • Hypoeutectic: C_0 = 50 wt% Sn
  • Eutectic: C_0 = 61.9 wt% Sn
  • Hypereutectic: (illustration only)

Intermetallic Compounds

  • Note: Intermetallic compound exists as a line on the diagram - not an area - because of stoichiometry (i.e. composition of a compound is a fixed value).

Eutectoid Diagram

  • Eutectoid - one solid phase transforms to two other solid phases (3 solid phases NO Liq.)

Peritectic Diagram (Ag - Pt)

  • Peritectic - liquid and one solid phase transform to a second solid phase

Phase Diagram Vocabulary

  • Phase Boundaries
  • Liquidus
  • Solidus
  • Solvus
  • Composition
  • Temperature
  • Alloy
  • Eutectic/Eutectoid as an adjective
  • Hypoeutectic alloy
  • Hypereutectic alloy
  • Proeutectic Phase

Eutectic, Eutectoid, and Peritectic Transformations

  • Eutectoid: One solid phase transforms to two other solid phases S2 \rightarrow S1 + S3 (e.g., \gamma \rightarrow \alpha + Fe_3C for Fe-C, 727°C, 0.76 wt% C).
  • Eutectic: Liquid transforms to two solid phases L \rightarrow \alpha + \beta (e.g., for Pb-Sn, 183°C, 61.9 wt% Sn).
  • Peritectic: Liquid and one solid phase transform to a second solid phase S1 + L \rightarrow S2 (e.g., \delta + L \rightarrow \gamma for Fe-C, 1493°C, 0.16 wt% C).

Cu-Zn Phase Diagram

  • Eutectoid transformation: \delta \rightarrow \gamma + \epsilon
  • Peritectic transformation: \gamma + L \rightarrow \delta

Iron-Carbon (Fe-C) Phases

  • α-Ferrite: BCC, 0°C – 910°C, Solubility is max 0.02 wt % at 750°C.
  • γ-Austenite: FCC, 910°C – 1410°C, Solubility is max 1.7 wt % at 1150°C.
  • δ-Ferrite: BCC, 1410°C – 1535°C, Solubility is max 0.1 wt % at 1493°C.
  • Cementite (Fe_3C): Complex Orthorhombic Crystal Structure 25 at% C = 6.7 wt% C.
  • Graphite

Iron-Carbon (Fe-C) Phase Diagram

  • Two important points:
    • Eutectoid (B): \gamma \rightarrow \alpha + Fe_3C
    • Eutectic (A): L \rightarrow \gamma + Fe_3C
  • Result: Pearlite = alternating layers of α and Fe_3C phases

Peritectic, Eutectic, and Eutectoid Reactions in Fe-C System

  • Peritectic: L + \delta \rightarrow \gamma
  • Eutectic: L \rightarrow \gamma + Fe_3C
  • Eutectoid: \gamma \rightarrow \alpha + Fe_3C

Hypoeutectoid Steel

  • Proeutectoid ferrite and pearlite

Lever Rule in Fe-C System

  • W_\alpha = s / (r + s)
  • W\gamma = (1 - W\alpha)

Hypereutectoid Steel

  • Proeutectoid Fe_3C and pearlite

Lever Rule for Hypereutectoid Steel

  • W{\text{pearlite}} = W\gamma = x / (v + x)
  • W{Fe3C} = (1 - W_\gamma)

Example Problem: 99.6 wt% Fe-0.40 wt% C Steel

  • For a 99.6 wt% Fe-0.40 wt% C steel at a temperature just below the eutectoid, determine the following:
    • a) The compositions of Fe_3C and ferrite (α).
    • b) The amount of cementite (in grams) that forms in 100 g of steel.
    • c) The amounts of pearlite and proeutectoid ferrite (α) in the 100 g.
  • Solution:
    • a) C\alpha = 0.022 wt% C, C{Fe_3C} = 6.70 wt% C
    • b) W{Fe3C} = (C0 - C\alpha) / (C{Fe3C} - C\alpha) = (0.40 - 0.022) / (6.70 - 0.022) = 0.057; Amount of Fe3C in 100 g = (100 g)(0.057) = 5.7 g
    • c) W{\text{pearlite}} = (C0 - C\alpha) / (C\gamma - C_\alpha) = (0.40 - 0.022) / (0.76 - 0.022) = 0.512; Amount of pearlite in 100 g = (100 g)(0.512) = 51.2 g

Summary

  • Phase diagrams are useful tools to determine:
    • The number and types of phases present.
    • The composition of each phase.
    • The weight fraction of each phase given the temperature and composition of the system.
  • The microstructure of an alloy depends on:
    • Its composition.
    • Whether or not cooling rate allows for maintenance of equilibrium.
  • Important phase diagram phase transformations include eutectic, eutectoid, and peritectic.