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 $T<em>A = 1320$°C: Only Liquid (L) present, $C</em>L = C_0 = 35$ wt% Ni
    • At $T<em>B = 1250$°C: Both α and L present, $C</em>L = C<em>{liquidus} = 32$ wt% Ni, $C</em>\alpha = C_{solidus} = 43$ wt% Ni
    • At $T<em>D = 1190$°C: Only Solid (α) present, $C</em>\alpha = C_0 = 35$ wt% Ni
• Rule 3: If we know T and $C_0$, then can determine the weight fraction of each phase.
  • Examples:
    • At $T<em>A$: Only Liquid (L) present, $W</em>L = 1.00, W_\alpha = 0$
    • At $T<em>D$: Only Solid (α) present, $W</em>L = 0, W_\alpha = 1.00$
    • 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).
• $M<em>L * S = M</em>\alpha * R$
• $W<em>L = (C</em>\alpha - C<em>0) / (C</em>\alpha - C_L)$
• $W<em>\alpha = (C</em>0 - C<em>L) / (C</em>\alpha - C_L)$

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
• $T<em>E$: No liquid below $T</em>E$: Composition at temperature $T_E$.
• $C<em>E$: Composition at temperature $T</em>E$.
• Eutectic reaction: $L(C<em>E) \rightarrow \alpha(C</em>{\alpha E}) + \beta(C_{\beta E})$
  • 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: $C<em>\alpha = 11$ wt% Sn, $C</em>\beta = 99$ wt% Sn
  • Relative amount of each phase:
    • $W<em>\alpha = (C</em>\beta - C<em>0) / (C</em>\beta - C_\alpha) = (99 - 40) / (99 - 11) = 0.67$
    • $W<em>\beta = (C</em>0 - C<em>\alpha) / (C</em>\beta - C_\alpha) = (40 - 11) / (99 - 11) = 0.33$
• For a 40 wt% Sn-60 wt% Pb alloy at 220°C:
  • Phases present: α + L
  • Phase compositions: $C<em>\alpha = 17$ wt% Sn, $C</em>L = 46$ wt% Sn
  • Relative amount of each phase:
    • $W<em>\alpha = (C</em>L - C<em>0) / (C</em>L - C_\alpha) = (46 - 40) / (46 - 17) = 0.21$
    • $W<em>L = (C</em>0 - C<em>\alpha) / (C</em>L - C_\alpha) = (40 - 17) / (46 - 17) = 0.79$

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 $C<em>0 = C</em>E$:
  • 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 $T<em>E$: $W</em>\alpha = R / (R + S) = 0.50, W<em>L = (1 - W</em>\alpha) = 0.50, C<em>\alpha = 18.3$ wt% Sn, $C</em>L = 61.9$ wt% Sn
  • Just below $T<em>E$: $C</em>\alpha = 18.3$ wt% Sn, $C<em>\beta = 97.8$ wt% Sn, $W</em>\alpha = S / (R + S) = 0.73, W_\beta = 0.27$

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

• $W_\alpha = s / (r + s)$
• $W<em>\gamma = (1 - W</em>\alpha)$

Hypereutectoid Steel

• Proeutectoid $Fe_3C$ and pearlite

Lever Rule for Hypereutectoid Steel

• $W<em>{\text{pearlite}} = W</em>\gamma = x / (v + x)$
• $W<em>{Fe</em>3C} = (1 - W_\gamma)$

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) $C<em>\alpha = 0.022$ wt% C, $C</em>{Fe_3C} = 6.70$ wt% C
  • b) $W<em>{Fe</em>3C} = (C<em>0 - C</em>\alpha) / (C<em>{Fe</em>3C} - C<em>\alpha) = (0.40 - 0.022) / (6.70 - 0.022) = 0.057$; Amount of $Fe</em>3C$ in 100 g = (100 g)(0.057) = 5.7 g
  • c) $W<em>{\text{pearlite}} = (C</em>0 - C<em>\alpha) / (C</em>\gamma - C_\alpha) = (0.40 - 0.022) / (0.76 - 0.022) = 0.512$; 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.