Chapter 9: Phase Diagrams

• When we combine two elements... what equilibrium state do we get? • In particular, if we specify... --a composition (e.g., wt% Cu - wt% Ni), and

--a temperature (T ) then... How many phases do we get? What is the composition of each phase? How much of each phase do we get?

Phase A Phase B

Nickel atom Copper atom Chapter 9 - 1 Definitions and basic concepts

• Component: pure metals and/or compounds of which an alloy is composed • System: a specific body of material under consideration • A phase: a homogeneous portion of a system that has uniform physical and chemical characteristics • Equilibrium: a system is at equilibrium if its free energy is at a minimum under some specified combination of temperature, pressure, and composition. • Phase equilibrium: minimum energy for a system with multiple phases • Phase diagram: information about phases as function of T, composition, and pressure

Chapter 9 - Solubility limit

• A maximum concentration of solute atoms may dissolve in the solvent

• Solubility increases with T

Chapter 9 - Phase Equilibria: Solubility Limit

Sucrose/Water Phase Diagram Question: What is the 10 0

solubility limit at 20C? Solubility Answer: 65wt% sugar. 8 0 Limit L If Co < 65wt% sugar: syrup (liquid) 6 0 If Co > 65wt% sugar: syrup L + + sugar. 4 0 (liquid solution S i.e., syrup) (solid 20 Temperature (°C) Temperature sugar) • Solubility limit increases with T:

0 20 40 6065 80 100

e.g., if T = 100C, solubility Co =Composition (wt% sugar) Sugar limit = 80wt% sugar. Pure Water Pure

Chapter 9 - 4 Effect of T & Composition (Co) • Changing T can change # of phases: path A to B. • Changing Co can change # of phases: path B to D. B (100°C,70) D (100°C,90) 1 phase 2 phases 100

80 L (liquid) water- 60 + sugar L S (liquid solution system 40 (solid i.e., syrup) sugar) 20 A (20°C,70) Temperature (°C) Temperature 2 phases Adapted from 0 Fig. 9.1, 0 20 40 60 70 80 100 Callister 7e. Co =Composition (wt% sugar) Chapter 9 - 5 Binary isomorphous systems

• Complete liquid and solid solubility of the two components • Cu-Ni solid solution • liquidus line • solidus line • melting temperatures • phase composition • phase relative amount: tie line and lever rule

Chapter 9 - Phase Diagrams

• Indicate phases as function of T, Co, and P. • For this course: -binary systems: just 2 components. -independent variables: T and Co (P = 1 atm is almost always used). T(°C) 1600 • Phase • 2 phases: L (liquid) Diagram 1500 for Cu-Ni L (liquid) α (FCC solid solution) system 1400 • 3 phase fields: L 1300 L + α 1200 α α Adapted from Fig. 9.3(a), Callister 7e. (Fig. 9.3(a) is adapted from Phase 1100 (FCC solid Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, solution) OH (1991). 1000 0 20 40 60 80 100 wt% Ni Chapter 9 - 7 Phase Diagrams: composition of phases • Rule 2: If we know T and Co, then we know: --the composition of each phase. Cu-Ni system • Examples: T(°C) A C o = 35 wt% Ni T A tie line At T A = 1320°C: 1300 L (liquid) Only Liquid (L) B CL = Co ( = 35 wt% Ni) T B α At T D = 1190°C: 1200 D (solid) Only Solid ( α ) T D Cα = Co ( = 35 wt% Ni) 20 3032 35 40 4 3 50 At T B = 1250°C: C LC o C α wt% Ni Both α and L Adapted from Fig. 9.3(b), Callister 7e. (Fig. 9.3(b) is adapted from Phase Diagrams C L = C liquidus ( = 32 wt% Ni here) of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.) C α = C solidus ( = 43 wt% Ni here)

Chapter 9 - 8 Phase Diagrams: weight fractions of phases • Rule 3: If we know T and Co, then we know: --the amount of each phase (given in wt%). Cu-Ni • Examples: T(°C) system C o = 35 wt% Ni A T A tie line At T A : Only Liquid (L) 1300 L (liquid) W = 100 wt%, W = 0 L α B At T D : Only Solid ( α ) T B R S α W L = 0, W α = 100 wt% 1200 D (solid) At T B : Both α and L T D S 20 30 40 50 W = 32 35 4 3 L R + S C LC o C α wt% Ni

Adapted from Fig. 9.3(b), Callister 7e. R (Fig. 9.3(b) is adapted from Phase Diagrams of W = = 27 wt% Binary Nickel Alloys, P. Nash (Ed.), ASM α R + S International, Materials Park, OH, 1991.) Chapter 9 - 9 The Lever Rule

• Tie line – connects the phases in equilibrium with each other - essentially an isotherm T(°C) How much of each phase? tie line Think of it as a lever (teeter-totter) 1300 L (liquid) M M B L α T B α 1200 (solid) R S R S

20 30 40 50 C L Co Cα wt% Ni Adapted from Fig. 9.3(b), Callister 7e.

Chapter 9 - 10 Ex: Cooling in a Cu-Ni Binary

• Phase diagram: T(°C) L (liquid) L: 35wt%Ni Cu-Ni system. Cu-Ni system • System is: 130 0 A --binary L: 35 wt% Ni α: 46 wt% Ni B i.e., 2 components: 35 46 32 C Cu and Ni. 43 --isomorphous D 24 36 L: 32 wt% Ni i.e., complete α: 43 wt% Ni solubility of one 120 0 E component in L: 24 wt% Ni α: 36 wt% Ni another; α phase α field extends from (solid) 0 to 100 wt% Ni. • Consider 110 0 Co = 35 wt%Ni. 20 3 0 35 4 0 5 0 Adapted from Fig. 9.4, C o wt% Ni Callister 7e. Chapter 9 - 11 Cored vs Equilibrium Phases

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

Uniform C : First α to solidify: α 46 wt% Ni 35 wt% Ni

Last α to solidify: < 35 wt% Ni

Chapter 9 - 12 Binary-Eutectic Systems has a special composition 2 components with a min. melting T. Cu-Ag T(°C) system Ex.: Cu-Ag system 1200 • 3 single phase regions L (liquid) (L, α, β ) 1000 • Limited solubility: α L + α 800 779°C L + β β α : mostly Cu TE 8.0 71.9 91.2 β: mostly Ag 600 • TE : No liquid below TE α + β 400 • CE : Min. melting TE composition 200 0 20 40 60 CE 80 100 • Eutectic transition Co , wt% Ag L(C ) (C ) + (C ) Adapted from Fig. 9.7, E α αE β βE Callister 7e.

Chapter 9 - 13 Binary eutectic systems (Example)

• Explain how spreading salt on ice that is at a temperature below 00C can cause the ice to melt.

Chapter 9 - EX: Pb-Sn Eutectic System (1) • For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find... --the phases present: α + β Pb-Sn --compositions of phases: T(°C) system

CO = 40 wt% Sn C = 11 wt% Sn 300 α L (liquid) Cβ = 99 wt% Sn --the relative amount L + α 200 α 183°C L + β β of each phase: 18.3 61.9 97.8 S Cβ - CO 150 W α = = R S R+S Cβ - Cα 100 α + β 99 - 40 59 = = = 67 wt% 99 - 11 88 0 11 20 40 60 80 99100 R CO - Cα W β = = Cα Co Cβ R+S Cβ - Cα C, wt% Sn Adapted from Fig. 9.8, 40 - 11 29 Callister 7e. = = = 33 wt% 99 - 11 88 Chapter 9 - 15 EX: Pb-Sn Eutectic System (2) • For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find... --the phases present: α + L Pb-Sn --compositions of phases: T(°C) system

CO = 40 wt% Sn C = 17 wt% Sn 300 α L (liquid) CL = 46 wt% Sn L + α --the relative amount 220 200 α R S L + β β of each phase: 183°C CL - CO 46 - 40 W = = α C - C 46 - 17 100 L α α + β 6 = = 21 wt% 29 0 17 20 40 46 60 80 100 Cα Co CL CO - Cα 23 C, wt% Sn W L = = = 79 wt% Adapted from Fig. 9.8, CL - Cα 29 Callister 7e.

Chapter 9 - 16