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Chapter 11: Phase Diagrams ISSUES to ADDRESS Chapter 11: Phase Diagrams ISSUES TO ADDRESS... • When we combine two elements... what is the resulting equilibrium state? • In particular, if we specify... -- the composition (e.g., wt% Cu - wt% Ni), and -- the temperature (T) then... How many phases form? What is the composition of each phase? What is the amount of each phase? Phase A Phase B Nickel atom Copper atom AMSE 205 Spring ‘2016 Chapter 11 - 1 Phase Equilibria: Solubility Limit • Solution – solid, liquid, or gas solutions, single phase • Mixture – more than one phase Adapted from Fig. 11.1, Callister & Rethwisch 9e. Sugar/Water Phase Diagram • Solubility Limit: 100 Maximum concentration for ) Solubility C which only a single phase o 80 Limit L solution exists. (liquid) 60 L + Question: What is the 40 (liquid solution S solubility limit for sugar in i.e., syrup) (solid o 20 water at 20 C ? ( Temperature sugar) Answer: 65 wt% sugar. 0 20 At 20°C, if C < 65 wt% sugar: syrup 40 6065 80 100 C = Composition (wt% sugar) At 20°C, if C > 65 wt% sugar: Sugar syrup + sugar Water AMSE 205 Spring ‘2016 Chapter 11 - 2 Components and Phases • Components: The elements or compounds which are present in the alloy (e.g., Al and Cu) • Phases: The physically and chemically distinct material regions that form (e.g., α and β). Aluminum- β (lighter Copper phase) Alloy α (darker Adapted from chapter- phase) opening photograph, Chapter 9, Callister, Materials Science & Engineering: An Introduction, 3e. AMSE 205 Spring ‘2016 Chapter 11 - 3 Effect of Temperature & Composition • Altering T can change # of phases: path A to B. • Altering C can change # of phases: path B to D. B (100 oC, C = 70) D (100 oC, C = 90) 1 phase 2 phases 100 L ) 80 C (liquid) o water- 60 + sugar L S (liquid solution system 40 (solid i.e., syrup) sugar) 20 A (20 oC, C = 70) Temperature ( Temperature 2 phases Fig. 11.1, Callister & 0 Rethwisch 9e. 0 20 40 60 70 80 100 C = Composition (wt% sugar) AMSE 205 Spring ‘2016 Chapter 11 - 4 Criteria for Solid Solubility Simple system (e.g., Ni-Cu solution) Crystal electroneg r (nm) Structure Ni FCC 1.9 0.1246 Cu FCC 1.8 0.1278 • Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume – Rothery rules) suggesting high mutual solubility. • Ni and Cu are totally soluble in one another for all proportions. AMSE 205 Spring ‘2016 Chapter 11 - 5 Phase Diagrams • Indicate phases as a function of T, C, and P. • For this course: - binary systems: just 2 components. - independent variables: T and C (P = 1 atm is almost always used). T(oC) Phase 1600 • 2 phases: Diagram L (liquid) 1500 for Cu-Ni L (liquid) α (FCC solid solution) system 1400 • 3 different phase fields: L 1300 L + α 1200 α α Fig. 11.3(a), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary (FCC solid Nickel Alloys, P. Nash, Editor, 1991. Reprinted 1100 by permission of ASM International, Materials solution) Park, OH.) 1000 0 20 40 60 80 100 wt% Ni AMSE 205 Spring ‘2016 Chapter 11 - 6 Isomorphous Binary Phase Diagram • Phase diagram: T(oC) Cu-Ni system. 1600 • System is: 1500 L (liquid) Cu-Ni -- binary 1400 phase i.e., 2 components: diagram Cu and Ni. -- isomorphous 1300 i.e., complete 1200 α solubility of one component in 1100 (FCC solid another; α phase solution) field extends from 1000 0 20 40 60 80 100 wt% Ni 0 to 100 wt% Ni. Fig. 11.3(a), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Park, OH.) AMSE 205 Spring ‘2016 Chapter 11 - 7 Phase Diagrams: Determination of phase(s) present • Rule 1: If we know T and Co, then we know: -- which phase(s) is (are) present. T(oC) • Examples: 1600 A(1100 oC, 60 wt% Ni): L (liquid) 1 phase: α 1500 Cu-Ni B(1250 oC, 35 wt% Ni): 1400 phase 2 phases: L + α diagram 1300 (1250ºC,35) α B (FCC solid 1200 Fig. 11.3(a), Callister & Rethwisch 9e. solution) (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, 1991. Reprinted 1100 A(1100ºC,60) by permission of ASM International, Materials Park, OH.) 1000 0 20 40 60 80 100 wt% Ni AMSE 205 Spring ‘2016 Chapter 11 - 8 Phase Diagrams: Determination of phase compositions • Rule 2: If we know T and C0, then we can determine: -- the composition of each phase. Cu-Ni system • Examples: T(oC) A Consider C0 = 35 wt% Ni TA tie line o 1300 L (liquid) At TA = 1320 C : Only Liquid (L) present B T CL = C0 ( = 35 wt% Ni) B o α At TD = 1190 C : 1200 D (solid) Only Solid (α) present TD Cα = C0 ( = 35 wt% Ni) 20 303235 4043 50 o At TB = 1250 C : CL C0 Cα wt% Ni Both α and L present Fig. 11.3(b), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary CL = Cliquidus ( = 32 wt% Ni) Nickel Alloys, P. Nash, Editor, 1991. Reprinted by permission of ASM International, Materials Cα = Csolidus ( = 43 wt% Ni) Park, OH.) AMSE 205 Spring ‘2016 Chapter 11 - 9 Phase Diagrams: Determination of phase weight fractions • Rule 3: If we know T and C0, then can determine: -- the weight fraction of each phase. Cu-Ni • Examples: T(oC) system Consider C = 35 wt% Ni A 0 TA tie line At TA : Only Liquid (L) present 1300 L (liquid) WL = 1.00, W = 0 B T At TD : Only Solid (α) present B R S α WL = 0, Wα = 1.00 1200 (solid) At T : Both α and L present D B TD S 43 35 20 303235 4043 50 WL 0.73 R +S 43 32 CL C0 Cα wt% Ni Fig. 11.3(b), Callister & Rethwisch 9e. (Adapted from Phase Diagrams of Binary R Nickel Alloys, P. Nash, Editor, 1991. Reprinted W = 0.27 by permission of ASM International, Materials α R +S Park, OH.) AMSE 205 Spring ‘2016 Chapter 11 - 10 The Lever Rule • Tie line – connects the phases in equilibrium with each other – also sometimes called an isotherm T(oC) What fraction of each phase? tie line Think of the tie line as a lever 1300 L (liquid) (teeter-totter) B T B ML Mα α 1200 (solid) R S 20 30C 40 50 R S L C0 C αAdapted from Fig. 11.3(b), wt% Ni Callister & Rethwisch 9e. AMSE 205 Spring ‘2016 Chapter 11 - 11 Ex: Cooling of a Cu-Ni Alloy o • Phase diagram: T( C) L (liquid) L: 35 wt%Ni Cu-Ni system. Cu-Ni system • Consider 1300 A L: 35 wt% Ni microstuctural α: 46 wt% Ni B 35 46 changes that 32 C 43 accompany the D L: 32 wt% Ni cooling of a 24 36 α: 43 wt% Ni C = 35 wt% Ni alloy 1200 E 0 L: 24 wt% Ni α α: 36 wt% Ni (solid) α: 35 wt% Ni 110 0 20 3035 40 50 wt% Ni Adapted from Fig. 11.4, C0 Callister & Rethwisch 9e. AMSE 205 Spring ‘2016 Chapter 11 - 12 Mechanical Properties: Cu-Ni System • Effect of solid solution strengthening on: -- Tensile strength (TS) -- Ductility (%EL) ) 60 %EL for pure Cu 400 EL 50 %EL for TS for pure Ni pure Ni 40 300 TS for pure Cu 30 200 Elongation (% 20 0 20406080100 020406080100 Tensile Strength (MPa) Tensile Cu Ni Cu Ni Composition, wt% Ni Composition, wt% Ni Adapted from Fig. 11.5(a), Adapted from Fig. 11.5(b), Callister & Rethwisch 9e. Callister & Rethwisch 9e. AMSE 205 Spring ‘2016 Chapter 11 - 13 Binary-Eutectic Systems has a special composition 2 components with a min. melting T. Cu-Ag T(oC) 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+ββ α: mostly Cu TE 8.0 71.9 91.2 β: mostly Ag 600 • TE : No liquid below TE α β 400 • CE : Composition at temperature TE 200 0 20 40 60 CE 80 100 • Eutectic reaction C, wt% Ag Fig. 11.6, Callister & Rethwisch 9e [Adapted from Binary Alloy Phase Diagrams, 2nd edition, L(CE) α(CαE) + β(CβE) Vol. 1, T. B. Massalski (Editor-in-Chief), 1990. Reprinted by permission of ASM International, Materials Park, OH.]. cooling heating AMSE 205 Spring ‘2016 Chapter 11 - 14 EX 1: Pb-Sn Eutectic System • For a 40 wt% Sn-60 wt% Pb alloy at 150 oC, determine: -- the phases present Pb-Sn Answer: α + β T(oC) system -- the phase compositions 300 Answer: Cα = 11 wt% Sn 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 Answer: 150 R S S Cβ - C0 100 Wα = = R+S Cβ - Cα α + β 99 - 40 59 = = = 0.67 99 - 11 88 0 11 2040 60 80 99100 C C C R C0 -Cα α 0 β W = = C, wt% Sn β C -C Fig. 11.7, Callister & Rethwisch 9e. R+S β α [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in- 40 - 11 29 Chief), 1990. Reprinted by permission of ASM = = = 0.33 International, Materials Park, OH.] 99 - 11 88 AMSE 205 Spring ‘2016 Chapter 11 - 15 EX 2: Pb-Sn Eutectic System • For a 40 wt% Sn-60 wt% Pb alloy at 220 oC, determine: -- the phases present: Pb-Sn Answer: α + L T (oC) system -- the phase compositions 300 Answer: Cα = 17 wt% Sn L (liquid) CL = 46 wt% Sn L+ α -- the relative amount 220 200 α R S L+β β of each phase 183°C Answer: 100 CL -C0 46 - 40 α + β Wα = = CL -Cα 46 - 17 6 17 100 = = 0.21 0 2040 46 60 80 29 Cα C0 CLC, wt% Sn Fig.
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