Chapter 11: Phase Diagrams. Phase Equilibria: Solubility Limit

Transcription

1 Temperature ( C) Water ugar 217/1/4 Chapter 11: Phase Diagrams IUE TO ADDRE... 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 Chapter 11-1 Phase Equilibria: olubility imit olution solid, liquid, or gas solutions, single phase Mixture more than one phase olubility imit: Maximum concentration for which only a single phase solution exists. Question: What is the solubility limit for sugar in water at 2 C? Answer: 65 wt% sugar. 1 At 2 C, if C < 65 wt% sugar: syrup At 2 C, if C > 65 wt% sugar: syrup + sugar ugar/water Phase Diagram olubility imit (liquid solution i.e., syrup) Adapted from Fig. 11.1, Callister & (liquid) + (solid sugar) C = Composition (wt% sugar) Chapter

2 Temperature ( C) 217/1/4 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., and ). Aluminum- Copper Alloy Adapted from chapteropening photograph, Chapter 9, Callister, Materials cience & Engineering: An Introduction, 3e. (lighter phase) (darker phase) 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. watersugar Fig. 11.1, Callister & (liquid solution i.e., syrup) B (1 C,C = 7) D (1 C,C = 9) 1 phase 2 phases (liquid) + (solid sugar) A (2 C,C = 7) 2 phases C = Composition (wt% sugar) Chapter

3 217/1/4 Criteria for olid olubility imple (e.g., Ni-Cu solution) Crystal tructure electroneg r (nm) Ni FCC Cu FCC 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. Chapter 11-5 Phase Diagrams Indicate phases as a function of T, C, and P. For this course: - binary s: just 2 components. - independent variables: T and C (P = 1 atm is almost always used). Phase Diagram for Cu-Ni (liquid) (FCC solid solution) phases: (liquid) (FCC solid solution) 3 different phase fields: + Fig. 11.3(a), Callister & (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, Reprinted by permission of AM International, Materials Park, OH.) wt% Ni Chapter

4 B (125ºC,35) 217/1/4 Isomorphous Binary Phase Diagram Phase diagram: Cu-Ni. ystem is: -- binary i.e., 2 components: Cu and Ni. -- isomorphous i.e., complete solubility of one component in another; phase field extends from to 1 wt% Ni (liquid) (FCC solid solution) Cu-Ni phase diagram wt% Ni Fig. 11.3(a), Callister & (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, Reprinted by permission of AM International, Materials Park, OH.) Chapter 11-7 Phase Diagrams: Determination of phase(s) present Rule 1: If we know T and C o, then we know: -- which phase(s) is (are) present. Examples: A(11 C, 6 wt% Ni): 1 phase: B(125 C, 35 wt% Ni): 2 phases: + Fig. 11.3(a), Callister & (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, Reprinted by permission of AM International, Materials Park, OH.) (liquid) (FCC solid solution) A(11ºC,6) Cu-Ni phase diagram wt% Ni Chapter

5 217/1/4 Phase Diagrams: Determination of phase compositions Rule 2: If we know T and C, then we can determine: -- the composition of each phase. Cu-Ni Examples: Consider C = 35 wt% Ni At T A = 132 C: Only iquid () present C = C ( = 35 wt% Ni) At T D = 119 C: Only olid () present C = C ( = 35 wt% Ni) At T B = 125 C: Both and present C = Cliquidus ( = 32 wt% Ni) C = Csolidus ( = 43 wt% Ni) T A 13 T B T D 2 (liquid) A B D 3235 tie line (solid) 4 C C C Fig. 11.3(b), Callister & (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, Reprinted by permission of AM International, Materials Park, OH.) wt% Ni Chapter 11-9 Phase Diagrams: Determination of phase weight fractions Rule 3: If we know T and C, then can determine: -- the weight fraction of each phase. Cu-Ni Examples: Consider C = 35 wt% Ni A At T A : Only iquid () present At T D : At T B : W = W = W = 1., W a = Only olid ( ) present W =, W = 1. Both and present R + R R = = =.27 T A 13 T B T D 2 (liquid) B R D tie line (solid) CC C Fig. 11.3(b), Callister & (Adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash, Editor, Reprinted by permission of AM International, Materials Park, OH.) wt% Ni Chapter

6 217/1/4 The ever Rule Tie line connects the phases in equilibrium with each other also sometimes called an isotherm 13 TB 2 (liquid) R B C C wt% Ni tie line (solid) C Adapted from Fig. 11.3(b), Callister & What fraction of each phase? Think of the tie line as a lever (teeter-totter) M R M Chapter Ex: Cooling of a Cu-Ni Alloy Phase diagram: Cu-Ni. Consider microstuctural changes that accompany the cooling of a C = 35 wt% Ni alloy (liquid) 13 : 35 wt% Ni : 46 wt% Ni (solid) A B C 24 D 36 E : 35 wt%ni Cu-Ni : 32 wt% Ni : 43 wt% Ni : 24 wt% Ni : 36 wt% Ni : 35 wt% Ni 11 2 Adapted from Fig. 11.4, Callister & C wt% Ni Chapter

7 Tensile trength (MPa) Elongation (%E) 217/1/4 Mechanical Properties: Cu-Ni ystem Effect of solid solution strengthening on: -- Tensile strength (T) -- Ductility (%E) 4 3 T for pure Cu Cu Ni Composition, wt% Ni Adapted from Fig. 11.5(a), Callister & T for pure Ni %E for pure Cu Cu Ni %E for pure Ni Composition, wt% Ni Adapted from Fig. 11.5(b), Callister & Chapter Binary-Eutectic ystems 2 components Ex.: Cu-Ag 3 single phase regions (,, ) imited solubility: : mostly Cu : mostly Ag T E : No liquid below T E C E : Composition at temperature T E has a special composition with a min. melting T Eutectic reaction (C E ) (C E ) + (C E ) cooling + (liquid) + Cu-Ag + T C E C E 8 1 C, wt% Ag Fig. 11.6, Callister & Rethwisch 9e [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 199. Reprinted by permission of AM International, Materials Park, OH.]. heating Chapter

8 217/1/4 EX 1: Pb-n Eutectic ystem For a 4 wt% n-6 wt% Pb alloy at 15 C, determine: -- the phases present Pb-n Answer: + -- the phase compositions 3 Answer: C = 11 wt% n (liquid) C = 99 wt% n -- the relative amount C + of each phase Answer: 15 R C W - C 1 = R+ = C - C + = W = = =.67 R R+ = 4-11 = C - C C - C = = C C C Fig. 11.7, Callister & [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in- Chief), 199. Reprinted by permission of AM International, Materials Park, OH.] C, wt% n Chapter EX 2: Pb-n Eutectic ystem For a 4 wt% n-6 wt% Pb alloy at 22 C, determine: -- the phases present: Pb-n Answer: + -- the phase compositions 3 Answer: C = 17 wt% n (liquid) C = 46 wt% n + -- the relative amount 22 2 R + of each phase 183 C Answer: 1 + W = C - C 46-4 = C - C = 6 29 =.21 W = C - C C - C = = C C C Fig. 11.7, Callister & [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in- Chief), 199. Reprinted by permission of AM International, Materials Park, OH.] C, wt% n Chapter

9 217/1/4 Microstructural Developments in Eutectic ystems I For alloys for which C < 2 wt% n Result: at room temperature -- polycrystalline with grains of phase having composition C T E : C wt% n : C wt% n + (Pb-n ystem) 1 + Fig. 11.1, Callister & 1 2 C 2 (room T solubility limit) 3 C, wt% n Chapter Microstructural Developments in Eutectic ystems II For alloys for which 2 wt% n < C < 18.3 wt% n Result: at temperatures in + range -- polycrystalline with grains and small -phase particles T E + : C wt% n : C wt% n 1 + Pb-n Fig , Callister & C (sol. limit at Troom) 18.3 (sol. limit at T E ) 3 C, wt% n Chapter

10 217/1/4 Microstructural Developments in Eutectic ystems III For alloy of composition C = C E Result: Eutectic microstructure (lamellar structure) -- alternating layers (lamellae) of and phases. 3 Pb-n 2 T E C : C wt% n + Micrograph of Pb-n eutectic microstructure 1 Fig , Callister & + : 97.8 wt% n : 18.3 wt%n C E C, wt% n 16μm Fig , Callister & (From Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, Reproduced by permission of AM International, Materials Park, OH.) Chapter amellar Eutectic tructure Figs & 11.14, Callister & (Fig from Metals Handbook, 9th edition, Vol. 9, Metallography and Microstructures, Reproduced by permission of AM International, Materials Park, OH.) Chapter

11 217/1/4 Microstructural Developments in Eutectic ystems IV For alloys for which 18.3 wt% n < C < 61.9 wt% n Result: phase particles and a eutectic microconstituent : C wt% n Just above T E : 3 C Pb-n 2 T E 1 + Fig , Callister & R R C, wt% n + primary eutectic eutectic C = 18.3 wt% n = 61.9 wt% n W = =.5 R + W = (1- W) =.5 Just below T E : C = 18.3 wt% n C = 97.8 wt% n W = =.73 R + W =.27 Chapter Hypoeutectic & Hypereutectic Fig. 11.7, Callister & Rethwisch 9e. [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 3, T. B. Massalski (Editor-in-Chief), 199. Reprinted by permission of AM International, Materials Park, OH.] 3 2 T E (Pb-n ystem) (Figs and from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, Reproduced by permission of AM International, Materials Park, OH.) C, wt% n eutectic hypoeutectic: C = 5 wt% n 61.9 hypereutectic: (illustration only) 175 μm Fig , Callister & eutectic: C = 61.9 wt% n 16 μm eutectic micro-constituent Fig , Callister & Adapted from Fig , Callister & (Illustration only) Chapter

12 217/1/4 Intermetallic Compounds Fig , Callister & [Adapted from Phase Diagrams of Binary Magnesium Alloys, A. A. Nayeb-Hashemi and J. B. Clark (Editors), Reprinted by permission of AM International, Materials Park, OH.] Mg 2 Pb 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). Chapter Eutectic, Eutectoid, & Peritectic Eutectic - liquid transforms to two solid phases cool + (For Pb-n, 183 C, 61.9 wt% n) heat Eutectoid one solid phase transforms to two other solid phases intermetallic compound cementite 3 cool γ + Fe 3 C (For Fe-C, 727 C,.76 wt% C) heat Peritectic - liquid and one solid phase transform to a second solid phase cool δ + γ (For Fe-C, 1493 C,.16 wt% C) heat Chapter

13 Fe 3 C (cementite) 217/1/4 Eutectoid & Peritectic Cu-Zn Phase diagram Peritectic transformation γ + δ Eutectoid transformation δ γ + e Fig. 11.2, Callister & [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 2, T. B. Massalski (Editor-in- Chief), 199. Reprinted by permission of AM International, Materials Park, OH.] Chapter Iron-Carbon (Fe-C) Phase Diagram 2 important points - Eutectic (A): γ + Fe 3 C - Eutectoid (B): γ +Fe 3 C δ (austenite) B 1148 C A γ+fe 3 C 727 C = Teutectoid +Fe 3 C +Fe 3 C 12 μm Result: Pearlite = alternating layers of and Fe 3 C phases Fig , Callister & (From Metals Handbook, Vol. 9, 9th ed., Metallography and Microstructures, Reproduced by permission of AM International, Materials Park, OH.) (Fe) C, wt% C Fe 3 C (cementite-hard) (ferrite-soft) Fig , Callister & [Adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 199. Reprinted by permission of AM International, Materials Park, OH.] Chapter

14 Fe 3 C (cementite) Fe 3 C (cementite) 217/1/4 γ γ γ γ δ a Hypoeutectoid teel + (austenite) 727 C (Fe) C C, wt% C pearlite 1148 C γ + Fe 3 C + Fe 3 C +Fe 3 C (Fe-C ystem) Adapted from Figs and 11.28, Callister & [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 199. Reprinted by permission of AM International, Materials Park, OH.] pearlite Adapted from Fig , Callister & (Photomicrograph courtesy of Republic teel Corporation.) 1 μm Hypoeutectoid steel proeutectoid ferrite Chapter γ γ W = s/(r + s) W γ =(1 - W ) δ 14 8 r s W pearlite = W γ W = /(R + ) R 6 Hypoeutectoid teel + (austenite) 727 C pearlite (Fe) C C, wt% C W Fe 3 C =(1 W ) pearlite 1148 C γ + Fe 3 C + Fe 3 C +Fe 3 C (Fe-C ystem) 1 μm Hypoeutectoid steel Adapted from Figs and 11.28, Callister & [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 199. Reprinted by permission of AM International, Materials Park, OH.] proeutectoid ferrite Adapted from Fig , Callister & (Photomicrograph courtesy of Republic teel Corporation.) Chapter

15 Fe 3 C (cementite) Fe 3 C (cementite) 217/1/4 Fe 3 C δ Hypereutectoid teel + (austenite) 727 C 4 1 C (Fe) C C, wt% C pearlite 1148 C γ + Fe 3 C + Fe 3 C +Fe 3 C (Fe-C ystem) Adapted from Figs and 11.31, Callister & [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 199. Reprinted by permission of AM International, Materials Park, OH.] pearlite 6 μm Hypereutectoid steel proeutectoid Fe 3 C Adapted from Fig , Callister & (Copyright 1971 by United tates teel Corporation.) Chapter Fe 3 C δ 14 W γ =x/(v + x) 8 W Fe 3 C =(1-W γ ) Hypereutectoid teel + (austenite) 727 C 6 + Fe 3 C pearlite 4 1 C (Fe) C C, wt% C W pearlite = W γ W = X/(V + X) W =(1 - W ) Fe 3 C V v pearlite 1148 C X γ + Fe 3 C x +Fe 3 C (Fe-C ystem) Adapted from Figs and 11.31, Callister & [Figure 9.24 adapted from Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 199. Reprinted by permission of AM International, Materials Park, OH.] 6 μm Hypereutectoid steel proeutectoid Fe 3 C Adapted from Fig , Callister & (Copyright 1971 by United tates teel Corporation.) Chapter

16 Fe 3 C (cementite) 217/1/4 Example Problem For a 99.6 wt% Fe-.4 wt% C steel at a temperature just below the eutectoid, determine the following: a) The compositions of Fe 3 C and ferrite (). b) The amount of cementite (in grams) that forms in 1 g of steel. c) The amounts of pearlite and proeutectoid ferrite () in the 1 g. Chapter olution to Example Problem a) Using the R tie line just below the eutectoid C =.22 wt% C C Fe3 C = 6.7 wt% C b) Using the lever rule with the tie line shown Amount of Fe 3 C in 1 g = (1 g)w Fe3 C = (1 g)(.57) = 5.7 g δ 14 + (austenite) 8 R 6 Fig , Callister & [From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 199. Reprinted by permission of AM International, Materials Park, OH.] 727 C C C C, wt% C C Fe 3 C 1148 C γ + Fe 3 C + Fe 3 C +Fe 3 C Chapter

17 TEutectoid (ºC) Ceutectoid (wt% C) Fe 3 C (cementite) 217/1/4 olution to Example Problem (cont.) c) Using the VX tie line just above the eutectoid and realizing that C =.4 wt% C C =.22 wt% C C pearlite = C γ =.76 wt% C Amount of pearlite in 1 g δ 14 + (austenite) 8 V X 6 Fig , Callister & [From Binary Alloy Phase Diagrams, 2nd edition, Vol. 1, T. B. Massalski (Editor-in-Chief), 199. Reprinted by permission of AM International, Materials Park, OH.] 727 C 1148 C γ + Fe 3 C + Fe 3 C +Fe 3 C = (1 g)w pearlite = (1 g)(.512) = 51.2 g C C C γ C, wt% C Chapter Alloying with Other Elements T eutectoid changes: C eutectoid changes: Ti Mo i W Ni Cr Ni Cr Mn Ti i Mo W Mn wt. % of alloying elements Fig , Callister & (From Edgar C. Bain, Functions of the Alloying Elements in teel, Reproduced by permission of AM International, Materials Park, OH.) wt. % of alloying elements Fig ,Callister & (From Edgar C. Bain, Functions of the Alloying Elements in teel, Reproduced by permission of AM International, Materials Park, OH.) Chapter

18 217/1/4 ummary Phase diagrams are useful tools to determine: -- the number and types of phases present, -- the composition of each phase, -- and the weight fraction of each phase given the temperature and composition of the. The microstructure of an alloy depends on -- its composition, and -- whether or not cooling rate allows for maintenance of equilibrium. Important phase diagram phase transformations include eutectic, eutectoid, and peritectic. Chapter