Phase Diagrams. Phases

Transcription

1 Phase Diagrams Reading: Callister Ch. 10 What is a phase? What is the equilibrium i state t when different elements are mixed? What phase diagrams tell us. How phases evolve with temperature and composition (microstructures). β (lighter phase) (darker phase) 1 Phases Phase: A homogeneous portion of a system that has uniform physical and chemical characteristics. 1. Different physical states: t vapor, liquid, id solid e.g. water 2 1

2 2. Different chemical compositions Phases e.g. sugar water Solubility imit: Max concentration for which only a solution occurs. Solubility limit increases with T: e.g., if T = 100C, solubility Question: What is the solubility limit at 20C? limit = 80wt% sugar. Answer: 65wt% sugar. If C o < 65wt% sugar: syrup If C o > 65wt% sugar: syrup + sugar. 3 Components: elements or compounds which are mixed initially (e.g., Al and Cu) Phases: physically ysca y and chemically ca ds distinct regions ego that result (e.g., and β). Aluminum- Copper Alloy β (lighter phase) (darker phase) Plain carbon steel 4 2

3 Temperature and composition B C A A: 2 phases present (liquid solution and solid sugar) B: Single phase C: back to 2 phases but at different composition Note: this is an equilibrium phase diagram (What does it mean to have phase equilibrium?) 5 Phase Equilibrium Equilibrium: minimum energy state for a given T, P, and composition (i.e. equilibrium state will persist indefinitely for a fixed T, P and composition). An equilibrium phase will stay constant over time. Phase diagrams tell us about equilibrium phases as a function of T, P and composition (here, we ll always keep P constant for simplicity). 6 3

4 Unary Systems Single component system Consider 2 elemental metals separately: Cu has melting T = 1085 o C Ni has melting T = 1453 o C T (at standard P = 1 atm) T liquid liquid 1453 o C 1085 o C solid solid Cu Ni What happens when Cu and Ni are mixed? 7 Binary Isomorphous Systems 2 components Complete liquid and solid solubility Expect T m of solution to lie in between T m of two pure components T T liquid liquid 1453 o C 1085 o C solid S solid Cu wt% Ni Ni For a pure component, complete melting occurs before T increases (sharp phase transition). But for multicomponent systems, there is usually a coexistence of and S. 8 4

5 Binary Isomorphous Systems What can we learn from this phase diagram? Solid-liquid coexistence region 1. Phase(s) present. A: solid () only B: solid and liquid 2. Composition of those phases A: 60 wt% Ni B: 35 wt% Ni overall (how about in and S separately?) 3. Amount of the phases. A: 100% phase B: % solid and % liquid? 9 Determining phase composition in 2-phase region: 1. Draw the tie line. 2. Note where the tie line intersects the liquidus and solidus lines (i.e. where the tie line crosses the phase boundaries). 3. Read off the composition at the boundaries: iquid is composed of C amount of Ni (31.5 wt% Ni). Solid is composed of C amount of Ni (42.5 wt% Ni). 10 5

6 ever Rule Determining phase amount in the 2- phase region: 1. Draw the tie line. 2. Determine the distance from the point of interest (B) to each of the phase boundaries. R = C o C S = C -C o 3. Mass fractions (wt%) of each phase: S C Co W = = iquid: = = 0.68 R + S C C R Co C Solid: W = = = = 0.32 R + S C C i.e. 68% of the mass is liquid and 32% of the mass is solid. 11 ever Rule: Derivation Since we have only 2 phases: W W = 1 (1) + Conservation of mass requires that: Amount of Ni in -phase + amount of Ni in liquid phase = total amount of Ni or W C + WC = Co (2) From 1 st condition, we have: W =1 W Sub-in to (2): (1 W ) C + W C = C o Solving for W and W gives : W C = C C C o W C = C o C C 12 6

7 ever Rule: Derivation A geometric interpretation: C Co C moment equilibrium: R S W R = W S W W 1 W solving gives ever Rule 13 Microstructures in Isomorphous Alloys Microstructures will vary on the cooling rate (i.e. processing conditions) 1. Equilibrium Cooling: Very slow cooling to allow phase equilibrium to be maintained during the cooling process. a (T>1260 o C): start as homogeneous liquid solution. b (T ~ 1260 o C): liquidus line reached. phase begins to nucleate. C = 46 wt% Ni; C = 35 wt% Ni c (T= 1250 o C): calculate composition and mass fraction of each phase. d (T~ 1220 o C): solidus line reached. Nearly complete solidification. C = 35 wt% Ni; C = 24 wt% Ni e (T<1220 o C): homogeneous solid solution with 35 wt% Ni. 14 7

8 Example problem 65 wt% Ni 35 wt% Cu alloy is heated to T within the + region. If -phase h contains 70 wt% Ni, determine: a. Temperature of the alloy. b. Composition of the liquid phase. c. Mass fraction of both phases. 15 Non-equilibrium cooling Fast cooling, but how fast? Fast w.r.t. diffusion Since diffusion rate is especially low in solids, consider case where: Cooling rate >> diffusion rate in solid Cooling rate << diffusion rate in liquid (equilibrium maintained in liquids phase) 16 8

9 Non-equilibrium cooling a (T>1260 o C): start as homogeneous liquid solution. b (T ~ 1260 o C): liquidus line reached. phase begins to nucleate. C = 46 wt% Ni; C = 35 wt% Ni c (T= 1250 o C): solids that formed at pt b remain with same composition (46wt%) and new solids with 42 wt% Ni form around the existing solids (Why around them?). d (T~ 1220 o C): solidus line reached. Nearly complete solidification. Previously solidified regions maintain original composition and further solidification occurs at 35 wt% Ni. e (T<1220 o C): Non-equilibrium solidification complete (with phase segregation). 17 Cored vs. Equilibrium phases C changes as we solidify. Cu-Ni case: First to solidify has C = 46wt%Ni. ast to solidify has C = 35wt%Ni. Fast rate of cooling: Cored structure Slow rate of cooling: Equilibrium structure First to solidfy: 46wt%Ni ast to solidfy: < 35wt%Ni Uniform C : 35wt%Ni 18 9

10 Binary Eutectic Systems Cu-Ag phase diagram easily melted Single phase regions: -phase (solid solution rich in Cu). β-phase (solid solution rich in Ag). -phase (liquid solution). 2-Phase coexistence regions: +β phases (limited solubility of Ag in Cu and vice versa lead to 2 different solid solution phases). + and β+ phase regions. Tie lines and ever Rule can be applied in the 2-phase regions. 19 Binary Eutectic System Cu-Ag phase diagram C =7wt% Ag C = 50wt% Ag B Tie line A Tie line Calculate composition and mass fractions of each phases at points A and B. C = 3wt% Ag C β = 97wt% Ag 20 10

11 Eutectic Point Eutectic point: Where 2 liquidus lines meet (pt. E). Sometimes also referred to as invariant point. Eutectic Reaction: cool (C E ) (C E ) + β(c βe ) heat similar to one component (pure) system except 2 solid phases. Cu-Ag phase diagram Eutectic Isotherm: Horizontal solidus line at T = T E. 21 Eutectic System: Example Pb-Sn phase diagram At 150 o C for a 40wt% Sn/ 60 wt% Pb alloy: 1. What phases are present? 2. What are the compositions of the phases present? 3. What are the mass fractions of the phases? 4. What are the volume fractions? For a 10wt% Sn/ 90wt% Pb alloy: 1. At what T, can a state with 50% liquid be achieved? 2. At 250 o C, how much Sn must be added to achieve the same state (50% liquid)? 22 11

12 Microstructures in Eutectic Alloys 1. One component rich composition. a: start with homogeneous liquid. b: -phase solids with liquid. Compositions and mass fractions can be found via tie lines and lever rule. c: -phase solid solution only. Net result: polycrystalline solid. Cooling at this composition is similar to binary isomorphous systems. Partial Pb-Sn phase diagram 23 Microstructures in Eutectic Alloys 2. One-component rich but cooling to + β coexistence. d: homogeneous liquid. e: + phase (same as previous but at different compositions and mass fractions). f: all -phase solid solution. g: + β phase (passing through solvus line leads to exceeding solubility limit and β phase precipitates out). Net result: polycrystalline -solid with fine β crystals

13 Microstructures in Eutectic Alloys 3. Cooling through eutectic point. h: homogeneous liquid. i: solidification via eutectic reaction cool (C E ) (C E ) + β(c βe ) heat note: and β phases have very different compositions than the original composition of the liquid (e.g. C = 18.3 wt% Sn; C β = 97.8 wt% Sn whereas C = 61.9 wt% Sn). Eutectic Structure: layered (lamellar) structure. Why does this structure form? 25 Eutectic Structure Pb-Sn Eutectic microstructure t Sn Pb β Pb rich Sn rich In order to achieve large homogeneous regions, long diffusion lengths are required. amellar structure forms because relatively short diffusion lengths are required

14 Microstructures in Eutectic Alloys Hypoeutectic 4. Cooling through eutectic isotherm. j: homogeneous liquid. k: + phases: use tie lines and lever rule. l: just above eutectic isotherm compositions given but what about mass fraction? m: remaining liquid transforms to eutectic structure upon crossing eutectic isotherm. Microconstituent: an element of a microstructure with identifiable and characteristic structure (at pt. m there are 2 microconstituents: primary 27 and eutectic structures) Just above the eutectic isotherm at C 4 : W P P + Q = All of this amount will turn into eutectic structure (W e ). Primary structure: Total (primary + eutectic): Total β (all in eutectic): W ' Q = P + Q W Q + R = P + Q + R P = P + Q + R Note: this is for equilibrium cooling. Non-equilibrium cooling will lead to: Cored primary phases & Increased fraction of eutectic microconstituent 28 W β 14

15 T( C) 300 Adapted from Fig. 9.7, Callister 6e. (Fig. 9.7 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in- Chief), ASM International, Materials Park, OH, 1990.) (Figs and 9.15 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) hypoeutectic: C 200 TE β Co hypoeutectic eutectic o =50wt%Sn eutectic: C β Co hypereutectic β o =61.9wt%Sn 175 μm 160 μm Adapted from Fig. 9.15, Callister 6e. eutectic micro-constituent Adapted from Fig. 9.12, Callister 6e. Co, wt% Sn (Pb-Sn System) hypereutectic: (illustration only) β β β β β Adapted from Fig. 9.15, Callister 6e. (Illustration only) β 29 Congruent phase transformation Congruent transformation: no change in composition upon phase transformation. Incongruent transformation: phase transformation where at least one of the phases go through composition change. At what overall composition does Mg 2 Pb melt congruently? 30 15

16 Intermetallic compounds Intermediate (intermetallic) compounds: discrete metal compounds rather than solutions (i.e. distinct chemical formula). A x B y : in solution x and y can vary in compounds x and y are fixed (always fixed composition of A and B). Mg-Pb phase diagram Can be considered as two phase diagrams put together back to back. Mg-Mg 2 Pb phase diagram Mg 2 Pb-Pb phase diagram 31 Intermetallic compounds: Example Within the + Mg 2 Pb coexistence region, each of the two phases is found to have 50% mass fraction. If the overall composition is 50wt%Pb, determine the composition of each of the phases and temperature

17 Intermediate phases Intermediate solid solutions (intermediate phases): Solid solutions that do not extend to pure components in the phase diagram. Cu-Zn Terminal solutions: and η. Intermediate solutions: β, γ, δ and ε. Tie lines and lever rule can be used to determine compositions and wt% of phases. e.g. at 800 o C with 70 wt% Zn C = 78 wt% Zn C γ = 67 wt% Zn 33 Eutectoid and peritectic reactions Eutectoid reaction: one solid phase turning into two other solid phases upon cooling cool e.g. δ γ + ε heat Peritectic reaction: one solid phase transforms into liquid and a different solid phases upon heating cool e.g. δ + ε 34 heat 17

18 Ceramic phase diagrams Al 2 O 3 -Cr 2 O 3 MgO-Al 2 O 3 35 Gibbs Phase Rule Number of phases present P + F = C + N Number of non-compositional variables (Temperature & Pressure) Degree of freedom (externally controllable parameters: i.e. T, P, and C) Number of components 36 18

19 P + F = C + N Gibbs Phase Rule e.g. Cu-Ag phase diagram Cu and Ag are the only components -> C = 2 Temperature is the only noncompositional variable here (i.e. fixed pressure). -> N = 1 (but in general N = 2) When 2 phases are present -> P = 2 which leads to F =C+NP=2+1 C+N-P = =1 When only 1 phase is present -> P = 1 which leads to F = 2 What does this mean? Why should you care? 37 Gibbs Phase Rule In the previous example of Cu-Ag phase diagram, when F = 1, only one parameter (T or C) needs to be specified to completely define the system. e.g. (for + region) If T is specified to be 1000 o C, compositions are already determined (C and C ). Or If composition of the a phase is specified to be C then both T and C are already determined. C C 38 19

20 Gibbs Phase Rule When F = 2, both T and C have to be specified to completely define the state of the system. e.g.(for region) If T is specified to be 800 o C, C can be any where between 0 to ~8 wt% Ag) Or If composition of the a phase is specified to be C = 3 wt%, then T and can be any where between ~600 to 1100 o C. C 39 Gibbs Phase Rule Where in the Cu Ag diagram, is there a 0 degree of freedom? (i.e. T, P, and C are all fixed) 40 20

21 Iron-Carbon System Typical metal (e.g. Cu) Iron T T( o C) iquid 1538 T m 1394 Solid 912 iquid -Fe(BCC) γ-fe (FCC) -Fe (BCC) 41 Iron-Carbon System Eutectic point Eutectoid Note: only goes out to 6.7 wt% C (100 wt% Fe 3 C intermediate compound) FCC γ-phase has highest C concentration (2.14 wt% C) whereas BCC - phase has low solubility (0.022 wt% C). Recall FCC is close packed (i.e. larger APF). Why is C more soluble in FCC? 42 21

22 Eutectoid cooling γ (0.76 wt% C) cool heat (0.022 wt% C) + Fe 3 C (6.7 wt% C) Pearlite structure ayered structure forms due to the same reason as eutectic structure formation. 43 Hypoeutectoid Alloys Cooling below eutectoid composition. c: homogeneous γ solid. d: + γ coexistence. -phase nucleate at the grain boundaries (Why?). e -> f: - crossing eutectoid isotherm will cause all remaining γ-phase into eutectoid structure. - -phase that formed prior to eutectoid isotherm are called proeutectoid ferrite. Fraction of pearlite = W C o p = Fraction of proeutectoid = W 0.76 C o ' = 22

23 Hypereutectoid Alloys Cooling above the eutectoid composition Compositions and wt% can be found similarly il l as hypoeutectoid t cooling. Instead of proeutectoid, proeutectoid cementite appears. 45 Example problem For 0.35 wt% C, at T just below eutectoid isotherm, determine: a) Fractions of total ferrite and cementite phases. b) Fractions of proeutectoid ferrite and pearlite. c) Fraction of eutectoid ferrite

24 Influence of other alloying elements Changes eutectoid T Changes eutectoid composition Useful processing info to control microstructure. 47 Concepts to remember Phases, physical states, chemical composition, phase equilibrium. Phase diagrams tell us about: Number and types of phases present. Composition of each phase. Mass fraction (wt%) of each phase. Binary isomorphous systems. Binary eutectic systems. Intermediate compounds. Intermediate phases. Fe-C phase diagram. Microstructure evolution. Phase diagrams help us to determine the equilibrium microstructures which in turn determines the properties of materials! 48 24