Phase Diagrams- Introduction ES-312 DRAFT 1 Phase Diagrams- Objectives • Identify the phases in a phase diagram • In binary phase diagrams, given composition and temperature, identify the phases present, composition of phases, weight fraction of phases • Read an Iron-carbon phase diagram and identify the phases and their respective quantities present • Development of microstructures in alloy systems • Prediction of properties of the final microstructure • Last but not least, design and process alloys DRAFT 2 What is a Solution ? DRAFT 3 Imperfections in Solids Conditions for substitutional solid solution: Hume – Rothery rules –1. –2. –3. –4. r (atomic radius) < 15% Proximity in periodic table (i.e., similar electronegativities) Same crystal structure for pure metals Valency equality All else being equal, a metal will have a greater tendency to dissolve a metal of higher valency than one of lower valency DRAFT 4 Alloys/Compounds/Intermediates Intermetallics Ni, Al FCC Ni-Al Alloy NiAl (Nickel Aluminide Intermetallic) Disordered at high temperature – Individual lattice is BCC - Room temperature shows Simple Cubic Crystal Lattice [Light, 5 X stronger than SS, Refractory: Has best of both metals & ceramics – but workability is less due to high brittleness] AKA Electron Compounds/Hume Rothery Compounds: SC [ORDERED ALLOYS ?????] DRAFT Motif: 1Ni + 1Al 5 Definition f Phase/Component/State Phases of Materials What is the difference between a phase and state ? A phase of matter is uniform (homogeneous) with respect to its physical and chemical properties – “Primary” phases of matter is also known as States of Matter – Solid, Liquid, Gas, Plasma & BEC (or degenerate matter) DRAFT 6 Components/States/Phases DRAFT 7 Difference between Purely Physical Terms: Components/States/Phases Phase of matter: A region of space in a material where intensive properties are uniform [Density/Composition/Temperature/Pressure] - Salt + Water = Salt solution (2C/1P) Salt + Water + Ice = Salt Solution+ Ice (2C/2P) Salt + Water + Olive Oil = Salt Solution+ Olive Oil (3C/2P) Salt + Water + Olive Oil+ Ice = Salt Solution+ Olive Oil + Ice (3C/3P) DRAFT 8 Phase Diagram of Pure Substances The lines themselves represent the combinations of T and P at which two states are in equilibrium. Invariant Triple Point DRAFT 9 Iron Same Line (Assume Transposed) DRAFT 10 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., a and b). b (lighter phase) Al-Cu Alloy a (darker phase) Adapted from chapteropening photograph, Chapter 9, Callister, Materials Science & Engineering: An Introduction, 3e. DRAFT 11 Titanium Room Temperature 1 atm DRAFT 12 Gibbs Phase Rule [DOF = (# of Components) – (# of Phases) + Environmental Variables] F=C-P+E For most processes, E= Temp, Pressure etc. For equilibrium diagrams, pressure is assumed constant at 1atm Gibbs’ Phase Rule: F=C-P+2 F=C-P+1 [Callister, Sec 9.17] For one component with p and T and variables DRAFT 13 Phase rule applied to a Pure Substance F=C-P+2 Degree of Freedom => Thermodynamic Variables that can independently be changed without changing phases present If any change is made in one of the variables, another one must change itself DRAFT 14 H2O Triple Point of H2O Where liquid water, water vapor and ice coexist, there is a 'triple point' where both the boiling point of water and melting point of ice are equal. A 'critical point' phase boundary (liquid/vapor) vanishes L/V phases become indistinguishable. Below critical temperature (Tc ) a vapor can be liquefied by increasing pressure only. liq. H2O is hot enough and gaseous H2O is under sufficient pressure that their densities are identical (0.322 g cm-3 ) Triple Point: 273.16K/4.58 mm Hg (0.00603659 atm) [−22 °C/2070 atm ] Critical Point: 373.99 oC/217.75 atm [liq H2O hot enough & H2O Vapor is under sufficient pressure that their densities are identical ] DRAFT 15 What is the Difference ??? DRAFT 16 Stillinger-Weber Potential Liquified noble gases (Ar, Kr, Xe) typify such systems, for which the famous Lennard-Jones (LJ) interaction is well defined/discussed No usual pair potential describes general structures involving tetrahedral coordination/diamond structure (Si) as UL-J stabilizes the close-packed crystals characteristic of the noble gases. Molecular Dynamics Simulation- using Stillinger-Weber Potential Quadruple Point The crystalline phases are BCC, β-tin, DC and SC DRAFT 18 Phase Equilibria: Solubility Limit Sugar/Water Phase Diagram Adapted from Fig. 9.1, Callister & Rethwisch 8e. • Solubility Limit: Answer: Approx 65 wt% sugar. At 20ºC, if C < 65 wt% sugar: syrup At 20ºC, if C > 65 wt% sugar: L2 (liquid) 60 40 20 0 L1 (liquid solution i.e., syrup) + s (solid sugar) 20 40 6065 80 100 C = Composition (wt% sugar) Sugar for sugar in water at 20ºC? Solubility Limit 80 Water Question: What is the solubility limit Temperature (ºC) Maximum concentration for which only a single phase solution exists. 100 syrup + sugar DRAFT 19 Moving the Phase Equilibria 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 (100ºC,C = 70) D (100ºC,C = 90) 100 Temperature (ºC) Water-Sugar System Adapted from Fig. 9.1, Callister & Rethwisch 8e. 1 phase 2 phases L 80 (liquid) 60 L (liquid solution 40 i.e., syrup) + S (solid sugar) A (20ºC,C = 70) 20 2 phases 0 0 20 40 60 70 80 100 C = Composition (wt% sugar) DRAFT 20 Criteria for Solid Solubility Simple system (e.g., Ni-Cu solution) Crystal Structure electroneg r (nm) Ni FCC 1.91 0.1246 Cu FCC 1.9 0.1278 • Both have the same crystal structure (FCC) and have similar electronegativity values and atomic radii (W. Hume – Rothery rules) suggesting high mutual solubility. • Ni and Cu are totally soluble in one another in all proportions. DRAFT 21 Cooling curves of Alloys Cu-Ni at different compositions time Flat region indicates latent heat of fusion (transformation of phase) 22 Binary Phase Diagrams • Indicate phases as a function of T, C, and P. • For this course: - only binary (2 components) systems: - independent variables: T and C (P = constant = 1 atm). T(ºC) • 2 phases: 1600 Phase Diagram Cu-Ni system 1500 L (liquid) 1400 1300 a (FCC solid solution) 1200 1100 1000 L (liquid) a (FCC solid solution) 0 20 40 60 DRAFT 80 • 3 different phase fields: L L+a a Adapted from Fig. 9.3(a), Callister & Rethwisch 8e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). 100 wt% Ni 23 Binary Isomorphous System • Phase diagram: Cu-Ni system. • System is: T(ºC) 1600 1500 L (liquid) -- 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. Cu-Ni phase diagram 1400 1300 a (FCC solid solution) 1200 1100 1000 0 20 40 60 80 100 Adapted from Fig. 9.3(a), Callister & Rethwisch 8e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). DRAFT wt% Ni 24 Phase Determination • Rule 1: If we know T and Co, then we know: -- which phase(s) is (are) present [Coordinates in phase diagram]. Point A(1100ºC, 60 wt% Ni): 1 phase: a B(~ 1250ºC, 35 wt% Ni): 2 phases: L + a 1600 L (liquid) B (1250ºC,35) • Examples: T(ºC) 1500 1400 1300 a (FCC solid solution) 1200 Adapted from Fig. 9.3(a), Callister & Rethwisch 8e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). Cu-Ni phase diagram A(1100ºC,60) 1100 1000 0 20 DRAFT 40 60 80 100 wt% Ni 25 Composition Determination: Tie Rule • Rule 2: We know T and C0, we can determine: Composition of each phase. T(ºC) • Examples: = 1320ºC: B TB a = 1190ºC: Only Solid (a) present Ca = C0 1200 TD ( = 35 wt% Ni) = 1250ºC: Both (solid) D 20 At TB tie line L (liquid) 1300 Only Liquid (L) present ( = 35 wt% Ni) CL = C0 At TD A TA Consider C0 = 35 wt% Ni At TA Cu-Ni system 30 40 32 35 a and L present CL C0 50 wt% Ni 43 Ca CL = C liquidus ( = 32 wt% Ni) Ca = C solidus ( = 43 wt% Ni) Comp. at “tie-line hitting solidus” 26 Between 33% to 42% (wt%) Ni and a fixed temp of 1250oC - the CL & Ca are fixed The Inverse-Lever Rule T(ºC) What fraction of each phase? Think of the tie line as a lever (teeter-totter) tie line 1300 L (liquid) TB a (solid) 1200 R 20 Ma ML B 30CL S R C0 40 Ca wt% Ni Ca C0 ML S WL ML Ma R S Ca CL 50 Adapted from Fig. 9.3(b), Callister & Rethwisch 8e. C0 CL R Wa R S Ca CL DRAFT S We can find both the weight fraction and the composition of SS formed at a given temperature for any alloy during solidification 27 Determination of weight fractions • Rule 3: Knowing T and C0, we can determine the weight fraction • Examples: Consider C0 = 35 wt% Ni T(ºC) S R +S Wa R = 0.27 R +S WL=Red/Red+Blue WS=Blue/Red+Blue tie line B R S TB 1200 TD D 20 30 35 32 C0 CL a (solid) 40 43 50 Ca wt% Ni Adapted from Fig. 9.3(a), Callister & Rethwisch 8e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). DRAFT S L (liquid) 1300 43 35 0.73 43 32 WL A TA At TA : Only Liquid (L) present WL = 1.00, Wa = 0 At TD : Only Solid ( a) present WL = 0, Wa = 1.00 At TB : Both a and L present L Cu-Ni system 28 Equilibrium Cooling • Phase diagram: Cu-Ni system. • Consider microstuctural changes that accompany the cooling of a C0 = 35 wt% Ni alloy T(ºC) L (liquid) 130 0 L: 35 wt% Ni a: 46 wt% Ni L: 35wt%Ni Compositional readjustments are accomplished by diffusional processes A 35 32 B C 46 43 D 24 L: 32 wt% Ni 36 120 0 a: 43 wt% Ni E L: 24 wt% Ni a: 36 wt% Ni a (solid) 110 0 20 30 Adapted from Fig. 9.4, Callister & Rethwisch 8e. Cu-Ni system Diffusion is slower with reducing temp 35 C0 DRAFT 40 50 wt% Ni 29 Non-equilibrium cooling ~1240 deg C – diffusion is slow and composition doesn’t reach 40% Nickel but rather is ~ 42% CL=29%Ni-71%Cu (Cα should be 40%) α-phase formed at b’ should lose Ni with lowering of temp but slow diffusion inhibits the process and hence has a higher value than predicted from tie-line As if the solidus line on the phase diagram has been shifted to higher Ni contents due to fast cooling 30 Cored vs Equilibrium Structures • Ca changes as we solidify. • Cu-Ni case: First a to solidify has Ca = 46 wt% Ni. • Slow rate of cooling: • Fast rate of cooling: Equilibrium structure Cored structure Uniform Ca: 35 wt% Ni First a to solidify: 46 wt% Ni DRAFT 31 Mechanical Properties: Cu-Ni System • Effect of solid solution strengthening on: -- Ductility (%EL) 400 TS for pure Ni 300 TS for pure Cu 200 0 20 40 Cu 60 80 100 Ni Composition, wt% Ni Adapted from Fig. 9.6(a), Callister & Rethwisch 8e. Elongation (%EL) Tensile Strength (MPa) -- Tensile strength (TS) 60 %EL for pure Cu %EL for pure Ni 50 40 30 20 0 20 Cu 40 60 80 100 Ni Composition, wt% Ni Adapted from Fig. 9.6(b), Callister & Rethwisch 8e. DRAFT 32 Binary-Eutectic Systems Cu-Ag & Pb-Sn systems DRAFT 33 Violation of Hume-Rothery Rules Crystal Structure electroneg r (nm) Ag FCC 1.93 0.172 Cu FCC 1.9 0.127 DRAFT 34 Copper-Silver: Binary-Eutectic Systems has a special composition with a min. melting T. Cu-Ag system 2 components Ex: Cu-Ag system T(ºC) 1200 • 3 single phase regions (L, a, b) L (liquid) 1000 a : mostly Cu or Cu-rich phase b : mostly Ag or Ag-rich phase TE 800 • TE : No liquid below TE • CE : Composition at temp TE • Eutectic reaction L+a cooling heating 779ºC 8.0 71.9 L+b b 91.2 600 ab 400 200 0 20 a(CaE) + b(CbE) L(CE) L(71.9 wt% Ag) a 40 60 CE 80 100 wt% Ag Adapted from Fig. 9.7, Callister & Rethwisch 8e. a (8.0 wt% Ag) b (91.2 wt% Ag) 35 Solder Solder Alloys Solder alloys can be elemental metals such as tin and indium; binary alloys such as Sn–Ag or Sn-Pb; complex ternary and quaternary compositions Of interest are the solidus temperature, at which melting begins, and the liquidus temperature, at and above which the alloy is fully liquid. The temperature range between the solidus and liquidus temperatures is identified as the pasty range or the mushy zone, in which the solder is a mixture of liquid and solid particles. P.T. Vianco, Y Feng, Electronic Packaging: Solder Mounting Technologies, Reference Module in Materials Science and Materials Engineering, Elsevier, 2016 DRAFT 36 Lead-Tin (Pb-Sn) System (solder) • For a 40% Sn (60% Pb) alloy at 150ºC, determine: Pb-Sn system -- the phases present and weight fractions T(ºC) Answer: a + b -- the phase compositions Answer: Ca = 11 wt% Sn Cb = 99 wt% Sn -- the relative amount of each phase Answer: S = W = R +S a 99 - 40 = 99 - 11 Wb = R = R+S 40 - 11 = 99 - 11 300 200 Cb - C0 Cb - Ca 59 = 0.67 88 C0 - Ca Cb - Ca = = 29 = 0.33 88 L (liquid) a L+ a 100 L+b b 183ºC 18.3 150 Tie-line Isotherm @150 degC 61.9 R 97.8 S a+b 0 11 20 Ca 40 C0 60 80 99100 Cb wt% Sn Adapted from Fig. 9.8, Callister & Rethwisch 8e. DRAFT 37 Example 2: Pb-Sn Eutectic System • For a 40 wt% Sn-60 wt% Pb alloy at 220ºC, determine: -- the phases and compositions: T(ºC) Answer: a + L -- the phase compositions Answer: Ca = 17 wt% Sn CL = 46 wt% Sn -- the relative amount of each phase Pb-Sn system 300 a 220 200 L (liquid) L+a R L+b S b 183ºC Answer: CL - C0 46 - 40 = Wa = CL - Ca 46 - 17 6 = = 0.21 29 C0 - C a 23 = = 0.79 WL = CL - C a 29 100 a+b 0 17 20 Ca 40 46 60 C0 CL 80 100 C, wt% Sn Adapted from Fig. 9.8, Callister & Rethwisch 8e. DRAFT 38 Microstructure Development Binary-Eutectic Systems DRAFT 39 Microstructural Developments Pb-Sn System Limited solubility – completely soluble upto 2% Sn – T(ºC) L: C0 wt% Sn 400 • For alloys where C0 < 2 wt% Sn L a L 300 200 TE • Result: at room temperature -- polycrystalline with grains of a-phase having composition C0 100 Adapted from Fig. 9.11, Callister & Rethwisch 8e. DRAFT a L+ a a: C0 wt% Sn (Pb-Sn System) a+ b 0 10 20 30 C0 C , wt% Sn 2 (room T solubility limit) 40 Microstructure Developments: Intermediate Composition L: C0 wt% Sn T(ºC) 400 • For alloys for which 2 wt% Sn < C0 < 18.3 wt% Sn L is low-Sn (High-Pb) L 300 • Result: at temperatures in a + b range -- polycrystalline with a grains and small b phase particles L +a a 200 TE a: C0 wt% Sn a b 100 Adapted from Fig. 9.12, Callister & Rethwisch 8e. L a a+ b 0 10 20 Pb-Sn system 30 C0 C, wt% 2 (sol. limit at T room ) 18.3 (sol. limit at TE) DRAFT Sn 41 Microstructural Developments: Eutectic Composition Eutectic Mixture • For alloy of composition C0 = CE • Result: Eutectic microstructure (lamellar structure) -- alternating layers (lamellae) of a and b phases. T(ºC) L : C0 wt% Sn 300 Pb-Sn system a 200 L+ a L Lb b 183ºC TE α - Pb-rich 100 ab β - Sn-rich Tie-line gives composition 0 20 18.3 Adapted from Fig. 9.13, Callister & Rethwisch 8e. Micrograph of Pb-Sn eutectic microstructure 40 b: 97.8 wt% Sn a: 18.3 wt%Sn 60 CE 61.9 80 Adapted from Fig. 9.14, Callister & Rethwisch 8e. 100 97.8 C, wt% Sn DRAFT 160 m 42 Lamellar Eutectic Structure Alternate layered system with two alloy types - α and β α and β phases have different compositions neither of which is the same as that of the liquid Redistribution is accomplished by atomic diffusion Interdiffusion of Pb & Sn occurs in the liquid just ahead of the “eutectic solid–liquid” interface. Lamellar configuration preferred as atomic diffusion of Pb and Sn need only occur over relatively short distances. Adapted from Figs. 9.14 & 9.15, Callister & Rethwisch 8e. DRAFT 43 Liquidus: Line separating pure Liquid (L) phase field from a mixture of {Liquid (L) + solid(α)} phase field Solidus: ? Solvus line: a line on a phase diagram that separates a solid phase (α) from a mixed solid (β) + solid(α) phase Two phase fields have different microstructures. DRAFT 44 Hypoeutectic & Hypereutectic 300 T(ºC) Adapted from Fig. 9.8, Callister & Rethwisch 8e. 200 (Fig. 10.8 adapted from TE Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in-Chief), 100 ASM International, Materials Park, OH, 1990.) a L+ a 20 a L+b b 40 hypoeutectic: C0 = 50 wt% Sn a Eutectic: Lamellar μ-structure Hypo: α-rich + eutectic Hyper: β-rich +eutectic a+b 0 (Figs. 9.14 and 9.17 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) (Pb-Sn System) L a 60 80 eutectic 61.9 C, wt% Sn hypereutectic: (illustration only) eutectic: C0 = 61.9 wt% Sn b a a b a 175 m 100 b b b b 160 m eutectic micro-constituent 45 Intermetallic Compounds Intermetallic compounds are defined as solid phases involving two or more metallic or semimetallic elements with an ordered structure Usually has well-defined and fixed stoichiometry Non-stoichiometric Intermetallics: Ordered but have a variable composition DRAFT 46 Intermetallic Compounds in Phase Diagrams Adapted from Fig. 9.20, Callister & Rethwisch 8e. Mg2Pb Note: intermetallics usually exists as a line on the diagram (i.e. composition of a compound is a fixed value). Non-stoichiometric Intermetallics usually have a variable composition Invariant Reactions Syntectic: L1+L2α Catatectic: β L+α DRAFT 48 Tin 63% / Lead 37% solder which melts and freezes at 183 °C. This melting point is much lower than the melting points of either pure metal which are 232 °C (Sn) and 327 °C (Pb) DRAFT 49 Eutectoid & Peritectic Peritectic transformation + L Cu-Zn Phase diagram (Brass) Eutectoid transformation + DRAFT Adapted from Fig. 9.21, Callister & Rethwisch 8e. 50 Allotropes of Iron and various phases DRAFT 51 Fe-C Phase Diagram: Invariant reactions Eutectic - liquid transforms to two solid phases L cool a + b (For Pb-Sn, 183ºC, 61.9 wt% Sn) heat • Eutectoid – one solid phase transforms to two other solid phases compound - cementite S2 S1+S3 cool a + Fe3C (For Fe-C, 727ºC, 0.76 wt% C) heat • Peritectic - liquid and one solid phase transform to a second solid phase S1 + S2 +L cool heat (For Fe-C, 1493ºC, 0.16 wt% C) DRAFT 52 Iron-Iron carbide (Fe-Fe3C) Phase Diagram L + Fe3C T(ºC) 1600 1200 - Eutectoid (B): a + Fe3C L 1400 +L (austenite) 1000 a 800 A 1148ºC +Fe3C 727ºC = T eutectoid B a+Fe3C 600 120 m Result: Pearlite = alternating layers of a and Fe3C phases (Adapted from Fig. 9.27, Callister & Rethwisch 8e.) 400 0 (Fe) L+Fe3C 1 0.76 2 3 4 4.30 5 6 Fe3C (cementite) • 2 important points - Eutectic (A): 6.7 C, wt% C Fe3C (cementite-hard) a (ferrite-soft) Adapted from Fig. 9.24, DRAFT Callister & Rethwisch 8e. 53 Voids in Octahedron/Tetrahedron FCC: 8 OV and 4 TV per unit cell BCC: 12 OV and 6TV Each of 8 OVs in FCC >> Each of 12 OVs in BCC DRAFT 54 Hypoeutectoid Steel T(ºC) 1600 L a a a +L 1200 (austenite) 1000 + Fe3C 800 727ºC a a + Fe3C 600 pearlite 1 0.76 a 400 0 (Fe)C0 L+Fe3C 1148ºC 2 3 4 5 6 6.7 Adapted from Figs. 9.24 and 9.29,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-inChief), ASM International, Materials Park, OH, 1990.) C, wt% C 100 m pearlite (Fe-C System) Fe3C (cementite) 1400 Hypoeutectoid steel proeutectoid ferrite Adapted from Fig. 9.30, Callister & Rethwisch 8e. DRAFT 55 Hypoeutectoid Steel T(ºC) 1600 L a +L 1200 (austenite) a a 1000 + Fe3C Wa = s/(r + s) 800 r s W =(1 - Wa) 727ºC aRS a + Fe3C 600 pearlite Wpearlite = W 400 0 (Fe)C0 1 0.76 a L+Fe3C 1148ºC Wa’ = S/(R + S) WFe3C =(1 – Wa’) pearlite 2 3 4 5 6 (Fe-C System) Fe3C (cementite) 1400 6.7 Adapted from Figs. 9.24 and 9.29,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-inChief), ASM International, Materials Park, OH, 1990.) C, wt% C 100 m Hypoeutectoid steel proeutectoid ferrite Adapted from Fig. 9.30, Callister & Rethwisch 8e. DRAFT 56 Hypereutectoid Steel T(ºC) 1600 L Fe3C +L 1200 (austenite) 1000 L+Fe3C 1148ºC +Fe3C 800 a a +Fe3C 600 400 0 (Fe) pearlite 0.76 (Fe-C System) 1 C0 2 3 4 5 6 Fe3C (cementite) 1400 Adapted from Figs. 9.24 and 9.32,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) 6.7 C, wt%C 60 mHypereutectoid steel pearlite proeutectoid Fe3C Adapted from Fig. 9.33, Callister & Rethwisch 8e. DRAFT 57 Hypereutectoid Steel T(ºC) 1600 L +L 1200 (austenite) W =x/(v + x) WFe3C =(1-W) +Fe3C v x 800 a V 600 pearlite 400 0 (Fe) Wpearlite = W X 1 C0 Wa = X/(V + X) WFe 3C’ L+Fe3C 1148ºC 1000 0.76 Fe3C (Fe-C System) a +Fe3C 2 3 4 5 6 Fe3C (cementite) 1400 Adapted from Figs. 9.24 and 9.32,Callister & Rethwisch 8e. (Fig. 9.24 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) 6.7 C, wt%C 60 mHypereutectoid steel =(1 - Wa) pearlite proeutectoid Fe3C Adapted from Fig. 9.33, Callister & Rethwisch 8e. DRAFT 58 Alloying with Other Elements Ti Mo Si W Cr Mn Ni • Ceutectoid changes: Ceutectoid (wt% C) TEutectoid (ºC) • Teutectoid changes: Ni Cr Si Ti Mo W Mn wt. % of alloying elements wt. % of alloying elements Adapted from Fig. 9.34,Callister & Rethwisch 8e. (Fig. 9.34 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) Adapted from Fig. 9.35,Callister & Rethwisch 8e. (Fig. 9.35 from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) DRAFT 59 Example Problem 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 Fe3C and ferrite (a). b) The amount of cementite (in grams) that forms in 100 g of steel. c) The amounts of pearlite and proeutectoid ferrite (a) in the 100 g. DRAFT 60 Solution to Example Problem a) Using the RS tie line just below the eutectoid Ca = 0.022 wt% C CFe3C = 6.70 wt% C WFe 3C = = R C - Ca = 0 R + S CFe 3C - Ca 0.40 - 0.022 = 0.057 6.70 - 0.022 1600 L 1400 T(ºC) 1200 +L 1000 + Fe3C 800 727ºC R Amount of Fe3C in 100 g S a + Fe3C 600 = (100 g)WFe3C 400 0 = (100 g)(0.057) = 5.7 g Ca C0 DRAFT L+Fe3C 1148ºC (austenite) Fe3C (cementite) b) Using the lever rule with the tie line shown 1 2 3 4 C, wt% C 5 6 6.7 CFe 3C 61 Solution to Example Problem (cont.) c) Using the VX tie line just above the eutectoid and realizing that C0 = 0.40 wt% C Ca = 0.022 wt% C Cpearlite = C = 0.76 wt% C 1600 Wpearlite = = V C - Ca = 0 V + X Cg - Ca 0.40 - 0.022 = 0.512 0.76 - 0.022 Amount of pearlite in 100 g = (100 g)Wpearlite L 1400 T(ºC) 1200 +L L+Fe3C 1148ºC (austenite) 1000 + Fe3C 800 727ºC VX a + Fe3C 600 400 0 1 Ca C0 C = (100 g)(0.512) = 51.2 g 2 3 4 5 6 Fe3C (cementite) • 6.7 C, wt% C DRAFT 62 Summary • 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 system. • 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. DRAFT 63