Chapter 9: Phase Diagrams ISSUES TO ADDRESS... • 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 Phase Equilibria: Solubility Limit Introduction – Solutions – solid solutions, single phase – Mixtures – more than one phase Adapted from Fig. 9.1, Callister 7e. Sucrose/Water Phase Diagram • Solubility Limit : 100 Max concentration for Solubility which only a single phase 80 Limit L solution occurs. (liquid) 60 L + Question: What is the 40 (liquid solution S solubility limit at 20°C? i.e., syrup) (solid 20 Temperature (°C) Temperature sugar) Answer: 65 wt% sugar . If C < 65 wt% sugar: syrup o 0 20 40 60 65 80 100 If Co > 65 wt% sugar: syrup + sugar. Co =Composition (wt% sugar) Sugar Pure Pure Water Chapter 9 - 2 Components and Phases • Components : The elements or compounds which are present in the mixture (e.g., Al and Cu) • Phases : The physically and chemically distinct material regions that result (e.g., α and β). Aluminum- β (lighter Copper phase) Alloy α (darker phase) Adapted from chapter-opening photograph, Chapter 9, Callister 3e. A phase maybe defined as a homogeneous portion of a system that has Chapter 9 - 3 uniform physical and chemical characteristics. 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) 2 phases Adapted from 0 Fig. 9.1, 0 20 40 60 7080 100 Callister 7e. Co =Composition (wt% sugar) Chapter 9 - 4 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). Phase Equilibrium: If there is more than 1 phase present, phase characteristics 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). Chapter 9 - 5 Unary Systems Triple point Chapter 9 - 6 Phase Equilibria Simple solution 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 miscible in all proportions. Chapter 9 - 7 Unary Systems Single component system Consider 2 metals: Cu has melting T = 1085 oC Ni has melting T = 1453 oC (at standard P = 1 atm) T T liquid liquid 1453 oC 1085 oC solid solid Cu Ni Chapter 9 - 8 What happens when Cu and Ni are mixed? 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 For a pure liquid component, complete melting occurs before T liquid L 1453 oC increases (sharp phase transition). But for 1085 oC multicomponent solid systems, there is S solid usually a coexistence of L 0 100 and S. Cu wt% Ni Ni Chapter 9 - 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 • 2 phases: • Phase L (liquid) Diagram 1500 for Cu-Ni L (liquid) α (FCC solid solution) system 1400 • 3 phase fields: s du L ui α 1300 liq + us L + α L id ol 1200 s 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 - 10 Phase Diagrams : # and types of phases • Rule 1: If we know T and Co, then we know: --the # and types of phases present. T(°C) • Examples: 1600 A(1100°C, 60): L (liquid) 1500 1 phase: α s Cu-Ni du ui iq phase B(1250°C, 35): 1400 l us lid 2 phases: L + α so diagram (1250°C,35) α 1300 + α B L (FCC solid 1200 solution) Adapted from Fig. 9.3(a), Callister 7e. (Fig. 9.3(a) is adapted from Phase 1100 A(1100°C,60) Diagrams of Binary Nickel Alloys , P. Nash (Ed.), ASM International, Materials Park, OH, 1991). 1000 0 20 40 60 80 100 wt% Ni Chapter 9 - 11 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 = 35 wt% Ni T tie line s o A idu iqu At TA = 1320°C: 1300 L (liquid) l + α Only Liquid ( L) L B us T olid CL = C o ( = 35 wt% Ni) B s α At TD = 1190°C: + α 1200 L D (solid) Only Solid ( α) TD Cα = Co ( = 35 wt% Ni ) 20 3032 35 4043 50 At TB = 1250°C: CLCo Cα wt% Ni Both α and L Adapted from Fig. 9.3(b), Callister 7e. (Fig. 9.3(b) is adapted from Phase Diagrams CL = Cliquidus ( = 32 wt% Ni here) of Binary Nickel Alloys , P. Nash (Ed.), ASM Cα = Csolidus ( = 43 wt% Ni here) International, Materials Park, OH, 1991.) Chapter 9 - 12 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: Liquid is composed of CL amount of Ni (31.5 wt% Ni). Solid is composed of Cααα amount of Ni (42.5 wt% Ni). Chapter 9 - 13 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 Co = 35 wt% Ni T A A tie line dus qui At TA: Only Liquid (L) 1300 L (liquid) li α W = 100 wt%, W = 0 L + L α B s idu At TD: Only Solid ( α) TB R S sol α WL = 0, Wα = 100 wt% + α 1200 L D (solid) At TB: Both α and L TD S 43 − 35 20 30 40 50 W = = = 73 wt % 32 35 43 L R +S 43 − 32 CLCo Cα wt% Ni Adapted from Fig. 9.3(b), Callister 7e. R (Fig. 9.3(b) is adapted from Phase Diagrams of Wα = = 27 wt% Binary Nickel Alloys , P. Nash (Ed.), ASM R +S International, Materials Park, OH, 1991.) Chapter 9 - 14 Lever Rule: Derivation Since we have only 2 phases: W W (1) L + α =1 Conservation of mass requires that: Amount of Ni in α-phase + amount of Ni in liquid phase = total amount of Ni or W C W C C (2) α α + L L = o From 1 st condition, we have: W W α =1− L W C W C C Sub-in to (2): 1( − L ) α + L L = o W W Solving for L and α gives : C − C C − C W = α o W = o L L C C α C C α − L α − L Chapter 9 - 15 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 dus qui Think of it as a lever (teeter-totter) 1300 L (liquid) li + α M M L L α B s idu TB sol α + α 1200 L (solid) R S R S 20 30CL C 40C 50 o ααα M α ⋅S = M L ⋅R wt% Ni Adapted from Fig. 9.3(b), Callister 7e. ML S Cα −C0 R C0 −CL WL = = = Wα = = ML + Mα R + S Cα −CL R + S Cα −CL Chapter 9 - 16 Ex: Cooling in a Cu-Ni Binary • Phase diagram: T(°C) L (liquid) L: 35wt%Ni Cu-Ni system. Cu-Ni system • System is: 1300 α A + -- binary L: 35 wt% Ni L 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 α: 43 wt% Ni 1200 + solubility of one L E component in L: 24 wt% Ni another; α phase α α: 36 wt% Ni field extends from (solid) 0 to 100 wt% Ni. • Consider 1100 Co = 35 wt%Ni . 20 3035 40 50 Adapted from Fig. 9.4, Co wt% Ni Callister 7e. Chapter 9 - 17 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 oC): start as homogeneous liquid solution. b (T ~ 1260 oC): liquidus line reached. α phase begins to nucleate. Cα = 46 wt% Ni; C L = 35 wt% Ni c (T= 1250 oC): calculate composition and mass fraction of each phase. Cα = 43 wt% Ni; CL = 32 wt% Ni d (T~ 1220 oC): solidus line reached.