Chapter 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? Chapter 9: Phase Diagrams Phase B Phase A Nickel atom Copper atom
Chapter Phase Equilibria: Solubility Limit Introduction –Solutions – solid solutions, single phase –Mixtures – more than one phase Solubility Limit: Max concentration for which only a single phase solution occurs. Question: What is the solubility limit at 20°C? Answer: 65 wt% sugar. If C o < 65 wt% sugar: syrup If C o > 65 wt% sugar: syrup + sugar. 65 Sucrose/Water Phase Diagram Pure Sugar Temperature (°C) CoCo =Composition (wt% sugar) L (liquid solution i.e., syrup) Solubility Limit L (liquid) + S (solid sugar) PureWater Adapted from Fig. 9.1, Callister 7e.
Chapter 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., and ). Aluminum- Copper Alloy Components and Phases (darker phase) (lighter phase) Adapted from chapter-opening photograph, Chapter 9, Callister 3e.
Chapter Effect of T & Composition (C o ) Changing T can change # of phases: Adapted from Fig. 9.1, Callister 7e. D (100°C,90) 2 phases B (100°C,70) 1 phase path A to B. Changing C o can change # of phases: path B to D. A (20°C,70) 2 phases Temperature (°C) CoCo =Composition (wt% sugar) L ( liquid solution i.e., syrup) L (liquid) + S (solid sugar) water- sugar system
Chapter Phase Equilibria Crystal Structure electroneg r (nm) NiFCC CuFCC Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume – Rothery rules) suggesting high mutual solubility. Simple solution system (e.g., Ni-Cu solution) Ni and Cu are totally miscible in all proportions.
Chapter Phase Diagrams Indicate phases as function of T, C o, and P. For this course: -binary systems: just 2 components. -independent variables: T and C o (P = 1 atm is almost always used). Phase Diagram for Cu-Ni system Adapted from Fig. 9.3(a), Callister 7e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). 2 phases: L (liquid) (FCC solid solution) 3 phase fields: L L + wt% Ni T(°C) L (liquid) (FCC solid solution) L + liquidus solidus
Chapter wt% Ni T(°C) L (liquid) (FCC solid solution) L + liquidus solidus Cu-Ni phase diagram Phase Diagrams : # and types of phases Rule 1: If we know T and C o, then we know: --the # and types of phases present. Examples: A(1100°C, 60): 1 phase: B(1250°C, 35): 2 phases: L + Adapted from Fig. 9.3(a), Callister 7e. (Fig. 9.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991). B (1250°C,35) A(1100°C,60)
Chapter wt% Ni T(°C) L (liquid) (solid) L + liquidus solidus L + Cu-Ni system Phase Diagrams : composition of phases Rule 2: If we know T and C o, then we know: --the composition of each phase. Examples: T A A 35 C o 32 C L At T A = 1320°C: Only Liquid (L) C L = C o ( = 35 wt% Ni) At T B = 1250°C: Both and L C L = C liquidus ( = 32 wt% Ni here) C = C solidus ( = 43 wt% Ni here) At T D = 1190°C: Only Solid ( ) C = C o ( = 35 wt% Ni) C o = 35 wt% Ni Adapted from Fig. 9.3(b), Callister 7e. (Fig. 9.3(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.) B T B D T D tie line 4 C 3
Chapter Rule 3: If we know T and C o, then we know: --the amount of each phase (given in wt%). Examples: At T A : Only Liquid (L) W L = 100 wt%, W = 0 At T D : Only Solid ( ) W L = 0, W = 100 wt% C o = 35 wt% Ni Adapted from Fig. 9.3(b), Callister 7e. (Fig. 9.3(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.) Phase Diagrams : weight fractions of phases wt% Ni T(°C) L (liquid) (solid) L + liquidus solidus L + Cu-Ni system T A A 35 C o 32 C L B T B D T D tie line 4 C 3 R S At T B : Both and L = 27 wt% WLWL S R+S WW R R+S
Chapter Tie line – connects the phases in equilibrium with each other - essentially an isotherm The Lever Rule How much of each phase? Think of it as a lever (teeter-totter) MLML MM RS wt% Ni T(°C) L (liquid) (solid) L + liquidus solidus L + B T B tie line C o C L C S R Adapted from Fig. 9.3(b), Callister 7e.
Chapter wt% Ni L (liquid) (solid) L + L + T(°C) A 35 C o L: 35wt%Ni Cu-Ni system Phase diagram: Cu-Ni system. System is: --binary i.e., 2 components: Cu and Ni. --isomorphous i.e., complete solubility of one component in another; phase field extends from 0 to 100 wt% Ni. Adapted from Fig. 9.4, Callister 7e. Consider C o = 35 wt%Ni. Ex: Cooling in a Cu-Ni Binary :43 wt% Ni L: 32 wt% Ni L: 24 wt% Ni :36 wt% Ni B : 46 wt% Ni L: 35 wt% Ni C D E 24 36
Chapter C changes as we solidify. Cu-Ni case: Fast rate of cooling: Cored structure Slow rate of cooling: Equilibrium structure First to solidify has C = 46 wt% Ni. Last to solidify has C = 35 wt% Ni. Cored vs Equilibrium Phases First to solidify: 46 wt% Ni Uniform C : 35 wt% Ni Last to solidify: < 35 wt% Ni
Chapter Mechanical Properties: Cu-Ni System Effect of solid solution strengthening on: --Tensile strength (TS)--Ductility (%EL,%AR) --Peak as a function of C o --Min. as a function of C o Adapted from Fig. 9.6(a), Callister 7e.Adapted from Fig. 9.6(b), Callister 7e. Tensile Strength (MPa) Composition, wt% Ni Cu Ni TS for pure Ni TS for pure Cu Elongation (%EL) Composition, wt% Ni Cu Ni %EL for pure Ni %EL for pure Cu
Chapter : Min. melting T E 2 components has a special composition with a min. melting T. Adapted from Fig. 9.7, Callister 7e. Binary-Eutectic Systems Eutectic transition L(C E ) (C E ) + (C E ) 3 single phase regions (L, ) Limited solubility: : mostly Cu : mostly Ag T E : No liquid below T E C E composition Ex.: Cu-Ag system Cu-Ag system L (liquid) L + L+ CoCo,wt% Ag T(°C) CECE TETE °C
Chapter L+ L+ + 200 T(°C) 18.3 C, wt% Sn L (liquid) 183°C For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find... --the phases present: Pb-Sn system EX: Pb-Sn Eutectic System (1) + --compositions of phases: C O = 40 wt% Sn --the relative amount of each phase: CoCo 11 CC 99 CC S R C = 11 wt% Sn C = 99 wt% Sn W = C - C O C - C = = = 67 wt% S R+SR+S = W = C O - C C - C = R R+SR+S = = 33 wt% = Adapted from Fig. 9.8, Callister 7e.
Chapter L+ + 200 T(°C) C, wt% Sn L (liquid) L+ 183°C For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find... --the phases present: Pb-Sn system Adapted from Fig. 9.8, Callister 7e. EX: Pb-Sn Eutectic System (2) + L --compositions of phases: C O = 40 wt% Sn --the relative amount of each phase: W = C L - C O C L - C = = 6 29 = 21 wt% W L = C O - C C L - C = = 79 wt% 40 CoCo 46 CLCL 17 CC 220 S R C = 17 wt% Sn C L = 46 wt% Sn
Chapter C o < 2 wt% Sn Result: --at extreme ends --polycrystal of grains i.e., only one solid phase. Adapted from Fig. 9.11, Callister 7e. Microstructures in Eutectic Systems: I 0 L + 200 T(°C) CoCo,wt% Sn CoCo L 30 + 400 (room T solubility limit) TETE (Pb-Sn System) L L: C o wt% Sn : C o wt% Sn
Chapter wt% Sn < C o < 18.3 wt% Sn Result: Initially liquid + then alone finally two phases polycrystal fine -phase inclusions Adapted from Fig. 9.12, Callister 7e. Microstructures in Eutectic Systems: II Pb-Sn system L + 200 T(°C) CoCo,wt% Sn CoCo L 30 + 400 (sol. limit at T E ) TETE 2 (sol. limit at T room ) L L: C o wt% Sn : C o wt% Sn
Chapter C o = C E Result: Eutectic microstructure (lamellar structure) --alternating layers (lamellae) of and crystals. Adapted from Fig. 9.13, Callister 7e. Microstructures in Eutectic Systems: III Adapted from Fig. 9.14, Callister 7e. 160 m Micrograph of Pb-Sn eutectic microstructure Pb-Sn system LL 200 T(°C) C, wt% Sn L L+ 183°C 40 TETE 18.3 : 18.3 wt%Sn 97.8 : 97.8 wt% Sn CECE 61.9 L: C o wt% Sn
Chapter Lamellar Eutectic Structure Adapted from Figs & 9.15, Callister 7e.
Chapter wt% Sn < C o < 61.9 wt% Sn Result: crystals and a eutectic microstructure Microstructures in Eutectic Systems: IV SR 97.8 S R primary eutectic WLWL = (1-W ) = 50 wt% C = 18.3 wt% Sn CLCL = 61.9 wt% Sn S R +S W = = 50 wt% Just above T E : Just below T E : C = 18.3 wt% Sn C = 97.8 wt% Sn S R +S W = = 73 wt% W = 27 wt% Adapted from Fig. 9.16, Callister 7e. Pb-Sn system L+ 200 T(°C) C o, wt% Sn L L+ 40 + TETE L: C o wt% Sn L L
Chapter L+ L+ + 200 C o, wt% Sn L TETE 40 (Pb-Sn System) Hypoeutectic & Hypereutectic Adapted from Fig. 9.8, Callister 7e. (Fig. 9.8 adapted from Binary Phase Diagrams, 2nd ed., Vol. 3, T.B. Massalski (Editor-in- Chief), ASM International, Materials Park, OH, 1990.) 160 m eutectic micro-constituent Adapted from Fig. 9.14, Callister 7e. hypereutectic: (illustration only) Adapted from Fig. 9.17, Callister 7e. (Illustration only) (Figs and 9.17 from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) 175 m hypoeutectic: C o = 50 wt% Sn Adapted from Fig. 9.17, Callister 7e. T(°C) 61.9 eutectic eutectic: C o = 61.9 wt% Sn
Chapter Eutectoid & Peritectic Eutectic - liquid in equilibrium with two solids L + cool heat intermetallic compound - cementite cool heat Eutectoid - solid phase in equation with two solid phases S 2 S 1 +S 3 + Fe 3 C (727ºC)
Chapter Iron-Carbon (Fe-C) Phase Diagram 2 important points -Eutectoid (B): + Fe 3 C -Eutectic (A): L + Fe 3 C Adapted from Fig. 9.24,Callister 7e. Fe 3 C (cementite) L (austenite) +L+L +Fe 3 C + L+Fe 3 C (Fe) C o, wt% C 1148°C T(°C) 727°C = T eutectoid A SR 4.30 Result: Pearlite = alternating layers of and Fe 3 C phases 120 m (Adapted from Fig. 9.27, Callister 7e.) RS 0.76 C eutectoid B Fe 3 C (cementite-hard) (ferrite-soft)
Chapter Hypoeutectoid Steel Adapted from Figs and 9.29,Callister 7e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) Fe 3 C (cementite) L (austenite) +L+L + Fe 3 C L+Fe 3 C (Fe) C o, wt% C 1148°C T(°C) 727°C (Fe-C System) C0C Adapted from Fig. 9.30,Callister 7e. proeutectoid ferrite pearlite 100 m Hypoeutectoid steel R S w =S/(R+S) w Fe 3 C =(1-w ) w pearlite =w rs w =s/(r+s) w =(1-w )
Chapter Hypereutectoid Steel Fe 3 C (cementite) L (austenite) +L+L +Fe 3 C L+Fe 3 C (Fe) C o, wt%C 1148°C T(°C) Adapted from Figs and 9.32,Callister 7e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) (Fe-C System) 0.76 CoCo Adapted from Fig. 9.33,Callister 7e. proeutectoid Fe 3 C 60 m Hypereutectoid steel pearlite RS w =S/(R+S) w Fe 3 C =(1-w ) w pearlite =w s r w Fe 3 C =r/(r+s) w =(1-w Fe 3 C )
Chapter Example: Phase Equilibria For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following a)composition of Fe 3 C and ferrite ( ) b)the amount of carbide (cementite) in grams that forms per 100 g of steel c)the amount of pearlite and proeutectoid ferrite ( )
Chapter Chapter 9 – Phase Equilibria Solution: b)the amount of carbide (cementite) in grams that forms per 100 g of steel a) composition of Fe 3 C and ferrite ( ) C O = 0.40 wt% C C = wt% C C Fe C = 6.70 wt% C 3 Fe 3 C (cementite) L (austenite) +L+L + Fe 3 C L+Fe 3 C C o, wt% C 1148°C T(°C) 727°C COCO R S C Fe C 3 CC
Chapter Chapter 9 – Phase Equilibria c.the amount of pearlite and proeutectoid ferrite ( ) note: amount of pearlite = amount of just above T E C o = 0.40 wt% C C = wt% C C pearlite = C = 0.76 wt% C pearlite = 51.2 g proeutectoid = 48.8 g Fe 3 C (cementite) L (austenite) +L+L + Fe 3 C L+Fe 3 C C o, wt% C 1148°C T(°C) 727°C COCO R S CC CC
Chapter Alloying Steel with More Elements T eutectoid changes: C eutectoid changes: Adapted from Fig. 9.34,Callister 7e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) Adapted from Fig. 9.35,Callister 7e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) T Eutectoid (°C) wt. % of alloying elements Ti Ni Mo Si W Cr Mn wt. % of alloying elements C eutectoid (wt%C) Ni Ti Cr Si Mn W Mo
Chapter Phase diagrams are useful tools to determine: --the number and types of phases, --the wt% of each phase, --and the composition of each phase for a given T and composition of the system. Alloying to produce a solid solution usually --increases the tensile strength (TS) --decreases the ductility. Binary eutectics and binary eutectoids allow for a range of microstructures. Summary