Chapter ISSUES TO ADDRESS... 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? Chapter 10: Phase Diagrams Phase B Phase A Nickel atom Copper atom
Chapter Phase Diagrams Indicate phases as a function of T, C, and P. For this course: - binary systems: just 2 components. - independent variables: T and C (P = 1 atm is almost always used). Phase Diagram for Cu-Ni system Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(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 different phase fields: L L + wt% Ni T(°C) L (liquid) (FCC solid solution) L + liquidus solidus
Chapter Cu-Ni phase diagram Isomorphous Binary Phase Diagram Phase diagram: Cu-Ni system. System is: Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). -- 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. wt% Ni T(°C) L (liquid) (FCC solid solution) L + liquidus solidus
Chapter 10 - wt% Ni T(°C) L (liquid) (FCC solid solution) L + liquidus solidus 4 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(1100°C, 60 wt% Ni): 1 phase: B(1250°C, 35 wt% Ni): 2 phases: L + B (1250°C,35) A(1100°C,60) Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).
Chapter wt% Ni T(°C) L (liquid) (solid) L + liquidus solidus L + Cu-Ni system Phase Diagrams : Determination of phase compositions Rule 2: If we know T and C 0, then we can determine: -- the composition of each phase. Examples: TATA A 35 C0C0 32 CLCL At T A = 1320°C: Only Liquid (L) present C L = C 0 ( = 35 wt% Ni) At T B = 1250°C: Both and L present CLCL = C liquidus ( = 32 wt% Ni) CC = C solidus ( = 43 wt% Ni) At T D = 1190°C: Only Solid ( ) present C = C 0 ( = 35 wt% Ni) Consider C 0 = 35 wt% Ni D TDTD tie line 4 CC 3 Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991). B TBTB
Chapter Rule 3: If we know T and C 0, then can determine: -- the weight fraction of each phase. Examples: At T A : Only Liquid (L) present W L = 1.00, W = 0 At T D : Only Solid ( ) present WL WL = 0, W = 1.00 Phase Diagrams : Determination of phase weight fractions wt% Ni T(°C) L (liquid) (solid) L + liquidus solidus L + Cu-Ni system TATA A 35 C0C0 32 CLCL B TBTB D TDTD tie line 4 CC 3 R S At T B : Both and L present = 0.27 WLWL S R+S WW R R+S Consider C 0 = 35 wt% Ni Adapted from Fig. 10.3(a), Callister & Rethwisch 3e. (Fig. 10.3(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).
Chapter Tie line – connects the phases in equilibrium with each other – also sometimes called an isotherm The Lever Rule What fraction of each phase? Think of the tie line as a lever (teeter-totter) MLML MM RS wt% Ni T(°C) L (liquid) (solid) L + liquidus solidus L + B T B tie line C0C0 CLCL CC S R Adapted from Fig. 10.3(b), Callister & Rethwisch 3e.
Chapter wt% Ni L (liquid) (solid) L + L + T(°C) A 35 C0C0 L: 35wt%Ni Cu-Ni system Phase diagram: Cu-Ni system. Adapted from Fig. 10.4, Callister & Rethwisch 3e. Consider microstuctural changes that accompany the cooling of a C 0 = 35 wt% Ni alloy Ex: Cooling of a Cu-Ni Alloy : 43 wt% Ni L: 32 wt% Ni B : 46 wt% Ni L: 35 wt% Ni C E L: 24 wt% Ni : 36 wt% Ni D
Chapter 10 - Slow rate of cooling: Equilibrium structure Fast rate of cooling: Cored structure First to solidify: 46 wt% Ni Last to solidify: < 35 wt% Ni 9 C changes as we solidify. Cu-Ni case: First to solidify has C = 46 wt% Ni. Last to solidify has C = 35 wt% Ni. Cored vs Equilibrium Structures Uniform C : 35 wt% Ni
Chapter Mechanical Properties: Cu-Ni System Effect of solid solution strengthening on: -- Tensile strength (TS)-- Ductility (%EL) Adapted from Fig. 10.6(a), Callister & Rethwisch 3e. 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 Adapted from Fig. 10.6(b), Callister & Rethwisch 3e.
Chapter components has a special composition with a min. melting T. Adapted from Fig. 10.7, Callister & Rethwisch 3e. Binary-Eutectic Systems 3 single phase regions (L, , ) Limited solubility: : mostly Cu : mostly Ag T E : No liquid below T E : Composition at temperature T E C E Ex.: Cu-Ag system Cu-Ag system L (liquid) L + L+ C,wt% Ag T(°C) CECE TETE °C cooling heating Eutectic reaction L(C E ) (C E ) + (C E )
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, determine: -- the phases present Pb-Sn system EX 1: Pb-Sn Eutectic System Answer: + -- the phase compositions -- the relative amount of each phase C0C0 11 CC 99 CC S R Answer: C = 11 wt% Sn C = 99 wt% Sn W = C - C 0 C - C = = = 0.67 S R+SR+S = W = C 0 - C C - C = R R+SR+S = = 0.33 = Answer: Adapted from Fig. 10.8, Callister & Rethwisch 3e.
Chapter Answer: C = 17 wt% Sn -- the phase compositions L+ + 200 T(°C) C, wt% Sn L (liquid) L+ 183°C For a 40 wt% Sn-60 wt% Pb alloy at 220°C, determine: -- the phases present: Pb-Sn system EX 2: Pb-Sn Eutectic System -- the relative amount of each phase W = C L - C 0 C L - C = = 6 29 = 0.21 WLWL = C 0 - C C L - C = = C0C0 46 CLCL 17 CC 220 S R Answer: + L C L = 46 wt% Sn Answer: Adapted from Fig. 10.8, Callister & Rethwisch 3e.
Chapter For alloys for which C 0 < 2 wt% Sn Result: at room temperature -- polycrystalline with grains of phase having composition C 0 Microstructural Developments in Eutectic Systems I 0 L + 200 T(°C) C,wt% Sn C0C L 30 + 400 (room T solubility limit) TETE (Pb-Sn System) L L: C 0 wt% Sn : C 0 wt% Sn Adapted from Fig , Callister & Rethwisch 3e.
Chapter For alloys for which 2 wt% Sn < C 0 < 18.3 wt% Sn Result: at temperatures in + range -- polycrystalline with grains and small -phase particles Adapted from Fig , Callister & Rethwisch 3e. Microstructural Developments in Eutectic Systems II Pb-Sn system L + 200 T(°C) C,wt% Sn C0C L 30 + 400 (sol. limit at T E ) TETE 2 (sol. limit at T room ) L L: C 0 wt% Sn : C 0 wt% Sn
Chapter For alloy of composition C 0 = C E Result: Eutectic microstructure (lamellar structure) -- alternating layers (lamellae) of and phases. Adapted from Fig , Callister & Rethwisch 3e. Microstructural Developments in Eutectic Systems III Adapted from Fig , Callister & Rethwisch 3e. 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 0 wt% Sn
Chapter Lamellar Eutectic Structure Adapted from Figs & 10.15, Callister & Rethwisch 3e.
Chapter For alloys for which 18.3 wt% Sn < C 0 < 61.9 wt% Sn Result: phase particles and a eutectic microconstituent Microstructural Developments in Eutectic Systems IV SR 97.8 S R primary eutectic WLWL = (1-W ) = 0.50 CC = 18.3 wt% Sn CLCL = 61.9 wt% Sn S R +S WW = = 0.50 Just above T E : Just below T E : C = 18.3 wt% Sn C = 97.8 wt% Sn S R +S W = = 0.73 W = 0.27 Adapted from Fig , Callister & Rethwisch 3e. Pb-Sn system L+ 200 T(°C) C, wt% Sn L L+ 40 + TETE L: C 0 wt% Sn L L
Chapter L+ L+ + 200 C, wt% Sn L TETE 40 (Pb-Sn System) Hypoeutectic & Hypereutectic Adapted from Fig. 10.8, Callister & Rethwisch 3e. (Fig 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 , Callister & Rethwisch 3e. hypereutectic: (illustration only) Adapted from Fig , Callister & Rethwisch 3e. (Illustration only) (Figs and from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.) 175 m hypoeutectic: C 0 = 50 wt% Sn Adapted from Fig , Callister & Rethwisch 3e. T(°C) 61.9 eutectic eutectic: C 0 = 61.9 wt% Sn
Chapter Intermetallic Compounds 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). Adapted from Fig , Callister & Rethwisch 3e.
Chapter Eutectoid – one solid phase transforms to two other solid phases S 2 S 1 +S 3 + Fe 3 C (For Fe-C, 727 C, 0.76 wt% C) intermetallic compound - cementite cool heat Eutectic, Eutectoid, & Peritectic Eutectic - liquid transforms to two solid phases L + (For Pb-Sn, 183 C, 61.9 wt% Sn) cool heat cool heat Peritectic - liquid and one solid phase transform to a second solid phase S 1 + L S 2 + L (For Fe-C, 1493°C, 0.16 wt% C)
Chapter Eutectoid & Peritectic Cu-Zn Phase diagram Adapted from Fig , Callister & Rethwisch 3e. Eutectoid transformation + Peritectic transformation + L
Chapter Ceramic Phase Diagrams MgO-Al 2 O 3 diagram: Adapted from Fig , Callister & Rethwisch 3e.
Chapter Iron-Carbon (Fe-C) Phase Diagram 2 important points - Eutectoid (B): + Fe 3 C - Eutectic (A): L + Fe 3 C Adapted from Fig , Callister & Rethwisch 3e. Fe 3 C (cementite) L (austenite) +L+L +Fe 3 C + (Fe) C, wt% C 1148°C T(°C) 727°C = T eutectoid 4.30 Result: Pearlite = alternating layers of and Fe 3 C phases 120 m (Adapted from Fig , Callister & Rethwisch 3e.) 0.76 B A L+Fe 3 C Fe 3 C (cementite-hard) (ferrite-soft)
Chapter Fe 3 C (cementite) L (austenite) +L+L + Fe 3 C L+Fe 3 C (Fe) C, wt% C 1148°C T(°C) 727°C (Fe-C System) C0C Hypoeutectoid Steel Adapted from Figs and 10.33,Callister & Rethwisch 3e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) Adapted from Fig , Callister & Rethwisch 3e. proeutectoid ferrite pearlite 100 m Hypoeutectoid steel pearlite
Chapter Fe 3 C (cementite) L (austenite) +L+L + Fe 3 C L+Fe 3 C (Fe) C, wt% C 1148°C T(°C) 727°C (Fe-C System) C0C Hypoeutectoid Steel sr W = s/(r + s) W =(1 - W ) R S pearlite W pearlite = W W ’ = S/(R + S) W =(1 – W ’ ) Fe 3 C Adapted from Figs and 10.33,Callister & Rethwisch 3e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.) Adapted from Fig , Callister & Rethwisch 3e. proeutectoid ferrite pearlite 100 m Hypoeutectoid steel
Chapter Hypereutectoid Steel Fe 3 C (cementite) L (austenite) +L+L +Fe 3 C L+Fe 3 C (Fe) C, wt%C 1148°C T(°C) Adapted from Figs and 10.36,Callister & Rethwisch 3e. (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 C0C0 Fe 3 C Adapted from Fig , Callister & Rethwisch 3e. proeutectoid Fe 3 C 60 m Hypereutectoid steel pearlite
Chapter Fe 3 C (cementite) L (austenite) +L+L +Fe 3 C L+Fe 3 C (Fe) C, wt%C 1148°C T(°C) Hypereutectoid Steel (Fe-C System) 0.76 C0C0 pearlite Fe 3 C x v VX W pearlite = W W = X/(V + X) W =(1 - W ) Fe 3 C’ W =(1-W ) W =x/(v + x) Fe 3 C Adapted from Fig , Callister & Rethwisch 3e. proeutectoid Fe 3 C 60 m Hypereutectoid steel pearlite Adapted from Figs and 10.36,Callister & Rethwisch 3e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in- Chief), ASM International, Materials Park, OH, 1990.)
Chapter 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 Fe 3 C and ferrite ( ). b)The amount of cementite (in grams) that forms in 100 g of steel. c)The amounts of pearlite and proeutectoid ferrite ( ) in the 100 g.
Chapter Solution to Example Problem b)Using the lever rule with the tie line shown a) Using the RS tie line just below the eutectoid C = wt% C C Fe 3 C = 6.70 wt% C Fe 3 C (cementite) L (austenite) +L+L + Fe 3 C L+Fe 3 C C, wt% C 1148°C T(°C) 727°C C0C0 R S C Fe C 3 CC Amount of Fe 3 C in 100 g = (100 g)W Fe 3 C = (100 g)(0.057) = 5.7 g
Chapter Solution to Example Problem (cont) c) Using the VX tie line just above the eutectoid and realizing that C 0 = 0.40 wt% C C = wt% C C pearlite = C = 0.76 wt% C Fe 3 C (cementite) L (austenite) +L+L + Fe 3 C L+Fe 3 C C, wt% C 1148°C T(°C) 727°C C0C0 V X CC CC Amount of pearlite in 100 g = (100 g)W pearlite = (100 g)(0.512) = 51.2 g
Chapter Alloying with Other Elements T eutectoid changes: Adapted from Fig ,Callister & Rethwisch 3e. (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 C eutectoid changes: Adapted from Fig ,Callister & Rethwisch 3e. (Fig from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) 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 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. Summary
Chapter Core Problems: Self-help Problems: ANNOUNCEMENTS Reading: