Introduction to Materials Science and Engineering Chapter 9: Phase Diagrams Textbook Chapter 11: Phase Diagrams
Content Introduction Solubility Limit Phase Diagrams Microstructure Evolution during Cooling Eutectic Systems Fe-C Alloy
Introduction 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?
Introduction ISSUES TO ADDRESS... Phase diagram is a graphical illustration of a thermodynamic equilibrium state of a material system which is defined by the composition, temperature, and pressure, namely, Gibbs free energy of a system, and illustrates, The number of phases at equilibrium The composition of each phases The amount of each phases at equilibrium.
Content Introduction Solubility Limit Phase Diagrams Microstructure Evolution during Cooling Eutectic Systems Fe-C Alloy
THE SOLUBILITY LIMIT • Solubility Limit: + L S • Ex: Phase Diagram: Max. concentration for which only a solution occurs. Pure Sugar Temperature (ºC) 20 4 6 8 10 C o = Composition (wt% sugar) L (liquid solution i.e., syrup) Solubility Limit (liquid) + S (solid sugar) 65 Water • Ex: Phase Diagram: Water-Sugar System Question: What is the solubility limit at 20C? Answer: 65wt% sugar. If Co < 65wt% sugar: sugar syrup If Co > 65wt% sugar: syrup + sugar. Adapted from Fig. 9.1, Callister 6e. • Solubility limit increases with T: e.g., if T = 100C, solubility limit = 80wt% sugar.
COMPONENTS AND PHASES • Components: • Phases: The elements or compounds which are mixed initially (e.g., Al and Cu) • Phases: The physically and chemically identical material regions that result (e.g., a and b). Aluminum- Copper Alloy Adapted from Fig. 9.0, Callister 3e.
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. 70 8 10 6 4 2 Temperature (ºC) C o =Composition (wt% sugar) L ( liquid solution i.e., syrup) A (20,70 ) 2 phases B (100,70) 1 phase 20 D (100,90) (liquid) + S (solid sugar) • water- sugar system Adapted from Fig. 9.1, Callister 6e.
Content Introduction Solubility Limit Phase Diagrams Microstructure Evolution during Cooling Eutectic Systems Fe-C Alloy
PHASE DIAGRAMS • Tell us about phases as function of T, Co, P. Gibbs Phase Rule: • Tell us about phases as function of T, Co, P. • For this course: --binary systems: just 2 components. --independent variables: T and Co (P = 1 atm is always used). P+F=C+2 2 phases: L (liquid) a (FCC solid solution) 3 phase fields: L + wt% Ni 20 4 6 8 10 100 110 120 130 140 150 160 T(C) L (liquid) (FCC solid solution) + liquidus solidus • Phase Diagram for Cu-Ni system Adapted from Fig. 9.2(a), Callister 6e. (Fig. 9.2(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH (1991).
PHASE DIAGRAMS: # and types of phases • Rule 1: If we know T and Co, then we know: --the # and types of phases present. wt% Ni 20 4 6 8 10 100 110 120 130 140 150 160 T(ºC) L (liquid) a (FCC solid solution) L + liquidus solidus A(1100,60) B (1250,35) • Examples: Cu-Ni phase diagram Adapted from Fig. 9.2(a), Callister 6e. (Fig. 9.2(a) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991).
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: wt% Ni 20 120 130 T(ºC) L (liquid) a (solid) L + liquidus solidus 3 4 5 T A D B tie line 35 32 C o C o = 35wt%Ni At T A : ? Only Liquid (L) C L = C o ( = 35wt% Ni) At T D : ? Only Solid ( a ) C a = C o ( = 35wt% Ni ) At T B : ? Both a and L Adapted from Fig. 9.2(b), Callister 6e. (Fig. 9.2(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.) C L = C liquidus ( = 32wt% Ni here) C a = C solidus ( = 43wt% Ni here)
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 system • Examples: T(ºC) C o = 38wt%Ni A T A At T tie line A : Only Liquid (L) L (liquid) 130 W L = 100wt%, W a = 0 a B + L At T a D : Only Solid ( ) T B R S W L = 0, W a = 100wt% a + a L At T : Both a B and L 120 D T (solid) D What would be WL and W? 20 3 32 38 4 4 4 5 W L = S R + C L C C o a wt% Ni W a = R + S
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 system • Examples: T(ºC) A T A tie line L (liquid) 130 a B + L T B R S a + a L 120 D T (solid) D What would be WL and W? 20 3 32 35 4 4 4 5 W L = S R + C L C C o a wt% Ni W a = R + S
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%). • Examples: Cu-Ni system wt% Ni 20 120 130 T(ºC) L (liquid) a (solid) L + liquidus solidus 3 4 5 T A D B tie line 35 32 C o R S What would be WL and W? W L = S R + = 27wt% W a = R + S Adapted from Fig. 9.2(b), Callister 6e. (Fig. 9.2(b) is adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.)
THE LEVER RULE: A PROOF • Sum of weight fractions: • Conservation of mass (Ni): • Combine above equations: • A geometric interpretation:
Content Introduction Solubility Limit Phase Diagrams Microstructure Evolution during Cooling Eutectic Systems Fe-C Alloy
EX: COOLING IN A Cu-Ni BINARY • Phase diagram: Cu-Ni system. T(ºC) L (liquid) L: 35wt%Ni Cu-Ni system • System is: --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 100wt% Ni. 130 a A + L: 35wt%Ni L a : 46wt%Ni B 35 46 C 32 43 D 24 L: 32wt%Ni 36 a a + : 43wt%Ni 120 L E L: 24wt%Ni a a : 36wt%Ni (solid) • Consider Co = 35wt%Ni. What would be the microstructure? 110 20 3 35 4 5 C wt% Ni o
EX: COOLING IN A Cu-Ni BINARY wt% Ni 20 120 130 3 4 5 110 L (liquid) a (solid) L + T(ºC) A D B 35 C o L: 35wt%Ni : 46wt%Ni E 46 43 32 24 36 43wt%Ni L: 32wt%Ni L: 24wt%Ni 36wt%Ni Cu-Ni system • Consider Co = 35wt%Ni. Adapted from Fig. 9.3, Callister 6e.
If the alloy (65%Cu-35%Ni) is cooled from A to D at 10oC/sec, what would be the details inside the solid? wt% Ni 20 120 130 T(ºC) L (liquid) a (solid) L + liquidus solidus 3 4 5 T A D B tie line 35 32 C o
CORED VS EQUILIBRIUM PHASES • Ca changes as we solidify. • Cu-Ni case: First a to solidify has Ca = 46wt%Ni. Last a to solidify has Ca = 35wt%Ni. • Fast rate of cooling: Cored structure • Slow rate of cooling: Equilibrium structure
MECHANICAL PROPERTIES: Cu-Ni System • Effect of solid solution strengthening on: --Tensile strength (TS) --Ductility (%EL,%AR) Elongation (%EL) Composition, wt%Ni Cu Ni 20 4 6 8 10 3 5 %EL for pure Ni for pure Cu Tensile Strength (MPa) 200 00 TS for TS for pure Cu ?
MECHANICAL PROPERTIES: Cu-Ni System • Effect of solid solution strengthening on: --Tensile strength (TS) --Ductility (%EL,%AR) --Peak as a function of Co --Min. as a function of Co
Content Introduction Solubility Limit Phase Diagrams Microstructure Evolution during Cooling Eutectic Systems Fe-C Alloy
BINARY-EUTECTIC SYSTEMS Ex.: Cu-Ag system Gibbs Phase Rule: P+F=C+2 3 single phase regions (L, a, b) Limited solubility: L (liquid) a L + b C o , wt% Ag 20 4 6 8 10 2 120 T(ºC) 00 E T 8.0 71 .9 91.2 779ºC a : mostly Cu b : mostly Ag TE: no liquid below TE CE: minimum melting Temp. composition
EX: Pb-Sn EUTECTIC SYSTEM (1) • For a 40wt%Sn-60wt%Pb alloy at 150C, find... --the phases present: a + b --the compositions of the phases: T(ºC) 3 00 L (liquid) L + a a L + b b 2 183ºC 18.3 61.9 97.8 150 a + b 1 00 20 4 6 8 10 C o C , wt% Sn o
EX: Pb-Sn EUTECTIC SYSTEM (2) • For a 40wt%Sn-60wt%Pb alloy at 150C, find... --the phases present: a + b --the compositions of the phases: Ca = ? Cb = ? --the relative amounts of each phase: T(ºC) 3 00 L (liquid) L + a a L + b b 2 183ºC 18.3 61.9 97.8 150 R S 1 00 Wa = ? wt% Wb = ? wt% a + b 11 20 4 6 8 99 10 C o C , wt% Sn o
EX: Pb-Sn EUTECTIC SYSTEM (2) • For a 40wt%Sn-60wt%Pb alloy at 150C, find... --the phases present: a + b --the compositions of the phases: Ca = 11wt%Sn Cb = 99wt%Sn --the relative amounts of each phase: T(ºC) 3 00 L (liquid) L + a a L + b b 2 183ºC 18.3 61.9 97.8 150 R S 1 00 a + b 11 20 4 6 8 99 10 C o C , wt% Sn o
MICROSTRUCTURES IN EUTECTIC SYSTEMS-I • Co < 2wt%Sn • Result: ? T(ºC) L: C o wt%Sn 4 00 L 3 00 L + a a 2 (Pb-Sn T E System) 1 00 a + b 1 2 3 C o C o , wt% Sn 2 (room T solubility limit)
MICROSTRUCTURES IN EUTECTIC SYSTEMS-I • Co < 2wt%Sn • Result: --polycrystal of a grains. L + a 2 T(ºC) C o , wt% Sn 1 3 00 L: C wt%Sn : C b 4 (room T solubility limit) T E
MICROSTRUCTURES IN EUTECTIC SYSTEMS-II • 2wt%Sn < Co < 18.3wt%Sn • Result:? T(ºC) L: C o wt%Sn 4 00 L L 3 00 a L + a a : C o wt%Sn a 2 T E a b 1 00 a + b Pb-Sn system 1 2 3 C C o , wt% Sn 2 o (sol. limit at T room ) 18.3 (sol. limit at T E )
MICROSTRUCTURES IN EUTECTIC SYSTEMS-II • 2wt%Sn < Co < 18.3wt%Sn • Result: --a polycrystal with fine b crystals. a : C o wt%Sn L + 2 T(ºC) , wt% Sn 1 18.3 3 00 L: C b 4 T E ) Pb-Sn system (sol. limit at T room (sol. limit at T E )
MICROSTRUCTURES IN EUTECTIC SYSTEMS-III • Co = CE • Result: ? L + a 2 T(ºC) C o , wt% Sn 4 3 00 1 6 L: C wt%Sn b T E : 18.3wt%Sn 8 100 18.3 97.8 61.9 183ºC : 97.8wt%Sn Pb-Sn system
MICROSTRUCTURES IN EUTECTIC SYSTEMS-III • Co = CE • Result: Eutectic microstructure --alternating layers of a and b crystals. T(ºC) L: C o wt%Sn 3 00 L Pb-Sn system L + a L + b 2 a b 183ºC T E 1 00 a + b b : 97.8wt%Sn a : 18.3wt%Sn 2 4 6 8 100 18.3 C 97.8 E C o , wt% Sn 61.9
Formation of Eutectic Lamellar Structure
MICROSTRUCTURES IN EUTECTIC SYSTEMS-IV • 18.3wt%Sn < Co < 61.9wt%Sn • Result: a crystals and an eutectic microstructure Just above TE: L + a 2 T(ºC) C o , wt% Sn 4 3 00 1 6 L: C wt%Sn b T E 8 100 18.3 61.9 97.8 S R C a = 18.3wt%Sn C L = 61.9wt%Sn S W a = =50wt% R + S W L = (1- W a ) =50wt% Just below TE: C a = 18.3wt%Sn C b = 97.8wt%Sn S W a = =73wt% R + S W b = 27wt%
MICROSTRUCTURES IN EUTECTIC SYSTEMS-IV • 18.3wt%Sn < Co < 61.9wt%Sn • Result: a crystals and an eutectic microstructure Just above TE : W L = (1- a ) =50wt% C = 18.3wt%Sn = 61.9wt%Sn S R + = L: C wt%Sn L a o T(ºC) L 3 00 L a L + a b 2 R S L + b a T E R S Just below TE : C a = 18.3wt%Sn b = 97.8wt%Sn S R + W = =73wt% = 27wt% 1 00 a + b primary a eutectic a eutectic b 2 4 6 8 100 18.3 C o 61.9 97.8 C o , wt% Sn
HYPOEUTECTIC & HYPEREUTECTIC T(ºC) 3 00 L L + a 2 a L + b (Pb-Sn T E b System) a + b 1 00 C o C o hypoeutectic hypereutectic C 2 4 6 8 100 o , wt% Sn eutectic 97.8 18.3 175 m a hypoeutectic: C o =50wt%Sn 61.9 hypereutectic: (illustration only) 160 m o =61.9wt%Sn eutectic: C b b b b b b eutectic micro-constituent
OTHER EXAMPLES
OTHER EXAMPLES
Content Introduction Solubility Limit Phase Diagrams Microstructure Evolution during Cooling Eutectic Systems Fe-C Alloy
IRON-CARBON (Fe-C) PHASE DIAGRAM 3 C (cementite) 1600 1400 1200 1000 800 6 00 4 1 2 5 6.7 L g ( austenite) +L +Fe C a + L+Fe d (Fe) o , wt% C 0.77 4.30 727ºC = T eutectoid 1148ºC T(ºC) A B S R Two important points -Eutectic (A): -Eutectoid (B): L + g + Fe 3 C g + a
IRON-CARBON (Fe-C) PHASE DIAGRAM 3 C (cementite) 1600 1400 1200 1000 800 6 00 4 1 2 5 6.7 L g ( austenite) +L +Fe C a + L+Fe d (Fe) o , wt% C 0.77 4.30 727ºC = T eutectoid 1148ºC T(ºC) A B S R C (cementite-hard) ferrite-soft) Two important points -Eutectic (A): -Eutectoid (B): L + g + Fe 3 C g + a Result: Pearlite = alternating layers of a and Fe3C phases. 120 m
Formation of Pearlite Lamellar Structure
HYPOEUTECTOID STEEL Fe 3 C (cementite) L g ( austenite) +L +Fe a L+Fe 1600 1400 1200 1000 800 6 00 4 1 2 5 6.7 L g ( austenite) +L +Fe a L+Fe d , wt% C 0.77 727ºC 1148ºC T(ºC) R S r s
HYPOEUTECTOID STEEL Fe 3 C (cementite) L g ( austenite) +L +Fe a L+Fe 1600 1400 1200 1000 800 6 00 4 1 2 5 6.7 L g ( austenite) +L +Fe a L+Fe d , wt% C 0.77 727ºC 1148ºC T(ºC) R S r s w = /( + ) (1- Fe3C pearlite 100 m
HYPEREUTECTOID STEEL a T(ºC) d L (Fe-C g g +L System) L+Fe C ( 3 1600 d L 1400 (Fe-C g g +L System) 1200 L+Fe C ( 1148ºC 3 austenite) 1000 g +Fe C Adapted from Figs. 9.21 and 9.29,Callister 6e. (Fig. 9.21 adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.) 3 C (cementite) 800 r s a R S 6 00 3 a +Fe 3 C Fe 4 00 1 0.77 C 2 3 4 5 6 6.7 o C o , wt% C
HYPEREUTECTOID STEEL (Fe-C System) Fe 3 C (cementite) L g ( austenite) 1600 1400 1200 1000 800 6 00 4 1 2 5 6.7 L g ( austenite) +L +Fe a L+Fe d , wt% C 0.77 1148ºC T(ºC) R S s w Fe3C = r /( + ) =(1- (1- pearlite m Hypereutectoid steel
SUMMARY • 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.