1 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? Lecture 17 Chapter 9: Phase Diagrams

2 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.

3 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

4 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)

5 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

6 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 WW  R R+S

7 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 MM 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.

8 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

9 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

10 : 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

11 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 CC 99 CC 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.

12 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 CC 220 S R C  = 17 wt% Sn C L = 46 wt% Sn

13 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

14 2 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

15 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 LL  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

16 Lamellar Eutectic Structure Adapted from Figs & 9.15, Callister 7e.

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 

18 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

19 Eutectoid 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)

20 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)

21 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  )              

22 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 )            

23 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 (  )

24 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 CC

25 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 CC CC

26 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

27 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