1. Phase Equilibrium Engr 2110 – Chapter 9 Dr. R. R. Lindeke

2. Topics for Study Definitions of Terms in Phase studies Binary Systems  Complete solubility systems  Multiphase systems Eutectics Eutectoids and Peritectics Intermetallic Compounds The Fe-C System

3. Definitions Component of a system: Pure metals and or compounds of which an alloy is composed, e.g. Cu and Ag. They are the solute(s) and solvent System: A body of engineering material under investigation. e.g. Ag – Cu system Solubility Limit: The maximum concentration of solute atoms that may dissolve in the Solvent to form a “solid solution” at some temperature.

4. Phases: A homogenous portion of a system that has uniform physical and chemical characteristics, e.g. pure material, solid solution, liquid solution, and gaseous solution, ice and water, syrup and sugar. Microstructure: A system’s microstructure is characterized by the number of phases present, their proportions, and the manner in which they are distributed or arranged. Factors affecting microstructure are: alloying elements present, their concentrations, and the heat treatment of the alloy. Single phase system = Homogeneous system Multi phase system = Heterogeneous system or mixtures Definitions, cont

5. Phase Equilibrium: A stable configuration with lowest free-energy (internal energy of a system, and also randomness or disorder of the atoms or molecules (entropy). Any change in Temp., Comp., and Pressure = increase in free energy and away from Equilibrium. And move to another state Equilibrium Phase Diagram: It is a diagram of the information about the control of microstructure or phase structure of a particular alloy system. The relationships between temperature and the compositions and the quantities of phases present at equilibrium are shown. Definition that focus on “Binary Systems” Binary Isomorphous Systems: An alloy system that contains two components that attain complete liquid and solid solubility of the components, e.g. Cu and Ni alloy. It is the simplest binary system to understand. Binary Eutectic Systems: An alloy system that contains two components that has a special composition with a minimum melting temperature. Definitions, cont

6. With these definitions in mind: ISSUES TO ADDRESS... When we combine two elements... what “equilibrium state” would we expect to get? In particular, if we specify... --a composition (e.g., wt% Cu - wt% Ni), and --a temperature (T ) and/or a Pressure (P) then... How many phases do we get? What is the composition of each phase? How much of each phase do we get? Phase B Phase A Nickel atom Copper atom

7. Lets consider a commonly observed System exhibiting:  Solutions – liquid (solid) regions that contain a single phase  Mixtures – liquid & solid regions contain more than one phase Solubility Limit: Max concentration for which only a single phase solution occurs (as we saw earlier). 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) 0 20 4060 80100 CoCo =Composition (wt% sugar) L (liquid solution i.e., syrup) Solubility Limit L (liquid) + S (solid sugar) 20 40 60 80 100 PureWater Adapted from Fig. 9.1, Callister 7e. A 1-phase region A 2-phase region

8. Effect of Temperature (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 See path A to B. Changing C o can change # of phases: See path B to D. A (20°C,70) 2 phases 70801006040200 Temperature (°C) CoCo =Composition (wt% sugar) L ( liquid solution i.e., syrup) 20 100 40 60 80 0 L (liquid) + S (solid sugar) water- sugar system

9. Gibbs Phase Rule: a tool to define the number of phases that can be found in a system at equilibrium For the system under study the rule determines if the system is at equilibrium For a given system, we can use it to predict how many phases can be expected Using this rule, for a given phase field, we can predict how many independent parameters we can specify Ex: Determine how many equilibrium phases that can exist in a ternary alloy? C = 3; N = 1 (temperature)  P + F = 3+1 = 4 So, if F = 0 (all 3 elements compositions and the temperature is set) we could have (at most) 4 phases present

10. Phase Diagrams: A chart based on a System’s Free Energy indicating “equilibuim” system structures Indicate ‘stable’ phases as function of T, C o, and P. We will focus on: -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 are possible: L (liquid)  (FCC solid solution) 3 phase fields are observed: L L +   wt% Ni 204060801000 1000 1100 1200 1300 1400 1500 1600 T(°C) L (liquid)  (FCC solid solution) L +  liquidus solidus An “Isomorphic” Phase System

11. wt% Ni 204060801000 1000 1100 1200 1300 1400 1500 1600 T(°C) L (liquid)  (FCC solid solution) L +  liquidus solidus Cu-Ni phase diagram Phase Diagrams: Rule 1: If we know T and C o then we know the # and types of all 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)

12. wt% Ni 20 1200 1300 T(°C) L (liquid)  (solid) L +  liquidus solidus 304050 L +  Cu-Ni system Phase Diagrams: Rule 2: If we know T and C o 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

13. 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: wt% Ni 20 1200 1300 T(°C) L (liquid)  (solid) L +  liquidus solidus 304050 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 Notice: as in a lever “the opposite leg” controls with a balance (fulcrum) at the ‘base composition’ and R+S = tie line length = difference in composition limiting phase boundary, at the temp of interest

14. Tie line – a line connecting the phases in equilibrium with each other – at a fixed temperature (a so-called Isotherm) The Lever Rule How much of each phase? We can Think of it as a lever! So to balance: MLML MM RS wt% Ni 20 1200 1300 T(°C) L (liquid)  (solid) L +  liquidus solidus 304050 L +  B T B tie line C o C L C  S R Adapted from Fig. 9.3(b), Callister 7e.

15. wt% Ni 20 1200 1300 304050 1100 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 46 35 43 32  :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

16. 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

17. Cored (Non-equilibrium) Cooling Notice: The Solidus line is “tilted” in this non- equilibrium cooled environment

18. : Min. melting T E 2 components has a special composition with a min. melting Temp. Adapted from Fig. 9.7, Callister 7e. Binary-Eutectic (PolyMorphic) 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 204060 80100 0 200 1200 T(°C) 400 600 800 1000 CECE TETE 8.071.991.2 779°C

19. L+  L+   +  200 T(°C) 18.3 C, wt% Sn 206080 100 0 300 100 L (liquid)  183°C 61.997.8  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 (lever rule) : 150 40 CoCo 11 CC 99 CC S R C  = 11 wt% Sn C  = 99 wt% Sn W  = C  - C O C  - C  = 99 - 40 99 - 11 = 59 88 = 67 wt% S R+SR+S = W  = C O - C  C  - C  = R R+SR+S = 29 88 = 33 wt% = 40 - 11 99 - 11 Adapted from Fig. 9.8, Callister 7e.

20. L+   +  200 T(°C) C, wt% Sn 206080 100 0 300 100 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  = 46 - 40 46 - 17 = 6 29 = 21 wt% W L = C O - C  C L - C  = 23 29 = 79 wt% 40 CoCo 46 CLCL 17 CC 220 S R C  = 17 wt% Sn C L = 46 wt% Sn

21. 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 10 2% 20 CoCo 300 100 L  30  +  400 (room Temp. solubility limit) TETE (Pb-Sn System)  L L: C o wt% Sn  : C o wt% Sn

22. 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 10 18.3 200 CoCo 300 100 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

23. 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 2060801000 300 100 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 45.1%  and 54.8% 

24. Lamellar Eutectic Structure Adapted from Figs. 9.14 & 9.15, Callister 7e.

25. 18.3 wt% Sn < C o < 61.9 wt% Sn Result:  crystals and a eutectic microstructure Microstructures in Eutectic Systems: IV 18.361.9 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  = = 72.6 wt% W  = 27.4 wt% Adapted from Fig. 9.16, Callister 7e. Pb-Sn system L+  200 T(°C) C o, wt% Sn 2060801000 300 100 L   L+  40  +  TETE L: C o wt% Sn L  L 

26. L+  L+   +  200 C o, wt% Sn 2060801000 300 100 L   TETE 40 (Pb-Sn System) Hypoeutectic & Hypereutectic Compositions 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. 9.14 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

27. “Intermetallic” Compounds Mg 2 Pb Note: an intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact. Adapted from Fig. 9.20, Callister 7e. An Intermetallic Compound is also an important part of the Fe-C system!

28. Eutectoid: solid phase in equilibrium with two solid phases S 2 S 1 +S 3   + Fe 3 C (727ºC) intermetallic compound - cementite cool heat Peritectic: liquid + solid 1 in equilibrium with a single solid 2 (Fig 9.21) S 1 + L S 2  + L  (1493ºC) cool heat Eutectic: a liquid in equilibrium with two solids L  +  Eutectoid & Peritectic – some definitions cool heat

29. Eutectoid & Peritectic Cu-Zn Phase diagram Adapted from Fig. 9.21, Callister 7e. Eutectoid transition   +  Peritectic transition  + L 

30. 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) 1600 1400 1200 1000 800 600 400 0 1234566.7 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) Max. C solubility in  iron = 2.11 wt%

31. Hypoeutectoid Steel Adapted from Figs. 9.24 and 9.29,Callister 7e. (Fig. 9.24 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) 1600 1400 1200 1000 800 600 400 0 1234566.7 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) C0C0 0.76 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  )              

32. Hypereutectoid Steel Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 1234566.7 L  (austenite)  +L+L  +Fe 3 C  L+Fe 3 C  (Fe) C o, wt%C 1148°C T(°C)  Adapted from Figs. 9.24 and 9.32,Callister 7e. (Fig. 9.24 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 )            

33. 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 (  )

34. 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  = 0.022 wt% C C Fe C = 6.70 wt% C 3 Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 1234566.7 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

35. Solution, cont: 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  = 0.022 wt% C C pearlite = C  = 0.76 wt% C pearlite = 51.2 g proeutectoid  = 48.8 g Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 1234566.7 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 Looking at the Pearlite: 11.1% Fe 3 C (.111*51.2 gm = 5.66 gm) & 88.9%  (.889*51.2gm = 45.5 gm)  total  = 45.5 + 48.8 = 94.3 gm

36. Alloying Steel with More Elements T eutectoid changes: C eutectoid changes: Adapted from Fig. 9.34,Callister 7e. (Fig. 9.34 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. 9.35 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

37. Looking at a Tertiary Diagram When looking at a Tertiary (3 element) P. diagram, like this image, the Mat. Engineer represents equilibrium phases by taking “slices at different 3 rd element content” from the 3 dimensional Fe-C-Cr phase equilibrium diagram What this graph shows are the increasing temperature of the euctectoid and the decreasing carbon content, and (indirectly) the shrinking  phase field From: G. Krauss, Principles of Heat Treatment of Steels, ASM International, 1990

38. 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 (cause) a range of microstructures. Summary