1. Chapter 9 - 1 ISSUES TO ADDRESS... When we mix 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 (relative amount)? Chapter 9: Phase Diagrams Phase  (A and B atoms) Phase  (A and B atoms) Nickel atom Copper atom Structure Properties Processing

2. Chapter 9 - 2 Phase Equilibria: Review: Solubility Limit Introduction -Solutions – solid solutions, single phase (e.g. syrup) -Mixtures – more than one phase (syrup and solid sugar) Phase: Homogeneous portion of a system that has uniform physical and chemical properties Solubility Limit: Max concentration for which only a single phase solution occurs. Q: 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

3. Chapter 9 - 3 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  ). Equilibrium No change with time – fixed composition for each component Al-Cu Alloy Components and Phases  (darker phase) Includes Al and Cu  (lighter phase) Includes Al and Cu Two phases Two Components

4. Chapter 9 - 4 Effect of T & Composition (C o ) Changing T can change # of phases: 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 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

5. Chapter 9 - 5 Phase Diagrams Types of phase diagrams depend on Solubility Isomorphous phase diagram (complete solubility) Eutectic phase diagram (partial solubility) Complex phase diagram (…….) How to obtain phase diagrams ? - From cooling curves for various starting solutions (Thermal method) Thermodynamics: 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: A Plot of T versus C indicating various boundaries between phases T Time ? Phase Change T composition Uses of Phase Diagrams: - No. of phases - Composition of each phase - Relative amount of each phase + Predict the microstructure

6. Chapter 9 - 6 Phase Equilibria - Solutions Crystal Structure electroneg r (nm) NiFCC1.90.1246 CuFCC1.80.1278 Both metals have Nearly the same atomic radius The same crystal structure (FCC) similar electronegativities (W. Hume – Rothery rules) suggesting high mutual solubility. Simple Solution System (e.g., Ni-Cu solution) Remember Ni and Cu are totally miscible (soluble) in all proportions (compositions). ONE PHASE ALWAYS – Isomorphous !

7. Chapter 9 - 7 Phase Diagrams- Metals 2 phases: L (liquid)  (FCC solid solution) 3 phase fields: L L +   Phase Diagram for Cu-Ni system 204060801000 wt% Ni 1000 1100 1200 1300 1400 1500 1600 T(°C) L (liquid)  (FCC solid Soln) L +  Liquidus line Solidus line Isomorphous Phase Diagram Complete solubility - one phase field extends from 0 to 100 wt% Ni.

8. Chapter 9 - 8 Phase Change Phase boundary: Lines between different phases e.g. –Liquid –solid one phase two phases …etc) Phase Change: –Solidification (or melting) –Phase Reaction: one phase gives DIRECTLY two phases e.g. Liquid  Solid phase 1 + Solid Phase 2 (both insoluble) This occurs at –Certain (fixed) temperature: T E –Certain (fixed) composition: C E Eutectic Phase Reaction: L   There are other types of phase reactions (Later)

9. Chapter 9 - 9 Cooling Curves There are various types of cooling curves: Depending on composition (characteristics) 1.Pure Element 2.Mixture of soluble 2 elements 3.Mixture of Partially Soluble 2 elements 4.Mixtures undergoing Phase Reactions 1.At the composition for phase reaction 2.Before the composition for phase reaction

10. Chapter 9 - 10 Types of Cooling Curves 2) Mixture of soluble 2 elements Time L L L   T 3) Phase Reaction at C E 3) Phase Reaction: C o > C E Time L L    T L L  T Time L L   A or B)  1) Pure Element T melting Time L L    T TETE

11. Chapter 9 - 11 Constructing Phase Diagrams Thermal Method Start with several different solutions Each solution at certain c o Prepare the solutions in liquid form (melting) Cool each solution while recording T versus time Plot the cooling curve for each solution At each phase change: there are changes in the slope –Latent Heat –Change in the properties (specific heat) –Phase reaction (later) Project cooling curves on T versus c o plot. Connect the points from projection creating boundaries. Label different obtained areas and boundaries.

12. Chapter 9 - 12 Constructing Isomorphous Phase Diagrams Cooling Curves at Various Initial C o Phase Diagram T verus C o T Composition Time

13. Chapter 9 - 13 wt% Ni 204060801000 1000 1100 1200 1300 1400 1500 1600 T(°C) L (liquid)  (FCC solid solution) L +  liquidus solidus Cu-Ni phase diagram Using Phase Diagrams : # and types of phases Rule 1: If we know T and C o, then we know: - Number and types of phases present. Examples: A(1100°C, 60): 1 phase:  B(1250°C, 35): 2 phases: L +  B (1250°C,35) A(1100°C,60)

14. Chapter 9 - 14 wt% Ni 20 1200 1300 T(°C) L (liquid)  (solid) L +  liquidus solidus 304050 L +  Cu-Ni system Using 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 B T B D T D tie line 4 C  3

15. Chapter 9 - 15 Rule 3: If we know T and C o, then we know: --the relative amount of each phase (given in wt%). Examples: At T A : Only Liquid (L) W L = 1.0, W =  = 0 At T D : Only Solid (  ) W L = 0, W  = 1.0 C o = 35 wt% Ni Using Phase Diagrams : weight fractions of phases 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 = 0.27 WLWL  S R+S WW  R R+S

16. Chapter 9 - 16 Tie line – connects the phases in equilibrium with each other - essentially an isotherm The Lever Rule Think How much of each phase? Think of it as a lever (teeter-totter) 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

17. Chapter 9 - 17 wt% Ni 20 1200 1300 304050 1100 L (liquid)  (solid) L +  L +  T(°C) A 35 C o L: 35wt%Ni Cu-Ni system Observe the microstructure while going down (cooling) at the same composition. Consider C o = 35 wt%Ni. Microstructure - 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

18. Chapter 9 - 18 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

19. Chapter 9 - 19 Mechanical Properties: Cu-Ni System Effect of solid solution strengthening on: --Tensile strength (TS) - Peak as a function of C o Tensile Strength (MPa) Composition, wt% Ni Cu Ni 020406080100 200 300 400 TS for pure Ni TS for pure Cu --Ductility (%EL,%AR) - Min. as a function of C o Elongation (%EL) Composition, wt% Ni Cu Ni 020406080100 20 30 40 50 60 %EL for pure Ni %EL for pure Cu

20. Chapter 9 - 20 Eutectic Phase Diagram With Eutectic Phase Reaction L(C E )  (C  E ) +  (C  E ) The phase diagram is characterized with: 3 single phase regions (L,  and  ) Partial Solubility – Solubility Limit of A in B forming  B in A forming  Occurs at L (liquid)  L +  L+    Wt% A: C o 100 0 T CECE TETE : Eutectic Composition C E T E : Eutectic Temperature Remember Phase Reaction at C E Time L L    T TETE

21. Chapter 9 - 21 Eutectic Phase Diagram – Cooling Curves L  L +  L+    Wt% A: C o 100 0 T(°C) T Time L L L   1) Pure A or B T melting 2) Giving  or  Time L L L   T L L    T L L  Time L L    T TETE What is the difference in this side

22. Chapter 9 - 22 Maximum solubility of A in B L  L +  L+    Wt% A: C o 100 0 T(°C) Liquidus Solidus Solvus Eutectic Reaction Eutectic Phase Diagram Maximum solubility of B in A Boundaries - lines: Liquidus Solidus Solvus Now, apply previous rules ? On real systems

23. Chapter 9 - 23

24. Chapter 9 - 24

25. Chapter 9 - 25 Example-Eutectic System  : mostly Cu  : mostly Ag T E : No liquid below T E Ex.: Cu-Ag system L (liquid)  L +  L+    CoCo,wt% Ag 204060 80100 0 200 1200 T(°C) 400 600 800 1000 CECE TETE 8.0% C E =71.9% 91.2% T E =779°C Given T and C Values are characteristics Of Cu-Ag system

26. Chapter 9 - 26 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 --compositions of phases: C O = 40 wt% Sn relative amount --the relative amount of each phase : 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 = 0.67 S R+SR+S = W  = C O - C  C  - C  = R R+SR+S = 29 88 = 0.33 (or 1 – 0.67) = 40 - 11 99 - 11

27. Chapter 9 - 27 L+   +  200 T(°C) C, wt% Sn 206080 100 0 300 100 L (liquid)   L+  183°C 200°C For a 40 wt% Sn-60 wt% Pb alloy at 200°C, find... --the phases present: Pb-Sn system EX: Pb-Sn Eutectic System (2)  + L compositions --compositions of phases: C O = 40 wt% Sn relative amount --the relative amount of each phase : W  = C L - C O C L - C  = 46 - 40 46 - 17 = 6 29 = 0.21 W L = C O - C  C L - C  = 23 29 = 0.79 40 CoCo 46 CLCL 17 CC 220 S R C  = 17 wt% Sn C L = 46 wt% Sn

28. Chapter 9 - 28 C o < 2 wt% Sn Result: --at extreme ends --polycrystal of  grains i.e., only one solid phase. Microstructures in Eutectic Systems: I 0 L +  200 T(°C) CoCo,wt% Sn 10 2 20 CoCo 300 100 L  30  +  400 (room T solubility limit) TETE (Pb-Sn System)  L L: C o wt% Sn  : C o wt% Sn Depends on the composition

29. Chapter 9 - 29 2 wt% Sn < C o < 18.3 wt% Sn Result:  Initially liquid +   then  alone   finally two phases   polycrystal  fine  -phase inclusions 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

30. Chapter 9 - 30 C o = C E Result: Eutectic microstructure (lamellar structure) --alternating layers (lamellae) of  and  crystals. Microstructures in Eutectic Systems: III 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

31. Chapter 9 - 31 Lamellar Eutectic Structure

32. Chapter 9 - 32 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   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 

33. Chapter 9 - 33 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  ) = 0.50 C  = 18.3 wt% Sn CLCL = 61.9 wt% Sn S R +S W  ‘ = = 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 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  What is the fraction of Eutectic  ? Pro-eutectic (primary) 

34. Chapter 9 - 34 L+  L+   +  200 C o, wt% Sn 2060801000 300 100 L   TETE 40 (Pb-Sn System) Hypoeutectic & Hypereutectic 160  m eutectic micro-constituent hypereutectic : (illustration only)       175  m       hypoeutectic : C o = 50 wt% Sn T(°C) 61.9 eutectic eutectic : C o = 61.9 wt% Sn

35. Chapter 9 - 35 Intermetallic Compounds Note: intermetallic compound forms a line - not an area like  ? Mg 2 Pb In Previous Phase diagrams α and β are terminal solid solutions Can we obtain other intermediate phases ? Now: Mg2Pb is an intermetallic compound with a ratio (2:1) (Mg:Pb)-Molar melts at a fixed temperature M Cooling curve ??? Because stoichiometry (i.e. composition) is exact: Calculate ? Calculate Wt% Pb = 207 /(207+2(24.3)) X 100 = 81 %

36. Chapter 9 - 36 Same Story ? No. and type of each phase Composition of each phase Relative amount or fraction of each phase Distinguish between –primary phase –Eutectic phase Microstructure Cooling Curves Fill the regions with types of phases?

37. Chapter 9 - 37 Eutectoid & Peritectic Phase Reactions Eutectic - liquid in equilibrium with two solids L  +  cool heat intermetallic compound - cementite cool heat Eutectoid - solid phase in equilibrium with two solid phases S 2 S 1 +S 3   + Fe 3 C (727ºC) cool heat Peritectic - liquid + solid 1  solid 2 (Fig 9.21) S 1 + L S 2  + L  (1493ºC)

38. Chapter 9 - 38 Complex Phase Diagrams α and η are terminal solid solutions - β, β’, γ, δ and ε are intermediate solid solutions. new phenomena exist: ……… reaction ………. reaction

39. Chapter 9 - 39 Eutectoid & Peritectic Cu-Zn Phase diagram Eutectoid transition   +  Peritectic transition  + L 

40. Chapter 9 - 40 Phase Diagram with Eutectoid Phase Reactions The eutectoid (eutectic-like in Greek) reaction is similar to the eutectic This phase diagram includes both: Eutectic Reaction Eutectoid Reaction Our Interest In Fe-C phase diagram

41. Chapter 9 - 41 Iron-Carbon (Fe-C) Phase Diagram 2 important points -Eutectoid (B):  + Fe 3 C -Eutectic (A): L  + Fe 3 C Fe 3 C (cementite) 1600 1400 1200 1000 800 600 400 0 123456 6.7 L  austenite (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    RS 0.76 C eutectoid B hard Fe 3 C (cementite-hard)  ferritesoft (ferrite-soft)

42. Chapter 9 - 42 Pearlite Development of Microstructure in Iron-Carbon alloys Pearlite Structure Diffusion By Diffusion ferritecemetitepearlite Note: remember the names! Austenite, ferrite, cemetite and pearlite

43. Chapter 9 - 43 Hypoeutectoid 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)  727°C (Fe-C System) C0C0 0.76 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  )              

44. Chapter 9 - 44 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)  (Fe-C System) 0.76 CoCo Fe 3 C 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 )            

45. Chapter 9 - 45 Phases in Fe–Fe 3 C Phase Diagram α -ferrite - solid solution of C in …….. Fe Stable form of iron at room temperature. The maximum solubility of C is 0.022 wt% (interstitial solubility) Soft and relatively easy to deform + Fe-C liquid solution γ -austenite - solid solution of C in …….. Fe The maximum solubility of C is 2.14 wt % at 1147°C. Interstitial lattice positions are much larger than ferrite (higher C%) Is not stable below the eutectic temperature (727 °C) unless cooled rapidly (Chapter 10). δ -ferrite solid solution of C in ……. Fe The same structure as α -ferrite Stable only at high T, above 1394 °C Also has low solubility for carbon (BCC) Fe3C (iron carbide or cementite) This intermetallic compound is metastable, it remains as a compound indefinitely at room T, but decomposes (very slowly, within several years) into α -Fe and C (graphite) at 650 - 700 °C

46. Chapter 9 - 46 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 (  )

47. Chapter 9 - 47 Example – Fe-C System b)the amount of 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

48. Chapter 9 - 48 Chapter 9 – Phase Equilibria c.the amount of pearlite and proeutectoid ferrite (  ) pearlite  just above 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 Pearlite include ? Fe3C Eutectoid Fe3C + ferrite  Eutectoid ferrite 

49. Chapter 9 - 49 Alloying Steel with More Elements T eutectoid changes: C eutectoid changes: 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

50. Chapter 9 - 50 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