Chapter 9: Phase Diagrams

# Chapter 9: Phase Diagrams

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## Chapter 9: Phase Diagrams

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1. Chapter 9: Phase Diagrams Phase b (A and B atoms) Phase a (A and B atoms) Nickel atom Copper atom ISSUES TO ADDRESS... • When we mix two elements... what equilibrium state do we get? Structure Properties Processing • 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)?

2. Sucrose/Water Phase Diagram 10 0 Solubility L Limit 8 0 (liquid) 6 0 + L S Temperature (°C) (liquid solution (solid 4 0 i.e., syrup) sugar) 20 0 20 40 60 80 100 65 Co =Composition (wt% sugar) Sugar Water Pure Pure 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 Co < 65 wt% sugar: syrup If Co > 65 wt% sugar: syrup + sugar.

3. Components and Phases • • 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., a and b). • Equilibrium • No change with time – fixed composition for each component Two phases Two Components b (lighter phase) Includes Al and Cu Al-Cu Alloy a (darker phase) Includes Al and Cu

4. B (100°C,70) 1 phase D (100°C,90) 2 phases 100 L 80 (liquid) + 60 L S Temperature (°C) ( liquid solution (solid 40 i.e., syrup) sugar) A (20°C,70) 2 phases 20 0 0 20 40 60 70 80 100 Co =Composition (wt% sugar) Effect of T & Composition (Co) path A to B. • Changing T can change # of phases: • Changing Co can change # of phases: path B to D. water- sugar system

5. T composition T ? Phase Change Time Phase Diagrams • Thermodynamics: Phase Diagrams Indicate phases as function of T, Co, and P. • For this course: -binary systems: just 2 components. -independent variables: T and Co (P = 1 atm is almost always used). Phase Diagram: A Plot of T versus C indicating various boundaries between phases 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) Uses of Phase Diagrams: - No. of phases - Composition of each phase - Relative amount of each phase + Predict the microstructure

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

7. Phase Diagram for Cu-Ni system 1600 1500 L (liquid) 1400 Liquidus line a T(°C) + 1300 Solidus line L 1200 a 1100 (FCC solid Soln) 1000 wt% Ni 0 20 40 60 80 100 Phase Diagrams- Metals Isomorphous Phase Diagram Complete solubility - one phase field extends from 0 to 100 wt% Ni. • 2 phases: L (liquid) a (FCC solid solution) • 3 phase fields: L a L + a

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: TE • Certain (fixed) composition: CE • Eutectic Phase Reaction: L a + b • There are other types of phase reactions (Later)

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

10. 1) Pure Element 2) Mixture of soluble 2 elements T L T T L L a (A or B) L a Tmelting a a Time Time L L T L a + b L a a + b TE L a + b a + b Time Time Types of Cooling Curves 3) Phase Reaction at CE 3) Phase Reaction: Co > CE

11. Constructing Phase Diagrams Thermal Method • Start with several different solutions • Each solution at certain co • 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 co plot. • Connect the points from projection creating boundaries. • Label different obtained areas and boundaries.

12. T Time Composition Constructing Isomorphous Phase Diagrams Phase Diagram T verus Co Cooling Curves at Various Initial Co

13. T(°C) 1600 L (liquid) 1500 a 1 phase: Cu-Ni phase diagram liquidus B (1250°C, 35): (1250°C,35) 1400 solidus a 2 phases: L + a a + 1300 L B (FCC solid 1200 solution) 1100 A(1100°C,60) 1000 wt% Ni 0 20 40 60 80 100 Using Phase Diagrams:# and types of phases Rule 1: If we know T and Co, then we know: - Number and types of phases present. • Examples: A(1100°C, 60):

14. Cu-Ni system T(°C) A T C = 35 wt% Ni A o tie line liquidus L (liquid) At T = 1320°C: 1300 A a + L Only Liquid (L) B T solidus B C = C ( = 35 wt% Ni) L o a a At T = 1190°C: + D L (solid) 1200 D a Only Solid ( ) T D C = C ( = 35 wt% Ni ) a o 32 35 4 3 20 30 40 50 At T = 1250°C: C C C a B L o wt% Ni a Both and L C = C ( = 32 wt% Ni here) L liquidus C = C ( = 43 wt% Ni here) a solidus Using Phase Diagrams:composition of phases Rule 2:If we know T and Co, then we know: - the composition of each phase. • Examples:

15. Using Phase Diagrams:weight fractions of phases Cu-Ni system T(°C) A T C = 35 wt% Ni A o tie line liquidus L (liquid) At T : Only Liquid (L) 1300 a A + L B W = 1.0 , W = = 0 a L T solidus B S R a At T : Only Solid ( ) D a a + W = 0, W = 1.0 L a L (solid) 1200 D T a D At T : Both and L B 32 35 4 3 20 3 0 4 0 5 0 S = WL C C C a L o wt% Ni R + S R = = 0.27 Wa R + S Rule 3: If we know T and Co, then we know: --the relative amount of each phase (given in wt%). • Examples:

16. T(°C) tie line liquidus L (liquid) 1300 a + M ML L B solidus T B a a + L (solid) 1200 R S S R 20 3 0 4 0 5 0 C C C a L o wt% Ni The Lever Rule • Tie line – connects the phases in equilibrium with each other - essentially an isotherm • How much of each phase?Think of it as a lever (teeter-totter)

17. Microstructure - Cooling in a Cu-Ni Binary T(°C) L: 35wt%Ni L (liquid) Cu-Ni system a 130 0 A + L L: 35 wt% Ni B a: 46 wt% Ni 35 46 C 32 43 D L: 32 wt% Ni 24 36 a a : 43 wt% Ni + 120 0 E L L: 24 wt% Ni a : 36 wt% Ni a (solid) 110 0 35 20 3 0 4 0 5 0 wt% Ni C o • Consider Co = 35 wt%Ni. Observe the microstructure while going down (cooling) at the same composition.

18. Cored vs Equilibrium Phases Uniform C : a a First to solidify: 35 wt% Ni 46 wt% Ni a Last to solidify: < 35 wt% Ni • Ca changes as we solidify. • Cu-Ni case: First a to solidify has Ca = 46 wt% Ni. Last a to solidify has Ca = 35 wt% Ni. • Fast rate of cooling: Cored structure • Slow rate of cooling: Equilibrium structure

19. --Tensile strength (TS) --Ductility (%EL,%AR) 60 %EL for pure Cu 400 %EL for 50 pure Ni TS for Elongation (%EL) 40 pure Ni Tensile Strength (MPa) 300 30 TS for pure Cu 200 20 0 20 40 60 80 100 0 20 40 60 80 100 Cu Ni Cu Ni Composition, wt% Ni Composition, wt% Ni - Peak as a function of Co - Min. as a function of Co Mechanical Properties:Cu-Ni System • Effect of solid solution strengthening on:

20. Eutectic Phase Diagram T a L + L • TE : Eutectic Temperature L a + b : Eutectic Composition • CE a + b TE Time With Eutectic Phase Reaction L(CE) (CE) + (CE) The phase diagram is characterized with: • 3 single phase regions (L, a and b) • Partial Solubility – Solubility Limit of A in B forming b B in A forming a L (liquid) Occurs at T a b L + b TE a + b CE 100 0 Wt% A: Co Remember Phase Reaction at CE

21. 1) Pure A or B 2) Giving aor b T L T T L T(°C) L a L L a Tmelting a a a L + a b L + Time b Time What is the difference in this side L L T L a + b a + b L a a + b TE 100 0 L a + b a + b Wt% A: Co Time Time Eutectic Phase Diagram – Cooling Curves

22. a L + Eutectic Phase Diagram • Boundaries - lines: • Liquidus • Solidus • Solvus • Now, apply previous rules ? • On real systems T(°C) L Liquidus Solidus a b L + b Eutectic Reaction Solvus a + b 100 0 Wt% A: Co Maximum solubility of A in B Maximum solubility of B in A

23. Example-Eutectic System T(°C) Ex.: Cu-Ag system 1200 L (liquid) a : mostly Cu 1000 b a : mostly Ag L + a b L + b 800 TE • TE : No liquid below TE TE=779°C 600 CE=71.9% Given T and C Values are characteristics Of Cu-Ag system 400 a + b 8.0% 91.2% 200 80 100 0 20 40 60 CE Co , wt% Ag

24. EX: Pb-Sn Eutectic System (1) T(°C) 300 L (liquid) a L + a b b L + 200 183°C 18.3 61.9 97.8 C- CO 150 S R S = W = a R+S C- C 100 a + b 99 - 40 59 = = = 0.67 99 - 11 88 100 0 11 20 60 80 99 40 CO - C R W C C Co = =  C, wt% Sn C - C R+S 40 - 11 29 = = 0.33 (or 1 – 0.67) = 99 - 11 88 • For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find... --the phases present: a + b Pb-Sn system --compositions of phases: CO = 40 wt% Sn Ca = 11 wt% Sn Cb = 99 wt% Sn --the relative amountof each phase:

25. EX: Pb-Sn Eutectic System (2) T(°C) CL - CO 46 - 40 = W = a CL - C 46 - 17 300 L (liquid) 6 a L + = = 0.21 29 220 a b b R L + S 200 183°C 100 a + b 100 17 46 0 20 40 60 80 C CL Co C, wt% Sn CO - C 23 = W = = 0.79 L CL - C 29 • For a 40 wt% Sn-60 wt% Pb alloy at200°C, find... --the phases present: a + L Pb-Sn system --compositions of phases: CO = 40 wt% Sn Ca = 17 wt% Sn CL = 46 wt% Sn --the relative amountof each phase:

26. Microstructures in Eutectic Systems: I T(°C) L: Cowt% Sn 400 L a L 300 L a + a 200 (Pb-Sn a: Cowt% Sn TE System) 100 b + a 0 10 20 30 Co , wt% Sn Co 2 (room T solubility limit) Depends on the composition • Co < 2 wt% Sn • Result: --at extreme ends --polycrystal of a grains i.e., only one solid phase.

27. Microstructures in Eutectic Systems: II L: Co wt% Sn T(°C) 400 L L 300 a L + a a: Cowt% Sn a 200 TE a b 100 b + a Pb-Sn system 0 10 20 30 Co , wt% Sn Co 2 (sol. limit at T ) 18.3 room (sol. limit at TE) • • 2 wt% Sn < Co < 18.3 wt% Sn • • Result: • Initially liquid +  • then  alone • finally two phases • a polycrystal • fine -phase inclusions

28. Microstructures in Eutectic Systems: III Micrograph of Pb-Sn T(°C) eutectic L: Co wt% Sn microstructure 300 L Pb-Sn system a L + a b L 200 183°C TE 100 160m a : 97.8 wt% Sn : 18.3 wt%Sn 0 20 40 60 80 100 97.8 18.3 CE C, wt% Sn 61.9 • Co = CE • Result: Eutectic microstructure (lamellar structure) --alternating layers (lamellae) of a and b crystals.

29. Lamellar Eutectic Structure

30. Microstructures in Eutectic Systems: IV L T(°C) L: Co wt% Sn a L a 300 L Pb-Sn system a L + a b b L + R S 200 TE S R a b 100 + a primary a eutectic b eutectic 0 20 40 60 80 100 18.3 61.9 97.8 Co, wt% Sn • 18.3 wt% Sn < Co < 61.9 wt% Sn • Result:a crystals and a eutectic microstructure

31. Microstructures in Eutectic Systems: IV • Just aboveTE : L T(°C) L: Co wt% Sn a C = 18.3 wt% Sn a L a CL = 61.9 wt% Sn 300 L Pb-Sn system S W a L + a = 0.50 ‘ = R + S a b WL = (1- W ) = 0.50 b L + a R S 200 TE S R • Just belowTE : C = 18.3 wt% Sn a a b 100 + a primary C = 97.8 wt% Sn b a eutectic S b eutectic W a = 0.73 = R + S 0 20 40 60 80 100 W = 0.27 b 18.3 61.9 97.8 Co, wt% Sn • 18.3 wt% Sn < Co < 61.9 wt% Sn • Result: a crystals and a eutectic microstructure What is the fraction of Eutectica ? Pro-eutectic (primary) a ?

32. Hypoeutectic & Hypereutectic hypoeutectic: Co = 50 wt% Sn hypereutectic: (illustration only) a b a b a a b b a b a b 175 mm 300 L T(°C) a L + a b b L + (Pb-Sn 200 TE System) a + b 100 Co, wt% Sn 0 20 40 60 80 100 eutectic 61.9 eutectic: Co=61.9wt% Sn 160 mm eutectic micro-constituent

33. Mg2Pb Intermetallic Compounds In Previous Phase diagrams α and β are terminalsolid 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 ??? Note: intermetallic compound forms a line - not an area like a ? Because stoichiometry (i.e. composition) is exact: Calculate ? Calculate Wt% Pb = 207 /(207+2(24.3)) X 100 = 81 %

34. 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?

35. cool cool cool heat heat heat • Eutectoid - solid phase in equilibrium with two solid phases • S2S1+S3 •  + Fe3C (727ºC) intermetallic compound - cementite • Peritectic - liquid + solid 1 solid 2 (Fig 9.21) • S1 + LS2 •  + L (1493ºC) Eutectoid & Peritectic Phase Reactions • Eutectic - liquid in equilibrium with two solids L + 

36. 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

37. Peritectic transition  + L Eutectoid transition  +  Eutectoid & Peritectic Cu-Zn Phase diagram

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

39. T(°C) 1600 d -Eutectic (A): L 1400 Þ g + L Fe3C g +L g A 1200 L+Fe3C 1148°C -Eutectoid (B): (austenite) R S g Þ a + Fe3C g g 1000 g +Fe3C g g a Fe3C (cementite) + 800 B g a 727°C = T eutectoid R S 600 a +Fe3C 400 0 1 2 3 4 5 6 6.7 4.30 0.76 Co, wt% C (Fe) 120 mm Fe3C (cementite-hard) Result: Pearlite = alternating layers of eutectoid a (ferrite-soft) a and Fe3C phases C Iron-Carbon (Fe-C) Phase Diagram • 2 important points

40. Development of Microstructure in Iron-Carbon alloysPearlite Structure By Diffusion Note: remember the names!Austenite, ferrite, cemetite and pearlite

41. T(°C) 1600 d L 1400 (Fe-C g +L g g g System) 1200 L+Fe3C 1148°C (austenite) g g g 1000 g g +Fe3C g g Fe3C (cementite) r s 800 a g g 727°C a a a g g R S 600 a +Fe3C w = s /( r + s ) a w = (1- w ) g a 400 0 1 2 3 4 5 6 6.7 a Co, wt% C (Fe) C0 0.76 pearlite w = w g pearlite Hypoeutectoid 100 mm w = S /( R + S ) a steel w = (1- w ) a Fe3C pearlite proeutectoid ferrite Hypoeutectoid Steel

42. T(°C) 1600 d L 1400 (Fe-C g +L g g g System) 1200 L+Fe3C 1148°C g g (austenite) g 1000 g g +Fe3C g g Fe3C (cementite) Fe3C s r 800 g g a g g R S 600 a +Fe3C w = r /( r + s ) Fe3C w =(1- w ) g Fe3C 400 0 1 2 3 4 5 6 6.7 Co, wt%C (Fe) pearlite w = w g pearlite w = S /( R + S ) a Hypereutectoid 60mm steel w = (1- w ) a Fe3C pearlite proeutectoid Fe3C Hypereutectoid Steel Co 0.76

43. Phases in Fe–Fe3C 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 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) + Fe-C liquid solution 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

44. Example: Phase Equilibria For a 99.6 wt% Fe-0.40 wt% C at a temperature just below the eutectoid, determine the following • composition of Fe3C and ferrite () • the amount of carbide (cementite) in grams that forms per 100 g of steel • the amount of pearlite and proeutectoid ferrite ()

45. CO = 0.40 wt% CCa = 0.022 wt% CCFe C = 6.70 wt% C 3 1600 d L 1400 T(°C) g +L g 1200 L+Fe3C 1148°C (austenite) 1000 g +Fe3C Fe3C (cementite) 800 727°C R S 600 a +Fe3C 400 0 1 2 3 4 5 6 6.7 CO C Co, wt% C CFe C 3 Example – Fe-C System a) composition of Fe3C and ferrite () • the amount of cementite in grams that forms per 100 g of steel

46. 1600 d L 1400 T(°C) g +L g 1200 L+Fe3C 1148°C (austenite) 1000 g +Fe3C Fe3C (cementite) 800 727°C R S pearlite = 51.2 gproeutectoid  = 48.8 g 600 a +Fe3C 400 CO 0 1 2 3 4 5 6 6.7 C C Co, wt% C Chapter 9 – Phase Equilibria • the amount of pearlite and proeutectoid ferrite () note: amount of pearlite = amount of g just aboveTE Co = 0.40 wt% CCa = 0.022 wt% CCpearlite = C = 0.76 wt% C Pearlite include ? Eutectoid Fe3C + Eutectoidferrite a

47. • Teutectoid changes: • Ceutectoid changes: (wt%C) Ti Si Mo (°C) Ni W Cr Cr eutectoid Eutectoid Si Mn W Mn Ti Mo C T Ni wt. % of alloying elements wt. % of alloying elements Alloying Steel with More Elements

48. 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.