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Phase Equilibrium

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

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Phase Equilibrium

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  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 System: A body of engineering material under investigation. e.g. Ag – Cu system 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 Solubility Limit:The maximum concentration of solute atoms that may dissolve in the Solvent to form a “solid solution” at some temperature.

  4. Definitions, cont 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. Single phase system = Homogeneous system Multi phase system = Heterogeneous system or mixtures 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.

  5. Definitions, cont 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.

  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. Sucrose/Water Phase Diagram 10 0 Solubility L Limit 8 0 (liquid) + 6 0 L Temperature (°C) (solid S (liquid solution sugar) 4 0 i.e., syrup) 20 0 20 40 60 80 100 65 Co =Composition (wt% sugar) Sugar Water Pure Pure • 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). A 1-phase region Question: What is the solubility limit at 20°C? A 2-phase region Answer: 65 wt% sugar. If Co < 65 wt% sugar: syrup If Co > 65 wt% sugar: syrup + sugar. Adapted from Fig. 9.1, Callister 7e.

  8. 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 Temperature (T) & Composition (Co) See path A to B. • Changing T can change # of phases: • Changing Co can change # of phases: See path B to D. water- sugar system Adapted from Fig. 9.1, Callister 7e.

  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. T(°C) • 2 phases are possible: 1600 L (liquid) 1500 L (liquid) a (FCC solid solution) • 3 phase fields are observed: 1400 L a liquidus + 1300 a L + L solidus a a 1200 (FCC solid 1100 solution) 1000 wt% Ni 0 20 40 60 80 100 Phase Diagrams: A chart based on a System’s Free Energy indicating “equilibuim” system structures • Indicate ‘stable’ phases as function of T, Co, and P. • We will focus on: -binary systems: just 2 components. -independent variables: T and Co (P = 1 atm is almost always used). •Phase Diagram for Cu-Ni system An “Isomorphic” Phase 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).

  11. 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 Phase Diagrams: • Rule 1: If we know T and Co then we know the # and types of all phases present. • Examples: A(1100°C, 60): 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).

  12. 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 Phase Diagrams: • Rule 2: If we know T and Co we know the composition of each phase • Examples: 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.)

  13. Phase Diagrams: 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 = 100 wt%, W = 0 a L T solidus B S R a At T : Only Solid ( ) D a a + W = 0, W = 100 wt% 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 = = 27 wt% Wa R + S • Rule 3: If we know T and Co then we know the amount of each phase (given in wt%) • Examples: 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.) 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. 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 • How much of each phase?We can Think of it as a lever! So to balance: • Tie line – a line connecting the phases in equilibrium with each other – at a fixed temperature (a so-called Isotherm) Adapted from Fig. 9.3(b), Callister 7e.

  15. Ex: 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 • 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; a phase field extends from 0 to 100 wt% Ni. • Consider Co = 35 wt%Ni. Adapted from Fig. 9.4, Callister 7e.

  16. 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

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

  18. Binary-Eutectic (PolyMorphic) Systems • Eutectic transition L(CE) (CE) + (CE) has a special composition with a min. melting Temp. 2 components Cu-Ag system T(°C) Ex.: Cu-Ag system 1200 • 3 single phase regions L (liquid) a, b (L, ) 1000 a L + a • Limited solubility: b L + 779°C b 800 TE a : mostly Cu 8.0 71.9 91.2 b : mostly Ag 600 • TE : No liquid below TE a + b 400 • CE : Min. melting TE composition 200 80 100 0 20 40 60 CE Co , wt% Ag Adapted from Fig. 9.7, Callister 7e.

  19. 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 = = = 67 wt% 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 = = 33 wt% = 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 amount of each phase (lever rule): Adapted from Fig. 9.8, Callister 7e.

  20. EX: Pb-Sn Eutectic System (2) T(°C) CL - CO 46 - 40 = W = a CL - C 46 - 17 300 L (liquid) 6 a L + = = 21 wt% 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 = = 79 wt% L CL - C 29 • For a 40 wt% Sn-60 wt% Pb alloy at 200°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 amount of each phase: Adapted from Fig. 9.8, Callister 7e.

  21. 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 Temp. solubility limit) • Co < 2 wt% Sn • Result: --at extreme ends --polycrystal of a grains i.e., only one solid phase. Adapted from Fig. 9.11, Callister 7e.

  22. 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 Adapted from Fig. 9.12, Callister 7e.

  23. 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 Adapted from Fig. 9.14, Callister 7e. : 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. 45.1%  and 54.8%  Adapted from Fig. 9.13, Callister 7e.

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

  25. Microstructures in Eutectic Systems: IV • Just above TE : 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 = 50 wt% = R + S a b WL = (1- W ) = 50 wt% b L + a R S 200 TE S R • Just below TE : C = 18.3 wt% Sn a a b 100 + a primary C = 97.8 wt% Sn b a eutectic S b eutectic W a = 72.6 wt% = R + S 0 20 40 60 80 100 W = 27.4 wt% b 18.3 61.9 97.8 Adapted from Fig. 9.16, Callister 7e. Co, wt% Sn • 18.3 wt% Sn < Co < 61.9 wt% Sn • Result:a crystals and a eutectic microstructure

  26. Hypoeutectic & Hypereutectic Compositions hypoeutectic: Co = 50 wt% Sn hypereutectic: (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.) a b a b a a b b a b a b 175 mm Adapted from Fig. 9.17, Callister 7e. Adapted from Fig. 9.17, Callister 7e. (Illustration only) 300 L T(°C) a L + a b b L + (Pb-Sn 200 TE System) 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.) 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 Adapted from Fig. 9.14, Callister 7e.

  27. “Intermetallic” Compounds Adapted from Fig. 9.20, Callister 7e. An Intermetallic Compound is also an important part of the Fe-C system! Mg2Pb Note: an intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact.

  28. 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 in equilibrium with a single solid 2 (Fig 9.21) S1 + LS2  + L (1493ºC) Eutectoid & Peritectic – some definitions • Eutectic: a liquid in equilibrium with two solids L + 

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

  30. 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 (Adapted from Fig. 9.27, Callister 7e.) Adapted from Fig. 9.24,Callister 7e. C Iron-Carbon (Fe-C) Phase Diagram Max. C solubility in  iron = 2.11 wt% • 2 important points

  31. T(°C) 1600 d L 1400 (Fe-C System) g +L g g g 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 Adapted from Fig. 9.30,Callister 7e. 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.)

  32. 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 Adapted from Fig. 9.33,Callister 7e. Hypereutectoid Steel 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.) Co 0.76

  33. 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 ()

  34. 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 Solution: a) composition of Fe3C and ferrite () • the amount of carbide (cementite) in grams that forms per 100 g of steel

  35. 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 Solution, cont: • the amount of pearlite and proeutectoid ferrite () note: amount of pearlite = amount of g just above TE Co = 0.40 wt% CCa = 0.022 wt% CCpearlite = C = 0.76 wt% C Looking at the Pearlite: 11.1% Fe3C (.111*51.2 gm = 5.66 gm) & 88.9%  (.889*51.2gm = 45.5 gm)  total  = 45.5 + 48.8 = 94.3 gm

  36. • 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 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.) Alloying Steel with More Elements

  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 3rd element content” from the 3 dimensional Fe-C-Cr phase equilibrium diagram From: G. Krauss, Principles of Heat Treatment of Steels, ASM International, 1990 What this graph shows are the increasing temperature of the euctectoid and the decreasing carbon content, and (indirectly) the shrinking  phase field

  38. 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 (cause) a range of microstructures.

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