1 / 42

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 (fully miscible) systems Multiphase systems Eutectics Eutectoids and Peritectics Intermetallic Compounds The Fe-C System.

sydney-munoz
Télécharger la présentation

Phase Equilibrium

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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 (fully miscible) systems • Multiphase systems • Eutectics • Eutectoids and Peritectics • Intermetallic Compounds • The Fe-C System

  3. Some Definitions System: A body of engineering material under investigation. e.g. Ag – Cu system, NiO-MgO system (or even sugar-milk system) Component of a system:Pure metals and or compounds of which an alloy is composed, e.g. Cu and Ag or Fe and Fe3C. 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. Figure 9.3(a) Schematic representation of the one-component phase diagram for H2O. (b) A projection of the phase diagram information at 1 atm generates a temperature scale labeled with the familiar transformation temperatures for H2O (melting at 0°C and boiling at 100°C).

  6. Figure 9.4(a) Schematic representation of the one-component phase diagram for pure iron. (b) A projection of the phase diagram information at 1 atm generates a temperature scale labeled with important transformation temperatures for iron. This projection will become one end of important binary diagrams, such as that shown in Figure 9.19

  7. 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 Temperature, Composition, and Pressure causes an increase in free energy and away from Equilibrium thus forcing a move to another ‘state’ Equilibrium Phase Diagram: It is a “map” of the information about the control of microstructure or phase structure of a particular material system. The relationships between temperature and the compositions and the quantities of phases present at equilibrium are represented. 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. Binary Eutectic Systems:An alloy system that contains two components that has a special composition with a minimum melting temperature.

  8. 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

  9. Gibb’s Phase Rule: a tool to define the number of phases and/or degrees of phase changes that can be found in a system at equilibrium • For any 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 (degrees of freedom) we can specify • Typically, N = 1 in most condensed systems – pressure is fixed!

  10. Looking at a simple “Phase Diagram” for Sugar – Water (or milk)

  11. B (100°C,70) 1 phase D (100°C,90) 2 phases A (20°C,70) 2 phases 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. 100 L 80 (liquid) + water- sugar system 60 L S Temperature (°C) (liquid solution (solid 40 i.e, syrup) sugar) 20 0 Adapted from Fig. 9.1, Callister 7e. 0 20 40 60 70 80 100 Co =Composition (wt% sugar)

  12. 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 Diagram: A Map based on a System’s Free Energy indicating “equilibuim” system structures – as predicted by Gibbs Rule •Indicate ‘stable’ phases as function of T, P & Composition, • 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

  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 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):

  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 Phase Diagrams: • Rule 2: If we know T and Co we know the composition of each phase • Examples: adapted from Phase Diagrams of Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.

  15. Figure 9.31 The lever rule is a mechanical analogy to the mass-balance calculation. The (a) tie line in the two-phase region is analogous to (b) a lever balanced on a fulcrum.

  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 • 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)

  17. Therefore we define 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: 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

  18. 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.

  19. 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

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

  21. 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

  22. 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 (by lever rule): Adapted from Fig. 9.8, Callister 7e.

  23. 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 CL CO - C 23 C Co = W = = 79 wt% L CL - C 29 C, wt% Sn • For a 40 wt% Sn-60 wt% Pb alloy at 220°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:

  24. 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.

  25. 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 liquid +  • then  alone • finally two solid phases • a polycrystal • fine -phase inclusions

  26. 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. 45.1%  and 54.8%  -- by Lever Rule

  27. Lamellar Eutectic Structure

  28. 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 Co, wt% Sn • 18.3 wt% Sn < Co < 61.9 wt% Sn • Result:a crystals and a eutectic microstructure

  29. Hypoeutectic & Hypereutectic Compositions hypoeutectic: Co = 50 wt% Sn hypereutectic: (illustration only) 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 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

  30. “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.

  31. 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 + 

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

  33. 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.) C Iron-Carbon (Fe-C) Phase Diagram Max. C solubility in  iron = 2.11 wt% • 2 important points

  34. 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 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.)

  35. 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 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

  36. 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 ()

  37. 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

  38. 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

  39. • 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 Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) Adapted from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.) Alloying Steel with More Elements

  40. 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

  41. Fe-C-Si Ternary phase diagram at about 2.5 wt% Si This is the “controlling Phase Diagram for the ‘Graphite‘ Cast Irons

  42. 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.

More Related