Presentation on theme: "Engr 2110 – Chapter 9 Dr. R. R. Lindeke"— Presentation transcript:
1 Engr 2110 – Chapter 9 Dr. R. R. Lindeke Phase EquilibriumEngr 2110 – Chapter 9Dr. R. R. Lindeke
2 Topics for Study Definitions of Terms in Phase studies Binary Systems Complete solubility (fully miscible) systemsMultiphase systemsEutecticsEutectoids and PeritecticsIntermetallic CompoundsThe Fe-C System
3 Some DefinitionsSystem: 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 solventSolubility Limit: The maximum concentration of solute atoms that may dissolve in the Solvent to form a “solid solution” at some temperature.
4 Definitions, contPhases: 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 systemMulti phase system = Heterogeneous system or mixturesMicrostructure: 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 (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 (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 Definition that focus on “Binary Systems” 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 BPhase ANickel atomCopper 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 equilibriumFor any system under study the rule determines if the system is at equilibriumFor a given system, we can use it to predict how many phases can be expectedUsing this rule, for a given phase field, we can predict how many independent parameters (degrees of freedom) we can specifyTypically, N = 1 in most condensed systems – pressure is fixed!
10 Looking at a simple “Phase Diagram” for Sugar – Water (or milk)
11 Effect of Temperature (T) & Composition (Co) • Changing T can change # of phases:See path A to B.Changing Co can change # of phases:See path B to D.B (100°C,70)1 phaseD (100°C,90)2 phasesAdapted from Fig. 9.1,Callister 7e.A (20°C,70)2 phases7080100604020Temperature (°C)Co=Composition (wt% sugar)L(liquid solutioni.e, syrup)(liquid)+S(solidsugar)water-sugarsystem
12 T(°C) L L (liquid) a + (FCC solid solution) • 2 phases are possible: 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).• 2 phases are possible:L(liquid)a(FCC solid solution)• 3 ‘phase fields’ are observed:L +wt% Ni204060801001000110012001300140015001600T(°C)L (liquid)(FCC solidsolution)+liquidussolidusPhase Diagram --for Cu-Ni systemAn “Isomorphic” Phase System
13 Phase Diagrams:• Rule 1: If we know T and Co then we know the # and types of all phases present.wt% Ni204060801001000110012001300140015001600T(°C)L (liquid)a(FCC solidsolution)L+liquidussolidusCu-Niphasediagram• Examples:A(1100°C, 60):1 phase:aB(1250°C,35)B(1250°C, 35):2 phases: L +aA(1100°C,60)
14 adapted from Phase Diagrams • Rule 2: If we know T and Co we know the composition of each phasewt% Ni2012001300T(°C)L (liquid)a(solid)L+liquidussolidus304050Cu-Ni system• Examples:TACo= 35 wt% Nitie line35CoAt TA= 1320°C:Only Liquid (L)CL= Co( = 35 wt% Ni)BT32CL4Ca3At TD= 1190°C:DTOnly Solid (a)C= Co( = 35 wt% NiAt TB= 1250°C:Bothaand Ladapted from Phase Diagramsof Binary Nickel Alloys, P. Nash (Ed.), ASM International, Materials Park, OH, 1991.CL= Cliquidus( = 32 wt% Ni here)Ca= Csolidus( = 43 wt% Ni here)
15 Figure 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 The Lever RuleTie line – a line connecting the phases in equilibrium with each other – at a fixed temperature (a so-called Isotherm)How much of each phase? We can Think of it as a lever! So to balance:wt% Ni2012001300T(°C)L (liquid)a(solid)L+liquidussolidus345BTtie lineCoSRMLMRS
17 Therefore we define• Rule 3: If we know T and Co then we know the amount of each phase (given in wt%)wt% Ni2012001300T(°C)L (liquid)a(solid)L+liquidussolidus345Cu-Ni systemTA35Co32BDtie lineRS• Examples:Co= 35 wt% NiAt TA: Only Liquid (L)WL= 100 wt%, Wa= 0At TD:Only Solid (a)WL= 0, W= 100 wt%At TB:Bothaand L= 27 wt%WL=SR+WaNotice: 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 • Phase diagram:Cu-Ni system.wt% Ni20120130345110L (liquid)a(solid)L+T(°C)A35CoL: 35wt%NiCu-Nisystem• System is:--binaryi.e., 2 components:Cu and Ni.--isomorphousi.e., completesolubility of onecomponent inanother; a phasefield extends from0 to 100 wt% Ni.a: 46 wt% NiL: 35 wt% NiB4635C4332a:43 wt% NiL: 32 wt% NiD2436L: 24 wt% Nia:36 wt% NiE• ConsiderCo = 35 wt%Ni.Adapted from Fig. 9.4, Callister 7e.
19 Cored vs Equilibrium Phases • 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 structureFirstato solidify:46 wt% NiUniform C:35 wt% NiLast< 35 wt% Ni
20 Cored (Non-equilibrium) Cooling Notice:The Solidus line is “tilted” in this non-equilibrium cooled environment
21 Binary-Eutectic (PolyMorphic) Systems has a ‘special’ composition with a min. melting Temp.2 componentsCu-AgsystemT(°C)Ex.: Cu-Ag system1200• 3 single phase regionsL (liquid)(L,a, b)1000a• Limited solubility:aL+L+b800779°CbaTE: mostly Cu8.071.991.2b: mostly Ag600• TE: No liquid below TEa+b• CE: Min. melting TE400composition200204060CE80100• Eutectic transitionL(CE) (CE) + (CE)Co,wt% Ag
22 EX: Pb-Sn Eutectic System (1) • For a 40 wt% Sn-60 wt% Pb alloy at 150°C, find...--the phases present:a + bPb-Snsystem--compositions of phases:L+ab200T(°C)18.3C, wt% Sn206080100300L (liquid)183°C61.997.8CO = 40 wt% SnCa = 11 wt% SnCb = 99 wt% Sn--the relative amount of each phase (by lever rule):Wa=C - COC - C5988= 67 wt%SR+S150R11C40CoS99CW=CO - CC - CRR+S2988= 33 wt%Adapted from Fig. 9.8,Callister 7e.
23 EX: Pb-Sn Eutectic System (2) • For a 40 wt% Sn-60 wt% Pb alloy at 220°C, find...--the phases present:a + LPb-Snsystem--compositions of phases:L+ba200T(°C)C, wt% Sn206080100300L (liquid)183°CCO = 40 wt% SnCa = 17 wt% SnCL = 46 wt% Sn--the relative amount of each phase:22017CR40CoS46CLWa=CL - COCL - C629= 21 wt%WL=CO - CCL - C2329= 79 wt%
24 Microstructures In Eutectic Systems: I • Co < 2 wt% Sn• Result:--at extreme ends--polycrystal of a grainsi.e., only one solid phase.L+a200T(°C)Co,wt% Sn102%2030010030b400(room Temp. solubility limit)TE(Pb-SnSystem)L: Co wt% SnaLa: Co wt% Sn
25 Microstructures in Eutectic Systems: II L: Co wt% Sn• 2 wt% Sn < Co < 18.3 wt% Sn• Result:Initially liquidThen liquid + then alonefinally two solid phasesa polycrystalfine -phase inclusionsPb-SnsystemL+a200T(°C)Co,wt% Sn1018.32030010030b400(sol. limit at TE)TE2(sol. limit at Troom)Laa: Co wt% Snab
26 Microstructures in Eutectic Systems: III • Co = CE• Result: Eutectic microstructure (lamellar structure)--alternating layers (lamellae) of a and b crystals.160 mMicrograph of Pb-SneutecticmicrostructurePb-SnsystemLa200T(°C)C, wt% Sn206080100300Lab+183°C40TEL: Co wt% SnCE61.918.3: 18.3 wt%Sn97.8: 97.8 wt% Sn45.1% and 54.8% -- by Lever Rule
28 Microstructures in Eutectic Systems: IV • wt% Sn < Co < 61.9 wt% Sn• Result: a crystals and a eutectic microstructureWL= (1-Wa)= 50 wt%C= 18.3 wt% SnCL= 61.9 wt% SnSR+=• Just above TE :Pb-SnsystemL+b200T(°C)Co, wt% Sn206080100300a40TEL: Cowt% SnLaLa18.361.9SR97.8SRprimaryaeutecticb• Just below TE :Ca= 18.3 wt% Snb= 97.8 wt% SnSR+W== 72.6 wt%= 27.4 wt%
29 Hypoeutectic & Hypereutectic Compositions 300LT(°C)L+aa200L+bb(Pb-SnTESystem)a+b100from Metals Handbook, 9th ed., Vol. 9, Metallography and Microstructures, American Society for Metals, Materials Park, OH, 1985.175 mmahypoeutectic: Co = 50 wt% Snhypereutectic: (illustration only)bCo, wt% Sn20406080100eutectic61.9eutectic: Co = 61.9 wt% Sn160 mmeutectic micro-constituent
30 “Intermetallic” Compounds Adapted fromFig. 9.20, Callister 7e.An Intermetallic Compound is also an important part of the Fe-C system!Mg2PbNote: an intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact.
31 Eutectoid & Peritectic – some definitions Eutectic: a liquid in equilibrium with two solidsL + coolheatEutectoid: solid phase in equilibrium with two solid phasesS S1+S3 + Fe3C (727ºC)intermetallic compound - cementitecoolheatPeritectic: liquid + solid 1 in equilibrium with a single solid 2 (Fig 9.21)S1 + L S2 + L (1493ºC)coolheat
33 Iron-Carbon (Fe-C) Phase Diagram Max. C solubility in iron = 2.11 wt%• 2 importantFe3C (cementite)16001400120010008006004001234566.7Lg(austenite)+L+Fe3Ca+L+Fe3Cd(Fe)Co, wt% C1148°CT(°C)727°C = Teutectoidpoints-Eutectic (A):LÞg+Fe3CASR4.30-Eutectoid (B):gÞa+Fe3CgResult: Pearlite =alternating layers ofaand Fe3C phases120 mm(Adapted from Fig. 9.27, Callister 7e.)0.76CeutectoidBRSFe3C (cementite-hard)a(ferrite-soft)
34 Hypoeutectoid Steel T(°C) d L +L g (austenite) Fe3C (cementite) a 16001400120010008006004001234566.7Lg(austenite)+L+ Fe3CaL+Fe3Cd(Fe)Co , wt% C1148°CT(°C)727°C(Fe-CSystem)C00.76ggAdapted from Figs and 9.29,Callister 7e. (Fig adapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.)rswa=/(+)g(1-proeutectoid ferritepearlite100 mmHypoeutectoidsteelRSaw=/(+)Fe3C(1-g
35 Hypereutectoid Steel Fe3C (cementite) L g (austenite) +L a d T(°C) g g 16001400120010008006004001234566.7Lg(austenite)+L+Fe3CaL+Fe3Cd(Fe)Co , wt%C1148°CT(°C)(Fe-CSystem)ggadapted from Binary Alloy Phase Diagrams, 2nd ed., Vol. 1, T.B. Massalski (Ed.-in-Chief), ASM International, Materials Park, OH, 1990.srwFe3C=/(+)g=(1-proeutectoid Fe3C60 mmHypereutectoidsteelpearliteRSwa=/(+)Fe3C(1-gCo0.76
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 steelthe amount of pearlite and proeutectoid ferrite ()
37 Solution: a) composition of Fe3C and ferrite () CO = 0.40 wt% C Ca = wt% C CFe C = 6.70 wt% C3a) composition of Fe3C and ferrite ()the amount of carbide (cementite) in grams that forms per 100 g of steelFe3C (cementite)16001400120010008006004001234566.7Lg(austenite)+L+ Fe3CaL+Fe3CdCo , wt% C1148°CT(°C)727°CCORSCFe C3C
38 Solution, cont: the amount of pearlite and proeutectoid ferrite () note: amount of pearlite = amount of g just above TECo = 0.40 wt% C Ca = wt% C Cpearlite = C = 0.76 wt% CFe3C (cementite)16001400120010008006004001234566.7Lg(austenite)+L+ Fe3CaL+Fe3CdCo , wt% C1148°CT(°C)727°CCOpearlite = 51.2 g proeutectoid = 48.8 gRSCCLooking at the Pearlite:11.1% Fe3C (.111*51.2 gm = 5.66 gm) & 88.9% (.889*51.2gm = 45.5 gm) total = = 94.3 gm
39 Alloying Steel with More Elements • Teutectoid changes:• Ceutectoid changes:Adapted from Edgar C. Bain, Functions of the Alloying Elements in Steel, American Society for Metals, 1939, p. 127.)TEutectoid(°C)wt. % of alloying elementsTiNiMoSiWCrMnCeutectoid(wt%C)
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 diagramFrom: G. Krauss, Principles of Heat Treatment of Steels, ASM International, 1990What 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 phasefor 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.