Phase Diagrams [9] Cu-Ag phase diagram Brazil’s map 1> Some important definitions… • Component - chemically recognizable species (Fe and C in carbon steel, H2O and NaCl in salted water). A binary alloy contains two components, a ternary alloy – three, etc. • Phase – a portion of a system that has uniform physical and chemical characteristics. Two distinct phases in a system have distinct physical or chemical characteristics (e.g. water and ice) and are separated from each other by definite phase boundaries. A phase may contain one or more components. • A single-phase system is called homogeneous, systems with two or more phases are mixtures or heterogeneous systems. • Solvent - host or major component in solution. • Solute - minor component Solubility Limit of a component in a phase is the maximum amount of the component that can be dissolved in it (e.g. alcohol has unlimited solubility in water, sugar has a limited solubility, oil is insoluble). The same concepts apply to solid phases: Cu and Ni are mutually soluble in any amount (unlimited solid solubility), while C has a limited solubility in Fe. 2> Some important definitions… ¾ Solubility Limit: after Callister & Rethwisch (2014) 3> Some important definitions… • Microstructure: The properties of an alloy depend not only on proportions of the phases but also on how they are arranged structurally at the microscopic level. Thus, the microstructure is specified by the number of phases, their proportions, and their arrangement in space. matrix graphite This is an alloy of Fe with 4 wt.% C. There are several phases. The long gray regions are flakes of graphite. The matrix is a fine mixture of BCC Fe and Fe3C compound. 4> Some important definitions… • Equilibrium – thermodynamic state where a system is stable at constant temperature, pressure and composition, not changing with time. 5> Gibb’s Phase Rule ¾ A tool to define the number of phases and/or degrees of phase changes that can be found in a system at equilibrium P + F = C + N P: # of phases present F: degrees of freedom (temp., pressure, composition) C: components or compounds N: noncompositional variables (usually 1) For the Cu-Ag system @ 1 atm for a single phase P: N=1 (temperature), C = 2 (Cu-Ag), P= 1 (a, b, L) F = 2 + 1 – 1= 2 This means that to characterize the alloy within a single phase field, 2 parameters must be given: temperature and composition. If 2 phases coexist, for example, α+L , β+L, α+β, then according to GPR, we have 1 degree of freedom: F = 2 + 1 – 2= 1. So, if we have temperature or composition, then we can completely define the system. If 3 phases exist (for a binary system), there are 0 degrees of freedom. This means the composition and temperature are fixed. This condition is met for a eutectic system by the eutectic isotherm. 6> Unary phase diagram water pure iron 7> Binary Systems ¾ 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 Crystal Electro- r (nm) Structure negativity Ni FCC 1.9 0.1246 Cu FCC 1.8 0.1278 • Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (W. Hume– Rothery rules) suggesting high mutual solubility. 8> Cu-Ni Isomorphous phase diagram Information: ¾ Phases, ¾ Compositions, ¾ Weight fractions. Lever Rule 9> Determination of phase(s) present ¾ Rule 1: If we know T and Co, then we know: - which phase(s) is (are) present. T(ºC) •Examples: 1600 A (1100ºC, 60 wt% Ni): L (liquid) 1 phase: α 1500 dus Cu-Ni ui 1400 liq phase B (1250ºC, 35 wt% Ni): us id 2 phases: L + α ol diagram α s 1300 (1250ºC,35) + α B L (FCC solid 1200 solution) 1100 A(1100ºC,60) 1000 0 20 40 60 80 100 wt% Ni 10> Determination of phase compositions ¾ Rule 2: If we know T and C0, then we can determine: - the composition of each phase. Cu-Ni system •Examples: T(ºC) T A Consider C0 = 35 wt% Ni A tie line dus qui 1300 L (liquid) li At TA = 1320ºC: + α Only Liquid (L) present L B us T olid CL = C0 ( = 35 wt% Ni) B s α α At TD = 1190ºC: + 1200 L D (solid) Only Solid (α) present TD Cα = C0 ( = 35 wt% Ni) 20 303235 4043 50 At TB = 1250ºC: CL C0 Cα wt% Ni Both α and L present CL = Cliquidus ( = 32 wt% Ni) Cα = Csolidus ( = 43 wt% Ni) 11> Determination of phase weight fractions ¾ Rule 3: If we know T and C0, then can determine: - the weight fraction of each phase. Cu-Ni •Examples: T(ºC) system Consider C = 35 wt% Ni A 0 T tie line s A idu iqu At TA : Only Liquid (L) present 1300 L (liquid) l + α WL = 1.00, Wα = 0 L B us T olid At TD : Only Solid ( α) present B R S s α WL = 0, Wα = 1.00 + α 1200 L (solid) At T :Both and L present D B α TD S 43− 35 20 303235 4043 50 WL = = 0= . 73 R +S 43− 32 CLC0 Cα wt% Ni R W = = 0.27 α R +S 12> The Lever Rule: A Proof ¾ Sum of weight fractions: W L + Wα = 1 ¾ Conservation of mass: C o = WLCL + WαCα ¾ Combine above equations: C − C S C − C R W = α o = W o L = L + α = + Cα − CL R S Cα − CL R S • A geometric interpretation: moment equilibrium: CL Co Cα W R = W S R S L α WL Wα 1 − Wα solving gives Lever Rule 13> Development of the microstructure ¾ Solidification in the solid + liquid phase occurs gradually upon cooling from the liquidus line. ¾ The equilibrium composition of the solid and the liquid change gradually during cooling (as can be determined by the tie-line method.) ¾ Nuclei of the solid phase form and they grow to consume all the liquid at the solidus line. 14> Development of the microstructure When cooling is too fast for atoms to diffuse and produce equilibrium conditions, nonequilibrium concentrations are produced. The nonuniform composition produced by nonequilibrium solidification is known as segregation. Microsegregation can cause hot shortness which is the melting of the material below the melting point of the equilibrium solidus. Homogenization, which involves heating the material just below the non-equilibrium solidus and holding it there for a few hours, reduces the microsegregation by enabling diffusion to bring the composition back to equilibrium. 15> Cored vs Equilibrium Structures • Cα changes as we solidify. • Cu-Ni case: First α to solidify has Cα = 46 wt% Ni. Last α to solidify has Cα = 35 wt% Ni. • Slow rate of cooling: • Fast rate of cooling: Equilibrium structure Cored structure Uniform Cα: 35 wt% Ni First α to solidify: 46 wt% Ni Last α to solidify: < 35 wt% Ni 16> Mechanical Properties of the Cu-Ni System ¾ Effect of solid solution strengthening on: 17> Three-phase Reactions in Phase Diagrams 18> Eutectoid & Peritectic Peritectic transition γ + L δ ¾ Cu-Zn Phase diagram: Eutectoid transition δγ+ ε 19> Intermetallic Compounds Mg2Pb Note: an intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact. 20> Microstructures in Eutectic Systems Pb-Sn system cooling ( 61 . 9 wt% Sn) L (1α 8 .3 wt%+ β Sn) ( 9 7.8 wt% Sn) heating 21> Lamellar Eutectic Structure ¾A 2-phase microstructure resulting from the solidification of a liquid having the eutectic composition where the phases exist as a lamellae Pb‐rich that alternate with one another. ¾Formation of eutectic Sn‐rich layered microstructure in the Pb-Sn system during solidification at the eutectic composition. Compositions of α and β phases are very different. Solidification involves redistribution of Pb and Sn atoms by atomic diffusion. 22> Hypoeutectic & Hypereutectic Compositions 300 L T(°C) L+α 200 α L+β β TE (Pb-Sn α + β System) 100 0 20 40 60 80 100 Co, wt% Sn eutectic hypoeutectic: Co = 50 wt% Sn 61.9 hypereutectic: (illustration only) eutectic: C =61.9wt% Sn α o β α β α α β β α β α β 175 μm 160 μm eutectic micro-constituent 23> Iron-Carbon (Fe-C) Phase Diagram • 2 important T(°C) Max. C solubility in γ iron = 2.11 wt% 1600 points δ -Eutectic (A): 1400 L L ⇒γ+ Fe C 3 γ γ+L 1200 A L+Fe C -Eutectoid (B): (austenite) 1148°C 3 R S γ⇒α+Fe3C 1000 γγ γ+Fe3C α γ γ + 800α γ B 727°C = Teutectoid R S C (cementite) 3 600 α+Fe C 3 Fe 400 0 1234566.7 (Fe) 0.76 4.30 120 μm Co, wt% C Result: Pearlite = Fe3C (cementite-hard) alternating layers of α (ferrite-soft) α and Fe3C phases eutectoid C 24> Iron-Carbon (Fe-C) Phase Diagram Hypoeutectoid Steel T(°C) 1600 δ 1400 L (Fe-C System) γγ γ γ+L γ γ 1200 1148°C L+Fe C (austenite) 3 γ γ 1000 γ γ γ +Fe3C α 800 r s αγ γ 727°C α γ γ R S C (cementite) α 3 wα =s/(r+s) 600 α+Fe C wγ =(1- wα) 3 Fe 400 0 1234566.7 α (Fe) C0 Co, wt% C pearlite 0.76 w pearlite = wγ Hypoeutectoid 100 μm wα =S/(R+S) steel w =(1-w ) Fe3C α pearlite proeutectoid ferrite 25> Iron-Carbon (Fe-C) Phase Diagram T(°C) Hypereutectoid Steel 1600 δ 1400 L (Fe-C System) γ γ γ γ+L 1200 1148°C L+Fe C γ γ (austenite) 3 γ γ 1000 γ γ γ +Fe3C Fe C 3 s γ γ 800 r γ γ C (cementite) α R S 3 600 wFe C =r/(r+s) α +Fe C 3 3 Fe wγ =(1-w Fe C) 3 400 0 1234566.7Co (Fe) C , wt%C pearlite 0.76 o w pearlite = wγ wα =S/(R+S) Hypereutectoid w =(1-w α) 60 μm Fe3C steel pearlite proeutectoid Fe3C 26> Application of the Lever Rule (I) Example: For a AISI 1040 carbon steel (0.40 wt% C) at a temperature just below the eutectoid, determine the following: a) composition of cementite (Fe3C) and ferrite (α) b) the amount of carbide (cementite) in grams that forms per 100 g of steel c) the amount of pearlite and proeutectoid ferrite (α) 27> Application of the Lever Rule (I) ¾ Solution: CO = 0.40 wt% C C = 0.022 wt% C a) composition of Fe C and ferrite (α) α 3 C = 6.70 wt% C Fe3 C b) the amount of carbide (cementite) in grams that forms per 100 g of steel 1600 δ Fe3 C CCo − α = x 100 1400 L Fe+ C αCC− T(°C) 3 Fe3 C α γ+L 1200 γ L+Fe C 0 .