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Engineering 45. Phase Diagrams (1). Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. Learning Goals – Phase Diagrams. When Two Elements Are Combined, Determine the Resulting MicroStructural Equilibrium State For Example Specify
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Engineering 45 PhaseDiagrams (1) Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu
Learning Goals – Phase Diagrams • When Two Elements Are Combined, Determine the Resulting MicroStructural Equilibrium State • For Example • Specify • a composition (e.g., wt%Cu - wt%Ni), and • a temperature (T) • a pressure (P) • almost ALWAYS assume ROOM Pressure • Determine Structure
Phase A Phase B Nickel atom Copper atom Learning Goals.2 – Phase Dia. • Cont: Determine Structure • HOW MANY phases Result • The COMPOSITION of each phase • Relative QUANTITY of each phase
Definitions – Phase Systems • Component Pure Constituent of a Compound • Typcially an ATOM, but can also be a Molecular Unit • Solvent/Solute • Solvent Majority Component in a Mixture • Solute Minority Component in a Mixture • System Possible Alloys Formed by Specific Components (e.g. C-Fe Sys)
Solubility Limit Max Concentration of Solute that will actually DISSOLVE in a Solvent to form a SOLUTION Sucrose/Water Phase Diagram 10 0 Solubility L Limit 8 0 (liquid) 6 0 + L Temperature (°C) S (liquid solution 4 0 i.e., syrup) (solid 20 sugar) 0 20 40 60 80 100 Co =Composition (wt% sugar) Sugar Water Pure Pure The Solid Solubility Limit • Example: Water-Sugar • Add Sugar (Solute) to Water (Solvent) • Initially ALL the Sugar Dissolves • But after a Certain Amount, SOLID Sugar Starts to Collect on the bottom of the Vessel
Sol-Sol Quantitative Example At What wt% Sugar does the Sugar NO Longer Dissolve for 20 °C 80 °C For 20 °C 10 0 Solubility L Limit 8 0 (liquid) 6 0 + L Temperature (°C) S (liquid solution 4 0 i.e., syrup) (solid 20 sugar) 0 20 40 60 80 100 Co =Composition (wt% sugar) Water Pure Pure The Solid Solubility Limit cont. 75 63 Sugar • Cast Right from 20C • Find Solid Sugar in Vessel at C0 = 63 wt% • For 80C, Again Cast Rt • Find Solid Sugar in Vessel at C0 = 75 wt% • INcreased Temp INcreases Sol-Sol Limit
Components The elements or compounds which are mixed initially (e.g., Al and Cu) Phases The PHYSICALLY and CHEMICALLY DISTINCT material regions that result from mixing (e.g., a and b below) Components & Phases • AluminumCopperAlloy
B D (100,70) (100,90) 1 phase 2 phases 10 0 L 8 0 (liquid) + 6 0 L S Temperature (°C) ( liquid solution (solid 4 0 i.e., syrup) sugar) A 20 (70, 2 0 ) 2 phases 0 0 2 0 4 0 6 0 70 8 0 10 0 C = Composition (wt% sugar) o Effect of T & Composition (C0) • Changing T can change No. of phases: path A to B. • Changing C0 can change No. of phases: path B to D • WaterSugarSystem
Consider the Cu-Ni Alloy System Phase Equilibria • Both have the same crystal structure (FCC) and have similar electronegativities and atomic radii (c.f. Hume – Rothery rules) suggesting high mutual solubility. • Copper and Nickel are, in fact, totally miscible in all Proportions
• 2 phases: 1600 L (liquid) 1500 L (liquid) a (FCC solid soln) • 3 phase fields: 1400 LL+ aa a liquidus T(°C) + 1300 L solidus a 1200 (FCC solid 1100 solution) 1000 wt% Ni 0 20 4 0 6 0 8 0 10 0 Phase Diagrams • Describes Phase Formation as a Function of T, C0, P • This Course Considers • binary systems: 2 components • independent variables: T & C0 (P = 1atm in all Cases) • The Cu-Ni Phase Diagram
Rule-1: Given T & C0 (for P = 1 atm) then Find NUMBER & TYPES of Phases Present T(°C) 1600 L (liquid) 1500 liquidus 1400 (1250,35) solidus a a + 1300 L B (FCC solid 1200 solution) 1100 A(1100,60) 1000 wt% Ni 0 20 4 0 6 0 8 0 10 0 Phase Dia.’s: Phase No.s & Types • Examples • Pt-A (1100C, 60wt-%) • 1 Phase → a; the FCC Solid Solution • Pt-B (1250,35) • 2 Phases → L+a • Cu-Ni PhaseDiagram
Rule-2: Given T & C0 (for P = 1 atm) then Find The COMPOSITION (wt% or at%) for EACH Phase T(°C) A T A tie line liquidus L (liquid) 1300 a + L B solidus T B a a + L (solid) 1200 D T D 20 3 0 32 35 4 0 4 3 5 0 CL C0 Ca wt% Ni Phase Dia.’s: Phase Composition • Example: C0 = 35 wt% Ni • At TA: • Only Liquid • CL = CO = 35 wt% Ni • Cu-Ni PhaseDiagram
T(°C) A T A tie line liquidus L (liquid) 1300 a + L B solidus T B a a + L (solid) 1200 D T D 20 3 0 32 35 4 0 4 3 5 0 CL C0 Ca wt% Ni Phase Dia.’s: Phase Comp. cont. • Example: C0 = 35 wt% Ni • At TD: • Only Solid (a-FCC) • Ca = C0 = 35 wt% Ni • At TB: • BOTH a and L • Ca = Csolidus • 43 wt% Ni • CL = Cliquidus • 32 wt% Ni • Cu-Ni PhaseDiagram • Note the Use of the IsoThermal “Tie Line” at TB to Find CL & Ca
T(°C) A T A tie line liquidus L (liquid) 1300 a + L B solidus T B a a + L (solid) 1200 D T D 20 3 0 32 35 4 0 4 3 5 0 CL C0 Ca wt% Ni Phase Dia.’s: Phase Wt Fractions • Rule-3: Given T & C0 (for P = 1 atm) then Find • The AMOUNT of EACH Phase in Wt-Fraction • Example: C0 = 35 wt% Ni • At TA: • Only Liquid • WL = 1.00 & Wa = 0.00 (wt Frac’s) • At TD: • Only Solid • WL = 0.00 & Wa = 1.00 (Frac’s) • Cu-Ni PhaseDiagram
S = WL + T(°C) R S A T A tie line liquidus L (liquid) 1300 a + S L B solidus T R B = 27wt% = Wa R a a + + R S L (solid) 1200 D T D 20 3 0 32 35 4 0 4 3 5 0 CL C0 Ca wt% Ni Phase Dia.’s: Wt Fractions cont. • Example: C0 = 35 wt% Ni • At TB: • BOTH a and L • Calc Wa,B & WL,B Using the INVERSE LEVER RULE • Cu-Ni PhaseDiagram
Sum of weight fractions: Lever Rule Proof • Conservation of mass (Ni): • Combine These Two Equations for WL & Wα • A Geometric Interpretation Balance massXdist at Tip-Pt
T(°C) L: 35wt%Ni L (liquid) a 1300 + A L L: 35wt%Ni a : 46wt%Ni B 35 46 C 32 43 D L: 32wt%Ni 24 36 a a : 43wt%Ni + L 1200 E L: 24wt%Ni a : 36wt%Ni a (solid) 1100 20 3 0 35 4 0 5 0 wt% Ni C0 Cooling Cu-Ni Binary Phase-Sys • Phase Diagram for Cu-Ni System → • System Characteristics: • BINARY → 2 components: Cu & Ni • ISOMORPHOUS → Complete Solubility of one Component in Another • At least One Solid Phase-Field Extends from 0 to 100 wt% Ni
T(°C) L: 35wt%Ni L (liquid) a 1300 + A L L: 35wt%Ni a : 46wt%Ni B 35 46 C 32 43 D L: 32wt%Ni 24 36 a a : 43wt%Ni + L 1200 E L: 24wt%Ni a : 36wt%Ni a (solid) 1100 20 3 0 35 4 0 5 0 wt% Ni C0 Ex: Cu-Ni Binary Cooling • Consider 35 wt% Ni Cooled: 1300 °C → Rm-Temp • Pt-A • 1.00 Liquid • 35 wt% Ni • Pt-B on Liquidus • Tiny Amount of solid-a in Liq. Suspension • Liq → 35 wt% Ni • a → 46 wt% Ni
T(°C) L: 35wt%Ni L (liquid) a 1300 + A L L: 35wt%Ni a : 46wt%Ni B 35 46 C 32 43 D L: 32wt%Ni 24 36 a a : 43wt%Ni + L 1200 E L: 24wt%Ni a : 36wt%Ni a (solid) 1100 20 3 0 35 4 0 5 0 wt% Ni C0 Ex: Cu-Ni Binary Cooling cont. • Pt-C in 2-Ph Region • (43-35)/(43-32) = 0.727 Liquid • Liq → 32 wt% Ni • a → 43 wt% Ni • Pt-D on Solidus • Small Liq Pockets in Solid Suspension • Liq → 24 wt% Ni • a → 36 wt% Ni • Pt E • 1.00 a, @ C0
NonEquilibrium Cooling • Phases Diagrams are Constructed Under the Assumption of ThermoDynamic Equilibrium • i.e., All Phases have Formed Sufficiently Slowly to allow for HOMOGENOUS (same) Concentrations WITHIN ALL Phases • In the Previous Example The Solid STARTS at 46 wt%-Ni (pt-B) and ENDS at 35 wt%-Ni (Pt-E) • Thus Solid particles that WERE 46Ni Had to CHANGE to 35Ni by SOLID STATE DIFFUSION • But Solid-State Diffusion Proceeds Slowly • Rapid Cooling Can result in NonUniform Comp.
Ca Changes Composition Upon Cooling First a to solidify has Ca = 46 wt%Ni Last a to solidify has Ca = 35 wt%Ni a First to solidfy: 46wt%Ni Last to solidfy: a < 35wt%Ni Uniform Ca 35wt%Ni NonEquil Cool → Cored Structure • Fast Cool Rate → Cored structure • Slow Cool Rate → Equil. Structure
Recall Solid-Solution Strengthening 60 %EL for pure Cu 400 %EL for 50 pure Ni TS for Elongation (%EL) 40 pure Ni 300 Tensile Strength (MPa) 30 TS for pure Cu 200 20 0 20 4 0 6 0 8 0 10 0 0 20 4 0 6 0 8 0 10 0 Cu Ni Cu Ni Composition, wt%Ni Composition, wt%Ni Mech Props → Cu-Ni System • Tensile Strength, TS • Ductility (%EL,%AR) • Max As Fcn of C0 • Min as Fcn of C0
WhiteBoard PPT Work • Problems 9.[5,6] • The Affect of PRESSURE on Phase Diagrams • Water Ice, Has at Least TEN, yes 10, Distinct Structural Phases • Phases form in Response to the PRESSURE Above The Ice
Ice is Nice – Problem 9.5 • Given Ice-I at −15C & 10atm → Find MELTING and SUBLIMATION PRESSURES • Note Typo in Book • Temperature needs to be –15 °C for this to work Starting Point
Ice is Nice P9.5a – Melt Temp • At −15C Cast UPward to the Solid-LIQUID Phase Boundary • Find that Ice-I, when held at −15C, MELTS at about 1000 atm (~15000 psi, ~100 Mpa) 1000
Ice is Nice P9.5b – Sublime Temp • At −15C Cast DOWNward to the Solid-VAPOR Phase Boundary • Find that Ice-I, when held at −15C, VAPORIZES at about 0.003 atm (~0.0002 psi, ~20 Pa) 0.003
Ice is Nice P9.6 P = 0.1 Atm • At 0.1 Atm Cast RIGHTward to intercept the Sol-Liq and Liq-Vap Phase-Boundaries • Ice-I MELTS at 2 °C • Water BOILS at 75 °C • i.e., the VAPOR PRESSURE of Water at 75 °C is10% of Atm 2.0 75