1 / 27

Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

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

kim-johnston
Download Presentation

Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

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. Engineering 45 PhaseDiagrams (1) Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  2. 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

  3. 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

  4. 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)

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. • 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

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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

  20. 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.

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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

  26. 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

  27. 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 is10% of Atm 2.0 75

More Related