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

Engineering 45. Phase Diagrams (2). Bruce Mayer, PE Licensed 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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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

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  1. Engineering 45 PhaseDiagrams (2) Bruce Mayer, PE Licensed 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) • 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. Binary → Two Components 3 Phases (A, B, Liq) Eutectic → Easily Melted Eutectic Point Composition & Temp at Which Pure-Liquid and Pure-Solid CoExisit The Low-Melt Temp Eutectic Point Binary Eutectic Systems

  5. 3 Phases: a, b, L LIMITED Solubility a → Mostly Cu 8.0 wt% Ag b → Mostly Ag 8.8 wt% Cu TE → NO Liquid Below 779 °C CE → Min. Melting-Temp Composition 71.9% Ag T(°C) 1200 L (liquid) 1000 a a L + b L + 779°C 800 b TE 8.0 71 .9 91.2 600 a + b 400 200 8 0 10 0 0 20 4 0 6 0 CE C0, wt% Ag Cu-Ag Binary Eutectic Sys

  6. At the Eutectic Composition there is NO “Mushy Phase” L(CE) (CE) + (CE) Eutectic Transition Cu-Ag system T(°C) 1200 L (liquid) 1000 a L + a b L + 779°C b 800 TE 8.0 71.9 91.2 • At CE the alloy when heated “flashes” to Liquid 600 a + b 400 200 • Eutectic transition Liquid and α&β 80 100 0 20 40 60 CE Co , wt% Ag

  7. For 40wt%Sn-60wt%Pb Alloy at 150 °C Find Phases Present Phase Compositions Phase Fractions At C0 = 40 wt% Sn @ 150C the Phases a b T(°C) 3 00 L (liquid) a L + a b b L + 2 0 0 183 °C 18.3 61.9 97.8 150 a + b 1 00 0 20 4 0 6 0 8 0 10 0 C o C , wt% Sn o Ex: Pb-Sn Binary Eutectic Sys Pb-Sn Phase Diagram

  8. Phase Composition Need to Cast Left for a, and Right for b Ca= 11 wt% Sn Cb= 99 wt% Sn 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 Ex: Pb-Sn Eutectic Sys cont. Pb-Sn system • For Phase Fractions Use LEVER Rule

  9. For 40wt%Sn-60wt%Pb at 200 °C T(°C) 300 L (liquid) a L + 220 a b b R L + S 200 183°C 100 a + b 100 17 46 0 20 40 60 80 C CL Co C, wt% Sn Ex: Pb-Sn Eutectic Sys cont. Pb-Sn system • Phases Present → L + α • Phase Compositions • Co = 40 wt% Sn • Cα = 17 wt% Sn • CL = 46 wt% Sn

  10. For 40wt%Sn-60wt%Pb at 200 °C 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 C CL Co C, wt% Sn CO - C 23 = W = = 79 wt% L CL - C 29 Ex: Pb-Sn Eutectic Sys cont. Pb-Sn system • The Relative Amounts of L & αby Lever-Law

  11. Consider C01 Wt% Sn Cooled from 350C First Liquid at C0 Then L+ a The a-Particles will Have “Cored” Structure with Very Slight Composition gradient Lastly a @C0 Grains Grow Out from Particles formed in the L+a Phase-Field T(°C) L: C wt%Sn o 4 00 L a L 3 00 a L + a (Pb-Sn) 2 0 0 T a E : C wt%Sn o 1 00 a b + 0 1 0 2 0 3 0 C C , wt% Sn o o 2 (room T solubility limit) 1-Phase Cooling MicroStructure-1 Pb-SnPhaseDiagram

  12. Consider 2wt%Sn < C0 < 18.3wt%Sn Cooled from 325C First Liquid at C0 Then L+a The a-Particles with “Cored” Structure with Significant Comp gradient Next aGrains at Net-C0 Lastly a+b (b Precipitate) b Particles Could have Cored Structure L: C wt%Sn T(°C) o 4 00 L L 3 00 a a L + a : C wt%Sn o a 2 0 0 T E a b 1 00 a b + 0 1 0 2 0 3 0 C , wt% Sn C o 2 o (sol. limit at T ) 18.3 room (sol. limit at T ) E 1-Phase Cooling MicroStructure-2 Pb-SnPhaseDiagram

  13. C0 = CE, Cooled From TE First Liquid at CE Then b+ain Solid State at TE–T In the Solid, Phases Form Compositions at the ENDS of the Eutectic IsoTherm T(°C) L: C wt%Sn o 3 00 L a L + b b L + a 2 0 0 183°C T E a b 1 00 + b : 97.8wt%Sn a : 18.3wt%Sn 0 8 0 2 0 4 0 6 0 100 0 97.8 18.3 C E C , wt% Sn o 61.9 1-Phase Cooling MicroStructure-3 Pb-Sn Phase Diagram • a = 18.3 wt% Sn • b = 97.8 wt% Sn

  14. Eutectic Cooling Forms Lamellar Structure Micrograph of Pb-SnEutecticMicroStructure 160 µm 1-Phase Cooling MicroStructure-4 DumpsPb 61.9 Sn 18.3-SnLEAD Rich 97.8-SnTIN Rich Dumps Sn • Composition Relaxation by Atomic Diffusion

  15. 18.3wt%Sn < C0 < 61.9wt%Sn Ca,max < C0 < CE Result → a-Crystals + eutectic-microstructure Calc Compositions and Phase-Fractions by Lever Rules Ref Levers: L L: C wt%Sn a o T(°C) L a L 3 00 a L + b b L + P Q a 2 0 0 T E P R 1 00 a b + a primary a eutectic b 0 eutectic 8 0 2 0 100 0 4 0 6 0 C 61.9 97.8 18.3 o C , wt% Sn o P R Q 1-Phase Cooling MicroStructure-5 Pb-Sn Phase Diagram

  16. L L: C wt%Sn a o T(°C) L a L 3 00 C a = 18.3wt%Sn a C a = 18.3wt%Sn L + C b = 97.8wt%Sn b b L + P Q a 2 0 0 C = 61.9wt%Sn R L T E W a = =73wt% P R Q P + R W a = = 50 wt% R + Q W b = 27wt% 1 00 a b + W = (1 - W a = 50 wt% ) L a primary a eutectic b 0 eutectic 8 0 2 0 100 0 4 0 6 0 C 61.9 97.8 18.3 o C , wt% Sn o P R Q 1-Phase Cooling MicroStructure-6 Pb-Sn Phase Diagram • Just ABOVE TE in L+a Field • Just BELOW TE

  17. T(°C) L 3 00 a L + b L + 2 0 0 a T b E a b + 1 00 C0 C0 hypoeutectic hypereutectic 0 C , wt% Sn o 8 0 2 0 4 0 6 0 100 0 eutectic 18.3 97.8 61.9 hypoeutectic: C0 = 50wt%Sn hypereutectic: (illustration only) eutectic: C0 = 61.9wt%Sn a b a b a a b b a b a b 175 µm 160 µm eutectic micro-constituent [HYPO/HYPER]-eutectic • HYPOeutectic →BELOW Eutecticcomposition • Yields Island-likea-regions with Lamellar Eutectic Structure • HYPEReutectic →ABOVE EutecticComposition • Yields lsland-likeb-regions withLamellar Eutectic Structure Pb-SnPhaseDiagram

  18. InterMetallic Compounds • An Intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact by a chemical reaction between the pure constituents Mg2Pb

  19. Eutectoid & Peritectic • Eutectic liquid in equilibrium with two solids • L  α + β • Eutectoid solid phase in equilibrium with two OTHER solid phases • S1 S2 + S3 • Example of Iron-Carbon @ 727 °C: g  α + Fe3C • Peritectic liquid + solid1  solid 2 • L + S1 S2 • Example of Iron-Carbon @ 1493 °C: L + δ g

  20. Peritectic transition  + L Eutectoid transition  +  Eutectoid & Peritectic Cu-Zn Phase diagram

  21. T(°C) 1600 d L 1400 g +L g A L+Fe3C 1200 1148°C (austenite) R S g g 1000 g+Fe3C g g (cementite) a + B 800 727°C = T a g eutectoid R S 6 00 a +Fe C 3 Fe3C 4 00 0 1 2 3 4 5 6 6.7 4.30 0.77 (Fe) C , wt% C o Fe3C (cementite-hard) Ceutectoid a ( ferrite-soft) 120 µm Iron-Carbon Phase Diagram Fe-C Phase Diagram • Two Significant (C0,T) points on the Fe-Fe3C Phase Diagram • Eutectic Point-A • L g + Fe3C • EutectoidPoint-B • ga + Fe3C Result: PEARLITE = Alternating Layers of a and Fe3C Phase

  22. T(°C) 1600 d L 1400 g +L g g g L+Fe C 1200 g 3 1148°C g ( austenite) g g 1000 g +Fe C g g 3 C (cementite) a g g r s 800 727°C a a a g g R S 3 6 00 w a = s /( r + s ) a +Fe C 3 Fe w g = (1- w a ) 4 00 0 1 2 3 4 5 6 6.7 C a o 0.77 a C , wt% C o pearlite a w = w g pearlite a w = S /( R + S ) w = (1- w a ) Fe3C HYPOeutectoid Steel Fe-C Phase Diagram • Cool From Solid Austenite @1460C • @1000C → Grains of g-Only • @~800C → Tiny Islands of a (ferrite) Form Along g-Grain Boundaries • @727+d °C → g + Ferrite in proportions as given by Lever Rule

  23. T(°C) 1600 d L 1400 g +L g g g L+Fe C 1200 g 3 1148°C g ( austenite) g g 1000 g +Fe C g g 3 C (cementite) a g g r s 800 727°C a a a g g R S 3 6 00 w a = s /( r + s ) a +Fe C 3 Fe w g = (1- w a ) 4 00 0 1 2 3 4 5 6 6.7 C a o 0.77 100 µm a C , wt% C o pearlite a w = w g pearlite a w = S /( R + S ) HypoEutectoidSteel w = (1- w a ) Fe3C HYPOeutectoid Steel cont Fe-C Phase Diagram • @727-d °C → ProEutectoid-FERRITE and Pearlite in proportions asgiven by the Lever Rule

  24. T(°C) 1600 d L 1400 g +L g g g L+Fe C 1200 g 1148°C 3 g (austenite) g g 1000 g g +Fe C g 3 C (cementite) Fe C 3 g s r g 800 a g g R S 3 6 00 w = r /( r + s ) a +Fe C Fe3C 3 Fe w g =(1- w ) Fe3C 4 00 0 1 2 3 4 5 6 6.7 C o 0.77 C , wt% C o pearlite w = w g pearlite w a = S /( R + S ) a w = (1- w ) Fe3C HYPEReutectoid Steel Fe-C Phase Diagram • Cool From Solid Austenite @1000C • @1000C → Grains of g-Only • @~860C → Tiny Islands of Cementite Form Along g-Grain Boundaries • @727+d °C → g+ Cementite in proportions as given by Lever Rule

  25. T(°C) 1600 d L 1400 g +L g g g L+Fe C 1200 g 1148°C 3 g (austenite) g g 1000 g g +Fe C g 3 C (cementite) Fe C 3 g s r g 800 a g g R S 3 6 00 w = r /( r + s ) a +Fe C Fe3C 3 Fe w g =(1- w ) Fe3C 4 00 0 1 2 3 4 5 6 6.7 C o 0.77 C , wt% C o pearlite w = w g pearlite w a = S /( R + S ) 60 µm a w = (1- w ) Fe3C HYPEReutectoid Steel cont. Fe-C Phase Diagram • @727–d °C → ProEutectoid-CEMENTITE and Pearlite in proportions asgiven by the Lever Rule HyperEutectoidSteel

  26. WhiteBoard PPT Work Starting Point • Given: 1.8 kg of 1.5 wt%-C Austenite is Cooled 1050C → 725C • Find • ProEutectoid Phase • TOTAL kg’s of Ferrite & Cementite • kg of MicroConstituents Pearlite & ProEu-Phase

  27. HyperEutectoid Steel • PROeutectoid Phase  The FINAL, or Persistent, Phase that is Present ABOVE the Euectoid Temperature • In this Case ProEutectoid Phase is: Fe3C, a.k.a. CEMENTITE

  28. DCcem = 6.7-1.5 DCa = 1.5-.022 Lever Rule for a + Cementite

  29. Total Lever Length DCtot From Lever Rule a +Fe3C • Now Apply Mass Fractions, W, against the Total Mass of 1.8 kg • Thus • Since 1.5 wt%C is Closer to the a Terminous, Then Expect more a

  30. Lever Rule for Pearlite + ProEu DCp = 6.7-1.5 DCFe3C = 1.5-0.76

  31. Total Lever Length DCtot From Lever Rule for Pearlite • Now Apply Mass Fractions, W, against the Total Mass of 1.8 kg • Now All g present at 727+d °C converts to PEARLITE • Since 1.5wt%C is closer to g-line than Fe3C, expect More PEARLITE than ProEutectoid Fe3C

  32. Cementite MicroConstituents • Quantities of the two forms of Cementite • ProEutectoid Cementite = 0.225 kg • Total Cementite = 0.398 kg • Thus Lamellar (Pearlite) Cementite = • Then the Cementite-Form Fractions • ProEutectoid = 225/398 = 56.5% • Pearlitic Cementite = 173/398 =43.5%

  33. Pearlite Mass Balance • Note at the ALL the α-Iron is in the MicroForm of PEARLITE • Recall Mα = 1.402 kg • So Total Pearlite by Phase Addition = • This is the SAME value for Mp as that Calculated by an independent LEVER Rule Calculation

  34. WhiteBoard Work None Today

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