1. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 1 Bruce Mayer, PE Engineering-45: Materials of Engineering Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 45 Phase Diagrams (2)

2. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 2 Bruce Mayer, PE Engineering-45: Materials of Engineering 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. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 3 Bruce Mayer, PE Engineering-45: Materials of Engineering Learning Goals.2 – Phase Dia. Cont: Determine Structure –HOW MANY phases Result –The COMPOSITION of each phase –Relative QUANTITY of each phase Phase A Phase B

4. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 4 Bruce Mayer, PE Engineering-45: Materials of Engineering Binary Eutectic Systems  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

5. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 5 Bruce Mayer, PE Engineering-45: Materials of Engineering Cu-Ag Binary Eutectic Sys  3 Phases: , , L  LIMITED Solubility  → Mostly Cu –8.0 wt% Ag  → Mostly Ag –8.8 wt% Cu  T E → NO Liquid Below 779 °C  C E → Min. Melting- Temp Composition 71.9% Ag

6. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 6 Bruce Mayer, PE Engineering-45: Materials of Engineering Eutectic Transition  At the Eutectic Composition there is NO “Mushy Phase” Cu-Ag system L (liquid)  L +  L+    CoCo,wt% Ag 204060 80100 0 200 1200 T(°C) 400 600 800 1000 CECE TETE 8.071.991.2 779°C  At C E the alloy when heated “flashes” to Liquid  Eutectic transition Liquid and α&β L(C E )  (C  E ) +  (C  E )

7. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 7 Bruce Mayer, PE Engineering-45: Materials of Engineering Ex: Pb-Sn Binary Eutectic Sys  For 40wt%Sn- 60wt%Pb Alloy at 150 °C Find Phases Present Phase Compositions Phase Fractions  At C 0 = 40 wt% Sn @ 150C the Phases   Pb-Sn Phase Diagram

8. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 8 Bruce Mayer, PE Engineering-45: Materials of Engineering Ex: Pb-Sn Eutectic Sys cont.  Phase Composition Need to Cast Left for , and Right for  C  = 11 wt% Sn C  = 99 wt% Sn L+  L+   +  200 T(°C) 18.3 C, wt% Sn 206080 100 0 300 100 L (liquid)  183°C 61.997.8  Pb-Sn system 150 40 CoCo 11 CC 99 CC S R  For Phase Fractions Use LEVER Rule W  = C  - C O C  - C  = 99 - 40 99 - 11 = 59 88 = 67 wt% S R+SR+S = W  = C O - C  C  - C  = R R+SR+S = 29 88 = 33 wt% = 40 - 11 99 - 11

9. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 9 Bruce Mayer, PE Engineering-45: Materials of Engineering Ex: Pb-Sn Eutectic Sys cont.  For 40wt%Sn- 60wt%Pb at 200 °C L+   +  200 T(°C) C, wt% Sn 206080 100 0 300 100 L (liquid)   L+  183°C Pb-Sn system 40 CoCo 46 CLCL 17 CC 220 S R  Phases Present → L + α  Phase Compositions C o = 40 wt% Sn C α = 17 wt% Sn C L = 46 wt% Sn

10. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 10 Bruce Mayer, PE Engineering-45: Materials of Engineering Ex: Pb-Sn Eutectic Sys cont.  For 40wt%Sn- 60wt%Pb at 200 °C L+   +  200 T(°C) C, wt% Sn 206080 100 0 300 100 L (liquid)   L+  183°C Pb-Sn system 40 CoCo 46 CLCL 17 CC 220 S R  The Relative Amounts of L & α by Lever-Law W  = C L - C O C L - C  = 46 - 40 46 - 17 = 6 29 = 21 wt% W L = C O - C  C L - C  = 23 29 = 79 wt%

11. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 11 Bruce Mayer, PE Engineering-45: Materials of Engineering 1-Phase Cooling MicroStructure-1  Consider C 0  1 Wt% Sn Cooled from 350C First Liquid at C 0 Then L+  –The  -Particles will Have “Cored” Structure with Very Slight Composition gradient Lastly  @C 0 –Grains Grow Out from Particles formed in the L+  Phase-Field Pb-Sn Phase Diagram

12. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 12 Bruce Mayer, PE Engineering-45: Materials of Engineering 1-Phase Cooling MicroStructure-2  Consider 2wt%Sn < C 0 < 18.3wt%Sn Cooled from 325C First Liquid at C 0 Then L+  –The  -Particles with “Cored” Structure with Significant Comp gradient Next  Grains at Net-C 0 Lastly  +  (  Precipitate) –  Particles Could have Cored Structure Pb-Sn Phase Diagram

13. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 13 Bruce Mayer, PE Engineering-45: Materials of Engineering 1-Phase Cooling MicroStructure-3  C 0 = C E, Cooled From T E First Liquid at C E Then  +  in Solid State at T E –  T  In the Solid, Phases Form Compositions at the ENDS of the Eutectic IsoTherm Pb-Sn Phase Diagram  = 18.3 wt% Sn  = 97.8 wt% Sn

14. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 14 Bruce Mayer, PE Engineering-45: Materials of Engineering 1-Phase Cooling MicroStructure-4  Eutectic Cooling Forms Lamellar Structure 160 µm Micrograph of Pb-Sn Eutectic MicroStructure Composition Relaxation by Atomic Diffusion 18.3-Sn LEAD Rich Dumps Sn 97.8-Sn TIN Rich 61.9 Sn Dumps Pb

15. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 15 Bruce Mayer, PE Engineering-45: Materials of Engineering 1-Phase Cooling MicroStructure-5  18.3wt%Sn < C 0 < 61.9wt%Sn C ,max < C 0 < C E  Result →  -Crystals + eutectic- microstructure  Calc Compositions and Phase-Fractions by Lever Rules Ref Levers: Pb-Sn Phase Diagram P Q R

16. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 16 Bruce Mayer, PE Engineering-45: Materials of Engineering 1-Phase Cooling MicroStructure-6 Pb-Sn Phase Diagram P Q R  Just BELOW T E  Just ABOVE T E in L+  Field W L = (1 -W  ) = 50 wt% C  = 18.3wt%Sn C L = 61.9wt%Sn Q R +Q W  = = 50 wt% C  = 18.3wt%Sn C  = 97.8wt%Sn R P +R W  = =73wt% W  = 27wt%

17. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 17 Bruce Mayer, PE Engineering-45: Materials of Engineering [HYPO/HYPER]-eutectic Pb-Sn Phase Diagram  HYPOeutectic → BELOW Eutectic composition Yields Island-like  -regions with Lamellar Eutectic Structure  HYPEReutectic → ABOVE Eutectic Composition Yields lsland-like  -regions with Lamellar Eutectic Structure

18. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 18 Bruce Mayer, PE Engineering-45: Materials of Engineering InterMetallic Compounds Mg 2 Pb  An Intermetallic compound forms a line - not an area - because stoichiometry (i.e. composition) is exact by a chemical reaction between the pure constituents

19. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 19 Bruce Mayer, PE Engineering-45: Materials of Engineering Eutectoid & Peritectic  Eutectic  liquid in equilibrium with two solids L  α + β  Eutectoid  solid phase in equilibrium with two OTHER solid phases S 1  S 2 + S 3 Example of Iron-Carbon @ 727 °C:  α + Fe 3 C  Peritectic  liquid + solid1  solid 2 L + S 1  S 2 Example of Iron-Carbon @ 1493 °C: L + δ 

20. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 20 Bruce Mayer, PE Engineering-45: Materials of Engineering Eutectoid & Peritectic Cu-Zn Phase diagram Eutectoid transition   +  Peritectic transition  + L 

21. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 21 Bruce Mayer, PE Engineering-45: Materials of Engineering Iron-Carbon Phase Diagram Fe-C Phase Diagram  Two Significant (C 0,T) points on the Fe-Fe 3 C Phase Diagram 1.Eutectic Point-A – L   + Fe 3 C 2.Eutect oid Point-B –    + Fe 3 C Result: PEARLITE = Alternating Layers of  and Fe 3 C Phase 120 µm

22. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 22 Bruce Mayer, PE Engineering-45: Materials of Engineering HYPOeutectoid Steel Fe-C Phase Diagram  Cool From Solid Austenite @1460C 1.@1000C → Grains of  -Only 2.@~800C → Tiny Islands of  (ferrite) Form Along  -Grain Boundaries 3.@727+  °C →  + Ferrite in proportions as given by Lever Rule

23. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 23 Bruce Mayer, PE Engineering-45: Materials of Engineering HYPOeutectoid Steel cont Fe-C Phase Diagram 4.@727-  °C → ProEutectoid- FERRITE and Pearlite in proportions as given by the Lever Rule 100 µm HypoEutectoid Steel

24. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 24 Bruce Mayer, PE Engineering-45: Materials of Engineering HYPEReutectoid Steel Fe-C Phase Diagram  Cool From Solid Austenite @1000C 1.@1000C → Grains of  -Only 2.@~860C → Tiny Islands of Cementite Form Along  -Grain Boundaries 3.@727+  °C →  + Cementite in proportions as given by Lever Rule

25. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 25 Bruce Mayer, PE Engineering-45: Materials of Engineering HYPEReutectoid Steel cont. Fe-C Phase Diagram 4.@727–  °C → ProEutectoid- CEMENTITE and Pearlite in proportions as given by the Lever Rule 60 µm HyperEutectoid Steel

26. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 26 Bruce Mayer, PE Engineering-45: Materials of Engineering WhiteBoard PPT Work  Given: 1.8 kg of 1.5 wt%-C Austenite is Cooled 1050C → 725C Find 1.ProEutectoid Phase 2.TOTAL kg’s of Ferrite & Cementite 3.kg of MicroConstituents Pearlite & ProEu-Phase Starting Point

27. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 27 Bruce Mayer, PE Engineering-45: Materials of Engineering HyperEutectoid Steel  PROeutectoid Phase  The FINAL, or Persistent, Phase that is Present ABOVE the Euectoid Temperature 1.In this Case ProEutectoid Phase is: Fe 3 C, a.k.a. CEMENTITE

28. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 28 Bruce Mayer, PE Engineering-45: Materials of Engineering Lever Rule for  + Cementite  C  = 1.5-.022  C cem = 6.7-1.5

29. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 29 Bruce Mayer, PE Engineering-45: Materials of Engineering From Lever Rule  +Fe 3 C  Total Lever Length  C tot  Since 1.5 wt%C is Closer to the  Terminous, Then Expect more   Now Apply Mass Fractions, W, against the Total Mass of 1.8 kg 2.Thus

30. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 30 Bruce Mayer, PE Engineering-45: Materials of Engineering Lever Rule for Pearlite + ProEu  C p = 6.7-1.5  C Fe3C = 1.5-0.76

31. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 31 Bruce Mayer, PE Engineering-45: Materials of Engineering From Lever Rule for Pearlite  Total Lever Length  C tot  Now All  present at 727+  °C converts to PEARLITE Since 1.5wt%C is closer to  -line than Fe 3 C, expect More PEARLITE than ProEutectoid Fe 3 C  Now Apply Mass Fractions, W, against the Total Mass of 1.8 kg

32. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 32 Bruce Mayer, PE Engineering-45: Materials of Engineering 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. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 33 Bruce Mayer, PE Engineering-45: Materials of Engineering Pearlite Mass Balance  Note at the ALL the α-Iron is in the MicroForm of PEARLITE Recall M α = 1.402 kg  So Total Pearlite by Phase Addition =  This is the SAME value for M p as that Calculated by an independent LEVER Rule Calculation

34. BMayer@ChabotCollege.edu ENGR-45_Lec-22_PhaseDia-2.ppt 34 Bruce Mayer, PE Engineering-45: Materials of Engineering WhiteBoard Work