1. L10B: Phases in equilibrium with one another: Composition and amount of each The temperature and pressure must be the same in each phase in equilibrium with one another. Since binary phase diagrams are usually at 1 atm, we draw a horizontal isotherm (tie line) to find the composition of the phases. The amount of each is found by a material balance: the total amount of each component in the mixture of phases must equal the sum of that contained in the individual phases. Most conveniently utilized as the “lever rule.” If the phase compositions are in weight %, then the lever rule gives the weight fraction of each phase. If the phase compositions are in atom%, then the lever rule gives the atom fraction of each phase. In a micrograph, you’d see the volume fractions. To convert from weight fraction to volume fraction you’d have to divide the volume fraction of each phase by the density of that phase. Keep careful track of the units! Last revised December 3, 2013 by W.R.Wilcox at Clarkson University.

2. wt% Ni 20 1200 1300 T( o C) L (liquid)  (solid) L +  liquidus solidus 304050 L +  Cu-Ni system Determination of phase compositions If we know T and C mix, we can determine the composition of each phase. Examples for Cu-Ni: TATA A 35 C mix 32 CLCL At T A = 1320 o C: Only liquid (L) present C L = C mix ( = 35 wt% Ni) At T B = 1250 o C: Both  and L present CLCL = C liquidus ( = 32 wt% Ni) CC = C solidus ( = 43 wt% Ni) At T D = 1190 o C: Only solid  present C  = C mix ( = 35 wt% Ni) Consider C mix = 35 wt% Ni D TDTD tie line 4 CC 3 B TBTB

3. If we know T and C mix, then can determine the weight fraction of each phase. Examples for Cu-Ni: At T A : Only liquid (L) present W L = 1.00, W  = 0 At T D : Only solid (  ) present WL WL = 0, W  = 1.00 Determination of the weight fraction of each phase wt% Ni 20 1200 1300 T( o C) L (liquid)  (solid) L +  liquidus solidus 304050 L +  Cu-Ni system TATA A 35 C mix 32 CLCL B TBTB D TDTD tie line 4 CC 3 R S At T B : Both  and L present = 0.27 WLWL  S R+S WW  R R+S Consider C mix = 35 wt% Ni

4. Tie line – connects the phases in equilibrium with each other – an isotherm for binary mixtures (only) The Lever Rule What fraction of each phase? Think of the tie line as a lever (teeter-totter or seesaw) MLML MM RS wt% Ni 20 1200 1300 T( o C) L (liquid)  (solid) L +  liquidus solidus 304050 L +  B T B tie line C mix CLCL CC S R

5. Another viewpoint C mix Distance to opposite phase Total distance For example: simple eutectic with no solid solubility. the total distance between phases A & B Check: the closer the mixture composition is to a phase the more of that phase must be present, in the limit 100%! Fraction of A equals the distance from the mixture composition to the opposite phase (B) divided by

6. wt% Ni 20 1200 1300 304050 1100 L (liquid)  (solid) L +  L +  T( o C) A 35 C mix L: 35wt%Ni Cu-Ni system Consider microstructural changes that accompany cooling of a 35 wt% Ni alloy. Slow cooling of a Cu-Ni Alloy (equilibrium) For equilibrium to be achieved at each stage, there must be sufficient diffusion in the solid for its composition to be uniform. This requires that the particles be small and the cooling rate very slow. In usual solidification, the cooling rate is not slow enough even if the particles are small. Illustrated in next 2 slides 46 35 43 32  : 43 wt% Ni L: 32 wt% Ni B  : 46 wt% Ni L: 35 wt% Ni C E L: 24 wt% Ni  : 36 wt% Ni 24 36 D

7. Equilibrium: very slow cooling Same as previous slide. With normal cooling rates, we can ignore diffusion inside the particles. Thus, the center of each particle is its original composition and only the surface of the particle is in equilibrium with the melt. The melt is gradually depleted in the component preferentially taken up by the solid, here nickel. This is “segregation.” Shown in the next slide.

8. Non-equilibrium solidification For Cu-Ni, the melt continues to be enriched in Cu as more Ni goes into the solid. Thus, as time progresses the melt composition continues moving down the liquidus. The last infinitesimal amount of liquid will be pure Cu! (The textbook is wrong) In casting of alloys, first get dendrites growing out from the walls. The composition of the melt between dendrites changes, becoming enriched in Cu. So ingot not uniform!

9. Directional solidification of eutectic metals When a melt of eutectic composition is solidified, the two solid phases often form side by side. A good example is Pb-Sn: Simulations using organic analog (CBr 4 -C 2 Cl 6 ) and numerical at Clarkson. See times 6:47, 7.16, 8.12 and 8.57 in the following video (which must be in the same file as this ppt):  Clarkson Simulation of eutectic solidification (On-line site) Clarkson Simulation of eutectic solidificationOn-line site

10. Microstructure from solidification of eutectics A.Freezing regular eutectics such as Pb-Sn give a lamellar structure. B.When the eutectic consists of a small amount of one phase in a large amount of the other, freezing gives a rod or fibrous eutectic. Al-Cu lamellar eutectic Mo-NiAl rod eutectic

11. Eutectic and off-eutectic solidification The faster the freezing rate, the thinner the lamella or rods, and the closer they are together. (Less time for diffusion in the adjacent melt.) When the melt deviates from the eutectic composition, primary dendrites form first, the entrapped melt changes until it reaches the eutectic composition, and then the eutectic forms between the dendrite; e.g., hypoeutectic solidification: Source of micrograph

12. Fraction of grains with eutectic structure Consider the red point. Rather than asking how much of A and B are present, we can ask what weight fraction of the grains is eutectic and what fraction is primary B. To do this, treat the eutectic as a compound. Then use the lever rule in the usual way to calculate the weight fraction of grains that have the eutectic microstructure. The fraction of eutectic is opposite/total. T A B Weight fraction of B Liquid B + L A + L opposite total

13. Eutectoid solidification: Pearlite Can also cool a solid with eutectoid composition to get a lamellar structure, e.g. pearlite in Fe-C. http://en.wikipedia.org/wiki/Pearlitehttp://en.wikipedia.org/wiki/Pearlite Get alternating layers of ferrite and cementite. Can be drawn into thin wires, and is then one of the strongest materials!