the Phase Diagrams Today s Topics

Transcription

1 MME 291: Lecture 04 Interpretation of the Phase Diagrams Prof. A.K.M.B. Rashid Department of MME BUET, Dhaka Today s Topics Interpretation of phase diagrams Development of microstructures during equilibrium i cooling Formation of cored structure and method of its elimination Properties of solid solution alloys Reference: 1. WD Callister, Jr. Materials Science and Engineering: An Introduction, 5 th Ed., Ch SH Avner. Introduction to physical metallurgy, 2 nd Ed., Ch. 6. Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 02 1

2 Working with Phase Diagrams Overall composition of the alloy Liquidus and solidus temperatures Limits of solid solubility Chemical composition of phases at any temperature Amount of phases at any temperature Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 03 Working with Phase Diagrams Phase diagram nomenclatures Liquidus Start of solidification (or, end of liquification) temperature Solidus End of solidification (or, start of liquification) temperature Not to be confused with Solvus point, which indicates the limit of solid solubility in a solid solution Concentration/Composition of alloy Relative amounts of each constituent It is the horizontal axis in all binary phase diagrams The scale can be in weight %, atomic % or mole % Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 04 2

3 Working with Phase Diagrams Phase diagram nomenclatures Chemical composition of Phases It is the chemical composition of each phase in the system. In a system having more than one phase, each phase will have a unique chemical composition which will be different from each other, and will also be different from the overall composition. Not to be confused with overall composition of the alloy. Relative amounts of Phases When a system contains more than one phases, then it is the amount of each phase relative to overall amount of the alloy. Depends on temperature and composition of the alloy. Not to be confused with composition of phases. Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 05 Working with Phase Diagrams Phase diagram nomenclatures Liquid Liquidus Freezing range Twophase Solidus Solid X % Y added Y Phase diagram with complete solubility of one component into another Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 06 3

4 Working with Phase Diagrams Phase diagram nomenclatures The Cooling Curve T X Alloy 1 Liquid Freezing range Twophase Solid Te emperature Liquid T X Pure metal X ARREST POINTS start of solidification Liquid + Solid end of solidification Solid Alloy 1 X % Y added Y Time Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 07 Predicting the nature of stable phases as function of T, P and C 0 Consider a simple binary system of Cu and Ni Independent variables: T and X ( always P=1 atm) one-phase area (L) two-phase area (L + α) one-phase area (α) Wt% Ni Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 08 4

5 Number and types of phases Phase diagram rule #1: If we know the alloy composition, C 0, then we can tell the number and type of phases present in the structure at any given temperature. Point A: (30%Ni, 1225 C) Phases are: L and α Point B: (60%Ni, 1100 C) Phases are: α only wt.%ni Cu-Ni Phase Diagram Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 09 Composition of each phase Phase diagram rule #2: Tie Line Rule If we know C 0, then we can tell the composition of each phase at any temperature. At T A : L phase only C L = 35%Ni tie line At T B: L and α phases C L = C liquidus = 32%Ni C α = C solidus = 43%Ni At T D : α phase only C α = 35%Ni 5

6 Relative amount of each phase Phase diagram rule #3: Lever Rule If we know C 0, then we can tell the weight fraction of each phase at any temperature. Consider C 0 = 35 wt.%ni At T A, only L phase W L = 100 wt.% At T D, only α phase W α = 100 wt.% D At T B, both L and α phases 100 S 100 (43-35) W L = = = 73 wt.% R+S R 100 (35-32) W α = = = 27 wt.% R+S α The Lever Rule: A Proof Sum of weight fractions: W L + W α = 1 Conservation of mass of Ni: C 0 = W L C L + W α C α Combining above equations gives: W L = C α C o Cα CL = S R + S W α = C o C L C α C L = R R + S A geometric interpretation C α R S W L C 0 C L W α The moment equilibrium: W L R = W α S (1 W α ) R = W α S Solving this gives Lever Rules 6

7 To summarise: Finding the composition in a two phase region: 1. Locate composition and temperature in diagram. 2. In the two phase region, draw a tie line at the given temperature. 3. The liquid composition will be on the liquidus line and the solid composition will be on the solidus line. Note the intersections with phase boundaries. Read compositions at the intersections. Finding the amounts of phases in a two phase region: 1. Locate composition and temperature in diagram. 2. In the two phase region, draw a tie line at the given temperature. 3. Fraction of a phase is determined by taking the length of the tie line to the phase boundary for the other phase, and dividing by the total length of tie line. Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 13 Problem One kilogram of an alloy of 70% Pb and 30% Sn is slowly cooled from 300ºC. Calculate the following: 1. Chemical composition of the liquid and α phase at 225ºC. 2. Weight % of liquid and α at 225ºC. Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 14 7

8 Problem One kilogram of an alloy of 70% Pb and 30% Sn is slowly cooled from 300ºC. Calculate the following: 1. Chemical composition of the liquid and α phase at 225ºC. 2. Weight % of liquid and α at 225ºC. C L = 45 wt.% Sn C α = 17 wt.% Sn W α = W L = 100 (45 30) (30 17) = 54 wt.% = 56 wt.% Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 15 Development of Microstructures During Equilibrium Cooling The microstructural development depends on the overall composition and the cooling rate Point E α: 35 wt.%ni Point A L: 35 wt.%ni Point B α: 46 wt.%ni L: 34 wt.%ni T ( C) L A B C D 36 E L+α Point D α: 36 wt.%ni Point C L: 24 wt.%ni C α: 43 wt.%ni 0 =35 L: 32 wt.%ni α C =35 wt.% Ni 8

9 Development of Microstructures During Equilibrium Cooling All microstructures at a glance Summary of Solidification Process Solidification begins at the liquidus line and continues upon cooling The composition and amount of the solid and the liquid change gradually during cooling Solid phase grows by consuming all the liquid and solidification ends at the solidus line Development of Microstructures During Non-equilibrium Cooling Faster cooling rate cored structure a.k.a. microsegregation Slower cooling rate equilibrium structure C α changes during solidification C α composition is equals to C 0 throughout Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 18 9

10 Development of Microstructures During Non-equilibrium Cooling L L + α Wt.% Ni α Complete solidification occurs at lower T and higher Ni content than equilibrium at that T % Ni 90% Ni Solid can t freeze fast enough: solidus line effectively shifted to higher Ni concentrations. Shift increases with faster cooling rates, slower diffusion 80% Ni 70% Ni Development of Microstructures During Non-equilibrium Cooling On re-heating, grain boundaries will melt first, as they are rich in lower melting-point constituent. This can lead to premature mechanical failure! Lighter phase with low Al content Darker phase with high Al content Homogenising the cored structure Cored structure in Al-Cu alloy Microsegregation or the cored structure can be eliminated by reheating the material at a higher temperature. Diffusion of inhomogeneous element takes place. Macrosegregation, however, cannot be eliminated by reheating. Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 20 10

11 Properties of Solid Solution Alloys Alloying element causes distortion in the crystal lattice of solvent metal, which results in an increase in strength, hardness and electrical resistivity but a decrease in ductility of the solid solution alloy. This phenomenal is called Solid Solution Strengthening. Rashid, DMME, BUET MME 291, Lec 04: Interpretation of the phase diagrams P 21 Next Class MME 291: Lecture 05 Eutectic Type Phase Diagrams 11