1. Binary phase diagrams

2. Binary phase diagrams and Gibbs free energy curves
Binary solutions with unlimited solubility
Relative proportion of phases (tie lines and the lever principle)
Development of microstructure in isomorphous alloys
Binary eutectic systems (limited solid solubility)
Solid state reactions (eutectoid, peritectoid reactions)
Binary systems with intermediate phases/compounds
The iron-carbon system (steel and cast iron)
Gibbs phase rule
Temperature dependence of solubility
Three-component (ternary) phase diagrams
Reading: Chapters – of Porter and Easterling,
Chapter 10 of Gaskell

3. Binary phase diagram and Gibbs free energy
A binary phase diagram is a temperature - composition map which indicates the equilibrium phases present at a given temperature and composition.
The equilibrium state can be found from the Gibbs free energy dependence on temperature and composition.
We have discussed the dependence of G of a one component system on T:

4. G =XAGA + XBGB + Hmix -TSmix
We have also discussed the
dependence of the Gibbs free
energy from composition at a
given T:
G =XAGA + XBGB + Hmix -TSmix

5. Binary solutions with unlimited solubility
Let’s construct a binary phase diagram for the simplest case: A and B components are mutually soluble in any amounts in both solid (isomorphous system) and liquid phases, and form ideal solutions.
We have 2 phases – liquid and solid. Let’s consider Gibbs free energy curves for the two phases at different T
T1 is above the equilibrium melting temperatures of both pure components: T1 > Tm(A) > Tm(B) → the liquid phase will be the stable phase for any composition.

7. Decreasing the temperature below T1 will have two effects:
GAliquid dan GBliquid will decrease more rapidly than GAsolid and GBsolid . WHY
The curvature of the G(XB) curves will decrease. WHY ?

8. Eventually we will reach T2 – the melting point pure component A, where GAliquid = GAsolid

9. For even lower temperature T3 < T2 = Tm(A) the Gibbs free energy curves for the liquid and solid phases will cross.
As we discussed before, the common tangent construction can be used to show that for compositions near cross-over of Gsolid and Gliquid, the total Gibbs free energy can be minimized by separation into two phases.

10. As temperature decreases below T3, GAliquid and GBliquid
continue to increase more rapidly than GAsolid and GBsolid
Therefore, the intersection of the Gibbs free energy curves, as well as points X1 and X2 are shifting to the right, until, at T4 = Tm(B) the curves will intersect at X1 = X2 = 1
At T4 and below this temperature the Gibbs free energy of the solid phase is lower than the G of the liquid phase in the whole range of compositions –the solid phase is the only stable phase.

11. Based on the Gibbs free energy curves we can now construct a phase diagram for a binary isomorphous systems

12. Example of isomorphous system: Cu-Ni (the complete solubility occurs because both Cu and Ni have the same crystal structure, FCC, similar radii, electronegativity and valence).
Liquidus line separates liquid from liquid + solid
Solidus line separates solid from liquid + solid

13. In one-component system melting occurs at a well-defined melting temperature.
In multi-component systems melting occurs over the range of temperatures, between the solidus and liquidus lines. Solid and liquid phases are in equilibrium in this temperature range.

15. Interpretation of Phase Diagrams
For a given temperature and composition we can use phase diagram to determine:
The phases that are present
Compositions of the phases
The relative fractions of the phases
Finding the composition in a two phase region:
1. Locate composition and temperature in diagram
2. In two phase region draw the tie line or isotherm
3. Note intersection with phase boundaries. Read compositions at the intersections.
The liquid and solid phases have these compositions.

16. Interpretation of Phase Diagrams: the Lever Rule
Finding the amounts of phases in a two phase region:
1. Locate composition and temperature in diagram
2. In two phase region draw the tie line or isotherm
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
The lever rule is a mechanical analogy to the mass balance calculation. The tie line in the two phase region is analogous to a lever balanced on a fulcrum.

17. Derivation of the lever rule:
All material must be in one phase or the other: Wα + WL = 1
2) Mass of a component that is present in both phases equal to the mass of the component in one phase + mass of the component in the second phase: WαCα + WβCβ = Co
3) Solution of these equations gives us the Lever rule.
Wβ = (C0 - Cα) / (Cβ- Cα) and Wα = (Cβ- C0) / (Cβ - Cα)

18. Composition/Concentration: weight fraction vs. molar fraction
Composition can be expressed in
Molar fraction, XB, or atom percent (at%) that is useful when trying to understand the material at the atomic level.
Atom percent (at %) is a number of moles (atoms) of a particular element relative to the total number of moles (atoms) in alloy.
For two -component system, concentration of element B in at. % is
Where nmA and nmB are numbers of moles of elements A and B in the system.

19. where mA and mB are the weights of the components in the system.
Weight percent (C, wt %) that is useful when making the solution. Weight percent is the weight of a particular component relative to the total alloy weight. For two component system, concentration of element B in wt. % is
where mA and mB are the weights of the components in the system.
where AA and AB are atomic
weights of elements A and B.

20. Composition Conversions
Weight % to Atomic %:
Atomic % to Weight %:

21. ML = (XBα - XB0)/(XBα - XBL) = (Catα - Cato) / (Catα- CatL)
Of course the lever rule can be formulated for any specification
of composition:
ML = (XBα - XB0)/(XBα - XBL) = (Catα - Cato) / (Catα- CatL)
Mα = (XB0 - XBL)/(XBα - XBL) = (Cat0 - CatL) / (Catα- CatL)
WL = (Cwtα - Cwto) / (Cwtα- CwtL)
WL = (Cwto - CwtL) / (Cwtα- CwtL)

22. Phase compositions and amounts. An example.
Co = 35 wt. %, CL = 31.5 wt. %, Cα = 42.5 wt. %
Mass fractions: WL = S / (R+S) = (Cα - Co) / (Cα- CL) = 0.68
Wα = R / (R+S) = (Co - CL) / (Cα- CL) = 0.32

23. Development of microstructure in isomorphous alloys Equilibrium (very slow) cooling

24. Development of microstructure in isomorphous alloys Equilibrium (very slow) cooling
Solidification in the solid + liquid phase occurs gradually upon cooling from the liquidus line.
The composition of the solid and the liquid change gradually during cooling (as can be determined by the tie-line method.)
Nuclei of the solid phase form and they grow to consume all the liquid at the solidus line.

25. Development of microstructure in isomorphous alloys
Non-equilibrium cooling

26. Development of microstructure in isomorphous alloys Non-equilibrium cooling
Compositional changes require diffusion in solid and liquid phases
Diffusion in the solid state is very slow.  The new layers that solidify on top of the existing grains have the equilibrium composition at that temperature but once they are solid their composition does not change.  Formation of layered (cored) grains and the invalidity of the tie-line method to determine the composition of the solid phase.
The tie-line method still works for the liquid phase, where diffusion is fast. Average Ni content of solid grains is higher.  Application of the lever rule gives us a greater proportion of liquid phase as compared to the one for equilibrium cooling at the same T.  Solidus line is shifted to the right (higher Ni contents), solidification is complete at lower T, the outer part of the grains are richer in the low-melting component (Cu).
Upon heating grain boundaries will melt first. This can lead to premature mechanical failure.

27. Binary solutions with a miscibility gap
Let’s consider a system in which the liquid phase is approximately ideal, but for the solid phase we have ∆Hmix > 0 ( A and B atoms dislike each other)
At low temperatures, there is a region where the solid solution is most stable as a mixture of two phases α1 and α2 with compositions X1 and X2. This region is called a miscibility gap.

28. Eutectic phase diagram
For an even larger ∆Hmix the miscibility gap can extend into the liquid phase region. In this case we have eutectic phase diagram.

29. Eutectic phase diagram with different crystal structures of pure phases
A similar eutectic phase diagram can result if pure A and B have
different crystal structures.

30. Eutectic systems - alloys with limited solubility (I)
Three single phase regions (α - solid solution of Ag in Cu matrix, β = solid solution of Cu in Ag matrix, L - liquid)
Three two-phase regions (α + L, β +L, α +β)
Solvus line separates one solid solution from a mixture of solid
solutions. Solvus line shows limit of solubility

31. Eutectic systems - alloys with limited solubility (II)
Eutectic or invariant point - Liquid and two solid phases coexist in equilibrium at the eutectic composition CE and the eutectic temperature TE.
Eutectic isotherm - the horizontal solidus line at TE.
Eutectic reaction – transition between liquid and mixture of two
solid phases, α + β at eutectic concentration CE.
The melting point of the eutectic alloy is lower than that of the components (eutectic = easy to melt in Greek).

32. Eutectic systems - alloys with limited solubility (III)
Compositions and relative amounts of phases are determined
from the same tie lines and lever rule, as for isomorphous alloys
For points A, B, and C calculate the compositions (wt. %) and relative amounts (mass fractions) of phases present.

33. L → α+L → α Development of microstructure in eutectic alloys (I)
Several different types of microstructure can be formed in slow cooling an different compositions. Let’s consider cooling of liquid lead – tin system as an example.
In the case of lead-rich alloy (0-2 wt. % of tin) solidification proceeds in the same manner as for isomorphous alloys (e.g. Cu-Ni) that we discussed earlier.
L → α+L → α

34. Development of microstructure in eutectic alloys (II)
At compositions between the room temperature solubility limit and the maximum solid solubility at the eutectic temperature, β phase nucleates as the α solid solubility is exceeded upon crossing the solvus line.

35. Development of microstructure in eutectic alloys (III)
Solidification at the eutectic composition
No changes above the eutectic temperature TE. At TE all the
liquid transforms to α and β phases (eutectic reaction).

36. Development of microstructure in eutectic alloys (IV)
Solidification at the eutectic composition
Compositions of α and β phases are very different → eutectic reaction involves redistribution of Pb and Sn atoms by atomic diffusion This simultaneous formation of α and β phases result in a layered (lamellar) microstructure that is called eutectic
structure.
Formation of the eutectic structure in the lead-tin system. In the micrograph, the dark layers are lead reach α phase, the light layers are the tin-reach β phase.

37. Development of microstructure in eutectic alloys (V)
Compositions other than eutectic but within the range of the eutectic isotherm
Primary α phase is formed in the α + L region, and the eutectic structure that includes layers of α and β phases (called eutectic α and eutectic β phases) is formed upon crossing the eutectic
isotherm.

38. Development of microstructure in eutectic alloys (VI)
Microconstituent – element of the microstructure having a
distinctive structure. In the case described in the previous page,
microstructure consists of two microconstituents, primary α
phase and the eutectic structure.
Although the eutectic structure consists of two phases, it is a microconstituent with distinct lamellar structure and fixed ratioof the two phases.

39. How to calculate relative amounts of microconstituents?
Eutectic microconstituent forms from liquid having eutectic composition (61.9 wt% Sn)
We can treat the eutectic as a separate phase and apply the lever rule to find the relative fractions of primary α phase (18.3 wt% Sn) and the eutectic structure (61.9 wt% Sn):
We = P / (P+Q) (eutectic)
Wα’ = Q / (P+Q) (primary)

40. How to calculate the total amount of α phase (both eutectic and primary)?
Fraction of α phase determined by application of the lever rule
across the entire α + β phase field:
Wα = (Q+R) / (P+Q+R) (α phase)
Wβ = P / (P+Q+R) (β phase)

41. Binary solutions with ∆Hmix < 0 - ordering
If ∆Hmix < 0 bonding becomes stronger upon mixing → melting point of the mixture will be higher than the ones of the pure
components.
For the solid phase strong interaction between
unlike atoms can lead to (partial) ordering → |∆Hmix| canbecome larger than |ΩXAXB| and the Gibbs free energy curve for the solid phase can become steeper than the one for liquid.
At low temperatures, strong attraction between unlike atoms can lead to the formation ofordered phase α`.

42. Binary solutions with ∆Hmix < 0 - intermediate phases
If attraction between unlike atoms is very strong, the ordered
phase may extend up to the liquid.
In simple eutectic systems, discussed above, there are only two
solid phases (α and β) that exist near the ends of phase
diagrams.
Phases that are separated from the composition extremes (0%
and 100%) are called intermediate phases. They can have
crystal structure different from structures of components A and B.

43. ∆Hmix<0 - tendency to form high-melting point intermediate phase

44. Phase diagrams with intermediate phases: example
Example of intermediate solid solution phases: in Cu-Zn, α and
η are terminal solid solutions, β, β’, γ, δ, ε are intermediate
solid solutions.

45. lPhase diagrams for systems containing compounds
For some sytems, instead of an intermediate phase, an
intermetallic compound of specific composition forms.
Compound is represented on the phase diagram as a vertical line,
since the composition is a specific value.
When using the lever rule, compound is treated like any other
phase, except they appear not as a wide region but as a vertical
line

46. This diagram can be thought of as two joined eutectic diagrams,
for Mg-Mg2Pb and Mg2Pb-Pb. In this case compound Mg2Pb(19%wt Mg and 81%wt Pb) can be considered as a component.
A sharp drop in the Gibbs free energy at the compound composition should be added to Gibbs free energy curves for the existing phases in the system

47. Eutectoid Reactions
The eutectoid (eutectic-like in Greek) reaction is similar to the eutectic reaction but occurs from one solid phase to two new solid phases.
Eutectoid structures are similar to eutectic structures but are much finer in scale (diffusion is much slower in the solid state).
Upon cooling, a solid phase transforms into two other solid phases (δ ↔ γ + ε in the example below)
Looks as V on top of a horizontal tie line (eutectoid isotherm) in
the phase diagram.

48. Eutectic and Eutectoid Reactions
The above phase diagram contains both an eutectic reaction and its solid-state analog, an eutectoid reaction

49. Peritectic Reactions
A peritectic reaction - solid phase and liquid phase will together form a second solid phase at a particular temperature and composition upon cooling - e.g. L + α ↔ β
These reactions are rather slow as the product phase will form at the boundary between the two reacting phases thus separating them, and slowing down any further reaction.
Peritectoid is a three-phase reaction similar to peritectic but occurs from two solid phases to one new solid phase (α + β = γ).

50. Example: The Iron–Iron Carbide (Fe–Fe3C) Phase Diagram
In their simplest form, steels are alloys of Iron (Fe) and Carbon (C). The Fe-C phase diagram is a fairly complex one, but we will only consider the steel part of the diagram, up to around 7% Carbon.

51. α-ferrite - solid solution of C in BCC Fe
Phases in Fe–Fe3C Phase Diagram
α-ferrite - solid solution of C in BCC Fe
• Stable form of iron at room temperature.
• The maximum solubility of C is wt%
• Transforms to FCC γ-austenite at 912 °C
γ-austenite - solid solution of C in FCC Fe
• The maximum solubility of C is 2.14 wt %.
• Transforms to BCC δ-ferrite at 1395 °C
• Is not stable below the eutectic temperature (727oC unless cooled rapidly
δ-ferrite solid solution of C in BCC Fe
• The same structure as α-ferrite
• Stable only at high T, above 1394 °C
• Melts at 1538 °C

52. Fe3C (iron carbide or cementite)
• This intermetallic compound is metastable, it remains
as a compound indefinitely at room T, but decomposes
(very slowly, within several years) into α-Fe and C
(graphite) at °C
Fe-C liquid solution

53. A few comments on Fe–Fe3C system
C is an interstitial impurity in Fe. It forms a solid solution with
α, γ, δ phases of iron
Maximum solubility in BCC α-ferrite is limited (max wt% at 727 °C) - BCC has relatively small interstitial positions
Maximum solubility in FCC austenite is 2.14 wt% at 1147 °C - FCC has larger interstitial positions
Mechanical properties: Cementite is very hard and brittle - can
strengthen steels. Mechanical properties also depend on the
microstructure, that is, how ferrite and cementite are mixed.
Magnetic properties: α -ferrite is magnetic below 768 °C,
austenite is non-magnetic

54. Classification. Three types of ferrous alloys:
• Iron: less than wt % C in α−ferrite at room T
• steels: wt % C (usually < 1 wt % ) α-ferrite + Fe3C at room T
• Cast iron: wt % (usually < 4.5 wt %)

55. Eutectic and eutectoid reactions in Fe–Fe3C
Eutectic: 4.30 wt% C, 1147 °C
Eutectic and eutectoid reactions are very important in heat treatment of steels

56. Development of Microstructure in Iron - Carbon alloys
Microstructure depends on composition (carbon content) and heat treatment. In the discussion below we consider slow cooling in which equilibrium is maintained.
Microstructure of eutectoid steel (II)
When alloy of eutectoid composition (0.76 wt % C) is cooled slowly it forms pearlite, a lamellar or layered structure of two phases: α-ferrite and cementite (Fe3C)

57. The layers of alternating phases in pearlite are formed for the same reason as layered structure of eutectic structures:
redistribution C atoms between ferrite (0.022 wt%) and cementite (6.7 wt%) by atomic diffusion.
Mechanically, pearlite has properties intermediate to soft, ductile
ferrite and hard, brittle cementite.
In the micrograph, the dark areas areFe3C layers, the light phase is α-ferrite

58. Microstructure of hypoeutectoid steel (I)
Compositions to the left of eutectoid ( wt % C) hypoeutectoid (less than eutectoid -Greek) alloys.

59. Microstructure of hypoeutectoid steel (II)
Hypoeutectoid alloys contain proeutectoid ferrite (formed above the eutectoid temperature) plus the eutectoid perlite that contain eutectoid ferrite and cementite.

60. Microstructure of hypereutectoid steel (I)
Compositions to the right of eutectoid ( wt % C)
hypereutectoid (more than eutectoid -Greek) alloys.

61. Microstructure of hypereutectoid steel

62. Microstructure of hypereutectoid steel (II)
Hypereutectoid alloys contain proeutectoid cementite (formed above the eutectoid temperature) plus pearlite that contain eutectoid ferrite and cementite.

63. How to calculate the relative amounts of proeutectoid phase (α or Fe3C) and pearlite?
Application of the lever rule with tie line that extends from the eutectoid composition (0.75 wt% C) to α – (α + Fe3C) boundary (0.022 wt% C) for hypoeutectoid alloys and to (α + Fe3C) – Fe3C boundary (6.7 wt% C) for hypereutectoid alloys.
Fraction of α phase is determined by application of the lever rule across the entire (α + Fe3C) phase field:

64. Example for hypereutectoid alloy with composition C1
Fraction of pearlite:
WP = X / (V+X) = (6.7 – C1) / (6.7 – 0.76)
Fraction of proeutectoid cementite:
WFe3C = V / (V+X) = (C1 – 0.76) / (6.7 – 0.76)

66. The Gibbs phase rule (I)
Let’s consider a simple one-component system.
In the areas where only one phase
is stable both pressure and
temperature can be independently
varied without upsetting the
equilibrium → there are 2 degrees
of freedom.
Along the lines where two phases coexist in equilibrium, only one variable can be independently varied without upsetting the
two-phase equilibrium (P and T are related by the Clapeyron equation) → there is only one degree of freedom.
At the triple point, where solid liquid and vapor coexist any change in P or T would upset the three-phase equilibrium →
there are no degrees of freedom.

67. In general, the number of degrees of freedom, F, in a system that contains C components and can have Ph phases is given by the Gibbs phase rule:
F = C - Ph + 2
Let’s now consider a multi-component system containing C components and having Ph phases
A thermodynamic state of each phase can be described by pressure P, temperature T, and C - 1 composition variables. The state of the system can be then described by Ph×(C-1+2) variables

68. But how many of them are independent?
The condition for Ph phases to be at equilibrium are:
T = T = T = ………….. Ph -1 equations
P = P = P = ………….. Ph -1 equations
A = A = A = ………….. Ph -1 equations
C sets of equation
B= B = B = ………….. Ph -1 equations
Therefore we have (Ph – 1)×(C + 2) equations that connect the variables in the system.

69. The number of degrees of freedom is the difference between the total number of variables in the system and the minimum number of equations among these variables that have to be satisfied in order to maintain the equilibrium.
F = Ph(C+1) – (Ph-1)(C+2) = C – Ph+2
F = C – Ph+2
Gibbs Phase Rule

70. The Gibbs phase rule – example (an eutectic systems)
F = C – Ph +2
P = constant
F = C – Ph +1
C = 2
F = 3 – Ph
In one-phase regions of the phase diagram T and XB can be changed independently.
In two-phase regions, F = 1. If the temperature is chosen independently, the compositions of both phases are fixed.
Three phases (L, α, β) may be in equilibrium only at a few points along the eutectic isotherm (F = 0).

71. Temperature dependence of solubility
Let’s consider a regular binary solution with limited solubilities of A in B and B in A.
The closer is the minimum of the Gibbs free energy curve Gα(XB) to the axes XB = 0, the smaller is the maximum possible concentration of B in phase α. Therefore, to discuss the temperature dependence of solubility let’s find the minimum of Gα(XB).

72. Greg = XAGA + XBGB + ΩXAXB + RT[XAlnXA + XBlnXB]
minimum of G(XB)
Greg = (1-XB)GA + XBGB + Ω(1-XB)XB + RT[(1-XB) ln(1-XB) + XBlnXB]
Solid solubility of B in α increases exponentially with temperature

73. Multicomponent systems (I)
The approach used for analysis of binary systems can be extended to multi component systems.
Representation of the composition in a ternary system (the Gibbs triangle). The total length of the red lines is 100% :
XA + XB + XC =1

74. The Gibbs free energy surfaces (instead of curves for a binary system) can be plotted for all the possible phases and for different temperatures.
The chemical potentials of A, B, and C of any phase in thissystem are given by the points where the tangential plane to the free energy surfaces intersects the A, B, and C axis.

75. A three-phase equilibrium in the ternary system for a given temperature can be derived by means of the tangential planeconstruction.
For two phases to be in equilibrium, the chemical potentials should be equal, that is the compositions of the two phases in equilibrium must be given by points connected by a common tangential plane (e.g. l and m).

76. The relative amounts of phases are given by the lever rule (e. g
The relative amounts of phases are given by the lever rule (e.g.using tie-line l-m).
A three phase triangle can result from a common tangential plane simultaneously touching the Gibbs free energies of three phases e.g. points x, y, and z).
Eutectic point: four-phase equilibrium between α, β, γ, and liquid

77. An example of ternary system
The ternary diagram of Ni-Cr-Fe. It includes Stainless Steel (wt.% of Cr > 11.5 %, wt.% of Fe > 50 %) and Inconeltm (Nickel based super alloys).
Inconel have very good corrosion resistance, but are more expensive and therefore used in corrosive environments where Stainless Steels are not sufficient (Piping onNuclear Reactors or Steam Generators).

78. Another example of ternary phase diagram:oil – water – surfactant system
Surfactants are surface-active molecules that can form interfaces between immiscible fluids (such as oil and water).
A large number of structurally different phases can be formed, such as droplet, rod-like, and bicontinuous microemulsions, along with hexagonal, lamellar, and cubic liquid crystalline phases.

79. Summary Make sure you understand language and concepts:
􀂾 Common tangent construction
􀂾 Separation into 2 phases
􀂾 Eutectic structure
􀂾 Composition of phases
􀂾 Weight and atom percent
􀂾 Miscibility gap
􀂾 Solubility dependence on T
􀂾 Intermediate solid solution
􀂾 Compound
􀂾 Isomorphous
􀂾 Tie line, Lever rule
􀂾 Liquidus & Solidus lines
􀂾 Microconstituent
􀂾 Primary phase
􀂾 Solvus line, Solubility limit
􀂾 Austenite, Cementite, Ferrite
􀂾 Pearlite
􀂾 Hypereutectoid alloy
􀂾 Hypoeutectoid alloy
􀂾 Ternery alloys
􀂾 Gibbs phase rule

80. Elements of phase diagrams:

81. END…….