Interactive Lecture
Understanding Material Phases and Transformations
by Dr Arnaldo Delli Carri
Phase diagrams are graphical representations that show the relationships between temperature, composition, and the phases present in a material system at equilibrium. They let you predict which phases form during processing and service.
A phase is a physically distinct and chemically homogeneous portion of a material system that has uniform physical and chemical characteristics.
The Gibbs Phase Rule determines the number of degrees of freedom in a system:
\(F = C - P + 2\)
Where:
For constant pressure (isobaric systems): \(F = C - P + 1\)
Binary phase diagrams show the phase relationships for two-component systems:
The Lever Rule calculates the relative amounts of phases in a two-phase region:
\(W_\alpha = \frac{C_0 - C_L}{C_\alpha - C_L} \times 100\%\)
\(W_L = \frac{C_\alpha - C_0}{C_\alpha - C_L} \times 100\%\)
Where:
The same rule works in any two-phase region — swap the liquid subscript for the second solid phase. In an \(\alpha + \beta\) field, use \(C_\beta\) and \(W_\beta\) in place of \(C_L\) and \(W_L\):
\(W_\alpha = \frac{C_\beta - C_0}{C_\beta - C_\alpha} \times 100\%, \qquad W_\beta = \frac{C_0 - C_\alpha}{C_\beta - C_\alpha} \times 100\%\)
Explore a binary phase diagram! Click or drag directly on the diagram to set temperature and composition, or use the sliders below. In two-phase regions, the tie-line and lever rule fractions are shown automatically.
Watch how microstructure evolves during cooling. Set the alloy composition, then click Start Cooling to see realistic grain formation with Voronoi grain boundaries and lamellar eutectic patterns.
Drag the three points along the composition axis to explore how the lever rule works as a balance beam. The fulcrum is at the overall composition C₀, and the beam tilts to show relative phase amounts.
Phase diagrams assume equilibrium conditions, but real processes involve non-equilibrium cooling. TTT (Time-Temperature-Transformation) and CCT (Continuous-Cooling-Transformation) diagrams bridge theory and the shop floor — they let you predict what microstructure you'll actually get.
| Concept | Equation | Description |
|---|---|---|
| Gibbs Phase Rule | \(F = C - P + 2\) | Degrees of freedom in phase equilibrium |
| Isobaric Phase Rule | \(F = C - P + 1\) | At constant pressure |
| Lever Rule (α phase) | \(W_\alpha = \frac{C_0 - C_L}{C_\alpha - C_L}\) | Weight fraction of solid phase |
| Lever Rule (L phase) | \(W_L = \frac{C_\alpha - C_0}{C_\alpha - C_L}\) | Weight fraction of liquid phase |
| Phase Conservation | \(W_\alpha + W_L = 1\) | Total mass conservation |
A Cu-Ni alloy with 35 wt% Ni is at 1250°C in a two-phase region. At this temperature, the liquidus composition is 32 wt% Ni and the solidus composition is 43 wt% Ni. Calculate:
Show Solution
Given: \(C_0 = 35\) wt% Ni, \(C_L = 32\) wt% Ni, \(C_\alpha = 43\) wt% Ni
a) Weight fraction of liquid:
\(W_L = \displaystyle \frac{C_\alpha - C_0}{C_\alpha - C_L} = \frac{43 - 35}{43 - 32} = \frac{8}{11} = 0.727\) (72.7%)
b) Weight fraction of solid:
\(W_\alpha = \displaystyle \frac{C_0 - C_L}{C_\alpha - C_L} = \frac{35 - 32}{43 - 32} = \frac{3}{11} = 0.273\) (27.3%)
Check: \(W_L + W_\alpha = 0.727 + 0.273 = 1.0\) ✓
Answer: \(W_L = 72.7\%\), \(W_\alpha = 27.3\%\).
For a binary alloy system (2 components) at constant pressure:
Show Solution
Using \(F = C - P + 1\) (constant pressure)
a) Single-phase region:
\(F = 2 - 1 + 1 = 2\) degrees of freedom
Both temperature and composition can be varied independently.
b) Eutectic point (3 phases):
\(F = 2 - 3 + 1 = 0\) degrees of freedom
The eutectic point has fixed temperature and composition.
c) Practical meaning:
In single-phase regions, you can control both T and composition. At the eutectic point, the system is invariant - you cannot change T or composition without losing one of the three phases.
Answer: a) \(F = 2\); b) \(F = 0\); c) single-phase regions allow independent control of both T and composition, whereas the eutectic point is invariant (T and composition are both fixed).
A Pb-Sn alloy with 40 wt% Sn is slowly cooled from 250°C to room temperature. The eutectic composition is 61.9 wt% Sn at 183°C. Describe the microstructure that forms and estimate the relative amounts of phases at room temperature. (Assume maximum solubilities: α ≈ 19 wt% Sn, β ≈ 97 wt% Sn)
Show Solution
Cooling sequence:
1. Above liquidus: Single liquid phase
2. Entering α + L region: Primary α grains start forming
3. At 183°C (eutectic temperature): Remaining liquid solidifies as eutectic (α + β mixture)
4. Below eutectic: Two-phase α + β microstructure
Final microstructure:
Primary α grains surrounded by eutectic (lamellar α + β)
Phase amounts at room temperature (using lever rule):
Given: \(C_0 = 40\) wt% Sn, \(C_\alpha \approx 19\) wt% Sn, \(C_\beta \approx 97\) wt% Sn
\(W_\alpha = \displaystyle \frac{C_\beta - C_0}{C_\beta - C_\alpha} = \frac{97 - 40}{97 - 19} = \frac{57}{78} = 0.731\) (73.1% α phase)
\(W_\beta = \displaystyle \frac{C_0 - C_\alpha}{C_\beta - C_\alpha} = \frac{40 - 19}{97 - 19} = \frac{21}{78} = 0.269\) (26.9% β phase)
Answer: Primary α grains surrounded by lamellar eutectic (α + β); at room temperature \(W_\alpha \approx 73.1\%\) and \(W_\beta \approx 26.9\%\).