Solidification, Phase Relations and Strengthening Mechanism

Solidification is the change of a metal from liquid to solid. It occurs in two steps: nucleation of tiny solid particles (nuclei) and growth of these into grains.

Cooling curve of a pure metal

 T
 |\
 | \
 Tm|--\___ ______  <- freezing at Tm (arrest)
 |      \/  <-- undercooling (recalescence)
 |         \____ solid cools
 +---------------------> time

A pure metal freezes at a constant temperature $T_m$. Liquid cools to slightly below $T_m$ (undercooling $\Delta T$), nuclei form, latent heat of fusion is released, temperature rises back to $T_m$ (recalescence), and stays constant until solidification is complete.

Nucleation

• Homogeneous nucleation: nuclei form in pure liquid by random clustering of atoms. A cluster of radius $r$ has free energy change $\Delta G = \tfrac43\pi r^3\Delta G_v + 4\pi r^2\gamma$. Only clusters with $r > r^*$ (critical radius) are stable. Large undercooling is needed.
• Heterogeneous nucleation: nuclei form on mould walls, impurities or added inoculants. Surface energy barrier is lower, so only a small undercooling (0.1-10 K) is needed. This is the common case in castings; grain refiners (Ti, B in aluminium) exploit it.

Grain growth and dendrites

Nuclei grow by attachment of atoms. With undercooled liquid ahead of the interface, small protrusions grow faster into the liquid than the flat front, giving a tree-like crystal called a dendrite (primary arm, secondary and tertiary arms). Arms thicken until they meet and the remaining liquid fills the gaps. Each dendrite becomes one grain, and grains meet at grain boundaries.

     |   /|\   dendrite arm
   --+--/-+-\--
     | /  |  \
     |/   |   \

Effect of cooling rate

Fast cooling gives more nuclei, hence fine grains, higher strength and toughness; slow cooling gives coarse grains.

Derivation

For a spherical nucleus of radius $r$, the total free-energy change is volume term (negative) plus surface term (positive):

$$\Delta G = \frac{4}{3}\pi r^3 \Delta G_v + 4\pi r^2 \gamma$$

where $\Delta G_v = -\dfrac{\Delta H_f\,\Delta T}{T_m}$ (per unit volume) and $\gamma$ is the solid-liquid surface energy. $\Delta G$ is maximum at $r^*$:

$$\frac{d\Delta G}{dr} = 4\pi r^2 \Delta G_v + 8\pi r\gamma = 0$$ $$r^{*} = \frac{-2\gamma}{\Delta G_v} = \frac{2\gamma T_m}{\Delta H_f\,\Delta T}$$

Substituting back gives the barrier

$$\Delta G^{*} = \frac{16\pi\gamma^3 T_m^2}{3\,\Delta H_f^2\,\Delta T^2}$$

A bigger undercooling lowers both $r^*$ and $\Delta G^*$.

Numerical

$$r^{*} = \frac{2 \times 0.177 \times 1356}{(1.628\times10^{9})(236)} = \frac{480.0}{3.842\times10^{11}} = 1.249\times10^{-9}\ \text{m} = 1.25\ \text{nm}$$ $$\Delta G^{*} = \frac{16\pi (0.177)^3 (1356)^2}{3 (1.628\times10^{9})^2 (236)^2} = 1.16\times10^{-18}\ \text{J}$$

Volume of nucleus:

$$V = \frac{4}{3}\pi (1.249)^3 = 8.17\ \text{nm}^3$$

Volume per atom in FCC copper $= \dfrac{a^3}{4} = \dfrac{(0.3615)^3}{4} = 0.0118\ \text{nm}^3$.

$$N = \frac{8.17}{0.0118} \approx 692\ \text{atoms}$$

Answer: $r^* = 1.25$ nm, $\Delta G^* = 1.16\times10^{-18}$ J, about 690 atoms in the critical nucleus.

A solid solution is a single-phase crystalline alloy in which solute atoms are dissolved in the lattice of the solvent (like sugar in water), with the solvent crystal structure retained.

Hume-Rothery rules (substitutional, extensive solubility)

1. Atomic sizes differ by less than about 15%.
2. Same crystal structure.
3. Similar electronegativity (so no compound forms).
4. Similar valency (a metal dissolves a higher-valency metal more readily than the reverse).

Strengthening

Solute atoms of different size distort the lattice and create stress fields around them. Dislocations interact with these fields, so more stress is needed to move them. Thus yield strength, tensile strength and hardness increase, while ductility and conductivity decrease. Strengthening rises with solute content and with the size mismatch. Example: 70/30 brass is stronger than pure copper; carbon in iron gives strong steel.

Gibbs phase rule

$$F = C - P + 2$$

where $F$ = degrees of freedom, $C$ = number of components, $P$ = number of phases. For a binary alloy at constant (atmospheric) pressure it becomes $F = C - P + 1$. In a two-phase region of a binary diagram, $F = 2 - 2 + 1 = 1$, so once the temperature is fixed, the compositions of both phases are fixed.

Lever rule

In a two-phase region, draw a horizontal tie line at the temperature. Its ends give the compositions of the two phases, and the mass fractions are inversely proportional to the distance of the alloy composition from each end:

$$W_{L} = \frac{C_\alpha - C_0}{C_\alpha - C_L}, \qquad W_\alpha = \frac{C_0 - C_L}{C_\alpha - C_L}$$

Calculation

Given $C_0 = 35$, $C_L = 31.5$, $C_\alpha = 42.5$ wt% Ni. Since 35 wt% Ni at 1250 °C lies between the liquidus and solidus, both liquid and $\alpha$ are present.

$$W_L = \frac{42.5 - 35}{42.5 - 31.5} = \frac{7.5}{11.0} = 0.682$$ $$W_\alpha = \frac{35 - 31.5}{42.5 - 31.5} = \frac{3.5}{11.0} = 0.318$$

Check: $0.682 + 0.318 = 1$.

Masses for 5 kg:

$$m_L = 0.682 \times 5 = 3.41\ \text{kg}, \qquad m_\alpha = 0.318 \times 5 = 1.59\ \text{kg}$$

Answer: liquid (31.5% Ni) = 68.2% = 3.41 kg; α (42.5% Ni) = 31.8% = 1.59 kg.

Cast structure (three zones)

 +---------------------------+
 | chill zone (fine, random) |
 |  +---------------------+  |
 |  | columnar zone       |  |
 |  |   +-------------+   |  |
 |  |   | equiaxed    |   |  |
 |  |   | central zone|   |  |
 |  |   +-------------+   |  |
 |  +---------------------+  |
 +---------------------------+

1. Chill zone: at the cold mould wall, rapid cooling gives many small randomly oriented grains.
2. Columnar zone: grains grow inwards, opposite to the heat flow, as long parallel dendrites.
3. Equiaxed zone: in the centre, grains nucleate in the undercooled melt and grow roughly equal in all directions.

Solidification defects

• Shrinkage cavity / pipe: metal contracts on solidifying (3-7%), leaving a cavity at the last-to-freeze region; reduced by risers.
• Porosity: gas (H$_2$ in aluminium, for example) is rejected on freezing and trapped as bubbles.
• Segregation: non-uniform composition, as in coring (dendrite cores differ from edges) and macrosegregation.
• Hot tears / cracks: restrained contraction cracks the weak, partly solid metal.
• Inclusions and cold shuts: oxide/slag particles, or incomplete fusion of two metal streams.