Materials Science · Processing routes and microstructural evolution

Solidification, nucleation and growth

Useful background: Processing, structure and useful properties; Energy, stability and environmental context. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Distinguish nucleation, growth and solute redistribution during freezing.
Use a stated surface–volume model to distinguish a critical nucleus from the zero of its formation energy.
Track heat removal and solute conservation through a finite freezing sequence.
Distinguish dendrite spacing, grain boundaries and composition evidence when comparing processing histories.

A new solid must first appear

Solidification involves creating stable solid regions and growing them into liquid. Interfaces carry an energy cost, so a favourable bulk free-energy change does not eliminate every nucleation barrier. Existing surfaces can assist nucleation.

Nucleation is the appearance of a region of the new phase that can persist and grow under the stated conditions. Growth is subsequent movement of its solid–liquid interface as atoms join the solid. An embryo can form and disappear before it becomes a successful nucleus. A liquid below its equilibrium freezing temperature can therefore remain liquid for an observation interval: a bulk preference for solid is not a promise of immediate transformation.

Homogeneous nucleation begins within the liquid without a special foreign surface. Heterogeneous nucleation uses a site such as a mould wall or particle. The new solid replaces some existing interfaces as well as creating others; favourable wetting can reduce the formation barrier. “Any added particle guarantees more grains” does not follow: its interfacial chemistry, geometry, survival in the melt and available sites matter. Growth from an existing crystal also differs from creating an independent new nucleus.

Small embryos pay a large interface cost per volume

Consider an original idealised spherical solid embryo of radius r in a large uniform liquid at fixed temperature and pressure. Let γ be the positive solid–liquid interfacial free energy in joules per square metre, J/m². Let Δg be the positive magnitude of the bulk free-energy advantage of solid over liquid in joules per cubic metre, J/m³. Thus forming solid lowers the bulk contribution by Δg times its volume. ΔG is the total formation free-energy change in joules relative to liquid without the embryo.

The sphere has area 4πr² and volume (4π/3)r³, giving ΔG(r) = 4πγr² − (4π/3)Δg r³. The interface penalty grows with r²; the bulk benefit grows with r³. Their ratio, interface penalty divided by bulk-benefit magnitude, is 3γ/(Δg r). Small embryos have relatively more interface. This continuum model assumes constant isotropic γ and Δg, a spherical shape and negligible elastic strain energy; it does not resolve individual atoms, a wall contact or local depletion of the liquid.

The critical radius r* = 2γ/Δg is a maximum of ΔG. One way to find it is to set the slope dΔG/dr = 4πr(2γ − Δg r) to zero for r > 0. Below r*, shrinking lowers the formation energy; above r*, further growth lowers it, provided atoms and heat can be transported. At r* itself the ideal embryo is unstable to changes in size, rather than sitting in a stable minimum. The formation barrier is ΔG* = 16πγ³/(3Δg²).

The nonzero crossing ΔG = 0 occurs later, at r = 3γ/Δg = 1.5r*. A slightly supercritical embryo can already lower its energy by growing even while its total ΔG remains positive. The sign of ΔG alone therefore does not determine the local growth direction. Thermal fluctuations and transport determine whether embryos form and cross the barrier; this static curve supplies neither a waiting time nor a nucleation rate.

For the chosen γ = 0.100 J/m² and Δg = 2.00 × 10⁷ J/m³, r* = 10.0 nm and ΔG* = 4.19 × 10⁻¹⁷ J. Here 1 nm = 10⁻⁹ m. At 5.00 nm the energy is 2.09 × 10⁻¹⁷ J; at 12.0 nm it is 3.62 × 10⁻¹⁷ J but is falling with increasing radius. The zero crossing is 15.0 nm. These supplied constants illustrate the distinction; they are not measured values for a named alloy.

Increasing undercooling, the difference between the relevant equilibrium freezing temperature and the liquid temperature, can increase the bulk driving force. However, atomic attachment and transport also change with temperature. A smaller calculated barrier by itself does not establish faster overall freezing, more final grains or better component performance. The equilibrium and kinetics lessons explain why thermodynamic preference and observed rate need separate evidence.

Original spherical embryo model with interfacial free energy 0.100 joule per square metre and bulk driving-force magnitude 20.0 million joules per cubic metre. Radius spans zero to 18 nanometres. Total formation energy rises to its maximum of 4.19 times ten to the minus 17 joule at radius 10.0 nanometres, then falls. It crosses zero at 15.0 nanometres. The square marks the critical maximum; the circle marks the later zero. The curve is formation energy versus radius, not radius versus time or a nucleation-rate prediction.

Growth establishes structure

Heat removal and solute transport influence interface motion, grain shape and segregation. Rapid growth can preserve composition gradients or nonequilibrium structures. A final casting therefore depends on cooling path and geometry as well as overall composition.

A concrete sequence is: a viable solid region appears; its interface advances; solidification releases latent heat and may partition the alloying elements; growing regions meet or run out of liquid. Latent heat is the energy released by the phase change itself, distinct from sensible heat released as temperature falls. Heat can leave through the mould and surrounding material, or first enter cooler liquid nearby. The interface temperature is not automatically the chamber or cooling-fluid setting.

In an alloy, the solid forming at an interface need not have the liquid’s composition. If that solid incorporates less of an element than the liquid contains, the remaining liquid becomes enriched. If mixing in the liquid and redistribution in the solid are too slow to eliminate gradients during the actual schedule, early and late solid regions can retain different compositions. This is segregation. A closed casting can conserve its overall composition while becoming locally nonuniform.

The interface need not remain flat. A small protrusion changes the surrounding heat and solute transport; under suitable conditions it grows preferentially and branches into a dendrite. Dendrite means a tree-like growth shape, not a separate phase. Several arms can belong to one crystal orientation and hence one grain. Grain boundaries instead separate differently oriented crystals. Counting visible arms is not a grain count.

Competition between available nuclei, growth directions and spatial thermal conditions can produce elongated columnar grains or more equiaxed grains, with similar extents in different directions. The words describe shape, not a universal strength ranking. Dendrite spacing, grain size, phase identity and segregation are distinct observations. An image with fine branches alone neither measures all four nor identifies a unique cooling history. The retained branched sketch below is a transport reminder, not a concentration map or a predicted grain shape.

A branched schematic solid grows into liquid. Growing interfaces and transport can leave spatial variation; the sketch is not a predicted dendrite or concentration map.

Conserve the element through successive increments

Use the following original finite bookkeeping example, not a phase diagram or an alloy recipe. A closed 100 g liquid initially contains 10.0 mass% B: 10.0 g of element B and 90.0 g of the other element. “Mass% B” means 100 times the mass of B divided by the total mass of that particular portion, not the fraction of the casting that is solid.

The first 20.0 g of solid is supplied to have an average of 5.00 mass% B. It contains 1.00 g B. The remaining 80.0 g liquid contains 9.00 g B, so its average composition is 11.25 mass% B. Next, another 30.0 g of solid forms with a supplied average of 6.00 mass% B. That increment takes 1.80 g B; 50.0 g liquid remains with 7.20 g B, or 14.4 mass% B.

The two solid increments together contain 2.80 g B in 50.0 g, so their combined average is 5.60 mass% B. They are not each 5.60%. If all remaining liquid freezes without element loss, its 50.0 g contributes 7.20 g B to the final solid, whose casting-wide average is still 10.0%. A bulk assay cannot distinguish these early and late compositions.

Assume the liquid mixes between increments, earlier solid does not exchange B with it further during this account, and each supplied composition is the mass-average of that entire solid increment. The account uses conservation to find remaining composition; it does not predict the supplied increments from a phase boundary or a constant partition coefficient. Actual solid can partly homogenise during or after freezing, and flow or loss of material can change the boundary. The diagram’s widths encode masses, not spatial distances or liquid volumes.

Original stipulated closed mass account. Each solid composition is the average over its whole increment. Percentages use the indicated portion’s mass; all B masses sum to 10.0 g.
Stage	Solid increments already formed	Remaining liquid	B in remaining liquid	Liquid B (mass%)
Start	None	100 g	10.0 g	10.0
After increment 1	20.0 g at 5.00% B	80.0 g	9.00 g	11.25
After increment 2	Plus 30.0 g at 6.00% B	50.0 g	7.20 g	14.4

Original closed 100 gram charge starts with 10.0 grams B, or 10.0 mass percent B. The first row is all liquid. After the first increment, 20.0 grams solid contains 1.00 gram B and 80.0 grams liquid contains 9.00 grams B, giving 11.25 mass percent B in the liquid. After the second increment, the first 20.0 gram solid portion remains, a second 30.0 gram solid portion contains 1.80 grams B, and 50.0 grams liquid contains 7.20 grams B, giving 14.4 mass percent B in the liquid. All three bars total 100 grams. Width represents portion mass, not volume, concentration or position in a casting. Solid 1 uses diagonal hatching, solid 2 uses dots, and liquid uses a plain fill.

Compare a defined freezing interval, then examine the structure

For a separate original heat account, take a 0.100 kg charge that fully freezes and decreases in temperature by 40.0 K over a defined interval. Supply a constant common specific heat capacity c = 500 J/(kg·K) for the sensible part and latent heat of solidification L = 240 kJ/kg. Neglect other phase changes, mechanical work, reactions and additional stored-energy changes. These simplifications make Q_out = mcΔT_drop + mL a useful total heat-removal account. Here m is charge mass in kilograms, ΔT_drop is the positive temperature decrease in kelvin, and Q_out is the positive heat energy leaving the charge in joules. The account does not require that freezing occur at one temperature or in a particular order.

Sensible heat is 2.00 kJ and latent heat is 24.0 kJ, so the total is 26.0 kJ. Removing this same amount during 40.0 s gives mean outward heat flow 650 W; during 10.0 s it gives 2.60 kW. Average power is total energy divided by interval duration. Neither value says that the instantaneous heat flow, interface speed or nucleation rate was constant. A temperature plateau or transient warming can accompany latent-heat release, depending on heat flow and measurement location.

To interpret a difference between two castings, record starting composition and melt history, geometry and position, relevant particles or surfaces, actual local temperature histories and the time interval. Examine orientation/grain boundaries separately from branch spacing, and measure early/late composition at stated spatial resolution. For example, smaller branch spacing plus a changed local cooling record supports a thermal-processing association; attributing it solely to more nuclei requires additional evidence about grains and sites.

Even a well-characterised final microstructure may not reveal one unique processing path. Flow, gradients, initial surfaces, nucleation and later redistribution can produce overlapping observations. Use changed-condition comparisons to test explanations, rather than treating a faster chamber setting or uniform bulk chemistry as a complete description. Solid-state diffusion and deformation-driven changes are the next lessons; they are not determined by the spherical nucleation curve or this mass ledger.

Worked examples

Work through the case

A casting has uniform overall chemistry but local compositional differences. Can that arise during freezing?

Solid and liquid may favour different compositions.
Growing solid redistributes solute.
Limited diffusion can preserve local segregation.

Yes. Overall composition does not imply local uniformity.

Positive formation energy can coexist with a growth tendency

For γ = 0.100 J/m² and Δg = 2.00 × 10⁷ J/m³, compare embryos of radius 5.00, 10.0 and 12.0 nm. Must the 12.0 nm embryo shrink because ΔG is positive?

Use metres: r* = 2(0.100)/(2.00 × 10⁷) = 1.00 × 10⁻⁸ m = 10.0 nm. The nonzero energy crossing is 1.5r* = 15.0 nm.
At 5.00 nm, 4πγr² = 3.14 × 10⁻¹⁷ J and (4π/3)Δg r³ = 1.05 × 10⁻¹⁷ J; subtracting gives 2.09 × 10⁻¹⁷ J. It lies below the critical radius, so reducing its radius lowers ΔG in the model.
At 10.0 nm the total is the maximum, 4.19 × 10⁻¹⁷ J. At 12.0 nm the total has fallen to 3.62 × 10⁻¹⁷ J. The slope factor 2γ − Δg r is then 0.200 − 0.240 = −0.0400 J/m².
The negative slope means increasing radius lowers ΔG, even before ΔG becomes zero. Availability of transport and fluctuations still matters; the calculation is a local energetic tendency, not a deterministic waiting-time prediction.

No. The 12.0 nm embryo is supercritical in this model. Its growth tendency follows the slope, while its positive total formation energy reflects the earlier barrier.

An enriched last liquid without changing the bulk chemistry

After the two supplied solid increments, what are the remaining liquid composition and the combined earlier-solid composition? What if the remaining liquid then freezes completely without loss?

Initial B inventory is (0.100)(100 g) = 10.0 g. The first solid removes (0.0500)(20.0 g) = 1.00 g B; the second removes (0.0600)(30.0 g) = 1.80 g B.
Liquid mass is 100 − 20 − 30 = 50.0 g. Its B mass is 10.0 − 1.00 − 1.80 = 7.20 g, so its composition is 100(7.20/50.0) = 14.4 mass% B.
Earlier solid contains 2.80 g B in 50.0 g, giving 5.60 mass% B. This is an average of the distinct 5.00% and 6.00% increments.
Complete loss-free freezing gives 2.80 + 7.20 = 10.0 g B in 100 g solid. Local composition can vary even though the final bulk composition remains 10.0 mass% B.

The remaining liquid averages 14.4 mass% B; the earlier solid averages 5.60 mass% B. Full loss-free freezing preserves the bulk 10.0 mass% B, not local uniformity.

A shorter interval changes average heat flow, not a measured grain count

For the stated 0.100 kg heat account, compare 40.0 s and 10.0 s intervals. Does four times the mean heat flow imply four times as many grains?

Convert L = 240 kJ/kg to 240,000 J/kg. Sensible heat is (0.100)(500)(40.0) = 2,000 J; latent heat is (0.100)(240,000) = 24,000 J.
The total is 26,000 J. Dividing by 40.0 s gives 650 W; dividing by 10.0 s gives 2,600 W. The latter mean is four times as large.
No equation in this account relates heat flow to grain number. Nucleation sites, local undercooling, interface growth and impingement also matter. Measure grain boundaries/orientations and local thermal records separately from branch spacing.

The mean outward heat flows are 650 W and 2.60 kW. The fourfold ratio alone establishes no grain-count ratio.

Try the reasoning

Does nucleation mean the whole liquid freezes instantly?
Show the answer and reasoning
No.
Growth must follow.
Why include heat removal?
Show the answer and reasoning
Latent heat must be transported.
Thermal conditions influence interface growth.
Starting from the spherical model’s supplied constants, case A doubles Δg and case B halves γ. Compare critical radii and barriers with the original. Do equal critical radii mean equal barriers?
Show the answer and reasoning
Both radii become 5.00 nm. A has one-quarter of the original barrier, 1.05 × 10⁻¹⁷ J; B has one-eighth, 5.24 × 10⁻¹⁸ J. Equal radius does not imply equal barrier.
r* scales as γ/Δg, whereas ΔG* scales as γ³/Δg². The two parameter changes have different powers. No nucleation-rate ratio follows without temperature, sites and transport.
An observer calls the 15.0 nm zero crossing the critical nucleus and says every smaller embryo must shrink. Give a counterexample from the supplied curve.
Show the answer and reasoning
A 12.0 nm embryo has ΔG > 0 but r > r* = 10.0 nm, so increasing its radius lowers ΔG.
The critical size is the maximum, found from the slope. The total energy’s zero occurs at 1.5r* and answers a different question. This remains a conditional continuum-model tendency.
After increment 2, remove 10.0 g of the well-mixed remaining liquid at 14.4 mass% B, without removing earlier solid. Find the average B composition of the 90.0 g left. Can you still demand 10.0%?
Show the answer and reasoning
The exported liquid carries 1.44 g B; 8.56 g B remains in 90.0 g, giving 9.51 mass% B. The remaining body is no longer a closed 100 g account.
Use the removed liquid’s composition: 0.144 × 10.0 = 1.44 g. The original 10.0 g inventory equals 8.56 + 1.44 g. Its conservation applies to the body plus exported liquid, not to the smaller body alone.
Keep the first 20.0 g at 5.00 mass% B, but supply the next 30.0 g at 5.00% rather than 6.00%. Find the liquid composition after that step. Does the mass balance tell you why the solid composition changed?
Show the answer and reasoning
The 50.0 g liquid contains 7.50 g B, so it averages 15.0 mass% B. Conservation alone does not identify the changed interfacial or processing mechanism.
The solid now removes 1.00 + 1.50 = 2.50 g B. Less B entering the solid leaves more in the liquid. A measured composition/history or a separate constitutive/phase relation would be needed to explain the supplied change.
A separate 0.200 kg charge fully freezes, drops by 20.0 K and obeys the same supplied c and L. Its defined interval lasts 25.0 s. Find total heat and mean outward heat flow.
Show the answer and reasoning
Sensible heat is 2.00 kJ; latent heat is 48.0 kJ; total heat is 50.0 kJ and mean heat flow is 2.00 kW.
Use Q = (0.200)(500)(20.0) + (0.200)(240,000) = 50,000 J. Divide by 25.0 s. Doubling mass does not double the original total because the temperature drop has also halved.
Two matched sections have the same bulk composition. Section X has finer dendrite branches, but no orientation map was made. A report claims that X has more grains and no segregation. Which observations would discriminate those claims?
Show the answer and reasoning
Map crystal orientations or identified grain boundaries to compare grains, and measure composition across early and late solid regions at a stated resolution to test segregation.
Several branches can belong to one grain. A bulk assay averages local differences, and branch spacing alone is not a concentration measurement. Keep actual thermal position/history and section preparation in the comparison.
Adding particles under a nominally identical chamber schedule gives more identified equiaxed grains. Does this prove a lower homogeneous nucleation barrier? What comparison would test the proposed surface-site account?
Show the answer and reasoning
No. The proposed mechanism is heterogeneous nucleation on suitable surfaces, and a chamber setting does not establish equal local conditions.
Compare actual local thermal histories, initial chemistry, geometry and surviving particle population; look for a reproducible association between candidate particles and grain initiation under matched conditions. More grains alone does not distinguish a new surface route from changes in growth or other sites.

What to carry forward

Separate nucleation from growth.
Follow heat and solute transport.
In the stated spherical model, critical radius is an energy maximum; the later energy zero is different.
A closed element balance can preserve overall composition while early and late solid differ.
Total heat divided by duration gives mean heat flow, not nucleation rate or grain count.
Use separate thermal, orientation and local-composition evidence to explain the observed structure.

References

Teaching resources and research supporting this lesson’s explanations and examples.