Melting and Solidification Simulation

Category: Thermal Analysis – Phase Change | Revised 2026-10-01
CAE visualization for melting solidification theory - technical simulation diagram
Melting and Solidification Simulation

Theory: a moving interface and latent heat

Overview

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How does melting/solidification differ from ordinary heat conduction?

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Latent heat is absorbed or released at the solid–liquid boundary. At the melting point large amounts of heat flow without changing temperature, so the interface advances only as fast as that heat is removed. Its position is part of the answer, so the mesh can't be fitted to it. The enthalpy method computes each cell's energy (enthalpy) and derives liquid fraction and temperature from it, avoiding explicit interface tracking and extending directly to multiple dimensions. The standard check is the Stefan problem, which has an exact solution.

Neumann solution (one-phase Stefan problem)

$$ \lambda\,e^{\lambda^2}\,\mathrm{erf}(\lambda) = \frac{St}{\sqrt{\pi}},\qquad St = \frac{c\,\Delta T}{L},\qquad s(t) = 2\lambda\sqrt{\alpha t} $$

The liquid is at the melting point and the wall held ΔT below it. c and α are the solid's specific heat and diffusivity, L latent heat. For small St, λ ≈ √(St/2) (quasi-steady approximation).

Enthalpy method

$$ \frac{\partial H}{\partial t} = \nabla\cdot(k\nabla T),\qquad H = \rho c\,(T - T_m) - \rho L\,(1 - f_l),\qquad f_l = \begin{cases}1 & H \ge 0\\ 1 + H/(\rho L) & -\rho L < H < 0\\ 0 & H \le -\rho L\end{cases} $$

f_l is liquid fraction. For a pure substance it jumps from 0 to 1 at the melting point; for alloys it changes over a temperature range, forming the mushy zone.

Coffee Break Trivia

From the thickness of Arctic ice

In 1889 Stefan formulated heat conduction with a moving interface while studying how fast polar sea ice thickens. Neumann is said to have presented the exact solution earlier in lectures, and it is now called the Neumann solution. Growth of the solid layer with the square root of time applies to sea ice, welding, casting and thermal storage alike, and it remains the first verification problem for numerical methods.

Worked examples

Example 1: time for the solid layer to grow

MaterialWall undercoolingStλ10 mm50 mm100 mm
Water (ice)10 K0.0630.176711 s4.9 h19.7 h
Aluminium300 K0.760.5541.0 s26 s105 s
Paraffin (PCM)10 K0.100.22057 min23.9 h95.6 h

Ten times thicker takes 100 times longer. Aluminium's diffusivity is about 70 times ice's and its undercooling larger, so it is orders of magnitude faster (real moulds slow it through their own thermal resistance).

Example 2: verifying the enthalpy method (ice, 400 cells)

Front positionEnthalpy methodNeumannDifference
10 mm710 s711 s−0.1%
50 mm17,762 s17,764 s0.0%

Example 3: quasi-steady approximation error

MaterialStλ errorTime to 100 mm (exact / approx.)
Water (ice)0.063+1.0%71,056 / 69,609 s
Paraffin0.10+1.6%344,302 / 333,333 s
Aluminium0.76+11.0%105 / 85 s
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So simple approximations fail when the Stefan number is large.

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Right. The quasi-steady approximation assumes a linear temperature profile in the solid. With large latent heat and a slow interface, the solid's temperature settles first and it works well. At large Stefan numbers the sensible heat to cool the solid matters, the profile departs from linear, and the approximation predicts solidification too fast — 19% short for aluminium cooled 300 K. In casting that means underestimating solidification time and mispredicting shrinkage porosity. Numerical methods like the enthalpy method include this automatically, but grid and time step should be checked once against exact solutions like these.

Simulation workflow

  1. Compute the Stefan number to see whether latent or sensible heat dominates.
  2. Give liquid fraction over a narrow band for pure substances, solidus to liquidus for alloys.
  3. Where liquid convection matters, include flow resistance in the mushy zone.
  4. Verify grid and time step against exact solutions such as Neumann's.
  5. Choose models handling volume change (shrinkage) or undercooling if important.
Coffee Break Trivia

“The storage tank only half melted”

A paraffin thermal store designed to melt fully overnight (10 hours) only half melted in practice. The design had used one enlarged conductivity for paraffin that included liquid convection. In reality the melt convected above the heat-transfer tubes but barely moved below them, where heat crossed at the solid's 0.24 W/mK. Melting 50 mm by conduction alone takes about 24 hours, roughly matching measurements. Adding fins to shorten conduction distances solved it.

Common mistakes

Mistakes and fixes

MistakeEffectFix
Quasi-steady at large StTime underestimatedNeumann or numerics
Liquid-fraction band too narrowTemperature oscillatesSpread over a few cells
Convection as uniform conductivityTop–bottom difference missedSolve the flow
Coarse gridFront advances in stepsCheck vs exact
Ignoring mould resistanceSolidification too fastInclude interface conductance
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Structural AnalysisFluid Analysis (CFD)Manufacturing & Process Simulation
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