Validation/Spray and evaporation
An evaporating droplet
The d-squared law is a statement about shape, not speed — and this case tests only the shape: linearity, independence of d₀ and of the time step, mass closure, and a saturated droplet that does not evaporate at all.
Why this case
The spray family had no reference of any kind behind it. Its solvers inject droplets, evaporate them and report a cloud, and every part of that happens whether or not the phase-change model is doing the right thing.
The classical statement about an evaporating droplet is not about how fast it goes. It is about the shape of the history: under quasi-steady evaporation the square of the diameter falls linearly with time, and the slope does not depend on how big the droplet started. That is a strong claim and a cheap one to test, and — importantly for what follows — it can be tested without knowing the value of the slope.
| Physics | Lagrangian spray, two-way coupled, liquidEvaporationBoil phase change |
|---|---|
| Droplet | 20 µm heptane (C7H16) at 320 K, 5 parcels, 3.1416 × 10⁻¹¹ kg of liquid in total |
| Atmosphere | Nitrogen at 1000 K and 101 325 Pa, at rest — 40 cm³ in 2560 cells |
| Held still | gravity removed from particleForces and the injection speed set to zero, so Re = 0 and Sh = 2 |
| Held isothermal | heatTransferModel none — the droplet stays at 320 K, measured spread 0.00 K |
| Time | To 2.5 ms at a fixed 8.3333 × 10⁻⁶ s, with 12 writes |
| Reference | The d-squared law for quasi-steady droplet evaporation, d² = d₀² − Kt |
| Agreement | d² falls in a straight line to R² = 0.99999955, worst point 0.0363 % of the fall; the slope moves 0.104 % when the step is quartered |
What is checked, and what is not
What is checked needs no such constant, and each part fails independently:
- The line is straight. Anything that makes the transfer coefficient drift through the run bends it — a droplet that moves, a droplet that cools, a cell that saturates — and all three are live in this case.
- The slope does not depend on d₀. That is the entire content of the d-squared law, and it is checked by running the same case at 20 and 10 µm.
- The slope does not depend on the time step. Otherwise what is being fitted is a property of the discretisation.
- Mass is conserved. What leaves the droplets appears in the gas as vapour. No reference is involved on either side.
- A droplet in a saturated atmosphere does not evaporate at all. The driving force is zero by construction, so the diameter cannot move.
The reference
A droplet whose surface is at a fixed vapour pressure, in a quiescent atmosphere, loses mass at a rate set by diffusion from a sphere. Writing the mass as ρl πd³/6 and the flux over the surface area πd², the diameter obeys
d(d^2)/dt = -K so d^2(t) = d0^2 - K t
with K constant. Two conditions make it constant, and both are enforced rather than hoped for. The Sherwood number must be 2, which is true only at zero relative velocity — so gravity is removed from particleForces and the injection velocity is set to zero, where Re is identically zero. And the surface vapour pressure must not move, which needs a constant droplet temperature — so heatTransferModel is none, and the measured temperature spread over the run is 0.00 K.
The third condition is that the surroundings stay far from saturation. The droplet evaporates into the 2.5 mm cell it sits in and the vapour barely moves over 2.5 ms, so the injected mass is kept tiny: the peak vapour mass fraction anywhere in the run is 0.0015, against a harness ceiling of 0.02. Above that, C∞ stops being negligible against Cs, K drifts, and the straight line would be measuring the size of the box.
Result
| Quantity | Measured | Harness limit |
|---|---|---|
| Fitted K | 7.403548 × 10⁻⁸ m²/s | not compared — see above |
| R² of d² against t | 0.99999955 | — |
| Worst departure from the line, as a fraction of the total fall in d² | 0.0363 % | 0.5 % |
| Diameter lost over the run | 26.76 % | at least 10 % |
| Points in the fit | 12 | at least 6 |
| Peak vapour mass fraction in any cell | 0.0015 | 0.02 |
| Droplet temperature spread | 0.00 K | 0.01 K |
| Vapour reaching the outlet face | 0.0 | 10⁻⁶ |
The slope is then asked to be indifferent to two things it must not depend on. Halving the droplet is the d-squared law itself; quartering the step is the numerics.
| Run | d₀ | Time step | K (m²/s) | Against the reference run |
|---|---|---|---|---|
| Measured | 20 µm | 8.3333 × 10⁻⁶ s | 7.403548 × 10⁻⁸ | — |
| Quarter step | 20 µm | 2.0833 × 10⁻⁶ s | 7.411281 × 10⁻⁸ | +0.104 % |
| Half diameter | 10 µm | 2.0833 × 10⁻⁶ s | 7.405417 × 10⁻⁸ | +0.025 % |
And the mass has to close, which is not a comparison with anything external:
| Quantity | Value |
|---|---|
| Liquid in the cloud at the start | 2.95799217 × 10⁻¹¹ kg |
| Liquid lost over the run | 1.72355317 × 10⁻¹¹ kg |
| Vapour appearing in the gas | 1.72359770 × 10⁻¹¹ kg |
| Closure | +0.0026 % |
The controls, and what the product's own defaults do
The saturated control is the one check on this page that isolates the phase-change model from everything else. Fill the box with pure vapour and the driving force (Cs − C∞) is zero or negative by construction, so the diameter cannot move. Measured, it does not move at all — the drift from d₀ is exactly 0.000000 %, and the fitted slope comes out as −0.0.
That control cannot be shown to work on the healthy run, because on the healthy run its answer is exactly zero and no band can be tightened onto it. So it is demonstrated on a fixture instead: the same control with its one distinguishing ingredient removed — the atmosphere left as the product stages it, pure nitrogen. The droplet then loses 26.76 % of its diameter in the run that is supposed to show it losing none, and the check fires.
| Fixture | What it leaves as staged | What happens |
|---|---|---|
| Uncoupled | coupled false in the cloud’s solution block | The droplets evaporate exactly as before — same straight line, K 7.415586 × 10⁻⁸, 0.163 % from the coupled run. The gas-phase C7H16 field is zero at every write and mass closure is −100 %. No other number in the run is unwell. |
| With gravity | gravity left in particleForces | The droplet falls, so it has a Reynolds number, so its Sherwood number is no longer exactly 2. K moves 1.083 % when d₀ is halved, against 0.025 % without it — the d-squared law’s own statement is the thing that breaks. |
| Shipped controls | The case verbatim: endTime 1 with writeInterval 100 | It injects 998 parcels, runs the full second, reaches End and exits 0 — and writes no solution at all. The first write it can reach is a hundred times past the end of the run. |
What this page does not establish
- The magnitude of K, as above. It is measured and reported; it is compared with nothing. Everything verified here is a statement about the structure of the law, not its value.
- Nothing about a moving droplet. The case is built at Re = 0 precisely so that Sh = 2 and the reference is exact. Real sprays are convective, and the Ranz–Marshall correction that governs them is exercised by the with-gravity fixture only as a demonstration that it changes the answer — its value is not checked against anything.
- Nothing about droplet heating.
heatTransferModel noneis what holds the droplet at 320 K and therefore what makes K constant. A real droplet cools to a wet-bulb temperature first, and that transient is outside this case entirely. - Nothing about breakup, collision or the boiling regime. A
ReitzDiwakarbreakup model is present in the staged case, but a stationary 20 µm droplet never triggers it, and the run stops at about half the droplet’s lifetime. - One liquid, one atmosphere, one temperature pair. Heptane at 320 K in nitrogen at 1000 K. No sweep over any of them.
Repeat this yourself
Every case here is set up from the worked examples in the product, with no hand-editing of solver files — so you can run it, and get the same numbers. The free tier runs real cases up to 250,000 cells of fluids, or 100,000 nodes of solid, with no account needed to download and no time limit.
Other cases
- A particle settling in still airA sphere stops accelerating when drag balances its buoyant weight. The particle is sized so that Schiller–Naumann and Stokes agree to 0.86 %, which makes the reference one an engineer can also reach on paper.
- Heat crossing a solid–fluid interfaceNeither check needs a reference table. If both regions start at one temperature nothing may move; and with no heat source anywhere, no temperature may leave the range its own data started in.
- How much of the answer is the meshEvery other page here asks how far the answer is from a reference. This one asks how much of it is the mesh — and puts an uncertainty on a result that has no reference at all.
All validation cases · Written by the team building SHD Sim.