[Paper Review] A Droplet Sends Almost Every Sound Wave Back — Acoustic Impedance and a Sharp-Interface Two-Phase Solver

The cavity drive of 2026 Pa is 17.5 times the 116 Pa Laplace jump, so the droplet cannot hold a circle for even one period.

A diver barely hears a shout from above the surface. About 0.1% of the acoustic energy enters the water. The rest goes back into the air. Shoot a wave at a droplet and the grid turns 99.9% of it around at the interface. That contrast is the hard part of a compressible calculation that contains liquid.

Set medium 2 to water and the yellow pressure wave comes back almost whole, with the amplitude doubling at the interface. Lower $\rho_2$ and raise $c_2$ by the same factor and the picture does not move.

medium 1

medium 2 →

ρ₂998 kg/m³c₂1500 m/sinterface

What to watch: set medium 1 = air, medium 2 = water. The yellow pressure pulse comes back almost intact and nearly doubles at the interface (T = 1.998), while the cyan velocity pulse flips sign and the liquid barely moves. Now press medium 2 = air — R collapses to 0 and the pulse walks straight through. Only the ratio Z₂/Z₁ matters: slide ρ₂ down and c₂ up so that the product stays put, and the picture does not change at all.

The cyan velocity wave flips sign, and the amplitude on the water side dies near $10^{-4}$ . Pressure doubles. Velocity is nearly zero. The interface is acting like a wall.

Paper: A. Urbano, M. Bibal, S. Tanguy, A semi implicit compressible solver for two-phase flows of real fluids, Journal of Computational Physics (2022). ISAE-SUPAERO / IMFT. Level set + ghost fluid sharp interface, primitive-variable semi-implicit projection, van der Waals equation of state.

Add a Liquid and the Acoustic Time Step Dies First#

A fully explicit compressible two-phase step is chained to the acoustic CFL limit.

\Delta t \le \frac{\Delta x}{\max(|u| + c)}

$c$ is the sound speed and $\Delta x$ the cell size. $c$ is different in each phase. Air is 347 m/s, water 1500 m/s. One liquid cell tightens the global step by more than a factor of four.

Material inside the liquid moves at about 1 m/s. In Mach number (flow speed over sound speed) that is $10^{-3}$ . The motion worth watching is the interface deforming. The step is set by sound waves nobody is following.

The old workaround treated the liquid as incompressible. The step opens up, and the sound inside the liquid disappears with it. So does the question of how a droplet answers a wave.

This paper splits the equations into advection and acoustics. Advection stays explicit. Acoustics go implicit through a projection. What remains is this.

\frac{1}{\Delta t} = \frac{1}{\Delta t_{\mathrm{conv}}} + \frac{1}{\Delta t_{\mu}} + \frac{1}{\Delta t_{\sigma}}

$\Delta t_{\mathrm{conv}} = \Delta x / \max\lVert \vec{u} \rVert$ is advection, $\Delta t_{\mu} = \rho \Delta x^{2} / 2\mu$ is viscosity, and $\Delta t_{\sigma} = \tfrac{1}{2}\sqrt{\max(\rho)\Delta x^{3}/\sigma}$ is surface tension. No $c$ anywhere.

What Stays Continuous Across the Interface, and What Breaks#

A sharp interface treats $\Gamma$ as a surface of zero thickness and imposes the conditions on it directly. The paper uses three.

[\vec{u}]_{\Gamma} = 0, \qquad [p]_{\Gamma} = \sigma \kappa + 2\left[\mu \frac{\partial u_n}{\partial n}\right]_{\Gamma} - \frac{2}{3}\left[\mu \nabla \cdot \vec{u}\right]_{\Gamma}, \qquad [-k \nabla T \cdot \vec{n}]_{\Gamma} = 0

$[\,\cdot\,]_{\Gamma}$ is the jump across the interface, $\sigma$ surface tension, $\kappa$ curvature, $\mu$ viscosity, $k$ thermal conductivity. Velocity is continuous, pressure jumps by the capillary part, heat flux is continuous. With no mass transfer, temperature is continuous too ( $[T]_{\Gamma}=0$ ).

Density is not on the list. Each phase's equation of state sets its own density. Nothing averages density near the interface, so no intermediate state that neither material owns ever appears. That is where the fake sound speed, later, gets blocked.

The closure is cubic.

\rho^{3} + a_{1}\rho^{2} + a_{2}\rho + a_{3} = 0

The coefficients $a_1, a_2, a_3$ depend on pressure, temperature, and the material. van der Waals, Peng-Robinson, and Redlich-Soave-Kwong all have this shape. One expression holds a gas and a liquid. A perfect gas cannot do a liquid. Tait cannot do a gas.

One Impedance Fixes the Whole Reflection#

How a wave splits at the interface comes down to acoustic impedance.

Z = \rho c

Going from medium 1 into medium 2, the pressure reflection and transmission coefficients are these.

R = \frac{Z_2 - Z_1}{Z_2 + Z_1}, \qquad T = 1 + R = \frac{2 Z_2}{Z_2 + Z_1}

$R$ is the reflected pressure-amplitude ratio, $T$ the transmitted one. Not density alone, not sound speed alone. Only the product.

Air and water give $Z_{\text{air}} = 416$ , $Z_{\text{water}} = 1.50 \times 10^{6}$ , a ratio of 3600. $R = 0.9994$ , so 99.89% of the energy comes back. $T = 1.999$ , and the interface pressure nearly doubles. That product is why sliding $\rho_2$ and $c_2$ in opposite directions, in the figure above, changed nothing.

The wave the interface sends back meets the outlet of the domain once more. Whether the outlet swallows it or returns it is set by the $\sigma$ in NSCBC.

A Smeared Interface Invents a Sound Speed That Does Not Exist#

Spread the interface over a few cells and you get the situation every diffuse-interface method lives with.

When one cell holds gas and liquid, density and compressibility average like this.

\rho_m = \alpha \rho_g + (1-\alpha)\rho_\ell, \qquad \frac{1}{\rho_m c_m^{2}} = \frac{\alpha}{\rho_g c_g^{2}} + \frac{1-\alpha}{\rho_\ell c_\ell^{2}}

$\alpha$ is the gas volume fraction and the subscripts $g$ / $\ell$ mean gas and liquid. Wood's relation. The mixture is as soft as the gas and as heavy as the liquid.

At $\alpha = 0.5$ , $c_m = 24$ m/s. One-fourteenth of air, one-sixtieth of water. No material in the problem sends a signal at that speed. The average made the number. The race that widens the smear by hand sits just after the printed table.

What Solving in Primitive Variables Costs#

The debatable choice is to advance primitive variables $(p, \vec{u}, T)$ rather than the conserved set $(\rho, \rho\vec{u}, \rho E)$ . The energy equation is rewritten as a pressure equation.

What is gained is agreement of the two pressures, and an implicit heat-conduction term. The pressure from the equation of state and the pressure from the semi-implicit correction match every step. In conserved variables those two are known to drift. Conduction can also be treated implicitly, which matters in a low-Mach problem where heat leads, such as natural convection.

What is lost is conservation of total energy. The solver therefore cuts its own range at subsonic flow. Capturing a shock needs a different formulation.

Pressure positivity is not guaranteed either. The authors read that as room, not as a defect. A metastable liquid in cavitation really does sit at negative pressure, and an equation of state like van der Waals can pair a negative pressure with a positive temperature.

Reflection and Wood's Speed, Printed#

The numbers behind the figures, printed directly.

# Reflection/transmission at an air-water interface, and Wood's mixture sound speed
from math import sqrt
 
AIR   = (1.2,   347.0)    # (rho, c)
WATER = (998.0, 1500.0)
 
def impedance_split(m1, m2):
    z1, z2 = m1[0] * m1[1], m2[0] * m2[1]
    R = (z2 - z1) / (z2 + z1)
    return z1, z2, R, 1.0 + R, R * R
 
def wood_speed(alpha, gas=AIR, liq=WATER):
    rho_m = alpha * gas[0] + (1.0 - alpha) * liq[0]
    inv   = alpha / (gas[0] * gas[1] ** 2) + (1.0 - alpha) / (liq[0] * liq[1] ** 2)
    return 1.0 / sqrt(rho_m * inv)
 
z1, z2, R, T, E = impedance_split(AIR, WATER)
print(f"Z_air = {z1:8.1f}   Z_water = {z2:10.1f}   Z2/Z1 = {z2/z1:6.0f}")
print(f"R = {R:.5f}   T = 1+R = {T:.5f}   reflected energy = {100*E:.3f} %")
print(f"velocity amp in water / incident = {(z1/z2)*T:.3e}")
 
print("\nalpha   c_wood [m/s]")
for a in (0.0, 0.01, 0.1, 0.5, 0.9, 0.99, 1.0):
    print(f"{a:5.2f}   {wood_speed(a):8.1f}")

Z_air =    416.4   Z_water =  1497000.0   Z2/Z1 =   3595
R = 0.99944   T = 1+R = 1.99944   reflected energy = 99.889 %
velocity amp in water / incident = 5.562e-04
 
alpha   c_wood [m/s]
 0.00     1500.0
 0.01      120.5
 0.10       40.1
 0.50       24.0
 0.90       39.9
 0.99      114.3
 1.00      347.0

smear n8 cellsρ_liquid998 kg/m³

What to watch: at n = 0 the two lanes are the same run. Pull n up to 8 cells — a width any capturing scheme reaches within a few hundred steps — and the cyan curve digs a hole down to ~24 m/s that neither material has, so the smeared signal falls behind and arrives late. Drop ρ_liquid and the hole fills in: the dip is driven by the density ratio, not by the sound speeds. The yellow line is the bill an explicit solver pays for that hole.

Raise the n slider from 0 to 8 cells and the cyan curve digs a valley that belongs to neither material. Lower $\rho_\ell$ and the valley fills in. The $\alpha = 0.01$ row of the print, 120.5 m/s, is the mouth of that valley. Cut the density contrast and the valley goes away even if the sound-speed contrast stays.

Reflected energy is $R^{2}$ . Squaring the printed $R = 0.99944$ gives 99.889%. What gets through is 0.111%, the opening tenth. The sound-speed ratio 1500/347 is 4.32, but the impedance ratio is 3595. Density ratio 998/1.2 times the sound-speed ratio is that same 3595. Density builds the wall.

At $\alpha = 0.01$ the 1500 m/s falls to 120.5, a factor of 12.4. From there to 24 m/s at $\alpha = 0.5$ is another factor near five. Most of the damage sits in the first 1%. An explicit step follows that cell's sound speed, so one smeared cell tightens it twelvefold.

The standing wave is 2026 Pa against a Laplace jump of 116 Pa, ratio 17.5. A velocity of 4.9 m/s is 1.4% of 347 m/s, Mach 0.014, five times the 1 m/s liquid motion. $f = 347/(2 \times 0.008) = 21688$ Hz, matching 21.7 kHz. An incompressible liquid keeps the 17.5-fold disturbance out of the drop. One smeared cell lets 120.5 m/s shove the node.

Read the $\alpha = 0.01$ row again. One percent of gas in water drops the sound speed from 1500 to 120 m/s. That is the value seen in bubbly water, and it is what the grid sees if the interface smears by a single cell.

A Droplet in a Standing Wave — the Paper's Last Test#

The last test puts a 1 mm droplet in a cylindrical cavity of height $L = 8$ mm and drives the top boundary with a pressure oscillation.

p(t) = p_0 + \Delta p \sin(\omega t)

$\omega = 2\pi f$ and $f = c_0/(2L) \approx 21.7$ kHz is the cavity's first acoustic mode. $\Delta p$ is 2% of $p_0$ , 2026 Pa. Gravity is zero.

Once the standing wave stands, a pressure node sits at $z = L/2$ . A pressure node is a velocity antinode. The velocity amplitude follows from the impedance.

\Delta v = \frac{\Delta p}{\rho_0 c_0} \approx \frac{2026}{416} \approx 4.9\ \text{m/s}

The droplet sits in the middle of that oscillating velocity field. Pressure still propagates inside the liquid, and the initial circular drop holds a Laplace jump $2\sigma/r = 116$ Pa. The gap between the 2026 Pa acoustic disturbance and the 116 Pa the surface tension is defending decides the result. Each period the drop stretches and squeezes along the axis, and a large enough amplitude ends in breakup.

That picture is why the formulation above was needed. An incompressible liquid deletes the pressure wave inside the drop. A smeared interface lets a 24 m/s fake sound speed in the nearby cells shift the node. A sharp interface, with both phases compressible, is what produces the picture.

The One-Cell Smear That Shoves the Node#

The 24 m/s at $\alpha = 0.5$ is the number a solver steps on if the interface is left smeared. A five-equation model that keeps that sound speed on purpose is the Wood speed in Murrone–Guillard, which carries the same 24 m/s along.

In the droplet's standing wave, does keeping that 24 m/s hold the node, or does the 21.7 kHz sit still only after the one-cell 120.5 m/s is removed?