How acoustic streaming in a standing wave stirs the liquid

A steady flow pushed from the wall boundary layer

Liquid cannot slip at the top and bottom walls, while the liquid a little further away sloshes back and forth with the sound. The layer in which the velocity climbs from zero to the field value is thin: the viscous penetration depth $\delta$ , only about 0.4 µm in water near 2 MHz and hundreds of times thinner than the channel is high.

Inside that layer the velocity changes steeply, and the second-order average momentum flux does not vanish. It drives a slow, steady flow of the liquid outside the layer: Rayleigh streaming. Its velocity scale $u_0$ is proportional to the energy density.

\delta=\sqrt{\frac{2\nu}{\omega}},\qquad u_0=\frac{4E_{\mathrm{ac}}}{\rho_0c_0}

The left expression is Eq. 1 of Settnes and Bruus, with

\nu=\eta/\rho_0

the kinematic viscosity. The right one is the streaming speed scale of Eq. 5 of Barnkob and co-workers.

Source: M. Settnes and H. Bruus · R. Barnkob

Four rolls and the flow on the centre plane

When the wavelength is much longer than the channel height and the height much larger than $\delta$ , Rayleigh’s solution gives the flow outside the boundary layers. The cross-section holds four rolls: near the centre plane the liquid flows towards the pressure node, and near the top and bottom walls it flows back to the side walls.

On the centre plane $\zeta=0$ the horizontal speed is exactly $\tfrac{3}{16}u_0\sin(2ky)$ , with the same shape and direction as the radiation drift of a positive-contrast particle. So on that plane the two drifts simply add and the exact solution of Chapter 5 still holds. Away from it the streaming lifts and lowers the particle, and only numerical integration will do.

A real rectangular channel is not a pair of parallel plates. Barnkob and co-workers estimate that the rectangular cross-section reduces the centre-plane streaming to 0.82 of the parallel-plate value, while the thermoviscous effect, viscosity following the temperature oscillation of the sound field, raises it by 1.26. Together the coefficient comes to about 0.194, almost Rayleigh’s $3/16=0.1875$ . The experiment uses Rayleigh’s parallel-plate form.

U_y=\frac{3}{16}u_0\sin(2ky)\,(1-3\zeta^2),\qquad U_z=-\frac{3}{16}u_0\,kh\cos(2ky)\,(\zeta-\zeta^3),\qquad \zeta=\frac{2z}{h}-1

Eqs. 6a–b of Barnkob and co-workers with the origin at the left wall. For

|\zeta|>1/\sqrt3

the horizontal flow reverses; these are the upper and lower halves of the rolls.

Half-wave resonance across a channel with hard side walls. Teal: pressure amplitude, largest at the walls and zero at the centre. Amber: velocity amplitude, zero at the walls and largest at the centre. Beads with

\Phi>0

collect on the pressure node. Original schematic.

Source: R. Barnkob · M. Settnes and H. Bruus

How small a particle loses to streaming

The radiation drift scales with $a^2$ , while the drag from streaming does not depend on particle size. Their ratio is left with the particle radius and one length scale $a_0$ . The diameter at which the ratio is one is the critical diameter.

A key consequence: both drifts are proportional to $E_{\mathrm{ac}}$ , so more power cannot focus small particles; it only spins them faster. The critical diameter comes down only with a higher frequency, a lower viscosity or a particle with a larger contrast factor.

\frac{u^{\mathrm{rad}}}{u^{\mathrm{str}}}=\frac{1}{s}\frac{a^2}{a_0^2},\qquad a_0=\delta\sqrt{\frac{3}{\Phi}},\qquad 2a_c=2a_0\sqrt{s}=\sqrt{\frac{24\nu s}{\omega\Phi}}

Eqs. 12–14 of Barnkob and co-workers.

s

is the centre-plane streaming coefficient,

3/16

between parallel plates.

Source: R. Barnkob · M. Settnes and H. Bruus

Worked example: the critical diameter for polystyrene beads

Take the 1.940 MHz working frequency of Barnkob and co-workers and water at 25 °C, $\nu=8.93\times10^{-7}\,\mathrm{m^2/s}$ . The viscous penetration depth is $\delta=0.383\,\mu\mathrm{m}$ , matching the 0.38 µm of their Table II. With $\Phi=0.17$ , $a_0=1.61\,\mu\mathrm{m}$ .

The parallel-plate coefficient $s=3/16$ gives $2a_c=1.39\,\mu\mathrm{m}$ ; their corrected $s=0.194$ gives 1.42 µm. The paper quotes “about 1.4 µm” at 2 MHz. The experiment, at 1.985 MHz and with the viscous contrast factor, finds 1.395 µm.

At the experiment’s default $E_{\mathrm{ac}}=20\,\mathrm{J/m^3}$ , $u_0=53.6\,\mu\mathrm{m/s}$ and the centre-plane streaming peaks at 10.1 µm/s. A 5 µm bead drifts at 129 µm/s from radiation, 12.8 times the streaming, so radiation decides its path almost entirely. A 0.6 µm particle drifts at only 1.85 µm/s, less than a fifth of the streaming, and just rides the rolls.

Source: R. Barnkob · M. Settnes and H. Bruus

Try it in the Lab

Find the line between focusing and circling

Choose “Tiny beads ride the rolls” and watch how 0.6 µm particles move across the section. Then turn off “Include acoustic streaming” and compare.
Choose “Near the crossover size”. In the “Radiation versus streaming by particle size” chart, find the diameter where radiation provides half the drift and compare it with 1.4 µm.
Raise the energy density from 20 to 80 and check whether the critical-diameter readout changes. Explain with this chapter’s formula.

Exercises with explained answers

If the energy density doubles, what happens to the critical diameter and to the time a particle takes to go once round a roll?

Show the answer

The critical diameter stays the same, because radiation drift and streaming are both proportional to $E_{\mathrm{ac}}$ and their ratio does not depend on it. All speeds double, so a lap takes half as long.

Raise the frequency from 1.940 MHz to 3.900 MHz (the other series of Barnkob and co-workers), with $s=0.194$ and $\Phi=0.17$ . What is the critical diameter?

Show the answer

$2a_c\propto\omega^{-1/2}$ , so $1.42\,\mu\mathrm{m}/\sqrt{3.900/1.940}=1.00\,\mu\mathrm{m}$ . Doubling the frequency shrinks the critical diameter by only about 30 %.

Why can streaming and radiation drift simply be added on the centre plane but not away from it?

Show the answer

On the centre plane $U_z=0$ , and both $U_y$ and the radiation drift are proportional to $\sin(2ky)$ , so the combined equation keeps the form $\dot y=A\sin(2ky)$ . Away from it, $U_z$ carries the particle off the plane and the horizontal streaming changes with height, so the two motions couple and the path must be integrated numerically.

Sources and what we recomputed

R. Barnkob, P. Augustsson, T. Laurell and H. Bruus, Acoustic radiation- and streaming-induced microparticle velocities determined by micro-PIV in an ultrasound symmetry plane, arXiv:1208.6534v1; Physical Review E 86, 056307 (2012)

Version 1, 31 August 2012, read in full. Eq. 1 (standing-wave field), Eq. 2 (radiation force), Eqs. 5–7 (streaming scale and Rayleigh streaming), Eqs. 9–10 (geometry and thermoviscous corrections), Eqs. 12–14 (velocity amplitude, velocity ratio, critical diameter), Eqs. 16–18 and Table I (wall drag), Table II (material data), Tables IV–V and Figs. 8–9 (measured energy densities, streaming coefficients and velocity ratios). We recompute the theory values; the measured data are quoted, not reproduced.

M. Settnes and H. Bruus, Forces acting on a small particle in an acoustical field in a viscous fluid, arXiv:1110.6037v1; Physical Review E 85, 016327 (2012)

Version 1, 27 October 2011, read in full. Eq. 1 (viscous penetration depth), Eqs. 4–5 (first-order pressure and compressibility), Eqs. 48a–e (Gor’kov potential with the viscous dipole coefficient) and Table I (Pyrex and cell parameters).