Chapter 1 Theory of Stationary Streaming in a sound Field

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High-Intensity Ultrasonic Fields III Acoustic Streaming

L.K. Zarembo

Three types of Streaming:

1. Schlichting
2. Rayleigh
3. Eckart

The theory of acoustic streaming is based on the hydrodynamic equations for a viscous compressible fluid.

In Euler coordinates the equation of continuity has the form

$$\frac{\partial \rho}{\partial t} + \nabla (\rho \mathbf{v}) = 0$$

where $$\rho$$ is the density of the medium and $$\mathbf{v}$$ is the velocity vector.

The equation of motion of a viscous compressible fluid has the form

$$\rho[\frac{\partial \mathbf{v}}{\partial t} + (v \nabla) v] = \rho {\frac{\partial \mathbf{v}}{\partial t} + \frac{1}{2} \nabla v^2 - [\mathbf{v}\times(\nabla\times\mathbf{v}]} = \mathbf{F}$$

1. General Theory of Stationary Acoustic Streaming in the Second Approximation (Slow Streaming)

We represent the density, pressure, and velocity in series form:

$$\rho = \rho_0 + \rho^\prime+\rho^{\prime\prime} + \cdots\ p = p_0 + p^\prime+p^{\prime\prime} + \cdots\ v = v^{\prime} + v^{\prime\prime}$$

一级近似方程

$$\frac{\partial \rho^\prime}{\partial t} + \rho_0 \nabla \cdot \mathbf{v}^\prime = 0$$

$$\rho_0 \frac{\partial \mathbf{v}^\prime}{\partial t} = -\nabla p^\prime + (\frac{4}{3}\eta + \eta^\prime)\nabla\nabla v^\prime - \eta \nabla \times (\nabla \times \mathbf{v}^\prime)$$

二级近似方程

用 velocity vortex¹ 来

取时间平均

$$\frac{\partial \Omega}{\partial t} - v \nabla^2 \Omega = - \nabla \times \mathbf{f}$$

2. Eckart Streaming

Tlhe flow generated under the influence of a well-collimated （高准直）traveling-wave beam was first investigated by Eckart.

Assuming that the first-approximation velocity field is irrational, one can solve the problem on the basis of 18. It is essential to explore the permissibility of representing a confined sound beam as a potential field. It is readily seen that for a place wave, $$\nabla \times \mathbf{v}^\prime$$ depends significantly on the distribution of the particle velocity over the cross section of the beam. In the case of a well-collimated sound beam with a sharp boundary, goes to infinity at the boundary and is equal to zero over the rest of the space; for a smooth distribution of the particle velocity over the beam with a sharp boundary, $$\nabla \times \mathbf{v}^\prime$$ goes to infinity at the boundary and is equal to zero over the rest of the space; for a smooth distribution of the particle velocity over the beam cross section, $$\nabla \times \mathbf{v}^\prime$$has a nonzero value over the entire domain cross occupied by the field. In the interval of frequencies dimensions much larger than the wavelength, clearly, it may be assumed that the volume of the vortex domain at the boundary of the sound beam is small in comparison with the volume occupied by the sound field, and these surface vortex sources introduce a considerably smaller contribution than the volume sources to the stationary streaming. Eckart's theory, within the limits of its applicability, as will be seen below, is in fully satisfactory agreement with the experimental result. (不太懂）

For the determination of the stationary streaming velocity, we need to find the solution of Eq. (18); for $$p^\prime=c_0^2 \rho^\prime$$the solution can also be represented in the form

$$\bar{\Delta \Omega} = \frac{b}{\eta \rho_0^2} \bar{\nabla \frac{\partial \rho^\prime}{\partial t} \times \nabla \rho^\prime}$$

The particle velocity is directed along the z axis in cylindrical coordinates (r, \phi, z ) and has the form

(28) $$v^\prime = U(r) e^{-\alpha_0 z} \sin (\omega t - k z)$$

where $$\alpha_0 = b \omega^2/ 2\rho_0 c_0^3$$ is the sound absorption coefficient, and the function U(r) gives the distribution of the particle velocity over the wave front. Substituting (28) into (27), we can find the solution for stationary streaming. In this solution has been found under the following conditions: 1) the stationary streaming has only a z component on a given interval; 2) it is permissible in expression (28) to assume $$e^{-\alpha_0 z} = 1$$; 3) sound propogates in a cylindrical tube whose end opposite the sound source is terminated in an absorber, so that a traveling wave propagates in the tube; the mass flow of fluid through all cross sections of the tube is zero. Under these conditions the streaming velocity has the form

$$\bar{v_z}^{\prime\prime}=$$

here $$r_0$$ is the radius of the tube. The solution assumes a simple form for a uniform distribution of the particle velocity over the beam cross section:

$$U(r) = v_0, r<r_1$$

$$U(r) = 0, r_1<r\le r_0$$

here is the radius of the sound beam. Now the transverse distribution of the streaming velocity has the form

$$\bar{v_z}^{\prime\prime}= \begin{cases} U_0 {(1 - x^2/y^2) - (1 - \frac{1}{2} y^2)(1-x^2) - \ln y} \ - U_0 {(1-\frac{1}{2} y^2) (1- x^2) + \ln x}\end{cases}$$

where $$x = r0/r$$ _and \$$y = r_1/r_0 $$.

the streaming velocity is a maximum on the axis of the sound beam and depends on $$r_0/r_1$$. For $$r_0/r_1 = 1$$ the velocity revert to zero. The maximum velocity increases slowly with $$r_0/r_1$$, maintaining the same order under typical experimental conditions.

It follows from (32) that the ration of the streaming velocity to the particle velocity amplitude in a medium with a small dilatational viscosity is of the order M(kr_1)^2. The scale of the streaming is considerably larger than the wavelength, and, according to (21), the solution is only applicable for acoustic Reynolds numbers much smaller than unity. Streaming similar to Eckart streaming but induced by a sawtooth wave of large intensity will be considered below.

It follows from (32) that the streaming velocity is proportional to the sound intensity, the frequency squared, and the ratio of dilatational to the shear viscosity. The latter consideration suggested the measurement of Eckart streaming velocity as a new method for measuring the dilatational viscosity. However, it is seen at once from (32) that the velocity is proportional to the sound absorption coefficient \alpha_0 and time-average kinetic energy density _E_k = \rho__0v__0^2/4

It subsequently became clear that Eckart streaming is caused by all types of losses in sound wave (thermal conducting losses, scattering losses, etc), whereas deceleration of the flow occurs only under the influence of viscous forces.

Equation (33) implies that this type of streaming can be interpreted as streaming under the influence of the radiation pressure gradient, which is proportional to 2\alpha_0 \bar{E}_k. _In [8], relying on the law of the conservation of momentum, it has been shown that the sum of radiation pressure and dynamic pressure of the resulting stationary streaming is independent of the distance from the sound source. This result cannot be accepted as one of sufficient rigor, but, as shown by experiments performed at megacycle frequencies, the total streaming plus radiation pressure is in fact independent of the distance from the distance from the sound source within the experimental error limits. We note that this fact affords a fairly simple means for determining the sound energy flux through the surface of a source from a measurement of the total pressure far from the the source. because in this plane the streaming velocity is hypothetically equal to zero. At the present time, the determination of the intensity is hypothetical directed on the surface of a radiator at frequencies on the order of several megacycles presents a rather complex experimental problem. The force acting on an absorbing receiving experimental problem.

The solution (31) applies to a well-collimated sound beam. In [11] a solution has been found for a problem similar to the Eckart case when the divergence of the sound beam is taken into account. The first approximation in this case may be represented in the form

$$\rho^\prime = A_0 \frac{\sin(kr - \omega t)}{r} e^{-\alpha_0 r} \frac{2 J_1(k a \sin \theta)}{ka \sin \theta}$$

where and $$J1$$ is a first-order Bessel function. For $$r \gg a$$ expression (34), of course, represents the sound field produced by a piston of radius $$a$$ embedded in an infinite plane baffle and vibrating with an amplitude $$v_0$$ . Introducing a flow function such that$$v_r =\frac{1}{r} \frac{\partial \psi}{\partial \theta} + \frac{\psi}{r} \cot \theta\ v\theta = - \frac{\partial \psi}{\partial r} - \frac{\psi}{r}$$

and substituting (34) into (27), we obtain

(35)$$\nabla^4 \psi = - A_0^2 \frac{b k \omega}{2\eta \rho_0^3} \frac{e^{-2\alpha_0 r}}{r^3} \frac{d}{d \theta} [\frac{2 J_1(k a \sin \theta)}{k a \sin \theta}]^2$$

The solution of eq. (35) has been found in [11] for in the domain \theta < \theta0 , r_0 < r < r__1,and for small value of \theta_0 has the form

The constants C1 and C2 are chosen to make the velocities v_r and v\\theta vanish for \theta = \theta_0. This solution is only applicable for small angles . In as much as the sound field (34) is nonuniform, i.e., as the amplitude \rho^\prime is a maximum on the axis and decreases with increasing \theta, the flow velocity, as shown in Fig.3 This solution has been refined in [12], in which the streaming velocity is also determined for the sound field (34), but without the restriction to small angles \theta_0_

It can be shown that a sound source in the form of a pulsating sphere cannot generate streaming. In fact, in this case the particle velocity in spherical coordinates has the form $$vr^\prime = U(r) \sin(\omega t - k r)$$;$$v\theta^\prime = v_\phi^\prime = 0$$ in as much as $$\nabla\times \mathbf{v}^\prime = 0$$ it is permissible to use Eqs. (18) or (27). The right hand side of the equation vanishes. Since in this case could have had only a radial component, which vanishes on the surface of the sphere, we find ; this result is not unexpected, because the onset of streaming in this case would violate the condition of continuity. For the oscillations of a spherical source of other than zero order, of course, streaming is possible.

In [13] the streaming that occurs in the intersection of two plane traveling waves of equal frequency at arbitrary angles is analyzed. The streaming in this case can only have its velocity directed along the bisector of the angle of intersection, and the magnitude of the velocity varies by a cosine lw in the perpendicular direction. This creates "layered" streaming, as shown in Fig. 4. The maximum velocity in each layer is ~b/2\sqrt{2} \eta \cdot (v0^2/c_0).

Because Eckart streaming is a large-scale phenomeno, the theory based on Eqs. (13) and (14) is only applicable for acoustic Reynolds numbers less than unity, In liquids having a relatively low viscosity (0.01 P), this imposes severe limitations on the sound pressure amplitude in the frequency range of several megacycles. It has been shown in a number of experimental studies (see PART II) that, with an increase in the sound intensity, a departure form theory is observed; thus for example, the streaming velocity is no longer proportional to intensity. This was originally ascribed t the onset of turbulence. It now apperas reasonably certain that this departure is due to the inapplicability of the theory in the domain Re >> 1. In this domain, as we know, nonlinear effects such as distortion of the traveling wave profile and the attendant increase in wave absorption begin to play an important part. For Re > 1 a sine wave gradually transforms into a sawtooth at a certain distance from the source. The streaming velocity, as the experimental results indicate, no longer satisfies the slow streaming condition.

The problem of the acoustic streaming elicited by a sawtooth wave has been solved in [14]; the flow velocity in this case wave assumed small relative to the particle velocity amplitude of the wave. i.e., the solution is also limited to slow streaming.

The theory of fast acoustic streaming has only rencently begun to develop. One constructive approach to the determination of the streaming velocity for a sawtooth wave is indicated in [5]. We shall briefly consider the methods that can be used to determine the streaming velocity in the case when streaming cannot be regarded as slow. The method of successive approximations, naturally, is inapplicable in this case. The parameters of the sound field $$\rho$$ $$v$$ and $$p$$ may be written in the form [15,5]

$$\rho = \rho_0(x,y,z) + \rho_A(x, y, z, t)\ p = p_0(x,y,z) + p_A(x,y,z,t)\ v = v_0(x,y,z) + v_A(x, y, z, t) $$

This division of the parameters is made so that for example, the velocity $$v_0$$will represent the time-average value of the velocity v:

where t_2-t_1 is a certain time interval equal to multiple of the period of the sound wave. It follows form (38) that the average of the variables carrying the subscript A overt this time interval is zero. Inserting (37) into Eq.(2), represented in the form

assuming that $$\rho_A \ll \rho_0$$ and taking the time average, we obtain an equation in the form [15,5]

where

The time average of the equation of coninutity (1) gives

Equations (39) and (41) are the initial equations for determining the velocity of fast streaming, as their derivation did not rely on the assumption of a small flow streaming, as their derivation did not rely on the assumption of a small flow velocity relative to the particle velocity of the sound wave. A logical assumption make in the derivation of the equations is $$\rho_A /\rho_0 \ll 1$$. It was remarked earlier that the time-average values of the density, pressure and velocity in strong sound fields can differ from the values of these variables in the unperturbed medium. Consequently, in these equations, in general $$\rho_0$$ and $$p_0$$ are not equal to the density and pressure in the absence of sound. The deviation of these variables from their unperturbed values, at any rate, for problems that have been solved in nonlinear acoustic, is of the order M^2. under the condition $$M\ll 1$$, the variables $$\rho_0 $$ and $$p_0$$are not equal to the density and pressure in the absence of sound. The deviation of these varibles from their unperturbed values, at any rate, for problems that have been solved in nonlinear acoustics, is of the order M^2 Under the condition M<< 1, the variables \rho_0 and p_0 may be assumed equal to their unperturbed values; then the condition \rho_A \ll \rho_0 also corresponds to M \ll 1, and Eqs (39) and (41) can be somewhat simplified. It follows from (41), in particular, that

and from (39) that

where F can be expressed by means of $$\rho_A = c_0^{-2} p_A$$ only as a function of . The solution of (42) and (43) requires knowlegde of the solution of the problem of sound propagation in the approximation in which (42) and (43) were originally deduced. In [5] a solution has been found for a problem similar to the Eckart case when v_a has the form

3. Streaming Outside the Boundary Layer

The Eckart type of streaming, analyzed in the preceding section, is also essentially streaming outside the boundary layer because the processes in the thin vortex layer at the boundary of sound beam have been disregarded. In this section we consider certain problems related to the onset of streaming when the sound field is confined by rigid walls. An example of this type of streaming is the Rayleight variety [18]. i.e. two-dimensional streaming elicited by a standing wave between two planes. The oscillations in this standing wave between two planes. The oscillations in this standing wave are directed along the axis Ox, and far from the walls have the form v'_1(x,t) = v_0 cos kx cos omega t. If one of the planes has the coordinate y = 0 and the other plane has the coordinate y = 2 y_1, then the x and y components of the velocity from (8) for 0 have the form

$$v_1 = v_0 \cos kx [-\cos \omega t + e^{-\mu} \cos (\omega t -\mu)]\ v_2 = v_0 \frac{k \delta}{\sqrt{2}} \sin kx [(1-\frac{\mu}{\mu_1}\cos(\omega t - \frac{\pi}{4}) - e^{-\mu} \cos(\omega t -\frac{\pi}{4} -\mu)]$$

where $$\delta = (\frac{2\mu}{\omega})^{1/2} $$ $$\mu = y/\delta$$ $$\mu_1 = y_1/\delta$$

The motion is symmetric about the plane $$y = y_1$$; by the substitution $$y=2y_1 - y^\prime$$ we obtain the velocity for $$y_1 \ll y^\prime \ll 2y_1$$. It was assumed in the derivation of (46) and (47) that $$k \delta \ll 1$$ and $$y_1 \gg \lambda$$. Rayleigh obtained an approximate solution for the flow in the case when the first-approximation velocity field may be regarded as nearly solenoidal, so that Eq.(19) is applicable. The solution has the form

$$\bar{v_1^{\prime\prime} }= - \frac{v_0^2}{4c_0} \sin 2 kx {\frac{1}{2} e^{-2\mu} + e^{-\mu} \cos \mu + 2e^{-\mu} \sin \mu + \frac{3}{4} - \frac{9}{4}(1 - \frac{\mu}{\mu_1})^2}$$

$$\bar{v_2^{\prime\prime}} = -\frac{3 v_0^2}{8c_0} k \delta \cos 2kx { \frac{1}{2} e^{-2\mu} + 3e^{-\mu}\cos \mu + e^{-\mu} \sin{\mu} + \frac{3}{2} \mu_1 [(1-\frac{\mu}{\mu_1}) - (1 -\frac{\mu}{\mu_1})^3]}$$

It only approximately satisfies the boundary conditions and, hence, fails to give the true value of the streaming velocity in the boundary layer (as \mu\to 0)

Far from the boundary, where the terms $$e^{-\mu}$$ become negligible, (48) implies

The behavior of the streaming according to (49) is illustrated in Fig. 5. The streaming pattern comprises vortices spaced at a distance $$\lambda /4$$ along the axis Ox. The axes of the vortices are found at points with coordinates $$x = [(2n-1)\lambda]/8$$, $$y=0.423y_1$$ and $$x = [(2n-1)\lambda]/8, y=1.577 y_1$$ where n is an integer. According to (49), the velocity is independent of the viscosity, although, as will become apparent from the ensuing discussion, the formation of the vortices is due to interaction with boundary-layer vortices.

The Rayleigh solution was refined in [3], in which Eq. (20) was solved for the same problem, i.e., no assumptions were made with regard to the solenoidal character of the first-approximation velocity. The condition $$k \delta \ll 1$$ permits one to assume that the derivatives with respect to $$y$$ in (20) are much larger than the derivatives with respect to $$x $$, while $$v_2^\prime \ll v_1^\prime$$ and $$\bar{v_2^{\prime\prime}} \ll \bar{v_1^{\prime\prime}}$$. This makes it possible to simplify Eq.(20) considerably and to find a solution which still does not qualitatively alter the nature of the vortex flow.

The problem of the acoustic streaming produced by a standing wave in a narrow cylindrical tube for a solenoidal first-approximation velocit field (which is equivalent to saying that the domain is smaller than the sound wavelength and k R \ll 1, where R is the radius of the tube) has been treated in [19]. For the first approximation the stokes-Kirchhoff solution is adopted

$$v_z^\prime = j v_0 e^{j\omega t} \cos kz [1 - e^{(1+j) \kappa}]\v_r^\prime = j v_0 \frac{k \delta}{1+j} e^{j\omega t} \sin kz [1 - e^{(1+j)\kappa}]$$

where $$\kappa = (R-r)/\delta$$ and is the velocity amplitude on the tube axis. The second approximation is found under the condition that the thickness of the boundary layer is small relative to the tube radius. Substituting (50) into (19), we obtain the following relations for the stationary streaming velocity:

This solution, like the Rayleigh solution, approximately satisfies the boundary conditions as, for $$r=R$$ , we have $$\bar{v_z^{\prime\prime}} ~\delta v_0^2/Rc_0$$ and $$\bar{v_r} = k \delta v_0^2 /c_0$$. Near the tube axis ($$\kappa \ll 1$$), Eq (51) implies the parabolic velocity distribution typical of Poiseuille flow.

The streaming velocity does not depend on the viscosity of the medium and is proportional to $$v_0^2/c_0$$. The streaming pattern has a spatial period of $$\lambda/2$$; for $$r=0.707 R$$, the z component of the streaming velocity goes to zero. The nature of the streaming is similar to the one between two planes.

We point out certain characteristic features of the streaming outside the boundary layer at a large distance from the boundary. As in the case of Eckart streaming, for slow streaming outside, the ratio of the vortex velocity to the particle velocity in the sound wave is ~ M\phi, where \phi is a dimensionless variable depending on the geometry of the sound field, the frequency, and the viscosity. Far from the boundary, as indicated by (49) and of the medium. The streaming scale L'' ~ \lambda , and it follows from (21) that far from the boundary layer the solution is applicable for acoustic Reynolds numbers smaller than unity. For large Re, as indicated by the experimental results [20], the nature of the streaming stays roughly the same, but the vortex velocity is appreciably larger than implied by (49).

4. Acoustic Streaming in the Boundary Layer

In the theory of boundary-layer acoustic streaming, the viscous wavelength $$\lambda^\prime = 2\pi \delta$$ assumes great significance, where $$\delta = (2\nu/\omega)^{1/2}$$, is the thickness of the acoustic boundary layer. These quantities, of course, determine the distances over which are propagated the perturbations elicited by the viscous force (shear or viscosity waves in case of the shear oscillations of a plane in a viscous medium or perturbations of the amplitude of a sound wave propagating near a surface). The ratio of the viscous wavelength to the acoustic wavelength to the acoustic wavelength may be written in the form

$$\frac{\lambda^\prime}{\lambda} = k\delta = (\frac{2\omega \eta}{\rho_0c_0^3})^{1/2}=(\frac{3}{2} \frac{M}{Re})^{1/2}$$

Consequently, the condition is equivalent to the stipulation of small viscous stresses ~\omega \eta in comparison with the stresses produced by the volume elasicity of the medium $$\rho_0 c_0^2$$ to small sound absorption per wavelength, or, finally, to acoustic Reynolds numbers $$Re \gg M$$. The condition $$\lambda^\prime \ll \lambda$$ holds for practically all the cases treated in the present chapter.

Let us consider for a moment the qualitative characteristics of the processes in an acoustic boundary layer. The velocity on a perfectly rigid surface must vanish; hence, the velocity gradient is large in the boundary layer. This means that the momentum of the sound wave varies sharply in the boundary layer and the forces that produce streaming are large. These forces for a sufficiently thin boundary layer greatly exceed the forces produced in a free sound field by absorption.

The streaming generated near obstacles may be analyzed on the basis of the boundary-layer equations. In the derivation of the equations for a plane boundary layer it is assumed, of course, that the velocity gradients in the direction of the normal to the boundary are far greater than the gradients in the direction of boundary. Inasmuch as the first of these gradients is determined in the case of interest by the viscous wavelength, while the second is determined by the boundary-layer equations is as follows: $$\lambda \gg \lambda^\prime$$. The Prandtl equations for the boundary layer near the plane $$y=0$$ have the form

$$\frac{\partial v_1}{\partial t} + v_1 \frac{\partial v_1}{\partial x} + v_2 \frac{\partial v_1}{\partial y} - \nu \frac{\partial^2 v_1}{\partial y^2} =\frac{\partial U}{\partial t} + U \frac{\partial U}{\partial x}\\frac{\partial v_1}{\partial x} + \frac{\partial v_2}{\partial y} = 0$$

where U(x,t) is the known flow velocity far from the boundary, In these equations no allowance is made for the dilatational viscosity or thermal conductivity. Using the perturbation method, we represent the velocity components in the form

$$v_i = v_i^\prime + v_i^{\prime\prime}$$

Now the boundary-layer equations are written:

in the first approximation

$$\frac{\partial v_1^\prime}{\partial t} - \nu \frac{\partial^2 v_1^\prime} {\partial y^2} = \frac{\partial U}{\partial t}\ \frac{\partial v_1^\prime}{\partial x} + \frac{\partial v_2^\prime}{\partial y} = 0$$

in the second approximation

$$\frac{\partial v_1^{\prime\prime}}{\partial t} - \nu \frac{\partial v_2^{\prime\prime}}{\partial y^2} = U \frac{\partial U} {\partial x} - v_1 \frac{\partial v_1^\prime} {\partial x} - v_2^\prime \frac{\partial v_1^\prime} {\partial y}\ \frac{\partial v_1^{\prime\prime}}{\partial x} + \frac{\partial v_2^{\prime\prime}}{\partial y} = 0$$

The boundary conditions for the solution of the problem for the half-space $$y>0$$ are $$v1|{y=0}=v2|{y=0} =0$$, and

The solution of these equations is simplified by the introduction of the $$\phi = \phi^\prime + \phi^{\prime\prime} + \cdots$$ such that

In Schlichting's paper these equations have been solved for the standing wave $$U(x,t) = U_0(x) \cos \omega t $$. The solution of Eqs. (56) has the form

$$\phi^\prime = \frac{U_0(x)\delta}{2} \text{Re}{\sigma^\prime(\mu)} e^{j\omega t }$$

where

and the dimensionless distance $$\mu = y/\delta$$ is incorporated. Form (59) the components of the first-approximation particle velocity vector have the form

$$v_1^\prime = U_0(x) {\cos \omega t - e^{-\mu} \cos(\omega t - \mu)}\ v_2^\prime = -\frac{d U_0} {dx} \frac{\delta}{\sqrt{2}} {\frac{2\mu - 1}{\sqrt{2}} \cos \omega t - \frac{1}{\sqrt{2}}\sin \omega t + e^{-\mu} \cos (\omega t - \mu - \frac{\pi}{4})}$$

The solution of Eqs.(57) in the second approximation is

where

The form of the function is shown in Fig. 6.

As apparent from (61), in the second approximation there exists in the acoustic boundary layer, besides the double-frequency osillations, a constant velocity component. The streaming velocity has the following components:

$$\bar{v_1^{\prime\prime}} = \frac{1}{2\omega} \frac{d U_0}{d x} \frac{d \sigma^{\prime\prime}}{d \mu}\ \bar{v_2^{\prime\prime}} = -\frac{\delta}{2\omega} [(\frac{d U_0}{d x})^2 + U_0 \frac{d^2 U_0}{d x^2}] \sigma^{\prime\prime}$$

in the case of a space-periodic wave U_0(x) = -v_0 cos(k x), we obtain for the streaming velocit components

Near the boundary( for \mu \ll 1) correct to terms ~ \mu^2, we obtain from (65)

$$\bar{v_1^{\prime\prime}} = -\frac{v_0^2}{4c_0} (\mu - \mu^2) \sin 2 kx \ \bar{v_2^{\prime\prime}} = \frac{v_0^2}{4c_0} k \delta \mu^2 \cos 2 kx$$

Let us note some of the characteristic features of this streaming: 1) the flow velocity at points where U_0 = 0 or d_U_0/dx = 0 has only a normal component with respect to the boundary; this means that, at particle velocity nodes or antinodes, the flow velocity is as the sign of the derivative d\sigma_2^{\prime\prime} /d\mu changes for \mu ~ 1(64) indicates that in this domain the sign of the x component of the flow velocity changes, 3) the y component of the velocity normal to the boundary, according to (64) changes isgn for \mu \approx 1.9 (for arbitrary x). This determines the dimensions of the boundary-layer vortices.

Henceforth, we shall refer to the quantity \nabla, i.e., the distance from the boundary to the position at which the normal component of the flow velocity vector with respect to the boundary vanishes, as the boundary-layer vortex thickness. For Schlichting streaming \nabla = 1.9 \delta.

Schlichtung streaming comprises a series of cortices mear a surface, their dimensions equal to $$\lambda/4 \times 1.9\delta$$. The nature of the streaming is illustrated in Fig.7. The transfer of the medium near the boundary surface proceeds from particle velocity nodes to antinodes; on the outer part of the boundary layer, transfer occurs in the opposite direction. It is important to note here that the direction of rotation of the boundary-layer vortices is opposite to the rotation of the cortices outside the boundary layer.

The results of the solution of the problem of an acoustic boundary layer near a plane surface can be used to determine the streaming over curved surfaces whose local radius of curvature is considerably greater than the viscous wavelength[24]. This approach has been used to investigate stationary streaming in the boundary layer of a circular cylinder situated at a velocity antinode of a standing wave, so that its axis is perpendicular to the direction of oscillation. The solution (64) can be used here , where U_0(x) = 2v_0 \sin(x/a) . Then the Schlichting solution for the stationary streaming velocity near the cylinder has the form

The character of the streaming is illustrated in Fig. 8. In each quadrant there are two vortices, one in a relatively thin boundary layer and the other outside the boundary layer. The angular velocity of the boundary-layer vortices is opposite to the angular velocity of the external vortices. In the case of flow past

¹. $$\Omega = \nabla \times \mathbf{v}$$ ↩