Rayleigh Acoustic Streaming Above a Vibrating Surface

Derives Proposition A66.1. At first order Equation (35.32) linearizes to \(\pp_{t}u_{1}=-\rho^{-1}\pp_{x}p_{1}+\nu\,\pp_{y}^{2}u_{1}\), the convective terms being of second order and the streamwise viscous term smaller than the transverse one by the square of the ratio of the layer thickness to the scale of \(U\). The outer field carries no shear, so it satisfies \(\pp_{t}u_{\infty}=-\rho^{-1}\pp_{x}p_{1}\); and by Equation (A66.1) applied to the pressure together with the thinness of the layer, \(p_{1}\) has the same value inside the layer as just outside it. Subtracting,

since \(\pp_{y}^{2}u_{\infty}=0\). Seek \(u_{1}-u_{\infty} =\Re\left[A(x)\,\ee^{\ii\omega t}\ee^{-\gamma y}\right]\): then \(\ii\omega=\nu\gamma^{2}\), so \(\gamma=\sqrt{\ii\omega/\nu}=\left(1+\ii\right)/\delta\) with \(\delta\) as in Equation (A66.4), the root with positive real part being the one that decays. No slip at \(y=0\) requires \(u_{1}=0\) there, so \(A=-U(x)\) and

and expanding the real part with \(\ee^{-\left(1+\ii\right)\eta}\ee^{\ii\omega t} =\ee^{-\eta}\ee^{\ii\left(\omega t-\eta\right)}\) gives Equation (A66.3).

For \(v_{1}\), integrate continuity \(\pp_{x}u_{1}+\pp_{y}v_{1}=0\) upward from the wall, where \(v_{1}=0\):

and

which is Equation (A66.5). Note that \(v_{1}\) carries the factor \(\dd U/\dd x\): a spatially uniform oscillation drives no normal motion, and it will drive no streaming either.

Derives Lemma A66.2. Take the streamwise component of Equation (35.32) at order \(\epsilon^{2}\) and average over a period. The unsteady term averages to zero in a state steady in the mean; the convective terms contribute \(\avg{u_{1}\pp_{x}u_{1}+v_{1}\pp_{y}u_{1}}\), since the second-order velocity's own convection is of fourth order; and of the viscous term only \(\nu\,\pp_{y}^{2}\avg{u_{2}}\) survives, by the same comparison of transverse with streamwise gradients used in Proposition A66.1. Hence

Now evaluate the same equation just outside the layer, where the field is the irrotational \(u_{\infty}\) and there is no mean shear. There

that is, the Reynolds stress of an irrotational oscillation is balanced entirely by the second-order mean pressure and drives nothing. The layer is thin, so \(\avg{p_{2}}\) is the same function of \(x\) inside it as immediately outside; subtracting Equation (A66.13) from Equation (A66.12) eliminates the pressure and leaves Equation (A66.10). By construction \(F\to0\) as \(\eta\to\infty\), and exponentially, since \(u_{1}\to u_{\infty}\) exponentially.

For the boundary conditions: no slip holds at every order, so \(\avg{u_{2}}(0)=0\); and the mean shear at the outer edge of the layer belongs to the outer streaming, which varies on the acoustic length scale rather than on \(\delta\) and is therefore negligible here. Then integrating Equation (A66.10) downward from infinity, \(\mu\,\pp_{y}\avg{u_{2}}(y)=-\int_{y}^{\infty}F\,\dd y'\), and integrating that upward from the wall,

the exchange of the order of integration being legitimate because \(F\) decays exponentially. That is Equation (A66.11).

Derives Theorem A66.3. Write \(g(\eta,t):=\Re\left[h\,\ee^{\ii\omega t}\right]\) and \(G(\eta,t):=\Re\left[H\,\ee^{\ii\omega t}\right]\), so that by Proposition A66.1

with \(U'=\dd U/\dd x\). The two Reynolds-stress terms are therefore

the factor \(\delta\) cancelling between \(v_{1}\) and \(\pp_{y}u_{1}\) — which is already the reason the answer will not contain the layer thickness. Since \(\avg{u_{\infty}\pp_{x}u_{\infty}}=UU'\avg{\cos^{2}\omega t} =\tfrac{1}{2}UU'\),

Evaluating \(\Psi\). For two quantities written as \(\Re\left[A\ee^{\ii\omega t}\right]\) and \(\Re\left[B\ee^{\ii\omega t}\right]\) the period average of the product is \(\tfrac{1}{2}\Re\left[A\bar{B}\right]\). With \(\pp_{\eta}g=\Re\left[h'\ee^{\ii\omega t}\right]\) and \(h'=\left(1+\ii\right)E\),

Now \(h=1-E\) with \(E=\ee^{-\eta}\ee^{-\ii\eta}\), so

and, using \(\overline{h'}=\left(1-\ii\right)\bar{E}\) together with \(\left(1-\ii\right)/\left(1+\ii\right)=-\ii\),

Taking real parts with \(\bar{E}=\ee^{-\eta} \left(\cos\eta+\ii\sin\eta\right)\),

the last term of Equation (A66.21) being purely imaginary. Assembling Equations (A66.20) and (A66.22),

The moment. By Equations (A66.11) and (A66.18), with \(y=\delta\eta\),

since \(\rho\delta^{2}/\mu=\delta^{2}/\nu=2/\omega\) by Equation (A66.4). The required integrals follow from

with \(\left(1-\ii\right)^{2}=-2\ii\) and \(\left(1-\ii\right)^{3}=-2-2\ii\). For \(n=1\) the right-hand side is \(1/\left(-2\ii\right)=\ii/2\), so

and for \(n=2\) it is \(2/\left(-2-2\ii\right)=-1/\left(1+\ii\right) =\left(-1+\ii\right)/2\), so

while \(\int_{0}^{\infty}\eta\,\ee^{-2\eta}\dd\eta=\tfrac{1}{4}\). Multiplying Equation (A66.23) by \(\eta\) and integrating term by term,

so \(\int_{0}^{\infty}\eta\Psi\,\dd\eta=\tfrac{3}{8}\) and Equation (A66.24) gives \(u_{s}=-\left(2/\omega\right)\left(3/8\right)UU'\), which is the first member of Equation (A66.15). The second member follows from \(\avg{u_{\infty}^{2}}=\tfrac{1}{2}U^{2}\), whose \(x\) derivative is \(UU'\).

Derives Proposition A66.4. From the proof of Lemma A66.2, \(\mu\left(\pp_{y}\avg{u_{2}}\right)_{y=0} =-\int_{0}^{\infty}F\,\dd y =-\rho UU'\delta\int_{0}^{\infty}\Psi\,\dd\eta\). The zeroth moments needed are \(\int_{0}^{\infty}\ee^{-\eta}\cos\eta\,\dd\eta =\int_{0}^{\infty}\ee^{-\eta}\sin\eta\,\dd\eta=\tfrac{1}{2}\), from Equation (A66.25) at \(n=0\), together with \(\int_{0}^{\infty}\ee^{-2\eta}\dd\eta=\tfrac{1}{2}\) and Equation (A66.26). Integrating Equation (A66.23),

so \(\int_{0}^{\infty}\Psi\,\dd\eta=-\tfrac{1}{4}\) and \(\mu\left(\pp_{y}\avg{u_{2}}\right)_{y=0} =\tfrac{1}{4}\rho\delta UU'\). Dividing by \(\mu=\rho\nu\) and using \(\nu=\omega\delta^{2}/2\) gives Equation (A66.29). Since \(\avg{u_{2}}\) starts from zero at the wall with a shear of one sign and ends at \(u_{s}\) of the other, it changes sign once, at a height that the explicit profile places within a small multiple of \(\delta\).

Derives Lemma A66.5. When \(k_{p}\gg\omega/c\) the compressible wave equation for the velocity potential, \(\nabla^{2}\phi=c^{-2}\pp_{t}^{2}\phi\), reduces within a bending wavelength to Equation (35.25), because the term on the right is smaller than \(\pp_{x}^{2}\phi\) by \(\left(\omega/ck_{p}\right)^{2}\). Separating with the \(x\) dependence imposed by the plate and requiring decay as \(y\to\infty\) gives \(\phi\propto\sin\left(k_{p}x\right)\ee^{-k_{p}y}\), and \(\nabla^{2}\phi=\left(-k_{p}^{2}+k_{p}^{2}\right)\phi=0\) confirms it. Fixing the constant by \(\pp_{y}\phi=v\left(x,0,t\right)\) at \(y=0\) gives Equation (A66.31), whence

Since \(k_{p}\delta\ll1\), the exponential is indistinguishable from unity across the Stokes layer, so the field Equation (A66.33) is what Proposition A66.1 sees as \(u_{\infty}\), and Equation (A66.32) follows.

The physical reading is worth one sentence, because the result is the opposite of the naive guess. Over an antinode the plate pushes air straight up and down and the horizontal motion vanishes by symmetry; over a nodal line the plate does not move at all, but the air expelled from the antinode on one side must cross to the other, and it is there that the horizontal velocity is greatest.

Derives Corollary A66.6. \(U^{2}=V_{0}^{2}\cos^{2}\left(k_{p}x\right)\), so \(\dd\left(U^{2}\right)/\dd x =-V_{0}^{2}k_{p}\sin\left(2k_{p}x\right)\) and \(u_{s}=-\left(3/8\omega\right)\dd\left(U^{2}\right)/\dd x\) is Equation (A66.34). Take a nodal line at \(x=0\) and the neighbouring antinode at \(x=\pi/2k_{p}\): on the interval between them \(\sin\left(2k_{p}x\right)>0\), so \(u_{s}>0\) and the flow is directed towards the antinode. By the reflection symmetry about the nodal line the same holds on the other side, so the nodal lines are lines of divergence of the slip flow and the antinodes are lines of convergence. Mass conservation then requires the fluid to rise above each antinode and to descend over each node, closing the circulation with an outward branch aloft.

The statement Equation (A66.15) makes about \(\abs{U}\) is general; what is special to the plate is Lemma A66.5, which places the maxima of \(\abs{U}\) over the nodes. The two together are what invert the figure.

Derives Corollary A66.8. In a standing wave the velocity amplitude is greatest at the displacement antinodes and zero at the nodes, and the wall is parallel to the axis, so the tangential field imposed on the Stokes layer is \(U_{0}\sin\left(kx\right)\) with \(U_{0}\) the velocity amplitude. Then \(\dd\left(U^{2}\right)/\dd x=U_{0}^{2}k\sin\left(2kx\right)\) and Equation (A66.15) gives Equation (A66.37), the second form using \(\omega=ck\). Between a node at \(x=0\) and the antinode at \(x=\pi/2k\), \(\sin\left(2kx\right)>0\) and \(u_{s}<0\): the flow runs from the antinode back to the node. So the displacement nodes are lines of convergence and the ridges form there.

Phenomenon 39.12 asserts this collection and uses the ridge spacing to measure a wavelength; the chapter takes the location of the ridges from the observation. It is worth noting that the mechanism is the same as the plate's and the apparent reversal between them is not in the streaming at all but in Lemma A66.5: in the tube the maxima of \(\abs{U}\) coincide with the visible antinodes of the wave, whereas over a plate they sit above its nodal lines. In both cases the powder goes where \(\abs{U}\) is least.