There was an interesting talk on a wave-energy extraction device given by Tian-Shiang Yang (National Cheng Kung University, Taiwan).The question considered here seeks to determine how to relate the pressure prescribed on the free surface with the movement of the side pinned walls.  These walls will of course be coupled to a mass-spring-damper system.

An area where I think that we can make an impact is creating a better map from the pressure at the top p(x,t) to the lateral forces impacting the wall.

Imagine that the fluid is inviscid and irrotational.  Then, we could use the AFM method to map the boundary pressure at the free surface to the lateral walls.  Of course, we may not be able to create a direct map to the pressure.  However, chance are that we will be able to map to the tangential velocity of the fluid at the boundary.  With the velocity in the normal direction prescribed (via the movement of the walls), and the tangential velocity determined (via these AFM maps), the hope would be to determine a method to generate \displaystyle \int_{\text{side wall}} p(x,z,t) dA.

This calculation would ultimately allow for the coupling of the device to the side walls.

Of course, it isn’t quite clear how this will proceed.  Basically, along each of the rigid walls, we will need both the velocity in the normal direction, and the velocity in the tangential direction.  This tangential velocity is the component that remains unknown to the problem.

Recall that for any harmonic function \varphi, we find the following relationship for the fluid:

\displaystyle \int_{0}^L\varphi_x T_3 + \int_{L}^{x_2}\varphi_x T_2 - \varphi_z N_2\:dx

= \displaystyle \int_{x_1}^{x_2}\varphi_xq+x - \varphi_z\eta_t\:dx + \varphi_x + \int_{L}^{x_1}\varphi_x T_1 - \varphi_z N_1\:dx

where T_j, N_j represent the normal and tangential velocities at each interface.  The idea would be that given that there are three flat walls, and that Laplace’s equation is invariant under rotation, we could simply create three separate integral relations such as the one above that knock out the velocity in the tangent direction at each boundary.  This is because we already know the velocity in the normal direction at each boundary.

Thus, with the three equations that we generate, we can eliminate the tangential velocity at each of the linear boundaries and therefore solve for them.  Once we know the tangential velocities at each of the boundaries, it’s simple to calculate the total pressure along them.  Since we have the Bernoulli equation

\displaystyle \phi_t +\frac{1}{2} \left(\phi_x^2 + \phi_z^2\right) + gz + \frac{p}{\rho} = 0

we can also note that along the boundaries, this can be rewritten as

\displaystyle \phi_t(x,f(x),t) + \frac{1}{2}\left(T^2 + \frac{1 - f_x^2}{1+f_x^2}N^2\right) + gf(x) + \frac{p(x,f(x))}{\rho} = 0

where we used the fact that

\displaystyle \begin{bmatrix}1&f_x\\-f_x&1\end{bmatrix}\begin{bmatrix}u\\v\end{bmatrix} = \begin{bmatrix}T\\N\end{bmatrix}

I’m not quite sure if this will all play out, nor am I sure how I will be able to eliminate \phi_t at each of the side walls.  And then of course there’s the contact-line problem.  But those are questions for when I have a chance to think about things.

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