Wave Scattering By A Vertical Circular Cylinder

2010-09-09T13:34:22Z

Javier Arcos: /* Problem */

{{Ocean Wave Interaction with Ships and Offshore Structures
| chapter title = Wave Scattering By A Vertical Circular Cylinder
| next chapter = [[Forward-Speed Ship Wave Flows]]
| previous chapter = [[Long Wavelength Approximations]]
}}

{{incomplete pages}}

== Introduction ==
This important flow accepts a closed-form analytical solution for arbitrary values of the wavelength <math>\lambda\,</math>. This was shown to be the case by [[McCamy and Fuchs 1954]] using separation of variables.

== Problem ==

The incident potential is given as

<center><math> \Phi_I = \mathrm{Re} \left\{\phi_I e^{i\omega t} \right \} \,</math></center>

<center><math> \phi_I = \frac{i g A}{\omega} \frac{\cosh k(z+h)}{\cosh k h} e^{-ikx} </math></center>

Let the diffraction potential be

<center><math> \phi_7 = \frac{i g A}{\omega} \frac{\cosh k(z+h)}{\cos k h} \psi(x,y) </math></center>

For <math>\phi_7\,</math> to satisfy the 3D Laplace equation, it is easy to show that <math>\psi\,</math> must satisfy the Helmholtz equation:

<center><math> \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + k^2 \right) \psi = 0\, </math></center>

In polar coordinates:

<center><math>
\begin{Bmatrix}
x=R\cos\theta \\
y=R\sin\theta
\end{Bmatrix} ; \quad \psi(R,\theta)
</math></center>

The Helmholtz equation takes the form:

<center><math> \left( \frac{\partial^2}{\partial R^2} + \frac{1}{R} \frac{\partial}{\partial R} + \frac{1}{R^2} \frac{\partial^2}{\partial\theta^2} + k^2 \right) \psi = 0 \, </math></center>

On the cylinder:

<center><math> \frac{\partial\phi_7}{\partial n} = - \frac{\partial\phi_I}{\partial n} \,</math></center>

or

<center><math> \frac{\partial\psi}{\partial R} = - \frac{\partial}{\partial R} \left( e^{-ikx} \right) = -\frac{\partial}{\partial R} \left( e^{-ikR\cos\theta} \right) </math></center>

Here we make use of the familiar identity:

<center><math> e^{-ikR\cos\theta} = \sum_{m=0}^{\infty} \epsilon_m J_m ( k R ) \cos m \theta </math></center>

<center><math> \epsilon_m = \begin{Bmatrix}
1, & m = 0 \\
2(-i)^m, & m > 0
\end{Bmatrix} </math></center>

== Solution ==
Try:

<center><math> \psi(R,\theta) = \sum_{m=0}^{\infty} A_m F_m ( k R ) \cos m \theta \, </math></center>

Upon substitution in Helmholtz's equation we obtain:

<center><math> \left( \frac{\partial^2}{\partial R^2} + \frac{1}{R} \frac{\partial}{\partial R} - \frac{m^2}{R^2} + k^2 \right) F_m ( k R ) = 0 </math></center>

This is the Bessel equation of order m accepting as solutions linear combinations of the Bessel functions

<center><math> \begin{Bmatrix}
J_m ( k R ) \\
Y_m ( k R )
\end{Bmatrix} </math></center>

The proper linear combination in the present problem is suggested by the radiation condition that <math> \psi\,</math> must satisfy:

As <math> R \to \infty\,</math>:

<center><math> \psi(R,\theta) \sim e^{-ikR + i\omega t} \,</math></center>

Also as <math> R \to \infty\, </math>:

<center><math> J_m ( k R ) \sim \left( \frac{2}{\pi k R} \right)^{1/2} \cos \left( k R - \frac{1}{2} m \pi - \frac{\pi}{4} \right) </math></center>

<center><math> Y_m ( k R ) \sim \left( \frac{2}{\pi k R} \right)^{1/2} \sin \left( k R - \frac{1}{2} m \pi - \frac{\pi}{4} \right) </math></center>

Hence the Hankel function:

<center><math> H_m^{(2)} ( k R ) = J_m ( k R ) - i Y_m ( k R ) \,</math></center>

<center><math> \sim \left( \frac{2}{\pi k R} \right)^{1/2} e^{-i \left( k R - \frac{1}{2} m \pi - \frac{\pi}{4} \right)} </math></center>

Satisfies the far field condition required by <math> \psi(R,\theta) \,</math>. So we set:

<center><math> \psi(r,\theta) = \sum_{m=0}^{\infty} \epsilon_m A_m H_m^{(2)} ( k R ) \cos m \theta </math></center>

with the constants <math> A_m \,</math> to be determined. The cylinder condition requires:

<center><math> \left. \frac{\partial\psi}{\partial R} \right|_{R=a} = - \frac{\partial}{\partial R} \sum_{m=0}^{\infty} \epsilon_m J_m ( k R ) \left.\cos m \theta \right|_{r=a} </math></center>

It follows that:

<center><math> A_m {H_m^{(2)}}^' (k a) = - J_m^' (k a) \,</math></center>

or:

<center><math> A_m = - \frac{J_m^' ( k a ) }{{H_m^{(2)}}^' (k a)} \,</math></center>

where <math> (')\,</math> denotes derivatives with respect to the argument. The solution for the total velocity potential follows in the form

<center><math> (\psi+x)(r,\theta) = \sum_{m=0}^{\infty} \epsilon_m \left[ J_m (k R) - \frac{J_m^'(k a)}{{H_m^{(2)}}^'(k a)} H_m^{(2)} (k a) \right] \cos m \theta </math></center>

And the total original potential follows:

<center><math> \phi = \phi_I + \phi_7 = \frac{i g A}{\omega} \frac{\cosh k (z+h)}{\cosh k h } (\psi+x) (r,\theta) </math></center>

In the limit as <math> h \to \infty\,, \quad \frac{\cosh k (z+h)}{k h} \to e^{k z} \,</math> and the series expansion solution survives.

The total complex potential, incident and scattered, was derived above.

The hydrodynamic pressure follows from Bernoulli:

<center><math> P = \mathrm{Re} \left\{ \mathbf{P} e^{i\omega t} \right\} \,</math></center>

<center><math> \mathbf{P} = - i\omega \rho \left( \phi_I + \phi_7 \right) \, </math></center>

== Surge exciting force ==
The surge exciting force is given by

<center><math> X_1 = \iint_{S_B} P n_1 \mathrm{d}S = \mathrm{Re} \left\{ \mathbf{X}_1 e^{i\omega t} \right\} </math></center>

<center><math> \mathbf{X}_1 = \rho \int_{-\infty}^0 \mathrm{d}z \int_0^{2\pi} a \mathrm{d}\theta \left( - i \omega \frac{i g A}{\omega} \right) e^{k z} n_1 (\psi + x)_{R=a} </math></center>

Simple algebra in this case of water of infinite depth leads to the expression.

[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]

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Wave Scattering By A Vertical Circular Cylinder