Wave Scattering By A Vertical Circular Cylinder

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Wave and Wave Body Interactions
Current Chapter Wave Scattering By A Vertical Circular Cylinder
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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

[math] \Phi_I = \mathrm{Re} \left\{\phi_I e^{i\omega t} \right \} \,[/math]
[math] \phi_I = \frac{i g A}{\omega} \frac{\cosh k(z+h)}{\cosh k h} e^{-ikx} [/math]

Let the diffraction potential be

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

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:

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

In polar coordinates:

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

The Helmholtz equation takes the form:

[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]

On the cylinder:

[math] \frac{\partial\phi_7}{\partial n} = - \frac{\partial\phi_I}{\partial n} \,[/math]

or

[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]

Here we make use of the familiar identity:

[math] e^{-ikR\cos\theta} = \sum_{m=0}^{\infty} \epsilon_m J_m ( k R ) \cos m \theta [/math]
[math] \epsilon_m = \begin{Bmatrix} 1, & m = 0 \\ 2(-i)^m, & m \gt 0 \end{Bmatrix} [/math]

Solution

Try:

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

Upon substitution in Helmholtz's equation we obtain:

[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]

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

[math] \begin{Bmatrix} J_m ( k R ) \\ Y_m ( k R ) \end{Bmatrix} [/math]

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]:

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

Also as [math] R \to \infty\, [/math]:

[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]
[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]

Hence the Hankel function:

[math] H_m^{(2)} ( k R ) = J_m ( k R ) - i Y_m ( k R ) \,[/math]
[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]

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

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

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

[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]

It follows that:

[math] A_m {H_m^{(2)}}^' (k a) = - J_m^' (k a) \,[/math]

or:

[math] A_m = - \frac{J_m^' ( k a ) }{{H_m^{(2)}}^' (k a)} \,[/math]

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

[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]

And the total original potential follows:

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

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:

[math] P = \mathrm{Re} \left\{ \mathbf{P} e^{i\omega t} \right\} \,[/math]
[math] \mathbf{P} = - i\omega \rho \left( \phi_I + \phi_7 \right) \, [/math]

Surge exciting force

The surge exciting force is given by

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

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


Ocean Wave Interaction with Ships and Offshore Energy Systems