Difference between revisions of "Wave Scattering By A Vertical Circular Cylinder"

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The total complex potential, incident and scattered was derived above. The hydrodynamic pressure follows from Bernoulli:
 
The total complex potential, incident and scattered was derived above. The hydrodynamic pressure follows from Bernoulli:
  
<center><math> P = \mathbf{Re} \left\{ \mathbf{P} e^{i\omega t} \right\} \,</math></center>
+
<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>
 
<center><math> \mathbf{P} = - i\omega \rho \left( \phi_I + \phi_7 \right) \, </math></center>
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The surge exciting force is given by
 
The surge exciting force is given by
  
<center><math> X_1 = \iint_{S_B} P n_1 dS = \mathbf{Re} \left\{ \mathbf{X}_1 e^{i\omega t} \right\} </math></center>
+
<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 dZ \int_0^{2\pi} a d\theta \left( - i \omega \frac{i g A}{\omega} \right) e^{K Z} n_1 (\psi + x)_{R=a} </math></center>
+
<center><math> \mathbf{X}_1 = \rho \int_{-\infty}^0 \mathrm{d}Z \int_0^{2\pi} a 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.
 
Simple algebra in this case of water of infinite depth leads to the expression.

Revision as of 15:53, 9 April 2009

Wave and Wave Body Interactions
Current Chapter Wave Scattering By A Vertical Circular Cylinder
Next Chapter Forward-Speed Ship Wave Flows
Previous Chapter Long Wavelength Approximations


McCamy-Fuchs analytical solution of the scattering of regular waves by a vertical circular cylinder

This important flow accepts a closed-form analytical solution for arbitrary values of the wavelength [math]\displaystyle{ \lambda\, }[/math]. This was shown to be the case by McCamy-Fuchs using separation of variables

[math]\displaystyle{ \Phi_I = \mathbf{Re} \left\{\phi_I e^{i\omega t} \right \} \, }[/math]
[math]\displaystyle{ \phi_I = \frac{i g A}{\omega} \frac{\cosh K(Z+H)}{\cosh K H} e^{-iKX} }[/math]

Let the diffraction potential be:

[math]\displaystyle{ \phi_7 = \frac{i g A}{\omega} \frac{\cosh K(Z+H)}{\cos K H} \psi(X,Y) }[/math]
  • For [math]\displaystyle{ \phi_7\, }[/math] to satisfy the 3D Laplace equation, it is easy to show that [math]\displaystyle{ \psi\, }[/math] must satisfy the Helmholtz equation:
[math]\displaystyle{ \left( \frac{\partial^2}{\partial X^2} + \frac{\partial^2}{\partial y^2} + K^2 \right) \psi = 0\, }[/math]

In polar coordinates:

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

The Helmholtz equation takes the form:

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

or

[math]\displaystyle{ \frac{\partial\psi}{\partial R} = - \frac{\partial}{\partial R} \left( e^{-iKX} \right) = -\frac{\partial}{\partial R} \left( e^{-iKE\cos R} \right) }[/math]

Here we make use of the familiar identity:

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

Try:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \psi\, }[/math] must satisfy:

As [math]\displaystyle{ R \to \infty\, }[/math]:

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

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

[math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ H_m^{(2)} ( K R ) = J_m ( K R ) - i Y_m ( K R ) \, }[/math]
[math]\displaystyle{ \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]\displaystyle{ \psi(R,\theta) \, }[/math]. So we set:

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

With the constants [math]\displaystyle{ A_m \, }[/math] to be determined. The cylinder condition requires:

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

or:

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

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

[math]\displaystyle{ (\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]\displaystyle{ \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]\displaystyle{ H \to \infty \quad \frac{\cosh K (Z+H)}{K H} \longrightarrow e^{K Z} \, }[/math] and the series expansion solution survives.

Surge exciting force

The total complex potential, incident and scattered was derived above. The hydrodynamic pressure follows from Bernoulli:

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

The surge exciting force is given by

[math]\displaystyle{ X_1 = \iint_{S_B} P n_1 \mathrm{d}S = \mathrm{Re} \left\{ \mathbf{X}_1 e^{i\omega t} \right\} }[/math]
[math]\displaystyle{ \mathbf{X}_1 = \rho \int_{-\infty}^0 \mathrm{d}Z \int_0^{2\pi} a 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