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

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* Mccamy-Fuchs analytical solution of the scattering of regular waves by a vertical circular cylinder.
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{{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]]
 +
}}
  
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-Fuchs using separation of variables
+
{{incomplete pages}}
  
<center><math> \Phi_I = \mathfrak{Re} \left\{\phi_I e^{i\omega t} \right \} \,</math></center>
+
== 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.
  
<center><math> \phi_I = \frac{i g A}{\omega} \frac{\cosh K(Z+H)}{\cosh K H} e^{-iKX} </math></center>
+
== Problem ==
  
Let the diffraction potential be:
+
The incident potential is given as
  
<center><math> \phi_7 = \frac{i g A}{\omega} \frac{\cosh K(Z+H)}{\cos K H} \psi(X,Y) </math></center>
+
<center><math> \Phi_I = \mathrm{Re} \left\{\phi_I e^{i\omega t} \right \} \,</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> \phi_I = \frac{i g A}{\omega} \frac{\cosh k(z+h)}{\cosh k h} e^{-ikx} </math></center>
  
<center><math> \left( \frac{\partial^2}{\partial X^2} + \frac{\partial^2}{\partial y^2} + K^2 \right) \psi = 0\, </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:
 
In polar coordinates:
Line 21: Line 32:
 
   x=R\cos\theta \\
 
   x=R\cos\theta \\
 
   y=R\sin\theta
 
   y=R\sin\theta
\end{Bmatrix} j \quad \psi(R,\theta)
+
\end{Bmatrix} ; \quad \psi(R,\theta)
 
</math></center>
 
</math></center>
  
 
The Helmholtz equation takes the form:
 
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>
+
<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:
 
On the cylinder:
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or
 
or
  
<center><math> \frac{\partial\psi}{\partial R} = - \frac{\partial}{\partial R} \left( e^{-iKX} \right) = -\frac{\partial}{\partial R} \left( e^{-iKE\cos R} \right) </math></center>
+
<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:
 
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> 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}
 
<center><math> \epsilon_m = \begin{Bmatrix}
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\end{Bmatrix} </math></center>
 
\end{Bmatrix} </math></center>
  
 +
== Solution ==
 
Try:
 
Try:
  
<center><math> \psi(R,\theta) = \sum_{m=0}^{infty} A_m F_m ( K R ) \cos m \theta \, </math></center>
+
<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:
 
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>
+
<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
 
This is the Bessel equation of order m accepting as solutions linear combinations of the Bessel functions
  
 
<center><math> \begin{Bmatrix}
 
<center><math> \begin{Bmatrix}
   J_m ( K R } \\
+
   J_m ( k R ) \\
   Y_m ( K R )
+
   Y_m ( k R )
 
\end{Bmatrix} </math></center>
 
\end{Bmatrix} </math></center>
  
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As <math> R \to \infty\,</math>:
 
As <math> R \to \infty\,</math>:
  
<center><math> \psi(R,\theta) \nsim e^{-iKR + i\omega t} \,</math></center>
+
<center><math> \psi(R,\theta) \sim e^{-ikR + i\omega t} \,</math></center>
  
 
Also as <math> R \to \infty\, </math>:
 
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> 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>
+
<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:
 
Hence the Hankel function:
  
<center><math> H_m^{(2)} ( K R } = J_m ( K R ) - i Y_m ( K R ) \,</math></center>
+
<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>
+
<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:
 
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>
+
<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:
+
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>
+
<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:
 
It follows that:
  
<center><math> A_m {H_m^{(2)}}^' (K a) = - J_m^' (K a) \,</math></center>
+
<center><math> A_m {H_m^{(2)}}^' (k a) = - J_m^' (k a) \,</math></center>
  
 
or:
 
or:
  
<center><math> A_m = - \frac{J_m^' ( K a ) }{{H_m^{(2)}}^' (K a)} \,</math></center>
+
<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
 
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>
+
<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:
 
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>
+
<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} \longrightarrow e^{K Z} \,</math> and the series expansion solution survives.
+
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.
  
<u>Surge exciting force</u>
+
The total complex potential, incident and scattered, was derived above.
  
The total complex potential, incident and scattered was derived above. The hydrodynamic pressure follows from Bernoulli:
+
The hydrodynamic pressure follows from Bernoulli:
  
<center><math> P = \mathfrak{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>
  
The Surge exciting force is given by
+
== Surge exciting force ==
 +
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 \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.
 
Simple algebra in this case of water of infinite depth leads to the expression.
 +
 +
 +
[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]

Latest revision as of 13:34, 9 September 2010

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



Introduction

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 and Fuchs 1954 using separation of variables.

Problem

The incident potential is given as

[math]\displaystyle{ \Phi_I = \mathrm{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} ; \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^{-ikR\cos\theta} \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]

Solution

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} \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]\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]

Surge exciting force

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 \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