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

From WikiWaves
Jump to navigationJump to search
 
(20 intermediate revisions by 2 users not shown)
Line 5: Line 5:
 
}}
 
}}
  
* McCamy-Fuchs analytical solution of the scattering of regular waves by a vertical circular cylinder.
+
{{incomplete pages}}
  
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
+
== 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 = \mathbf{Re} \left\{\phi_I e^{i\omega t} \right \} \,</math></center>
+
== Problem ==
  
<center><math> \phi_I = \frac{i g A}{\omega} \frac{\cosh K(Z+H)}{\cosh K H} e^{-iKX} </math></center>
+
The incident potential is given as
  
Let the diffraction potential be:
+
<center><math> \Phi_I = \mathrm{Re} \left\{\phi_I e^{i\omega t} \right \} \,</math></center>
  
<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 = \frac{i g A}{\omega} \frac{\cosh k(z+h)}{\cosh k h} e^{-ikx} </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:
+
Let the diffraction potential be
  
<center><math> \left( \frac{\partial^2}{\partial X^2} + \frac{\partial^2}{\partial y^2} + K^2 \right) \psi = 0\, </math></center>
+
<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 27: 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:
Line 40: Line 45:
 
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}
Line 51: Line 56:
 
\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>
  
Line 70: Line 76:
 
As <math> R \to \infty\,</math>:
 
As <math> R \to \infty\,</math>:
  
<center><math> \psi(R,\theta) \sim 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 = \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>
  
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.

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