Difference between revisions of "Wave Scattering By A Vertical Circular Cylinder"
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− | + | {{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 | + | == 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> \ | + | <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> \left( \frac{\partial^2}{\partial | + | <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: | ||
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x=R\cos\theta \\ | x=R\cos\theta \\ | ||
y=R\sin\theta | y=R\sin\theta | ||
− | \end{Bmatrix} | + | \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} + | + | <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^{- | + | <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^{- | + | <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 ( | + | <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} + | + | <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 ( | + | J_m ( k R ) \\ |
− | Y_m ( | + | 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) \sim e^{- | + | <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 ( | + | <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 ( | + | <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)} ( | + | <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 | + | <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)} ( | + | <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 ( | + | <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)}}^' ( | + | <center><math> A_m {H_m^{(2)}}^' (k a) = - J_m^' (k a) \,</math></center> |
or: | or: | ||
− | <center><math> A_m = - \frac{J_m^' ( | + | <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 ( | + | <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 | + | <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> | + | 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 = \ | + | <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 == |
+ | The surge exciting force is given by | ||
− | <center><math> X_1 = \iint_{S_B} P n_1 | + | <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 | + | <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 | |
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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
Let the diffraction potential be
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:
In polar coordinates:
The Helmholtz equation takes the form:
On the cylinder:
or
Here we make use of the familiar identity:
Solution
Try:
Upon substitution in Helmholtz's equation we obtain:
This is the Bessel equation of order m accepting as solutions linear combinations of the Bessel functions
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]:
Also as [math]\displaystyle{ R \to \infty\, }[/math]:
Hence the Hankel function:
Satisfies the far field condition required by [math]\displaystyle{ \psi(R,\theta) \, }[/math]. So we set:
with the constants [math]\displaystyle{ A_m \, }[/math] to be determined. The cylinder condition requires:
It follows that:
or:
where [math]\displaystyle{ (')\, }[/math] denotes derivatives with respect to the argument. The solution for the total velocity potential follows in the form
And the total original potential follows:
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:
Surge exciting force
The surge exciting force is given by
Simple algebra in this case of water of infinite depth leads to the expression.
Ocean Wave Interaction with Ships and Offshore Energy Systems