Difference between revisions of "Long Wavelength Approximations"

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= Applications of GI Taylor's formula in wave-body interactions =
 
= Applications of GI Taylor's formula in wave-body interactions =
  
== Archimedean hydrostaticS ==
+
== Archimedean hydrostatics ==
  
 
<center><math> P=-\rho g Z, \quad \frac{\partial P}{\partial Z} = - \rho g \,</math></center>
 
<center><math> P=-\rho g Z, \quad \frac{\partial P}{\partial Z} = - \rho g \,</math></center>
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<center><math> \phi: \ \mbox{no added mass since there is no flow} </math></center>
 
<center><math> \phi: \ \mbox{no added mass since there is no flow} </math></center>
  
* So Archimedes' formula is a special case of GI Taylor when there
+
So Archimedes' formula is a special case of GI Taylor when there
 
is no flow. This offers an intuitive meaning to the term that includes the body displacement.
 
is no flow. This offers an intuitive meaning to the term that includes the body displacement.
  

Revision as of 20:39, 28 June 2007

Introduction

Very frequently the length of ambient waves [math]\displaystyle{ \lambda \, }[/math] is large compared to the dimension of floating bodies. For example the length of a wave with period [math]\displaystyle{ T=10 \ \mbox{sec}\, }[/math] is [math]\displaystyle{ \lambda \simeq T^2 + \frac{T^2}{2} \simeq 150\mbox{m} \, }[/math]. The beam of a ship with length [math]\displaystyle{ L=100\mbox{m}\, }[/math] can be [math]\displaystyle{ 20\mbox{m}\, }[/math] as is the case for the diameter of the leg of an offshore platform.


GI Taylor's formula

Consider a flow field given by

[math]\displaystyle{ U(X,t):\ \mbox{Velocity of ambient unidirectional flow} \, }[/math]

[math]\displaystyle{ P(X,t):\ \mbox{Pressure corresponding to} \ U(X,t) \, }[/math]

[math]\displaystyle{ \lambda \sim \frac{|U|}{|\nabla U|} \gg B \ = \ \mbox{Body characteristic dimension} \, }[/math]

In the absence of viscous effects and to leading order for [math]\displaystyle{ \lambda \gg B \, }[/math]:

[math]\displaystyle{ F_X = - \left( \forall + \frac{A_{11}}{\rho} \right) \left. \frac{\partial P}{\partial x} \right|_{X=0} }[/math]

where

[math]\displaystyle{ \ F_X: \ \mbox{Force in X-direction} \, }[/math]
[math]\displaystyle{ \ \forall: \ \mbox{Body displacement}\, }[/math]
[math]\displaystyle{ \ A_{11}: \ \mbox{Surge added mass} \, }[/math]

Derivation using Euler's equations

An alternative form of GI Taylor's formula for a fixed body follows from Euler's equations:

[math]\displaystyle{ \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial X} \simeq - \frac{1}{\rho} \frac{\partial P}{\partial X} }[/math]

Thus:

[math]\displaystyle{ F_X = \left( \rho \forall + A_{11} \right) + \left( \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial X} \right)_{X=0} }[/math]

If the body is also translating in the X-direction with displacement [math]\displaystyle{ X_1(t)\, }[/math] then the total force becomes

[math]\displaystyle{ \bullet \ F_X = \left( \rho\forall+A_{11} \right) \left( \frac{\partial U}{\partial t} + U \frac{\partial U}{\partial X} \right) - A_{11} \frac{d^2X_1(t)}{dt^2} }[/math]

Often, when the ambient velocity [math]\displaystyle{ U\, }[/math] is arising from plane progressive waves, [math]\displaystyle{ \left| U \frac{\partial U}{\partial X} \right| = 0(A^2) \, }[/math] and is omitted. Note that [math]\displaystyle{ U\, }[/math] does not include disturbance effects due to the body.

Applications of GI Taylor's formula in wave-body interactions

Archimedean hydrostatics

[math]\displaystyle{ P=-\rho g Z, \quad \frac{\partial P}{\partial Z} = - \rho g \, }[/math]
[math]\displaystyle{ F_Z = - ( \forall + \phi ) \frac{\partial P}{\partial Z} = \rho g \forall }[/math]
[math]\displaystyle{ \phi: \ \mbox{no added mass since there is no flow} }[/math]

So Archimedes' formula is a special case of GI Taylor when there is no flow. This offers an intuitive meaning to the term that includes the body displacement.

Regular waves over a circle fixed under the free surface

[math]\displaystyle{ \Phi_I = \mathfrak{Re} \left\{ \frac{i g A}{\omega} e^{KZ-iKX+i\omega t} \right\}, \quad K=\frac{\omega^2}{g} \, }[/math]
[math]\displaystyle{ u=\frac{\partial \Phi_I}{\partial X} = \mathfrak{Re} \left\{ \frac{i g A}{\omega} (-i K) e^{K Z - i K X + i \omega t } \right \} }[/math]
[math]\displaystyle{ \mathfrak{Re} \left\{ \omega A e^{ - K d +i \omega t} \right\}_{X=0,Z=-d} }[/math]

So the horizontal force on the circle is:

[math]\displaystyle{ F_X = \left( \forall + \frac{a_{11}}{\rho} \right) \frac{\partial u}{\partial t} + O \left( Z^2 \right) }[/math]
[math]\displaystyle{ \forall =\pi a^2, \quad a_{11} = \pi \rho a^2 \, }[/math]
[math]\displaystyle{ \frac{\partial u}{\partial t} = \mathfrak{Re} \left\{ i\omega^2 e^{-K d + i \omega t} \right\} }[/math]

Thus:

[math]\displaystyle{ F_X = - 2 \pi a^2 \omega^2 A e^{-K d} \sin \omega t \, }[/math]
  • Derive the vertical force along very similar lines. It is simply [math]\displaystyle{ 90^\circ\, }[/math] out of phase relative to [math]\displaystyle{ F_X\, }[/math] with the same modulus.

Horizontal force on a fixed circular cylinder of draft [math]\displaystyle{ T\, }[/math]

This case arises frequently in wave interactions with floating offshore platforms.

Here we will evaluate [math]\displaystyle{ \frac{\partial u}{\partial t} \, }[/math] on the axis of the platform and use a strip wise integration to evaluate the total hydrodynamic force.

[math]\displaystyle{ u = \frac{\partial \Phi}{\partial X} = \mathfrak{Re} \left\{ \frac{i g A}{\omega} (- i K) e^{KZ-i K X + i\omega t} \right\} }[/math]
[math]\displaystyle{ = \mathfrak{Re} \left\{ \omega A e^{KZ+i\omega t} \right\}_{X=0} \, }[/math]
[math]\displaystyle{ \frac{\partial u}{\partial t} (Z) = \mathfrak{Re} \left\{ \omega A ( i \omega) e^{KZ+i\omega t} \right\} }[/math]
[math]\displaystyle{ = - \omega^2 A e^{KZ} \sin \omega t \, }[/math]

The differential horizontal force over a strip [math]\displaystyle{ d Z \, }[/math] at a depth [math]\displaystyle{ Z \, }[/math] becomes:

[math]\displaystyle{ dF_Z = \rho ( \forall + a_{11} ) \frac{\partial u}{\partial t} d Z \, }[/math]
[math]\displaystyle{ \rho ( \pi a^2 + \pi a^2 ) \frac{\partial u}{\partial t} d Z \, }[/math]
[math]\displaystyle{ 2 \pi \rho a^2 \left( - \omega^2 A e^{KZ} \right) \sin \omega t d Z }[/math]

The total horizontal force over a truncated cylinder of draft [math]\displaystyle{ T\, }[/math] becomes:

[math]\displaystyle{ F_X = \int_{-T}^{0} dZ dF = -2\pi\rho a^2 \omega^2 A \sin \omega t \int_{-T}^0 e^{KZ} dZ }[/math]
[math]\displaystyle{ X_1 \equiv F_X = - 2 \pi \rho a^2 \omega^2 A \sin \omega t \cdot \frac{1-e^{-KT}}{K} }[/math]
  • This is a very useful and practical result. It provides an

estimate of the surge exciting force on one leg of a possibly multi-leg platform

  • As [math]\displaystyle{ T \to \infty; \quad \frac{1-e^{-KT}}{K} \to \frac{1}{K} \, }[/math]

Horizontal force on multiple vertical cylinders in any arrangement:

The proof is essentially based on a phasing argument. Relative to the reference frame:

[math]\displaystyle{ \Phi_I = \mathfrak{Re} \left\{ \frac{i g A}{\omega} e^{KZ-iKX + i\omega t} \right\} \, }[/math]
  • Express the incident wave relative to the local frames by

introducing the phase factors:

[math]\displaystyle{ \mathbf{P}_i = e^{-iKX_i} }[/math]

Let:

[math]\displaystyle{ X+X_i + \xi_i \, }[/math]

Then relative to the i-th leg:

[math]\displaystyle{ \Phi_I^{(i)} = \mathfrak{Re} \left\{ \frac{ i g A}{\omega} e^{KZ - iK\xi_i + i\omega t} \mathbf{P}_i \right\} \quad i=1,\cdots,N }[/math]

Ignoring interactions between legs, which is a good approximation in long waves, the total exciting force on an n-cylinder platform is:

[math]\displaystyle{ \mathbf{X}_1^N = \sum_{i=1}^N \mathbf{P}_i \mathbf{X}_1 \, }[/math]

The above expression gives the complex amplitude of the force with [math]\displaystyle{ \mathbf{X}_1\, }[/math] given in the single cylinder case.

  • The above technique may be easily extended to estimate the Sway force and Yaw moment on n-cylinders with little extra effort.

Surge exciting force on a 2D section

[math]\displaystyle{ \Phi_I = \mathfrak{Re} \left\{ \frac{ i g A}{\omega} e^{KZ-iKX+i\omega t} \right\} \, }[/math]
[math]\displaystyle{ u=\mathfrak{Re} \left\{ \frac{ i g A}{\omega} (- i K ) e^{KZ-iKX+i\omega t} \right\} \, }[/math]
[math]\displaystyle{ \frac{\partial u}{\partial t} = \mathfrak{Re} \left\{ \frac{ i g A}{\omega} \left(- i \frac{\omega^2}{g} \right) (i\omega) e^{i\omega t} \right\}_{X=0, Z=0} \, }[/math]
[math]\displaystyle{ = \mathfrak{Re} \left\{ i \omega^2 A e^{i\omega t} \right\} = -\omega^2 A \sin \omega t \, }[/math]
[math]\displaystyle{ \mathbf{X}_1 = \left( \rho \forall + A_{11} \right) \frac{\partial u}{\partial t} = - \omega^2 A \sin \omega t ( \rho \forall + A_{11} ) \, }[/math]
  • If the body section is a circle with radius [math]\displaystyle{ a\, }[/math]:
[math]\displaystyle{ \rho \forall = A_{11} = \pi\rho \frac{a^2}{2} \, }[/math]

So in long waves, the surge exciting force is equally divided between the Froude-Krylov and the diffraction components. This is not the case for Heave!

Heave exciting force on a surface piercing section

In long waves, the leading order effect in the exciting force is the hydrostatic contribution:

[math]\displaystyle{ \mathbf{X}_i \sim \rho g A_w A \, }[/math]

where [math]\displaystyle{ A_w\, }[/math] is the body water plane area in 2D or 3D. [math]\displaystyle{ A\, }[/math] is the wave amplitude. This can be shown to be the leading order contribution from the Froude-Krylov force

[math]\displaystyle{ \mathbf{X}_3^{FK} = \rho g A \iint_{S_B} e^{KZ-iKX} n_3 dS \, }[/math]

Using the Taylor series expansion:

[math]\displaystyle{ e^{KZ-iKX} = 1 + ( KZ - iKX ) + O ( KB )^2 \, }[/math]

It is easy to verify that: [math]\displaystyle{ \mathbf{X}_3 \to \rho g A A_w \, }[/math].

The scattering contribution is of order [math]\displaystyle{ KB\, }[/math]. For submerged bodies: [math]\displaystyle{ \mathbf{X}_3^{FK}=O(KB)\, }[/math].


Ocean Wave Interaction with Ships and Offshore Energy Systems