Difference between revisions of "Long Wavelength Approximations"
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| chapter title = Long Wavelength Approximations | | chapter title = Long Wavelength Approximations | ||
| next chapter = [[Wave Scattering By A Vertical Circular Cylinder]] | | next chapter = [[Wave Scattering By A Vertical Circular Cylinder]] | ||
− | | previous chapter = [[ | + | | previous chapter = [[Linear Wave-Body Interaction]] |
}} | }} | ||
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=== Regular waves over a circle fixed under the free surface === | === Regular waves over a circle fixed under the free surface === | ||
− | <center><math> \Phi_I = \mathrm{Re} \left\{ \frac{i g A}{\omega} e^{ | + | <center><math> \Phi_I = \mathrm{Re} \left\{ \frac{i g A}{\omega} e^{kz-ikx+i\omega t} \right\}, \quad k=\frac{\omega^2}{g} \, </math></center> |
− | <center><math>u=\frac{\partial \Phi_I}{\partial x} = \mathrm{Re} \left\{ \frac{i g A}{\omega} (-i | + | <center><math>u=\frac{\partial \Phi_I}{\partial x} = \mathrm{Re} \left\{ \frac{i g A}{\omega} (-i k) e^{k z - i k x + i \omega t } \right \} </math></center> |
− | <center><math> \mathrm{Re} \left\{ \omega A e^{ - | + | <center><math> \mathrm{Re} \left\{ \omega A e^{ - k h +i \omega t} \right\}_{x=0,z=-h} </math></center> |
So the horizontal force on the circle is: | So the horizontal force on the circle is: | ||
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<center><math> \forall =\pi a^2, \quad A_{11} = \pi \rho a^2 \,</math></center> | <center><math> \forall =\pi a^2, \quad A_{11} = \pi \rho a^2 \,</math></center> | ||
− | <center><math> \frac{\partial u}{\partial t} = \mathrm{Re} \left\{ i\omega^2 e^{- | + | <center><math> \frac{\partial u}{\partial t} = \mathrm{Re} \left\{ i\omega^2 e^{-kh + i \omega t} \right\} </math></center> |
Thus: | Thus: | ||
− | <center><math> F_x = - 2 \pi a^2 \omega^2 A e^{- | + | <center><math> F_x = - 2 \pi a^2 \omega^2 A e^{-k h} \sin \omega t \,</math></center> |
We can derive the vertical force along very similar lines. It is simply <math>90^\circ\,</math> out of phase relative to <math>F_x\,</math> with the same modulus. | We can derive the vertical force along very similar lines. It is simply <math>90^\circ\,</math> out of phase relative to <math>F_x\,</math> with the same modulus. | ||
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Here we will evaluate <math> \frac{\partial u}{\partial t} \,</math> on the axis of the platform and use a strip wise integration to evaluate the total hydrodynamic force. | Here we will evaluate <math> \frac{\partial u}{\partial t} \,</math> on the axis of the platform and use a strip wise integration to evaluate the total hydrodynamic force. | ||
− | <center><math> u = \frac{\partial \Phi}{\partial x} = \mathrm{Re} \left\{ \frac{i g A}{\omega} (- i | + | <center><math> u = \frac{\partial \Phi}{\partial x} = \mathrm{Re} \left\{ \frac{i g A}{\omega} (- i k) e^{kz-i k x + i\omega t} \right\} </math></center> |
− | <center><math> = \mathrm{Re} \left\{ \omega A e^{ | + | <center><math> = \mathrm{Re} \left\{ \omega A e^{kz+i\omega t} \right\}_{x=0} \,</math></center> |
− | <center><math> \frac{\partial u}{\partial t} (z) = \mathrm{Re} \left\{ \omega A ( i \omega) e^{ | + | <center><math> \frac{\partial u}{\partial t} (z) = \mathrm{Re} \left\{ \omega A ( i \omega) e^{kz+i\omega t} \right\} </math></center> |
− | <center><math> = - \omega^2 A e^{ | + | <center><math> = - \omega^2 A e^{kz} \sin \omega t \,</math></center> |
The differential horizontal force over a strip <math> \mathrm{d} z \,</math> at a depth <math> z \,</math> becomes: | The differential horizontal force over a strip <math> \mathrm{d} z \,</math> at a depth <math> z \,</math> becomes: | ||
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<center><math> \rho ( \pi a^2 + \pi a^2 ) \frac{\partial u}{\partial t} \mathrm{d} z \,</math></center> | <center><math> \rho ( \pi a^2 + \pi a^2 ) \frac{\partial u}{\partial t} \mathrm{d} z \,</math></center> | ||
− | <center><math> 2 \pi \rho a^2 \left( - \omega^2 A e^{ | + | <center><math> 2 \pi \rho a^2 \left( - \omega^2 A e^{kz} \right) \sin \omega t \mathrm{d} z </math></center> |
The total horizontal force over a truncated cylinder of draft <math>T\,</math> becomes: | The total horizontal force over a truncated cylinder of draft <math>T\,</math> becomes: | ||
− | <center><math> F_x = \int_{-T}^{0} \mathrm{d}Z \mathrm{d}F = -2\pi\rho a^2 \omega^2 A \sin \omega t \int_{-T}^0 e^{ | + | <center><math> F_x = \int_{-T}^{0} \mathrm{d}Z \mathrm{d}F = -2\pi\rho a^2 \omega^2 A \sin \omega t \int_{-T}^0 e^{kz} \mathrm{d}z </math></center> |
− | <center><math> X_1 \equiv F_x = - 2 \pi \rho a^2 \omega^2 A \sin \omega t \cdot \frac{1-e^{- | + | <center><math> X_1 \equiv F_x = - 2 \pi \rho a^2 \omega^2 A \sin \omega t \cdot \frac{1-e^{-kT}}{k} </math></center> |
− | 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> T \to \infty; \quad \frac{1-e^{- | + | 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> T \to \infty; \quad \frac{1-e^{-kT}}{k} \to \frac{1}{k}\,</math> |
=== Horizontal force on multiple vertical cylinders in any arrangement === | === Horizontal force on multiple vertical cylinders in any arrangement === | ||
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The proof is essentially based on a phasing argument. Relative to the reference frame, | The proof is essentially based on a phasing argument. Relative to the reference frame, | ||
− | <center><math> \Phi_I = \mathrm{Re} \left\{ \frac{i g A}{\omega} e^{ | + | <center><math> \Phi_I = \mathrm{Re} \left\{ \frac{i g A}{\omega} e^{kz-ikx + i\omega t} \right\} \,</math></center> |
Expressing the incident wave relative to the local frames by introducing the phase factors, | Expressing the incident wave relative to the local frames by introducing the phase factors, | ||
− | <center><math> \mathbf{P}_i = e^{- | + | <center><math> \mathbf{P}_i = e^{-ikx_i} </math></center> |
and letting | and letting | ||
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Then relative to the i-th leg, | Then relative to the i-th leg, | ||
− | <center><math> \Phi_I^{(i)} = \mathrm{Re} \left\{ \frac{ i g A}{\omega} e^{ | + | <center><math> \Phi_I^{(i)} = \mathrm{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></center> |
Ignoring interactions between legs, which is a good approximation in long waves, the total exciting force on an n-cylinder platform is | Ignoring interactions between legs, which is a good approximation in long waves, the total exciting force on an n-cylinder platform is | ||
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=== Surge exciting force on a 2D section === | === Surge exciting force on a 2D section === | ||
− | <center><math> \Phi_I = \mathrm{Re} \left\{ \frac{ i g A}{\omega} e^{ | + | <center><math> \Phi_I = \mathrm{Re} \left\{ \frac{ i g A}{\omega} e^{kz-ikx+i\omega t} \right\} \,</math></center> |
− | <center><math> u=\mathrm{Re} \left\{ \frac{ i g A}{\omega} (- i | + | <center><math> u=\mathrm{Re} \left\{ \frac{ i g A}{\omega} (- i k ) e^{kz-ikx+i\omega t} \right\} \,</math></center> |
<center><math> \frac{\partial u}{\partial t} = \mathrm{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></center> | <center><math> \frac{\partial u}{\partial t} = \mathrm{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></center> | ||
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where <math>A_w\,</math> is the body water plane area in 2D or 3D. <math>A\,</math> is the wave amplitude. This can be shown to be the leading order contribution from the Froude-Krylov force: | where <math>A_w\,</math> is the body water plane area in 2D or 3D. <math>A\,</math> is the wave amplitude. This can be shown to be the leading order contribution from the Froude-Krylov force: | ||
− | <center><math> \mathbf{X}_3^{FK} = \rho g A \iint_{S_B} e^{ | + | <center><math> \mathbf{X}_3^{FK} = \rho g A \iint_{S_B} e^{kz-ikx} n_3 \mathrm{d}S \,</math></center> |
Using the Taylor series expansion, | Using the Taylor series expansion, | ||
− | <center><math> e^{ | + | <center><math> e^{kz-ikx} = 1 + ( kz - ikx ) + O ( kB )^2 \,</math></center> |
It is easy to verify that <math>\mathbf{X}_3 \to \rho g A A_w \,</math>. | It is easy to verify that <math>\mathbf{X}_3 \to \rho g A A_w \,</math>. | ||
− | The scattering contribution is of order <math> | + | The scattering contribution is of order <math> kB\,</math>. For submerged bodies, <math> \mathbf{X}_3^{FK}=O(kB)\,</math>. |
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Latest revision as of 01:55, 12 February 2010
Wave and Wave Body Interactions | |
---|---|
Current Chapter | Long Wavelength Approximations |
Next Chapter | Wave Scattering By A Vertical Circular Cylinder |
Previous Chapter | Linear Wave-Body Interaction |
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{s}\, }[/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]
In the absence of viscous effects and to leading order for [math]\displaystyle{ \lambda \gg B \, }[/math]:
where
Derivation using Euler's equations
An alternative form of GI Taylor's formula for a fixed body follows from Euler's equations:
Thus:
If the body is also translating in the x-direction with displacement [math]\displaystyle{ x_1(t)\, }[/math] then the total force becomes
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
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
So the horizontal force on the circle is:
Thus:
We can 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.
The differential horizontal force over a strip [math]\displaystyle{ \mathrm{d} z \, }[/math] at a depth [math]\displaystyle{ z \, }[/math] becomes:
The total horizontal force over a truncated cylinder of draft [math]\displaystyle{ T\, }[/math] becomes:
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,
Expressing the incident wave relative to the local frames by introducing the phase factors,
and letting
Then relative to the i-th leg,
Ignoring interactions between legs, which is a good approximation in long waves, the total exciting force on an n-cylinder platform is
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
If the body section is a circle with radius [math]\displaystyle{ a\, }[/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
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:
Using the Taylor series expansion,
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].
This article is based on the MIT open course notes and the original article can be found here
Ocean Wave Interaction with Ships and Offshore Energy Systems