Long Wavelength Approximations - Revision history

Meylan at 01:55, 12 February 2010

2010-02-12T01:55:21Z

← Older revision		Revision as of 01:55, 12 February 2010
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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 = [[~~Added~~-~~Mass, Damping Coefficients And Exciting Forces~~]]		\| previous chapter = [[Linear Wave-Body Interaction]]
	}}		}}

Adi Kurniawan at 11:56, 11 December 2009

2009-12-11T11:56:15Z

Adi Kurniawan: /* Horizontal force on a fixed circular cylinder of draft T\, */

2009-12-02T10:17:40Z

Horizontal force on a fixed circular cylinder of draft T\,

Adi Kurniawan: /* Regular waves over a circle fixed under the free surface */

2009-12-02T10:16:51Z

Regular waves over a circle fixed under the free surface

← Older revision		Revision as of 10:16, 2 December 2009
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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^{-~~K h~~ + i \omega t} \right\} </math></center>		<center><math> \frac{\partial u}{\partial t} = \mathrm{Re} \left\{ i\omega^2 e^{-kh + i \omega t} \right\} </math></center>

	Thus:		Thus:

Adi Kurniawan: /* Regular waves over a circle fixed under the free surface */

2009-12-02T10:16:33Z

Regular waves over a circle fixed under the free surface

← Older revision		Revision as of 10:16, 2 December 2009
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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^{Kz-~~iKx~~+i\omega t} \right\}, \quad K=\frac{\omega^2}{g} \, </math></center>		<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 K) e^{K z - i K x + i \omega t } \right \} </math></center>		<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^{ - K h +i \omega t} \right\}_{x=0,z=-h} </math></center>		<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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	Thus:		Thus:

	<center><math> F_x = - 2 \pi a^2 \omega^2 A e^{-K h} \sin \omega t \,</math></center>		<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.

Adi Kurniawan at 09:56, 20 October 2009

2009-10-20T09:56:58Z

← Older revision		Revision as of 09:56, 20 October 2009
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	This article is based on the MIT open course notes and the original article can be found		This article is based on the MIT open course notes and the original article can be found
	[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/~~0B7683D3~~-~~9B31~~-~~453E~~-~~B98F~~-~~9F71A3C36C58~~/0/~~lecture2~~.pdf here]		[http://ocw.mit.edu/NR/rdonlyres/Mechanical-Engineering/2-24Spring-2002/D28E25F7-9600-48C1-A444-D1F0FA4F0A41/0/lecture11.pdf here]

	[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]		[[Ocean Wave Interaction with Ships and Offshore Energy Systems]]

	[[Category:Linear Water-Wave Theory]]		[[Category:Linear Water-Wave Theory]]

Meylan at 23:57, 16 October 2009

2009-10-16T23:57:06Z

← Older revision		Revision as of 23:57, 16 October 2009
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	\| previous chapter = [[Added-Mass, Damping Coefficients And Exciting Forces]]		\| previous chapter = [[Added-Mass, Damping Coefficients And Exciting Forces]]
	}}		}}

			{{incomplete pages}}

	== Introduction ==		== Introduction ==

Adi Kurniawan: /* Surge exciting force on a 2D section */

2009-04-10T12:02:28Z

Surge exciting force on a 2D section

← Older revision		Revision as of 12:02, 10 April 2009
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	<center><math> \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></center>		<center><math> \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></center>

	If the body section is a circle with radius <math> a\,</math>:		If the body section is a circle with radius <math> a\,</math>,

	<center><math> \rho \forall = A_{11} = \pi\rho \frac{a^2}{2} \,</math></center>		<center><math> \rho \forall = A_{11} = \pi\rho \frac{a^2}{2} \,</math></center>

Adi Kurniawan: /* Heave exciting force on a surface piercing section */

2009-04-10T12:01:37Z

Heave exciting force on a surface piercing section

← Older revision		Revision as of 12:01, 10 April 2009
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	=== Heave exciting force on a surface piercing section ===		=== Heave exciting force on a surface piercing section ===

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

	<center><math>\mathbf{X}_i \sim \rho g A_w A \,</math></center>		<center><math>\mathbf{X}_i \sim \rho g A_w A \,</math></center>

	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^{Kz-iKx} n_3 \mathrm{d}S \,</math></center>		<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^{Kz-iKx} = 1 + ( Kz - iKx ) + O ( KB )^2 \,</math></center>		<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> KB\,</math>. For submerged bodies: <math> \mathbf{X}_3^{FK}=O(KB)\,</math>.		The scattering contribution is of order <math> KB\,</math>. For submerged bodies, <math> \mathbf{X}_3^{FK}=O(KB)\,</math>.

	-----		-----

Adi Kurniawan: /* Surge exciting force on a 2D section */

2009-04-10T11:59:19Z

Surge exciting force on a 2D section

← Older revision		Revision as of 11:59, 10 April 2009
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	<center><math> \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></center>		<center><math> \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></center>

	* If the body section is a circle with radius <math> a\,</math>:		If the body section is a circle with radius <math> a\,</math>:

	<center><math> \rho \forall = A_{11} = \pi\rho \frac{a^2}{2} \,</math></center>		<center><math> \rho \forall = A_{11} = \pi\rho \frac{a^2}{2} \,</math></center>

← Older revision		Revision as of 11:56, 11 December 2009
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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^{Kz-~~iKx~~ + i\omega t} \right\} \,</math></center>		<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^{-~~iKx_i~~} </math></center>		<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^{Kz - iK\xi_i + i\omega t} \mathbf{P}_i \right\} \quad i=1,\cdots,N </math></center>		<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^{Kz-~~iKx~~+i\omega t} \right\} \,</math></center>		<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 K ) e^{Kz-~~iKx~~+i\omega t} \right\} \,</math></center>		<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^{Kz-~~iKx~~} n_3 \mathrm{d}S \,</math></center>		<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^{Kz-~~iKx~~} = 1 + ( Kz - ~~iKx~~ ) + O ( KB )^2 \,</math></center>		<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> KB\,</math>. For submerged bodies, <math> \mathbf{X}_3^{FK}=O(KB)\,</math>.		The scattering contribution is of order <math> kB\,</math>. For submerged bodies, <math> \mathbf{X}_3^{FK}=O(kB)\,</math>.

	-----		-----

← Older revision		Revision as of 10:17, 2 December 2009
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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 K) e^{Kz-i K x + i\omega t} \right\} </math></center>		<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^{Kz+i\omega t} \right\}_{x=0} \,</math></center>		<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^{Kz+i\omega t} \right\} </math></center>		<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^{Kz} \sin \omega t \,</math></center>		<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^{Kz} \right) \sin \omega t \mathrm{d} z </math></center>		<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^{Kz} \mathrm{d}z </math></center>		<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^{-KT}}{K} </math></center>		<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^{-KT}}{K} \to \frac{1}{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> 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 ===