Ship Kelvin Wake - Revision history

Adi Kurniawan: /* Kelvin wake */

2010-11-06T11:37:03Z

Kelvin wake

← Older revision		Revision as of 11:37, 6 November 2010
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	This implies the following		This implies the following
	* The phase velocity of the waves in the kelvin wake propagating in direction <math>\theta\,</math> must be equal to <math> U\cos\theta\,</math>, otherwise they cannot be stationary relative to the ship.		* The phase velocity of the waves in the kelvin wake propagating in direction <math>\theta\,</math> must be equal to <math> U\cos\theta\,</math>, otherwise they cannot be stationary relative to the ship.
	* Relative to the earth system the frequency of a local system propagating in direction <math> \theta \, </math> is given by the relation		* Relative to the earth system the frequency of a local system propagating in direction <math> \theta \, </math> is given by the relation <math> \omega = kU \cos \theta \, </math>
	~~<center>~~<math> \omega = kU \cos \theta \, </math~~></center~~>

	Relative to the earth system the [[Infinite Depth]] [[Dispersion Relation for a Free Surface]] states		Relative to the earth system the [[Infinite Depth]] [[Dispersion Relation for a Free Surface]] states

Meylan: /* Linear free surface conditions */

2010-04-06T08:50:32Z

Linear free surface conditions

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 <center><math> \frac{\mathrm{d}\Phi}{\mathrm{d}t} = -U \frac{\partial\bar{\Phi}}{\partial x} \,</math></center>
 or, the ship wake is stationary relative to the ship but not relative to an observed on the beach.
 == Kelvin wake ==

Meylan: /* Kelvin wake */

2010-04-04T10:26:38Z

Kelvin wake

← Older revision		Revision as of 10:26, 4 April 2010
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	Local view of Kelvin wake consists approximately of a [[Linear Plane Progressive Regular Waves\| plane progressive wave group]] propagating in direction <math> \theta\,\!</math>.		Local view of Kelvin wake consists approximately of a [[Linear Plane Progressive Regular Waves\| plane progressive wave group]] propagating in direction <math> \theta\,\!</math>.
	As noted above surface wave systems of general form always consist of combinations of plane progressive waves of different frequencies and directions. The same model will apply to the ship kelvin wake.		As noted above surface wave systems of general form always consist of combinations of plane progressive waves of different frequencies and directions. The same model will apply to the ship kelvin wake.
	Relative to the earth frame, the local plane ~~progressive~~ wave takes the form:		Relative to the earth frame, the local plane wave in [[Infinite Depth]] takes the form

	<center><math> \Phi = \frac{~~igA~~}{\omega} e^{KZ-iK(X\cos\theta+Y\sin\theta)+i\omega t} </math></center>		<center><math> \Phi = \frac{\mathrm{i}gA}{\omega} e^{kz-\mathrm{i}k(x\cos\theta+y\sin\theta)+ \mathrm{i}\omega t} </math></center>

	~~Substitute~~		Relative to the ship frame
			<center><math>bar{\Phi} = \frac{igA}{\omega} e^{kz-\mathrm{i}k(x\cos\theta+y\sin\theta)-\mathrm{i}(kU\cos\theta-\omega)t} </math></center>
	~~<center><math> X = x + Ut, Y=y, Z=z. \,</math></center>~~

	~~and use the results from [[Translating Coordinate Systems]] to give us, relative~~ to the ship frame

	<center><math> \~~phi~~ = \frac{igA}{\omega} e^{KZ-iK(x\cos\theta+y\sin\theta)-i(kU\cos\theta-\omega)t} </math></center>

	But relative to the ship frame waves are stationary, so we must have:		But relative to the ship frame waves are stationary, so we must have:

	<center><math> KU\cos\theta = \omega \,</math></center>		<center><math> kU\cos\theta = \omega \,</math></center>
	or		or
	<center><math> \frac{\omega}{K} = ~~V_P~~ = U \cos \theta \, </math></center>		<center><math> \frac{\omega}{k} = C_p = U \cos \theta \, </math></center>
	This implies the following		This implies the following
	* The phase velocity of the waves in the kelvin wake propagating in direction <math>\theta\,</math> must be equal to <math> U\cos\theta\,</math>, otherwise they cannot be stationary relative to the ship.		* The phase velocity of the waves in the kelvin wake propagating in direction <math>\theta\,</math> must be equal to <math> U\cos\theta\,</math>, otherwise they cannot be stationary relative to the ship.
	* Relative to the earth system the frequency of a local system propagating in direction <math> \theta \, </math> is given by the relation		* Relative to the earth system the frequency of a local system propagating in direction <math> \theta \, </math> is given by the relation
	<center><math> \omega = ~~K U~~ \cos \theta \, ~~</math></center>~~		<center><math> \omega = kU \cos \theta \, </math></center>
	~~where <math> K = \frac{2\pi}{\lambda}\, </math>.~~

	~~Relative to the earth system the deep water dispersion relation states:~~

	~~<center><math> \omega^2 = g K \, </math></center>~~

	or

	~~<center><math> K^2 U^2 \cos^2 \theta = g K \,</math></center>~~

	~~<center><math> \Longrightarrow K = \frac{g}{U^2 \cos^2 \theta} = \frac{2\pi}{\lambda(\theta)}~~ </math></center>

	<center><math> \~~Longrightarrow~~ \lambda(\theta) = \frac{2\pi U^2 \cos^2 \theta}{g} \, </math></center>		Relative to the earth system the [[Infinite Depth]] [[Dispersion Relation for a Free Surface]] states
			<center><math> \omega^2 = g k \, </math></center>
			so that
			<center><math> \lambda(\theta) = \frac{2\pi U^2 \cos^2 \theta}{g} \, </math></center>

	This is the wavelength of waves in a ~~kelvin~~ wake propagating in direction <math> \theta \, </math> ~~and being~~ stationary relative to the ship.		This is the wavelength of waves in a Kelvin wake propagating in direction <math> \theta \, </math> which
			are stationary relative to the ship.

	== Application of the Group velocity ==		== Application of the Group velocity ==

Meylan: /* Linear free surface conditions */

2010-04-04T10:16:36Z

Linear free surface conditions

← Older revision		Revision as of 10:16, 4 April 2010
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	The linear free surface conditions are as follows,		The linear free surface conditions are as follows,

	<center><math> \frac{\mathrm{d}^2\Phi}{\mathrm{d}t^2} + g \frac{\mathrm{d}\Phi}{\mathrm{d}Z} = 0, \quad Z=0 </math></center>		<center><math> \frac{\mathrm{d}^2\Phi}{\mathrm{d}t^2} + g \frac{\mathrm{d}\Phi}{\mathrm{d}z} = 0, \quad z=0 </math></center>

	We can calculate the second derivative above as		We can calculate the second derivative above as
	<center><math~~> \frac{\mathrm{d}\Phi}{\mathrm{d}t} = \frac{\partial\phi}{\partial t} - U \frac{\partial\phi}{\partial x} </math></center~~>		<center><math>
	~~and~~		\frac{\mathrm{d}^2\Phi}{\mathrm{d}t^2} = \frac{\partial^2 \bar{\Phi}}{\partial t^2}
	~~<center><math>~~ \frac{\mathrm{d}^2\Phi}{\mathrm{d}t^2} = \frac{\partial^2 \~~phi~~}{\partial t^2} - 2 U \frac{\partial^2\~~phi~~}{\partial x \partial t} + U^2 \frac{\partial^ \~~phi}~~{\~~partial x^2~~} ~~</math></center>~~		- 2 U \frac{\partial^2\bar{\Phi}}{\partial x \partial t}
			+ U^2 \frac{\partial^ \bar{\Phi}}{\partial x^2}
	~~This is the free surface condition governing the forward-speed linear seakeeping problem.~~		</math></center>

	~~When no ambient waves are present <math> \frac{\partial\phi}{\partial t} = 0 \, </math> and we obtain the free surface condition for the steady Kelvin ship wave problem~~

	~~<center><math> U^2 \frac{\partial^2 \phi~~}{\partial x^2} ~~+ g \frac{\partial\phi}{\partial z} = 0, \qquad z=0~~ </math></center>

			When no ambient waves are present <math> \frac{\partial\bar{\Phi}}{\partial t} = 0 \, </math> and we obtain the free surface condition for the steady Kelvin ship wave problem
			<center><math> U^2 \frac{\partial^2 \bar{\Phi}}{\partial x^2}
			+ g \frac{\partial\bar{\Phi}}{\partial z} = 0, \qquad z=0 </math></center>
	This is the famous Neumann-Kelvin free-surface condition governing the linear steady wave pattern generated by a translating ship.		This is the famous Neumann-Kelvin free-surface condition governing the linear steady wave pattern generated by a translating ship.

	Bernoulli equation:		Bernoulli equation:

	<center><math> P = - \rho \left( \frac{\mathrm{d}\Phi}{\mathrm{d}t} + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi + gZ \right) = - \rho \left[ \left( \frac{\partial\~~phi~~}{\partial t} - U \frac{\partial\~~phi~~}{\partial x} \right) + \frac{1}{2} \nabla \phi \cdot \nabla \phi + g Z \right] </math></center>		<center><math> P = - \rho \left( \frac{\mathrm{d}\Phi}{\mathrm{d}t} + \frac{1}{2} \nabla\Phi \cdot \nabla\Phi + gz \right) = - \rho \left[ \left( \frac{\partial\bar{\Phi}}{\partial t} - U \frac{\partial\bar{\Phi}}{\partial x} \right) + \frac{1}{2} \nabla \phi \cdot \nabla \phi + g z \right] </math></center>

	Free surface elevation:		Free surface elevation:

	<center><math> \zeta = - \left. \frac{1}{g} \frac{\mathrm{d}\Phi}{\mathrm{d}t} \right\|_{z=0} = -\frac{1}{g} \left. \left( \frac{\partial\~~phi~~}{\partial t} - U \frac{\partial\~~phi~~}{\partial x} \right)\right\|_{z=0} </math></center>		<center><math> \zeta = - \left. \frac{1}{g} \frac{\mathrm{d}\Phi}{\mathrm{d}t} \right\|_{z=0} = -\frac{1}{g} \left. \left( \frac{\partial\bar{\Phi}}{\partial t} - U \frac{\partial\bar{\Phi}}{\partial x} \right)\right\|_{z=0} </math></center>

	In the Kelvin ship wave problem		In the Kelvin ship wave problem

	<center><math> \frac{\partial\phi}{\partial t} = 0 </math></center>		<center><math> \frac{\partial\phi}{\partial t} = 0 </math></center>

	Hence		Hence
			<center><math> \zeta = \frac{U}{g} \frac{\partial\bar{\Phi}}{\partial x}, \qquad z=0 </math></center>
	<center><math> \zeta = \frac{U}{g} \frac{\partial\~~phi~~}{\partial x}, \qquad z=0 </math></center>

	~~So if the velocity potential <math> \phi(\overrightarrow{x}) \,</math> is available in some form relative to the translating frame, the wave pattern follows from the above expression.~~

	== Kelvin wake ==		== Kelvin wake ==

Meylan: /* Translating Coordinate System */

2010-04-04T10:09:21Z

Translating Coordinate System

← Older revision		Revision as of 10:09, 4 April 2010
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	The same potential expressed relative to the ship frame is <math> \bar{\Phi}(\mathbf{\bar{x}},t) \, </math>.		The same potential expressed relative to the ship frame is <math> \bar{\Phi}(\mathbf{\bar{x}},t) \, </math>.
	The relation between the two potentials is given by the identity		The relation between the two potentials is given by the identity
	<center><math> \Phi(x,y,z,t) = \bar{\~~phi~~}(\bar{x}-Ut,Y,Z,t) \,</math></center>		<center><math> \Phi(x,y,z,t) = \bar{\Phi}(\bar{x},y,z,t)
			= \bar{\Phi}(x-Ut,y,z,t) \,</math></center>
	where the relation between the coordinates of the two coordinate systems has been introduced.		where the relation between the coordinates of the two coordinate systems has been introduced.
			Note the time dependence occurs in two places in <math> \bar{\Phi} \,</math> and in one place in <math> \Phi \,</math>. The governing equations are always derived relative to the earth coordinate system and time derivatives are initially taken on <math> \Phi \, </math>.
			Therefore
			<center><math> \frac{\mathrm{d}\Phi}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \bar{\phi} ( x-Ut,y,z,t)
			= \frac{\partial\phi}{\partial t} - U \frac{\partial\phi}{\partial x} </math></center>

	Note the time dependence in the arguments of the two potentials. It occurs in two places in <math> \phi \,</math> and in one place in <math> \Phi \,</math>. The governing equations are always derived relative to the earth coordinate system and time derivatives are initially taken on <math> \Phi \, </math>:		All time derivatives of the earth fixed velocity potential <math> \Phi \, </math> which appear in the free surface condition and the Bernoulli equation can be expressed in terms of derivatives of <math> \bar{\Phi} \, </math> using the Galilean transformation derived above.

	<center><math> \frac{\mathrm{d}\Phi}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t} \phi ( X-Ut,y,z,t) = \frac{\partial\phi}{\partial x} \frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial\phi}{\partial t} = \frac{\partial\phi}{\partial t} - U \frac{\partial\phi}{\partial x} </math></center>

	~~The main result to follow is:~~

	~~<center><math> \frac{\mathrm{d}\Phi}{\mathrm{d}t} = \frac{\partial\phi}{\partial t} - U \frac{\partial\phi}{\partial x}. </math></center>~~

	All time derivatives of the earth fixed velocity potential <math> \Phi \, </math> which appear in the free surface condition and the Bernoulli equation can be expressed in terms of derivatives of <math> \~~phi~~ \, </math> using the Galilean transformation derived above.

	If the flow is steady relative to the ship fixed coordinate system		If the flow is steady relative to the ship fixed coordinate system
	<center><math> \frac{\partial\~~phi~~}{\partial t} = 0 \, </math></center>		<center><math> \frac{\partial\bar{\Phi}}{\partial t} = 0 \, </math></center>
	but		but
	<center><math> \frac{\mathrm{d}\Phi}{\mathrm{d}t} = -U \frac{\partial\~~phi~~}{\partial x} \,</math></center>		<center><math> \frac{\mathrm{d}\Phi}{\mathrm{d}t} = -U \frac{\partial\bar{\Phi}}{\partial x} \,</math></center>
	or, the ship wake is stationary relative to the ship but not relative to an observed on the beach.		or, the ship wake is stationary relative to the ship but not relative to an observed on the beach.

Meylan: /* Translating Coordinate System */

2010-04-04T08:49:43Z

Translating Coordinate System

← Older revision		Revision as of 08:49, 4 April 2010
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	== Translating Coordinate System ==		== Translating Coordinate System ==

	~~[[Image:Moving_coordinate~~.~~jpg\|thumb\|right\|600px\|Moving~~ coordinate ~~system]]~~		We have a the standard fixed coordinate system <math>x, y, z</math> and a moving coordinate systems
			which is moving in the <math>x</math> direction with speed <math>U</math>. We denote the moving coordinate
			systems in the <math>x</math> direction by
			<center><math>
			\bar{x} = x + U t
			</math></center>

	~~<math> \overrightarrow{X}: </math> Eulerian coordinate fixed and time independent relative to earth frame~~		Let <math> \Phi(\mathbf{x},t) \,</math> be the velocity potential describing the potential flow
			generated by the ship relative to the earth frame.
	<math> \overrightarrow{x}: </math> Cartesian coordinate of Eulerian point relative to ship frame. Must be time-dependent if <math> \overrightarrow{X} </math> is fixed. Its time dependence must be accounted for in taking time derivatives in formulation of governing laws and free surface conditions		The same potential expressed relative to the ship frame is <math> \bar{\Phi}(\mathbf{\bar{x}},t) \, </math>.

	~~<center><math> \begin{cases}~~
	~~x = X - U t \\~~
	~~y = Y \\~~
	~~z = Z~~
	~~\end{cases} </math></center>~~

	Let <math> \Phi(\~~overrightarrow~~{X},t) \,</math> be the velocity potential describing the potential flow generated by the ship relative to the earth frame.
	The same potential expressed relative to the ship frame is <math> \~~phi~~(\~~overrightarrow~~{x},t) \, </math>.
	The relation between the two potentials is given by the identity		The relation between the two potentials is given by the identity
	<center><math> \Phi~~(X,Y,Z,t) = \phi~~(x,y,z,t) = \phi(X-Ut,Y,Z,t) \,</math></center>		<center><math> \Phi(x,y,z,t) = \bar{\phi}(\bar{x}-Ut,Y,Z,t) \,</math></center>
	where the relation between the coordinates of the two coordinate systems has been introduced.		where the relation between the coordinates of the two coordinate systems has been introduced.

Meylan at 02:15, 4 April 2010

2010-04-04T02:15:59Z

← Older revision		Revision as of 02:15, 4 April 2010
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	\| chapter title = Ship Kelvin Wake		\| chapter title = Ship Kelvin Wake
	\| next chapter = [[Linear Wave-Body Interaction]]		\| next chapter = [[Linear Wave-Body Interaction]]
	\| previous chapter = [[~~Translating Coordinate Systems~~]]		\| previous chapter = [[Wavemaker Theory]]
	}}		}}

Meylan: /* Kelvin wake */

2010-04-04T02:12:30Z

Kelvin wake

@@ Line 13: / Line 13: @@
 A ship moving over the surface of undisturbed water sets up waves emanating from the bow and stern of the ship. The waves created by the ship consist of divergent and transverse waves. The divergent wave are observed as the [http://en.wikipedia.org/wiki/Wake wake] of a ship with a series of diagonal or oblique crests moving outwardly from the point of disturbance. These wave were first studied by [http://en.wikipedia.org/wiki/Lord_Kelvin Lord Kelvin]
 = Kelvin wake =
 [[Image:Kelvin_wake.jpg|thumb|right|600px|Diagram of the Kelvin Wake]]

Meylan at 01:30, 8 September 2009

2009-09-08T01:30:04Z

← Older revision		Revision as of 01:30, 8 September 2009
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	\| previous chapter = [[Translating Coordinate Systems]]		\| previous chapter = [[Translating Coordinate Systems]]
	}}		}}

			{{incomplete pages}}

	= Introduction =		= Introduction =

Meylan: /* Introduction */

2009-09-08T01:29:37Z

Introduction

← Older revision		Revision as of 01:29, 8 September 2009
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	= Introduction =		= Introduction =

	[[Image:Wake.avon.gorge.arp.750pix.jpg\|thumb\|right\|~~600px~~\|Wake created behind a ship]]		[[Image:Wake.avon.gorge.arp.750pix.jpg\|thumb\|right\|450px\|Wake created behind a ship]]

	A ship moving over the surface of undisturbed water sets up waves emanating from the bow and stern of the ship. The waves created by the ship consist of divergent and transverse waves. The divergent wave are observed as the [http://en.wikipedia.org/wiki/Wake wake] of a ship with a series of diagonal or oblique crests moving outwardly from the point of disturbance. These wave were first studied by [http://en.wikipedia.org/wiki/Lord_Kelvin Lord Kelvin]		A ship moving over the surface of undisturbed water sets up waves emanating from the bow and stern of the ship. The waves created by the ship consist of divergent and transverse waves. The divergent wave are observed as the [http://en.wikipedia.org/wiki/Wake wake] of a ship with a series of diagonal or oblique crests moving outwardly from the point of disturbance. These wave were first studied by [http://en.wikipedia.org/wiki/Lord_Kelvin Lord Kelvin]