Part Linear Wave Theory BRN

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CE 769Coastal and Ocean Environment

Dr. BALAJI RamakrishnanAssistant Professor

Department of Civil Engineering, IIT Bombay.email: [email protected]

Wave generation by Wind

Dr. BALAJI Ramakrishnan, Assistant Professor, Dept. of Civil Engg., IIT Bombay. email: [email protected]

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Why wind generated waves are important?

Distribution of energy as a function of wave periods and types(Fig. 9.4, Garrison, 2001).



Development of waves Due to the energy transfer from the wind to water surface, the waterparticles motion is initiated and initially small capillary waves of few mmhigh are formed.

When the wind is continuously blowing over the water surface, the wavesgrow and become gravity waves.

First waves grow in both length & height, later, when they reach heightlimit, they grow predominantly in length, if more energy is available.

The size/characteristics of the waves resulting from the energy transferare governed by;

velocity of the wind (W);

fetch (F) or distance over which the wind blows;

the duration (D) of time that the wind blows.

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Development of wavesDefinition of Fetch



Limited wavesFetch limited

If the wind duration exceeds the time required for waves to propagate theentire fetch length.

Duration limited

If the wind duration less, then the wave growth reaches OAC.

Fetch, F

Fully developed sea-state If both the fetch and duration aresufficiently large, a fully developed sea hasbeen generated for given wind velocity.

The state of the sea, in which the input ofenergy to the waves from the wind is inbalance with the transfer of energy amongthe different wave components, and with thedissipation of energy by wave breaking.

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Seas

Swell


Next...

Small amplitude wave theory

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General There are variety of wave types exist;

-Storm waves, flood waves in river

-oscillations in harbour basins, tidal bores

-hydraulic jump in estuaries

-waves generated by ships

-tsunami waves by landslides/earthquake/underwater explosions

A general solution is not possible, frommathematical point of view, as the boundaryconditions vary.

Assumptions made in various wave theoriesdefine the limits of validity.

The primary difficulty in water wave theories isone of the boundaries, free surface elevation, is oneof the unknowns !.



Assumptions The motion is assumed to irrotational;

Hence, a single-valued velocity potential function (x,y,z,t) exists &has to be found from continuity & momentum eqns.

The fluid is incompressible therefore, the density is a constant.and the continuity equation exists as,

The continuity eqn is expressed in terms of , as;

The momentum eqn is given as;

Laplace equationLaplace equation

Bernoulli equationBernoulli equation

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Assumptions

The fluid is homogeneous, ideal and inviscid;

Surface tension is neglected (surface waves are longer than the length where surface

tension effects are important);

Pressure at the free surface is uniform and constant (no pressure is exerted by

the wind and the aerostatic pressure difference between the wave crest and trough is negligible);

The particular wave being considered does not interact with any other

water motions;

The bottom is stationary, impermeable, and horizontal (bottom is not adding or

removing energy from the flow or reflecting wave energy);

The waves are long-crested and the wave amplitude is small.



Definition of domain

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Boundary conditions (BC)Bottom BC

The seabed is assumed to be horizontal, fixed and impermeable boundary,hence, the velocity normal to it is zero.

Kinematic Free Surface BC

The difficulty in water wave theory is one of the boundary, free surfaceelevation, is unknown.

The free surface is given as; z=(x,y,t) and its derivative with respect to t is;

Where;



Boundary conditions (BC)The kinematic free surface BC becomes;

Dynamic Free Surface BC

This BC is given by Bernoulli equation, in which the pressure is consideredas constant (=atmospheric pressure).

Periodic Lateral BC

This BC is given by;

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d Periodic lateral BC

Periodic lateral BC

2 =0



Boundary conditions (BC)

and are obtained by the solution of 2 =0, with application of boundaryconditions.



Linearization the Boundary conditions(1) Linearization of DFSBC: The nonlinear DFSBC is given as;

- The convective inertia term is neglected.

- At the free surface, z=, the above equation becomes;

or

- with assumption of small amplitude waves;

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Linearization the Boundary conditions(1) Linearization of KFSBC: The nonlinear KFSBC is given as;

- The convective terms are neglected.

- The slope components, /x & /y , are assumed to be small &neglected;

- Finally, the normal component of the fluid velocity at the free surface ()is assumed to be equal to the normal velocity of the surface itself.

- So, the approximated KFSBC is;



Solution of the linear water wave problemThe solution of Laplace equation is obtained through separation of

variables, in which the solution is expressed as product independentvariable terms;

As the solution must be periodic, we specify T(t) =sin t, where, is angularfrequency of the wave.

To find the , we use the lateral periodic BC;sin t =sin (t+T) (or) sin t =sin t cos T + cos t sin T

Which is true for T=2, or; =2/T

Now the velocity potential becomes;

Function that depends only on x

Function that depends only on z

Function that depends only on t

Angular frequency

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Solution of the linear water wave problemSubstituting the equation into the Laplace equation & dividing through ;

The x can vary independently of z and vice versa.

The only way in which the two functions will be equal is if both parts of theequation are equal to a same constant, for a different sign;

& where, k can be real or imaginary.

This is an ordinary differential equation & can be solved separately.

The value of k, can be real, k=0 or imaginary.



Solution of the linear water wave problemPossible solutions to Laplace equation, based on the separation of variables.

The solution is spacially periodic & physically meaningful, only if k is real &nonzero.

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Application of BCNow, we have the velocity potential as;

Applying the periodic BC;

Which is satisfied for cos kL=1 and since sin kL=0, leading to kL=2 or;

Keeping only A terms; as, & apply bottom BC;

on z=-d

Wave number



Application of BCFor this equation to be true for x & t;

Applying this in and dividing with D terms; we get,

Or,

where;

Applying the velocity potential into the DFSBC;

The terms within the brackets are constant; therefore, is given as aconstant times periodic terms in space and time plus a function of time.Now, can be written as;

now; G becomes as;

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Application of BCThe velocity potential now takes as;

Is this corresponding to a moving wave ?.

The pictorial representation of the is given as;

-At t = /2, the wave form is zero for all x,at t = 0, it has a cosine shape and atother times, the same cosine shape withdifferent magnitudes.

-This wave form is obviously a standingwave, as it does not propagate in anydirection.

-At positions kx = /2, and 3 /2, and soon, nodes exist; that is, there is no motionof the free surface at these points.



Application of BCConsidering another solution of Laplace equation, the & will be;

The differs from the previous solution in that the x and t terms are 900 outof phase. The associated water surface displacement is, as given above;

As the Laplace equation is linear & superposition is valid, we can add orsubtract solutions to the linearized boundary value problem to generatenew solutions. If we subtract the above from the previous , we obtain,

And the corresponding will be;

for progressive waves

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Next...

Properties of Waves

Properties of linear waves


Dispersion relationThe , after applying DFSBC, takes form;

Substituting the equations of & into KFSBC,

We get,

Applying the values of =2/T and k= 2/L in the above equation;

and rearranging the terms;

Similar algebraic manipulation gives another relationship;

Dispersion relation

Celerity / speed of wave

Wave length

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Wave length, LIn the wave length equation; or

In addition, the wave length equation is a iterative equation;

The wavelength equation is a iterative type and one can start with L0 asinitial value.

where,



Classification of water waves by depthThe gravity water waves are classified into three categories, depending upon

their water depth to wave length ratio, (d/L);

cosh(kd)sinh(kd)tanh(kd)

The hyperbolic functions present inthe equations have asymptotes thatleads to simplified forms ofequations.

Function Large kd Small kd

cosh kd ekd/2 1

sinh kd ekd/2 kd

tanh kd 1 kd


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Classification of water waves by depthShallow water: The dispersion relation,

reduces to;

So, in shallow water, the speed of the wave is solely depends on waterdepth.

Deep water: The dispersion relation reduces to,

Where, and

Function Large kd Small kdcosh kd ekd/2 1

sinh kd ekd/2 kd

tanh kd 1 kd



Water particle velocitiesThe horizontal and vertical components of water particle velocity (u and w,

respectively) can be determined from the velocity potential; as,

surface deep water particle speed

particle velocity variation over the vertical water column at a given location

& for particle velocity variation caused by the wave moving from deep to shallow water

phasing term dependent on position in the wave and time


Where, =(kx-t)

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Water particle accelerationsThe horizontal and vertical water particle

accelerations (ax and az, respectively) can bedetermined from the velocities; as,

w

ax

az


Where, =(kx-t)


Water particle displacementsThe displacement components of the water particle are obtained by

integrating the velocity w.r.t time;

Assuming the non-time varying parts as A & B in the above two equations;

Squaring and adding them together will lead to;

Which is a equation of ellipse with semi-axis A & B in x-z directions.

&

Shallow water Deep water


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Water particle displacements

dL0

dd

(a) deep water, d/L>0.5

(b) Intermediate water0.5>d/L

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EnergyThe total mechanical energy in a surface gravity wave is the sum of the

kinetic and potential energies.

The kinetic energy for a unit width of wave crest and for one wave length isequal to the integral over one wave length and the water depth of one-halftimes the mass of a differential element times the velocity of that elementsquared, as;

Apply u & w



EnergyIf we subtract the potential energy of a mass of still water (with respect to the bottom) from the potential energy of the wave form, the potential energy due solely to the wave form will be obtained. Potential energy per unit wave crest width and for one wave length is;

d

Thus, the kinetic and potential energies are equal and the total energy in a wave per unit crest width E is;

the avg. energy per unit surface area is;

Also called as energy density or specific energy


i

ii

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Summary of linear wave characteristics



Summary of linear wave characteristics


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Problem 1Given: A wave with a period T = 10 sec is propagating over a uniformlysloping sea from a depth d = 200 m.

Find: Wavelength, L and Celerity, C at the following depths;(a) d1 = 150 m (b) d2=50m (c) d3=5m

Solution:

For d1=150m;d1/L0= 0.9615 > 0.5 & hence, d1 is deepwater.

So, L1=L0=156m &

For d2=50m;d2/L0= 0.3205 < 0.5 & >0.05, hence, d2 is intermediate water.

L2=L0tanh(kd) =151.17m & C2=L2/T = 15.12 m/s

Problems related to linear waves


Problem 1For d3=5m;

d1/L0= 0.032 < 0.05 & hence, d3 is shallow water.

L3=L0tanh(kd) = 67.63m & C3=L3/T = 6.76 m/s

Summary:


Depth Description Wave length, L (m)

Celerity, C (m/s)

d1=150m Deep water 156 15.6

d2=50m Intermediate water 151.17 15.1

d3=5m Shallow water 67.63 6.76

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Problem 2Given: A wave with a period T = 8 sec, in a water depth d = 15 m, and aheight H = 5.5 m.

Find: The local horizontal and vertical velocities u and w, and accelerationsx and z when = (kx-t)=600, at following elevations below the SWL,

(a) z = -5 m (b) z= -9 (c) z=-14

Solution:For z=-5m;

Get the values of u & w and ax & az for various z values and tabulated them.


&


Problem 3Given: A wave in a depth d = 12 m, height H = 3 m, and a period T = 10 sec.The corresponding deepwater wave height is H0 = 3.13 m.

Find: (a) The maximum horizontal and vertical displacement of a waterparticle from its mean position when z = 0 and z = -d.

(b) The maximum water particle displacement at an elevation z = -7.5m when the wave is in infinitely deep water.

(c) For the deepwater conditions of (b) above, show that the particledisplacements are small relative to the wave height when z = -L0 /2.

Solution: (a) for z=0m;

Step 1: Calculate L0 & L

Step 2: Calculate A & B


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Problem 4Given: An average maximum pressure p = 124 kN/m2 is measured by asubsurface pressure gauge located in sea 0.6m above the bed in a waterdepth d = 12 m . The average wave period is 15sec.

Find: The height of the wave, H assuming that linear theory applies.

Solution: Step 1: Calculate L0 & L

Step 2: Rearrange the pressure formula !!

We know that,

and also know that;

Apply in p and find H.


Part Linear Wave Theory BRN

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