basics of wave motion

8/11/2019 Basics of Wave Motion

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Basics of wave motion

Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India

1

BASICS OF WAVE MOTION

Prof. V. Sundar,

Department of Ocean Engineering, Indian Institute of Technology Madras,

INDIA2.1 GENERAL:

Waves are periodic undulations of the sea surface, the complexity of which is most

challenging to those working in the oceans. They impose highly variable and fatigue type

loading on offshore and exposed coastal structures, and they may adversely affect coastlines

and harbor facilities and induce violent motions in moored ships and floating structures. The

designer of offshore structures must have a good understanding of the basics of wave

generation, its characteristics and behavior before evaluating forces and other potential effects.

Turbulent pressures fluctuations and variations in wind velocity cause an initial

disturbance of the sea surface, thus precipitating wave growth. The initial movement of the sea

surface is perpendicular to that of the wind, but it quickly aligns itself approximately with the

wind direction or some oblique angle to the wind. A light air movement of only two knots or

less will cause small ripples on the sea surface, which disappear immediately when the

generating wind stops. These ripples are associated with capillary force and are in conflict with

surface tension. As the wind speed increases, and if, it is of any duration, larger Gravity

waves begin to develop as the wave height builds up, and a more complex pressure

distribution forms at the surface. Normal stresses as well as tangential stresses now act on the

surface profile to further wave development. The exact way this growth begins is still not

completely understood. These are, however, many semiemperical relationships that describe

the growth of wind waves reasonably well. The ultimate state of wave growth depends

primarily on three parameters basic to wave forecasting: the fetch (F) or the distance over

which the wind blows, the wind velocity (V) and the duration (t) of time for which the wind

blows. Thus, for a given steady wind speed, the development of waves may be limited by the

fetch, or the duration. If however, the wind blows over a sufficient distance for a sufficient

length of time, a more or less steady state condition, where the average wave heights do not

change, will occur. This condition is called a fully developed sea (FDS). Waves moving out of

the generating area and are no longer subjected to significant wind action are known as swell.


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It is important to distinguish between various types of water waves that may be

generated and propagated. Ocean waves are classified according to apparent shape, relative

water depth and origin. According to apparent shape, waves can be classified as progressive

and standing waves. Progressive waves may be oscillatory or solitary. According to relative

water depth two types namely small amplitude waves and finite amplitude waves exist. Finite

amplitude waves may be further classified as intermediate depth waves (Stokes wave) and

shallow water waves (cnoidal waves). Fig.2.1 shows the classification of ocean waves.

2

WAVES

According

To Apparent shape According to relativewater depth

Progressive

waves

Standing waves

(Clapotis)

Small amplitude

waves (Airys theory

is used)

Finite Amplitude

waves

Oscillatory

waves Solitary waves Intermediate depthwaves (Stokes

waves)

Shallow waterwaves

According to origin

b. Ultragravity

waves (Surface

tension and gravityT upto 0.1 sec

a. Capillary

wave (due to

surface tension)

T upto 0.1 sec

d. Infra gravity waves

(stroms)

T = 30 sec to 5 min

c. Gravity wave

(wind)

T upto 30 sec

g. transtidal waves

(stroms, Tsunamis

due to under sea

explosives)

f. Ordinary tidal

waves (attraction of

astronomical bodies)

T = 12 hrs to 24 hrs

e. Long period waves

(stroms)T = 5 mins to 12 hrs

Fig.2.1 Classification of Ocean Waves


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3

The most fundamental description of a simple sinusoidal oscillatory wave consists of its

length, L or period, T and height, H. Small amplitude wave theory and some finite amplitude

wave theories can be developed by the introduction of a velocity potential (x,z,t). The

velocity potential, Laplaces equation and Bernoullis dynamic equation together with the

appropriate boundary conditions provide the necessary information needed in establishing the

small amplitude wave formulas.

The assumptions in deriving the expression for the velocity potential due to propagating ocean

waves are:

(a) Flow is said to be irrotational.

(b) Fluid is ideal.

(c) Surface tension is neglected.

(d) Pressure at the free surface is uniform and constant.

(e) The seabed is rigid, horizontal and impermeable.

(f) Wave height is small compared to its length.

(g) Potential flow theory is applicable.

(h) A velocity potential exists and the velocity components u and in the x and z directions

can be obtained as .z

andx

2.2 DERIVATION FOR VELOCITY POTENTIAL:

The governing equation is the Laplace Equation given by

0 (2.1)2 =

The continuity equation and Bernoullis equation given by equations (2.2) and (2.3) are used in

the solution procedure

0z

w

y

v

x

u=

+

+

(2.2)

( ) 0gzpwvu2

1

t

222 =+

++++

(2.3)


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4

2.2.1 BOUNDARY CONDITIONS:

(1)The equation (2.1) is to be satisfied in the region


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5

2.2.2 SOLUTION TO THE LAPLACE EQUATION:

0zx 2

2

2

2

=

+

(2.6)

Method of separable is used to obtain the solution to Equ.(2.6).

Let us assume

( ) ( ) ( ) ( )tTzZxXt,z,x = (2.7)

Substituting Eq.(2.7) in Eq.(2.6) we get

0T"ZXTZ"X =+

Where each prime denotes differentiation once with respect to the particular independent

variable.

Z

Z

X

XgivesTZ

""

=X Dividing both sides of the above

Let this be a constant = -k2.

0Xk"X 2 =+Then (2.8)

0Zk"Z 2 = (2.9)

kxsinBkxcosAX +=

kzkz eDeCZ +=

( ) ( )

Hence,

( ( )tTDeCekxsinBkxcosAt,z,xkzkz

++=

The solutions to are simple harmonic in time requiring T (t) be expressed as cos(t) or

sin(t),thus leading to four forms of solutions to , such that


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6

tsin.kxcoskzDekzCe4A4

tcos.kxsinkzDekzCe3A3

tsin.kxsin

kz

De

kz

Ce2A2

tcos.kxcoskzDekzCe1A1

+=

+=

+=

+=

2.2.3. DETERMINATION OF THE CONSTANTS:

The constants are determined by using the dynamic free surface boundary condition

and the kinematic bottom boundary condition.

Considering 2

tsin.kxsinkzDekzCe2A2

+= (2.10)

Applying, the kinematic bottom boundary condition,

dzat0z

==

i.e.,

0tsin.kxsinkzDkekzCke2Adzz

2 =

==

A20, sinkx . sint 0 [since velocity potential exists]

Substituting for C in eq.2.10 and simplifying,

( ) ( )tsin.kxsin

2

zdkezdkekdDe2A22

+++=

( ) tsin.kxsinzdkcoshDeA2 kd22 += (2.11)

and ( )tcos.kxsin.kdcosheDA2tkd

20z

2

=

=


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7

On assuming,

getwetg

1condition

boundarysurfacefreetheapplyingand2

Hamplitudewavea,where,tcos.kxsina

0z2

=

===

=

asinkx.cost =g

eDA2 kd2 coshkd. Sinkx. Cos t

2A2Dekd=

kdcosh

1.

ag

Substituting in eq.(2.11), we get

(2.12)

tcos.kxsinkzDekzCe3A3 +=

( )tsin.kxsin.

zdkcosh.

ag +kdcosh

2 =

Let us consider 3

(2.13)

Applying the kinematic bottom boundary condition

[ ] 0tcos.kxsinDkeCkeAz

kdkd3dz

3 ==

=

A30, sinkx.cost 0

C = De2kd

Substituting for C in eq. (2.13)

( ) tcos.SinkxzdkcoshDeA2 kd33 += (2.14)

and tsin.kxsin.kdcosheDA2t

kd30z =

=

On assuming and applying free surface boundary conditiontsin.xsina =

= kdcosh1

.

ag

DeA2kd

3


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8

Substituting this in eq.(2.14), we get

(2.15)( )

tcos.kxsin.kdcosh

zdkcoshag3

+

=

( tsin.kxcosDeCeA kzkz44 += (2.16)Applying the kinematic bottom boundary condition

( )

kd2

kdkd4dz

4

DeC

tsin.kxcosDeCeAz

=

=

=

Substituting for C in eq.(2.16)

( ) tsin.kxcos.zdkcoshDeA2 kd44 += (2.17)

and

( ) tsin.kxcos.zdkcosh.De.A2t

kd40z

4 +=

=

)5.2(.eqapplyingandtcos.kxcosa =

Assuming

We get 2A4Dekd

=kdcosh

1.

ag

Substituting this is eq.(2.17), we get

(2.18)

( tcos.kxcosDeCeA kzkz11 +=

( )tsin.kxcos.

zdkcoshag4

kdcosh

+

=

(2.19)

Applying the kinematic bottom boundary condition

kd2

dz1

DeC

0z

=

=

=

Substituting for C in eq.(2.19)

( ) tcos.kxcos.zdkcosh.De.A2 kd11 += (2.20)

Assuming = acoskx.sint and applying eq.(2.5)


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9

kdcosh

agDeA2 kd1

=

Substituting this in eq.(2.20)

(2.21)( ) tcos.kxcos.kdcosh

zdkcosh.

ag1

+

=

If += 2- 1

( )[ ]tsin.kxsintcos.kxcos.

kdcosh

zdkcosh.

ag+

+

=

(2.22)

)t

This is the expression for the velocity potential for a propagating wave in a constant water

depth.

Since 0zt.g

1

=

=

)tkxsin(.kdcosh

)zd(kcosh.

ag.

g

1

+

=

Hence

(2.23)

is periodic in x and t. If we locate a point and traverse along the wave, such that, at all time

t our position relative to the wave form remains fixed then the phase difference is zero or

(kx - t) = constant.

And the speed with which we must move to accomplish this is given by kx = +t Constant.

=dt

dxk

or C

T

L

2

L.

T

2

kdt

dx==

=

=

kxcos(.kdcosh

.)zd(kcoshag

+

=

)tkxsin(a =


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10

==T

LC CELERTY or Speed of the wave (2.24)

Wave moving in ve x direction

12 +=

=( )

)tkxcos(.kdcosh

tkxkcosh.

ag+

+

0z|t

.g1 ==

( )

++

= ))tkxsin(.(

kdcosh

zdkcoshag

g

1

= a sin (kx + t)

To obtain the celerity of the wave we have

kx + t = Constant

CT

L

kdt

dx=

=

= (2.25)

2.3 DISPERSION RELATIONSHIP :

The relationship between wavelength, period and water depth is obtained as given below. The

main assumption while establishing the relationship is that, since, we are dealing with small

amplitude waves, meaning that the slope of the wave profile are small so that

dt

dcan be

approximately said as equal to the vertical component velocity, w. This is,


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11

t

x.

xtdt

dw

+

=

=

Wave slope being small by setting

= 0x

tw

=

Butz

w

=

Hence,zt

=

(2.26)

Differentiating Eq. (2.22) we get

0z2

2

tg

1

t =

=

Hence,

( )tkxcos.kdcoshg

A

t

2

=

(2.27)

Wherekdcosh

1.

g

2 .

HA=

kAz

w = =sinhkd. cos (kx-t) (2.28)

Using the relation of Eq.(2.26), equating Eq. (2.27) to Eq. (2.28)

We get

=

)tkxcos(.kdcoshg

A 2A k.sinhkd. cos (kx-t)


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12

kdcosh

kdsinhk

g

2=

or the dispersion equation can be written as

tanhkd.gk2 = (2.29)

: Wave angular frequency =T

2and k: wave number

L

2=

The above equation can be written as

kdtanh.2

gL

T

L

kdtanh.L

2g

T

2

2

2

=

=

(2.30)

The speed at which a wave moves in its direction of propagation as a function of water depth is

given by Eq. (2.30)

T

LC=Since , from the above equation we get

kdtanh.2

gLC

=

kdtanh.k

g2C=

(2.31)

or kdtanh.2

gTL

2

= (2.32)


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13

Since the unknown L occurs on both sides (Implicit Eq.) of Eq.(2.32), it has to be solved by

trial and error.

2.4 CELERITY IN DIFFERENT WATER DEPTH CONDITIONS:

Classification of waves according to water depth is made with respect to the magnitude of d/L

and the resulting limiting values taken by the function tanh (kd) given in Table 2.1.

Table 2.1 Classification of ocean waves according to water depth.

Classification d/L

L

d2

L

d2tanh

Deep waters

Intermediate Waters

Shallow waters

>1/2

2

1to

20

1

>

20

1 1 for long period waves

N < 1 for short period waves

N = 1 for linear waves.

The pressure distribution under a progressive wave is given in Fig.2.7.

2.8 GROUP CELERITY:


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25

When a group of waves or a wave train travels, its speed is generally not identical to the speed

with individual waves within the group travel. If any two wave trains of the same amplitude,

but, slightly different wavelengths or periods progress in the same direction, the resultant

surface disturbance can be represented as the sum of the individual disturbances. For waves

propagating in deep or transitional waters, the group velocity is determined as follows.

)txksin(a)txksin(a 221121T +=+= (2.60)

+

+

= t

2x

2

kksin.t

2x

2

kkcosa2 21212121T

This is a form of a series of sine waves the amplitude of which varies slowly from 0 to 2a

according to the cosine factor.

The points of zero amplitude (nodes) of the wave envelope Tare located by finding the zeros

of the cosine factor.

i.e.,

2)1m2(t

2x

2

kk

whenoccurs0

2121

maxT

+=

=

( )

In other words, the nodes will occur on x axis at distances as follows:

tkkkk

1m2X

21

21

21node

+

+

=

Since the position of all the nodes is a function of time, they are not stationary. At t=0, there

will be nodes at

,......3,2,1,0mat.e.i.etc,kk

5,

kk

3,

kk 212121=

The distances between the nodes are given by

12

21

21LL

LL

kk

2x

=

= (2.61)


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26

The speed of propagation of the nodes and hence the speed of propagation of the wave group

is called the Group Velocityand is given by:

+=+==

=

==

=

=

=

dL

dk

1

dL

dC.k

CdK

dC

.kCdk

)KC(d

C

T

2

T

L.

L

2.C.KBut

dk

d

kk

CVelocityGroupWavedt

dx

G

1

21

Gnode

Since,L

2k

=

+=

2

G

L

2

1

dL

dC.

L

2CC

dL

dC.LC CG =

)kdtanh(.k

g2 =C Since

Substituting and on simplification we get

(2.62) +

kd

kd21

1G

==2sinh2

nC

C

For Deep waters, zeroiskd2sinh

kd2

Hence, oG C2

1C =

oo

G C2

1

T

L

2

1C ==


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27

(2.63)

The Group Celerity is one half of the phase velocity in deep waters. Further, it should be noted

that variables if associated with a suffix o refer to deep-water conditions. For example, Cois

deep-water celerity.

Table. 2.2 Variation of Asymptotic Functions

AsymptotesFunction

Shallow waters Deep waters

sinhkd

coshkd

tanhkd

kd

1

kd

2

ekd

2

ekd

1

In shallow waters, since sinh 2kd = 2kd

gdCCG == (2.64)

Hence, in shallow waters the group and phase velocities are same and is a function of only

depth of water and in deep waters, the CGis a function of wave length. Because of this, in

deep waters, the longer waves (long L) travel faster and produce the small phase differences

resulting in wave groups. These waves are said to be dispersive or propagating in a dispersive

medium, i.e., in a medium, where, their Celerity is dependent on wave length.

2.9 WAVE ENERGY

Total Energy = Potential energy + Kinetic energy.


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28

In order to determine the total energy under progressive waves, the potential energy of the

wave above z=-d with a wave form present is determined from which, the potential energy of

the water in the absence of a wave form is subtracted. Refer Fig.2.8for definitions.

The potential energy (with respect to z=-d) of a small column of water (d+

) high, dx

long and 1 m wide is

++=

=

2

d)d(.dx

xAdPE1

( ) dx.2

d2

+ (2.65)

The average potential energy per unit surface area (sometimes called the average potential

energy density) is

+

=++ Lx

x

2Tt

t1 dt.dx.)d(

T

1

L

1

2PE

The integration is from a time t over a wave period, T and from a certain distance x over a

distance x + L.

+

=++ Lx

x

2Tt

t1 dt.dx.)d(

LT2PE

becomes)66.2(.eq),tkxsin(a

(2.66)

Using =

++

=++ Tx

x

222Tt

t1 dt.dx)]tkx(sina)tkxsin(ad2d[(

LT2PE

On simplification


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29

4

a

2

dPE

22

1

+

= (2.67)

Which is the average potential energy per unit surface area of all the water above z = -d.

The potential energy in the absence of a wave would be

=

=++ Lx

x

22Tt

t2 .2/ddt.dxd

LT2PE (2.68)

The average potential energy density, PE which is attributable to the presence of the

progressive wave on the free surface, is

2

d

4

a

2

d

EnergyPotentialAveragePEPEPE

222

21

+

=

==

4

aPE

2= (2.69)

Kinetic Energy

2

1The kinetic energy, KE = mv2, where m is the mass of the fluid and v is the resultant

velocity. For a 2-D wave flow (Refer Fig.2.7for definitions)

( ) dx.dzwu2

1

dMwu2

1)KE(d

22

222

+=

+=

The average K.E. per unit of surface area is then given by


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30

+

=

++

d

22Lx

x

Tt

t

dt.dx.dz)wu(LT2

KE

with )tkxsin(a =

[ ] +++

=

++ 0

d

2222

22

2Lx

x

Tt

t

dt.dx.dz)tkx(cos).zd(ksinh)tkx(sin)zd(kcoshkdcosh

)agk(

LT2KE

Using the trignometrical identities

[ ]

[ ]

1)tkx(sin)tkx(cos

)tkx(cos)tkx(sin)tkx(cos

)zd(k2cosh12

1

)zd(ksinh

)zd(k2cosh12

1)zd(kcosh

2

222

2

2

=+

=

+

=+

++=+

sinh2kd = 2sinhd.coskd

and = gk tanhkd2

it can be shown

4

aKE

2= (2.70)

KE+PE Total Energy E =

2

2a E = (2.71)

The average total energy per unit surface area is the sum of the average potential and kineticenergy densities, often called as specific energy or energy density.

2.10 WAVE POWER:


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31

Wave energy flux is the rate at which energy is transmitted in the direction of wave

propagation across a vertical plane perpendicular to the direction of the wave advance and

extending down the entire depth. The average energy flux per unit wave crest width

transmitted across a plane perpendicular to wave advance is

P = Wave Power = Average energy flux per unit wave crest width

gCEncEP == (2.72)

where n =

+kd2sinh

kd21

2

1

For Deep WatersoG

C2

1Cand0

kd2sinh

kd2==

n =2

1

or ooo CE2

1P = (2.73)

For Shallow Waters

gCECEP == (since sinh2kd = 2kd)

Assume the wave propagates from deepwater towards the shore. The ocean bottom slope is

gradual and there are no undulations and has parallel bottom slope contours. Accordingly to

the conservation of energy, equating the power in the shallow waters (Eq.2.72) to that in deep

waters (Eq.2.73) we get

2C.

8HC.

8H o2o

G2 =

On substituting for CGand on simplification we obtain

+=

kd2sinh

kd21

1

C

C

H

H o2

o

sKn2

1.

C

oC

oH

H=

or


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32

(2.74)

where ,

+=kd2sinh

kd21

2

1n

The above equation giving the ratio between wave height at any water depth in shallower

waters and the deep-water height. This relationship obtained without considering the irregular

variation in the sea bottom contours is called as shoaling coefficient. The variation of the

different properties of small amplitude waves are shown in Fig.2.9.

2.11 MASS TRANSPORT VELOCITY

When waves are in motion, the particles upon completion of each nearly an elliptical or

circular motion would have advanced a short distance in the direction of propagation


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33

(Fig.2.10).Consequently there is a mass transport in the direction of progress of the wave. The

mass transport velocity at any depth z below S.W.L is given as

kdsinh

)zd(k2cosh.

2

C

L)z(u

2

2 +

= (2.75)

The mass transport speed is appreciable for high steep waves and is very small for waves of

long period.

2.12. IMPORTANT WEB ADDRESSES

www.coastal.udel.edu/faculty/rad/wavetheory.html

http://bigfoot.wes.army.mil/cem001.html

WORKED OUT EXAMPLES on WAVE MECHANICS

Problem 1

A wave flume is filled with fresh water to a depth of 5m. A wave of height 1m and period, 4sec. is generated. Calculate the wave celerity, group celerity, energy and power.

Solution: -
http://www.coastal.udel.edu/faculty/rad/wavetheory.htmlhttp://www.coastal.udel.edu/faculty/rad/wavetheory.html


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34

m21.22L

2003.0L

dtoingcorrespond2251.0

L

d

'TablesWave'From

2003.096.24

5Ld

m96.24)4(56.1T56.1L

kdtanhL

d

L

d

o

o

22o

o

=

==

==

===

=

Celerity= sec/m55.54

21.22

T

L=

Group Celerity=

+

kd2sinhkd21

2C

=

+436.8

83.21

2

52.

= 3.71 m/sec.

Energy = m/kg1258

2)1(1000

8

2H==

Power = sec/kg75.46371.3125G8

2H

=

C

PROBLEM 2

Oscillatory surface waves were observed in deep water and the wave period was found to be6.7 sec.

(i) At what bottom depth would the phase velocity begin to be changed by thedecreasing water depth.

(ii) What is the phase velocity at a bottom depth of 15.3 m and 3.06 m.

(iii) Compute the ratio of Celerity at the above water depths to the deep water celerities.

SOLUTION: -

(i) m03.707.656.1T56.1L 22o ===

The celerity will change, when water depth is less than

m01.352

03.70

2

Lo ==

(ii) At d = 15.3 m


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35

2185.003.70

3.15

L

d

o

==

From Appendix A (Wave tables) corresponding 2408.0L

d =

Hence L = m54.632408.0

3.15=

Celerity = sec/m48.97.6

54.63

T

L==

At d = 3.06 m

0437.003.70

06.3

L

d

o

==

Corresponding 0871.0L

d=

Hence m13.350871.0

06.3L ==

.sec/m24.57.6

13.35

T

L

==Celerity =

907.07.656.1

48.9

T56.1

48.9

C

C

o

=

=(iii) At d = 15.3, ratio

At d = 3.06, ratio 501.07.656.1

24.5

T56.1

24.5

C

C

o

=

=

=

PROBLEM- 3

A wave flume is filled with fresh water to a depth of 5m. A deep-water wave height 2m and

time period, 4 sec. is generated. For a given n=0.6689, calculate the wave celerity, groupcelerity, energy and power.

From tables, for n = 0.6689


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36

mH

H

H

Cg

TLC

T

TL

LL

L

d

L

d

o

o

o

83.12*9179.0

9179.0

72.3

57.5/

sec4

56.1

29.2212.25

2243.01990.0

2

0

==

=

=

==

=

=

==

==

Energy = mkgH

/6.4188

)83.1(1000

8

22

=

=

Power = sec/22.155772.3*6.4188

2kgH G ==C

PROBLEM 4

A wave of height 3m and period, 6 sec. is generated. Calculate the wave celerity, group

celerity, energy and power in water depths, 2m, 5m and 10m.

Solution: -

d=5 m

089.001313

''

089.016.56

5

16.56)6(56.156.1

tanh

22

==

==

===

=

o

o

o

o

L

dtoingcorrespond

L

d

TablesWaveFrom

L

d

mTL

kdL

d

L

d

n=0.8290 L=38.08

sec/34.64

08.38m

T

L==Celerity=

Group Celerity= C * n

= 6.34*0.8260

= 5.26 m/sec.

d=2m


37/55



37

0356.07748.0

''

0356.16.56

2

16.56)6(56.156.1

tanh

22

==

==

===

=

o

o

o

o

L

dtoingcorrespond

L

d

TablesWaveFrom

L

d

mTL

kdL

d

L

d

n=0.8290 L=25.81 m

Celerity= sec/30.44

81.25m

T

L==


= 4.30 * 0.9289

= 3.99m/sec.

d=10m

2066.001780.0

''

1780.016.56

10

16.56)6(56.156.1

tanh

22

==

==

===

=

o

o

o

o

L

dtoingcorrespond

L

d

TablesWaveFrom

L

d

mTL

kdL

d

L

d

n=0.6946 L=48.40

sec/06.8440.48 m

TL ==Celerity=


= 8.06*0.6946

= 5.59m/sec.


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38

PROBLEM 5

Consider a particle initially 3.06m below the SWL and 12.24 m above the sea bed. After the

wave motion is established, what is the size and character of the orbit of the particle. Repeat

the calculations for the particle at the surface and the other at the sea bed. T = 6.7 sec. and Ho

= 3.06 m.

Solution: -

Water depth, d = 3.06 + 12.24

= 15.3m

Lo= 1.56 T

2

= 70.03 m

2185.003.70

3.15

L

d

o

==

.2

1

L

d

20

1


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39

m4117.1kdsinh

kdsinh

2

HB ==

At the sea bed, z = -15.3 m

m6536.016.2

1*

2

8235.2D ==

00*2

HB ==

The water displacements at the three different levels are illustrated in the Fig. below.

PROBLEM 6

The semi major axis and semi minor axis are 1.1m and 1.0m at z = -2m. Calculate the water

depth, time period, wave height corresponding to this, as well as deep water wave height when

k = 0.14 and also find the displacement at z=-5 m, at sea bed and at free surface.

Solution: -

4222.0

4180.0

,14.0

=

=

=

L

d

Ld

tablefrom

kfor

o

H/HO=0.9795

LK

2=

L=44.87

4222.0=Ld


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40

d = 19m

4180.0=oL

d

Lo=45.45 m

T=5.45 sec

At z = -2

Cosh k(d+z) = 5.44, sinh kd = 7.113

mkd

zdkHD 1.1

sinh

)(cosh

2=

+=

1.1113.7

44.5*

2=

H

H =2.87m

(we can also use the semi minor axis to find the H value)

H0= H/0.9795

H0=2.93m

At the, sea bed z = 0

mkd

zdkHD 44.1

sinh

)(cosh

2=

+=

mkd

zdkHB 435.1

sinh

)(sinh

2=

+=

At the surface, z = -19 m

mD 20.0113.7

1*

2

87.2==

00*2

HB ==

Z=-5 m

Cosh k(d+z) = 3.62 sinh k(d+z) = 3.47

Semi Major axis (D) =

kdsinh

)zd(kcosh

2

H +


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41

= 0.730m

Semi minor axis (B) =kdsinh

)zd(ksinh

2

H +

= 0.700 m.

PROBLEM 7

Consider a particle initially 5m below the SWL and 20 m above the sea bed. After the wave

motion is established, what is the size and character of the orbit of the particle. Repeat the

calculations for the particle at the surface and the other at the sea bed. L=33m and a=2 m

Solution: -

Water depth, d = 5+ 20

= 25m

7575.033

25 ==L

d

.2

1>L

dThe particles will move in circular orbit.The wave is in deep water depth condition

7600.09994.075745.0 ===oo L

dfor

H

Hand

L

dFrom Appendix A,

i.e., H = a*2 = 4m

190.02

=

L

: kd = 4.75L = 33m ; k =

Z=-5 m

kdsinh

)zd(kcosh

2

H +Semi Major axis (D) =

= 0.774m

Semi minor axis (B) = 0.774 m.(since deep water)

At the surface, z = 0

mkd

kdHD 2

sinh

cosh

2==

(since deep water)mB 2=

At the sea bed, z = -25 m

mD 034.079.57

1

*2

4

==


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42

00*2

HB ==

Problem 8

For a wave of height 2m and period 7 secs, plot the variation of orbital velocity and

acceleration in the vertical and horizontal directions of a particle at a position 4m below SWL

and 20 m above the sea bed. Estimate the maximum velocities at this position, at SWL and at

the sea bed.

Solution:

( )

m96.73L,325.0L

dingcorrespond

314.07*56.1

24

T*56.1

24

L

d

m2420d

22o

==

===

=+=

At z = -4.0 m

Kd = 2.039, cosh k(d+z) = 2.826, cosh (kd) = 3.907Sinh k(d +z) = 2.643, sinh (kd) = 3.776

Substituting the values in the expressions for u and w (eq.2.39 & 2.40)

.sec/m628.0776.3

643.2

7

2w

.sec/m672.0776.3

826.2

7

2u

max

max

=

=

=

=

2

2

2

max

o

2

2

2

max

o

s/m564.0kdsinh

)zd(ksinh

T

H2w

s/m603.0kdsinh

)zd(kcosh

T

H2u

=+

=

=+

=


43/55



43

At z = 0.0

Coshk(d + z) =3.907, sinh k(d + z) = 3.776

.sec/m898776.3

776.3

7

2w

.sec/m929.0776.3907.3

72u

max

max

=

=

==

2max

o

2max

o

s/m805.0w

s/m833.0u

=

=

At z = -24.0

Coshk(d +z) = 1.0, sinh k(d +z) = 0.0

0maxw

2s/m213.0maxu

.sec/m238.0776.3

1

7

2umax

=

=

=

=

o

o

.sec/m0.00.07

2w max =

=

0 90 180 270 360

-1.0

-0.5

0.0

0.5

1.0

0 90 180 270 360

-1.0

-0.5

0.0

0.5

1.0

u

m/sec

0 90 180 270 360

-1.0

-0.5

0.0

0.5

1.0

w

m/sec

0 90 180 270 360

-1.0

-0.5

0.0

0.5

1.0

U

m/sec

0 90 180 270 360

-1.0

-0.5

0.0

0.5

1.0

Wm

/sec

2

o

(m)


44/55



44

The phase variation of u, w, are shown in the figure below.wanduoo

Problem 9

Determine the maximum orbital velocities and accelerations in the horizontal and verticaldirections of the particle at a position (i) 3.06m below SWL and 12.24 above the sea floor (ii)

at SWL and at the sea bed for H = 2.82 m and T = 6.7 sec.

Solution:

kdsinh

)zd(kcosh.

T

Hu max

+=

kdsinh

)zd(ksinh.

T

Hw max

+=

kdsinh

)zd(kcosh.

T

H2u

2

2. +=

kdsinh

)zd(ksinh.

T

H2w

2

2. +=

d = 3.06 +12.24 = 15.3 mL = 63.587m, kd = 1.512, sinh kd = 2.157

Using the appropriate expressions

For z = -3.06m: cosh k (d+z) = 1.825, sinh k (d+z) = 1.527

.sec/m049.1157.2

527.1

7.6

82.2w

.sec/m119.1157.2

825.1

7.6

82.2u

max

max

=

=

=

=

2max

.

sec/m049.1157.2

825.1

7.67.6

82.22u =

=

2max

.

sec/m878.0157.2

527.1

7.67.6

82.22w =

=


45/55



45

Similarly

at z = 0.0: cosh k (d+z) =2.378, sinh k(d+z) = 2.157

sec/m322.1wsec,/m457.1u maxmax ==

2max

.2

max

.

sec/m24.1w,sec/m367.1u ==

at z =15.3: cosh k(d+z) = 1.0, sinh k(d+z) = 0.0

0.0w,sec/m613.0u maxmax ==

0.0w,sec/m57.0u max.

2max

.

==

Problem -10

A wave with a height, 5.5m and period, 8 secs propagates in a water depth of 15m. Determinethe local horizontal and vertical velocities at a depth 3m below the SWL when phase angle is

60o.

Solution :

z =-3m, d = 15m, T = 8 sec., kd = 1.152, L = 81.79,

Cosh k(d+z) = 1.456, sinh kd = 1.425

Sinh k(d+z) = 1.058

+

= sin.kdsinh

)zd(kcosh.

T

Hu

+

= sin.kdsinh

)zd(ksinh.

T

Hw

Subsituting the corresponding values in the above expressions at

o60=

911.160sin425.1

456.1

8

5.5u =

= m/s


46/55



46

where as,

.sec/m21.2umax=

.sec/m8.060cos425.1058.1

85.5w ==

where as,

.sec/m604.1w max =

Problem 11

A wave of height 5.5m and wave length of 81.79m propagates in a water depth of 15m.Determine the local horizontal and vertical velocities at depth 3m below the SWL at a position

one fifth ahead of the wave crest.

Solution:

For L = 81.79 m, d=15m, kd = 1.152

= m358.1679.815

1L

5

1Phase angle at

oooo 162907236079.81

358.16 =+==

At z = -3m,

Cosh k(d+z) = 1.456, sinh kd = 1.425Sinh k(d+z) = 1.058

Using the expressions for u and w

.s/m68.0)162(sin425.1

456.1

8

5.5u =

=

Similarly for w

( ) .sec/m5216.1162cos425.1

058.1

8

5.5w +=

=


47/55



47

Problem 12

A wave of height H = 3m and wave period T = 10s, propagates in a water depth of12m, the corresponding deep-water wave height Ho= 3.5m. Estimate,

(a) The horizontal and vertical displacements from its mean position at z = 0 andZ=-d.

(b)The maximum water particle displacements at a depth of 7.50m below SWL,

Where the wave is in deep water.(c) For the deep water conditions of above show that the water particle

Displacements are small relative to the wave height at2

oLz =

(d)Also, compare the water particle displacements in deep water conditions for thecorresponding deep-water wave height, Ho= 3.5m and wave period T = 15s at z =

-7.5m

Solution:-

Given DataWave height, H = 3m

Wave period, T = 10s

Deep water wave height, Ho= 3.5m(a) The maximum horizontal (D) and vertical (B) displacements of water particle.

Lo= 1.56 T2= 1.56 (10)

2= 156m.

07692.0

156

12

Lo

d==

m58.991205.0

12

1205.0

dL1205.0

L

d====From table

0631.058.99

2

L

2K =

=

=

tanh kd = tanh (0.0631 x 12) = 0.6394sinh kd = sinh (0.0631 x 12) = 0.8316

@ z = 0

( )kdsinh

zdkcosh

2

HD

+=

( )

.m34.2

6394.0

1

2

3

kdtanh

1

2

H

kdsinh2

kdcosh

2

H0zD

=

====


48/55



( )2

0.3

2

H

kdsinh

kdsinh

2

H

kdsinh

zdksinh

2

HB ===

+=

= 1.5m.

@ z = -d = -12m

m80.18316.0

1

2

3

kdsinh

1

2

HD =

==

B = 0

B=1.50m

B = 0D = 180m

Z = -d

D=2.34m

Z = 0

(b) In deep water conditions, at z = -7.5m

( )( )m2938.1

5.70402.0e2

5.3zoke2o

HBD

0402.0156

2

oL

2oK

=

===

=

=

=

2

oL(c) In deep water conditions at z =

48

17D

D

B

17B

Z=-7.5m

Z=-78m= 20L

( )[ ]

.m07562.0

780402.0e2

5.3zoke2

oHBD

m5.3oH

m782

156z

=

===

=

=

=

So, it is inferred that for deep-water conditions the particle displacements are very small at

Compared with z = -7.5m.

m782

156z =

=

D171092.1707562.0

2938.1=


49/55



49

(d) T = 15s, Ho= 3.5m.

Lo= 1.56 T2= 1.56 (15)

2= 351m

Ko= .0179.0351

2

oL

2 =

At z = m5.175

2

351

2

oL =

=

( )( )5.1750179.0e2

5.3zoke2

oH = D = B =

= 0.07563m.

at Z = -78m

( )( )780179.0e2

5.3zoke2

oH =

( )

D = B =

= 0.43318m.

07563.0

43318.0

m5.1752

oLZat

ntsDisplacemeverticalandHorizontal

m78Zat

ntsDisplacemeverticalandHorizontal

=

==

=

= 5.72 6Hence, the displacements of the water particle at Z = -78m are approximately six times

the displacements at Z = .m5.1752

oL =

If z=-0.2depth, for T=15secs and 10secs, B=D=1.67m for the former and B=D=1.37m for thelater, indicating that the displacement for longer period wave, the displacement is more.

Problem 13

Aerial photographs of a coastal line displayed the presence of two wave systems. One with

crests 60m apart and another with crests at 12m spacing. Timing of major breaking on the

beach in the same period indicated the wave period to be10 sec., for the longer wave. What


50/55



50

was the depth of water in the zone of wave observation and what was the period of the minor

wave system?

Solution:

Lmajor = 60m : Lminor= 12m

Tmajor= 10 sec.

)kd(tanh2

2gTL

=

)kd(tanh.210

*2

g60

=

tanh (kd) = 0.3840

Corresponding to this value, from the wave tables d/L = 0.06478

Since Lmajor= 60 m

d = 0.06478 * 60 = 3.89m

= d*L

2tanh

2

gTL

2ormin

ormin

= 98.3*12

2tanh*

2

2ormin

gT12

sec8.297.0*8.9

24orminT ==

Problem 14

Ocean waves measure 90m from crest to crest when travelling at a point at a speed of 32

km/hr. Find the depth in the ocean at this point and the period of waves. If the waves were

fully grown and their steepness, H/L = 1/23, what is the wave height?

Solution:

L = 90m

hr/km32T

L=


51/55



.sec1.1032000

60*60*90

32

LT ===

sec/m89.818/5*32hr/km32C ===

89.8kdtanh2

gLC =

=

i.e. 012.792)89.8(2

kdtanhgL==

Therefore 5629.090*8.9

*2*012.79kdtanh =

=

Corresponding kd from wave tables = 0.6366

m1186.92

90*6366.0

d ==

23

1

L

H=

m913.323

90

23

LH ===

Problem 15

If a pressure sensing instrument is set up at 4m below SWL in a water depth of 20m, determine

the phase distribution of pressure head this instrument would record. Plot this pressure head

against phase and compare this result to the phase variation of hydrostatic pressure. The wave

height is 2m and period is 10 sec. and = 1020 kg/m3.

Solution:

The pressure head under a progressive wave is

zpKsin2

Hp

=

For T = 10 sec, d = 20 m

L is calculated as 121.24m

k = 0.0519 kd = 2/L * d = 1.037

51


52/55



52

859.0587.1

364.1

)20*k(cosh

))420(k(cosh

kdcosh

)zd(kcoshpK ==

=

+=

Dynamic pressure head , =

sin*859.0*

2

0.2p

2m/Kg8761020x1x859.0x2

0.2p,ssureDynamicpre ==

= 0.876 m of water column.

Total Pressure , p = 2m/kg18.49561020x4859.0sin2

2=

+

Hydrostatic pressure head = 0.5*H*sin- z

For T = 5 sec., d = 20 m, d/L = 0.5, L = Lo= 39

k = 2/39 = 0.1611, kd = 3.22

( )[ ]( )

( )52.0

54.12

6.6

)222.3(cosh

578.2cosh

20*1611.0cosh

4201611.0coshpK ===

=

2m/kg73.5381020xsin5282.0*2

0.2PessurePrTotal =

=

Note:

For a wave period of 5 secs the total pressure is less when compared to the wave period of 10

sec for the other conditions NOT being changed.

So, For longer waves the pressure is more

Problem 16

A subsurface pressure type recorder is installed at a depth of 6m at the point where water depthis 8m. The average maximum pressure and the period registered by the recorder are 3060

kg/m3and 9.2 sec respectively. Compute , = 1020 kg/m3.

Solution:

k = 0.0823


53/55



53

[ ][ ]

8276.08*0823.0cosh

)68(0823.0coshpK =

=

zpKp

=

)6(8276.0*1020

3060=

8276.0

1.6

1020

3060

=

= - 3.62 m

Note: Problem incorrect due to wave breaking condition

Problem 17

An average maximum pressure of 12500 kg/m2is measured by a sub surface pressure recorder

located at 0.6m above the sea bed in a water depth of 12m. The average wave frequency is

0.0666 cycles/sec. and = 1025 kg/m3. Determine the wave height.

Solution:

zpKp

=

The maximum pressure would occur at

= H/2 (Crest of the wave)

pmax = 12500, = 1025, z = - (12 0.6) = -11.4 m

m6.157L.,sec150666.0

1T ===

8952.01174.1

0003.1

12*6.157

2cosh

4.1112(*6.157

2cosh

pK ==

=

Substituting for the variables in the above formula


54/55



54

( )4.112

H*8952.0

1025

12500=

Hence H=1.78 m

Problem: 18

Water depth,d =12m, Wave period, T=10sec and Wave height, H=1.0m

(a) Calculate mass transport velocity, u(z) for z=0 to d and find out the effect of z/d on u(z)

(b) d=12m, T=10sec and H=0.5 to 9m at 0.5m interval. Find the effect of H/L on u(z) at z=0.0

(c) d=12m, H=2.0m and T=5 to 15sec at every 1sec interval. Find the effect of d/L on u(z) at

z=0.0

Solution:

kd2

sinh

)zd(k2cosh

2

C2

L

H +

Mass transport velocity u(z) = m/sec

(i) Effect of z/d on u(z):

Fig.1shows the variation of u(z) with respect to the various values of z/d. It is

observed that the mass transport velocity decreases with increase in z/d values. i.e the mass

transport velocity decrease from the free surface towards the sea bed.

(ii) Effect of H/L on u(z):

Fig.2shows the variation of u(z) with respect to the various of H/L. From the plot, it is clear

that u(z) increases with increase in H/L values. This means that, as the wave height increases

the mass transport velocity also increases.

(iii) Effect of d/L on u(z):

Fig.3shows the variation of u(z) for various values of d/L. The plot shows that with increase

in d/L values u(z) also increases. If wave period increases, then the mass transport decreases.


55/55



0.04 0.08 0.12

d/L

0.00

0.10

0.20

U(z)

m/sec

0.0 0.1 0.1

H/L

0.0

1.0

2.0

U(z)m/sec

0.0 0.5 1.0

z/d

0.000

0.010

0.020

U(z)m/sec

Fig.1 Variation of U(z) with z/d

Fig.2 Variation of U(z) with H/LFig.3 Variation of U(z) with d/L

d=12m, T=10sec , H=1.0mz=0.0 to d @ 1m interval

d=12.0m, T=12sec, z=0.0H=0.5 to 9m @ 0.5m interval

d=12.0m, H=2.0m, z=0.0T=5 to 15sec @ 1 sec interval

For Clarifications and comments if any contact the author at

[email protected]

basics of wave motion

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