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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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Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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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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2.2.1 BOUNDARY CONDITIONS:
(1)The equation (2.1) is to be satisfied in the region
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Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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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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Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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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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Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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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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Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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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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Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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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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Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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==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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Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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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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Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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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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Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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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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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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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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(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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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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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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+
=
++
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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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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(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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(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: -
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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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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
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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==
Group Celerity= C * n
= 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=
Group Celerity= C * n
= 8.06*0.6946
= 5.59m/sec.
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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
=+
=
=+
=
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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)
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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 =
=
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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
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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 +=
=
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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
=
====
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( )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=
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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
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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=
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.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
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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
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Basics of wave motion
Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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[ ][ ]
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
8/11/2019 Basics of Wave Motion
54/55
Basics of wave motion
Prof.V.Sundar, Dept.of Ocean Engg, I.I.T.Madras, India
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.
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
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