Chapter 3 linear wave theory and wave propagation

CHAPTER 3: WAVE THEORY AND
WAVE PROPAGATION
DR. MOHSIN SIDDIQUE
ASSISTANT PROFESSOR
1
0401444 - Intro. To Coastal Eng.
University of Sharjah
Dept. of Civil and Env. Engg.

WAVE THEORIES
2
Figure: Ranges of applicability of various
wave theories
Wave theories are
approximations to reality.
1. Linear wave theories
• Small amplitude wave theory
2. Non-linear wave theories
• Stokes finite amplitude wave
theory
3. Other theories
• Nonlinear shallow water eave
theories
• KdV and Boussinesq Eqs
• Cnoidal wave theory
• Stream function theory
• Fourier approximation

WAVE THEORIES
Figure: Wave profile of different progressive gravity waves

4
SMALL AMPLITUDE WAVE THEORY
The earliest mathematical description of periodic progressive waves is that
attributed to Airy in 1845.
Airy wave theory is strictly only applicable to conditions in which the wave
height is small compared to the wavelength and the water depth.
It is commonly referred to as linear or first order wave theory, because
of the simplifying assumptions made in its derivation.

5
L= wave length
T=wave period
σ=angular frequency=2π/T
C=wave speed (celerity)
H=wave height
a= amplitude of wave (H/2)=horizontal
excursion of water particle
η=instantaneous surface water elevation
d= local water depth
Figure: Definition of progressive surface wave parameters
u and w =The horizontal and
vertical components of the
water particle velocity at any
instant.
ς and ε = The horizontal and
vertical coordinates of a water
particle at any instant.

6
Assumptions:
The basic assumptions for linear wave theory are as follows:
1. Water is homogeneous and incompressible, wave lengths are greater
than 3m so that capillary effects may be ignored
2. Flow is irrotational, i.e. no shear stress present anywhere. Thus the
velocity potential, Φ, must satisfy the Laplace equation:
Laplace equation is actually continuity equation:
Where,

7
Assumptions:
3. The bottom is not moving and is impermeable and horizontal, i.e. no
energy transfer through the bed. This lead to:
4. The pressure along the air-sea interface is constant, i.e. no effect of
weather related pressure differences
5. The wave amplitude is small compared to the wave length and water
depth.
Assuming
η is very small

Where, k is wave number (=2 π /L) and
σ is angular frequency(=2 π /T)
Since the velocity potential, Φ, should be cyclic with horizontal position and
time, and it should vary with depth, we can assume that:
Substituting Eq.(2) into (1), one can get:
8
Basic Equations:
The water surface profile is given as:

9
The general solution to this partial differential equation is:
Where A an B are arbitrary constants. Putting this value of the function f(z)
in Eq.(2), we get:
Two boundary conditions are required to find A and B.

10
(a) The vertical velocity, w, at the bottom must be zero (Assumption 3):

11
(b) Second boundary condition (on the surface) may be derived from
Bernoulli’s equation for time-varying flow in two dimensions, i.e.

12

13
Dispersion of water waves generally refers to frequency dispersion, which
means that waves of different wavelengths travel at different phase speeds

14

Summary:
k =(2 π /L) & σ= (2 π /T)
&

16
Wave classification by relative depth (d/L)
1. Deep water:
When the relative depth, d/L, is greater that 0.5, tanh(2πd/L) ≈ 1; and
Eq.(7), (8) and (9) transform to the following, respectively:
This condition is called deep water condition and denoted by the subscript zero.
In this condition it may be noted
that the wave celerity is
independent of the water depth

17
2. Transitional water
In this case 0.5>d/L>0.05 & tanh(kh) < 1, and Eq.(7), (8) and (9) are used.

18
3. Shallow water:
Another extreme condition may be found when the relative depth, d/L, is
less than 0.05, tanh(2πd/L) ≈ 2πd/L. In this condition, Eq.(7) or (8)
becomes:
It is obvious that in this
situation the wave celerity
depends only on the water
depth.

Example: A wave with a period T=10s is propagated shoreward over a
uniformly sloping shelf from a depth of 200m to 1m. Calculate wave
celebrities, C, and lengths, L, to depths 200m, 3m & 1m.
Solution:
( )
mL
L
g
L
L
dgT
L
13.156
2002
tanh
2
10
2
tanh
2
2
2
=






=






=
π
π
π
π
5.028.1
13.156
200
>==
L
d
Since
So, the wave is deep water wave
For d=200m
( )
mL
L
g
L
L
dgT
L
16.53
32
tanh
2
10
2
tanh
2
2
2
=






=






=
π
π
π
π
05.0056.0
16.53
3
>==
L
d
Since
So, the wave is transition water wave
For d=3m
sm
T
L
C /6.15
10
13.156
===
mLL o 13.156==
sm
T
L
C /32.5
10
16.53
===
mL 16.53=
Using dispersion relationship Using dispersion relationship

Example: A wave with a period T=10s is propagated shoreward over a
uniformly sloping shelf from a depth of 200m to 1m. Calculate wave
celebrities, C, and lengths, L, to depths 200m, 3m & 1m.
Solution:
( )
mL
L
g
L
L
dgT
L
11.31
12
tanh
2
10
2
tanh
2
2
2
=






=






=
π
π
π
π
05.0032.0
11.31
1
<==
L
d
Since
So, the wave is shallow water wave
For d=1m
sm
T
L
C /11.3
10
11.31
===
mL 11.31=
Using dispersion relationship

Example: A wave tank is 193m long, 4.57m wide and 6.1m deep. The
tank is filled to a depth of 5m with fresh water and a 1m high, 4sec
period wave is generated. (a) Calculate the wave celerity and length
using small amplitude wave theory. (b) Calculate the corresponding
deep water wave length and celerity.
Deep water
Using dispersion relationship

Wave Kinematics: The horizontal and vertical components of water
particle velocity u and w may be determined as
respectively.

• η and u are in phase
• η and w are out of phase (phase
shift of 90o)
• η and ax are out of phase (phase
shift 90o)
• η and az are out of phase (Phase
shift is 180o)
Figure: profiles of particle velocity and
acceleration by Airy theory in relation
to the surface elevation (CEM, 2008)

Example: A wave with a period, T, =8sec, in a water depth, d, =15m
and a wave height, H,= 5.5m is propagating. Find local horizonal and
vertical velocities , u & w, and acceleration ax, & az at an elevation of
z=-5m below the still water level (SWL) when kx-σt=π/3 (60 degrees)
Solution:
Given: T=8s; d=15m; H=5.5
Find: u, w, ax, az =? @ z=-5m
First of all find L, using dispersion relationship
( )
mL
L
g
L
L
dgT
L
8.81
152
tanh
2
8
2
tanh
2
2
2
=






=






=
π
π
π
π
18.0
8.81
15
==
L
d
Since
So, the wave transitional water wave

Solution: velocities
L
kwhere
π2
, =
( )
( )
smu
u
/99.0
3/cos
15
8.81
2
sinh
515
8.81
2
cosh
8
5.5
=












−






= π
π
π
π
( )
( )
smw
w
/11.1
3/sin
15
8.81
2
sinh
515
8.81
2
sinh
8
5.5
=












−






= π
π
π
π
L
kwhere
π2
, =

Solution: Acceleration
L
kwhere
π2
, =
( )
( )
2
2
2
/35.1
3/sin
15
8.81
2
sinh
515
8.81
2
cosh
8
5.52
sma
a
x
x
=












−








= π
π
π
π
( )
( )
2
2
2
/504.0
3/cos
15
8.81
2
sinh
515
8.81
2
sinh
8
5.52
sma
a
z
z
−=












−








−= π
π
π
π

Particle displacement: The horizontal and vertical ordinates of the
particle displacement (ς and ε ) are related to the components of
particle velocity by:
Since

For deep water:
For Shallow water:

Figure. Water particle displacements from mean position for shallow-water and deep water
waves (CEM, 2008)

Pressure Field: In order to derive a relationship for pressure we can
use Bernoulli’s equation as follows:
In deep water, the dynamic pressure reduces to near zero at z =- L/2. A pressure
gauge (located above L/2) can be used as a wave gauge.

Wave Energy:

Example: Calculate the wave energy (total, Kinetic, and potential)
and power for the a wave if its height is 1m, wave period = 4s,
water depth=5m and wave length =22.2m.
Solution:
Given: H=1m; T=4s; d=5m; L=22.2m
Find: E, P=?

Wave power: Wave power, P, is the wave energy per unit time
transmitted in the direction of wave propagation, which is the product of
the force acting on a vertical plane normal to the direction of wave
propagation times the particle flow velocity across this plane.
Mathematically,

The n varies with the relative depth; where n increases from 0.5 in deep
water to 1.0 in shallow water.
For a wave train, considering the reflected or dissipated energy to be
negligible, the conservation of energy per unit time requires that:
If we draw lines orthogonal to the wave crests, and the spacing of the
orthogonals is B, we can write the energy between two orthogonals as
BE and therefore:

Commonly, waves are predicted for some deep water location and then
must be transformed to some intermediate or shallow water depth
nearshore using Eq. (38). For this, Eq. (38) becomes
(39)

Example: A wave in water 100 m deep has a period of 10 s and a height
of 2 m. When it has propagated into a water depth of 10 m without
refracting and assuming energy gains and losses can be ignored.
Determine the wave height, power and the water particle velocity
and pressure at a point 1 m below the still water level under the
wave crest.
Solution: Given: d=100m: H=2m & T=10s
Calculate L at d=10m
L = 93.3 m
d/L =10/ 93.3=0.11
k = 2π/93.3 = 0.0673m-1
L = 156 m
d/L = 100/156= 0.64 >0.5 (deep water)
Hence, Lo = 156 m & H=Ho=2m






=
L
dgT
L
π
π
2
tanh
2
2
Calculate L,
Using dispersion
relationship

Example:
where
Find H at d=10m Assuming no refraction i.e. B/Bo=1
Power:
wattsP 51.38806
8
3.9397.181.91000
10
874.0 2
=







 ×××
=

Velocity:
( )( )
( )
smw
smu
/0
/01.11
100673.0sinh
1100679.0cosh
10
97.1
=
=
×
−






=
π
Pressure:

Group Celerity: Considering two harmonic waves with same height
and slightly different wave lengths and periods, superimposed on each
other, the resulting wave profile is:
Since the wave lengths of the both waves, L1 and L2, have been assumed slightly
different, for some values of x the two components will be in phase (resultant wave
height will be 2H) with each other while some other values they will be completely
out of phase (the resultant wave height will be zero).

Figure: Characteristics of a wave group formed by the addition of
sinusoids with different periods (CEM, 2008)
(40)
(40)
Group velocity: It is the velocity of propagation of the envelope
curve

Note:
2π/L=k
2π/T= σ
θ=kx-σt

Mass Transport and Wave Set-up:
According to the small amplitude wave theory, the water particle paths are
closed and the particles oscillate about a mean position. In reality, the viscosity
is present and therefore the water particles do not follow exactly the path
depicted by the small amplitude wave theory.
More practical approaches like Stokes wave theory shows that the particle
paths are not closed and with the passage of each wave the particles
experience a small amount of translation. This fact is shown schematically in
Fig.
Figure: Water particle paths

The results of computation by Longuet-Higgins and Stewart (1963)
show that the mean sea level falls below the still water level from
offshore to the breaking point; a condition called as wave set-down.
In the surf zone (from breaking point to shoreline) the mean sea level
rises above the still water level; a condition named as wave set-up
(Fig.).
Figure: Schematic for wave set-up and set-down

Standing Waves and Wave Reflection: Changes in the bottom depth,
convergence of wave tank walls (or expansion) or presence of an
obstacle cause the reflection of waves.
If the wave is fully reflected, i.e. (Incident wave amplitude, ai, =
Reflected wave amplitude, ar,) the water surface shows a standing
wave called as Clapotis.
In a clapotis, nodes are the points on the water surface that remain
stationary whereas, the points called antinodes oscillate in vertical
direction.
If the wave is partially reflected the resulting standing wave is called as
partial clapotis.

For a general case:

Figure: Standing wave particle motion and surface profile envelope.
(a) Cr = 1.0, (b) Cr < 1.0.

Formulas for standing wave

Wave breaking: Reduction in wave energy and height in the surf zone
due to limited water depth.
Miche (1944) has given the limiting condition for wave breaking in any
water depth as:
Here H/L= wave steepness.
In deep water, tanh(2πd/L) ≅ 1, therefore:
In shallow water, tanh(2πd/L) ≅ 2πd/L, therefore:
Eq. (56) shows that wave height is limited by water depth.

Types of breaking waves:

Spilling breaker: the wave crest
becomes unstable, and cascades down
the shoreward face of the wave,
producing a foamy water surface (Fig.a).
Plunging breaker: The crest curls over
the shoreward face of the wave and falls
into the base of the wave, resulting in a
high splash (Fig.b).
Collapsing breaker: The crest remains
unbroken while the lower part of the
shoreward face steepens and the falls,
producing an irregular turbulent water
surface (Fig. c).
Surging breaker: The crest remains
unbroken and the front face of the wave
advances up the beach with minor
breaking

Breaker type may be correlated to the surf similarity parameter, ξo,
defined as:
Under mild beach slope and small wave steepness the spilling breaker
occurs.
For a relatively steeper beach slope and higher wave steepness,
plunging breaker can be observed.
Under extremely steep slopes of the beach the surging breaker is
observed

Although Eq.(56), gives an idea about the wave height at breaking, for
design purposes it is not enough to use this relationship. A number of
laboratory and field investigations have been carried out to study the
wave breaking phenomenon at various slopes. Some of this information
is provided in Figs. below.
Figure. Dimensionless breaker height
versus deep water wave steepness
Figure. Dimensionless breaker depth versus
breaker steepness for various beach slopes
Remember

Given the beach slope, the unrefracted deep water wave height, Ho’,
and the wave period one can calculate the deep water wave steepness
and then determine the breaker height from Figure below.
Figure: Dimensionless breaker height and class versus bottom slope and deep water
steepness. (Modified from U.S. Army Coastal Engineering Research Center, 1984.)

Figure: Dimensionless breaker depth versus bottom slope and breaker steepness.
(Modified from U.S. Army Coastal Engineering Research Center, 1984.)

Example: What is the wave height and water depth at breaking for
1-m high 4-sec period wave shoaling on a 1:10 slope in a wave tank
with water depth of 5m before the slope? What type of breaker would
you expect?
Using dispersion relation, calculate L;
Calculate deep water wave length, Lo;
Calculate unrefracted deep water wave height, Ho’;
Calculate breaking wave height, Hb;

Calculate breaking depth, db,

Wave Runup: After the wave breaking, the remaining energy causes
the water to run up the face of the beach or sloping surface of a
structure. The definition of the wave runup (R) is shown schematically
in Fig.
Figure (next slide) shows the experimental data for smooth, planar and
impermeable slopes with monochromatic waves and with ds/Ho’
between 1 and 3.
Figure: Schematic for wave runup (Sorensen, 1997)
r is determined from table and run-up on a smooth, impermeable surface is determined
from figure
α= slope

Therefore, the wave runup on a given surface can be found by first
determining R from Fig. below and then multiplying it with r from Table
for the given surface.
Figure. Dimensionless runup on smooth impermeable slopes versus bottom slope and
incident deep water wave steepness; 1 < ds/H0‘<3.
(Modified from U.S. Army Coastal Engineering Research Center, 1984.)

The effect of surface conditions of the slope may be introduced using
the data given in Table below. In this table,

Example: Consider the deep water wave (depth=100, wave period =10
s and wave height = 2 m.) propagating toward the shore without
refracting. The wave breaks and runs up on a 1:10 grass covered slope
having a toe depth of 4 m. Determine the breaking wave height and
the wave runup elevation on the grass-covered slope.
Solution:
For a deep water unrefracted wave height of 2 m (Ho= Ho’) and a period
of 10s, we have:
Then, from figure for m=0.1;

For breaking wave height

Runup:
From Figure , since ds/Ho’= 4/2 = 2, at cot (α) = 10

ravg=0.875

Since; R/Ho’=0.85
on smooth surface
on grass

WAVE PROPAGATION
Conceptual drawing of cross‐shore sediment processes in the near‐shore region (Zandan, 2016)
Shoaling: Increase in wave height due to decrease in water depth
Breaking: Reduction in wave energy and height in the surf zone due to limited water depth
(Collapse of wave crest over its front face)
Wave transformation: Change in wave energy due to action of physical process (i.e., change in
wave form as it propagates)
69

WAVE TRANSFORMATION
• Reflection: Reflection involves a
change in direction of waves when
they bounce off a barrier
• Refraction: Refraction of waves
involves a change in the direction of
waves as they pass from one medium
to another. It is accompanied by a
change in speed and wavelength of
the waves
• Diffraction: Diffraction involves a
change in direction of waves as they
pass through an opening or around a
barrier in their path.
70

WAVE PROPAGATION
71
Refraction: The refraction occurs in transitional (d/L=0.05~0.5) and
shallow water (d/L<0.05). As the wave celerity decreases with the water
depth, the waves in shallow portion travel slower than those in deeper
region. As a result the bending of wave crests occurs so that they approach
the orientation of bottom contours.

WAVE PROPAGATION
72
Where, refraction coefficient =
Refraction analyses were initially done by the manual construction of refraction
diagrams. Now, they are mostly done by numerical/computer analysis.
If a wave having a deep water height, Ho, refracts into a nearshore location where the
refraction coefficient is Krthen the unrefracted deep water height, Ho’, to be used in
these diagrams is
Ho’= KrHo.

WAVE PROPAGATION
73
Construction of wave refraction
diagrams: A popular method is
based on Snell’s law, generally
called as the Orthogonal Method.
Snell’s law can be derived
considering Fig.
A train of waves travels over a step (neglect the wave reflection by the step) where
the depth instantaneously decreases from d1 to d2 causing the wave celerity and
length to decrease from C1 and L1 to C2 and L2, respectively. From Fig. we can
observe that:
Eq. (60) expresses Snell’s law.

WAVE PROPAGATION
74
When waves shoal over near-shore contours that are essentially straight
and parallel as shown in Fig., it may be written that:
If we choose B0 and B1 so that the
orthogonal lengths equal L0 and L1 as
shown, then:
Equations (61) and (62) allow us to estimate the effects of wave refraction at a shoreline
with uniform near-shore bathymetry.

WAVE PROPAGATION
75
Example: A wave train is observed approaching a coast that has straight
parallel near-shore contours in the north-south direction. Where the depth is
5m, the wave length is 85m and the wave crest forms an angle of 9o with
the shore (waves from south-west). What is the incident wave direction in
deep water?
Shore line

WAVE PROPAGATION
76
Deep water
wave celerity

WAVE PROPAGATION
77
Wave diffraction: When a train of waves passes an impermeable
structure, there will be a transfer of energy (wave) along the wave crest into
the lee of the structure a shown in Fig. As a result, the wave height in the
region inside the dashed line will be affected.
Figure. Wave diffraction behind an
obstruction
It is found that the diffraction
coefficient is a function of the following
parameters as defined in Fig.

WAVE PROPAGATION
78
Wiegel (1962) used the exact solution presented by Penny and Price
(1952) and calculated diffraction coefficient as a function of selected values
of θ,β, and r/L.

WAVE PROPAGATION
79
Example: A train of 6-sec waves approaching an impermeable, non-
reflecting breakwater at an angle θ=60o . If the water depth is constant at 10
m, what would be the wave height at β=30o and r = 96.6 m from the
breakwater tip, when the incident wave height is 1m.

WAVE PROPAGATION
80
r/L=2; β=30o; θ=60o
KD=0.28

WAVE PROPAGATION
81
The theory for diffraction of water waves approaching in a direction normal
to a long straight structure and passing through a single gap in that
structure was also developed by Penny and Price (1952). A typical variation
of KD in the form of contours is shown in Fig. plotted by Johnson (1952).
Figure: Diffraction coefficient at barrier gap; gap width = 2L

WAVE PROPAGATION
82
Johnson showed that these diagrams could be used if the angle of wave
incidence is other than 90o by using a projected imaging gap width as
shown in Fig. (previous slide). If the gap width is greater that 5L , the
diffraction on each side of the gap is independent from the other. Then the
diffraction can be determined at each side by considering it a single barrier
(From Table).
Figure: Oblique wave incident to a barrier gap

WAVE PROPAGATION
83
Wave Reflection: A graphical method to determine the reflected crest
pattern is shown in Figure below. The incident wave crest is extended past
the reflecting barrier. The mirror image of the imaginary incident wave crest
extension is the reflected wave crest. The reflection coefficient, Cr, is given
as the ratio of the wave height of reflected wave to the wave height of the
incident wave, i.e.
Cr=Hr/Hi
Figure : Wave reflection analysis

WAVE PROPAGATION
84
To calculate the wave height that has
been transformed by diffraction as well
as reflection, consider Figure (on right),
where a straight crested wave passes a
barrier, diffracts into the lee of that
barrier, and then reflects off a second
barrier. The water depth is constant so
the wave only undergoes diffraction
and reflection.
The imaginary diffracted wave crest
pattern is carried to point A’ where the
imaginary wave height can be
determined from the incident height
and direction along with the values of r
and β as previously discussed. The
reflected wave height at point A would
equal the imaginary diffracted height at
point A’ times the reflection coefficient.

Chapter 3 linear wave theory and wave propagation

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Chapter 3 linear wave theory and wave propagation