Download - ANSWERS (WAVES EXAMPLES) AUTUMN 2021
Hydraulics 3 Answers (Waves Examples) -1 Dr David Apsley
ANSWERS (WAVES EXAMPLES) AUTUMN 2021
Q1.
Given:
β = 20 m
π΄ = 4 m (π» = 8 m)
π = 13 s
Then,
Ο =2Ο
π = 0.4833 rad sβ1
Dispersion relation:
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
0.4762 = πβ tanh πβ
The LHS is small, so iterate as
πβ =1
2(πβ +
0.4762
tanh πβ)
to get
πβ = 0.7499
π =0.7499
β = 0.03750 mβ1
πΏ =2Ο
π = 167.6 m
π =Ο
π (or
πΏ
π) = 12.89 m sβ1
Answer: wavelength 168 m, phase speed 12.9 m sβ1.
(b) By formula:
π’ =πππ΄
Ο
cosh π(β + π§)
cosh πβcos (ππ₯ β Οπ‘)
ππ₯ =ππ’
ππ‘+ non-linear terms = πππ΄
cosh π(β + π§)
cosh πβsin (ππ₯ β Οπ‘)
Hence
πx,max = 1.137 cosh π(β + π§)
Then:
surface (π§ = 0): πx,max = 1.137 cosh πβ = 1.472 m sβ2
mid-depth (π§ = β β 2β ): πx,max = 1.137 cosh(πβ/2) = 1.218 m sβ2
Hydraulics 3 Answers (Waves Examples) -2 Dr David Apsley
bottom: (π§ = ββ): πx,max = 1.137 m sβ2
Answer: accelerations (1.47, 1.22, 1.14) m sβ2.
Hydraulics 3 Answers (Waves Examples) -3 Dr David Apsley
Q2.
Waves A and B (no current)
In the absence of current the dispersion relation is
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
This may be iterated as either
πβ =Ο2β πβ
tanh πβ or πβ =
1
2(πβ +
Ο2β πβ
tanh πβ)
Wave A Wave B
β 20 m 3 m
π 10 s 12 s
Ο =2π
π 0.6283 rad sβ1 0.5236 rad sβ1
Ο2β
π 0.8048 0.08384
Iteration: πβ =1
2(πβ +
0.8048
tanh πβ) πβ =
1
2(πβ +
0.08384
tanh πβ)
πβ 1.036 0.2937
π 0.0518 mβ1 0.0979 mβ1
π =Ο
π 12.13 m sβ1 5.348
Type: Intermediate (Ο
10< πβ < Ο) Shallow (πβ <
Ο
10)
Wave C (with current)
ππ = 6 π
β = 24 m
π = β0.8 m sβ1
Then
Οπ =2π
ππ = 1.047 rad sβ1
Hydraulics 3 Answers (Waves Examples) -4 Dr David Apsley
With current, the dispersion relation is
(Οπ β ππ0)2 = Οπ2 = ππ tanh πβ
Rearrange for iteration as
π =(Οπ β ππ0)2
π tanh πβ
i.e.
π =(1.047 + 0.8π)2
9.81 tanh(24π)
Solve:
π = 0.1367 mβ1
πβ = 3.2808
The speed relative to the fixed-position sensor is the absolute speed:
ππ =Οπ
π =
1.047
0.1367 = 7.659 m sβ1
This is a deep-water wave, since πβ > Ο.
Answer: A: intermediate, 12.1 m sβ1; B: shallow, 5.348 m sβ1; C: deep, 7.66 m sβ1.
Hydraulics 3 Answers (Waves Examples) -5 Dr David Apsley
Q3.
Given:
β = 15 m
π = 0.1 Hz
πrange = 9500 N mβ2
Then,
Ο =2Ο
π = 2Οπ = 0.6283 rad sβ1
Dispersion relation:
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
0.6036 = πβ tanh πβ
The LHS is small, so iterate as
πβ =1
2(πβ +
0.6036
tanh πβ)
to get
πβ = 0.8642
π =0.8642
β = 0.05761 mβ1
πΏ =2Ο
π = 109.1 m
By formula, the wave part of the pressure is
π = Οππ΄cosh π(β + π§)
cosh πβcos (ππ₯ β Οπ‘)
and hence the range (max β min) at the seabed sensor (π§ = ββ) is
πrange =2Οππ΄
cosh πβ =
Οππ»
cosh πβ
Hence,
π» = πrange Γcosh πβ
Οπ = 9500 Γ
cosh 0.8642
1025 Γ 9.81 = 1.32 m
Answer: height = 1.32 m.
Hydraulics 3 Answers (Waves Examples) -6 Dr David Apsley
Q4.
On the seabed (π§ = ββ) the gauge pressure distribution is
π = Οπβ +Οππ΄
cosh πβcos(ππ₯ β Οππ‘)
The average and the fluctuation of this will give the water depth and wave amplitude,
respectively:
οΏ½Μ οΏ½ = Οπβ , Ξπ =Οππ΄
cosh πβ
i.e.
β =οΏ½Μ οΏ½
Οπ , π΄ =
Ξπ
Οgcosh πβ
Here,
οΏ½Μ οΏ½ =98.8 + 122.6
2Γ 1000 = 110400 Pa
Ξπ =122.6 β 98.8
2Γ 1000 = 11900 Pa
Hence,
β =110400
1025 Γ 9.81 = 10.98 m
For the amplitude, we require also πβ, which must be determined, via the dispersion relation,
from the period. From, e.g., the difference between the peaks of 5 cycles,
π =49.2 β 6.8
5 = 8.48 s
Οπ =2Ο
π = 0.7409 rad sβ1
(a) No current.
The dispersion relation rearranges as
Οπ2β
π= πβ tanh πβ
0.6144 = πβ tanh πβ
Rearrange for iteration. Since the LHS < 1 the most effective form is
πβ =1
2(πβ +
0.6144
tanh πβ)
Iteration from, e.g., πβ = 1, gives
πβ = 0.8737
Hydraulics 3 Answers (Waves Examples) -7 Dr David Apsley
Then the amplitude is
π΄ =Ξπ
Οgcosh πβ =
11900
1025 Γ 9.81Γ cosh 0.8737 = 1.665 m
Corresponding to a wave height
π» = 2π΄ = 3.33 m
Finally, for the wavelength,
π =πβ
β =
0.8737
10.98 = 0.07957 mβ1
πΏ =2Ο
π = 78.96 m
Answer: water depth = 11.0 m; wave height = 3.33 m; wavelength = 79.0 m.
(b) With current π = 2 m sβ1
The dispersion relation is
(Οπ β ππ)2 = ππ tanh πβ
Rearrange as
π =(Οπ β ππ)2
π tanh πβ
Here,
π =(0.7409 β 2π)2
9.81 tanh(10.98π)
This does not converge very easily. An alternative using under-relaxation is
π =1
2[π +
(0.7409 β 2π)2
9.81 tanh(10.98π)]
Iteration from, e.g., π = 0.1 mβ1, gives
π = 0.06362 mβ1
and, with β = 10.98 m),
πβ = 0.6985
Then the amplitude is
π΄ =Ξπ
Οgcosh πβ =
11900
1025 Γ 9.81Γ cosh 0.6985 = 1.484 m
corresponding to a wave height
π» = 2π΄ = 2.968 m
Hydraulics 3 Answers (Waves Examples) -8 Dr David Apsley
Finally, for the wavelength,
πΏ =2Ο
π = 98.76 m
Answer: water depth = 11.0 m; wave height = 2.97 m; wavelength = 98.8 m.
Hydraulics 3 Answers (Waves Examples) -9 Dr David Apsley
Q5.
There are three depths to be considered:
deep / transducer (22 m) / shallow (8 m)
The shoreward rate of energy transfer is constant; i.e.
(π»2ππ)deep = (π»2ππ)transducer = (π»2ππ)shallow
We can find π and π from the dispersion relation at all locations, and the height π» from the
transducer pressure measurements.
Given:
π = 12 s
Then
Ο =2Ο
π = 0.5236 rad sβ1
Dispersion relation:
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
Solve for πβ, and thence π by iterating either
πβ =Ο2β πβ
tanh πβ or (better here) πβ =
1
2(πβ +
Ο2β πβ
tanh πβ)
together with
π =Ο
π , π =
1
2[1 +
2πβ
sinh 2πβ]
In shallow water (β = 8 m), Ο2β π = 0.2236β , and hence:
πβ = 0.4912 , π = 0.06140 mβ1, π = 8.528 m sβ1 , π = 0.9278
At the transducer (β = 22 m), Ο2β π = 0.6148β , and hence:
πβ = 0.8740 , π = 0.03973 mβ1, π = 13.18 m sβ1 , π = 0.8139
In deep water, tanh πβ β 1 and so
Ο2 = ππ
whence
π =Ο2
π= 0.02795
The phase speed is
π =Ο
π =
π
Ο = 18.74 m sβ1
Hydraulics 3 Answers (Waves Examples) -10 Dr David Apsley
and, in deep water,
π =1
2
At the pressure transducer we have
Οππ΄cosh π(β + π§)
cosh πβ= 10000
with Ο = 1025 kg mβ3 (seawater) and π§ = β20 m in water of depth β = 22 m. Hence,
π΄ =10000 Γ cosh 0.8740
1025 Γ 9.81 Γ cosh[0.03973 Γ 2] = 1.395 m
and hence
π»transducer = 2π΄ = 2.790 m
Then, from the shoaling equation:
π»deep = π»transducerβ(ππ)transducer
(ππ)deep = 2.790 Γ β
0.8139 Γ 13.18
0.5 Γ 18.74 = 2.985 m
π»shallow = π»transducerβ(ππ)transducer
(ππ)shallow = 2.790 Γ β
0.8139 Γ 13.18
0.9278 Γ 8.528 = 3.249 m
Answer: wave heights (a) nearshore: 3.25 m; (b) deep water: 2.99 m.
Hydraulics 3 Answers (Waves Examples) -11 Dr David Apsley
Q6.
Given:
β = 1 m
π = 0.1 m
π = 0.05 m
First deduce πβ, and hence the wavenumber, from the ratio of the semi-axes of the particle
orbits:
dπ
dπ‘= π’ β
π΄ππ
Ο
cosh π(β + π§0)
cosh πβcos(ππ₯0 β Οπ‘)
π = π₯0 βπ΄ππ
Ο2
cosh π(β + π§0)
cosh πβsin(ππ₯0 β Οπ‘)
π = π₯0 β π΄cosh π(β + π§0)
sinh πβsin(ππ₯0 β Οπ‘)
dπ
dπ‘= π€ β
π΄ππ
Ο
sinh π(β + π§0)
cosh πβsin(ππ₯0 β Οπ‘)
π = π§0 +π΄ππ
Ο2
sinh π(β + π§0)
cosh πβcos(ππ₯0 β Οπ‘)
π = π§0 + π΄sinh π(β + π§0)
sinh πβcos(ππ₯0 β Οπ‘)
Where, in both directions we have used the dispersion relation Ο2 = ππ tanh πβ to simplify.
Hence,
π = π΄cosh(πβ/2)
sinh πβ = 0.1
π = π΄sinh(πβ/2)
sinh πβ = 0.05
Then
π
π= tanh(πβ/2) = 0.5
πβ = 2 tanhβ1(0.5) = 1.099
π =1.099
1 = 1.099 mβ1
Then wave height can be deduced from, e.g., the expression for π:
π΄ = πsinh πβ
cosh(πβ/2) = 0.1 Γ
sinh 1.099
cosh(1.099/2) = 0.1155
Hydraulics 3 Answers (Waves Examples) -12 Dr David Apsley
π» = 2π΄ = 0.2310 m
From the dispersion relation:
Ο2 = ππ tanh πβ β Ο = 2.937 rad sβ1
Finally,
π =2Ο
Ο = 2.139 s
πΏ =2Ο
π = 5.717 m
Answer: height = 0.0268 m; period = 2.14 s; wavelength = 5.72 m.
Hydraulics 3 Answers (Waves Examples) -13 Dr David Apsley
Q7.
Given
β = 20 m
π΄ = 1 m (π» = 2 m)
(a) Here,
π = +1 m sβ1
ππ = 3 s
Οπ =2Ο
ππ = 2.094 rad sβ1
Dispersion relation:
(Οπ β ππ)2 = Οπ2 = ππ tanh πβ
Iterate as
π =(Οπ β ππ)2
π tanh πβ
i.e.
π =(2.094 β π)2
9.81 tanh 20π
to get
π = 0.3206
πΏ =2Ο
π = 19.60 m
The maximum particle velocity occurs at the surface and is the wave-relative maximum
velocity plus the current, i.e. from the wave-induced particle velocity
π’π =π΄ππ
Οπ
cosh π(β + π§)
cosh πβcos(ππ₯ β Οπ‘)
we have
Οπ = Οπ β ππ = 2.094 β 0.3206 Γ 1 = 1.773 rad sβ1
As the wave is travelling in the same direction as the current:
|π’|max =π΄ππ
Οπ+ π =
1 Γ 9.81 Γ 0.3206
1.773+ 1 = 2.774 m sβ1
Answer: wavelength 19.6 m; maximum particle speed 2.77 m sβ1.
(b) Now,
π = β1 m sβ1
ππ = 7 s
Hydraulics 3 Answers (Waves Examples) -14 Dr David Apsley
Οπ =2Ο
ππ = 0.8976 rad sβ1
Dispersion relation is rearranged for iteration as above:
π =(0.8976 + π)2
9.81 tanh 20π
to get
π = 0.1056
πΏ =2Ο
π = 59.50 m
Wave-relative frequency:
Οπ = Οπ β ππ = 0.8976 + 0.1056 Γ 1 = 1.003 rad sβ1
This time, as the current is opposing, the maximum speed is the magnitude of the backward
velocity:
|π’|max =π΄ππ
Οπ+ |π| =
1 Γ 9.81 Γ 0.1056
1.003+ 1 = 2.033 m sβ1
Answer: wavelength 59.5 m; maximum particle speed 2.03 m sβ1.
Hydraulics 3 Answers (Waves Examples) -15 Dr David Apsley
Q8.
As the waves move from deep to shallow water they refract (change angle ΞΈ) according to
π0 sin ΞΈ0 = π1 sin ΞΈ1
and undergo shoaling (change height, π») according to
(π»2ππ cos ΞΈ)0 = (π»2ππ cos ΞΈ)1
where subscript 0 indicates deep and 1 indicates the measuring station. Period is unchanged.
Given:
π = 5.5 s
then
Ο =2Ο
π = 1.142 rad sβ1
In deep water,
π0 =Ο2
π = 0.1329
π0 =1
2
π0 =ππ
2Ο (or
Ο
π) = 8.587 m sβ1
In the measured depth β = 6 m the wave height π»1 = 0.8 m. The dispersion relation is
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
0.7977 = πβ tanh πβ
Iterate as
πβ =0.7977
tanh πβ or (better here) πβ =
1
2(πβ +
0.7977
tanh πβ)
to get (adding subscript 1):
π1β = 1.030
π1 =1.030
β = 0.1717 mβ1
π1 =Ο
π1 = 6.651 m sβ1
π1 =1
2[1 +
2π1β
sinh 2π1β] = 0.7669
Refraction
Hydraulics 3 Answers (Waves Examples) -16 Dr David Apsley
π0 sin ΞΈ0 = π1 sin ΞΈ1
Hence:
sin ΞΈ0 =π1
π0sin ΞΈ1 =
0.1717
0.1329sin 47Β° = 0.9449
ΞΈ0 = 70.89Β°
Shoaling
(π»2ππ cos ΞΈ)0 = (π»2ππ cos ΞΈ)1
Hence:
π»0 = π»1β(ππ cos ΞΈ)1
(ππ cos ΞΈ)0 = 0.8 Γ β
0.7669 Γ 6.651 Γ cos 47Β°
0.5 Γ 8.587 Γ cos 70.89Β° = 1.259 m
Answer: deep-water wave height = 1.26 m; angle = 70.9Β°.
Hydraulics 3 Answers (Waves Examples) -17 Dr David Apsley
Q9.
(a)
Deep water
πΏ0 = 300 m
π»0 = 2 m
From the wavelength,
π0 =2Ο
πΏ0 = 0.02094 mβ1
From the dispersion relation with tanh πβ = 1:
Ο2 = ππ0
whence
Ο = βππ0 = 0.4532 rad sβ1
This stays the same as we move into shallower water.
π0 =Ο
π0 = 21.64 m sβ1
π0 = 0.5
Depth β = 30 m
The dispersion relation is
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
0.6281 = πβ tanh πβ
Iterate as
πβ =0.6281
tanh πβ or (better here) πβ =
1
2(πβ +
0.6281
tanh πβ)
to get (adding subscript 1):
πβ = 0.8856
π =0.8856
β = 0.02952 mβ1
πΏ =2Ο
π = 212.8 m
π =Ο
π = 15.35 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ] = 0.8103
Hydraulics 3 Answers (Waves Examples) -18 Dr David Apsley
ππ = ππ = 12.44 m sβ1
From the shoaling relationship
(π»2ππ)0 = π»2ππ
Hence,
π» = π»0β
(ππ)0
ππ = 2 Γ β
0.5 Γ 21.64
0.8103 Γ 15.35 = 1.865 m
Answer: wavelength = 213 m; height = 1.87 m; group velocity = 12.4 m sβ1.
(b)
πΈ =1
2Οππ΄2 =
1
8Οππ»2 =
1
8Γ 1025 Γ 9.81 Γ 1.8652 = 4372 J mβ2
Answer: energy = 4370 J mβ2.
(c) If the wave crests were obliquely oriented then the wavelength and group velocity would
not change (because they are fixed by period and depth). However, the direction would change
(by refraction) and the height would change (due to shoaling).
Refraction:
π0 sin ΞΈ0 = π sin ΞΈ
Hence:
sin ΞΈ =π0
πsin ΞΈ0 =
0.02094
0.02952sin 60Β° = 0.6143
ΞΈ0 = 37.90Β°
Shoaling:
(π»2ππ cos ΞΈ)0 = π»2ππ cos ΞΈ
Hence:
π» = π»0β
(ππ cos ΞΈ)0
ππ cos ΞΈ = 2 Γ β
0.5 Γ 21.64 Γ cos 60Β°
0.8103 Γ 15.35 Γ cos 37.90Β° = 1.485 m
Answer: wavelength and group velocity unchanged; height = 1.48 m.
Hydraulics 3 Answers (Waves Examples) -19 Dr David Apsley
Q10.
(a)
π = 8 s
Ο =2Ο
π = 0.7854 rad sβ1
Shoaling from 10 m depth to 3 m depth. Need wave properties at these two depths.
The dispersion relation is
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
This may be iterated as either
πβ =Ο2β πβ
tanh πβ or πβ =
1
2(πβ +
Ο2β πβ
tanh πβ)
β = 3 m β = 10 m
Ο2β
π 0.1886 0.6288
Iteration: πβ =1
2(πβ +
0.1886
tanh πβ) πβ =
1
2(πβ +
0.6288
tanh πβ)
πβ 0.4484 0.8862
π 0.1495 mβ1 0.08862 mβ1
π =Ο
π 5.254 m sβ1 8.863 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ] 0.9388 0.8101
π» ? 1 m
From the shoaling equation:
(π»2ππ)3 m = (π»2ππ)10 m
Hence:
π»3m = π»10mβ(ππ)10 m
(ππ)3 m = 1 Γ β
0.8101 Γ 8.863
0.9388 Γ 5.254 = 1.207 m
Answer: wave height = 1.21 m.
Hydraulics 3 Answers (Waves Examples) -20 Dr David Apsley
(b) The wave continues to shoal:
(π»2ππ)π = (π»2ππ)3 m = 7.186
(in m-s units). Assuming that it breaks as a shallow-water wave, then
ππ = 1
ππ = βπβπ
and we are given, from the breaker depth index:
π»π = 0.78βπ
Substituting in the shoaling equation,
(0.78βπ)2βπβπ = 7.186
1.906βπ5 2β
= 7.186
giving
βπ = 1.700 m
π»π = 0.78βπ = 1.326 m
Answer: breaking wave height 1.33 m in water depth 1.70 m.
Hydraulics 3 Answers (Waves Examples) -21 Dr David Apsley
Q11.
Narrow-band spectrum means that a Rayleigh distribution is appropriate, for which
π(wave height > π») = exp [β (π»
π»rms)
2
]
Central frequency of 0.2 Hz corresponds to a period of 5 seconds; i.e. 12 waves per minute. In
8 minutes there are (on average) 8 Γ 12 = 96 waves, so the question data says basically that
π(wave height > 2 m) =1
96
Comparing with the Rayleigh distribution with π» = 2 m:
exp [β (2
π»rms)
2
] =1
96
exp [(2
π»rms)
2
] = 96
4
π»rms2
= ln 96
π»rms = β4
ln 96 = 0.9361 m
(a)
π(wave height > 3 m) = exp [β (3
π»πππ )
2
] = 3.463 Γ 10β5 =1
28877
This corresponds to once in every 28877 waves, or, at 12 waves per minute,
28877
12= 2406 min = 40.1 hours
Answer: about once every 40 hours.
(b) The median wave height, π»med, is such that
π(wave height > π»med) =1
2
Hence
exp [β (π»med
π»rms)
2
] =1
2
Hydraulics 3 Answers (Waves Examples) -22 Dr David Apsley
exp [(π»med
π»rms)
2
] = 2
(π»med
π»rms)
2
= ln 2
π»med = π»rmsβln 2 = 0.9361βln 2 = 0.7794 m
Answer: median height = 0.779 m.
Hydraulics 3 Answers (Waves Examples) -23 Dr David Apsley
Q12.
(a) Narrow-banded, so a Rayleigh distribution is appropriate. Then
π»rms =π»π
1.416 =
2
1.416 = 1.412 m
(b)
π»1 10β = 1.800 π»rms = 1.800 Γ 1.412 = 2.542 m
(c)
π(4 m < height < 5 m) = π(height > 4) β π(height > 5)
= exp [β (4
1.412)
2
] β exp [β (5
1.412)
2
]
= 3.235 Γ 10β4
or about once in every 3090 waves.
Answer: (a) π»πππ = 1.41 m; (b) π»1 10β = 2.54 m; (c) probability = 3.2410β4.
Hydraulics 3 Answers (Waves Examples) -24 Dr David Apsley
Q13.
The Bretschneider spectrum is
π(π) =5
16π»π
2ππ
4
π5exp (β
5
4
ππ4
π4)
Here we have
π»π = 2.5 m
ππ =1
ππ =
1
6 Hz
Amplitudes
For a single component the energy (per unit weight) is
1
2π2 = π(π)Ξπ
Hence,
ππ = β2π(π)Ξπ
In this instance, Ξπ = 0.25ππ, so that, numerically,
ππ = β0.9766
(ππ ππβ )5 exp [β
1.25
(ππ ππβ )4]
These are computed (using Excel) in a column of the table below.
Wavenumbers
For deep-water waves,
Ο2 = ππ
or, since Ο = 2Οπ
ππ =4Ο2ππ
2
π
Numerically here:
ππ = 0.1118 (ππ
ππ)
2
These are computed (using Excel) in a column of the table below.
Velocity
The maximum particle velocity for one component is
Hydraulics 3 Answers (Waves Examples) -25 Dr David Apsley
π’π =πππππ
ππ
cosh ππ(β + π§)
cosh ππβ
or, since the water is deep and hence cosh π ~1
2eπ:
π’π =πππππ
2πππexp πππ§
Numerically here, with π§ = β5 m:
π’π =9.368ππππ
ππ/ππexp(β5ππ)
These are computed (using Excel) in a column of the table below.
ππ/ππ ππ ππ π’π
0.75 0.281410 0.062888 0.161410
1 0.528962 0.111800 0.316769
1.25 0.437929 0.174688 0.239372
1.5 0.316967 0.251550 0.141566
1.75 0.228204 0.342388 0.075503
2 0.168004 0.447200 0.037614
Sum: 1.961475 0.972235
From the table, with all components instantaneously in phase
Ξ·πππ₯ = β ππ = 1.961 m
π’max = β π’π = 0.9722 m sβ1
Answers: (a), (b) [ππ] and [ππ] given in the table; (c) Ξ·max = 1.96 m, π’max = 0.972 m sβ1.
Hydraulics 3 Answers (Waves Examples) -26 Dr David Apsley
Q14.
(a) Waves are duration-limited if the wind has not blown for sufficient time for wave energy
to propagate across the entire fetch. Otherwise they are fetch-limited.
(b) Given
π = 13.5 m sβ1
πΉ = 64000 m
π‘ = 3 Γ 60 Γ 60 = 10800 s
then the relevant non-dimensional fetch is
οΏ½ΜοΏ½ β‘ππΉ
π2 = 3445
The minimum non-dimensional time required for fetch-limited waves is found from
οΏ½ΜοΏ½min β‘ (ππ‘
π)
min = 68.8οΏ½ΜοΏ½2 3β = 15693
but the actual non-dimensional time for which the wind has blown is
οΏ½ΜοΏ½ β‘ππ‘
π = 7848
which is less. Hence, the waves are duration-limited and in the predictive curves we must use
an effective fetch πΉeff given by
68.8οΏ½ΜοΏ½eff2/3
= 7848
whence
οΏ½ΜοΏ½eff = 1218
Then
ππ»π
π2β‘ οΏ½ΜοΏ½π = 0.0016οΏ½ΜοΏ½eff
1/2 = 0.05584
πππ
πβ‘ οΏ½ΜοΏ½π = 0.286οΏ½ΜοΏ½eff
1/3 = 3.054
from which the corresponding significant wave height and peak period are
π»π = 1.037 m
ππ = 4.203 s
The significant wave period is estimated as
ππ = 0.945ππ = 3.972 s
Answer: duration-limited; significant wave height = 1.04 m and period 3.97 s.
Hydraulics 3 Answers (Waves Examples) -27 Dr David Apsley
Q15.
Wave Properties
Given:
β = 6 m
π = 6 s
Then,
Ο =2Ο
π = 1.047 rad sβ1
Dispersion relation:
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
0.6705 = πβ tanh πβ
Iterate as
πβ =0.6705
tanh πβ or (better here) πβ =
1
2(πβ +
0.6705
tanh πβ)
to get
πβ = 0.9223
π =0.9223
β = 0.1537 mβ1
Forces and Moments
The height of the fully reflected wave is 2π», where π» is the height of the incident wave. Thus
the maximum crest height above SWL is π» = 1.5 m. Since the SWL depth is 6 m, the
maximum crest height does not overtop the caisson.
The modelled pressure distribution is as shown, consisting of
hydrostatic (from π1 = 0 at the crest to π2 = Οππ» at the SWL);
linear (from π2 at the SWL to the wave value π3 at the base);
linear underneath (from π3 at the front of the caisson to 0 at the rear).
To facilitate the computation of moments, the linear distribution on the front face can be
decomposed into a constant distribution (π3) and a triangular distribution with top π2 β π3.
Hydraulics 3 Answers (Waves Examples) -28 Dr David Apsley
The relevant pressures are:
π1 = 0
π2 = Οππ» = 1025 Γ 9.81 Γ 1.5 = 15080 Pa
π3 =Οππ»
cosh πβ =
1025 Γ 9.81 Γ 1.5
cosh(0.1537 Γ 6) = 10360 Pa
This yields the equivalent forces and points of application shown in the diagram below. Sums
of forces and moments (per metre width) are given in the table.
Region Force, πΉπ₯ or πΉπ§ Clockwise turning moment about heel
1 πΉπ₯1 =1
2π2 Γ π» = 11310 N/m πΉπ₯1 Γ (β +
1
3π») = 73520 N m mβ
2 πΉπ₯2 = π3 Γ β = 62160 N/m πΉπ₯2 Γ1
2β = 186480 N m mβ
3 πΉπ₯3 =1
2(π2 β π3) Γ β = 14160 N/m πΉπ₯3 Γ
2
3β = 56640 N m mβ
4 πΉπ§4 =1
2π3 Γ π€ = 20720 N /m πΉπ§4 Γ
2
3π€ = 55250 N m mβ
The sum of the horizontal forces
πΉπ₯1 + πΉπ₯2 + πΉπ₯3 = 87630 N/m
The sum of all clockwise moments
371900 N m / m
Answer: per metre of caisson: horizontal force = 87.6 kN; overturning moment = 372 kN m.
p1
3p
3p
1
2
3
4
2p
heel
Fx
My
Fz
Hydraulics 3 Answers (Waves Examples) -29 Dr David Apsley
Q16.
Period:
π =1
π =
1
0.1 = 10 s
The remaining quantities require solution of the dispersion relationship.
Ο =2Ο
π = 0.6283 rad sβ1
Rearrange the dispersion relation Ο2 = ππ tanh πβ to give
Ο2β
π= πβ tanh πβ
Here, with β = 18 m,
0.7243 = πβ tanh πβ
Since the LHS < 1, rearrange for iteration as
πβ =1
2[πβ +
0.7243
tanh πβ]
Iteration from, e.g., πβ = 1 produces
πβ = 0.9683
π =0.9683
β = 0.05379 mβ1
The wavelength is then
πΏ =2Ο
π = 116.8 m
and the phase speed and ratio of group to phase velocities are
π =Ο
π (or
πΏ
π) = 11.68 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ] = 0.7852
For the wave power,
power (per m) =1
8Οππ»2ππ =
1
8Γ 1025 Γ 9.81 Γ 2.12 Γ 0.7852 Γ 11.68
= 50840 W mβ1
Answer: period = 10 s; wavelength = 117 m; phase speed = 11.68 m sβ1;
power (per metre of crest) = 50.8 kW mβ1.
(b) First need new values of π, π and π in 6 m depth.
Ο2β
π= πβ tanh πβ
Hydraulics 3 Answers (Waves Examples) -30 Dr David Apsley
Here, with β = 6 m,
0.2414 = πβ tanh πβ
Since the LHS < 1, rearrange for iteration as
πβ =1
2[πβ +
0.2414
tanh πβ]
Iteration from, e.g., πβ = 1 produces
πβ = 0.5120
π =0.5120
β = 0.08533 mβ1
π =Ο
π = 7.363 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ] = 0.9222
For direction, use Snellβs Law:
(π sin ΞΈ)18 m = (π sin ΞΈ)6 m
whence
(sin ΞΈ)6 m =(π sin ΞΈ)18 m
π6 m =
0.05379 Γ sin 25Β°
0.08533 = 0.2664
and hence the wave direction in 6 m depth is 15.45Β°.
For wave height, use the shoaling equation:
(π»2ππ cos ΞΈ)6 m = (π»2ππ cos ΞΈ)18 m
Hence,
π»6 m = π»18 mβ(ππ cos ΞΈ)18 m
(ππ cos ΞΈ)6 m = 2.1β
0.7852 Γ 11.68 Γ cos 25Β°
0.9222 Γ 7.363 Γ cos 15.45Β° = 2.367 m
Answer: wave height = 2.37 m; wave direction = 15.5Β°.
(c) The Miche breaking criterion gives
(π»
πΏ)
π= 0.14 tanh(πβ)π
If we are to assume shallow-water behaviour, then tanh πβ ~πβ and so
(π»π
2Ο)
π= 0.14(πβ)π
whence
Hydraulics 3 Answers (Waves Examples) -31 Dr David Apsley
(π»
β)
π= 2Ο Γ 0.14 = 0.8796
or
π»π = 0.8796βπ (*)
This is used to eliminate wave height on the LHS of the shoaling equation
(π»2ππ cos ΞΈ)π = (π»2ππ cos ΞΈ)6 m
For refraction, Snellβs Law in the form (sin ΞΈ)/π = ππππ π‘πππ‘, together with the shallow-water
phase speed at breaking, gives
(sin ΞΈ
βπβ)
π
= (sin ΞΈ
π)
6 m =
sin 15.45Β°
7.363 = 0.03618
whence
sin ΞΈπ = 0.1133ββπ
cos ΞΈπ = β1 β sin2 ΞΈπ = β1 β 0.01284βπ
With the shallow water approximations π = 1, π = βπβ, and the relation (*), the shoaling
equation becomes
(0.8796βπ)2βπβπβ1 β 0.01284βπ = 36.67
or
2.423βπ5/2(1 β 0.01284βπ)1/2 = 36.67
This rearranges for iteration:
βπ = 2.965(1 β 0.01284βπ)β1/5
to give
βπ = 2.988 m
Answer: breaks in water of depth 2.99 m.
Hydraulics 3 Answers (Waves Examples) -32 Dr David Apsley
Q17.
(a)
π = 35 m sβ1
πΉ = 150000 m
From these,
οΏ½ΜοΏ½ =ππΉ
π2 = 1201
From the JONSWAP curves,
οΏ½ΜοΏ½min = 68.8οΏ½ΜοΏ½2 3β = 7774
(i) For π‘ = 4 hr = 14400 s,
οΏ½ΜοΏ½ =ππ‘
π = 4036
This is less than 7774. We conclude that there is insufficient time for wave energy to have
propagated right across the fetch; the waves are duration limited. Hence, instead of οΏ½ΜοΏ½, we use
an effective dimensionless fetch οΏ½ΜοΏ½eff such that
68.8οΏ½ΜοΏ½eff2/3
= 4036
οΏ½ΜοΏ½eff = 449.3
Then
οΏ½ΜοΏ½π = 0.0016οΏ½ΜοΏ½eff1/2
= 0.03391
οΏ½ΜοΏ½π = 0.2857οΏ½ΜοΏ½eff1/3
= 2.188
The dimensional significant wave height and peak wave period are
π»π =οΏ½ΜοΏ½π π2
π = 4.234 m
ππ =οΏ½ΜοΏ½ππ
π = 7.806 s
Answer: significant wave height = 4.23 m; peak period = 7.81 s.
(ii) For π‘ = 8 hr = 28800 s,
οΏ½ΜοΏ½ =ππ‘
π = 8072
This is greater than 7774, so the wave statistics are fetch limited and we can use οΏ½ΜοΏ½ = 1201.
Then
οΏ½ΜοΏ½π = 0.0016οΏ½ΜοΏ½1/2 = 0.05545
οΏ½ΜοΏ½π = 0.2857οΏ½ΜοΏ½1/3 = 3.037
The dimensional significant height and peak wave period are
Hydraulics 3 Answers (Waves Examples) -33 Dr David Apsley
π»π =οΏ½ΜοΏ½π π2
π = 6.924 m
ππ =οΏ½ΜοΏ½ππ
π = 10.84 s
Answer: significant wave height = 6.92 m; period = 10.8 s.
(b) (i) The wave spectrum is narrow-banded, so it is appropriate to use the Rayleigh probability
distribution for wave heights.
In deep water π»rms is known: π»rms = 1.8 m. For a single wave:
π(π» > 3 m) = exp[β(3/1.8)2] = 0.06218
This is once in every
1
π= 16.08 waves
or, with wave period 9 s, once every 144.7 s.
Answer: every 145 s.
(ii) In this part the wave height is π» = 3 m at depth 10 m, but π»rms is given in deep water. So
we either need to transform π» to deep water or π»rms to the 10 m depth. Weβll do the former.
Use the shoaling equation (for normal incidence):
(π»2ππ)0 = (π»2ππ)10 m
First find π and π in 10 m depth:
π = 9 s
Ο =2Ο
π = 0.6981 rad sβ1
Rearrange the dispersion relation Ο2 = ππ tanh πβ to give
Ο2β
π= πβ tanh πβ
Here, with β = 10 m,
0.4968 = πβ tanh πβ
Since the LHS < 1, rearrange for iteration as
πβ =1
2[πβ +
0.4968
tanh πβ]
Iteration from, e.g., πβ = 1 produces
πβ = 0.7688
Hydraulics 3 Answers (Waves Examples) -34 Dr David Apsley
Then,
π =0.7688
β = 0.07688 mβ1
π =Ο
π = 9.080 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ] = 0.8464
In deep water:
π0 =ππ
2Ο = 14.05 m sβ1
π0 =1
2
Hence, when the inshore wave height π»10 = 3 m, the deep-water wave height, from the
shoaling equation, is
π»0 = π»10β(ππ)10
(ππ)0 = 3 Γ β
0.8464 Γ 9.080
0.5 Γ 14.05 = 3.138 m
(If we had transformed π»rms to the 10 m depth instead we would have got 1.721 m.)
Now use the Rayleigh distribution to determine wave probabilities. For a single wave:
π(π»10 > 3 m) = π(π»0 > 3.138 m) = exp[β(3.138/1.8)2] = 0.04787
This is once in every
1
π= 20.89 waves
or, with wave period 9 s, once every 188.0 s.
Answer: every 188 s.
Hydraulics 3 Answers (Waves Examples) -35 Dr David Apsley
Q18.
(a)
(i) Refraction β change of direction as oblique waves move into shallower water;
(ii) Diffraction β spreading of waves into a region of shadow;
(iii) Shoaling β change of height as waves move into shallower water.
(b)
Ο =2Ο
π =
2Ο
12 = 0.5236 rad sβ1
Deep water
Dispersion relation
Ο2 = ππ0
Hence:
π0 =Ο2
π = 0.02795 mβ1
πΏ0 =2Ο
π0 = 224.8 m
π0 =Ο
π0 (or
πΏ0
π) = 18.73 m sβ1
π0 =1
2
Shallow water
Dispersion relation:
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
0.1397 = πβ tanh πβ
Iterate as either
πβ =0.1397
tanh πβ or (better here) πβ =
1
2(πβ +
0.1397
tanh πβ)
to get
πβ = 0.3827
π =0.3827
β = 0.07654 mβ1
πΏ =2Ο
π = 82.09 m
Hydraulics 3 Answers (Waves Examples) -36 Dr David Apsley
π =Ο
π (or
πΏ
π) = 6.841 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ ] = 0.9543
Refraction
π sin ΞΈ = π0 sin ΞΈ0
Hence,
sin ΞΈ =π0 sin ΞΈ0
π =
0.02795 Γ sin 20Β°
0.07654 = 0.1249
ΞΈ = 7.175Β°
Shoaling
(π»2ππ cos ΞΈ)5 m = (π»2ππ cos ΞΈ)0
Hence:
π»5 m = π»0β(ππ cos ΞΈ)0
(ππ cos ΞΈ)5 m = 3 Γ β
0.5 Γ 18.73 Γ cos 20Β°
0.9543 Γ 6.841 Γ cos 7.175Β° = 3.497 m
Answer: wavelength = 82.1 m; direction = 7.18Β°; height = 3.50 m.
(c) The breaker height index is
π»π
π»0= 0.56 (
π»0
πΏ0)
β1 5β
= 0.56 Γ (3
224.8)
β1 5β
= 1.328
Hence,
π»π = 1.328 π»0 = 3.984 m
The breaker depth index is
Ξ³π β‘ (π»
β)
π= π β π
π»π
ππ2
where, with a beach slope π = 1 20β = 0.05:
π = 43.8(1 β eβ19π) = 26.86
π =1.56
1 + eβ19.5π = 1.133
Hence,
Ξ³π = 1.133 β 26.86 Γ3.984
9.81Γ122 =1.058
giving:
Hydraulics 3 Answers (Waves Examples) -37 Dr David Apsley
βπ =π»π
Ξ³π = 3.766 m
Answer: breaker height = 3.98 m; breaking depth = 3.77 m.
Hydraulics 3 Answers (Waves Examples) -38 Dr David Apsley
Q19.
(a)
(i) βNarrow-banded sea stateβ β narrow range of frequencies.
(ii) βSignificant wave heightβ β average height of the highest one-third of waves
or, from the wave spectrum,
π»π0 = 4( π2Μ Μ Μ )1 2β
= 4π0
(iii) βEnergy spectrumβ: distribution of wave energy with frequency; specifically,
π(π) dπ
is the wave energy (divided by Οπ) in the small interval (π, π + dπ).
(iv) βDuration-limitedβ β condition of the sea state when the storm has blown for
insufficient time for wave energy to propagate across the entire fetch.
(b) If the period is π = 10 s then the number of waves per hour is
π =3600
π = 360
(i) For a narrow-banded sea state a Rayleigh distribution is appropriate:
π(height > π») = eβ(π» π»rmsβ )2
Hence,
π(height > 3.5) = eβ(3.5 2.5β )2 = 0.1409
For 360 waves, the expected number exceeding this is
360 Γ 0.1409 = 50.72
Answer: 51 waves.
(ii)
π(height > 5) = eβ(5 2.5β )2 = 0.01832
Corresponding to once in every
1
0.01832= 54.59 waves
With a period of 10 s, this represents a time of
10 Γ 54.59 = 545.9 s
(Alternatively, this could be expressed as 6.59 times per hour.)
Answer: 546 s (about 9.1 min) or, equivalently, 6.59 times per hour.
Hydraulics 3 Answers (Waves Examples) -39 Dr David Apsley
(iii) In deep water the dispersion relation reduces to:
Ο2 = ππ
or
(2Ο
π)
2
= π (2Ο
πΏ)
Hence,
πΏ =ππ2
2Ο
With π = 10 s,
πΏ = 156.1 m
π =πΏ
π = 15.61 m sβ1
and, in deep water
ππ =1
2π = 7.805 m sβ1
Answer: wavelength = 156.1 m; celerity = 15.6 m sβ1; group velocity = 7.81 m sβ1.
(c) Given
π = 20 m sβ1
πΉ = 105 m
π‘ = 6 hours = 21600 s
Then
οΏ½ΜοΏ½ β‘ππΉ
π2 = 2453
οΏ½ΜοΏ½min β‘ππ‘min
π = 68.8πΉ2 3β = 12513
But the non-dimensional time of the storm is
οΏ½ΜοΏ½ =ππ‘
π = 10590
This is less than οΏ½ΜοΏ½min, hence the sea state is duration-limited.
Hence, we must calculate an effective fetch by working back from οΏ½ΜοΏ½:
οΏ½ΜοΏ½eff = (οΏ½ΜοΏ½
68.8)
3 2β
= 1910
The non-dimensional wave height and peak period then follow from the JONSWAP formulae:
οΏ½ΜοΏ½π β‘ππ»π
π2 = 0.0016οΏ½ΜοΏ½eff
1 2β = 0.06993
Hydraulics 3 Answers (Waves Examples) -40 Dr David Apsley
οΏ½ΜοΏ½π β‘πππ
π = 0.2857οΏ½ΜοΏ½eff
1 3β = 3.545
whence, extracting the dimensional variables:
π»π =π2
ποΏ½ΜοΏ½π = 2.851 m
ππ =π
ποΏ½ΜοΏ½π = 7.227 s
Answer: duration-limited; significant wave height = 2.85 m; peak period = 7.23 s.
Hydraulics 3 Answers (Waves Examples) -41 Dr David Apsley
Q20.
(a)
(i) Two of:
spilling breakers: steep waves and/or mild beach slopes;
plunging breakers: moderately steep waves and moderate beach slopes;
collapsing breakers: long waves and/or steep beach slopes.
(ii)
π = 7 s
Ο =2Ο
π = 0.8976 rad sβ1
Shoaling from 100 m depth to 12 m depth. Need wave properties at these two depths.
The dispersion relation is
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
This may be iterated as either
πβ =Ο2β πβ
tanh πβ or πβ =
1
2(πβ +
Ο2β πβ
tanh πβ)
β = 12 m β = 100 m
Ο2β
π 0.9855 8.213
Iteration: πβ =0.9855
tanh πβ πβ =
8.213
tanh πβ
πβ 1.188 8.213
π 0.099 mβ1 0.08213 mβ1
π =Ο
π 9.067 m sβ1 10.93 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ] 0.7227 0.5000
ΞΈ 0ΒΊ (necessarily) 0ΒΊ
π» ? 1.8 m
From the shoaling equation:
Hydraulics 3 Answers (Waves Examples) -42 Dr David Apsley
(π»2ππ)12 m = (π»2ππ)100 m
Hence:
π»12 m = π»100 mβ(ππ)100 m
(ππ)12 m = 1.8 Γ β
0.5 Γ 10.93
0.7227 Γ 9.067 = 1.644 m
Answer: wave height = 1.64 m.
(iii)
The relevant pressures are:
π1 = 0
π2 = Οππ» = 1025 Γ 9.81 Γ 1.644 = 16530 Pa
π3 =Οππ»
cosh πβ =
1025 Γ 9.81 Γ 1.644
cosh(1.188) = 9221 Pa
This yields the equivalent forces and points of application shown in the diagram.
Region Force, πΉπ₯ or πΉπ§ Moment arm
1 πΉπ₯1 =1
2π2 Γ π» = 13590 N/m π§1 = β +
1
3π» = 12.55 m
2 πΉπ₯2 = π3 Γ β = 110700 N/m π§2 =1
2β = 6 m
3 πΉπ₯3 =1
2(π2 β π3) Γ β = 43850 N/m π§3 =
2
3β = 8 m
The sum of all clockwise moments about the heel:
πΉπ₯1π§1 + πΉπ₯2π§2 + πΉπ₯3π§3 = 1186000 N m/m
Answer: overturning moment = 1186 kN m per metre of breakwater.
(b) New waves have revised properties:
π = 13 s
p1
3p
1
2
3
2p
heel
Fx
My
Fz
Hydraulics 3 Answers (Waves Examples) -43 Dr David Apsley
Ο =2Ο
π = 0.4833 rad sβ1
β = 12 m β = 100 m
Ο2β
π 0.2857 2.381
Iteration: πβ =1
2(πβ +
0.2857
tanh πβ) πβ =
2.381
tanh πβ
πβ 0.5613 2.419
π 0.04678 mβ1 0.02419 mβ1
π =Ο
π 10.33 m sβ1 19.98 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ] 0.9086 0.5383
ΞΈ ? 30ΒΊ
π» ? 1.3 m
To estimate travel time of wave groups consider the group velocity.
The previous waves had group velocity
ππ = ππ = 5.465 m sβ1
These longer-period waves have group velocity
ππ = ππ = 10.76 m sβ1
The latter have a larger group velocity and hence have a shorter travel time.
(ii) Refraction:
(π sin ΞΈ)12 m = (π sin ΞΈ)100 m
0.04678 sin ΞΈ = 0.02419 sin 30Β°
Hence,
ΞΈ = 14.98Β°
From the shoaling equation:
(π»2ππ cos ΞΈ)12 m = (π»2ππ cos ΞΈ)100 m
Hence:
Hydraulics 3 Answers (Waves Examples) -44 Dr David Apsley
π»12m = π»100mβ(ππ cos ΞΈ)100 m
(ππ cos ΞΈ)12 m= 1.3 Γ β
0.5383 Γ 19.98 Γ cos 30Β°
0.9086 Γ 10.33 Γ cos 14.98Β°= 1.318 m
The maximum freeboard for a combination of the two waves, allowing for reflection, is twice
the sum of the amplitudes, i.e. the sum of the heights.
Hence, the freeboard must be
1.644 + 1.318 = 2.962 m
or height from the bed:
12 + 2.962 = 14.96 m
Answer: caisson height 14.96 m (from bed level), or 2.96 m (freeboard).
Hydraulics 3 Answers (Waves Examples) -45 Dr David Apsley
Q21.
(a)
(b) Given:
β = 8 m
π = 5 s
Then,
Ο =2Ο
π = 1.257 rad sβ1
Dispersion relation:
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
1.289 = πβ tanh πβ
Iterate as
πβ =1.289
tanh πβ
to get
πβ = 1.442
π =1.442
β = 0.1803 mβ1
π
10< πβ < π, so this is an intermediate-depth wave.
Answer: πβ = 1.44; intermediate-depth wave.
(c)
(i) The velocity potential is
π =π΄π
Ο
cosh π(β + π§)
cosh πβsin (ππ₯ β Οπ‘)
pSWL
bed
shallow water
z
deep water
Hydraulics 3 Answers (Waves Examples) -46 Dr David Apsley
Then
π’ β‘ππ
ππ₯ =
π΄ππ
Ο
cosh π(β + π§)
cosh πβcos (ππ₯ β Οπ‘)
At the SWL (π§ = 0) the amplitude of the horizontal velocity is
π΄ππ
Ο
(ii) At the sensor height (π§ = β6 m) the amplitude of horizontal velocity is
π΄ππ
Ο
cosh π(β + π§)
cosh πβ
and hence, from the given data,
π΄ Γ 9.81 Γ 0.1803
1.257Γ
cosh[0.1803 Γ 2]
cosh 1.442 = 0.34
Hence,
π΄ = 0.5062
π» = 2π΄ = 1.012 m
Answer: wave height = 1.01 m.
(iii) From
π’ =π΄ππ
Ο
cosh π(β + π§)
cosh πβcos (ππ₯ β Οπ‘)
we have
ππ₯ =ππ’
ππ‘+ non β linear terms = π΄ππ
cosh π(β + π§)
cosh πβsin (ππ₯ β Οπ‘)
The maximum horizontal acceleration at the bed (π§ = ββ) is
π΄ππ
cosh πβ=
0.5062 Γ 9.81 Γ 0.1803
cosh 1.442 = 0.4010 m sβ2
Answer: maximum acceleration at the bed = 0.401 m sβ2.
Hydraulics 3 Answers (Waves Examples) -47 Dr David Apsley
Q22.
(a) βSignificant wave heightβ is the average of the highest 1/3 of waves. For irregular waves,
π»π = 4βΞ·2Μ Μ Μ
whilst for a regular wave
Ξ·2Μ Μ Μ =1
2π΄2
Hence
π»π = 4βπ΄2
2 = 2β2π΄ = 2β2 Γ 0.8 = 2.263 m
Answer: significant wave height = 2.26 m.
(b) βShoalingβ is the change in wave height as a wave moves into shallower water. Linear
theory can be applied provided the height-to-depth and height-to-wavelength ratios remain
βsmallβ and, in practice, up to the point of breaking.
(c) (i) No current.
Given:
β = 35 m
π΄ = 0.8 m (π» = 1.6 m)
π = 9 s
Then,
Ο =2Ο
π = 0.6981 rad sβ1
Dispersion relation:
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
1.739 = πβ tanh πβ
Iterate as
πβ =1.739
tanh πβ
to get
πβ = 1.831
Hydraulics 3 Answers (Waves Examples) -48 Dr David Apsley
π =1.831
β = 0.05231 mβ1
πΏ =2Ο
π = 120.1 m
π =Ο
π (or
πΏ
π ) = 13.35 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ] = 0.5941
π =1
8Οππ»2(ππ) =
1
8Γ 1025 Γ 9.81 Γ 1.62 Γ (0.5941 Γ 13.35) = 25520 W mβ1
Answer: wavelength = 120 m; power = 25.5 kW per metre of crest.
(ii) With current β0.5 m sβ1.
Οπ = 0.6981 rad sβ1
Dispersion relation:
(Οπ β ππ)2 = Οπ2 = ππ tanh πβ
Rearrange as
π =(Οπ β ππ)2
π tanh πβ
or here:
π =(0.6981 + 0.5π)2
9.81 tanh 35π
to get
π = 0.05592 mβ1
πβ = 1.957
πΏ =2Ο
π = 112.4 m
Οπ = Οπ β ππ = 0.6981 + 0.05592 Γ 0.5 = 0.7261 rad sβ1
ππ =ππ
π = 12.98 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ] = 0.5782
π =1
8Οππ»2(πππ) =
1
8Γ 1025 Γ 9.81 Γ 1.62 Γ (0.5782 Γ 12.98) = 24150 W mβ1
(Since power comes from integrating the product of wave fluctuating properties ππ’ to get rate
of working, the group velocity here is that in a frame moving with the current; i.e. the βrβ suffix).
Answer: wavelength = 112 m; power = 24.2 kW per metre of crest.
Hydraulics 3 Answers (Waves Examples) -49 Dr David Apsley
(d) For the heading change (refraction), find the new wavenumber and then apply Snellβs Law.
For the wavenumber at β = 5 m depth, Ο as in part c(i), but
Ο2β
π= 0.2484 = πβ tanh πβ
This is small, so iterate as
πβ =1
2(πβ +
0.2484
tanh πβ)
to get
πβ = 0.5200
π =0.5200
β = 0.104 mβ1
From Snellβs Law:
π sin ΞΈ = π1 sin ΞΈ1
Hence:
sin ΞΈ =π1
πsin ΞΈ1 =
0.05231
0.104sin 25Β° = 0.2126
ΞΈ = 12.27Β°
For the shoaling, the shoreward component of power is constant; i.e.
π cos ΞΈ = π1 cos ΞΈ1
Hence,
π = π1
cos ΞΈ1
cos ΞΈ = 25520
cos 25Β°
cos 12.27Β° = 23020 W mβ1
Answer: heading = 12.3Β°; power = 23.0 kW per metre of crest.
Hydraulics 3 Answers (Waves Examples) -50 Dr David Apsley
Q23.
(a) Given
β = 24 m
π = 6 s
Then,
Ο =2Ο
π = 1.047 rad sβ1
Dispersion relation:
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
2.682 = πβ tanh πβ
Iterate as
πβ =2.682
tanh πβ
to get
πβ = 2.706
π =2.706
β = 0.1128 mβ1
πΏ =2Ο
π = 55.70 m
π =Ο
π (or
πΏ
π) = 9.282 m sβ1
Answer: wavelength 55.7 m, phase speed 9.28 m sβ1.
(b)
(i) From the velocity potential
π =π΄π
Ο
cosh π(β + π§)
cosh πβsin (ππ₯ β Οπ‘)
then the dynamic pressure is
π = βΟππ
ππ‘ = Οππ΄
cosh π(β + π§)
cosh πβcos (ππ₯ β Οπ‘)
(ii) At any particular depth the amplitude of the pressure variation is
Οππ΄cosh π(β + π§)
cosh πβ
Hydraulics 3 Answers (Waves Examples) -51 Dr David Apsley
and hence, from the given conditions at π§ = β22.5 m,
1025 Γ 9.81π΄ Γcosh(0.1128 Γ 1.5)
cosh 2.706= 1220
Hence,
π΄ = 0.8993 m
π» = 2π΄ = 1.799 m
Answer: wave height = 1.80 m.
(iii) From the velocity potential
π =π΄π
Ο
cosh π(β + π§)
cosh πβsin (ππ₯ β Οπ‘)
the horizontal velocity is
π’ =ππ
ππ₯ =
π΄ππ
Ο
cosh π(β + π§)
cosh πβcos (ππ₯ β Οπ‘)
The maximum horizontal velocity occurs at the surface (π§ = 0), and is
π’max =π΄ππ
Ο =
0.8993 Γ 9.81 Γ 0.1128
1.047 = 0.9505 m sβ1
Answer: maximum horizontal velocity = 0.950 m sβ1.
(c) Kinematic boundary condition β the boundary is a material surface (i.e. always composed
of the same particles), or, equivalently, there is no flow through the boundary.
Dynamic boundary condition β stress (here, pressure) is continuous across the boundary.
(d) Determine the wave period that would result in the same absolute period if this were
recorded in the presence of a uniform flow of 0.6 m sβ1 in the wave direction.
With current 0.6 m sβ1.
Οπ = 1.047 rad sβ1
Dispersion relation:
(Οπ β ππ)2 = Οπ2 = ππ tanh πβ
Rearrange as
π =(Οπ β ππ)2
π tanh πβ
or here:
Hydraulics 3 Answers (Waves Examples) -52 Dr David Apsley
π =(1.047 β 0.6π)2
9.81 tanh 24π
to get
π = 0.1008 mβ1
Οπ = Οπ β ππ = 1.047 β 0.1008 Γ 0.6 = 0.9865 rad sβ1
ππ =2Ο
Οπ = 6.369 s
Answer: wave (relative) period = 6.37 s.
Hydraulics 3 Answers (Waves Examples) -53 Dr David Apsley
Q24.
(a) Given:
β = 1.2 m
π = 1.5 s
Then,
Ο =2Ο
π = 4.189 rad sβ1
Dispersion relation:
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
2.147 = πβ tanh πβ
Iterate as
πβ =2.147
tanh πβ
to get
πβ = 2.200
These are intermediate-depth waves β particle trajectories sketched below.
(b) From
π€ =π΄ππ
Ο
sinh π(β + π§)
cosh πβsin (ππ₯ β Οπ‘)
we have
ππ§ =ππ€
ππ‘+ non β linear terms = βπ΄ππ
sinh π(β + π§)
cosh πβcos (ππ₯ β Οπ‘)
At the free surface (π§ = 0) the amplitude of the vertical acceleration is
ππ§,max = π΄ππ tanh πβ
Substituting the dispersion relation Ο2 = ππ tanh πβ,
ππ§,max = π΄Ο2
Here,
intermediate depth
Hydraulics 3 Answers (Waves Examples) -54 Dr David Apsley
π =2.200
β = 1.833 mβ1
Hence
π΄ =ππ§,max
Ο2 =
0.88
4.1892 = 0.05015 m
π» = 2π΄ = 0.1003
Also
πΏ =2Ο
π = 3.428 m
π =Ο
π = 2.285 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ] = 0.5540
π =1
8Οππ»2ππ =
1
8Γ 1025 Γ 9.81 Γ 0.10032 Γ 0.5540 Γ 2.285 = 16.01 W mβ1
Answer: wavelength = 3.43 m; wave height = 0.100 m; power = 16.0 W per metre crest.
(c)
With current β0.3 m sβ1.
Οπ = 4.189 rad sβ1
Dispersion relation:
(Οπ β ππ)2 = Οπ2 = ππ tanh πβ
Rearrange as
π =(Οπ β ππ)2
π tanh πβ
or here:
π =(4.189 + 0.3π)2
9.81 tanh(1.2π)
to get
π = 2.499 mβ1
πΏ =2Ο
π = 2.514 m
Answer: wavelength = 2.51 m.
Hydraulics 3 Answers (Waves Examples) -55 Dr David Apsley
(d) Spilling breakers occur for steep waves and/or mildly-sloped beaches. Waves gradually
dissipate energy as foam spills down the front faces.
Miche criterion in deep water (tanh πβ β 1):
π»π
πΏ= 0.14
where, in deep water, and with period π = 1 s:
πΏ =ππ2
2Ο = 1.561 m
Hence,
π»π = 0.14 Γ 1.561 = 0.2185 m
Answer: breaking height = 0.219 m.
Hydraulics 3 Answers (Waves Examples) -56 Dr David Apsley
Q25.
(a) Period is unchanged:
π = 5.5 s
Deep-water wavelength
πΏ0 =ππ2
2Ο = 47.23 m
Answer: period = 5.5 s; wavelength = 47.2 m.
(b)
Wave properties:
β = 4 m
Ο =2Ο
π = 1.142 rad sβ1
The dispersion relation is
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
0.5318 = πβ tanh πβ
Since the LHS is small this may be iterated as
πβ =1
2(πβ +
0.5318
tanh πβ)
to give
πβ = 0.8005
Hydrodynamic pressure:
Crest: π1 = 0
Still-water line: π2 = Οππ» = 1025 Γ 9.81 Γ 0.75 = 7541 Pa
Bed: π3 =Οππ»
cosh πβ =
π2
cosh 0.8005 = 5637 Pa
(c) Decompose the pressure forces on the breakwater as shown. Relevant dimensions are:
β = 4 m
π» = 0.75 m
π = 3 m
Hydraulics 3 Answers (Waves Examples) -57 Dr David Apsley
Region Force, πΉπ₯ or πΉπ§ Moment arm
1 πΉπ₯1 =1
2π2 Γ π» = 2828 N/m π§1 = β +
1
3π» = 4.25 m
2 πΉπ₯2 = π3 Γ β = 22548 N/m π§2 =1
2β = 2 m
3 πΉπ₯3 =1
2(π2 β π3) Γ β = 3808 N/m π§3 =
2
3β = 2.667 m
4 πΉπ§4 =1
2π3 Γ π = 8456 N/m π§4 =
2
3π = 2 m
The sum of all clockwise moments about the heel:
πΉπ₯1π§1 + πΉπ₯2π§2 + πΉπ₯3π§3 + πΉπ§4π₯4 = 84180 N m/m
Answer: overturning moment = 84.2 kN m per metre of breakwater.
p1
3p
3p
1
2
3
4
2p
heel
Fx
My
Fz
Hydraulics 3 Answers (Waves Examples) -58 Dr David Apsley
Q26.
(a)
β = 3 m
For deep water:
πβ > Ο
π >Ο
β = 0.1366 mβ1
πΏ <2π
0.1366 = 46.00 m
Also, from the dispersion relation:
Ο2 = ππ tanh πβ
Ο > β9.81 Γ 0.1366 Γ tanh Ο = 1.155 rad sβ1
π <2Ο
1.155 = 5.440 s
Answer: largest wavelength = 46 m; largest period = 5.44 s.
(b)
(i) Given
β = 23 m
π = 9 s
Then,
Ο =2Ο
π = 0.6981 rad sβ1
Dispersion relation:
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
1.143 = πβ tanh πβ
Iterate as
πβ =1.143
tanh πβ
to get
πβ = 1.319
π =1.319
β = 0.05735 mβ1
Hydraulics 3 Answers (Waves Examples) -59 Dr David Apsley
πΏ =2π
π = 109.6 m
Answer: wavenumber = 0.0573 mβ1; wavelength = 110 m.
(ii) From the velocity potential
π =π΄π
Ο
cosh π(β + π§)
cosh πβsin (ππ₯ β Οπ‘)
the horizontal velocity is
π’ =ππ
ππ₯ =
π΄ππ
Ο
cosh π(β + π§)
cosh πβcos (ππ₯ β Οπ‘)
Hence, the amplitude of the horizontal particle velocity at height π§ is
π’max =π΄ππ
Ο
cosh π(β + π§)
cosh πβ
From the given data, π’πππ₯ = 0.31 m sβ1 when π§ = β20 m,
π΄ =Οπ’max
ππ
cosh πβ
cosh π(β + π§)=
0.6981 Γ 0.31
9.81 Γ 0.05735Γ
cosh 1.319
cosh[0.05735 Γ 3]= 0.7594 m
π» = 2π΄ = 1.519 m
Answer: wave height = 1.52 m.
(iii) At π§ = 0,
π’max =π΄ππ
Ο =
0.7594 Γ 9.81 Γ 0.05735
0.6981 = 0.6120 m sβ1
The wave speed is
π =Ο
π =
0.6981
0.05735 = 12.17 m sβ1
Answer: particle velocity = 0.612 m sβ1, much less than the wave speed.
Hydraulics 3 Answers (Waves Examples) -60 Dr David Apsley
Q27.
Given
π»π = 1.5 m
π = 14 m sβ1
π‘ = 6 Γ 3600 = 21600 s
then
οΏ½ΜοΏ½ β‘ππ‘
π = 15140
If duration-limited then this would imply an effective fetch related to time π‘ by the non-
dimensional relation
οΏ½ΜοΏ½ = 68.8οΏ½ΜοΏ½eff2/3
or
οΏ½ΜοΏ½eff = (οΏ½ΜοΏ½
68.8)
3 2β
= 3264
and consequent wave height
ππ»π
π2β‘ οΏ½ΜοΏ½π = 0.0016οΏ½ΜοΏ½eff
1/2 = 0.09141
π»π = 0.09141 Γπ2
π = 1.826
But the actual wave height is less than this, so it is limited by fetch, not duration.
Hydraulics 3 Answers (Waves Examples) -61 Dr David Apsley
Q28.
(a) Kinematic boundary condition β the boundary is a material surface (i.e. always composed
of the same particles), or, equivalently, there is no flow through the boundary.
Dynamic boundary condition β stress (here, pressure) is continuous across the boundary.
(b)
(i) Given:
β = 20 m
π = 8 s
Then,
Ο =2Ο
π = 0.7854 rad sβ1
Dispersion relation:
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
1.258 = πβ tanh πβ
Iterate as
πβ =1.258
tanh πβ
to get
πβ = 1.416
π =1.416
β = 0.07080 mβ1
πΏ =2Ο
π = 88.75 m
π =Ο
π (or
πΏ
π) = 11.09 m sβ1
Answer: wavelength = 88.7 m; speed = 11.1 m sβ1.
(ii) From the velocity potential
π =π΄π
Ο
cosh π(β + π§)
cosh πβsin (ππ₯ β Οπ‘)
then the dynamic pressure is
π = βΟππ
ππ‘ = Οππ΄
cosh π(β + π§)
cosh πβcos (ππ₯ β Οπ‘)
Hydraulics 3 Answers (Waves Examples) -62 Dr David Apsley
From the given magnitude of the pressure fluctuations at the bed (π§ = ββ):
Οππ΄
cosh πβ= 6470
Hence
π΄ =6470 Γ cosh πβ
Οπ =
6470 Γ cosh 1.416
1025 Γ 9.81 = 1.404 m
π» = 2π΄ = 2.808 m
Answer: wave height = 2.81 m.
(c)
(i) For π < 5 s,
Ο =2Ο
π > 1.257 rad sβ1
At the limiting value on the RHS,
Ο2β
π= 3.221
Solving the dispersion relationship in the form
Ο2β
π= πβ tanh πβ
by iteration:
πβ =3.221
tanh πβ
produces
πβ = 3.220
This is a deep-water wave (πβ > Ο), hence negligible wave dynamic pressure is felt at the bed.
(ii) From the velocity potential
π =π΄π
Ο
cosh π(β + π§)
cosh πβsin (ππ₯ β Οπ‘)
the horizontal velocity is
π’ =ππ
ππ₯ =
π΄ππ
Ο
cosh π(β + π§)
cosh πβcos (ππ₯ β Οπ‘)
and the particle acceleration is
ππ₯ =ππ’
ππ‘+ non β linear terms = π΄ππ
cosh π(β + π§)
cosh πβsin (ππ₯ β Οπ‘)
and the amplitude of horizontal acceleration at π§ = 0 is
π΄ππ
Hydraulics 3 Answers (Waves Examples) -63 Dr David Apsley
Q29.
(a) The irrotationality condition (given in that yearβs exam paper) is
πππ¦
ππ₯β
πππ₯
ππ¦= 0
Wave behaviour is the same all the way along the coast; hence π ππ¦β = 0 for all variables. This
leaves
πππ¦
ππ₯= 0
But
ππ¦ = π sin ΞΈ
(see diagram). Hence
π sin ΞΈ = constant
or
(π sin ΞΈ)1 = (π sin ΞΈ)2
for two locations on a wave ray.
(b)
π = 7 s
Hence
Ο =2Ο
π = 0.8976 rad sβ1
The dispersion relation is
Ο2 = ππ tanh πβ
Ο2β
π= πβ tanh πβ
This may be iterated as either
πβ =Ο2β πβ
tanh πβ or πβ =
1
2(πβ +
Ο2β πβ
tanh πβ)
β = 5 m β = 28 m
Ο2β
π 0.4106 2.300
Iteration: πβ =1
2(πβ +
0.4106
tanh πβ) πβ =
2.300
tanh πβ
πβ 0.6881 2.343
π 0.1376 mβ1 0.08368 mβ1
k
kx
ky
x
y
coast
Hydraulics 3 Answers (Waves Examples) -64 Dr David Apsley
π =Ο
π 6.523 m sβ1 10.73 m sβ1
π =1
2[1 +
2πβ
sinh 2πβ] 0.8712 0.5432
ΞΈ ? 35ΒΊ
π» ? 1.2 m
Refraction:
(π sin ΞΈ)5 m = (π sin ΞΈ)28 m
0.1376 sin ΞΈ = 0.08368 sin 35Β°
Hence,
ΞΈ = 20.41Β°
From the shoaling equation:
(π»2ππ cos ΞΈ)5 m = (π»2ππ cos ΞΈ)28 m
Hence:
π»5 m = π»28 mβ(ππ cos ΞΈ)28 m
(ππ cos ΞΈ)5 m= 1.2 Γ β
0.5432 Γ 10.73 Γ cos 35Β°
0.8712 Γ 6.523 Γ cos 20.41Β°= 1.136 m
The power per metre of crest at depth 5 m is
π =1
8Οππ»2(ππ) =
1
8Γ 1025 Γ 9.81 Γ 1.1362 Γ (0.8712 Γ 6.523) = 9218 W mβ1
Answer: direction 20.4Β°; wave height 1.14 m; power = 9.22 kW mβ1.
(c) Given
π = 7 s
and at depth 5 m:
π» = 2.8 m
ΞΈ = 0Β°
Breaker Height
Breaker height index (on formula sheet):
π»π = 0.56π»0 (π»0
πΏ0)
β1 5β
Hydraulics 3 Answers (Waves Examples) -65 Dr David Apsley
First find deep-water conditions π»0 and πΏ0. By shoaling (with no refraction):
(π»2ππ)0 = (π»2ππ)5 m
In deep water,
π0 =1
2
π0 =ππ
2π = 10.93 m sβ1
whilst π and π at depth 5 m come from part (b). Hence,
π»02 Γ 0.5 Γ 10.93 = 2.82 Γ 0.8712 Γ 6.523
whence
π»0 = 2.855 m
Also in deep water,
πΏ0 =ππ2
2Ο = 76.50 m
Then from the formula for breaker height:
π»π = 0.56π»0 (π»0
πΏ0)
β1 5β
= 3.086 m
Breaking Depth
From the formula sheet the breaker depth index is
Ξ³π β‘ (π»
β)
π= π β π
π»π
ππ2
where, with a beach slope π = 1 40β = 0.025:
π = 43.8(1 β eβ19π) = 43.8(1 β eβ19Γ0.025) = 16.56
π =1.56
1 + eβ19.5π =
1.56
1 + eβ19.5Γ0.025 = 0.9664
Hence, from the breaker depth index:
3.086
βπ= 0.9664 β 16.56 Γ
3.086
9.81 Γ 72
giving depth of water at breaking:
βπ = 3.588 m
Answer: breaker height = 3.09 m; breaking depth = 3.59 m.