answers (waves examples) autumn 2021
TRANSCRIPT
Hydraulics 3 Answers (Waves Examples) -1 Dr David Apsley
ANSWERS (WAVES EXAMPLES) AUTUMN 2021
Q1.
Given:
â = 20 m
ðŽ = 4 m (ð» = 8 m)
ð = 13 s
Then,
Ï =2Ï
ð = 0.4833 rad sâ1
Dispersion relation:
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
0.4762 = ðâ tanh ðâ
The LHS is small, so iterate as
ðâ =1
2(ðâ +
0.4762
tanh ðâ)
to get
ðâ = 0.7499
ð =0.7499
â = 0.03750 mâ1
ð¿ =2Ï
ð = 167.6 m
ð =Ï
ð (or
ð¿
ð) = 12.89 m sâ1
Answer: wavelength 168 m, phase speed 12.9 m sâ1.
(b) By formula:
ð¢ =ðððŽ
Ï
cosh ð(â + ð§)
cosh ðâcos (ðð¥ â Ïð¡)
ðð¥ =ðð¢
ðð¡+ non-linear terms = ðððŽ
cosh ð(â + ð§)
cosh ðâsin (ðð¥ â Ïð¡)
Hence
ðx,max = 1.137 cosh ð(â + ð§)
Then:
surface (ð§ = 0): ðx,max = 1.137 cosh ðâ = 1.472 m sâ2
mid-depth (ð§ = â â 2â ): ðx,max = 1.137 cosh(ðâ/2) = 1.218 m sâ2
Hydraulics 3 Answers (Waves Examples) -2 Dr David Apsley
bottom: (ð§ = ââ): ðx,max = 1.137 m sâ2
Answer: accelerations (1.47, 1.22, 1.14) m sâ2.
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
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
This may be iterated as either
ðâ =Ï2â ðâ
tanh ðâ or ðâ =
1
2(ðâ +
Ï2â ðâ
tanh ðâ)
Wave A Wave B
â 20 m 3 m
ð 10 s 12 s
Ï =2ð
ð 0.6283 rad sâ1 0.5236 rad sâ1
Ï2â
ð 0.8048 0.08384
Iteration: ðâ =1
2(ðâ +
0.8048
tanh ðâ) ðâ =
1
2(ðâ +
0.08384
tanh ðâ)
ðâ 1.036 0.2937
ð 0.0518 mâ1 0.0979 mâ1
ð =Ï
ð 12.13 m sâ1 5.348
Type: Intermediate (Ï
10< ðâ < Ï) Shallow (ðâ <
Ï
10)
Wave C (with current)
ðð = 6 ð
â = 24 m
ð = â0.8 m sâ1
Then
Ïð =2ð
ðð = 1.047 rad sâ1
Hydraulics 3 Answers (Waves Examples) -4 Dr David Apsley
With current, the dispersion relation is
(Ïð â ðð0)2 = Ïð2 = ðð tanh ðâ
Rearrange for iteration as
ð =(Ïð â ðð0)2
ð tanh ðâ
i.e.
ð =(1.047 + 0.8ð)2
9.81 tanh(24ð)
Solve:
ð = 0.1367 mâ1
ðâ = 3.2808
The speed relative to the fixed-position sensor is the absolute speed:
ðð =Ïð
ð =
1.047
0.1367 = 7.659 m sâ1
This is a deep-water wave, since ðâ > Ï.
Answer: A: intermediate, 12.1 m sâ1; B: shallow, 5.348 m sâ1; C: deep, 7.66 m sâ1.
Hydraulics 3 Answers (Waves Examples) -5 Dr David Apsley
Q3.
Given:
â = 15 m
ð = 0.1 Hz
ðrange = 9500 N mâ2
Then,
Ï =2Ï
ð = 2Ïð = 0.6283 rad sâ1
Dispersion relation:
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
0.6036 = ðâ tanh ðâ
The LHS is small, so iterate as
ðâ =1
2(ðâ +
0.6036
tanh ðâ)
to get
ðâ = 0.8642
ð =0.8642
â = 0.05761 mâ1
ð¿ =2Ï
ð = 109.1 m
By formula, the wave part of the pressure is
ð = ÏððŽcosh ð(â + ð§)
cosh ðâcos (ðð¥ â Ïð¡)
and hence the range (max â min) at the seabed sensor (ð§ = ââ) is
ðrange =2ÏððŽ
cosh ðâ =
Ïðð»
cosh ðâ
Hence,
ð» = ðrange Ãcosh ðâ
Ïð = 9500 Ã
cosh 0.8642
1025 Ã 9.81 = 1.32 m
Answer: height = 1.32 m.
Hydraulics 3 Answers (Waves Examples) -6 Dr David Apsley
Q4.
On the seabed (ð§ = ââ) the gauge pressure distribution is
ð = Ïðâ +ÏððŽ
cosh ðâcos(ðð¥ â Ïðð¡)
The average and the fluctuation of this will give the water depth and wave amplitude,
respectively:
ï¿œÌ ï¿œ = Ïðâ , Îð =ÏððŽ
cosh ðâ
i.e.
â =ï¿œÌ ï¿œ
Ïð , ðŽ =
Îð
Ïgcosh ðâ
Here,
ï¿œÌ ï¿œ =98.8 + 122.6
2Ã 1000 = 110400 Pa
Îð =122.6 â 98.8
2Ã 1000 = 11900 Pa
Hence,
â =110400
1025 Ã 9.81 = 10.98 m
For the amplitude, we require also ðâ, which must be determined, via the dispersion relation,
from the period. From, e.g., the difference between the peaks of 5 cycles,
ð =49.2 â 6.8
5 = 8.48 s
Ïð =2Ï
ð = 0.7409 rad sâ1
(a) No current.
The dispersion relation rearranges as
Ïð2â
ð= ðâ tanh ðâ
0.6144 = ðâ tanh ðâ
Rearrange for iteration. Since the LHS < 1 the most effective form is
ðâ =1
2(ðâ +
0.6144
tanh ðâ)
Iteration from, e.g., ðâ = 1, gives
ðâ = 0.8737
Hydraulics 3 Answers (Waves Examples) -7 Dr David Apsley
Then the amplitude is
ðŽ =Îð
Ïgcosh ðâ =
11900
1025 Ã 9.81Ã cosh 0.8737 = 1.665 m
Corresponding to a wave height
ð» = 2ðŽ = 3.33 m
Finally, for the wavelength,
ð =ðâ
â =
0.8737
10.98 = 0.07957 mâ1
ð¿ =2Ï
ð = 78.96 m
Answer: water depth = 11.0 m; wave height = 3.33 m; wavelength = 79.0 m.
(b) With current ð = 2 m sâ1
The dispersion relation is
(Ïð â ðð)2 = ðð tanh ðâ
Rearrange as
ð =(Ïð â ðð)2
ð tanh ðâ
Here,
ð =(0.7409 â 2ð)2
9.81 tanh(10.98ð)
This does not converge very easily. An alternative using under-relaxation is
ð =1
2[ð +
(0.7409 â 2ð)2
9.81 tanh(10.98ð)]
Iteration from, e.g., ð = 0.1 mâ1, gives
ð = 0.06362 mâ1
and, with â = 10.98 m),
ðâ = 0.6985
Then the amplitude is
ðŽ =Îð
Ïgcosh ðâ =
11900
1025 Ã 9.81Ã cosh 0.6985 = 1.484 m
corresponding to a wave height
ð» = 2ðŽ = 2.968 m
Hydraulics 3 Answers (Waves Examples) -8 Dr David Apsley
Finally, for the wavelength,
ð¿ =2Ï
ð = 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.
(ð»2ðð)deep = (ð»2ðð)transducer = (ð»2ðð)shallow
We can find ð and ð from the dispersion relation at all locations, and the height ð» from the
transducer pressure measurements.
Given:
ð = 12 s
Then
Ï =2Ï
ð = 0.5236 rad sâ1
Dispersion relation:
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
Solve for ðâ, and thence ð by iterating either
ðâ =Ï2â ðâ
tanh ðâ or (better here) ðâ =
1
2(ðâ +
Ï2â ðâ
tanh ðâ)
together with
ð =Ï
ð , ð =
1
2[1 +
2ðâ
sinh 2ðâ]
In shallow water (â = 8 m), Ï2â ð = 0.2236â , and hence:
ðâ = 0.4912 , ð = 0.06140 mâ1, ð = 8.528 m sâ1 , ð = 0.9278
At the transducer (â = 22 m), Ï2â ð = 0.6148â , and hence:
ðâ = 0.8740 , ð = 0.03973 mâ1, ð = 13.18 m sâ1 , ð = 0.8139
In deep water, tanh ðâ â 1 and so
Ï2 = ðð
whence
ð =Ï2
ð= 0.02795
The phase speed is
ð =Ï
ð =
ð
Ï = 18.74 m sâ1
Hydraulics 3 Answers (Waves Examples) -10 Dr David Apsley
and, in deep water,
ð =1
2
At the pressure transducer we have
ÏððŽcosh ð(â + ð§)
cosh ðâ= 10000
with Ï = 1025 kg mâ3 (seawater) and ð§ = â20 m in water of depth â = 22 m. Hence,
ðŽ =10000 à cosh 0.8740
1025 Ã 9.81 Ã cosh[0.03973 Ã 2] = 1.395 m
and hence
ð»transducer = 2ðŽ = 2.790 m
Then, from the shoaling equation:
ð»deep = ð»transducerâ(ðð)transducer
(ðð)deep = 2.790 Ã â
0.8139 Ã 13.18
0.5 Ã 18.74 = 2.985 m
ð»shallow = ð»transducerâ(ðð)transducer
(ðð)shallow = 2.790 Ã â
0.8139 Ã 13.18
0.9278 Ã 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:
â = 1 m
ð = 0.1 m
ð = 0.05 m
First deduce ðâ, and hence the wavenumber, from the ratio of the semi-axes of the particle
orbits:
dð
dð¡= ð¢ â
ðŽðð
Ï
cosh ð(â + ð§0)
cosh ðâcos(ðð¥0 â Ïð¡)
ð = ð¥0 âðŽðð
Ï2
cosh ð(â + ð§0)
cosh ðâsin(ðð¥0 â Ïð¡)
ð = ð¥0 â ðŽcosh ð(â + ð§0)
sinh ðâsin(ðð¥0 â Ïð¡)
dð
dð¡= ð€ â
ðŽðð
Ï
sinh ð(â + ð§0)
cosh ðâsin(ðð¥0 â Ïð¡)
ð = ð§0 +ðŽðð
Ï2
sinh ð(â + ð§0)
cosh ðâcos(ðð¥0 â Ïð¡)
ð = ð§0 + ðŽsinh ð(â + ð§0)
sinh ðâcos(ðð¥0 â Ïð¡)
Where, in both directions we have used the dispersion relation Ï2 = ðð tanh ðâ to simplify.
Hence,
ð = ðŽcosh(ðâ/2)
sinh ðâ = 0.1
ð = ðŽsinh(ðâ/2)
sinh ðâ = 0.05
Then
ð
ð= tanh(ðâ/2) = 0.5
ðâ = 2 tanhâ1(0.5) = 1.099
ð =1.099
1 = 1.099 mâ1
Then wave height can be deduced from, e.g., the expression for ð:
ðŽ = ðsinh ðâ
cosh(ðâ/2) = 0.1 Ã
sinh 1.099
cosh(1.099/2) = 0.1155
Hydraulics 3 Answers (Waves Examples) -12 Dr David Apsley
ð» = 2ðŽ = 0.2310 m
From the dispersion relation:
Ï2 = ðð tanh ðâ â Ï = 2.937 rad sâ1
Finally,
ð =2Ï
Ï = 2.139 s
ð¿ =2Ï
ð = 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
â = 20 m
ðŽ = 1 m (ð» = 2 m)
(a) Here,
ð = +1 m sâ1
ðð = 3 s
Ïð =2Ï
ðð = 2.094 rad sâ1
Dispersion relation:
(Ïð â ðð)2 = Ïð2 = ðð tanh ðâ
Iterate as
ð =(Ïð â ðð)2
ð tanh ðâ
i.e.
ð =(2.094 â ð)2
9.81 tanh 20ð
to get
ð = 0.3206
ð¿ =2Ï
ð = 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
ð¢ð =ðŽðð
Ïð
cosh ð(â + ð§)
cosh ðâcos(ðð¥ â Ïð¡)
we have
Ïð = Ïð â ðð = 2.094 â 0.3206 Ã 1 = 1.773 rad sâ1
As the wave is travelling in the same direction as the current:
|ð¢|max =ðŽðð
Ïð+ ð =
1 Ã 9.81 Ã 0.3206
1.773+ 1 = 2.774 m sâ1
Answer: wavelength 19.6 m; maximum particle speed 2.77 m sâ1.
(b) Now,
ð = â1 m sâ1
ðð = 7 s
Hydraulics 3 Answers (Waves Examples) -14 Dr David Apsley
Ïð =2Ï
ðð = 0.8976 rad sâ1
Dispersion relation is rearranged for iteration as above:
ð =(0.8976 + ð)2
9.81 tanh 20ð
to get
ð = 0.1056
ð¿ =2Ï
ð = 59.50 m
Wave-relative frequency:
Ïð = Ïð â ðð = 0.8976 + 0.1056 Ã 1 = 1.003 rad sâ1
This time, as the current is opposing, the maximum speed is the magnitude of the backward
velocity:
|ð¢|max =ðŽðð
Ïð+ |ð| =
1 Ã 9.81 Ã 0.1056
1.003+ 1 = 2.033 m sâ1
Answer: wavelength 59.5 m; maximum particle speed 2.03 m sâ1.
Hydraulics 3 Answers (Waves Examples) -15 Dr David Apsley
Q8.
As the waves move from deep to shallow water they refract (change angle Ξ) according to
ð0 sin Ξ0 = ð1 sin Ξ1
and undergo shoaling (change height, ð») according to
(ð»2ðð cos Ξ)0 = (ð»2ðð cos Ξ)1
where subscript 0 indicates deep and 1 indicates the measuring station. Period is unchanged.
Given:
ð = 5.5 s
then
Ï =2Ï
ð = 1.142 rad sâ1
In deep water,
ð0 =Ï2
ð = 0.1329
ð0 =1
2
ð0 =ðð
2Ï (or
Ï
ð) = 8.587 m sâ1
In the measured depth â = 6 m the wave height ð»1 = 0.8 m. The dispersion relation is
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
0.7977 = ðâ tanh ðâ
Iterate as
ðâ =0.7977
tanh ðâ or (better here) ðâ =
1
2(ðâ +
0.7977
tanh ðâ)
to get (adding subscript 1):
ð1â = 1.030
ð1 =1.030
â = 0.1717 mâ1
ð1 =Ï
ð1 = 6.651 m sâ1
ð1 =1
2[1 +
2ð1â
sinh 2ð1â] = 0.7669
Refraction
Hydraulics 3 Answers (Waves Examples) -16 Dr David Apsley
ð0 sin Ξ0 = ð1 sin Ξ1
Hence:
sin Ξ0 =ð1
ð0sin Ξ1 =
0.1717
0.1329sin 47° = 0.9449
Ξ0 = 70.89°
Shoaling
(ð»2ðð cos Ξ)0 = (ð»2ðð cos Ξ)1
Hence:
ð»0 = ð»1â(ðð cos Ξ)1
(ðð cos Ξ)0 = 0.8 à â
0.7669 à 6.651 à cos 47°
0.5 à 8.587 à cos 70.89° = 1.259 m
Answer: deep-water wave height = 1.26 m; angle = 70.9°.
Hydraulics 3 Answers (Waves Examples) -17 Dr David Apsley
Q9.
(a)
Deep water
ð¿0 = 300 m
ð»0 = 2 m
From the wavelength,
ð0 =2Ï
ð¿0 = 0.02094 mâ1
From the dispersion relation with tanh ðâ = 1:
Ï2 = ðð0
whence
Ï = âðð0 = 0.4532 rad sâ1
This stays the same as we move into shallower water.
ð0 =Ï
ð0 = 21.64 m sâ1
ð0 = 0.5
Depth â = 30 m
The dispersion relation is
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
0.6281 = ðâ tanh ðâ
Iterate as
ðâ =0.6281
tanh ðâ or (better here) ðâ =
1
2(ðâ +
0.6281
tanh ðâ)
to get (adding subscript 1):
ðâ = 0.8856
ð =0.8856
â = 0.02952 mâ1
ð¿ =2Ï
ð = 212.8 m
ð =Ï
ð = 15.35 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ] = 0.8103
Hydraulics 3 Answers (Waves Examples) -18 Dr David Apsley
ðð = ðð = 12.44 m sâ1
From the shoaling relationship
(ð»2ðð)0 = ð»2ðð
Hence,
ð» = ð»0â
(ðð)0
ðð = 2 Ã â
0.5 Ã 21.64
0.8103 Ã 15.35 = 1.865 m
Answer: wavelength = 213 m; height = 1.87 m; group velocity = 12.4 m sâ1.
(b)
ðž =1
2ÏððŽ2 =
1
8Ïðð»2 =
1
8Ã 1025 Ã 9.81 Ã 1.8652 = 4372 J mâ2
Answer: energy = 4370 J mâ2.
(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:
ð0 sin Ξ0 = ð sin Ξ
Hence:
sin Ξ =ð0
ðsin Ξ0 =
0.02094
0.02952sin 60° = 0.6143
Ξ0 = 37.90°
Shoaling:
(ð»2ðð cos Ξ)0 = ð»2ðð cos Ξ
Hence:
ð» = ð»0â
(ðð cos Ξ)0
ðð cos Ξ = 2 à â
0.5 à 21.64 à cos 60°
0.8103 à 15.35 à cos 37.90° = 1.485 m
Answer: wavelength and group velocity unchanged; height = 1.48 m.
Hydraulics 3 Answers (Waves Examples) -19 Dr David Apsley
Q10.
(a)
ð = 8 s
Ï =2Ï
ð = 0.7854 rad sâ1
Shoaling from 10 m depth to 3 m depth. Need wave properties at these two depths.
The dispersion relation is
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
This may be iterated as either
ðâ =Ï2â ðâ
tanh ðâ or ðâ =
1
2(ðâ +
Ï2â ðâ
tanh ðâ)
â = 3 m â = 10 m
Ï2â
ð 0.1886 0.6288
Iteration: ðâ =1
2(ðâ +
0.1886
tanh ðâ) ðâ =
1
2(ðâ +
0.6288
tanh ðâ)
ðâ 0.4484 0.8862
ð 0.1495 mâ1 0.08862 mâ1
ð =Ï
ð 5.254 m sâ1 8.863 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ] 0.9388 0.8101
ð» ? 1 m
From the shoaling equation:
(ð»2ðð)3 m = (ð»2ðð)10 m
Hence:
ð»3m = ð»10mâ(ðð)10 m
(ðð)3 m = 1 Ã â
0.8101 Ã 8.863
0.9388 Ã 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:
(ð»2ðð)ð = (ð»2ðð)3 m = 7.186
(in m-s units). Assuming that it breaks as a shallow-water wave, then
ðð = 1
ðð = âðâð
and we are given, from the breaker depth index:
ð»ð = 0.78âð
Substituting in the shoaling equation,
(0.78âð)2âðâð = 7.186
1.906âð5 2â
= 7.186
giving
âð = 1.700 m
ð»ð = 0.78âð = 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
ð(wave height > ð») = exp [â (ð»
ð»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 Ã 12 = 96 waves, so the question data says basically that
ð(wave height > 2 m) =1
96
Comparing with the Rayleigh distribution with ð» = 2 m:
exp [â (2
ð»rms)
2
] =1
96
exp [(2
ð»rms)
2
] = 96
4
ð»rms2
= ln 96
ð»rms = â4
ln 96 = 0.9361 m
(a)
ð(wave height > 3 m) = exp [â (3
ð»ððð )
2
] = 3.463 Ã 10â5 =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, ð»med, is such that
ð(wave height > ð»med) =1
2
Hence
exp [â (ð»med
ð»rms)
2
] =1
2
Hydraulics 3 Answers (Waves Examples) -22 Dr David Apsley
exp [(ð»med
ð»rms)
2
] = 2
(ð»med
ð»rms)
2
= ln 2
ð»med = ð»rmsâln 2 = 0.9361âln 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
ð»rms =ð»ð
1.416 =
2
1.416 = 1.412 m
(b)
ð»1 10â = 1.800 ð»rms = 1.800 à 1.412 = 2.542 m
(c)
ð(4 m < height < 5 m) = ð(height > 4) â ð(height > 5)
= exp [â (4
1.412)
2
] â exp [â (5
1.412)
2
]
= 3.235 Ã 10â4
or about once in every 3090 waves.
Answer: (a) ð»ððð = 1.41 m; (b) ð»1 10â = 2.54 m; (c) probability = 3.2410â4.
Hydraulics 3 Answers (Waves Examples) -24 Dr David Apsley
Q13.
The Bretschneider spectrum is
ð(ð) =5
16ð»ð
2ðð
4
ð5exp (â
5
4
ðð4
ð4)
Here we have
ð»ð = 2.5 m
ðð =1
ðð =
1
6 Hz
Amplitudes
For a single component the energy (per unit weight) is
1
2ð2 = ð(ð)Îð
Hence,
ðð = â2ð(ð)Îð
In this instance, Îð = 0.25ðð, so that, numerically,
ðð = â0.9766
(ðð ððâ )5 exp [â
1.25
(ðð ððâ )4]
These are computed (using Excel) in a column of the table below.
Wavenumbers
For deep-water waves,
Ï2 = ðð
or, since Ï = 2Ïð
ðð =4Ï2ðð
2
ð
Numerically here:
ðð = 0.1118 (ðð
ðð)
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
ð¢ð =ððððð
ðð
cosh ðð(â + ð§)
cosh ððâ
or, since the water is deep and hence cosh ð ~1
2eð:
ð¢ð =ððððð
2ðððexp ððð§
Numerically here, with ð§ = â5 m:
ð¢ð =9.368ðððð
ðð/ððexp(â5ðð)
These are computed (using Excel) in a column of the table below.
ðð/ðð ðð ðð ð¢ð
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
ηððð¥ = â ðð = 1.961 m
ð¢max = â ð¢ð = 0.9722 m sâ1
Answers: (a), (b) [ðð] and [ðð] given in the table; (c) ηmax = 1.96 m, ð¢max = 0.972 m sâ1.
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
ð = 13.5 m sâ1
ð¹ = 64000 m
ð¡ = 3 Ã 60 Ã 60 = 10800 s
then the relevant non-dimensional fetch is
ï¿œÌï¿œ â¡ðð¹
ð2 = 3445
The minimum non-dimensional time required for fetch-limited waves is found from
ï¿œÌï¿œmin â¡ (ðð¡
ð)
min = 68.8ï¿œÌï¿œ2 3â = 15693
but the actual non-dimensional time for which the wind has blown is
ï¿œÌï¿œ â¡ðð¡
ð = 7848
which is less. Hence, the waves are duration-limited and in the predictive curves we must use
an effective fetch ð¹eff given by
68.8ï¿œÌï¿œeff2/3
= 7848
whence
ï¿œÌï¿œeff = 1218
Then
ðð»ð
ð2â¡ ï¿œÌï¿œð = 0.0016ï¿œÌï¿œeff
1/2 = 0.05584
ððð
ðâ¡ ï¿œÌï¿œð = 0.286ï¿œÌï¿œeff
1/3 = 3.054
from which the corresponding significant wave height and peak period are
ð»ð = 1.037 m
ðð = 4.203 s
The significant wave period is estimated as
ðð = 0.945ðð = 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:
â = 6 m
ð = 6 s
Then,
Ï =2Ï
ð = 1.047 rad sâ1
Dispersion relation:
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
0.6705 = ðâ tanh ðâ
Iterate as
ðâ =0.6705
tanh ðâ or (better here) ðâ =
1
2(ðâ +
0.6705
tanh ðâ)
to get
ðâ = 0.9223
ð =0.9223
â = 0.1537 mâ1
Forces and Moments
The height of the fully reflected wave is 2ð», where ð» is the height of the incident wave. Thus
the maximum crest height above SWL is ð» = 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 ð1 = 0 at the crest to ð2 = Ïðð» at the SWL);
linear (from ð2 at the SWL to the wave value ð3 at the base);
linear underneath (from ð3 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 (ð3) and a triangular distribution with top ð2 â ð3.
Hydraulics 3 Answers (Waves Examples) -28 Dr David Apsley
The relevant pressures are:
ð1 = 0
ð2 = Ïðð» = 1025 Ã 9.81 Ã 1.5 = 15080 Pa
ð3 =Ïðð»
cosh ðâ =
1025 Ã 9.81 Ã 1.5
cosh(0.1537 Ã 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, ð¹ð¥ or ð¹ð§ Clockwise turning moment about heel
1 ð¹ð¥1 =1
2ð2 à ð» = 11310 N/m ð¹ð¥1 à (â +
1
3ð») = 73520 N m mâ
2 ð¹ð¥2 = ð3 à â = 62160 N/m ð¹ð¥2 Ã1
2â = 186480 N m mâ
3 ð¹ð¥3 =1
2(ð2 â ð3) à â = 14160 N/m ð¹ð¥3 Ã
2
3â = 56640 N m mâ
4 ð¹ð§4 =1
2ð3 à ð€ = 20720 N /m ð¹ð§4 Ã
2
3ð€ = 55250 N m mâ
The sum of the horizontal forces
ð¹ð¥1 + ð¹ð¥2 + ð¹ð¥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:
ð =1
ð =
1
0.1 = 10 s
The remaining quantities require solution of the dispersion relationship.
Ï =2Ï
ð = 0.6283 rad sâ1
Rearrange the dispersion relation Ï2 = ðð tanh ðâ to give
Ï2â
ð= ðâ tanh ðâ
Here, with â = 18 m,
0.7243 = ðâ tanh ðâ
Since the LHS < 1, rearrange for iteration as
ðâ =1
2[ðâ +
0.7243
tanh ðâ]
Iteration from, e.g., ðâ = 1 produces
ðâ = 0.9683
ð =0.9683
â = 0.05379 mâ1
The wavelength is then
ð¿ =2Ï
ð = 116.8 m
and the phase speed and ratio of group to phase velocities are
ð =Ï
ð (or
ð¿
ð) = 11.68 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ] = 0.7852
For the wave power,
power (per m) =1
8Ïðð»2ðð =
1
8Ã 1025 Ã 9.81 Ã 2.12 Ã 0.7852 Ã 11.68
= 50840 W mâ1
Answer: period = 10 s; wavelength = 117 m; phase speed = 11.68 m sâ1;
power (per metre of crest) = 50.8 kW mâ1.
(b) First need new values of ð, ð and ð in 6 m depth.
Ï2â
ð= ðâ tanh ðâ
Hydraulics 3 Answers (Waves Examples) -30 Dr David Apsley
Here, with â = 6 m,
0.2414 = ðâ tanh ðâ
Since the LHS < 1, rearrange for iteration as
ðâ =1
2[ðâ +
0.2414
tanh ðâ]
Iteration from, e.g., ðâ = 1 produces
ðâ = 0.5120
ð =0.5120
â = 0.08533 mâ1
ð =Ï
ð = 7.363 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ] = 0.9222
For direction, use Snellâs Law:
(ð sin Ξ)18 m = (ð sin Ξ)6 m
whence
(sin Ξ)6 m =(ð sin Ξ)18 m
ð6 m =
0.05379 à sin 25°
0.08533 = 0.2664
and hence the wave direction in 6 m depth is 15.45°.
For wave height, use the shoaling equation:
(ð»2ðð cos Ξ)6 m = (ð»2ðð cos Ξ)18 m
Hence,
ð»6 m = ð»18 mâ(ðð cos Ξ)18 m
(ðð cos Ξ)6 m = 2.1â
0.7852 à 11.68 à cos 25°
0.9222 à 7.363 à cos 15.45° = 2.367 m
Answer: wave height = 2.37 m; wave direction = 15.5°.
(c) The Miche breaking criterion gives
(ð»
ð¿)
ð= 0.14 tanh(ðâ)ð
If we are to assume shallow-water behaviour, then tanh ðâ ~ðâ and so
(ð»ð
2Ï)
ð= 0.14(ðâ)ð
whence
Hydraulics 3 Answers (Waves Examples) -31 Dr David Apsley
(ð»
â)
ð= 2Ï Ã 0.14 = 0.8796
or
ð»ð = 0.8796âð (*)
This is used to eliminate wave height on the LHS of the shoaling equation
(ð»2ðð cos Ξ)ð = (ð»2ðð cos Ξ)6 m
For refraction, Snellâs Law in the form (sin Ξ)/ð = ðððð ð¡ððð¡, together with the shallow-water
phase speed at breaking, gives
(sin Ξ
âðâ)
ð
= (sin Ξ
ð)
6 m =
sin 15.45°
7.363 = 0.03618
whence
sin Ξð = 0.1133ââð
cos Ξð = â1 â sin2 Ξð = â1 â 0.01284âð
With the shallow water approximations ð = 1, ð = âðâ, and the relation (*), the shoaling
equation becomes
(0.8796âð)2âðâðâ1 â 0.01284âð = 36.67
or
2.423âð5/2(1 â 0.01284âð)1/2 = 36.67
This rearranges for iteration:
âð = 2.965(1 â 0.01284âð)â1/5
to give
âð = 2.988 m
Answer: breaks in water of depth 2.99 m.
Hydraulics 3 Answers (Waves Examples) -32 Dr David Apsley
Q17.
(a)
ð = 35 m sâ1
ð¹ = 150000 m
From these,
ï¿œÌï¿œ =ðð¹
ð2 = 1201
From the JONSWAP curves,
ï¿œÌï¿œmin = 68.8ï¿œÌï¿œ2 3â = 7774
(i) For ð¡ = 4 hr = 14400 s,
ï¿œÌï¿œ =ðð¡
ð = 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 ï¿œÌï¿œ, we use
an effective dimensionless fetch ï¿œÌï¿œeff such that
68.8ï¿œÌï¿œeff2/3
= 4036
ï¿œÌï¿œeff = 449.3
Then
ï¿œÌï¿œð = 0.0016ï¿œÌï¿œeff1/2
= 0.03391
ï¿œÌï¿œð = 0.2857ï¿œÌï¿œeff1/3
= 2.188
The dimensional significant wave height and peak wave period are
ð»ð =ï¿œÌï¿œð ð2
ð = 4.234 m
ðð =ï¿œÌï¿œðð
ð = 7.806 s
Answer: significant wave height = 4.23 m; peak period = 7.81 s.
(ii) For ð¡ = 8 hr = 28800 s,
ï¿œÌï¿œ =ðð¡
ð = 8072
This is greater than 7774, so the wave statistics are fetch limited and we can use ï¿œÌï¿œ = 1201.
Then
ï¿œÌï¿œð = 0.0016ï¿œÌï¿œ1/2 = 0.05545
ï¿œÌï¿œð = 0.2857ï¿œÌï¿œ1/3 = 3.037
The dimensional significant height and peak wave period are
Hydraulics 3 Answers (Waves Examples) -33 Dr David Apsley
ð»ð =ï¿œÌï¿œð ð2
ð = 6.924 m
ðð =ï¿œÌï¿œðð
ð = 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 ð»rms is known: ð»rms = 1.8 m. For a single wave:
ð(ð» > 3 m) = exp[â(3/1.8)2] = 0.06218
This is once in every
1
ð= 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 ð» = 3 m at depth 10 m, but ð»rms is given in deep water. So
we either need to transform ð» to deep water or ð»rms to the 10 m depth. Weâll do the former.
Use the shoaling equation (for normal incidence):
(ð»2ðð)0 = (ð»2ðð)10 m
First find ð and ð in 10 m depth:
ð = 9 s
Ï =2Ï
ð = 0.6981 rad sâ1
Rearrange the dispersion relation Ï2 = ðð tanh ðâ to give
Ï2â
ð= ðâ tanh ðâ
Here, with â = 10 m,
0.4968 = ðâ tanh ðâ
Since the LHS < 1, rearrange for iteration as
ðâ =1
2[ðâ +
0.4968
tanh ðâ]
Iteration from, e.g., ðâ = 1 produces
ðâ = 0.7688
Hydraulics 3 Answers (Waves Examples) -34 Dr David Apsley
Then,
ð =0.7688
â = 0.07688 mâ1
ð =Ï
ð = 9.080 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ] = 0.8464
In deep water:
ð0 =ðð
2Ï = 14.05 m sâ1
ð0 =1
2
Hence, when the inshore wave height ð»10 = 3 m, the deep-water wave height, from the
shoaling equation, is
ð»0 = ð»10â(ðð)10
(ðð)0 = 3 Ã â
0.8464 Ã 9.080
0.5 Ã 14.05 = 3.138 m
(If we had transformed ð»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:
ð(ð»10 > 3 m) = ð(ð»0 > 3.138 m) = exp[â(3.138/1.8)2] = 0.04787
This is once in every
1
ð= 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 â change of direction as oblique waves move into shallower water;
(ii) Diffraction â spreading of waves into a region of shadow;
(iii) Shoaling â change of height as waves move into shallower water.
(b)
Ï =2Ï
ð =
2Ï
12 = 0.5236 rad sâ1
Deep water
Dispersion relation
Ï2 = ðð0
Hence:
ð0 =Ï2
ð = 0.02795 mâ1
ð¿0 =2Ï
ð0 = 224.8 m
ð0 =Ï
ð0 (or
ð¿0
ð) = 18.73 m sâ1
ð0 =1
2
Shallow water
Dispersion relation:
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
0.1397 = ðâ tanh ðâ
Iterate as either
ðâ =0.1397
tanh ðâ or (better here) ðâ =
1
2(ðâ +
0.1397
tanh ðâ)
to get
ðâ = 0.3827
ð =0.3827
â = 0.07654 mâ1
ð¿ =2Ï
ð = 82.09 m
Hydraulics 3 Answers (Waves Examples) -36 Dr David Apsley
ð =Ï
ð (or
ð¿
ð) = 6.841 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ ] = 0.9543
Refraction
ð sin Ξ = ð0 sin Ξ0
Hence,
sin Ξ =ð0 sin Ξ0
ð =
0.02795 à sin 20°
0.07654 = 0.1249
Ξ = 7.175°
Shoaling
(ð»2ðð cos Ξ)5 m = (ð»2ðð cos Ξ)0
Hence:
ð»5 m = ð»0â(ðð cos Ξ)0
(ðð cos Ξ)5 m = 3 à â
0.5 à 18.73 à cos 20°
0.9543 à 6.841 à cos 7.175° = 3.497 m
Answer: wavelength = 82.1 m; direction = 7.18°; height = 3.50 m.
(c) The breaker height index is
ð»ð
ð»0= 0.56 (
ð»0
ð¿0)
â1 5â
= 0.56 Ã (3
224.8)
â1 5â
= 1.328
Hence,
ð»ð = 1.328 ð»0 = 3.984 m
The breaker depth index is
γð â¡ (ð»
â)
ð= ð â ð
ð»ð
ðð2
where, with a beach slope ð = 1 20â = 0.05:
ð = 43.8(1 â eâ19ð) = 26.86
ð =1.56
1 + eâ19.5ð = 1.133
Hence,
γð = 1.133 â 26.86 Ã3.984
9.81Ã122 =1.058
giving:
Hydraulics 3 Answers (Waves Examples) -37 Dr David Apsley
âð =ð»ð
γð = 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) âNarrow-banded sea stateâ â narrow range of frequencies.
(ii) âSignificant wave heightâ â average height of the highest one-third of waves
or, from the wave spectrum,
ð»ð0 = 4( ð2Ì Ì Ì )1 2â
= 4ð0
(iii) âEnergy spectrumâ: distribution of wave energy with frequency; specifically,
ð(ð) dð
is the wave energy (divided by Ïð) in the small interval (ð, ð + dð).
(iv) âDuration-limitedâ â 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 ð = 10 s then the number of waves per hour is
ð =3600
ð = 360
(i) For a narrow-banded sea state a Rayleigh distribution is appropriate:
ð(height > ð») = eâ(ð» ð»rmsâ )2
Hence,
ð(height > 3.5) = eâ(3.5 2.5â )2 = 0.1409
For 360 waves, the expected number exceeding this is
360 Ã 0.1409 = 50.72
Answer: 51 waves.
(ii)
ð(height > 5) = eâ(5 2.5â )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 Ã 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:
Ï2 = ðð
or
(2Ï
ð)
2
= ð (2Ï
ð¿)
Hence,
ð¿ =ðð2
2Ï
With ð = 10 s,
ð¿ = 156.1 m
ð =ð¿
ð = 15.61 m sâ1
and, in deep water
ðð =1
2ð = 7.805 m sâ1
Answer: wavelength = 156.1 m; celerity = 15.6 m sâ1; group velocity = 7.81 m sâ1.
(c) Given
ð = 20 m sâ1
ð¹ = 105 m
ð¡ = 6 hours = 21600 s
Then
ï¿œÌï¿œ â¡ðð¹
ð2 = 2453
ï¿œÌï¿œmin â¡ðð¡min
ð = 68.8ð¹2 3â = 12513
But the non-dimensional time of the storm is
ï¿œÌï¿œ =ðð¡
ð = 10590
This is less than ï¿œÌï¿œmin, hence the sea state is duration-limited.
Hence, we must calculate an effective fetch by working back from ï¿œÌï¿œ:
ï¿œÌï¿œeff = (ï¿œÌï¿œ
68.8)
3 2â
= 1910
The non-dimensional wave height and peak period then follow from the JONSWAP formulae:
ï¿œÌï¿œð â¡ðð»ð
ð2 = 0.0016ï¿œÌï¿œeff
1 2â = 0.06993
Hydraulics 3 Answers (Waves Examples) -40 Dr David Apsley
ï¿œÌï¿œð â¡ððð
ð = 0.2857ï¿œÌï¿œeff
1 3â = 3.545
whence, extracting the dimensional variables:
ð»ð =ð2
ðï¿œÌï¿œð = 2.851 m
ðð =ð
ðï¿œÌï¿œð = 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)
ð = 7 s
Ï =2Ï
ð = 0.8976 rad sâ1
Shoaling from 100 m depth to 12 m depth. Need wave properties at these two depths.
The dispersion relation is
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
This may be iterated as either
ðâ =Ï2â ðâ
tanh ðâ or ðâ =
1
2(ðâ +
Ï2â ðâ
tanh ðâ)
â = 12 m â = 100 m
Ï2â
ð 0.9855 8.213
Iteration: ðâ =0.9855
tanh ðâ ðâ =
8.213
tanh ðâ
ðâ 1.188 8.213
ð 0.099 mâ1 0.08213 mâ1
ð =Ï
ð 9.067 m sâ1 10.93 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ] 0.7227 0.5000
Ξ 0º (necessarily) 0º
ð» ? 1.8 m
From the shoaling equation:
Hydraulics 3 Answers (Waves Examples) -42 Dr David Apsley
(ð»2ðð)12 m = (ð»2ðð)100 m
Hence:
ð»12 m = ð»100 mâ(ðð)100 m
(ðð)12 m = 1.8 Ã â
0.5 Ã 10.93
0.7227 Ã 9.067 = 1.644 m
Answer: wave height = 1.64 m.
(iii)
The relevant pressures are:
ð1 = 0
ð2 = Ïðð» = 1025 Ã 9.81 Ã 1.644 = 16530 Pa
ð3 =Ïðð»
cosh ðâ =
1025 Ã 9.81 Ã 1.644
cosh(1.188) = 9221 Pa
This yields the equivalent forces and points of application shown in the diagram.
Region Force, ð¹ð¥ or ð¹ð§ Moment arm
1 ð¹ð¥1 =1
2ð2 à ð» = 13590 N/m ð§1 = â +
1
3ð» = 12.55 m
2 ð¹ð¥2 = ð3 à â = 110700 N/m ð§2 =1
2â = 6 m
3 ð¹ð¥3 =1
2(ð2 â ð3) à â = 43850 N/m ð§3 =
2
3â = 8 m
The sum of all clockwise moments about the heel:
ð¹ð¥1ð§1 + ð¹ð¥2ð§2 + ð¹ð¥3ð§3 = 1186000 N m/m
Answer: overturning moment = 1186 kN m per metre of breakwater.
(b) New waves have revised properties:
ð = 13 s
p1
3p
1
2
3
2p
heel
Fx
My
Fz
Hydraulics 3 Answers (Waves Examples) -43 Dr David Apsley
Ï =2Ï
ð = 0.4833 rad sâ1
â = 12 m â = 100 m
Ï2â
ð 0.2857 2.381
Iteration: ðâ =1
2(ðâ +
0.2857
tanh ðâ) ðâ =
2.381
tanh ðâ
ðâ 0.5613 2.419
ð 0.04678 mâ1 0.02419 mâ1
ð =Ï
ð 10.33 m sâ1 19.98 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ] 0.9086 0.5383
Ξ ? 30º
ð» ? 1.3 m
To estimate travel time of wave groups consider the group velocity.
The previous waves had group velocity
ðð = ðð = 5.465 m sâ1
These longer-period waves have group velocity
ðð = ðð = 10.76 m sâ1
The latter have a larger group velocity and hence have a shorter travel time.
(ii) Refraction:
(ð sin Ξ)12 m = (ð sin Ξ)100 m
0.04678 sin Ξ = 0.02419 sin 30°
Hence,
Ξ = 14.98°
From the shoaling equation:
(ð»2ðð cos Ξ)12 m = (ð»2ðð cos Ξ)100 m
Hence:
Hydraulics 3 Answers (Waves Examples) -44 Dr David Apsley
ð»12m = ð»100mâ(ðð cos Ξ)100 m
(ðð cos Ξ)12 m= 1.3 à â
0.5383 à 19.98 à cos 30°
0.9086 à 10.33 à cos 14.98°= 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:
â = 8 m
ð = 5 s
Then,
Ï =2Ï
ð = 1.257 rad sâ1
Dispersion relation:
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
1.289 = ðâ tanh ðâ
Iterate as
ðâ =1.289
tanh ðâ
to get
ðâ = 1.442
ð =1.442
â = 0.1803 mâ1
ð
10< ðâ < ð, so this is an intermediate-depth wave.
Answer: ðâ = 1.44; intermediate-depth wave.
(c)
(i) The velocity potential is
ð =ðŽð
Ï
cosh ð(â + ð§)
cosh ðâsin (ðð¥ â Ïð¡)
pSWL
bed
shallow water
z
deep water
Hydraulics 3 Answers (Waves Examples) -46 Dr David Apsley
Then
ð¢ â¡ðð
ðð¥ =
ðŽðð
Ï
cosh ð(â + ð§)
cosh ðâcos (ðð¥ â Ïð¡)
At the SWL (ð§ = 0) the amplitude of the horizontal velocity is
ðŽðð
Ï
(ii) At the sensor height (ð§ = â6 m) the amplitude of horizontal velocity is
ðŽðð
Ï
cosh ð(â + ð§)
cosh ðâ
and hence, from the given data,
ðŽ à 9.81 à 0.1803
1.257Ã
cosh[0.1803 Ã 2]
cosh 1.442 = 0.34
Hence,
ðŽ = 0.5062
ð» = 2ðŽ = 1.012 m
Answer: wave height = 1.01 m.
(iii) From
ð¢ =ðŽðð
Ï
cosh ð(â + ð§)
cosh ðâcos (ðð¥ â Ïð¡)
we have
ðð¥ =ðð¢
ðð¡+ non â linear terms = ðŽðð
cosh ð(â + ð§)
cosh ðâsin (ðð¥ â Ïð¡)
The maximum horizontal acceleration at the bed (ð§ = ââ) is
ðŽðð
cosh ðâ=
0.5062 Ã 9.81 Ã 0.1803
cosh 1.442 = 0.4010 m sâ2
Answer: maximum acceleration at the bed = 0.401 m sâ2.
Hydraulics 3 Answers (Waves Examples) -47 Dr David Apsley
Q22.
(a) âSignificant wave heightâ is the average of the highest 1/3 of waves. For irregular waves,
ð»ð = 4âη2Ì Ì Ì
whilst for a regular wave
η2Ì Ì Ì =1
2ðŽ2
Hence
ð»ð = 4âðŽ2
2 = 2â2ðŽ = 2â2 à 0.8 = 2.263 m
Answer: significant wave height = 2.26 m.
(b) âShoalingâ 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
âsmallâ and, in practice, up to the point of breaking.
(c) (i) No current.
Given:
â = 35 m
ðŽ = 0.8 m (ð» = 1.6 m)
ð = 9 s
Then,
Ï =2Ï
ð = 0.6981 rad sâ1
Dispersion relation:
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
1.739 = ðâ tanh ðâ
Iterate as
ðâ =1.739
tanh ðâ
to get
ðâ = 1.831
Hydraulics 3 Answers (Waves Examples) -48 Dr David Apsley
ð =1.831
â = 0.05231 mâ1
ð¿ =2Ï
ð = 120.1 m
ð =Ï
ð (or
ð¿
ð ) = 13.35 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ] = 0.5941
ð =1
8Ïðð»2(ðð) =
1
8Ã 1025 Ã 9.81 Ã 1.62 Ã (0.5941 Ã 13.35) = 25520 W mâ1
Answer: wavelength = 120 m; power = 25.5 kW per metre of crest.
(ii) With current â0.5 m sâ1.
Ïð = 0.6981 rad sâ1
Dispersion relation:
(Ïð â ðð)2 = Ïð2 = ðð tanh ðâ
Rearrange as
ð =(Ïð â ðð)2
ð tanh ðâ
or here:
ð =(0.6981 + 0.5ð)2
9.81 tanh 35ð
to get
ð = 0.05592 mâ1
ðâ = 1.957
ð¿ =2Ï
ð = 112.4 m
Ïð = Ïð â ðð = 0.6981 + 0.05592 Ã 0.5 = 0.7261 rad sâ1
ðð =ðð
ð = 12.98 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ] = 0.5782
ð =1
8Ïðð»2(ððð) =
1
8Ã 1025 Ã 9.81 Ã 1.62 Ã (0.5782 Ã 12.98) = 24150 W mâ1
(Since power comes from integrating the product of wave fluctuating properties ðð¢ to get rate
of working, the group velocity here is that in a frame moving with the current; i.e. the ârâ 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 Snellâs Law.
For the wavenumber at â = 5 m depth, Ï as in part c(i), but
Ï2â
ð= 0.2484 = ðâ tanh ðâ
This is small, so iterate as
ðâ =1
2(ðâ +
0.2484
tanh ðâ)
to get
ðâ = 0.5200
ð =0.5200
â = 0.104 mâ1
From Snellâs Law:
ð sin Ξ = ð1 sin Ξ1
Hence:
sin Ξ =ð1
ðsin Ξ1 =
0.05231
0.104sin 25° = 0.2126
Ξ = 12.27°
For the shoaling, the shoreward component of power is constant; i.e.
ð cos Ξ = ð1 cos Ξ1
Hence,
ð = ð1
cos Ξ1
cos Ξ = 25520
cos 25°
cos 12.27° = 23020 W mâ1
Answer: heading = 12.3°; power = 23.0 kW per metre of crest.
Hydraulics 3 Answers (Waves Examples) -50 Dr David Apsley
Q23.
(a) Given
â = 24 m
ð = 6 s
Then,
Ï =2Ï
ð = 1.047 rad sâ1
Dispersion relation:
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
2.682 = ðâ tanh ðâ
Iterate as
ðâ =2.682
tanh ðâ
to get
ðâ = 2.706
ð =2.706
â = 0.1128 mâ1
ð¿ =2Ï
ð = 55.70 m
ð =Ï
ð (or
ð¿
ð) = 9.282 m sâ1
Answer: wavelength 55.7 m, phase speed 9.28 m sâ1.
(b)
(i) From the velocity potential
ð =ðŽð
Ï
cosh ð(â + ð§)
cosh ðâsin (ðð¥ â Ïð¡)
then the dynamic pressure is
ð = âÏðð
ðð¡ = ÏððŽ
cosh ð(â + ð§)
cosh ðâcos (ðð¥ â Ïð¡)
(ii) At any particular depth the amplitude of the pressure variation is
ÏððŽcosh ð(â + ð§)
cosh ðâ
Hydraulics 3 Answers (Waves Examples) -51 Dr David Apsley
and hence, from the given conditions at ð§ = â22.5 m,
1025 à 9.81ðŽ Ãcosh(0.1128 à 1.5)
cosh 2.706= 1220
Hence,
ðŽ = 0.8993 m
ð» = 2ðŽ = 1.799 m
Answer: wave height = 1.80 m.
(iii) From the velocity potential
ð =ðŽð
Ï
cosh ð(â + ð§)
cosh ðâsin (ðð¥ â Ïð¡)
the horizontal velocity is
ð¢ =ðð
ðð¥ =
ðŽðð
Ï
cosh ð(â + ð§)
cosh ðâcos (ðð¥ â Ïð¡)
The maximum horizontal velocity occurs at the surface (ð§ = 0), and is
ð¢max =ðŽðð
Ï =
0.8993 Ã 9.81 Ã 0.1128
1.047 = 0.9505 m sâ1
Answer: maximum horizontal velocity = 0.950 m sâ1.
(c) Kinematic boundary condition â 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 â 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 sâ1 in the wave direction.
With current 0.6 m sâ1.
Ïð = 1.047 rad sâ1
Dispersion relation:
(Ïð â ðð)2 = Ïð2 = ðð tanh ðâ
Rearrange as
ð =(Ïð â ðð)2
ð tanh ðâ
or here:
Hydraulics 3 Answers (Waves Examples) -52 Dr David Apsley
ð =(1.047 â 0.6ð)2
9.81 tanh 24ð
to get
ð = 0.1008 mâ1
Ïð = Ïð â ðð = 1.047 â 0.1008 Ã 0.6 = 0.9865 rad sâ1
ðð =2Ï
Ïð = 6.369 s
Answer: wave (relative) period = 6.37 s.
Hydraulics 3 Answers (Waves Examples) -53 Dr David Apsley
Q24.
(a) Given:
â = 1.2 m
ð = 1.5 s
Then,
Ï =2Ï
ð = 4.189 rad sâ1
Dispersion relation:
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
2.147 = ðâ tanh ðâ
Iterate as
ðâ =2.147
tanh ðâ
to get
ðâ = 2.200
These are intermediate-depth waves â particle trajectories sketched below.
(b) From
ð€ =ðŽðð
Ï
sinh ð(â + ð§)
cosh ðâsin (ðð¥ â Ïð¡)
we have
ðð§ =ðð€
ðð¡+ non â linear terms = âðŽðð
sinh ð(â + ð§)
cosh ðâcos (ðð¥ â Ïð¡)
At the free surface (ð§ = 0) the amplitude of the vertical acceleration is
ðð§,max = ðŽðð tanh ðâ
Substituting the dispersion relation Ï2 = ðð tanh ðâ,
ðð§,max = ðŽÏ2
Here,
intermediate depth
Hydraulics 3 Answers (Waves Examples) -54 Dr David Apsley
ð =2.200
â = 1.833 mâ1
Hence
ðŽ =ðð§,max
Ï2 =
0.88
4.1892 = 0.05015 m
ð» = 2ðŽ = 0.1003
Also
ð¿ =2Ï
ð = 3.428 m
ð =Ï
ð = 2.285 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ] = 0.5540
ð =1
8Ïðð»2ðð =
1
8Ã 1025 Ã 9.81 Ã 0.10032 Ã 0.5540 Ã 2.285 = 16.01 W mâ1
Answer: wavelength = 3.43 m; wave height = 0.100 m; power = 16.0 W per metre crest.
(c)
With current â0.3 m sâ1.
Ïð = 4.189 rad sâ1
Dispersion relation:
(Ïð â ðð)2 = Ïð2 = ðð tanh ðâ
Rearrange as
ð =(Ïð â ðð)2
ð tanh ðâ
or here:
ð =(4.189 + 0.3ð)2
9.81 tanh(1.2ð)
to get
ð = 2.499 mâ1
ð¿ =2Ï
ð = 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 ðâ â 1):
ð»ð
ð¿= 0.14
where, in deep water, and with period ð = 1 s:
ð¿ =ðð2
2Ï = 1.561 m
Hence,
ð»ð = 0.14 à 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:
ð = 5.5 s
Deep-water wavelength
ð¿0 =ðð2
2Ï = 47.23 m
Answer: period = 5.5 s; wavelength = 47.2 m.
(b)
Wave properties:
â = 4 m
Ï =2Ï
ð = 1.142 rad sâ1
The dispersion relation is
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
0.5318 = ðâ tanh ðâ
Since the LHS is small this may be iterated as
ðâ =1
2(ðâ +
0.5318
tanh ðâ)
to give
ðâ = 0.8005
Hydrodynamic pressure:
Crest: ð1 = 0
Still-water line: ð2 = Ïðð» = 1025 Ã 9.81 Ã 0.75 = 7541 Pa
Bed: ð3 =Ïðð»
cosh ðâ =
ð2
cosh 0.8005 = 5637 Pa
(c) Decompose the pressure forces on the breakwater as shown. Relevant dimensions are:
â = 4 m
ð» = 0.75 m
ð = 3 m
Hydraulics 3 Answers (Waves Examples) -57 Dr David Apsley
Region Force, ð¹ð¥ or ð¹ð§ Moment arm
1 ð¹ð¥1 =1
2ð2 à ð» = 2828 N/m ð§1 = â +
1
3ð» = 4.25 m
2 ð¹ð¥2 = ð3 à â = 22548 N/m ð§2 =1
2â = 2 m
3 ð¹ð¥3 =1
2(ð2 â ð3) à â = 3808 N/m ð§3 =
2
3â = 2.667 m
4 ð¹ð§4 =1
2ð3 à ð = 8456 N/m ð§4 =
2
3ð = 2 m
The sum of all clockwise moments about the heel:
ð¹ð¥1ð§1 + ð¹ð¥2ð§2 + ð¹ð¥3ð§3 + ð¹ð§4ð¥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)
â = 3 m
For deep water:
ðâ > Ï
ð >Ï
â = 0.1366 mâ1
ð¿ <2ð
0.1366 = 46.00 m
Also, from the dispersion relation:
Ï2 = ðð tanh ðâ
Ï > â9.81 Ã 0.1366 Ã tanh Ï = 1.155 rad sâ1
ð <2Ï
1.155 = 5.440 s
Answer: largest wavelength = 46 m; largest period = 5.44 s.
(b)
(i) Given
â = 23 m
ð = 9 s
Then,
Ï =2Ï
ð = 0.6981 rad sâ1
Dispersion relation:
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
1.143 = ðâ tanh ðâ
Iterate as
ðâ =1.143
tanh ðâ
to get
ðâ = 1.319
ð =1.319
â = 0.05735 mâ1
Hydraulics 3 Answers (Waves Examples) -59 Dr David Apsley
ð¿ =2ð
ð = 109.6 m
Answer: wavenumber = 0.0573 mâ1; wavelength = 110 m.
(ii) From the velocity potential
ð =ðŽð
Ï
cosh ð(â + ð§)
cosh ðâsin (ðð¥ â Ïð¡)
the horizontal velocity is
ð¢ =ðð
ðð¥ =
ðŽðð
Ï
cosh ð(â + ð§)
cosh ðâcos (ðð¥ â Ïð¡)
Hence, the amplitude of the horizontal particle velocity at height ð§ is
ð¢max =ðŽðð
Ï
cosh ð(â + ð§)
cosh ðâ
From the given data, ð¢ððð¥ = 0.31 m sâ1 when ð§ = â20 m,
ðŽ =Ïð¢max
ðð
cosh ðâ
cosh ð(â + ð§)=
0.6981 Ã 0.31
9.81 Ã 0.05735Ã
cosh 1.319
cosh[0.05735 Ã 3]= 0.7594 m
ð» = 2ðŽ = 1.519 m
Answer: wave height = 1.52 m.
(iii) At ð§ = 0,
ð¢max =ðŽðð
Ï =
0.7594 Ã 9.81 Ã 0.05735
0.6981 = 0.6120 m sâ1
The wave speed is
ð =Ï
ð =
0.6981
0.05735 = 12.17 m sâ1
Answer: particle velocity = 0.612 m sâ1, much less than the wave speed.
Hydraulics 3 Answers (Waves Examples) -60 Dr David Apsley
Q27.
Given
ð»ð = 1.5 m
ð = 14 m sâ1
ð¡ = 6 Ã 3600 = 21600 s
then
ï¿œÌï¿œ â¡ðð¡
ð = 15140
If duration-limited then this would imply an effective fetch related to time ð¡ by the non-
dimensional relation
ï¿œÌï¿œ = 68.8ï¿œÌï¿œeff2/3
or
ï¿œÌï¿œeff = (ï¿œÌï¿œ
68.8)
3 2â
= 3264
and consequent wave height
ðð»ð
ð2â¡ ï¿œÌï¿œð = 0.0016ï¿œÌï¿œeff
1/2 = 0.09141
ð»ð = 0.09141 Ãð2
ð = 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 â 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 â stress (here, pressure) is continuous across the boundary.
(b)
(i) Given:
â = 20 m
ð = 8 s
Then,
Ï =2Ï
ð = 0.7854 rad sâ1
Dispersion relation:
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
1.258 = ðâ tanh ðâ
Iterate as
ðâ =1.258
tanh ðâ
to get
ðâ = 1.416
ð =1.416
â = 0.07080 mâ1
ð¿ =2Ï
ð = 88.75 m
ð =Ï
ð (or
ð¿
ð) = 11.09 m sâ1
Answer: wavelength = 88.7 m; speed = 11.1 m sâ1.
(ii) From the velocity potential
ð =ðŽð
Ï
cosh ð(â + ð§)
cosh ðâsin (ðð¥ â Ïð¡)
then the dynamic pressure is
ð = âÏðð
ðð¡ = ÏððŽ
cosh ð(â + ð§)
cosh ðâcos (ðð¥ â Ïð¡)
Hydraulics 3 Answers (Waves Examples) -62 Dr David Apsley
From the given magnitude of the pressure fluctuations at the bed (ð§ = ââ):
ÏððŽ
cosh ðâ= 6470
Hence
ðŽ =6470 à cosh ðâ
Ïð =
6470 Ã cosh 1.416
1025 Ã 9.81 = 1.404 m
ð» = 2ðŽ = 2.808 m
Answer: wave height = 2.81 m.
(c)
(i) For ð < 5 s,
Ï =2Ï
ð > 1.257 rad sâ1
At the limiting value on the RHS,
Ï2â
ð= 3.221
Solving the dispersion relationship in the form
Ï2â
ð= ðâ tanh ðâ
by iteration:
ðâ =3.221
tanh ðâ
produces
ðâ = 3.220
This is a deep-water wave (ðâ > Ï), hence negligible wave dynamic pressure is felt at the bed.
(ii) From the velocity potential
ð =ðŽð
Ï
cosh ð(â + ð§)
cosh ðâsin (ðð¥ â Ïð¡)
the horizontal velocity is
ð¢ =ðð
ðð¥ =
ðŽðð
Ï
cosh ð(â + ð§)
cosh ðâcos (ðð¥ â Ïð¡)
and the particle acceleration is
ðð¥ =ðð¢
ðð¡+ non â linear terms = ðŽðð
cosh ð(â + ð§)
cosh ðâsin (ðð¥ â Ïð¡)
and the amplitude of horizontal acceleration at ð§ = 0 is
ðŽðð
Hydraulics 3 Answers (Waves Examples) -63 Dr David Apsley
Q29.
(a) The irrotationality condition (given in that yearâs exam paper) is
ðððŠ
ðð¥â
ððð¥
ððŠ= 0
Wave behaviour is the same all the way along the coast; hence ð ððŠâ = 0 for all variables. This
leaves
ðððŠ
ðð¥= 0
But
ððŠ = ð sin Ξ
(see diagram). Hence
ð sin Ξ = constant
or
(ð sin Ξ)1 = (ð sin Ξ)2
for two locations on a wave ray.
(b)
ð = 7 s
Hence
Ï =2Ï
ð = 0.8976 rad sâ1
The dispersion relation is
Ï2 = ðð tanh ðâ
Ï2â
ð= ðâ tanh ðâ
This may be iterated as either
ðâ =Ï2â ðâ
tanh ðâ or ðâ =
1
2(ðâ +
Ï2â ðâ
tanh ðâ)
â = 5 m â = 28 m
Ï2â
ð 0.4106 2.300
Iteration: ðâ =1
2(ðâ +
0.4106
tanh ðâ) ðâ =
2.300
tanh ðâ
ðâ 0.6881 2.343
ð 0.1376 mâ1 0.08368 mâ1
k
kx
ky
x
y
coast
Hydraulics 3 Answers (Waves Examples) -64 Dr David Apsley
ð =Ï
ð 6.523 m sâ1 10.73 m sâ1
ð =1
2[1 +
2ðâ
sinh 2ðâ] 0.8712 0.5432
Ξ ? 35º
ð» ? 1.2 m
Refraction:
(ð sin Ξ)5 m = (ð sin Ξ)28 m
0.1376 sin Ξ = 0.08368 sin 35°
Hence,
Ξ = 20.41°
From the shoaling equation:
(ð»2ðð cos Ξ)5 m = (ð»2ðð cos Ξ)28 m
Hence:
ð»5 m = ð»28 mâ(ðð cos Ξ)28 m
(ðð cos Ξ)5 m= 1.2 à â
0.5432 à 10.73 à cos 35°
0.8712 à 6.523 à cos 20.41°= 1.136 m
The power per metre of crest at depth 5 m is
ð =1
8Ïðð»2(ðð) =
1
8Ã 1025 Ã 9.81 Ã 1.1362 Ã (0.8712 Ã 6.523) = 9218 W mâ1
Answer: direction 20.4°; wave height 1.14 m; power = 9.22 kW mâ1.
(c) Given
ð = 7 s
and at depth 5 m:
ð» = 2.8 m
Ξ = 0°
Breaker Height
Breaker height index (on formula sheet):
ð»ð = 0.56ð»0 (ð»0
ð¿0)
â1 5â
Hydraulics 3 Answers (Waves Examples) -65 Dr David Apsley
First find deep-water conditions ð»0 and ð¿0. By shoaling (with no refraction):
(ð»2ðð)0 = (ð»2ðð)5 m
In deep water,
ð0 =1
2
ð0 =ðð
2ð = 10.93 m sâ1
whilst ð and ð at depth 5 m come from part (b). Hence,
ð»02 à 0.5 à 10.93 = 2.82 à 0.8712 à 6.523
whence
ð»0 = 2.855 m
Also in deep water,
ð¿0 =ðð2
2Ï = 76.50 m
Then from the formula for breaker height:
ð»ð = 0.56ð»0 (ð»0
ð¿0)
â1 5â
= 3.086 m
Breaking Depth
From the formula sheet the breaker depth index is
γð â¡ (ð»
â)
ð= ð â ð
ð»ð
ðð2
where, with a beach slope ð = 1 40â = 0.025:
ð = 43.8(1 â eâ19ð) = 43.8(1 â eâ19Ã0.025) = 16.56
ð =1.56
1 + eâ19.5ð =
1.56
1 + eâ19.5Ã0.025 = 0.9664
Hence, from the breaker depth index:
3.086
âð= 0.9664 â 16.56 Ã
3.086
9.81 Ã 72
giving depth of water at breaking:
âð = 3.588 m
Answer: breaker height = 3.09 m; breaking depth = 3.59 m.