Homework #1 Coastal Engineering

Surawut Nimtim 5310501355 E 21-3

Let: wave Period (T) = 8s, Wave height (H) = 1.5m, Water depth (d) = 6 m

a) Wave length in 6 m of water Assuming that the waves in the deep water

𝐿𝐿0 =𝑔𝑔𝑇𝑇2

2𝜋𝜋=

9.81(𝑚𝑚 𝑠𝑠2⁄ ) × �8(𝑠𝑠)�2

2𝜋𝜋= 99.923 𝑚𝑚

Check the value 𝑑𝑑 L⁄ = 699.923

= 0.06 Unusable

Assuming that the waves in the Shallow water

𝐶𝐶 = �𝑔𝑔𝑑𝑑 = √9.81 × 6 = 7.6720

𝐿𝐿 = 𝐶𝐶𝑇𝑇 = 7.6720 × 8 = 61.37

𝑑𝑑 𝐿𝐿⁄ = 6 61.37⁄ = 0.09 Available

Wave length (L) = 61.37 m.

b) Wave number (k)

𝑘𝑘 = 2𝜋𝜋𝐿𝐿

= 2𝜋𝜋

61.37= 0.10

c) Velocity of propagation (c)

𝐶𝐶 = �𝑔𝑔𝐿𝐿2𝜋𝜋

tanh𝑘𝑘𝑑𝑑 = �9.81 × 61.372𝜋𝜋

tanh(0.10 × 6) = 7.17

d) Group velocity (CG) 𝐶𝐶𝐺𝐺 = 𝐶𝐶 = 7.17

e) Energy density (Density of sea water=1025)

𝐸𝐸 = 18𝜌𝜌𝑔𝑔𝐻𝐻2 =

18

× 1025 × 9.81 × 1.52 = 2828.039 𝑗𝑗/𝑚𝑚2

f) Wave power 𝑃𝑃 = 𝐸𝐸𝐶𝐶 = 2828.039 × 7.17 = 20277.040 𝑤𝑤/𝑚𝑚

g) Horizontal component of orbital velocity at bottom (at bottom z = 0)

𝑢𝑢 =𝜋𝜋𝐻𝐻𝑇𝑇

cosh𝑘𝑘(𝑧𝑧 + 𝑑𝑑)sinh𝑘𝑘𝑑𝑑

cosh(𝑘𝑘𝑘𝑘 − 𝜔𝜔𝑡𝑡)

=𝜋𝜋 × 1.5

8×

cosh 0.10(0 + 6)sinh(0.10 × 6) cosh(0.10 × 0 − 0.785 × 8) = 293.67

h) Amplitude of the orbital motion at bottom (at bottom z = 0)

𝐴𝐴 =𝐻𝐻2

cosh𝑘𝑘(𝑧𝑧 + 𝑑𝑑)sinh𝑘𝑘𝑑𝑑

= 1.52

cosh 0.10(0 + 6)sinh(0.10 × 6) = 1.3965 𝑚𝑚