Question 15.2: At a depth of 20 m, waves of height 1.31 m and period 7 s ar......

At the place of observation, H = 1.31 m, d = 20 m and T = 7 s. From Fig. 14.4, L = 70  m, and from Fig. 14.3, c = 10.2 {m}{s}^{-1} .

L_0=\frac{g T^2}{2 \pi}=76.5 \, m .

For d / L_0=20 / 76.5=0.26, \quad H / H_0=0.95 (Fig. 14.8) and so  \textstyle H_{\mathrm{0}} = 1.31/0.95  = 1.38 m.

WAVE BREAKING

\frac{H_{ b }}{1.38}=0.38\left(\frac{1.38}{76.5}\right)^{-1 / 3}                   (equation (14.40) \left(H_{\mathrm{b}} / H_0\right)=0.38\left(H_0 / L_0\right)^{-1 / 3} )

RUN-UP

1. First trial. Assume a run-up of 2 m above SWL. The horizontal distance from the wave breaking section to the run-up, s = 10.0 + 5 + 6 + 5.6 = 26.6 m. The vertical distance is 2.0 + 2.56 = 4.56 m. The average slope is 4.56/26.6 = 0.17.

\frac{R_{ u }}{H_0}=\frac{R_{ u }}{1.38}=1.016 \times 0.17 \times\left(\frac{1.38}{76.5}\right)^{-0.5}                   (equation (15.18) R_{\mathrm{u}} / H_0=1.016 \tan \beta\left(H_0 / L_0\right)^{-0.5})
= 1.28
R_{ u }=1.28 \times 1.38=1.8\, m

2. Second trial. Assume that R_{\mathrm{u}}= 1.8 m. The average slope is

(1.8+2.56) /(9.0+5+6+5.6)=4.36 / 25.6 \approx 0.17 .

\frac{R_{ u }}{1.38}=1.016 \times 0.17(1.38 / 76.5)^{-1 / 2}=1.78 \,m                         (equation (15.18))