EXAMPLE PROBLEM II$-3-1$

FIND:

Wave height $H$ and angle $$ at water depths of $200,100,90,80,70,60,50,40,30,20,10,16,14,12,10,8,$

$6,$ and $4m$ for deepwater wave angles of $0,15,$ and $45.$

GIVEN:

A wave $1m$ high and $15-sec$ period in $500m$ of water, with a plane, sloping beach.

SOLUTION:

Routine solutions for a plane beach can be obtained using the ACES wave transformation code, by direct

calculation, or graphically using Figure II$-3-6.$

Table II$-3-1$ provides the results obtained by directly using the ACES code. On a personal computer with a

$486-$level microprocessor, the results may be obtained in seconds.

For a wave with a depth of $10m$ and an initial wave angle of $45$ deg, wave height and angle are calculated as

follows:

Since the deepwater wave length of a $15-sec$ wave is

$L_0=1\mathrm{.}56T^2=1\mathrm{.}56(15{)}^2=351m$

and since $500m$ is greater than $\frac{L}{2},$ the given initial wave is a deepwater condition. The wave length of the wave

in $10m$ must be estimated from

$L=\frac{g{T}^2}{2}tanh(\frac{2d}{L})$

and is $144m($see Problem II$-1-1).$

The shoaling coefficient $K,$ can be estimated from

$K_s=(\frac{C_{g0}}{C_{gl}}{)}^{\frac{1}{2}}$

In deep water $C_{k0}$ for a $15-$sec wave is

$\frac{1}{2}{C}_0=\frac{1}{2}(1\mathrm{.}56T)=\frac{23\mathrm{.}4}{2}=11\mathrm{.}7\frac{m}{s}$

The group velocity is given by

$C_g=nC=\frac{1}{2}(1+\frac{4\frac{d}{L}}{sinh(4\frac{d}{L})})\frac{gT}{2}tanh(\frac{2d}{L})$

Substitution of $d=10m,L=144m,T=15sec,$ and $g=9\mathrm{.}8\frac{m}{se{c}^2}$ yields $9\mathrm{.}05\frac{m}{s}.$

$K_s=(\frac{11\mathrm{.}70}{9\mathrm{.}05}{)}^{\frac{1}{2}}=1\mathrm{.}14$

Solution for $K,$ involves

$K_r=(\frac{1-si{n}^2{}_0}{1-si{n}^2{}_1}{)}^{\frac{1}{4}}$

$($Continued$)$