36
Coastal Engineering: Theory and Practice
(a) For d = 10 m, T = 10 sec, Lo = 156, d/LQ = 0.0641, sinh kd = 0.74
ttmax( d'jT _ ___ 7F___ _ TV = £
H
“ sinh (^) ~ 0.74
„ umax(-d)T
0.1365 x 10
■'■H=----- 44
44
(b) T = 8, H = 2, d = ?
^max(-d)T
H
01365 x 8- = 0.546
2
we know,
^max(-cT)T
H
TV
sinh kd
0.546 =
7F
sinh kd
sinh kd = 5.75
Corresponding
« 0.3840
Lo
Lo = 1.56 x 82 = 99.84
d ~ 40 m
(c) H = l, d = 20m, T — ?
^max(-cT)T
H
^max(-d)
TV
sinh kd
9D50
0.5
= 0.135
umax.T
umax(—d> )'L
——— or -----------—
H
H
T = 6.6s
^max(—d) = 0.18m/s
which is somewhat larger than the threshold velocity, For T = 5,
umax(-d)T
- 0.25, itmax(_d) == 0.05 m/s
which is much less than the required threshold.
Hence, by trial and error T = 6.6 s.