EXAMPLE PROBLEM III-2-9

EM 1110-2-1100 (Part III)

30 Apr 02

FIND:

(a) The time it will take for the structure to fill to half its length, if its length is 600 m.

(b) The distance seaward the shoreline will extend at the structure (y at x = 0) after 1 week (t =

604,800 sec) of continuous wave activity, and;

(c) the distance seaward the shoreline will extend at 500 m updrift from the structure (y at

x = 500 m) after 1 week (t = 604,800 sec) of continuous wave climate.

GIVEN:

A long terminal groin extending a few surf zone widths at the end of a project reach adjacent to an

inlet has been built to prevent sand from being lost from the beach into the inlet shoal system. Waves

approach the inlet from the updrift side with a breaking angle αb = 5 deg. Breaking wave height Hb =

2m, and wave period T = 10 sec. The structure is initially expected to block all sediment (i.e., no

bypassing). Sediment density to water density ratio ρs/ρ = 2.65, and porosity n = 0.4. Assume dB + dc =

6 m, K = 0.77, and K = 1.

SOLUTION:

Equation 2-26 gives

0.77 (2)2 9.81 @ 2

m2

1

@

' 0.287

ε'

sec

8

2.65&1

(1&0.4)

6

(a) Express Equation 2-28 for the shoreline location y at the structure x = 0:

y

1

0.564

'

2 ε t tan (αb)

π

and solve for y = 1/2 the length of the structure = 300 m, where αb = 5 deg:

300

0.564

0

2 0.287 t tan (5 )

Solve this expression for time t and find t = 3.22 x 107 sec (372 days).

(b) Use the expression for Equation 2-28 at the shoreline x = 0, and compute y for t = 1 week:

y

1

'

2 0.287 (604800) tan (5 0 )

π

and find y = 41.1 m (at x = 0).

(c) Solve Equation 2-28 for x = 500 m and t = 1 week.

x

500

' 0.60 ; 2 ε t tan(αb ) ' 72.9 m

'

2 εt

2 0.287 (604800)

From Figure III-2-33, erfc(0.60) 0.42, so that Equation 2-28 is solved directly (or by

Figure III-2-31) as

y

1

exp (&0.6 2 ) & 0.6 (0.42) ' 0.14

2 ε t tan(αb)

π

and

y = (0.14) (72.9 m) = 10.2 m

Longshore Sediment Transport

III-2-65