EXAMPLE PROBLEM II-7-8

FIND:

The wind drag for a Beaufort 5 fresh breeze of 20 knots (10.31 m/s) at an angle of 30 deg to the wind.

Assume gustiness can be accounted for by applying a factor of 1.20 to the wind speed. Beaufort 5 is usually the

limiting wind for port operations of loading and unloading.

GIVEN:

A C9 containership is moored at a pier with an asymmetrical container distribution of 5 high in the bow half

of the vessel and none in the aft portion. The useable length of the ship for cargo handling is 246 m. The height

above the water is 10.9 m. The containers are 2.7 m tall.

SOLUTION:

The forward and aft areas exposed to the wind are

246

10.9 ' 1,340 m 2

2

The net effective area is then

The wind speed, including a gustiness factor, is

Substituting these values into Equation II-7-31, we get

1

(1.2)(1.0)(2,170)(12.4)2 ' 200 kN

2

where we have assumed ρa = 1.2 kg/m3 at 20 EC and *C*D = 1.0 as a reasonable value since the mean drag coefficient

for bow and transverse winds varies between 0.82 to 1.11.

(h) Longitudinal current loads on ships are taken from procedures by NAVFAC Design Manual DM26.5.

The total longitudinal current load *F*c,tot is composed of form drag *F*c,form, skin friction *F*c,fric, and propeller

(II-7-33)

(i) Form drag is due to the flow of water past the vessel's cross-sectional area and is defined as

1

2

(II-7-34)

ρw Cc , *form *B T Vc cos θc

2

Harbor Hydrodynamics

II-7-71

Integrated Publishing, Inc. |