The wave profile changes the distribution of buoyancy along the ship. In still water the buoyancy is distributed according to the hull form, but in a seaway the waterline is no longer level: when a wave crest is amidships (hogging condition) the buoyancy is increased amidships and reduced at the ends, and when a wave trough is amidships (sagging condition) the buoyancy is reduced amidships and increased at the ends. This changes the net load (weight - buoyancy) distribution, which in turn changes the shear force and bending moment curves. The maximum hogging and sagging bending moments occur when the wave length is about equal to the ship length and the crest or trough is amidships. The wave-induced bending moment is added to the still-water bending moment to give the total bending moment, and the hull girder must be designed to withstand the maximum combined value. The shear force is also increased at the quarter points by the wave profile.

Wetted surface area of a container ship = 5946 m2. At service speed the effective power required is 11,250 kW, with frictional resistance 74% of the total resistance and specific fuel consumption 0.22 kg/kWh. To conserve fuel the speed is reduced by 10%; the daily fuel consumption is then 83.0 t. Frictional coefficient in sea water = 1.432, speed in m/s with index n = 1.825. Propulsive coefficient constant at 0.6.

(i) Service speed.

Frictional resistance Rf = 1.432 x S x V^1.825 = 1.432 x 5946 x V^1.825.

Total resistance = Rf/0.74.

Effective power = R_total x V = 11,250,000 W.

1.432 x 5946 x V^1.825/0.74 x V = 11,250,000.

1.432 x 5946/0.74 = 11,506. So 11,506 x V^2.825 = 11,250,000.

V^2.825 = 977.8, so V = 977.8^(1/2.825) = 11.44 m/s = 11.44 x 1.944 = 22.2 knots.

Answer: the service speed is about 22.2 knots.

(ii) Percentage increase in specific fuel consumption at reduced speed.

Reduced speed = 0.9 x 11.44 = 10.30 m/s.

Frictional resistance at reduced speed = 1.432 x 5946 x 10.30^1.825. 10.30^1.825 = 68.0, so Rf = 1.432 x 5946 x 68.0 = 579,000 N. Total resistance = 579,000/0.74 = 782,400 N.

Effective power at reduced speed = 782,400 x 10.30 = 8,058,000 W = 8058 kW.

Daily fuel at reduced speed = 83 t = 83,000 kg. Specific fuel consumption at reduced speed = 83,000/(8058 x 24) = 83,000/193,392 = 0.429 kg/kWh.

At service speed the SFC = 0.22 kg/kWh. Percentage increase = (0.429 - 0.22)/0.22 x 100 = 95%.

Answer: the specific fuel consumption increases by about 95% at the reduced speed.