Fuel consumption per unit time is proportional to the power developed, and power is approximately proportional to the cube of the speed (P proportional to V^3). If the ship is run above its design (operating) speed, the fuel consumption per hour rises very steeply (as V^3), so the specific fuel consumption per tonne-mile worsens and the voyage cost rises sharply. If the ship is run well below the operating range, the engine operates at a poor point on its specific fuel consumption curve (high SFC at low load), the propeller may be inefficient, and although the fuel per hour falls, the fuel per tonne-mile may not improve as much as expected; also the voyage time increases. Hence the most economical speed is near the design point where the SFC is minimum and the propulsive efficiency is good; running outside this range increases fuel consumption per unit of useful work (per tonne-mile) and per unit time at high speed.

Oil tanker 160 m long, 22 m beam, floats at a draught of 9 m in sea water. Cw = 0.865. The midship section is a rectangle with 1.2 m radius at the bilges. A midship tank 10.5 m long has twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5 m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

Waterplane area Aw = Cw x L x B = 0.865 x 160 x 22 = 3044.8 m2.

Midship section area (rectangle with bilge radius r=1.2 m): Ams = B x d - (4 - pi) r^2 = 22 x 9 - 0.858 x 1.44 = 198 - 1.236 = 196.76 m2.

The tank is 10.5 m long and spans the full beam (twin longitudinal bulkheads divide it, but it is holed for the whole transverse section). Volume of the tank below the original waterline = Ams x 10.5 = 196.76 x 10.5 = 2066 m3.

The tank contains oil of density rho_o = 1/1.4 = 0.714 t/m3. When holed, sea water (1.025 t/m3) replaces the oil, so the net loss of buoyancy is the volume times the relative density difference:

Vlost = 2066 x (1 - 0.714/1.025) = 2066 x (1 - 0.697) = 2066 x 0.303 = 626 m3.

The flooded tank provides no increase of buoyancy, so the effective sinking waterplane = Aw - (10.5 x 22) = 3044.8 - 231 = 2813.8 m2.

Sinkage = Vlost/effective waterplane = 626/2813.8 = 0.222 m.

New draught = 9 + 0.22 = 9.22 m.

Answer: the new draught is about 9.2 m.