Solution Naval Numerical 10

Numerical 10: An oil tanker 160m long and 22m beam floats at a draught of 9m in seawater. Cw is 0.865. The midships section is in the form of a rectangle with 1.2m radius at the bilges. A midships tank 10.5m longhas twin longitudinal bulkheads and contains oil of 1.4 m3/t to a depth of 11.5m. The tank is holed to the sea for the whole of its transverse section. Find the new draught.

Solution: A

Complete water plane area = 160 x 22 x 0.865 = 3045.8  

Intact water plane area = 3045.8 — 10.5 x 22 = 2814.8

It may be assumed that the whole of the mass of the oil is taken from the ship and that all the buoyancy of the compartment is lost.

Cross-sectional area of oil=

=

= 2.26 + 23.52 + 226.6

= 252.38

Immersed cross-sectional area = 252.38 — 22 x 2.5 = 197.38  

Mass of oil in compartment = 252.38 x 10.5 /1.4 = 1892.85 tonne

Mass of buoyancy lost = 197.38 x 10.5 x 1.025 = 2124.30 tonne

Nett loss in buoyancy = 2124.30 — 1892.85 = 231.55 tonne

Equivalent volume = 231.55/1.025 = 225.9

Increase in draught =nett volume of lost buoyancy / area of intact water plane =225.9/2814.8 = 0.0802 m

New draught = 9.08 m