201.NA October 2022 Q.6(b)

 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 long has twin longitudinal bulkheads and contains oil of 1.4m3/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

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Given

L = 160 m
B = 22 m
d = 9 m
Cw = 0.865
For the tank
Lt = 10.5 m
dt = 11.5 m
r = 1.2 m

 To find

1.New draught

Solution

Cw = Aw /Lx B
Area of water plane, Aw = Cw x L x B
                                        = 0.865 x 160 x 22
                                        = 3045.8 m2

It is given that the tank is holed.
Therefore area lost = Lt x B
                               = 10.5 x 22
                               = 231 m2

Intact water plane area = Aw - Area lost
                                     = 3045.8 - 231
                                     = 2814.8 m2

We need to find out area of space where oil is occupied.
So first we can consider the bilge part. Both bilge part together form a half circle.(Section (3))
Therefore Area of circle = π x r2

Since it is a half circle Area = (1/2) x π x r2
                                             = (1/2) x 3.14 x 1.22

              Area of section (3) = 2.26 m2 -------------------------------------------(1)

Consider section (2)
Bredth = 22 - (1.2 + 1.2)
            = 19.6 m
depth = 1.2 m
So area of section (2) = 19.6 x 1.2
                                   = 23.52 m2--------------------------------------------------(2)

Consider section (1)
Breadth = 22 m
Depth = 11.5 - 1.2
           = 10.3 m2
So area of section (1) = 22 x 10.3
                                   = 226.6 m2---------------------------------------------------(3)

Total area of oil = (1) + (2) + (3)
                          = 226.6 + 23.52 + 2.26
                          = 252.38 m2

Area of immersion = Total area of oil - (Breadth x Depth of non immersion)
                               = 252.38 - (22 x (11.5 - 9))
                               = 197.38 m2

As we know that density of oil = 1.4 m3/t
Density = Volume  / mass (Because the unit is given as m3/t)

Total mass of oil = Volume of oil / Density
                            = Area x Lt / Density
                            = 252.38 x 10.5 / 1.4
                            = 1892.85 t

So the compartment is holed we can assume that buoyancy is lost.

Mass of buoyancy lost = Area of immersion x Lt x SW density
                                     = 197.38 x 10.5 x 1.025
                                     = 2124.30 t

Net loss in buoyancy = Mass of buoyancy lost - Total mass of oil
                                   = 2124.30 - 1892.85
                                   = 231.55 t

Equivalent volume comparing to SW = Mass / density
                                                             = 231.55 / 1.025
                                                             = 225.9 m3

Increase in draught = Volume lost in buoyancy / Area of intact water plane
                                = 225.9 / 2814.8
                                = 0.0802 m

Increase in draught = 9 + 0.0802
                                = 9.08 m