100.NA October 2020 Q.9(b)

 A ship of 5000 t displacement has three regular double bottom tanks A - 12 m long and 16 m wide : Tank B - 14 m long and 15 m wide ; Tank C - 14m long and 16 m wide.

Calculate the free surface effect for any tank and state in which order the tank should be filled when making use of them for stability corrections.

IF YOU FIND ANY ERRORS KINDLY COMMENT BELOW

Solution

It is stated that The double bottom tanks in rectangular shape.

So Moment of inertia of a rectangular body I = (length x Width³) / 12

So Free surface effect  FSC = (Moment of inertia x Density of liquid) / Displacement

Consider tank A

Ia = (La x Ba³) / 12

    = (12 x 16³) / 12

    = 4096 m4

FSCa = (Ia x 1.025) / 5000

          = (4096 x 1.025) / 5000

          = 0.8396 m

Consider tank B

Ib = (Lb x Bb³) / 12

    = (14 x 15³) / 12

    = 3937.5 m4

FSCb = (Ib x 1.025) / 5000

          = (3937.5 x 1.025) / 5000

          = 0.8071 m

 

Consider tank C

Ic = (Lc x Bc³) / 12

    = (14 x 16³) / 12

    = 4778.6 m4

FSCc = (Ic x 1.025) / 5000

          = (4778.6 x 1.025) / 5000

          = 0.9796 m

 

Tank B has low free surface effect compared to A and C. So Tank B should fill first , then Tank A and finally tank C.