49.NA June 2019 Q.8

A box shaped vessel is 80 m long 12 m wide and float at a draft of 4 m. A full width mid ship compartment 15 m long is bilged and this result in a draft increasing to 4.5 m. Calculate each of the following.
1.Permeability of the compartment
2.Change in meta centric height due to bilging.

                           IF NOTICING ANY ERROR KINDLY COMMENT BELOW

Given

L = 80 m

B = 12 m

d1 = 4 m

l = 15 m

d2 = 4.5 m

To find

1.Permiability of compartment

2.The change in metacentric height

Solution

As we know that ,

Increase in draft = Volume of lost buoyancy / Area of intact water plane

                           = (µ x l x d1) / (L - µ x l)

                     0.5 = (µ x 15 x 4) / (80 - µ x 15)

1) Permiability µ = 0.592

2) KG is same in both conditions.

     KB1 = d1/2 = 2 m

     

     KB2 = d2/2 = 2.25 m

     

     BM1 = I₁ / Displacement                          (I – moment of inertia)         

              = (L x B³) / (12 X Displacement)

              = (80 x 12³) / (12 x 80 x 12 x 4)

              = 3 m

I_new = I₁ – I₂

        = [(L x B³) / 12] – [(µ x l x B³) / 12

        =  [(80 x 12³) / 12] – [(0.592 x 15 x 12³) / 12

        = 10241.3 m⁴

BM2 = 10241.3 / (80 x 12 x 4)

         = 2.67 m

GM1 = KB1 +BM1 –KG

          = 2 + 3 – KG

          = 5 – KG

GM2 = KB2 + BM2 – KG

          = 2.25 + 2.67 – KG

          = 4.92 – KG

Change in metacentric height = GM1 –GM2

                                                = 5 – KG – 4.92 + KG

                                                = 0.08 m