Chapter 4

1. A box-shaped vessel 120m long and 15m wide has alight draft of 4m and load draft of 9.8m in SW. find her light displacement, load displacement and DWT.

Solution :


Area of box shape vessel = (L x B) = 120m x 15m Light draft = 4m

Light displacement = (u/w volume) x ( density) = (L x b x d )x (1.025) = (120 x 15 x 4) x (1.025) = 7380 t

Load draft = 9.8m

So load displacement = (u/w volume) x (1.025) = ( L x b x d) x (1.025) = (120 x 15 x 9.8) x (1.025) = 18081t

Hence, DWT = (Load displacement – Light displacement) = ( 18081 – 7380 ) = 10, 701t

2. A box-shaped vessel 100m long and 14m wide is floating in SW at a draft of 7.6m. Her light draft is 3.6 m and load draft 8.5m. Find her present displacement, DWT aboard and DWT available.

Solution:


Area of box shape vessel = (L x B) = (100m x 14m)

Present draft = 7.6m Present displacement = (Lx b x d) x (density) = (100 x 14 x 7.6) x (1.025) =10906 t Light draft = 3.6m

Light displacement = (L x B x D) x(1.025) = (100 x 14 x 3.6 ) x (1.025) = 5166t

When Load draft is = 8.5m

Load displacement = ( L x b x d )x (1.025) = (100 x 14 x 8.5) x (1.025) = 12,197.5t

DWT aboard = (present displacement – Light displacement) = (10906 t – 5166 t ) = 5740 t

DWT available = (Load displacement – present displacement) = ( 12,197.5 t – 10,906 t ) = 1291.5t.

3. A ship is 200m long 20m wide at the waterline. If the coefficient of fineness of the water-plane is 0.8, find her TPC in SW, FW and DW of RD 1.015.

Solution :


Given : (L x B) = (200m x20) , Cw = 0.8 Cw = Area of water plane / ( L x B ) 0.8 = Area of water plane / (200 x 20) (0.8 x 200 x 20) = Water plane area

A = 3200m2

TPC in SW = (A/ 100 x 1.025) = (3200 / 100 x 1.025) = 32.8 t/cm

TPC in FW = (A / 100x 1 ) = (3200/100) = 32 t/cm

TPC in RD of 1.015 =( A /100 x 1.015) = (3200 /100 x 1.015) = 32.48 t/cm.

Note : here “A” refers to water plane area.

4. A double bottom tank 20m x 10.5m x 1.0m has a block a coefficient of 0.82. Calculate how much fuel oil of RD 0.9, it can hold .

Solution :


Given : – (L x B x H) of double bottom tank =( 20 x 10 .5 x 1) Cb = 0.82 , RD = 0.95

Cb = (volume of tank) /( L x B x depth) 0.82 = (volume of tank) /(20 x 10.5×1) volume of tank = 0.82 x ( 20 x 10.5x 1) = 172.2 m3

Weight of oil can be calculated as :- (volume of tank) x(density) = (172.2 x 0.95) = 163.59 t

5. A ship floating in SW at a draft of 8m is 110m long and 14m wide at the water line . If her block coefficient is 0,72, find her displacement. If her load displacement is 12000 t, find the DWT available .

Solution:


Given: – (L x B) = 110m x 14m, Present draft = 8m Cb = 0.72

Load displacement = 12000 t ,

Cb = (u/w volume ) / (Lx B x D) 0.72 = (u/w volume) / (110 x 14 x8)

u/w volume = 0.72 x ( 110 x 14 x 8 ) = 8870.4 t

DWT available = (Load displacement – present displacement) = (12000 – 8870.4) = 3129.6t.

6. A vessel of 14000t displacement is 160m long and 20m wide at the water line. If she is floating in SW at a draft of 6.1m, find her block coefficient.

Solution:


Given : – W = 14000 t , (L x B) of waterline = (160m x 20m), Present draft = 6.1m

Weight = (u/w volume )x (density of displaced water) 14000 = (u/w volume) x 1.025 u/w volume = 14000/ 1.025 = 13658.536 t

Cb = (u/w volume )/( Lx B x D) Cb = 13658.536 /(160 x 20 x6.1) = 0.699m = 0.7m

7. A box-shaped vessel 18m x 5m x2m floats in SW at a draft of 1.4m. Calculate her RB %.

Solution :