
A boxshaped 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

A boxshaped 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
We know that
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.

A ship is 200m long 20m wide at the waterline. If the coefficient of fineness of the waterplane 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
As we know that :
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 = 3200m^{2}
We can calculate :
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.

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
We know that:
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 m^{3}
Weight of oil can be calculated as : (volume of tank) x(density)
= (172.2 x 0.95)
= 163.59 t

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 ,
We know that :
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.

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

A boxshaped vessel 18m x 5m x2m floats in SW at a draft of 1.4m. Calculate her RB %.
Solution :
Given : (L x B x H) of box shape vessel =(18m x 5m x 2m), Present draft = 6.1m
We know that :
RB % = (Above water volume )/(total volume ) x 100
Total volume can be calculated as = (L x B x D)
= 18 x 5x 2m
= 180m^{3}
U/w volume = (L x B x present draft)
= 18 x 5 x 1.4
= 126m^{3}
Hence, Above water volume = (180 – 126)
= 54m3
RB % = (Above water volume) / (total volume ) x 100
Hence, RB % = (543/180 x 100)
= 30%
8. A box –shaped of 2000t displacement is 50m x 10m x 7m. Calculate her RB% in FW
Solution :
Given : W = 2000t,
(L x B x H) of box shaped vessel = (50m x 10m x 7m)
Total volume = 3500m^{3}
W = ( u/w volume) x (density)
2000 = (u/w volume) x (1.00)
u/w volume = 2000 m^{3}
Above water volume = (Total volume) – (u/w volume )
= (3500 – 2000)
= 1500 m^{3}
RB % = (Above water volume) / (total volume ) x 100
RB % = (1500 /3500)x 100
= 42.857%
9. The TPC of a ship in SW is 30. Calculate her TPC in FW and in DW of RD1.018.
Solution:
We know that:
TPC = (A / 100) x (density)
30 = (A / 100) x (1.025)
A = (30 x 100)/ 1.025
A = 2926.83 m^{2}
TPC in FW can be determined as = ( 2926.83/ 100)x(1)
=29.268 t/cm.
TPC in DW can be determined as = (2926.83 / 100) x(1.018)
= 29.79 t/cm.
10. A ship is a floating a draft of 8.2m in DW of RD 1.010. If her TPC in SW is 40, find how much cargo she can load to bring her draft in DW to 8.4m.
Solution :
Given:
RD of DW = 1.010,
Present Draft = 8.2m
Cargo can be load up to draft = 8.4m,
So, Sinkage = 0.2m = 20cm
TPC in SW = (A / 100) x (1.025)
40 = (40/100) x (1.025)
A = (40 x100)/ (1.025)
= 3902.44m^{2}
Now, TPC in DW = (A/100) x (1.010)
= (3902.44 / 100) x(1.010)
=39.41 t/cm.
Cargo can be load = (39.41 x 20)
= 788.2 t
question 5 is wrong. you forget to take the u/w vol and multiply by the sw density.
sir exercise no 4 problem no 7 draft you had taken is wrong it is not 4.1 it is 6.1 mtr