# Chapter 4 – Some important terms – displacement; deadweight; form coefficient; reserve buoyancy; Tonnes per centimeter.

1.

A box-shaped vessel 120m long and 15m wide has a light draft of 4m and a 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

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

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.

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

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   =  3200m2

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 waterplane 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

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 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 ,

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.

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

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 also know that:-

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 :

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
= 180m3

U/w volume = (L x B x present draft)
= 18 x 5 x 1.4
= 126m3

Hence, Above water volume = (180 – 126)
= 54m3

RB % = (Above water volume) / (total volume ) x 100
Hence, RB % = (543/180 x 100)
=  30%

###### 9.

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 = 3500m3
W = ( u/w volume) x (density)
2000 = (u/w volume) x (1.00)
u/w volume = 2000 m3

Above water volume = (Total volume) – (u/w volume )
= (3500 – 2000)
=  1500 m3

RB % = (Above water volume) / (total volume ) x 100
RB % = (1500 /3500)x 100
= 42.857%

###### 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)
= 3902.44m2

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