# 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 :

Given :-** (L x B x H) of box shape vessel =(18m x 5m x 2m), Present draft = 6.1mRB % = (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%

**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 = 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%

**9. The TPC of a ship in SW is 30. Calculate her TPC in FW and in DW of RD1.018.**

**Solution:**

** TPC = (A / 100) x (density)**
30 = (A / 100) x (1.025)
A = (30 x 100)/ 1.025
A = 2926.83 m2

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) = 3902.44m2

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