**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 :**

**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 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:**

**Solution:**

*A**rea* 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 water-plane is 0.8, find her TPC in SW, FW and DW of RD 1.015.**

**Solution :**

**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 :**

**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:*

*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:*

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

**Solution :**

**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**

**8. A box –shaped of 2000t displacement is 50m x 10m x 7m. Calculate her RB% in FW**

**Solution :**

**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.**

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

**Solution:**

**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.**

**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 :**

**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**

sir exercise no 4 problem no 7 draft you had taken is wrong it is not 4.1 it is 6.1 mtr