A vessel’s statutory freeboard is 2.0m. She is loading in DW of RDD 1.015 and her freeboard is 2.1m . TPC =24. FWA = 200mm. find the DWT available .

Solution:

Statutory freeboard is = 2.0m
DW RD = 1.015 & Freeboard = 2.1m
TPC = 24, FWA = 200mm = 20cm

As we know :

Change in draft = 

(Change in RD )x(FWA)
             0.025

= (1.015 – 1.025 ) x 20 /0.025
= ( 0.010 x 20) /0 .025
= 8cm

Hence , actual freeboard available =( 2.1 + 0.08)m
= 2.18m

Sinkage available = ( 2.18 – 2.00)
= 0.18m
= 18cm

TPC = 24 ( Given)

24  =  (A / 100) x(1.025)
A = (24 x 100)/ 1.025
A   = (24 / 1.025)
= 23.414 m2

Now, TPC for RD 1.015 = (A/ 100) x(1.015)
= (23.414 x 1.015)
= 23.76 t/cm

DWT available = (sinkage x TPC)
= (18 x 23.76 )
= 427.68t.

A vessel is floating at 7.8m draft in DW of RD 1.010. TPC is 18 and FWA is 250mm. the maximum permissible draft.

Solution:

Present draft = 7.8m,
RD of DW = 1.010 & TPC = 18 t/cm
FWA = 250mm = 25cm
Maximum permissible draft = 8.0m

As we know :

Change in draft = 

(Change in RD )x(FWA)
             0.025

= (1.010 – 1.020 ) x FWA /0 .025
=(0.015 x 25) /0 .025
= 15cm

When vessel enters SW,  the draft will rise by 15 cm,

Hence ,actual draft in SW = (7.80 -0. 15)
=7.65m

Sinkage available = (8.00 – 7.65)
=0.35m

TPC = 18 ( given)

We know that :

TPC = (A/100)  x ( density)
18 = (A/100) x 1.025
So, A = (18 x 1.025)/ 100
= 17.56 m2

Now , TPC at RD 1.010 = A / 100 x 1.010
= 17.56 x 1.010
= 17.73 t/cm.

DWT available =( sinkage x TPC)
= (35 x 17.73)
= 620.5t.

A vessel floating in DW of RD 1.005 has the upper edge of her summer loadline in the water line to starboard and 50mm above the waterline to port. If her FWA is 180mm and TPC is 24 , find the amount of cargo which the vessel can load to bring her to her permissible draft.

Solution:

RD = 1.0
Starboard side = 0cm above by below summer draft
Port side = 50mm (5cm above the water line to summer draft)

Mean draft   =    (0+5 ) / 2
= 5/ 2
= 2.5cm above the water line

FWA = 180mm
= 18cm,

TPC = 24

As we know :

Change in draft = 

(Change in RD )x(FWA)
             0.025

= (1.025 – 1.005 ) x 18 /0 .025
= (0.2 x 18) /0 .025
= 14.4cm

Total sinkage = (14.4 + 2.5) cm
=   16.9cm

TPC is SW = 24

TPC = (A / 100) x 1.025
= (24 x100)/ 1.025
A = 23.41 t/cm

Again ,

TPC in RD 1.005 = ( A / 100) x 1.005
= (23.41 x 1.005)
= 23.53t/cm

Cargo to load = sinkage x TPC
=   23.53 x 16.9
= 397.657 t

A vessel of 12000 t displacement arrives at the mouth of a river, drawing 10.0 m in SW. how much cargo must she discharge so that her draft in an upriver port of RD 1.012 would be 10m.

Solution:

Displacement (W)= 12000t ,
Depth = 10m & RD = 1.025

Let W1 be the displacement at RD 1.012

Now, W =( L x B x draft ) x( 1.025)
W1 = (L x B x d1) x (1.012)

According to question ,

Draft  =  d1
W/( L x B x 1.025)  = W1 / (L x B x 1.012)
W/ (L x B x 1.025)  = W1/ ( L x B x 1.025)

W1 = (12000 x 1.012) / 1.025
W1 = 11847.8 t

Hence, Cargo she has to discharge = (12000 – 11847.8) t
= 152.2 t

A vessel displaces 14500 tonnes, if floating in SW up to her winter load- line. If she is in a dock of RD 1.010 , with her winter load- line on the surface water, find how much cargo she can load, so that she would float at her winter load-line in SW.

Solution:

Displacement (W )= 14500t
RD = 1.025

Displacement (W) =( u/w volume ) x (density)
= (L x B x draft)  x  (1.025)

Draft = W /( L x B x 1.025)

Let W1 be the displacement at RD of 1.010

So W1 = (u/w volume )x( density)
= (L x B x d1 ) x  (1.010)
Hence, d1 = W1 / (L x B x 1.010)

According to question, Draft  = d1

So, W / ( L x B x 1.025)  =  W1 / (L x B x 1.010)
14500 / (L x B x 1.025) =  W1 / (L x B x 1.010)
W1   = (14500 x 1.010) / 1.025
= 14287.8 t

Hence, Cargo to load = (14500 – 14287.8)
= 212.2t.

A ship 100m long and 20m wide, block coefficient 0.8m, floats in SW at a mean draft of 8.0 m. Calculate the difference in displacement when floating at the same draft in FW .

Solution:

Area of Waterplane = (L X B)
= (100m x 20m)

Cb = 0.8, Depth = 8m

We know that :

Displacement = (u/w volume ) x (density)
= ( Lx B x draft ) x (0.8)
= (100 x 20 x 8) x(0.8)
= 12800m3

Displacement (W) = (u/w volume)x (density)
= (12800 x 1.025)
= 13120t ,

Draft = W / ( L x B x Cb x 1.025)

Let W1 be the displacement in FW
= (u/w volume) x (density)
=(L x B x d1 x Cb ) x ( density)

d1 = (W1 /( L x B x 1 )x (Cb)

According  to  question, Draft = d1

W /( L x B x Cb x 1.025) = ( W1 /( L x B x Cb)
W1 = ( 13120 x 1) /( 1.025)
= 12800t

So, Difference is displacement = (13120 – 12800)
= 320 t .

A vessel displaces 4500 t of FW at a certain draft . Find her displacement at the same draft in water of RD 1.020 .

Solution:

Displacement (W) = 4500t,

Displacement = (u/w volume ) x (density)
= (L x B x draft) x 1.025

Draft = W / (L x B x 1.025) m

Let W1 be the displacement at RD 1.020

So W1 =( L x B x d )x (1.020)
d = (W1 / (Lx B x 1.020)

according to question,  Draft  =  d

W /( L x B x 1.025)  =  W1 /( L x B x 1.020)
W1 = (4500 x 1.020) / 1.025
= 4478.04t.

The hydrostatic particular of a ship indicate that her displacement in SW at a draft of 5m is 3000t. Find her displacement when floating at 5m draft in water of RD 1.018.

Solution:

Displacement(W) = 3000t,
Draft= 5m & RD = 1.018

W = (u/w volume) x (density)
= (L x B x draft) x(1.025)
Draft = W/ (L x B x 1.025)

Let W1 be the displacement at RD = 1.018

W1 = ( L x B x d) x (1.018 )
d1 = W1/( Lx B x 1.018)

But according to question ,

Draft = d1
W/ (L x B x1.025 )= W1 /(L x B x 1.018)
3000 / 1.025 = W1 / 1.018
W1 = 3000 x (1.018 / 1.025)
= 2979.51 t

A box-shaped vessel 18 x5 x2 m floats in DW of RD1.000 at a draft 1.4m. Calculate her percentage reserve buoyancy when she enters the SW .

Solution:

Volume of box shape vessel = (L x B x H )
= (18m x 5m x 2m)

RD of DW = 1.000 & Depth = 1.4m

Displacement at DW of RD 1.000 (W)
= (u/w volume)x (density)
=( Lx B x 1.4) x(1)

Displacement at SW (W1) = (u/w volume) x (density)
=(L x B x Draft) x (1.025)

Since, Displacement of ship will remain constant with change of density as displacement of vessel is refer as MASS.

W = W1
(L x B x 1.4) x (1) = (L x B x Draft) x (1.025)

Hence, Draft = (1.4/1.025)
= 1.365m

Total volume of the ship = (L x B x H)
= (18 x 5 x 2)
= 180m3

Again, U/w volume of SW = (L x B x draft)
=(18 x 5 x 1.365)
= 122.85m3

Above water volume = (180 -122.85)
= 57.15m3

Hence, RB % = (Above water volume/ Total volume) x 100
= (57.15/ 180) x 100
= 31.75%.

A box-shaped vessel 20 x 4 x2 m has mean draft of 1.05m in SW. Calculate her draft in DW of RD 1.012.

Solution:

Volume of box shape vessel = (Lx B x H)
= (20 X 4 X 2)

Mean draft = 1.05m
Displacement  = (u/w volume )x density

Again, displacement can be calculated as (W)
=(L x B x D) x(density)
= (20 x 4 x 1.05) x( 1.025)
=86.1t

Let ‘X’ be the displacement at RD of 1.012
So, X = (L x B x D) x ( 1.012)

Since, Displacement of ship will remain constant with change of density as displacement of vessel is refer as MASS.
(20 x 4 x 1.05) x(1.025) = (20 x 4 x d) x(1.012)
d = (1.05 x 1.025) / 1.012
= 1.06m