# Stability 1 – Chapter 5 Effect of Density Solutions

Stability 1 – Chapter 5 Effect of Density Solutions > Exercise 5: Effect of density on draft and displacement – FWA ; DWA ; loadliness of ship.

1. A ship of 16000t displacement and TPC is 20 floating in SW at a draft of 8.0m . Find her draft in FW .
###### Solution:

Displacement (W) = 16000 t
TPC = 20 & SW draft = 8.0m

We know that:

Displacement when in SW = ( L X B x draft) x  1.025
Displacement when in FW  =( L X B X D) x 1

It is understood that displacement of ship will remain constant , as displacement is independent of change in density ,(is referred as MASS).

So, (L x B x 8) x 1.025 =  (L x B x draft) x 1
Hence draft  = ( 8 x 1.025)
= 8.2 m

###### 2nd  Method :

As we know that

FWA = (W/40TPC)
= 16000/(40 x 20)
= 20cm

We can calculated:

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.025 – 1) x 20 / 0.025
= 20c
=0.2m

So new draft of ship in FW = (8.0 +0.2)
= 8.2 m

1. A ship goes from water of RD 1.008 to SW. Find the change in  draft , if her FWA is 180mm, and state whether  it would be sinkage or rise.
###### Solution :

FWA = 180mm = 18cm

We can calculate :

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.008 – 1.025 ) x 18 / 0.025
=(0.017 x 18)/0.025
= 12.24c
= 0.12m

Here , Change in draft is  0.122m and it will be rise.

1. A vessel goes from water of RD 1.010 to FW. If her FWA is 160mm, State whether she would sink or rise and by how much.
###### Solution :

FWA = 160mm = 16cm

We can calculate:

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.010 – 1)x 16 /0.025
= 0.01 x 16 /0.025
= 6.4 cm
= 0.064 m

Here change in draft would lead to sinkage.

1. A ship of FWA 175mm goes from water of RD 1.006 to water 0f RD 1.018 . Find the amount of sinkage or rise.
###### Solution:

FWA = 175mm = 17.5cm

We can calculate:

Change in draft =

(Change in RD )x(FWA)
0.025

= (1.018 – 1.006) x 17.5 /(0 .025)
= (0.012 x 17.5) /(0 .025)
= 8.4cm

Here change in draft would cause rise of vessel.

1. A ship’s stability data book gives her load displacement to be 18000 t and TPC to be 25. If she is now loading in DW of RD 1.018, by how much may her be loadline be immerse so that she would not be over loaded.
###### Solution :

TPC =25.

We know that:
FWA = W/(40 TPC)
= 18000/( 40 x 25 )
=18cm.

We can calculate:

Change in draft =

(Change in RD )x(FWA)
0.025

=(1.025 – 1.018 ) x 18 /0.025
=5.04cm
= .05m

Since, Her load line should immersed to 0.05 m so that she will not be loaded.

1. 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