Stability 1 Chapter 2 – Water Pressure

7. A collision bulkhead is triangular shaped, having a breadth of 14m at the tank top and a height of 12m. As a result of a collision, the forepeak tank gets ruptured and SW enters the tank to a sounding of 9m. Calculate the thrust on the bulkhead.

Solution :

B = 14m, H =12m

As we know that :
Pressure = (depth x density)
Thrust   = (pressure x area)

An right angle triangle PBC , PB =7m
Since, Height divide AB  equally into two parts.
Triangle PBC  and   FEC are similar
So by law of similar angle triangle

PB/ FE = PC / FC
7 / FE  =12 / 9

Hence, FE  =  (63 / 12)
              = 5.25m

In triangle DEC
DF = FE
DE = (2 X FE)
       = 2X 5.25
       =10.5m

Area of triangle DEC = (1/2 x10.5×9)
                                    =47.25m²

Pressure = (depth x density)
                  = (9 x 1/3 x 1.025)
                 = 3.075 t/m²

Hence Thrust experienced  = (P X A)
                                            = (3.075 x 47.25)
                                            =145. 29 t

8. A tank has a triangular bulkhead, apex upwards. Its base is 14m and its sides, 15m each. It has a circular inspection hole of radius 0.5m the centre of a manhole 0.8m above the base and 1.6 from one corner. Find the thrust on the manhole cover when the tank contain oil of RD 0.95 to a sounding of 10m.(Assume π to be 3.1416).

Solution:

Base = 14m , Sides = 15m , Depth of the oil = 10m

Pressure = (depth x density)
             =  (10 – 0.8 x 0.95)
            = 8.74 t/m²

Area of the circle = π r²
                 = (3.1416 x 0.5 x 0.5)
                 =0.7854m²

Thrust = (P X A)
          =( 8.74 x 0.7854)
          = 6.864 t.

9. A rectangular deep tank is 22m x 20m x10m. Above the crown of the tank is a rectangular trunkway 0.2m high, 5m long and 4m wide. Find the thrust on the tank lid when the tank is pressed up with SW to a head of 2.64m above the crown of the tank .

Solution :

Rectangular deep tank  = 22m x 20m x10m
Trunk way = (5m x 4m x 0.2m)

Pressure = (depth x density)
              = (2.64m – 0.2) x1.025

Since , Depth is (2.64m – 0.2 )  and we have the calculate the thrust   act  on the trunkway and the water level above the trunkway will act as thrust )

If this trunkway would have been placed inside the tank then the depth will be   (10m + 2.64m – 0.2/2m ).

Area of the trunkway = (L X B)
                                = 4×5
                                 =20 m²

Thrust experienced = (P X  A)
                           = (2.501 x 20 )
                          = 50.02tonnes.

10. A double bottom tank measures 25m x20m x 2m. Find the thrust on the tank top when pressed upto a head 16m of SW. Also find the resultant thrust on the tank bottom, and the direction that it acts, if the ship’s draft in SW is 10m.

Solution :

Volume of  tank  = (L X B X H)
                         = (25m x 20m x 2m)

Area    =     (25 x20)
          =   500m2

Pressure = (depth x density)
                = (16 x 1.025)
                =16.4  t/m²

Thrust on tank top  = (PX A)
                                      = (16.4 X 500)
                                     = 8200 t

Pressure on the tank bottom = (depth x density)
                                                            = (18 x1.025)              = 8.45t/m²

Thrust   act  on the tank bottom = (P X A)
                                                               =   (500 x 18 .45)
                                                               = 9225t

Pressure act on outside of the tank bottom = (depth x density)
                                                                            = (10 x 1.025)
                                                                           = 10.25t

Hence, Thrust acting = (P X A)
                                 = (10.25 X500)
                                 = 5125 tonnes.

Finally, Resultant thrust = (9225 -5125)
                                    = 4100 tonnes.

Also read Stability Chapter 1 – Density and relative density Solutions.