# Stability 1 Chapter 2 – Water Pressure

**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 :*

*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)**** =47.25m ^{2}**

**Pressure = (depth x density)**** = (9 x 1/3 x 1.025)**** = 3.075 t/m ^{2}**

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

**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/m ^{2}**

**Area of the circle = π r ^{2}**

**= (3.1416 x 0.5 x 0.5)**

**=0.7854m**^{2}**Thrust = (P X A)**** =( 8.74 x 0.7854)**** = 6.864 t.**

**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 m ^{2}**

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

**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/m ^{2}**

**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/m ^{2}**

**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.**