Question

125.0% OUNOATONNVA UNIVERSITY(SKKU) [3][25 points A submarine of mass 10,000 kg is submerged in the ocean at a depth of 100 m. submarine has a total displacement volume of 12 . Part of the displacement volume is a ballast tank of 3 m that can . The Water Balast ank ballast ank filled be or air Valve with water Assume for the density of water 1 g/c,and g m/s2 The shown geoetry of the submarine is for ilkastration only 10 submarine mass The 10.000 kg is measured when its ballast is empty. (a)[5 points] Assuming an atmospher ic pressure of 10 Pa. what is the pressure at the depth of the submar ine? (b) [5 points] What is the buoyant force B of the submarine? (b (c)[5 points] To level the submar ine at its present depth (neutral (b buoyancy) hoW Buch water needs to be in the ballast tank ? (d)l 10 points] Assune the submar ine wants to surface and hence reduce the mass of the water in the ballast tanks by 100 kg. What air pressure in the ballast tank is needed so that this amount of water can be reaoved in 10 seconds through a valve with cross sect ion area 1 car connect ing the ballast tank to the sea. The valve is at the bot ton of the ballast tank (assune at 100 depth) and the ballast tank has a swater level of 2 which does not change significant ly dur ing this process. pr (c)

Answer 1

(a) The Pressure will be given by,

P =Pa+ pgh

Substituting values we get,

BP 101345(1000) (9.81) (100) 1082325 Pa

(b) The Buoyant force will be equal to the force generated due to the amount of water displaced, therefore we can write, (assuming that the ballast tank is filled with water, so the total displacement of water is 12-3)

F Vpg F, = (12 - 3) (1000) (9.81) = 88290 N

(c) For neutral buoyancy, let us say x amount of ballast should be filled with water. Then we can say that the buoyant force and downward force of the submarine should be same. This gives, (downward force will have weight of water + submarine)

mg+Vupg V pg (10000) (9.81)+x(1000) (9.81) = (12-x)(1000) (9.81)

10+ 12-a

$\Rightarrow x = 1$

So 1 m^3 ballast has to be filled with water for neutral buoyancy.

(d) Assuming that the submarine remains at the same depth, for 10 seconds until water is pumped out. The volume of water needed to be pumped out is,

100 0.1 m Vu 1000

The water has to be pumped out in 10 seconds through a valve of cross section area 1 cm^2 at a depth of 100 m. Hence the speed of water is,

V = Av =

0.1/10 100 m/s v= 1 x 10-4

Now, from Bernoulli's equation we can write, (at motor there is pressure P, outside the submarine there is pressure due to depth h, and there is velocity in water coming out)

P (Pa pgh)u

P (101325 (1000)(9.81) (100))(1000) (100)2

P6082325 Pa

This is the required pressure for the pump.