EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 13, Problem 92P

(a)

To determine

The fraction of the submarines volume is above the water surface when the tanks are filled with air.

(a)

Expert Solution
Check Mark

Answer to Problem 92P

  f= 2.5%

Explanation of Solution

Given:

Total mass of a submarine (including crew and equipment) is 2.4×106 kg

The volume of a pressure hull is 2.0×103 m3 ,

The volume of a ballast tank is 4×102 m3 ,

Total volume of the submarine is, V= 2.0×103 m3 + 4.0×102 m3 = 2.4×103 m3

Formula used:

The fraction of the submarine’s volume above the surface when the tanks are filled with air can be written as,

  f=VV'V=1V'V  (1)

Where V is the volume of the submarine.

  V' is the volume of seawater displaced by submarine when it is on the surface.

Apply Fy=0 to the submarine when its tanks are filled with air

  Bw=0  (2)

Where B is buoyant force on the submarine

  w is weight of the submarine.

Using Archimedes’ principle, the buoyant force B on the submarine can be expressed in terms of the volume of the displaced water as,

  B=ρswV'g

Where, ρsw is the density of sea water, ρsw=1.025×103 kg/m3

  g is acceleration due to gravity, g=9.81 m/s2

Weight of the submarine is, w=mg

Where, m is mass of the submarine.

Substituting for B and w in equation (2) ,

  ρswV'gmg=0ρswV'g=mgV'=mρ sw

Substituting for V' in equation (1)

  f=1mρswV  (3)

Calculation:

Substituting the numerical values in equation (3)

  f=12 .4×106 kg( 1 .025×10 3  kg/m 3 )( 2 .4×10 3  m 3 )f= 2.5%

Conclusion:

The fraction of the submarines volume is above the water surface when the tanks are filled with air is 2.5 .

(b)

To determine

The quantity of water must be admitted in to the tanks to give the submarine a neutral buoyancy.

(b)

Expert Solution
Check Mark

Answer to Problem 92P

  Vsw= 60 m3

Explanation of Solution

Given:

Total mass of a submarine (including crew and equipment) is 2.4×106 kg

The volume of a pressure hull is 2.0×103 m3 ,

The volume of a ballast tank is 4×102 m3 ,

Total volume of the submarine is, V= 2.0×103 m3 + 4.0×102 m3 = 2.4×103 m3

Neglect the mass of any air in the tanks,

Specific gravity of sea water is 1.025

Formula used:

The volume of the sea water in terms of its mass and density is, Vsw=mswρsw  (4)

Applying the condition for neutral buoyancy, Fy=0 to the submarine,

  Bwsubwsw=0  (5)

Where, B is buoyant force on the submarine,

  wsub is weight of the submarine,

  wsw is weight of the seawater,

Using Archimedes’ principle, the buoyant force B on the submarine can be expressed in terms of the volume of the displaced water as,

  B=ρswV'g

Where, ρsw is the density of sea water, ρsw=1.025×103 kg/m3

  g is acceleration due to gravity, g=9.81 m/s2

Weight of the submarine is, wsub=msubg

Where, msub is mass of the submarine.

Weight of the sea water is, wsw=mswg

Where, msw is mass of the submarine.

Substituting for B , wsub , and wsw in equation (5) ,

  ρswV'gmsubgmswg=0

  ρswV'msubmsw=0

  msw=ρswV'msub

Substituting for msw in equation (4) ,

  Vsw=ρ swV'm subρ swVsw=Vm subρ sw(6)

Calculation:

Substitute the numerical values in equation (6) ,

  Vsw=2.4×103m3-2 .4×106kg1 .025×103 kg/m3Vsw=60 m3

Conclusion:

The quantity of water must be admitted in to the tanks to give the submarine a neutral buoyancy is 60 m3 .

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