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.

Answer to Problem 92P

  $f=\text{ 2}\text{.5\%}$

Explanation of Solution

Given:

Total mass of a submarine (including crew and equipment) is $2.4\times {10}^6\text{ kg}$

The volume of a pressure hull is $2.0\times {10}^3{\text{ m}}^3$ ,

The volume of a ballast tank is $4\times {10}^2{\text{ m}}^3$ ,

Total volume of the submarine is, $\text{V}=\text{ 2}\text{.0}\times {10}^3{\text{ m}}^3\text{ + 4}\text{.0}\times {\text{10}}^2{\text{ m}}^3\text{ = 2}\text{.4}\times {\text{10}}^3{\text{ m}}^3$

Formula used:

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

  $f=\frac{V-V\text{'}}{V}=1-\frac{V\text{'}}{V}$  $\left(1\right)$

Where $V$ is the volume of the submarine.

  $V\text{'}$ is the volume of seawater displaced by submarine when it is on the surface.

Apply $\sum F_y=0$ to the submarine when its tanks are filled with air

  $B-w=0$  $\left(2\right)$

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,

  $B={\rho }_{sw}V\text{'}g$

Where, ${\rho }_{sw}$ is the density of sea water, ${\rho }_{sw}=1.025\times {10}^3{\text{ kg/m}}^3$

  $g$ is acceleration due to gravity, $g=9.81{\text{ m/s}}^2$

Weight of the submarine is, $w=mg$

Where, $m$ is mass of the submarine.

Substituting for $B$ and $w$ in equation $\left(2\right)$ ,

  $\begin{array}{l}{\rho }_{sw}V\text{'}g-mg=0\\ \Rightarrow {\rho }_{sw}V\text{'}g=mg\\ \Rightarrow V\text{'}=\frac{m}{{\rho }_{sw}}\end{array}$

Substituting for $V\text{'}$ in equation $\left(1\right)$

  $f=1-\frac{m}{{\rho }_{sw}V}$  $\left(3\right)$

Calculation:

Substituting the numerical values in equation $\left(3\right)$

  $\begin{array}{l}f=1-\frac{\text{2}{\text{.4×10}}^{\text{6}}\text{ kg}}{\left(\text{1}{\text{.025×10}}^{\text{3}}{\text{ kg/m}}^{\text{3}}\right)\left(\text{2}{\text{.4×10}}^{\text{3}}{\text{ m}}^{\text{3}}\right)}\\ f=2.\text{5\%}\end{array}$

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.

Answer to Problem 92P

  $V_{sw}={\text{ 60 m}}^3$

Explanation of Solution

Given:

Total mass of a submarine (including crew and equipment) is $2.4\times {10}^6\text{ kg}$

The volume of a pressure hull is $2.0\times {10}^3{\text{ m}}^3$ ,

The volume of a ballast tank is $4\times {10}^2{\text{ m}}^3$ ,

Total volume of the submarine is, $\text{V}=\text{ 2}\text{.0}\times {10}^3{\text{ m}}^3\text{ + 4}\text{.0}\times {\text{10}}^2{\text{ m}}^3\text{ = 2}\text{.4}\times {\text{10}}^3{\text{ m}}^3$

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, $V_{sw}=\frac{m_{sw}}{{\rho }_{sw}}$  $\left(4\right)$

Applying the condition for neutral buoyancy, $\sum F_y=0$ to the submarine,

  $B-w_{sub}-w_{sw}=0$  $\left(5\right)$

Where, $B$ is buoyant force on the submarine,

  $w_{sub}$ is weight of the submarine,

  $w_{sw}$ 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,

  $B={\rho }_{sw}V\text{'}g$

Where, ${\rho }_{sw}$ is the density of sea water, ${\rho }_{sw}=1.025\times {10}^3{\text{ kg/m}}^3$

  $g$ is acceleration due to gravity, $g=9.81{\text{ m/s}}^2$

Weight of the submarine is, $w_{sub}=m_{sub}g$

Where, $m_{sub}$ is mass of the submarine.

Weight of the sea water is, $w_{sw}=m_{sw}g$

Where, $m_{sw}$ is mass of the submarine.

Substituting for $B$ , $w_{sub}$ , and $w_{sw}$ in equation $\left(5\right)$ ,

  ${\rho }_{sw}V\text{'}g-m_{sub}g-m_{sw}g=0$

  $\Rightarrow {\rho }_{sw}V\text{'}-m_{sub}-m_{sw}=0$

  $\Rightarrow m_{sw}={\rho }_{sw}V\text{'}-m_{sub}$

Substituting for $m_{sw}$ in equation $\left(4\right)$ ,

  $\begin{array}{l}{V}_{sw}=\frac{{\rho }_{sw}V\text{'}-m_{sub}}{{\rho }_{sw}}\\ V_{sw}=V-\frac{m_{sub}}{{\rho }_{sw}}\left(6\right)\end{array}$

Calculation:

Substitute the numerical values in equation $\left(6\right)$ ,

  $\begin{array}{l}{V}_{sw}=2.4\times {\text{10}}^{\text{3}}{\text{m}}^{\text{3}}\text{-}\frac{\text{2}{\text{.4×10}}^{\text{6}}\text{kg}}{\text{1}{\text{.025×10}}^{\text{3}}{\text{kg/m}}^{\text{3}}}\\ V_{sw}=60{\text{ m}}^3\end{array}$

Conclusion:

The quantity of water must be admitted in to the tanks to give the submarine a neutral buoyancy is $60{\text{ m}}^3$ .