Question

As shown, a round viewing window of diameter D = 0.8 m is situated in a large tank of seawater (specific gravity S = 1.03). The top of the window is 1.2 m below the water surface, and the window is angled at 60° with respect to the horizontal. Find the hydrostatic force acting on the window and locate the corresponding center of pressure (CP).

Answer 1

Concepts and reason

This question is based on the concepts hydrostatic force and center of pressure.

Hydrostatic Force: A fluid always exerts a pressure on the submerged bodies. This pressure causes a force on the submerged bodies known as hydrostatic force.

Center of pressure: On a submerged body pressure force acts through a point, this particular point is known as the center of pressure.

Specific gravity: Specific gravity of a fluid is defined as the ratio of the density of fluid to density of the reference fluid. Most of the times water is considered as the reference fluid.

Calculate the vertical distance of the centroid of the window from the free surface. Use the expression for the hydrostatic force applied on a submerged object in order to calculate the value of hydrostatic force on the window.

Calculate the second area moment of inertia of the window about the axis passing through the centroid of the window. Use the expression for the center of pressure in order to calculate the distance of center of pressure of the window.

Fundamentals

The expression of hydrostatic force is,

$F = \rho gA\bar h$

Here, $F$ is the hydrostatic force that acts on the object , ${\rho _s}$ is the fluid, $g$ is the acceleration due to gravity and $\bar h$ is the center of gravity of window.

The expression of the center of pressure of an object is written as,

${h^*} = \bar h + \frac{{{I_G}{{\sin }^2}\theta }}{{A\bar h}}$

Here, ${h^*}$ is the center of pressure, ${I_G}$ is the moment of inertia of object about center of gravity, and $\theta$ is the inclination angle.

The expression of moment of inertia of circular section about center of gravity is written as,

${I_G} = \frac{{\pi {D^4}}}{{64}}$

Here, $D$ is the diameter of the window.

The area of circular section is expressed as,

$A = \frac{{\pi {D^2}}}{4}$

Here, $A$ is an area of the window.

The density of seawater is written as,

${\rho _s} = \rho S$

Here, $\rho$ is the density of water and ${\rho _s}$ is density of the sea water.

Calculate the area of the window.

$A = \frac{{\pi {D^2}}}{4}$

Substitute $0.80{\rm{ m}}$ for $D$ .

$\begin{array}{c}\\A = \frac{{\pi {{\left( {0.80{\rm{ m}}} \right)}^2}}}{4}\\\\ = 0.5026{\rm{ }}{{\rm{m}}^{\rm{2}}}\\\end{array}$

Write the expression for the vertical distance of centroid of window from free surface.

$\bar h = h + \frac{D}{2}\sin 60^\circ$

Here, $\bar h$ is the vertical distance of the centroid of the window from the free surface of water.

Substitute $1.20{\rm{ m}}$ for $h$ and $0.80{\rm{ m}}$ for $D$ .

$\begin{array}{c}\\\bar h = \left( {1.20{\rm{ m}}} \right) + \frac{{\left( {0.80{\rm{ m}}} \right)}}{2}\sin 60^\circ \\\\ = 1.5464{\rm{ m}}\\\end{array}$

Calculate the density of the sea water.

${\rho _s} = \rho S$

Here, ${\rho _s}$ is the density of the sea water, $\rho$ is density of fresh water.

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Substitute $1000{\rm{ kg/}}{{\rm{m}}^{\rm{3}}}$ for $\rho$ and $1.03$ for $S$ .

$\begin{array}{c}\\{\rho _s} = \left( {1000{\rm{ kg/}}{{\rm{m}}^{\rm{3}}}} \right)\left( {1.03} \right)\\\\ = 1030{\rm{ kg/}}{{\rm{m}}^3}\\\end{array}$

Calculate the hydrostatic force on the window.

$F = {\rho _s}gA\bar h$

Here, $F$ is the hydrostatic force that acts on the window, ${\rho _s}$ is the density of seawater and $g$ is the acceleration due to gravity.

Substitute $1030{\rm{ kg/}}{{\rm{m}}^3}$ for ${\rho _s}$ , $9.81{\rm{ m/}}{{\rm{s}}^{\rm{2}}}$ for $g$ , $1.5464{\rm{ m}}$ for $\bar h$ and $0.5026{\rm{ }}{{\rm{m}}^{\rm{2}}}$ for $A$ .

$\begin{array}{c}\\F = \left( {1030{\rm{ kg/}}{{\rm{m}}^{\rm{3}}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {0.5026{\rm{ }}{{\rm{m}}^{\rm{2}}}} \right)\left( {1.5464{\rm{ m}}} \right)\\\\ = 7853.60{\rm{ N}}\left( {\frac{{1{\rm{ kN}}}}{{{{10}^3}{\rm{ N}}}}} \right)\\\\{\rm{ = }}7.854{\rm{ kN}}\\\end{array}$

Write the expression for the moment of inertia of the object.

${I_G} = \frac{{\pi {D^4}}}{{64}}$

Here, ${I_G}$ is a moment of inertia.

Substitute $0.80{\rm{ m}}$ for $D$ .

$\begin{array}{c}\\{I_G} = \frac{{\pi {{\left( {0.80{\rm{ m}}} \right)}^4}}}{{64}}\\\\ = 0.020{\rm{ }}{{\rm{m}}^{\rm{4}}}\\\end{array}$

Write the expression for the center of pressure for the window.

${h^*} = \bar h + \frac{{{I_G}{{\sin }^2}\theta }}{{A\bar h}}$

Here, ${h^*}$ is center of pressure of the window.

Substitute $60^\circ$ for $\theta$ , $0.020{\rm{ }}{{\rm{m}}^{\rm{4}}}$ for ${I_G}$ , $1.5464{\rm{ m}}$ for $\bar h$ and $0.5026{\rm{ }}{{\rm{m}}^{\rm{2}}}$ for $A$ .

$\begin{array}{c}\\{h^*} = \left( {1.5464{\rm{ m}}} \right) + \frac{{\left( {0.020{\rm{ }}{{\rm{m}}^{\rm{4}}}} \right){{\sin }^2}60^\circ }}{{\left( {0.5026{\rm{ }}{{\rm{m}}^{\rm{2}}}} \right)\left( {1.5464{\rm{ m}}} \right)}}\\\\ = 1.565{\rm{ m}}\\\end{array}$

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Ans:

The hydrostatic force that acts on the window is $7.854{\rm{ kN}}$ .

The center of pressure of window is $1.565{\rm{ m}}$ from top of the surface of water.