91Ó°ÊÓ

When dealing with fluids in a tank, hydrostatic pressure plays a crucial role. It is the pressure exerted by a fluid at equilibrium due to the force of gravity acting on it. The deeper you dive into the liquid, the higher the pressure becomes.
For a tank filled with ethyl alcohol to a height of 1.5 meters above the top of the tank, we calculate hydrostatic pressure using the formula:

• \( P = \gamma \cdot h \)

where \( \gamma \) is the specific weight of the fluid (in this case, ethyl alcohol) and \( h \) is the vertical distance from the surface of the liquid to the point where we need the pressure.
In the exercise, the ethyl alcohol exerts a pressure at the bottom of an additional 1-meter radius semicircular cylindrical tank part, totalling 2.5 meters. Plugging this into the equation:

• \( P = 7915 \times 2.5 = 19787.5 \text{ N/m}^2 \)

This pressure is crucial for determining the resultant force on the tank, as it indicates how much force is being exerted over the surface area.

The force resulting from hydrostatic pressure over a given area is what we refer to as the resultant force. It acts perpendicular to the surface and is critical in designing storage tanks, pipes, and more.
To find this force in our cylindrical tank example, multiply the pressure by the area of the circle that forms the end of the tank. The formula used is:

• \( F = P \cdot A \)

The area of the tank's end is calculated using the formula for the area of a circle, \( A = \pi r^2 \).
Given the radius is 1 meter, the area becomes:

• \( A = \pi (1)^2 = 3.1416 \text{ m}^2 \)

Therefore, the resultant force is:

• \( F = 19787.5 \cdot 3.1416 = 62138.47 \text{ N} \)

The significance of this force is to understand how much pressure the tank's walls must withstand, ensuring that they remain intact and don’t yield.

The center of pressure is the point on a submerged surface at which the resultant hydrostatic pressure force acts. It is essential because it tells you where the sum of all the pressures can be concentrated to replicate the actual physical scenario.
To find the center of pressure in this exercise, the formula used is:

• \( h_{cp} = \frac{I_g}{A \cdot \bar{h}} + \bar{h} \)

\( I_g \) is the moment of inertia of the figure's area about its horizontal axis, and \( \bar{h} \) is the centroid's average depth from the liquid's surface.
In this case, \( I_g \) is calculated by:

• \( I_g = \frac{\pi \times (1)^4}{4} = 0.7854 \text{ m}^4 \)

Thus, the center of pressure can be found:

• \( h_{cp} = \frac{0.7854}{3.1416 \times 1.25} + 1.25 = 1.5 \text{ m} \)

This outcome indicates the uniform distribution of the force across the tank's surface, which is crucial in ensuring that the tank's design accommodates even pressure distribution.