A tank with vertical sides contains water to a depth of \(1.2 \mathrm{~m}\) and a layer of oil \(800 \mathrm{~mm}\) deep which rests on top of the water. The relative density of the oil is \(0.85\) and above the oil is air at atmospheric pressure. In one side of the tank, extending through its full height, is a protrusion in the form of a segment of a vertical circular cylinder. This is of radius \(700 \mathrm{~mm}\) and is riveted across an opening \(500 \mathrm{~mm}\) wide in the plane wall of the tank. Calculate the total horizontal thrust tending to force the protrusion away from the rest of the tank and the height of the line of action of this thrust above the base of the tank.

Step by step solution

01

Convert All Measurements to Consistent Units

First, convert all the depths and lengths to meters: - Depth of oil layer: \(800 \mathrm{~mm} = 0.8 \mathrm{~m}\)- Radius of cylinder: \(700 \mathrm{~mm} = 0.7 \mathrm{~m}\)- Width of opening: \(500 \mathrm{~mm} = 0.5 \mathrm{~m}\)Using meters will ensure consistency in calculations.

02

Calculate the Pressure Due to Water

The pressure due to the water at the interface between oil and water is given by \(P_w = \rho_w g h_w\), where - \(\rho_w\) is the density of water (\(1000 \mathrm{~kg/m^3}\)) - \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)) - \(h_w\) is the depth of water (\(1.2 \mathrm{~m}\))Calculate the pressure: \[P_w = 1000 \times 9.81 \times 1.2 = 11772 \mathrm{~Pa}\]

03

Calculate the Pressure Due to Oil

The pressure due to the oil at the interface is given by \(P_o = \rho_o g h_o + P_w\), where - \(\rho_o\) is the density of the oil (\(\rho_o = 0.85 \times 1000 = 850 \mathrm{~kg/m^3}\)) - \(h_o\) is the depth of the oil (\(0.8 \mathrm{~m}\))Calculate the pressure: \[P_o = 850 \times 9.81 \times 0.8 + 11772 = 6666.6 + 11772 = 18438.6 \mathrm{~Pa}\]

04

Calculate the Total Force Due to Water

The total horizontal force due to water, \(F_w\), is calculated using the formula: \(F_w = \frac{1}{2} \rho_w g h_w^2 A\), where - \(A\) is the area of the opening (\(0.5 \mathrm{~m}\) wide and \(1.2 \mathrm{~m}\) deep)Calculate the force: \[A = 0.5 \mathrm{~m} \times 1.2 \mathrm{~m} = 0.6 \mathrm{~m^2}\]\[F_w = \frac{1}{2} \times 1000 \times 9.81 \times 1.2^2 \times 0.6 = 4234.32 \mathrm{~N}\]

05

Calculate the Total Force Due to Oil

The total horizontal force due to oil, \(F_o\), is calculated using the formula: \(F_o = \frac{1}{2} \rho_o g h_o^2 A\), where - \(A\) is the area of the opening (\(0.5 \mathrm{~m}\) wide and \(0.8 \mathrm{~m}\) deep)Calculate the force: \[A = 0.5 \mathrm{~m} \times 0.8 \mathrm{~m} = 0.4 \mathrm{~m^2}\]\[F_o = \frac{1}{2} \times 850 \times 9.81 \times 0.8^2 \times 0.4 = 1068.16 \mathrm{~N}\]

06

Calculate the Total Horizontal Thrust

Add the forces due to water and oil to find the total horizontal thrust: \[F_{total} = F_w + F_o = 4234.32 + 1068.16 = 5302.48 \mathrm{~N}\]

07

Calculate the Height of Line of Action

The height of the line of action, \(h\), is given by the formula:\(\overline{h} = \frac{\sum F_i h_i}{\sum F_i}\) where \(\sum F_i\) is the total force and \(h_i\) the height of the centroid of each force.Since the forces are trapezoidal, the height is likely decomposable. Since water pressure increases linearly with depth, 2/3 down the depth is calculated simply by evaluating the height for each component.For Water:\(\overline{h_w} = \frac{1.2}{3} = 0.4\)For Oil:\(\frac{0.8}{6} = 0.1333 + 1.2 = 1.333\)\(\frac{5302.48}{1.7333} ≈ 3.06\)

08

Summarize

Therefore we have total thrust and line location, solving to find equity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fluid Statics

Fluid statics, also known as hydrostatics, deals with fluids at rest. In such situations, the primary concern is understanding how different layers of liquids exert pressure due to their weight. Consider a tank with water and oil layers. Each layer generates pressure based on its depth and density. This concept helps in solving problems related to tanks, dams, and even atmospheric pressure effects.

Pressure Calculations

Pressure is defined as the force applied per unit area. In fluid mechanics, pressure increases with depth due to the weight of the fluid above. For a given depth, the pressure in a fluid is calculated by the formula: \( P = \rho gh \), where \( \rho \) is fluid density, \( g \) is gravitational acceleration, and \( h \) is the height of the fluid column. In our exercise, we calculate pressure at the oil-water interface and the oil layer's top pressure.

Hydrostatic Force

The hydrostatic force is the force exerted by a static fluid on a surface due to pressure. When dealing with vertical surfaces, like the tank's side, this force is distributed linearly with depth. The total horizontal hydrostatic force due to a fluid layer is given by: \( F = \frac{1}{2} \rho g h^2 A \), where \( A \) is the surface area. Summing forces from different fluid layers gives us the total horizontal thrust on the protrusion.

Buoyancy

Buoyancy is the upward force experienced by objects submerged in a fluid, causing them to float or rise. It is governed by Archimedes' principle, which states that the buoyant force is equal to the weight of the displaced fluid. Although not directly tackled in the exercise, the buoyancy concept is essential in understanding how objects interact with different fluid layers and their stability within those fluids.