A tank \(3.5 \mathrm{~m}\) long and \(2.5 \mathrm{~m}\) wide contains alcohol of relative density \(0.82\) to a depth of \(3 \mathrm{~m}\). A \(50 \mathrm{~mm}\) diameter pipe leads from the bottom of the tank. What will be the reading on a gauge calibrated in Pa connected at a point (a) \(150 \mathrm{~mm}\) above the bottom of the tank; \((b)\) in the \(50 \mathrm{~mm}\) diameter pipe, \(2 \mathrm{~m}\) below the bottom of the tank; \((c)\) at the upper end of a \(25 \mathrm{~mm}\) diameter pipe, connected to the \(50 \mathrm{~mm}\) pipe \(2 \mathrm{~m}\) below the bottom of the tank, sloping upwards at \(30^{\circ}\) to the horizontal for \(1.2 \mathrm{~m}\) and then rising vertically for \(600 \mathrm{~mm}\) ? What is the load on the bottom of the tank?

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at a given point within the fluid, due to the force of gravity. This pressure increases with the depth of the fluid. To calculate hydrostatic pressure (\text{P}), we use the formula:
\[ P = \rho g h \]
Here:

• \(\rho\) is the density of the fluid
• \(g\) is the acceleration due to gravity (9.81 \text{m/s}^2)
• \(h\) is the depth of the fluid

In the exercise, the fluid is alcohol with a relative density (specific gravity) of 0.82. Relative density is the ratio of the density of a substance to the density of water. Therefore, the density of alcohol (\(\rho\)) is calculated as:
\[ \rho_{alcohol} = 0.82 \times 1000 \text{kg/m}^3 = 820 \text{kg/m}^3 \]
In part (a), we determine the pressure at a point 150 mm (0.15 m) above the bottom. The effective fluid column height (\(h\)) is 2.85 m, and the pressure at this depth is:
\[ P = 820 \times 9.81 \times 2.85 = 22907.7 \text{Pa} \]
This equation helps you understand how fluid pressure increases with depth.

Density is a physical property that defines mass per unit volume of a substance. It is usually denoted by \(\rho\) and is calculated as:
\[ \rho = \frac{m}{V} \]
Where:

• \(m\) is the mass
• \(V\) is the volume

Relative density, or specific gravity, is the ratio of the density of a fluid compared to the density of water (which is 1000 \text{kg/m}^3 for fresh water). It is dimensionless and often makes calculations easier. In the given problem, alcohol has a relative density of 0.82, so:
\[ \rho_{alcohol} = 0.82 \times \rho_{water} \]
Thus:
\[ \rho_{alcohol} = 0.82 \times 1000 = 820 \text{kg/m}^3 \]
This concept is essential because it directly affects the pressure calculation as shown in our solution steps. Relative density helps simplify calculations by comparing the fluid’s density to that of water.

In fluid systems, pressure distribution and variation are crucial for understanding system behavior. Pressure in fluids at rest is isotropic, meaning it acts equally in all directions and is calculated using hydrostatic principles. In differentiating locations within such a system, calculating pressure levels is critical as shown below:

• At a point 2 m below the tank, the fluid column height sums to 5 m.
• Using \(P = \rho g h\) for 5 m:
  \[ P = 820 \times 9.81 \times 5 = 40110 \text{Pa} \]

Likewise, for complex piping systems, pressure at specific points must account for vertical and sloped components. For a pipe extending 2 m below, then sloping upwards 1.2 m at 30 degrees and rise vertically 0.6 m:

• Height due to slope component:
  \[ h = 1.2 \times \sin(30^\circ) = 0.6 \text{m} \]
• Total height to point in pipe: 5 + 0.6 + 0.6 = 6.2 m
• Calculating final pressure:
  \[ P = 820 \times 9.81 \times 6.2 = 49731.6 \text{Pa} \]

Calculating the load on the bottom of a tank involves finding the pressure exerted by the fluid column and then determining the total force applied over the area. The pressure at the bottom is calculated using the depth of the fluid:
\[ P = \rho g h \]
Here:
\[ P = 820 \times 9.81 \times 3 = 24102 \text{Pa} \]
The area of the tank’s base can be calculated from its dimensions:
\[ A = \text{length} \times \text{width} = 3.5 \text{m} \times 2.5 \text{m} = 8.75 \text{m}^2 \]
The total load (or force) is the product of the pressure and the area:
\[ F = P \times A = 24102 \times 8.75 = 210892.5 \text{N} \]
Understanding this concept ensures accurate overall pressure and load calculations, such as the total force exerted on the tank base in varied fluid mechanics and engineering scenarios.