Question

Determine the total force, in \(\mathrm{kN}\), on the bottom of a \(100 \times 50 \mathrm{~m}\) swimming pool. The depth of the pool varies linearly along its length from \(1 \mathrm{~m}\) to \(4 \mathrm{~m}\). Also, determine the pressure on the floor at the center of the pool, in \(\mathrm{kPa}\). The atmospheric pressure is \(0.98\) bar, the density of the water is \(998.2 \mathrm{~kg} / \mathrm{m}^{3}\), and the local acceleration of gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\).

Short answer

Total force: 122,277.5 kN; Pressure at center: 122.46 kPa

Step-by-step solution

Calculate the average depth of the pool

The depth of the pool varies linearly from 1 m to 4 m. Therefore, the average depth (\bar{h}) can be calculated as:\[ \bar{h} = \frac{1 \text{ m} + 4 \text{ m}}{2} = 2.5 \text{ m} \]

Calculate the volume of the pool

The dimensions of the pool are 100 m by 50 m. Using these dimensions and the average depth, the volume (V) of the pool is:\[ V = \text{length} \times \text{width} \times \bar{h} = 100 \text{ m} \times 50 \text{ m} \times 2.5 \text{ m} = 12,500 \text{ m}^3 \]

Calculate the total force on the bottom of the pool

The force (F) on the bottom of the pool can be calculated using the hydrostatic pressure formula. The pressure at the bottom due to the water is given by:\[ P = \rho g h \]where \( \rho = 998.2 \text{ kg/m}^3 \) is the density of water, \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity, and \( h = 2.5 \text{ m} \) is the average depth.Therefore, the pressure on the bottom is:\[ P = 998.2 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \times 2.5 \text{ m} = 24,455.5 \text{ Pa} \]The area (A) of the bottom is:\[ A = 100 \text{ m} \times 50 \text{ m} = 5,000 \text{ m}^2 \]Thus, the total force is:\[ F = P \times A = 24,455.5 \text{ Pa} \times 5,000 \text{ m}^2 = 122,277,500 \text{ N} = 122,277.5 \text{ kN} \]

Calculate the atmospheric pressure in pascals

The atmospheric pressure is given as 0.98 bar. To convert this pressure to pascals, use the conversion factor 1 bar = 100,000 Pa:\[ P_{\text{atm}} = 0.98 \text{ bar} = 0.98 \times 100,000 \text{ Pa} = 98,000 \text{ Pa} \]

Calculate the pressure at the center of the pool

The center of the pool has a depth of 2.5 m. Thus, the pressure due to the water at the center of the pool is the same as the one calculated in Step 3, which is 24,455.5 Pa.The total pressure at the center, including the atmospheric pressure, is:\[ P_{\text{total}} = P + P_{\text{atm}} = 24,455.5 \text{ Pa} + 98,000 \text{ Pa} = 122,455.5 \text{ Pa} \]Converting this to kilopascals (kPa):\[ P_{\text{center}} = \frac{122,455.5 \text{ Pa}}{1,000} = 122.46 \text{ kPa} \]

Key concepts

Fluid Statics

Fluid statics, also known as hydrostatics, is the branch of fluid mechanics that studies fluids at rest. Understanding fluid statics is critical when dealing with problems involving submerged surfaces such as the bottom of a swimming pool. In this context, key concepts include pressure variation with depth and the impact of gravity on fluid pressure. The fundamental principle is that pressure increases with the depth of the fluid, which is why fluids exert a force on the container walls based on the fluid's weight above each point.
The key equation often used is the hydrostatic pressure formula:\[ P = \rho g h \]where:

• \(P\) is the pressure at a particular depth,
• \(\rho\) is the fluid density,
• \(g\) is the acceleration due to gravity,
• \(h\) is the depth of the fluid.

This formula helps in determining the pressure at any given depth of the fluid, which is essential for calculating the force exerted on the bottom or sides of a container.

Pressure Calculation

Pressure calculation is a vital concept in fluid mechanics and fluid statics. Accurate calculation of pressure at various points helps in determining the force exerted by the fluid on different surfaces. For example, in our swimming pool problem, the pressure at the bottom of the pool was calculated using the hydrostatic pressure formula:\[ P = 998.2 \ \text{kg/m}^3 \times 9.8 \ \text{m/s}^2 \times 2.5 \ \text{m} = 24,455.5 \ \text{Pa} \]This calculated pressure was then used to find the total force on the bottom of the pool. Moreover, to get the greatest accuracy, it's crucial to include the atmospheric pressure. This ensures the total pressure exerted on submerged objects also accounts for air pressure, making the calculation more precise:\[ P_{\text{total}} = P + P_{\text{atm}} \]Considering these calculations can simplify complex problems and lead to accurate results, as we saw in the case of determining pressure at the center of the swimming pool.

Force Distribution

Force distribution across a fluid container's surface is another essential aspect of fluid statics. The force exerted by a fluid in a container is determined by the pressure at various depths and the area over which the pressure acts. In our exercise, we first calculated the pressure at the bottom of the pool and then found the area to determine the force:\[ F = P \times A \]where:

• \( F \) is the force on the surface,
• \( P \) is the pressure at a given depth,
• \( A \) is the area over which the pressure acts.

Recognizing that pressure varies linearly with depth allows us to estimate an average pressure for uniformly varying depths, leading to a more manageable calculation. Understanding these concepts ensures we can predict and design structures that can withstand the forces exerted by fluids.

Thermodynamics Applications

Thermodynamics applications in fluid mechanics often deal with the interaction between thermal properties and fluid behaviors. While our exercise primarily focused on hydrostatic pressure, it's essential to understand that temperature can affect fluid density and pressure distribution. In practical applications like HVAC systems, fluid behavior under varying temperatures is crucial. Although not directly involved in the swimming pool problem, knowing how temperature influences fluid mechanics is beneficial. For instance, heated pools might experience changes in water density at different depths, leading to variations in pressure distribution that need to be accounted for. Therefore, integrating thermodynamic principles helps in designing and maintaining systems that operate efficiently despite changing thermal conditions.