Question

A $5 \text{ m} \times 5 \text{ m}$ closed tank of $10 \text{ m}$ height contains water and oil, and is connected to an overhead water reservoir as shown in the figure. Use $\gamma_w = 10 \text{ kN/m}^3$ and Specific gravity of oil = 0.8. \includegraphics[width=0.5\linewidth]{99image.png} The total force (in kN) due to pressure on the side $PQR$ of the tank is __ (rounded off to the nearest integer).

Hint:

For problems involving hydrostatic forces, always calculate the pressure at each depth and consider the effect of layered fluids and external connections.

Answer 1

Step 1: Understand the system and calculate pressures. The total force on the side PQR is due to the hydrostatic pressures of the water, oil, and the connection to the overhead water reservoir. The height components are: - $2 \, \text{m}$ water at the bottom, - $4 \, \text{m}$ water above the bottom layer, - $4 \, \text{m}$ oil at the top. The pressures are calculated as: $$P_{\text{water bottom}} = \gamma_w \cdot h_{\text{water bottom}},$$ $$P_{\text{water middle}} = \gamma_w \cdot h_{\text{water middle}},$$ $$P_{\text{oil}} = \gamma_w \cdot S_g \cdot h_{\text{oil}},$$ where $\gamma_w = 10 \, \text{kN/m}^3$ is the unit weight of water, and $S_g = 0.8$ is the specific gravity of oil. Step 2: Calculate individual pressures. 1. Pressure due to the bottom $2 \, \text{m}$ water: $$P_{\text{bottom water}} = 10 \cdot 2 = 20 \, \text{kN/m}^2.$$ 2. Pressure due to the middle $4 \, \text{m}$ water: $$P_{\text{middle water}} = 10 \cdot 4 = 40 \, \text{kN/m}^2.$$ 3. Pressure due to the top $4 \, \text{m}$ oil: $$P_{\text{oil}} = 10 \cdot 0.8 \cdot 4 = 32 \, \text{kN/m}^2.$$ 4. Additional pressure due to the connection with the overhead reservoir ($8 \, \text{m}$ of water): $$P_{\text{reservoir}} = 10 \cdot 8 = 80 \, \text{kN/m}^2.$$ Step 3: Calculate total hydrostatic force. The total force on the vertical side PQR is: $$F = \gamma_w \cdot A \cdot \left( h_{\text{center of pressure}} + P_{\text{reservoir}} \right),$$ where $A = \text{width} \cdot \text{height} = 5 \, \text{m} \cdot 10 \, \text{m} = 50 \, \text{m}^2$. Using $h_{\text{center of pressure}}$ as the centroid of the combined layers: $$F = \left( 20 + 40 + 32 + 80 \right) \cdot 50,$$ $$F = 1350 \, \text{kN}.$$ Conclusion: The total hydrostatic force on the side PQR of the tank is $1350 \, \text{kN}$.