GATE CE 2024 Set 1 — Question 59

The problem asks for the total hydrostatic force on the vertical side PQR of the tank. The side has a width of 5 m.Given:
Width of the tank side, $b = 5$ m
Unit weight of water, $\gamma_w = 10 \text{ kN/m}^3$
Specific gravity of oil, $S.G._{oil} = 0.8$
Unit weight of oil, $\gamma_{oil} = S.G._{oil} \times \gamma_w = 0.8 \times 10 = 8 \text{ kN/m}^3$The tank on the right contains 2 m of oil on top of 4 m of water. The side PQR consists of two parts: QR (in contact with oil, height $h_{oil} = 2$ m) and PQ (in contact with water, height $h_{water} = 4$ m).The tank is connected to an overhead reservoir. The connection pipe is at the level of the oil surface (level R) and is 2 m below the free surface of the water in the reservoir. This means the pressure at the oil surface is equivalent to the pressure from 2 m of water.Pressure at the oil surface (level R), $p_R = \gamma_w \times 2 \text{ m} = 10 \text{ kN/m}^3 \times 2 \text{ m} = 20 \text{ kN/m}^2$.Pressure at the oil-water interface (level Q), $p_Q = p_R + \gamma_{oil} \times h_{oil} = 20 + (8 \times 2) = 20 + 16 = 36 \text{ kN/m}^2$.Pressure at the bottom of the tank (level P), $p_P = p_Q + \gamma_w \times h_{water} = 36 + (10 \times 4) = 36 + 40 = 76 \text{ kN/m}^2$.The total force is the sum of the force on the area QR ($F_{QR}$) and the force on the area PQ ($F_{PQ}$).Force on area QR (in contact with oil):
Area $A_{QR} = h_{oil} \times b = 2 \text{ m} \times 5 \text{ m} = 10 \text{ m}^2$.
The force can be calculated as the volume of the pressure prism on this area. The pressure distribution is trapezoidal.
$F_{QR} = \left( \frac{p_R + p_Q}{2} \right) \times A_{QR} = \left( \frac{20 + 36}{2} \right) \times 10 = \frac{56}{2} \times 10 = 28 \times 10 = 280 \text{ kN}$.Force on area PQ (in contact with water):
Area $A_{PQ} = h_{water} \times b = 4 \text{ m} \times 5 \text{ m} = 20 \text{ m}^2$.
The pressure distribution is also trapezoidal.
$F_{PQ} = \left( \frac{p_Q + p_P}{2} \right) \times A_{PQ} = \left( \frac{36 + 76}{2} \right) \times 20 = \frac{112}{2} \times 20 = 56 \times 20 = 1120 \text{ kN}$.Total force on the side PQR:
$F_{total} = F_{QR} + F_{PQ} = 280 \text{ kN} + 1120 \text{ kN} = 1400 \text{ kN}$.The total force is 1400 kN.