Problem Statement
A circular opening, 3 m diameter, in a vertical side of a tank is closed by a disc of 3 m diameter which can rotate about a horizontal diameter. Calculate: (i) the force on the disc, and (ii) the torque required to maintain the disc in equilibrium in the vertical position when the head of water above the horizontal diameter is 6 m.
Given Data
- Diameter of disc, \( D = 3 \, \text{m} \)
- Head of water above horizontal diameter (pivot), \( \bar{h} = 6 \, \text{m} \)
- Density of Water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to Gravity, \( g = 9.81 \, \text{m/s}^2 \)
Solution
(i) Force on the Disc (\(F\))
The total force on the disc is the hydrostatic pressure at its centroid multiplied by its area.
(ii) Torque Required (\(T\))
The torque is the product of the force (\(F\)) and the vertical distance between the pivot point (the centroid) and the centre of pressure (\(h^*\)). This distance is the lever arm.
First, we find the depth of the centre of pressure, \(h^*\).
The lever arm is the distance between the centre of pressure and the pivot (centroid).
Finally, we calculate the torque.
(i) Force on the disc: \( F \approx 416.05 \, \text{kN} \).
(ii) Torque required: \( T \approx 39.0 \, \text{kN-m} \).
Explanation of Concepts
Hydrostatic Force: The total force on the disc is calculated based on the pressure at its geometric center (centroid). Since the disc is vertical, its centroid is at a depth of 6 m, as given in the problem.
Torque and Centre of Pressure: The hydrostatic force does not act at the centroid but at a lower point called the centre of pressure. This is because the pressure is greater on the bottom half of the disc than on the top half. The torque required to keep the disc vertical is the moment created by this force acting at the centre of pressure about the pivot point (the horizontal diameter/centroid). The lever arm for this torque is the vertical distance between these two points.
