A penstock made up by a pipe of 2 m diameter contains a circular disc of the same diameter to act as a valve which controls the discharge passing through it. It can rotate about a horizontal diameter. If the head of water above its centre is 20 m, find the total force acting on the disc and the torque required to maintain it in the vertical position.

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Penstock Valve Torque Problem

Problem Statement

A penstock made up by a pipe of 2 m diameter contains a circular disc of the same diameter to act as a valve which controls the discharge passing through it. It can rotate about a horizontal diameter. If the head of water above its centre is 20 m, find the total force acting on the disc and the torque required to maintain it in the vertical position.

Given Data

  • Diameter of disc, \( D = 2 \, \text{m} \)
  • Pivot: Horizontal diameter (centroidal axis)
  • Head of water above pivot, \( \bar{h} = 20 \, \text{m} \)
  • Fluid is water, \( \rho = 1000 \, \text{kg/m}^3 \)

Solution

(i) Total Force on the Disc (\(F\))

First, calculate the area of the disc.

$$ A = \frac{\pi}{4} D^2 $$ $$ A = \frac{\pi}{4} (2)^2 $$ $$ A = \pi \approx 3.1416 \, \text{m}^2 $$

Now, calculate the total force on the disc using the depth to the centroid.

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 3.1416 \times 20 $$ $$ F \approx 616387 \, \text{N} $$

(ii) Torque required to maintain equilibrium (\(T\))

The torque is created because the hydrostatic force \(F\) acts at the centre of pressure, which is below the pivot point (the centroid). The lever arm is the distance between these two points.

First, find the depth of the centre of pressure (\(h^*\)).

$$ I_G = \frac{\pi D^4}{64} $$ $$ I_G = \frac{\pi \times (2)^4}{64} $$ $$ I_G = \frac{16\pi}{64} = \frac{\pi}{4} \approx 0.7854 \, \text{m}^4 $$
$$ h^* = \frac{I_G}{A \bar{h}} + \bar{h} $$ $$ h^* = \frac{0.7854}{3.1416 \times 20} + 20 $$ $$ h^* = \frac{0.7854}{62.832} + 20 $$ $$ h^* \approx 0.0125 + 20 = 20.0125 \, \text{m} $$

The lever arm is the vertical distance between the centre of pressure and the centroid (pivot).

$$ \text{Lever Arm} = h^* - \bar{h} $$ $$ \text{Lever Arm} = 20.0125 - 20.0 $$ $$ \text{Lever Arm} = 0.0125 \, \text{m} $$

Finally, calculate the torque.

$$ T = F \times (\text{Lever Arm}) $$ $$ T = 616387 \times 0.0125 $$ $$ T \approx 7705 \, \text{N-m} $$
Final Results:

(i) Total Force on the disc: \( F \approx 616.39 \, \text{kN} \).

(ii) Torque required: \( T \approx 7.71 \, \text{kN-m} \).

Explanation of Concepts

Total Force: This is the total hydrostatic force exerted by the water on the valve. It is calculated by finding the pressure at the valve's geometric center (centroid) and multiplying it by the total area of the valve.

Torque and Centre of Pressure: The valve is pivoted at its centroid, but the total hydrostatic force acts at a slightly lower point called the centre of pressure. This is because the water pressure is greater on the bottom half of the valve than at the top. The offset between the point of force application (centre of pressure) and the pivot point (centroid) creates a lever arm. The torque required to prevent rotation is the moment produced by the hydrostatic force acting over this lever arm (\( T = F \times \text{Lever Arm} \)). When the head of water is very large compared to the diameter of the valve, this lever arm becomes very small, as seen in the calculation.

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