The pressure at the centre of a pipe of diameter 3 m is 29.43 N/cm². The pipe contains oil of sp. gr. 0.87 and is fitted with a gate valve. Find the force exerted by the oil on the gate and the position of the centre of pressure.

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Pressurized Pipe Gate Problem

Problem Statement

The pressure at the centre of a pipe of diameter 3 m is 29.43 N/cm². The pipe contains oil of sp. gr. 0.87 and is fitted with a gate valve. Find the force exerted by the oil on the gate and the position of the centre of pressure.

Given Data

  • Diameter of pipe/gate, \( D = 3 \, \text{m} \)
  • Pressure at centre, \( p = 29.43 \, \text{N/cm}^2 = 294300 \, \text{N/m}^2 \)
  • Specific Gravity of oil, \( S.G. = 0.87 \)
  • Density of oil, \( \rho = 0.87 \times 1000 = 870 \, \text{kg/m}^3 \)

Solution

(i) Force on the Gate (\(F\))

The total force on the gate is the pressure at its centroid multiplied by the area of the gate.

$$ A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (3)^2 \approx 7.0686 \, \text{m}^2 $$ $$ F = p \times A $$ $$ F = 294300 \times 7.0686 $$ $$ F \approx 208009 \, \text{N} $$

(ii) Position of Centre of Pressure (\(h^*\))

For a pressurized pipe, we first convert the pressure at the centroid into an equivalent pressure head (\(\bar{h}\)). This represents the height of an imaginary free surface above the pipe's center.

$$ \bar{h} = \frac{p}{\rho g} $$ $$ \bar{h} = \frac{294300}{870 \times 9.81} \approx 34.50 \, \text{m} $$

Now, we can find the depth of the centre of pressure, \(h^*\), using the standard formula.

$$ I_G = \frac{\pi D^4}{64} = \frac{\pi (3)^4}{64} \approx 3.976 \, \text{m}^4 $$ $$ h^* = \frac{I_G}{A\bar{h}} + \bar{h} $$ $$ h^* = \frac{3.976}{7.0686 \times 34.50} + 34.50 $$ $$ h^* \approx 0.0163 + 34.50 = 34.5163 \, \text{m} $$
Final Results:

Force on the gate: \( F \approx 208.01 \, \text{kN} \).

Position of centre of pressure: \( h^* \approx 34.516 \, \text{m} \) below the imaginary free surface, or 0.016 m below the pipe's centerline.

Explanation of Concepts

Pressurized Systems: When a fluid is under pressure (not open to the atmosphere), there isn't a physical free surface. To solve for hydrostatic forces, we create an imaginary one. The height of this imaginary surface above the point of interest is called the "pressure head" or "equivalent head" and is calculated as \(h = p / (\rho g)\).

Force Calculation: Once the pressure at the centroid is known, the total force is simply this pressure multiplied by the area. This is because the average pressure over the entire surface is the pressure at its centroid.

Centre of Pressure in Pressurized Pipes: The centre of pressure is found using the same formula as for unpressurized liquids, but the depth of the centroid (\(\bar{h}\)) is the calculated pressure head. When the pressure head is very large compared to the dimensions of the gate (as in this case, 34.5 m vs 3 m), the centre of pressure will be very close to the centroid.

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