Problem Statement
A pipeline which is 4 m in diameter contains a gate valve. The pressure at the centre of the pipe is 19.6 N/cm². If the pipe is filled with oil of specific gravity 0.87, find the force exerted by the oil upon the gate and the position of the centre of pressure.
Given Data
- Diameter of pipe, \(d = 4 \, \text{m}\)
- Area of gate, \(A = \frac{\pi}{4}d^2 = \frac{\pi}{4}(4)^2 = 4\pi \approx 12.566 \, \text{m}^2\)
- Pressure at centre, \(p = 19.6 \, \text{N/cm}^2 = 19.6 \times 10^4 \, \text{N/m}^2\)
- Specific gravity of oil, \(S = 0.87\)
- Density of oil, \(\rho_{oil} = 0.87 \times 1000 = 870 \, \text{kg/m}^3\)
Diagram
Visual representation of the gate valve inside the pipeline.
Solution
(i) Force on the Gate Valve
First, we convert the pressure at the center of the pipe to an equivalent pressure head (\(\bar{h}\)) of oil. This represents the height of an imaginary free surface of oil that would exert the same pressure.
Now, we can find the total force on the gate using this pressure head as the depth of the centroid:
Alternatively, since \(p = \rho_{oil} g \bar{h}\), we can simplify this to \(F = p A\):
(ii) Position of Centre of Pressure
The depth of the centre of pressure (\(h^*\)) from the imaginary free surface is found using:
Substitute the known values:
Force on the gate: \( F \approx 2.463 \, \text{MN} \)
Position of Centre of Pressure: \( h^* \approx 23.0315 \, \text{m} \) from the equivalent free surface
Explanation of Concepts
Pressure Head (\(\bar{h}\)): In problems involving pressurized pipes where there is no real "free surface", it is convenient to calculate an equivalent one. The pressure head is the height of a column of the specific fluid that would be required to produce the given pressure at that point. This allows us to use standard hydrostatic formulas.
Force on the Gate (\(F\)): This is the total force the gate must withstand. Since the pressure in a large-diameter pipe is substantial, the force on the gate is immense. In this case, it's over 2.4 million Newtons, equivalent to the weight of about 250 metric tons.
Centre of Pressure (\(h^*\)): This is the point on the gate where the entire hydrostatic force can be considered to act. It is crucial for the mechanical design of the gate and its operating mechanism.
Physical Meaning
The result shows that the centre of pressure (\(h^* \approx 23.03 \, \text{m}\)) is very close to the centre of the pipe (\(\bar{h} \approx 22.99 \, \text{m}\)). The difference is only about 4.35 cm.
This happens because the initial pressure in the pipe is very high, corresponding to a pressure head of nearly 23 meters. The additional pressure variation from the top of the 4-meter pipe to the bottom is small in comparison. Therefore, the pressure distribution across the gate is almost uniform. When pressure is uniform, the centre of pressure coincides with the geometric centre (centroid). The slight difference shows that there is still a marginally higher pressure at the bottom of the gate, but for many engineering purposes in high-pressure systems, the force can be assumed to act at the center without significant error.
