Problem Statement
A trapezoidal channel with side slopes of 1 to 1 has to be designed to convey 9 m³/s at a velocity of 1.5 m/s so that the amount of concrete lining for the bed and sides is the minimum. Calculate the area of lining required for one metre length of canal.
Given Data & Constants
- Side slope = 1 Horizontal to 1 Vertical, so \(n = 1\).
- Discharge, \(Q = 9 \, \text{m}^3/\text{s}\)
- Velocity, \(V = 1.5 \, \text{m/s}\)
Solution
1. Calculate the Required Flow Area (A)
2. Find the Dimensions for the Most Economical Section
For a trapezoidal channel to be most economical, the condition is that half of the top width equals the length of one sloping side.
Now substitute this relationship into the formula for the area.
3. Calculate the Area of Lining Required
The area of lining required for one meter length of the canal is equal to the wetted perimeter (P) of the channel.
The area of lining required for one metre length of the canal is approximately \(6.62 \, \text{m}^2\).
Explanation of "Most Economical Section"
The "most economical" or "most efficient" channel section is the one that can pass the maximum discharge for a given cross-sectional area. This is achieved by minimizing the wetted perimeter (\(P\)). A smaller wetted perimeter means less contact area between the water and the channel lining, which results in less frictional resistance and requires the least amount of lining material, thus minimizing construction cost.
For a trapezoidal channel, this optimal shape is achieved when the three wetted sides (the bed and the two sloping sides) are tangential to a semicircle with its center on the water's surface. This geometric condition leads to the specific relationships between width, depth, and side slope used in the calculation.
