Problem Statement
A rectangular channel carries water at the rate of 500 litres/s when bed slope is 1 in 3000. Find the most economical dimensions of the channel if C = 60.
Given Data & Constants
- Discharge, \(Q = 500 \, \text{L/s} = 0.5 \, \text{m}^3/\text{s}\)
- Bed slope, \(i = 1 \text{ in } 3000 = \frac{1}{3000}\)
- Chezy's constant, \(C = 60\)
Solution
1. Conditions for a Most Economical Rectangular Section
For a rectangular channel to be most economical (i.e., have the minimum wetted perimeter for a given area), two conditions must be met:
- The width is twice the depth: \(B = 2d\)
- The hydraulic mean depth is half the depth: \(m = d/2\)
2. Express Area and Hydraulic Mean Depth in Terms of Depth (d)
3. Use Chezy's Formula to Solve for Depth (d)
We start with the discharge equation and substitute the expressions from the previous step.
4. Calculate the Width of the Channel (B)
Using the condition for the most economical section:
The most economical dimensions of the channel are:
Width (B) \( \approx 1.26 \, \text{m} \), Depth (d) \( \approx 0.63 \, \text{m} \)
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, slope, and roughness. This is achieved by minimizing the wetted perimeter (\(P\)).
A smaller wetted perimeter means less contact area between the water and the channel walls, which in turn means less frictional resistance to flow. For a rectangular channel, the optimal shape to minimize this friction occurs when the width is exactly twice the depth. This condition maximizes the hydraulic mean depth (\(m = A/P\)), which is a key factor in the velocity calculation.
