Problem Statement
Find the discharge through a rectangular channel 3 m wide, having depth of water 2 m and bed slope as 1 in 1500. Take the value of N = 0.012 in Manning's formula.
Given Data & Constants
- Width of channel, \(B = 3 \, \text{m}\)
- Depth of water, \(d = 2 \, \text{m}\)
- Bed slope, \(i = 1 \text{ in } 1500 = \frac{1}{1500}\)
- Manning's roughness coefficient, \(N = 0.012\)
Solution
1. Calculate Geometric Properties
$$ \text{Area of flow, } A = B \times d = 3 \times 2 = 6 \, \text{m}^2 $$
$$ \text{Wetted Perimeter, } P = B + 2d = 3 + 2 \times 2 = 7 \, \text{m} $$
$$ \text{Hydraulic Mean Depth, } m = \frac{A}{P} = \frac{6}{7} \approx 0.857 \, \text{m} $$
2. Calculate Velocity (V) using Manning's Formula
The metric version of Manning's formula is used to find the flow velocity.
$$ V = \frac{1}{N} m^{2/3} i^{1/2} $$
$$ V = \frac{1}{0.012} \times (0.857)^{2/3} \times \left(\frac{1}{1500}\right)^{1/2} $$
$$ V = 83.333 \times 0.902 \times 0.0258 $$
$$ V \approx 1.94 \, \text{m/s} $$
3. Calculate Discharge (Q)
The discharge is the area multiplied by the velocity.
$$ Q = A \times V = 6 \, \text{m}^2 \times 1.94 \, \text{m/s} \approx 11.64 \, \text{m}^3/\text{s} $$
Final Result:
The discharge through the rectangular channel is approximately \(11.64 \, \text{m}^3/\text{s}\).
Explanation of Manning's Formula
Manning's formula is another widely used empirical equation for calculating the velocity of uniform flow in an open channel. It is often preferred over Bazin's formula in modern engineering practice.
The formula, \(V = \frac{1}{N} m^{2/3} i^{1/2}\), relates the velocity to:
- Manning's Roughness Coefficient (N): An empirically derived coefficient that represents the roughness of the channel lining. A lower value of N means a smoother channel and a higher velocity.
- Hydraulic Mean Depth (m): The ratio of the flow area to the wetted perimeter, representing the channel's geometric efficiency.
- Bed Slope (i): The gradient of the channel, which provides the gravitational force to drive the flow.
