Problem Statement
Water is flowing through a circular channel at the rate of 500 litres/s. The depth of water in the channel is 0.7 times the diameter and the slope of the bed of the channel is 1 in 8000. Find the diameter of the circular channel if the value of Manning's, N = 0.015.
Given Data & Constants
- Discharge, \(Q = 500 \, \text{L/s} = 0.5 \, \text{m}^3/\text{s}\)
- Depth of flow, \(d = 0.7 D\)
- Bed slope, \(i = 1 \text{ in } 8000 = \frac{1}{8000}\)
- Manning's roughness coefficient, \(N = 0.015\)
Solution
1. Define Geometric Properties in Terms of Diameter (D)
First, we find the half-angle (\(\alpha\)) subtended by the water surface at the pipe's center. Let \(R = D/2\).
$$ \cos(\alpha) = \frac{R - d}{R} = \frac{0.5D - 0.7D}{0.5D} = \frac{-0.2D}{0.5D} = -0.4 $$
$$ \alpha = \arccos(-0.4) \approx 1.9823 \, \text{radians} $$
Now we express the wetted area (A) and hydraulic mean depth (m) as functions of D.
$$ \text{Area, } A = R^2 (\alpha - \sin\alpha \cos\alpha) = \left(\frac{D}{2}\right)^2 (1.9823 - \sin(1.9823) \times (-0.4)) $$
$$ A = \frac{D^2}{4} (1.9823 - (0.982 \times -0.4)) = \frac{D^2}{4} (2.3751) \approx 0.5938 D^2 $$
$$ \text{Wetted Perimeter, } P = 2 R \alpha = 2 \left(\frac{D}{2}\right) \times 1.9823 = 1.9823 D $$
$$ \text{Hydraulic Mean Depth, } m = \frac{A}{P} = \frac{0.5938 D^2}{1.9823 D} \approx 0.2995 D $$
2. Apply Manning's Formula and Solve for D
We substitute these geometric expressions into the Manning's discharge formula.
$$ Q = A \times \frac{1}{N} m^{2/3} i^{1/2} $$
$$ 0.5 = (0.5938 D^2) \times \frac{1}{0.015} \times (0.2995 D)^{2/3} \times \left(\frac{1}{8000}\right)^{1/2} $$
$$ 0.5 = (0.5938 D^2) \times 66.67 \times (0.2995^{2/3} D^{2/3}) \times (0.01118) $$
$$ 0.5 = (0.5938 \times 66.67 \times 0.447 \times 0.01118) \times D^{2 + 2/3} $$
$$ 0.5 \approx 0.1978 \times D^{8/3} $$
Finally, we solve for the diameter D.
$$ D^{8/3} = \frac{0.5}{0.1978} \approx 2.5278 $$
$$ D = (2.5278)^{3/8} \approx 1.43 \, \text{m} $$
Final Result:
The required diameter of the circular channel is approximately \(1.43 \, \text{m}\).
