Answer :
To determine the necessary width of a rectangular channel carrying a discharge of 9 m³/s with a normal depth of 1.5 m, laid on a slope of 0.0005 with smooth concrete walls, we can use the Manning's equation for open channel flow.
The Manning's equation is given by:
[tex]Q = \frac{1}{n} A R^{2/3} S^{1/2}[/tex]
Where:
- [tex]Q[/tex] = discharge (9 m³/s)
- [tex]n[/tex] = Manning's roughness coefficient (0.013 for smooth concrete)
- [tex]A[/tex] = cross-sectional area of flow (m²)
- [tex]R[/tex] = hydraulic radius (m)
- [tex]S[/tex] = channel slope (0.0005)
The cross-sectional area [tex]A[/tex] of a rectangular channel is:
[tex]A = b \times y[/tex]
Where:
- [tex]b[/tex] = width of the channel (m)
- [tex]y[/tex] = depth of flow (1.5 m)
The hydraulic radius [tex]R[/tex] is calculated as:
[tex]R = \frac{A}{P}[/tex]
Where [tex]P[/tex] is the wetted perimeter. For a rectangular channel:
[tex]P = b + 2y[/tex]
Substituting [tex]A[/tex] and [tex]P[/tex] into the hydraulic radius:
[tex]R = \frac{b \times y}{b + 2y}[/tex]
Now, substitute the expressions for [tex]A[/tex] and [tex]R[/tex] into Manning’s equation:
[tex]9 = \frac{1}{0.013} \times (b \times 1.5) \left( \frac{1.5b}{b + 3} \right)^{2/3} (0.0005)^{1/2}[/tex]
Simplifying the equation, we need to solve for [tex]b[/tex]. Calculations can be somewhat complex analytically, so numerical methods or iterative approaches like trial and error can be used to find the optimal width [tex]b[/tex].
After solving this equation iteratively, we find that the channel width [tex]b[/tex] is approximately 2.5 m.
Therefore, a rectangular channel with a width of 2.5 meters is necessary to carry a discharge of 9 m³/s with a normal depth of 1.5 m over a 0.0005 slope with a Manning's coefficient of 0.013.
