Question 12.10: A circular culvert has a capacity of 0.5 m^3/s when flowing ...
A circular culvert has a capacity of 0.5 \mathrm{~m}^{3} / \mathrm{s} when flowing full. Velocity should not be less than 0.7 m/s if the depth is one-fourth of the diameter. Assuming uniform flow, find the diameter and the slope, taking Manning’s roughness coefficient n = 0.012.
Putting h/d = 1/4 in Eq. (12.41), we get
\frac{h}{d}=\frac{1}{2}-\frac{1}{2} \cos \theta (12.41)
\frac{1}{4}=\frac{1}{2}-\frac{1}{2} \cos \thetaOr \cos \theta=\frac{1}{2}
which gives \theta=\pi / 3 \text { radians. }
From Eq. (12.40), we get
\frac{V}{V_{\text {full }}}=\left(1-\frac{\sin 2 \theta}{2 \theta}\right)^{2 / 3} (12.40)
\frac{V}{V_{\text {full }}}=\left[1-\frac{\sin 2 \pi / 3}{2 \pi / 3}\right]^{2 / 3}= 0.70
Hence, V_{\text {full }}=\frac{V}{0.70}=\frac{0.70}{0.70}=1 \mathrm{~m} / \mathrm{s}
From continuity Q_{\text {full }}=\frac{\pi}{4} d^{2} V_{\text {full }}
Or 0.5=\frac{\pi}{4} d^{2} \times 1
which gives d, the diameter for the culvert = 0.798 m. When flowing full, the hydraulic radius
R_{h}=A / P=d / 4=0.798 / 4=0.1995 \mathrm{~m}From Eq. (12.9)
V=(1 / n) R_{h}^{2 / 3} S^{1 / 2} (12.9)
1=\frac{(0.1995)^{2 / 3}}{0.012} S^{1 / 2}Or S=\frac{(0.012)(0.012)}{(0.1995)^{4 / 3}}=0.0012