Question 12.10: A circular culvert has a capacity of 0.5 m^3/s when flowing ...

Putting h/d = 1/4 in Eq. (12.41), we get

\frac{h}{d}=\frac{1}{2}-\frac{1}{2} \cos \theta                 (12.41)

Or     \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)

= 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

From Eq. (12.9)

V=(1 / n) R_{h}^{2 / 3} S^{1 / 2}                           (12.9)

Or     S=\frac{(0.012)(0.012)}{(0.1995)^{4 / 3}}=0.0012