Discussion
A trapezoidal grassed waterway is constructed along a longitudinal gradient of 4%. If the cross-sectional area of flow is 1.52 m2, wetted perimeter is 12.5 m and Manning’s n for the waterway is 0.04 m−1/3 s, the flow through the waterway in m3 s−1 is
Answer: A
The flow is about 1.9 m3 s−1.
Use Manning's equation with the hydraulic radius \(R = A/P\) and the slope as a fraction:
\[ \begin{aligned} R &= 1.52 / 12.5 = 0.1216\ \text{m} \\ V &= \tfrac{1}{n} R^{2/3} S^{1/2} \\ &= \tfrac{1}{0.04}(0.2455)(0.2) \\ &= 1.228\ \text{m/s} \\ Q &= AV = 1.52 \times 1.228 = 1.87 \end{aligned} \]
So the discharge rounds to 1.9 m3 s−1. The larger options would need a larger hydraulic radius; a common slip is to use the slope 4 instead of 0.04 under the square root, or to forget the 2/3 power on R, both of which give values far from any option.
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