Grassed Waterways Questions and Answers

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.

Answer: B

The most appropriate dimensions are a top width of 6 m and a depth of 0.4 m.

The flow area needed at the permissible velocity is \(A = Q/V = 2.85/1.78 = 1.60\ \text{m}^{2}\). Manning’s equation fixes the hydraulic radius:

\[ \begin{aligned} R^{2/3} &= \frac{Vn}{\sqrt{S}} = \frac{1.78 \times 0.04}{\sqrt{0.03}} = 0.411 \\ R &= 0.264\ \text{m} \end{aligned} \]

For a parabolic section with top width \(T\) and depth \(d\), \(A = \tfrac{2}{3}Td\) and \(R \approx T^{2}d/(1.5T^{2} + 4d^{2})\). With \(T = 6\) and \(d = 0.4\):

\[ \begin{aligned} A &= \tfrac{2}{3}(6)(0.4) = 1.60\ \text{m}^{2} \\ R &= \frac{36 \times 0.4}{54 + 0.64} = 0.264\ \text{m} \end{aligned} \]

Both match. The pair 4 and 0.6 gives the right area but \(R = 0.38\) m, too deep a section, so the velocity would exceed the permissible value. The pairs 7 and 0.5 and 5.5 and 0.6 give areas of 2.33 and 2.2 m², larger than needed.