Question 11 of 13: Wide Shallow River Under Flood Conditions

Question 11: Wide Shallow River Under Flood Conditions (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Wide, shallow river of constant width $w$ and bed slope $S$; flood depth $y_{flood}=2y_{normal}$; Manning-type uniform-flow relations apply.

Find. How velocity and flow rate scale between normal and flood conditions, with theoretical justification.

For a WIDE shallow channel, the hydraulic radius $R=A/P$ is closely approximated by the depth itself, $R\approx y$, because the wetted perimeter is dominated by the (fixed, large) bottom width and the two side-wall contributions are comparatively negligible. Manning's equation for uniform flow, $V=\tfrac{1}{n}R^{2/3}S^{1/2}$, then gives velocity scaling as $V\propto y^{2/3}$ for fixed slope $S$ and roughness $n$. Doubling the depth therefore scales the velocity by $$\frac{V_{flood}}{V_{normal}}=\left(\frac{2y}{y}\right)^{2/3}=2^{2/3}=1.587$$

(a) Velocity: since $1.587\gt1$, the flood velocity is greater than the normal flow velocity (but, notably, NOT double — the two-thirds power law means the increase is sub-linear in depth).

Flow rate is $Q=VA=V(wy)$, and with fixed width $w$ this scales as $$\frac{Q_{flood}}{Q_{normal}}=\frac{V_{flood}}{V_{normal}}\times\frac{y_{flood}}{y_{normal}}=2^{2/3}\times2=2^{5/3}=3.175$$

(b) Flow rate: since $3.175\gt2$, the flood flow rate is greater than two times the normal flow rate — the combination of a higher velocity AND a doubled cross-sectional area compounds to more than double the discharge.

(c) Justification. Both results follow directly from Manning's uniform-flow equation applied to a wide channel where $R\approx y$: velocity depends on depth to the $2/3$ power (from the $R^{2/3}$ term, since $S$ and $n$ are unchanged), while flow rate depends on depth to the $2/3+1=5/3$ power (the extra power of 1 coming from the cross-sectional area $A=wy$ itself, which is exactly proportional to depth at fixed width). Because $5/3\gt1$, discharge grows faster than depth — a small rise in flood stage on a wide river therefore produces a disproportionately larger rise in discharge, which is exactly the nonlinearity that makes flood-flow forecasting from stage measurements a genuinely nonlinear (rating-curve) problem rather than a simple linear scale-up.