Distance to within 5% of normal depth

Given: wide rectangular channel, discharge taken as $q = 5\ \text{m}^2/\text{s}$ per metre width (a wide channel is analysed per unit width), $S_0 = 1/3600$, $n = 0.02$, $y = 3.5$ m. Wide channel: $R \approx y$.

Depths

$$y_n = \left(\frac{nq}{\sqrt{S_0}}\right)^{3/5} = \left(\frac{0.02 \times 5}{1/60}\right)^{0.6} = 6^{0.6} = 2.930\ \text{m},\qquad y_c = \left(\frac{5^2}{9.81}\right)^{1/3} = 1.366\ \text{m}$$

Profile type: $y_n > y_c$ (mild) and $y = 3.5 > y_n$ (zone 1), so the profile is M1 (backwater). The depth falls going upstream towards $y_n$. The target depth is within 5% of $y_n$: $1.05\,y_n = 3.077$ m, which lies upstream of the 3.5 m section.

Direct step method (2 equal steps, $\Delta y = 0.2117$ m), with $S_f = n^2q^2/y^{10/3}$:

$$L = 1859 + 2990 \approx 4850\ \text{m upstream}$$

(A finer calculation gives about 4.9 km.)

Answer: M1 profile; the depth comes within 5% of the normal depth about 4.85 km upstream of the 3.5 m section.