Direct step, 16 m wide channel, 40 m³/s

Given: $Q = 40\ \text{m}^3/\text{s}$, $b = 16\ \text{m}$, $S_0 = 1/5000 = 0.0002$, $n = 0.024$, $y = 1.5$ m at a section. Find the distance to $y = 2.5$ m.

Depths

$$q = 2.5\ \text{m}^2/\text{s},\quad y_c = \left(\frac{2.5^2}{9.81}\right)^{1/3} = 0.860\ \text{m},\quad y_n = 2.671\ \text{m}\ (\text{Manning, } R = A/P)$$

Since $y_n > y_c$ the slope is mild, and $y_n > y > y_c$ for both 1.5 m and 2.5 m. The profile is M2 (drawdown). For M2 the depth increases upstream, so the 2.5 m depth is upstream of the 1.5 m section.

Direct step method (3 equal steps, $\Delta y = 0.333$ m):

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$ $$L = 409.2 + 1012.9 + 2870.7 \approx 4293\ \text{m upstream}$$

Because the last step approaches $y_n$, where the profile is asymptotic, three steps give a lower value than a finer calculation (about 5.15 km).

Answer: M2 profile; the depth is 2.5 m about 4.3 km upstream of the section with 1.5 m depth (3-step direct step).