GVF in 5 m channel, depth 0.25 m to 0.40 m

Given: $b = 5$ m, $Q = 1.85\ \text{m}^3/\text{s}$, $S_0 = 0.00076$, $n = 0.0165$, $\alpha = 1.1$, $y = 0.25$ m at a section. Find the distance to $y = 0.40$ m with $\Delta y = 0.05$ m.

Depths

$$q = 0.37\ \text{m}^2/\text{s},\quad y_c = \left(\frac{\alpha q^2}{g}\right)^{1/3} = \left(\frac{1.1 \times 0.37^2}{9.81}\right)^{1/3} = 0.2485\ \text{m}$$

By Manning's equation (trial, $R = A/P$): $y_n = 0.431$ m.

Since $y_n > y_c$ the slope is mild. The depth 0.25 m satisfies $y_n > y > y_c$, so the profile is M2 (drawdown). The depth rises upstream, so 0.40 m is upstream of the 0.25 m section.

Direct step method

$$E = y + \alpha\frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$ $$L = 4.8 + 23.3 + 77.4 \approx 105.6\ \text{m upstream}$$

The last step is close to $y_n$ where the profile flattens, so a finer calculation gives about 126 m.

Answer: M2 profile; the depth is 0.40 m about 106 m upstream of the section where it is 0.25 m.