GVF in 12 m wide channel, 2 steps

Given: $Q = 30\ \text{m}^3/\text{s}$, $b = 12$ m, $S_0 = 1/6000$, $n = 0.025$, $y = 1.5$ m at a section. Find the distance to $y = 2$ m with 2 steps.

Depths

$$q = 2.5\ \text{m}^2/\text{s},\quad y_c = \left(\frac{2.5^2}{9.81}\right)^{1/3} = 0.860\ \text{m},\quad y_n = 3.034\ \text{m}\ (\text{Manning, } R = A/P)$$

Profile type: $y_n > y_c$ (mild), and $y_n > y > y_c$ (zone 2), so the profile is M2 (drawdown). The depth rises upstream, so the 2 m section is upstream of the 1.5 m section.

Direct step method (2 steps, $\Delta y = 0.25$ m):

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$ $$L = 226.1 + 415.4 \approx 641.6\ \text{m upstream}$$

(A finer calculation gives about 676 m.)

Answer: M2 profile; the depth is 2 m about 640 m upstream of the section where it is 1.5 m.