Slope break 0.01 to 0.0064: profile and length

Given: $b = 10$ m, $Q = 125\ \text{m}^3/\text{s}$ ($q = 12.5\ \text{m}^2/\text{s}$), $n = 0.015$, slope changes from $S_{01} = 0.01$ to $S_{02} = 0.0064$.

Depths

$$y_c = \left(\frac{12.5^2}{9.81}\right)^{1/3} = 2.516\ \text{m}$$

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

Both slopes are steep, and the flow is supercritical throughout. The flow arrives at the break with $y = 1.633$ m, which is below the new normal depth 1.896 m, so on the downstream reach $y < y_{n2} < y_c$. The profile is S3 (rising depth, asymptotic to $y_{n2}$). No jump occurs.

Length of S3 profile. The curve reaches $y_{n2}$ only at infinity, so its length is computed up to $0.99\,y_{n2} = 1.877$ m, using the direct step method with 4 equal steps and $S_0 = 0.0064$.

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$ $$L = 48.9 + 61.1 + 86.9 + 175.0 \approx 372\ \text{m}$$

A finer calculation gives about 407 m.

Answer: S3 profile on the downstream slope; length about 370 m (to within 1% of the new normal depth).