Backwater distance within 100 mm of normal depth

Given: $Q = 8.75\ \text{m}^3/\text{s}$, $b = 5\ \text{m}$, $S_0 = 1/5000 = 0.0002$, $n = 0.015$, depth at the dam $y = 2.5$ m.

Depths

$$q = 1.75\ \text{m}^2/\text{s},\quad y_c = \left(\frac{1.75^2}{9.81}\right)^{1/3} = 0.678\ \text{m}$$

By Manning's equation (trial, $R = A/P$): $y_n = 1.80\ \text{m}$.

Since $y = 2.5 > y_n > y_c$ on a mild slope, the profile is M1 (backwater). Going upstream, the depth falls towards $y_n$. We need the distance upstream to the point where $y = y_n + 0.1 = 1.90$ m.

Direct step method, 4 equal steps from 2.5 m to 1.90 m ($\Delta y = 0.15$ m):

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$ $$L = 1323 + 1552 + 2003 + 3270 \approx 8150\ \text{m}$$

The negative sign means upstream. A finer integration gives about 8.26 km.

Answer: The depth is within 100 mm of the normal depth about 8.15 km (roughly 8.2 km) upstream of the dam (M1 backwater curve).