Backwater distance into reservoir, 6 m channel

Given: $b = 6$ m, $n = 0.020$, $S_0 = 0.005236$, $Q = 30\ \text{m}^3/\text{s}$. The depth is 3 m at the reservoir entry and 2 m at the upstream end of the reach. Find the length $L$ between them. (Assumed reading of the figure: the 2 m depth is upstream, at distance $L$ from the entry.)

Depths ($q = 5\ \text{m}^2/\text{s}$):

$$y_c = \left(\frac{5^2}{9.81}\right)^{1/3} = 1.366\ \text{m},\qquad y_n = 1.417\ \text{m}\ (\text{Manning, } R = A/P)$$

$y_n > y_c$, so the slope is (just) mild. Depths of 2 m to 3 m are above $y_n$, so the profile is M1 (backwater from the reservoir). The depth falls going upstream.

Direct step method, 4 steps of 0.25 m, from $y = 3.0$ m to $y = 2.0$ m:

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}$$

(Steps are taken from the upstream depth 2 m towards the reservoir, so $\Delta x$ is measured downstream.)

$$L = 51.6 + 50.5 + 49.9 + 49.5 \approx 201.6\ \text{m}$$

A finer integration gives 201.0 m, which agrees.

Answer: $L \approx 202\ \text{m}$ (M1 backwater profile).