Slope break: jump depth and jump length by step

Given: $b = 4$ m, $Q = 16\ \text{m}^3/\text{s}$ ($q = 4\ \text{m}^2/\text{s}$), $n = 0.0165$, downstream slope $S_{02} = 0.0024$, upstream uniform depth $Y_{01} = 0.4$ m (steep reach).

$$y_c = \left(\frac{4^2}{9.81}\right)^{1/3} = 1.177\ \text{m}$$

a) Depth in the downstream channel

For the jump to end in uniform flow, the depth after the jump must equal the normal depth of the downstream channel. From Manning's equation with $R = A/P$ (trial):

$$Y_{02} = y_n = 1.495\ \text{m}\quad (> y_c,\ \text{so the slope is mild; } V = 2.68\ \text{m/s})$$

For comparison, the sequent depth of $Y_{01} = 0.4$ m is:

$$V_{01} = 10\ \text{m/s},\ Fr = 5.05,\quad y_2 = \frac{0.4}{2}\left(\sqrt{1 + 8(5.05)^2} - 1\right) = 2.66\ \text{m}$$

Since $2.66 > 1.495$ m, the tailwater is too low for the jump to form at the break. The supercritical flow continues on the mild slope as an M3 curve until its depth $y_j$ has a sequent depth equal to 1.495 m:

$$y_j:\ \frac{y_j}{2}\left(\sqrt{1 + 8\frac{q^2}{g\,y_j^3}} - 1\right) = 1.495 \;\Rightarrow\; y_j = 0.908\ \text{m}$$

b) Length of the M3 profile up to the jump (3 steps)

Depth from 0.40 m to 0.908 m, with $\Delta y = 0.169$ m:

$$E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_{02} - \bar S_f}$$ $$L = 31.6 + 32.1 + 28.9 \approx 92.5\ \text{m}$$

The jump itself then raises the depth from 0.908 m to 1.495 m over a length of about $6(y_2 - y_1) = 6(1.495 - 0.908) \approx 3.5$ m.

Answer: (a) $Y_{02} = 1.495$ m; (b) the toe of the jump is about 92 m downstream of the slope change (M3 profile), and the jump is about 3.5 m long.