Does a jump occur at slope change 0.065 to 0.0085

Given: $b = 1.1$ m, $Q = 7.2\ \text{m}^3/\text{s}$ ($q = 6.545\ \text{m}^2/\text{s}$), $n = 0.025$, $S_{01} = 0.065$ changing to $S_{02} = 0.0085$.

$$y_c = \left(\frac{6.545^2}{9.81}\right)^{1/3} = 1.635\ \text{m}$$

Normal depths (Manning's equation with $R = A/P$, by trial):

Flow on the upstream reach is supercritical ($y_{n1} < y_c$) and on the downstream reach it is subcritical ($y_{n2} > y_c$). The change from supercritical to subcritical flow can occur only through a hydraulic jump, so a jump occurs.

Location

Sequent depth of $y_{n1} = 1.224$ m:

$$y_2 = \frac{1.224}{2}\left(\sqrt{1 + 8(1.54)^2} - 1\right) = 2.128\ \text{m}$$

Because $y_2 = 2.128\ \text{m} < y_{n2} = 2.962$ m, the downstream depth is higher than needed. The jump is pushed upstream of the break, onto the steep slope. After the jump the depth rises from 2.128 m to 2.962 m along an S1 curve ending at the break. The jump forms where the S1 depth equals 2.128 m.

Direct step on the S1 curve (from 2.962 m at the break to 2.128 m, 4 steps), with $S_0 = 0.065$:

$$L = 3.04 + 2.95 + 2.81 + 2.58 \approx 11.4\ \text{m}$$

Answer: A hydraulic jump occurs on the steep reach about 11.4 m upstream of the slope change (depths 1.224 m to 2.128 m), followed by an S1 curve up to the break.