Profile for abrupt slope change 0.0016 to 0.0150

Given: $b = 3\ \text{m}$, $Q = 9.6\ \text{m}^3/\text{s}$ ($q = 3.2\ \text{m}^2/\text{s}$), $n = 0.014$, $S_{01} = 0.0016$, $S_{02} = 0.0150$.

Depths

$$y_c = \left(\frac{3.2^2}{9.81}\right)^{1/3} = 1.014\ \text{m}$$

From Manning's equation, $R = A/P$:

Profile

1. Upstream (mild) reach: the flow is subcritical and uniform at $y_n = 1.392$ m far upstream. The abrupt change to the steep slope forces critical depth at the break. The surface falls from the NDL to $y_c = 1.014$ m as an M2 curve, asymptotic to the NDL far upstream and meeting the CDL at the break.
2. Downstream (steep) reach: the flow starts at $y_c$ at the break and continues as an S2 curve, falling and becoming asymptotic to the uniform supercritical depth $y_n = 0.629$ m.
3. No hydraulic jump occurs because the flow goes from subcritical to supercritical.

 MILD  (yn=1.392)       |  STEEP (yn=0.629)
 ---- NDL ......___      |
              M2   ''--._|
 CDL - - - - - - - - - - *  (yc = 1.014, break)
                         |'-._  S2
 NDL(steep) - - - - - - -|- - -''--- yn = 0.629

Answer: M2 on the upstream mild slope (1.392 m falling to 1.014 m at the break); S2 on the downstream steep slope (1.014 m falling to 0.629 m).