Contraction and bed raise, 16 m³/s

Given: $Q = 16\ \text{m}^3/\text{s}$, $b_1 = 4\ \text{m}$, $y_1 = 2\ \text{m}$; at the downstream section $b_2 = 3.5\ \text{m}$ and the bed is raised by $\Delta z = 0.35\ \text{m}$.

Upstream:

$$V_1 = \frac{16}{4 \times 2} = 2\ \text{m/s},\quad Fr_1 = \frac{2}{\sqrt{9.81 \times 2}} = 0.45,\quad E_1 = 2 + \frac{2^2}{2g} = 2.204\ \text{m}$$

At the contracted, raised section (check for choking):

$$q_2 = \frac{16}{3.5} = 4.571\ \text{m}^2/\text{s},\quad y_c = \left(\frac{4.571^2}{9.81}\right)^{1/3} = 1.287\ \text{m},\quad E_c = 1.930\ \text{m}$$

Minimum upstream energy needed $= E_c + \Delta z = 1.930 + 0.35 = 2.280\ \text{m}$.

Since $2.280 > E_1 = 2.204\ \text{m}$, the approach flow has too little energy. The flow is choked: critical depth forms at the section and the upstream depth must rise.

New upstream depth:

$$y_1' + \frac{(16/4)^2}{2g\,y_1'^2} = 2.280 \;\Rightarrow\; y_1' = 2.094\ \text{m}$$

Effect on the water surface:

Answer: The surface is affected: it rises about 0.09 m upstream (to 2.09 m depth) and drops to critical depth 1.29 m over the raised, narrowed section.