Contraction and bed rise, 3.5 m channel (1.2 m depth)

Given data

$b_1=3.5$ m, $Q=4$ m³/s, $y_1=1.2$ m; contracted $b_2=2.0$ m, bed raised $\Delta z=0.15$ m.

$$q_1=\frac{4}{3.5}=1.1429\ \text{m}^2/\text{s},\quad V_1=\frac{1.1429}{1.2}=0.952\ \text{m/s},\quad E_1=1.2+\frac{0.952^2}{19.62}=1.2462\ \text{m}$$

Critical conditions at the contracted section

$$q_2=\frac{4}{2}=2\ \text{m}^2/\text{s},\quad y_{c2}=\sqrt[3]{\frac{2^2}{9.81}}=0.7415\ \text{m},\quad E_{min2}=1.5y_{c2}=1.1123\ \text{m}$$

Energy available at the contracted section if the upstream depth is unchanged:

$$E_2=E_1-\Delta z=1.2462-0.15=1.0962\ \text{m}<E_{min2}=1.1123\ \text{m}$$

The available energy is less than the minimum needed, so the flow is choked: critical flow occurs in the reduced section and the upstream depth must rise.

Depth at the reduced section

$$y_2=y_{c2}=0.742\ \text{m}$$

Upstream depth

Upstream energy must be $E_1'=E_{min2}+\Delta z=1.1123+0.15=1.2623$ m.

$$y_1'+\frac{q_1^2}{2gy_1'^2}=y_1'+\frac{0.06657}{y_1'^2}=1.2623\;\Rightarrow\;y_1'=1.217\ \text{m}$$

The upstream depth rises from 1.20 m to 1.217 m.

Answer: depth at the reduced section $=0.74$ m (critical); depth upstream $=1.22$ m.