Contraction with step down, 4.25 m³/s

Given: $Q = 4.25\ \text{m}^3/\text{s}$, $b_1 = 3\ \text{m}$, $y_1 = 1.15\ \text{m}$, $E_1 = 1.2\ \text{m}$, $b_2 = 1.5\ \text{m}$, step down $\Delta z = 0.6\ \text{m}$. Losses are neglected.

Step 1: Upstream flow type

$$\begin{aligned} V_1 &= \frac{4.25}{3 \times 1.15} = 1.232\ \text{m/s} \\ Fr_1 &= \frac{1.232}{\sqrt{9.81 \times 1.15}} = 0.367 < 1 \end{aligned}$$

The approach flow is subcritical, so the downstream flow is also taken on the subcritical branch.

Step 2: Critical conditions in the contracted section

$$\begin{aligned} q_2 &= \frac{4.25}{1.5} = 2.833\ \text{m}^2/\text{s} \\ y_c &= \left(\frac{q_2^2}{g}\right)^{1/3} = \left(\frac{2.833^2}{9.81}\right)^{1/3} = 0.935\ \text{m} \\ E_c &= 1.5\,y_c = 1.403\ \text{m} \end{aligned}$$

Step 3: Energy at the downstream section

The bed falls by 0.6 m, so the specific energy increases by 0.6 m:

$$E_2 = E_1 + \Delta z = 1.2 + 0.6 = 1.8\ \text{m} > E_c = 1.403\ \text{m}$$

So no choking occurs and the upstream depth is unchanged.

Step 4: Solve for $y_2$

$$y_2 + \frac{q_2^2}{2 g y_2^2} = 1.8 \quad\Rightarrow\quad y_2 + \frac{0.4092}{y_2^2} = 1.8$$

Trial: $y_2 = 1.65$ gives $1.65 + 0.150 = 1.800$. Checked by bisection: $y_2 = 1.650\ \text{m}$ (subcritical; $V_2 = 1.72\ \text{m/s}$).

Answer: Downstream depth $y_2 \approx 1.65\ \text{m}$. (Using the computed $E_1 = 1.227\ \text{m}$ instead of the rounded 1.2 m gives 1.68 m.)