Step and width reduction to produce critical flow

Critical flow at the transition requires $E_2=E_{min}=\tfrac32y_c$ with $y_c=\sqrt[3]{q_2^2/g}$ at that section.

Given data

$b=50$ m, $Q=250$ m³/s, $y_1=5$ m. $q_1=\dfrac{250}{50}=5$ m²/s, $V_1=1$ m/s.

$$E_1=5+\frac{1^2}{19.62}=5.0510\ \text{m},\qquad y_{c1}=\sqrt[3]{\frac{25}{9.81}}=1.366\ \text{m},\quad E_{min1}=2.049\ \text{m}$$

The approach flow is subcritical ($Fr_1=0.143$).

(i) Step in the bed (width constant)

$$E_1=E_{min1}+\Delta z\;\Rightarrow\;\Delta z=5.0510-2.0489=3.002\ \text{m}$$

Height of step $=3.00$ m.

(ii) Width reduction (bed unchanged)

Critical flow at the contraction with $E_2=E_1$:

$$y_c=\frac23E_1=3.3673\ \text{m},\qquad q_2=\sqrt{gy_c^3}=\sqrt{9.81\times3.3673^3}=19.354\ \text{m}^2/\text{s}$$ $$b_2=\frac{Q}{q_2}=\frac{250}{19.354}=12.92\ \text{m}$$

Reduction in width $=50-12.92=37.08$ m.

(iii) Combination of step and width reduction

Many combinations are possible. Take a step of $\Delta z=1.5$ m (assumed). Then energy at the contraction:

$$E_2=E_1-1.5=3.5510\ \text{m},\qquad y_c=\frac23E_2=2.3673\ \text{m}$$ $$q_2=\sqrt{9.81\times2.3673^3}=11.408\ \text{m}^2/\text{s},\qquad b_2=\frac{250}{11.408}=21.91\ \text{m}$$

Reduction in width $=50-21.91=28.09$ m.

Answer: (i) step $=3.00$ m; (ii) width reduced by 37.08 m to 12.92 m; (iii) e.g. step 1.5 m with width reduced by 28.09 m to 21.91 m.