Flow rate from drop in surface over hump

Given data

$b_1=3$ m, $y_1=2.05$ m; $b_2=2$ m; floor raised $\Delta z=0.30$ m; surface drops $0.20$ m at the contracted section. No loss assumed.

Depth at the contracted section

Water surface level at the contraction $=2.05-0.20=1.85$ m (above the approach bed). The floor is 0.30 m higher, so

$$y_2=1.85-0.30=1.55\ \text{m}$$

Energy equation (datum: approach bed)

$$y_1+\frac{V_1^2}{2g}=\Delta z+y_2+\frac{V_2^2}{2g},\qquad V_1=\frac{Q}{3\times2.05}=\frac{Q}{6.15},\quad V_2=\frac{Q}{2\times1.55}=\frac{Q}{3.10}$$ $$2.05+\frac{Q^2}{2g(6.15)^2}=0.30+1.55+\frac{Q^2}{2g(3.10)^2}$$ $$\frac{Q^2}{2g}\left(\frac{1}{3.10^2}-\frac{1}{6.15^2}\right)=2.05-1.85=0.20$$ $$\frac{Q^2}{19.62}\left(0.10406-0.026439\right)=0.20\;\Rightarrow\;Q^2=\frac{0.20\times19.62}{0.07762}=50.555$$ $$Q=7.110\ \text{m}^3/\text{s}$$

Check: $V_1=1.156$ m/s, $V_2=2.294$ m/s; $E_1=2.05+0.0681=2.1181$ m and $E_2$ (with datum) $=0.30+1.55+0.2681=2.1181$ m ✓. Flow is subcritical at the contraction ($Fr_2=0.588$).

Answer: discharge $Q\approx7.11$ m³/s.