Flood discharge in rectangular channel

The slope-area method is used: discharge from Manning's equation with friction slope found by the energy equation, solved by trial.

Data

$b = 10$ m, $y_1 = 3.0$ m, $y_2 = 2.9$ m, $L = 200$ m, fall $\Delta Z = 0.12$ m, $n = 0.025$.

Section properties and conveyance

$$\begin{aligned} A_1 &= 30\ \text{m}^2,\; P_1 = 16\ \text{m},\; R_1 = 1.875\ \text{m} \\ A_2 &= 29\ \text{m}^2,\; P_2 = 15.8\ \text{m},\; R_2 = 1.835\ \text{m} \\ K_1 &= \frac{1}{n}A_1R_1^{2/3} = \frac{30 \times 1.875^{2/3}}{0.025} = 1824.7 \\ K_2 &= \frac{29 \times 1.835^{2/3}}{0.025} = 1738.9 \\ K &= \sqrt{K_1K_2} = 1781.3 \end{aligned}$$

Trial solution

Area decreases downstream (contracting reach), so eddy loss is taken as zero. Then

$$h_f = \Delta Z + \frac{V_1^2 - V_2^2}{2g}, \qquad Q = K\sqrt{h_f/L}$$

Sample (trial 1): $Q = 1781.3\sqrt{0.12/200} = 1781.3 \times 0.02449 = 43.63$ m³/s.

Answer: Flood discharge ≈ 42.3 m³/s (taking $K_e = 0$ for the contracting reach).