Flood discharge in a 5-m rectangular channel

Given: rectangular channel $b=5$ m, $y_1=2.0$ m, $y_2=1.8$ m, $L=500$ m, fall $=0.25$ m, $n=0.025$, $h_e=0$, $\alpha=1$.

Section 1: $A_1=5\times2=10$ m$^2$, $P_1=5+2\times2=9$ m. Section 2: $A_2=5\times1.8=9$ m$^2$, $P_2=5+2\times1.8=8.6$ m.

Method: Manning's equation gives conveyance $K=\dfrac{1}{n}AR^{2/3}$ and $Q = K\sqrt{S_f}$. The friction slope is found from the energy equation between the two sections, so the discharge is found by trial.

Step 1: Conveyance

$$R_1=\frac{10}{9}=1.1111\ \text{m},\qquad R_2=\frac{9}{8.6}=1.0465\ \text{m}$$ $$K_1=\frac{1}{0.025}\times10\times1.1111^{2/3}=429.1,\qquad K_2=\frac{1}{0.025}\times9\times1.0465^{2/3}=371.1$$ $$K_m=\sqrt{K_1K_2}=399.0$$

Step 2: First trial (no velocity head)

$$S_f=\frac{0.25}{500},\qquad Q_1=K_m\sqrt{S_f}=8.923\ \text{m}^3/\text{s}$$

Step 3: Correct for velocity head and eddy loss

$$h_f=(h_1-h_2)+\left(\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right)-h_e,\qquad h_e=k_e\left|\frac{\alpha_1v_1^2}{2g}-\frac{\alpha_2v_2^2}{2g}\right|$$

With $h_e=0$: $h_f = 0.25 + (h_{v1}-h_{v2})$.

The trials converge to $h_f$ and $Q$ shown in the last row.

Answer: Discharge $Q \approx 8.8\ \text{m}^3/\text{s}$.