Slope-area flood discharge, 7 km reach

Given: $A_1=45$ m$^2$, $P_1=26$ m, $\alpha_1=1.18$; $A_2=50$ m$^2$, $P_2=28$ m, $\alpha_2=1.23$; fall $=1.15$ m; $L=7000$ m; $n=0.03$. The area increases downstream (expansion), so $k_e=0.3$.

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

$$R_1=\frac{45}{26}=1.7308\ \text{m},\qquad R_2=\frac{50}{28}=1.7857\ \text{m}$$ $$K_1=\frac{1}{0.03}\times45\times1.7308^{2/3}=2162.3,\qquad K_2=\frac{1}{0.03}\times50\times1.7857^{2/3}=2453.1$$ $$K_m=\sqrt{K_1K_2}=2303.1$$

Step 2: First trial (no velocity head)

$$S_f=\frac{1.15}{7000},\qquad Q_1=K_m\sqrt{S_f}=29.520\ \text{m}^3/\text{s}$$

Step 3: Correct for velocity head and eddy loss

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

Here $h_{v1}>h_{v2}$ (expansion), so $h_e = 0.3(h_{v1}-h_{v2})$.

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

Answer: Discharge $Q \approx 29.6\ \text{m}^3/\text{s}$.