Slope-area flood discharge, 6 km reach

Given: $A_1=60$ m$^2$, $P_1=18$ m, $A_2=45$ m$^2$, $P_2=14$ m, $L=6$ km, $n=0.025$, $k_e=0.15$. The bed levels differ by 0.45 m. Assuming the flow depth is the same at both sections (uniform depth reach), the water-surface fall equals the bed fall $=0.45$ m. $\alpha=1$.

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

$$R_1=\frac{60}{18}=3.3333\ \text{m},\qquad R_2=\frac{45}{14}=3.2143\ \text{m}$$ $$K_1=\frac{1}{0.025}\times60\times3.3333^{2/3}=5355.5,\qquad K_2=\frac{1}{0.025}\times45\times3.2143^{2/3}=3920.4$$ $$K_m=\sqrt{K_1K_2}=4582.1$$

Step 2: First trial (no velocity head)

$$S_f=\frac{0.45}{6000},\qquad Q_1=K_m\sqrt{S_f}=39.682\ \text{m}^3/\text{s}$$

Step 3: Correct for velocity head and eddy loss

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

The area decreases downstream (contraction), so $h_e = 0.15(h_{v2}-h_{v1})$.

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

Answer: Discharge $Q \approx 38.8\ \text{m}^3/\text{s}$.