Slope-area discharge, 1300 m reach

Given: $A_1=3500$ m$^2$, $P_1=650$ m, $\alpha_1=1.17$; $A_2=3250$ m$^2$, $P_2=621$ m, $\alpha_2=1.21$; fall $=0.4$ m, $L=1300$ m, $n=0.03$. No eddy-loss coefficient is given, so $h_e=0$ is assumed.

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

$$R_1=\frac{3500}{650}=5.3846\ \text{m},\qquad R_2=\frac{3250}{621}=5.2335\ \text{m}$$ $$K_1=\frac{1}{0.03}\times3500\times5.3846^{2/3}=358412.6,\qquad K_2=\frac{1}{0.03}\times3250\times5.2335^{2/3}=326555.2$$ $$K_m=\sqrt{K_1K_2}=342113.3$$

Step 2: First trial (no velocity head)

$$S_f=\frac{0.4}{1300},\qquad Q_1=K_m\sqrt{S_f}=6001.064\ \text{m}^3/\text{s}$$

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 $h_e=0$: $h_f = 0.4 + (h_{v1}-h_{v2})$.

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

Answer: Discharge $Q \approx 5754.8\ \text{m}^3/\text{s}$.

(If a contraction loss of $k_e=0.1$ were included, $Q$ would be about 5732 m$^3$/s, a change of 0.4 %.)