Slope-area discharge, 10 km reach

Given: $A_1=73.3$ m$^2$, $P_1=26.80$ m, $A_2=93.4$ m$^2$, $P_2=30.23$ m, $n=0.02$, $L=10$ km, fall $h_1-h_2 = 306.920-306.650 = 0.27$ m. Velocity-head coefficients are taken as $\alpha_1=\alpha_2=1$. The area increases downstream, so the reach is an expansion and $k_e = 0.3$.

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

$$R_1=\frac{73.3}{26.8}=2.7351\ \text{m},\qquad R_2=\frac{93.4}{30.23}=3.0896\ \text{m}$$ $$K_1=\frac{1}{0.02}\times73.3\times2.7351^{2/3}=7167.8,\qquad K_2=\frac{1}{0.02}\times93.4\times3.0896^{2/3}=9906.6$$ $$K_m=\sqrt{K_1K_2}=8426.6$$

Step 2: First trial (no velocity head)

$$S_f=\frac{0.27}{10000},\qquad Q_1=K_m\sqrt{S_f}=43.786\ \text{m}^3/\text{s}$$

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 $v_2<v_1$ (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 $Q \approx 44.2\ \text{m}^3/\text{s}$.