Slope-area discharge, 10 km reach, 104.77 m

Given: $A_A=73.293$ m$^2$, $R_A=2.733$ m, $A_B=93.375$ m$^2$, $R_B=3.089$ m, $n=0.02$, $L=10$ km, fall $=104.77-104.50=0.27$ m. Velocity-head coefficients are taken as 1. Section B has the larger area, so the reach expands and $k_e=0.3$. The wetted perimeter is found from $P=A/R$ ($P_A=26.82$ m, $P_B=30.23$ 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

$$R_1=\frac{73.293}{26.8178}=2.7330\ \text{m},\qquad R_2=\frac{93.375}{30.2282}=3.0890\ \text{m}$$ $$K_1=\frac{1}{0.02}\times73.293\times2.7330^{2/3}=7163.5,\qquad K_2=\frac{1}{0.02}\times93.375\times3.0890^{2/3}=9902.5$$ $$K_m=\sqrt{K_1K_2}=8422.4$$

Step 2: First trial (no velocity head)

$$S_f=\frac{0.27}{10000},\qquad Q_1=K_m\sqrt{S_f}=43.764\ \text{m}^3/\text{s}$$

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 $v_B<v_A$ (expansion), so $h_e = 0.3(h_{vA}-h_{vB})$.

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

Answer: Discharge $Q \approx 44.2\ \text{m}^3/\text{s}$.