Slope-area discharge with three sections

Given: reach length $L=260$ m between the u/s and d/s sections; fall $=315.50-315.15=0.35$ m; contraction coefficient $k_c=0.1$ (area decreasing downstream); $\alpha=1$. Roughness differs at each section ($n = 0.025, 0.027, 0.029$).

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

With three sections of different roughness, the conveyance of each section is computed with its own $n$:

The mean conveyance of the reach is taken as the geometric mean of the three sections:

$$K_m=(K_1K_2K_3)^{1/3}=5310.8$$

The middle section has no gauge reading, so the fall is that between the end sections, and the eddy loss is applied between the end sections.

Step 2: First trial (no velocity head)

$$S_f=\frac{0.35}{260},\qquad Q_1=K_m\sqrt{S_f}=194.853\ \text{m}^3/\text{s}$$

Step 3: Correct for velocity head and eddy loss

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

Velocity heads are taken at the u/s and d/s sections, and $h_e=k_c|h_{v1}-h_{v2}|$ with $k_c = 0.1$.

The trials converge to $h_f$ and $Q$ shown in the last row (trials 3 to 5 are not shown).

Answer: Discharge $Q \approx 185.2\ \text{m}^3/\text{s}$.