Slope-area flood discharge, Marin River at Kusuntar

Given: $A_1=175$ m$^2$, $P_1=75$ m, $A_2=224$ m$^2$, $P_2=166$ m, $L=120$ m, fall $=1.46$ m, $n=0.1$. No velocity-head or eddy-loss coefficient is given, so $\alpha=1$ and $h_e=0$ are 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

$$R_1=\frac{175}{75}=2.3333\ \text{m},\qquad R_2=\frac{224}{166}=1.3494\ \text{m}$$ $$K_1=\frac{1}{0.1}\times175\times2.3333^{2/3}=3078.6,\qquad K_2=\frac{1}{0.1}\times224\times1.3494^{2/3}=2735.3$$ $$K_m=\sqrt{K_1K_2}=2901.9$$

Step 2: First trial (no velocity head)

$$S_f=\frac{1.46}{120},\qquad Q_1=K_m\sqrt{S_f}=320.087\ \text{m}^3/\text{s}$$

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 $h_e=0$: $h_f = 1.46 + (h_{v1}-h_{v2})$.

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

Answer: Discharge $Q \approx 327.6\ \text{m}^3/\text{s}$.