Flood discharge in 10-m rectangular channel

Given: rectangular channel $b=10$ m, $y_1=3.0$ m, $y_2=2.7$ m, $L=200$ m, fall $=0.12$ m, $n=0.025$, $h_e = 0$, $\alpha=1$.

Section 1: $A_1=10\times3.0=30$ m$^2$, $P_1=10+6=16$ m. Section 2: $A_2=10\times2.7=27$ m$^2$, $P_2=10+5.4=15.4$ 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

$$R_1=\frac{30}{16}=1.8750\ \text{m},\qquad R_2=\frac{27}{15.4}=1.7532\ \text{m}$$ $$K_1=\frac{1}{0.025}\times30\times1.8750^{2/3}=1824.7,\qquad K_2=\frac{1}{0.025}\times27\times1.7532^{2/3}=1570.3$$ $$K_m=\sqrt{K_1K_2}=1692.7$$

Step 2: First trial (no velocity head)

$$S_f=\frac{0.12}{200},\qquad Q_1=K_m\sqrt{S_f}=41.463\ \text{m}^3/\text{s}$$

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 $h_e=0$: $h_f = 0.12 + (h_{v1}-h_{v2})$.

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

Answer: Discharge $Q \approx 38.0\ \text{m}^3/\text{s}$.

Parameters to be measured in the channel

To apply Manning's equation and the energy equation for stream flow computation the following are measured in the field:

1. Cross-sections at the ends (and middle) of the reach: width, depth, hence area $A$ and wetted perimeter $P$.
2. Length of the reach $L$ between the sections.
3. Elevations of the water surface at the ends of the reach (high-water marks), giving the fall $h$ and the slope.
4. Roughness coefficient $n$ from the bed and bank material and vegetation (estimated).
5. Velocity-head coefficients $\alpha$ and the eddy-loss coefficient $k_e$ (from the shape of the reach).