Crest level of fall, Q 30 m3/s

For a rectangular broad-crested weir with free flow, the discharge is $Q = 1.705\,L\,H^{3/2}$, where $H$ is the total head over the crest including the velocity head of approach. The crest level is the u/s energy level minus this head.

Data

$Q = 30$ m³/s, canal FSL 115.00 m, bed level 112.00 m (depth $D = 3.0$ m), bed width $B = 15$ m, side slope 2:1, crest length $L = 10$ m.

Step 1. Approach velocity

$$\begin{aligned} A &= (B + mD)D = (15 + 2\times3)\times3 = 63\ \text{m}^2 \\ v &= \frac{Q}{A} = \frac{30}{63} = 0.476\ \text{m/s}, \qquad h_a = \frac{v^2}{2g} = 0.0116\ \text{m} \end{aligned}$$

Step 2. Head over the crest

$$H = \left(\frac{Q}{1.705\,L}\right)^{2/3} = \left(\frac{30}{1.705\times10}\right)^{2/3} = (1.7595)^{2/3} = 1.457\ \text{m}$$

Step 3. Crest level

$$\text{Crest level} = FSL + h_a - H = 115.000 + 0.0116 - 1.457 = 113.55\ \text{m}$$

The crest is 1.55 m above the canal bed. Check: the head over the crest measured from the FSL is $115 - 113.554 = 1.446$ m; adding the velocity head of 0.012 m gives $H = 1.457$ m.

Answer: crest level $= 113.55$ m (head over crest $H = 1.46$ m).

 FSL 115.0 ~~~~~~~~~~~~~~~~~~~~~
                   ____ crest 113.55 m
 bed 112.0 _______|    |________
                  L = 10 m, broad crest