Pipe outlet minimum water level, 50 lps

A free-discharging pipe outlet needs a head equal to the sum of the entry loss, the friction loss and the velocity head at exit. The minimum water level in the distributary is the outlet level plus this head. The pipe delivers into the water course at its FSL 100.0 m; the pipe centre at the exit is taken at 100.0 m.

 distributary ~~~~ H.W.L (min) ~~~~
                     H = ?
               ---- pipe, L = 10 m, d = 20 cm --> free fall
                                     water course FSL 100.0 m

Data

$Q = 50$ l/s $= 0.05$ m³/s, $L = 10$ m, $d = 0.20$ m, Manning's $n = 0.016$, entry loss coefficient 0.5 (assumed).

Calculation

$$\begin{aligned} A &= \frac{\pi}{4}(0.2)^2 = 0.03142\ \text{m}^2, \qquad v = \frac{Q}{A} = 1.592\ \text{m/s} \\ \frac{v^2}{2g} &= 0.1291\ \text{m}, \qquad R = \frac{d}{4} = 0.050\ \text{m} \\ h_f &= \frac{n^2 v^2 L}{R^{4/3}} = \frac{0.016^2\times1.592^2\times10}{0.050^{4/3}} = 0.3520\ \text{m} \\ h_e &= 0.5\frac{v^2}{2g} = 0.0646\ \text{m} \end{aligned}$$

Total head required (entry loss + friction + velocity head at free discharge):

$$H = h_e + h_f + \frac{v^2}{2g} = 0.0646 + 0.3520 + 0.1291 = 0.546\ \text{m}$$

Minimum water level in the distributary $= 100.0 + 0.546 = 100.546$ m.

Answer: minimum water level in the distributary $\approx 100.55$ m (head required 0.55 m).