Siphon with 75 mm pipes and summit height

Data. $D = 0.075$ m, $H = 15$ m, $L_1 = 20$ m (to summit), $L = 50$ m, $f = 0.03$, minor losses ignored, $Q \ge 50$ l/s, summit pressure $\ge 25$ kPa absolute. Atmospheric pressure is taken as 101.3 kPa, $p_a/\gamma = 10.326$ m of water.

(a) Number of pipes

Level difference = friction head loss (no minor losses, so the exit velocity head is also ignored):

$$15 = f\frac{L}{D}\frac{V^2}{2g}, \qquad f\frac{L}{D} = 0.03\left(\frac{50}{0.075}\right) = 20.0$$ $$\frac{V^2}{2g} = \frac{15}{20.0} = 0.750\ \text{m},\quad V = 3.836\ \text{m/s}$$

Discharge per pipe $q = \dfrac{\pi}{4}(0.075)^2(3.836) = 0.01695$ m³/s $= 16.95$ l/s.

$$n = \frac{50}{16.95} = 2.95 \;\Rightarrow\; \textbf{3 pipes}$$

(b) Maximum height of the summit

Minimum absolute pressure head at the summit: $\dfrac{25}{9.81} = 2.548$ m. Energy equation from the main canal surface to the summit (velocity head and friction over 20 m):

$$\frac{p_a}{\gamma} = z + \frac{p_s}{\gamma} + \frac{V^2}{2g}\left(1 + f\frac{L_1}{D}\right), \qquad f\frac{L_1}{D} = 0.03\left(\frac{20}{0.075}\right) = 8.0$$ $$z = 10.326 - 2.548 - 0.750(1 + 8.0) = 1.03\ \text{m}$$

Answer: (a) 3 pipes; (b) the summit can be at most about 1.03 m above the water level of the main canal.