Siphon from main canal to branch canal

Data. $D = 0.1$ m, $L_{total} = 90$ m, $L_{summit} = 30$ m, level difference $H = 10$ m, $f = 0.025$, $Q = 0.05$ m³/s. All losses: entrance $0.5\dfrac{V^2}{2g}$, exit $1.0\dfrac{V^2}{2g}$ (pipe discharges below the branch canal level), friction.

(a) Number of pipelines

Energy equation between the two canal surfaces (both at atmospheric pressure, velocity zero):

$$H = \left(0.5 + f\frac{L}{D} + 1.0\right)\frac{V^2}{2g}, \qquad f\frac{L}{D} = 0.025\left(\frac{90}{0.1}\right) = 22.5$$ $$10 = 24.0\,\frac{V^2}{2g} \;\Rightarrow\; \frac{V^2}{2g} = 0.4167\ \text{m},\quad V = 2.859\ \text{m/s}$$

Discharge per pipe: $q = \dfrac{\pi}{4}(0.1)^2(2.859) = 0.02246$ m³/s $= 22.46$ l/s.

$$n = \frac{0.05}{0.02246} = 2.23 \;\Rightarrow\; \textbf{3 pipelines}$$

(b) Maximum height of the summit above the main canal

Let the summit be $z$ above the main canal surface. Atmospheric pressure head = 10 m of water (given), and the minimum absolute pressure at the summit is 20 kPa, i.e. $\dfrac{p_s}{\gamma} = \dfrac{20}{9.81} = 2.039$ m (absolute).

Energy equation from the main canal surface to the summit (losses over the first 30 m, including entrance; velocity head at the summit):

$$10 = z + 2.039 + \frac{V^2}{2g}\left(1 + 0.5 + f\frac{L_s}{D}\right), \qquad f\frac{L_s}{D} = 0.025\left(\frac{30}{0.1}\right) = 7.5$$ $$z = 10 - 2.039 - 0.4167(9.0) = 4.21\ \text{m}$$

(The velocity in each pipe is fixed by the 10 m head, so $V = 2.859$ m/s is used; the extra capacity of three pipes is controlled by a valve.)

Answer: (a) 3 pipelines; (b) the summit may be at most about 4.21 m above the water level of the main canal.