Siphon with 90 mm pipe, number and length

Data. $D = 0.09$ m, $L_1 = 25$ m (up to the summit), summit 2 m above the main canal, $p_s = 20$ kN/m² absolute, level difference $H = 10$ m, $f = 0.03$, $Q = 0.06$ m³/s. Minor losses: entrance $0.5\dfrac{V^2}{2g}$ and exit $1.0\dfrac{V^2}{2g}$. Atmospheric pressure is taken as 101.3 kPa, i.e. $p_a/\gamma = 10.326$ m of water.

(i) Number of pipelines

Absolute pressure head at the summit: $\dfrac{p_s}{\gamma} = \dfrac{20}{9.81} = 2.039$ m.

Energy equation from the main canal surface to the summit (velocity head at the summit, entrance loss and friction over 25 m):

$$\frac{p_a}{\gamma} = 2 + \frac{p_s}{\gamma} + \frac{V^2}{2g}\left(1 + 0.5 + f\frac{L_1}{D}\right), \qquad f\frac{L_1}{D} = 0.03\left(\frac{25}{0.09}\right) = 8.333$$ $$10.326 - 2 - 2.039 = 6.287 = 9.833\,\frac{V^2}{2g} \;\Rightarrow\; \frac{V^2}{2g} = 0.6394\ \text{m},\quad V = 3.542\ \text{m/s}$$

Discharge per pipe: $q = \dfrac{\pi}{4}(0.09)^2(3.542) = 0.02253$ m³/s $= 22.53$ l/s.

$$n = \frac{0.06}{0.02253} = 2.66 \;\Rightarrow\; \textbf{3 pipelines}$$

(ii) Total length of pipe

For the same velocity, the total head of 10 m between the canal surfaces is used up in the entrance, friction over the full length and exit:

$$10 = \left(0.5 + f\frac{L}{D} + 1.0\right)\frac{V^2}{2g} \;\Rightarrow\; f\frac{L}{D} = \frac{10}{0.6394} - 1.5 = 14.140$$ $$L = 14.140\times\frac{0.09}{0.03} = 42.4\ \text{m}$$

So the length beyond the summit is $42.4 - 25 = 17.4$ m.

Answer: (i) 3 pipelines; (ii) total pipe length $\approx 42.4$ m.