--- title: "Siphon from main canal to branch canal" subject: "Hydraulics" chapter: 2 key: "semester_4/hydraulics:siphon-canal-branch-number-of-pipes" source: https://www.bctengineer.com/q/semester_4/hydraulics/siphon-canal-branch-number-of-pipes site: BCT Engineering --- # Siphon from main canal to branch canal Hydraulics · Chapter 2: Simple pipe flow problems and solution - Asked in: 2072 Magh, 4+4 marks; 2071 Magh, 8 marks (asked 2 times) - Video: [What is Syphon.](https://www.youtube.com/watch?v=Dft-CcwkZPI) — Unique Series - Civil Mantraa - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/siphon-canal-branch-number-of-pipes - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2 ## Question Water from a main canal is siphoned to a branch canal over an embankment by means of a wrought iron pipe of 100 mm diameter. The length of the pipeline up to the summit is 30 m and the total length is 90 m. Water surface elevation in the branch canal is 10 m below that of main canal. Take f = 0.025 and consider all losses. (a) If the total quantity of water required to be conveyed is 0.05 m³/s, how many pipelines are needed? (b) What is the maximum permissible height of the summit above the water level in the main canal so that the water pressure at the summit may not fall below 20 kPa absolute, the barometer reading being 10 m of water? ## Short answer Each 100 mm pipe discharges 0.02246 m³/s under the 10 m head; thus 3 pipelines are needed. The summit must not exceed 4.21 m above the main canal water level. ## Answer **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Between the two canal surfaces, the 10 m head is used for: A. Friction only B. Entrance 0.5, friction f L/D and exit 1.0 velocity heads, giving 10 = 24.0 V^2/2g C. Entrance and exit losses only D. Friction over the first 30 m only Answer: B. With all losses the head equals (0.5 + 22.5 + 1.0) V^2/2g, so V^2/2g = 0.4167 m and V = 2.859 m/s. (Key point: Energy equation between canals) 2. How is the number of pipelines decided? A. Round 2.23 down to 2 B. Divide total length by pipe length C. Round 0.02246/0.05 up D. Divide required Q by the discharge of one pipe, 2.23, and round up to 3 Answer: D. A pipe carries 22.46 l/s, so 2 pipes would give too little (44.9 l/s); 3 pipes are needed. (Key point: Number of pipelines) 3. In (b), the energy equation from main canal to summit uses which losses? A. Velocity head, entrance loss and friction over the first 30 m B. Friction over the full 90 m C. Exit loss only D. No losses Answer: A. The bracket is 1 + 0.5 + f L_s/D = 9.0, giving z = 10 - 2.039 - 0.4167(9.0) = 4.21 m. (Key point: Summit pressure equation) 4. Why is the summit height limited? A. To keep the velocity low B. So that the absolute pressure at the summit does not fall below 20 kPa and the water column does not break C. To increase discharge D. To reduce friction Answer: B. The summit has the lowest pressure; too high a summit lets pressure fall to vapour pressure, causing cavitation or air release. (Key point: Meaning of the summit limit) ## Related in Hydraulics - [Siphon with 90 mm pipe, number and length](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Asiphon-90mm-pipe-total-length) — Hydraulics, chapter 2 - [Siphon with 75 mm pipes and summit height](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Asiphon-75mm-pipes-summit-height) — Hydraulics, chapter 2