--- title: "Siphon with 75 mm pipes and summit height" subject: "Hydraulics" chapter: 2 key: "semester_4/hydraulics:siphon-75mm-pipes-summit-height" source: https://www.bctengineer.com/q/semester_4/hydraulics/siphon-75mm-pipes-summit-height site: BCT Engineering --- # Siphon with 75 mm pipes and summit height Hydraulics · Chapter 2: Simple pipe flow problems and solution - Asked in: 2075 Baisakh, 4+4 marks - Video: [Problem on Syphon/ Flow through pipes](https://www.youtube.com/watch?v=Dj1tqNzSfUg) — Edutech Guru Engineering Learning-By Priyanka ma'am - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/siphon-75mm-pipes-summit-height - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2 ## Question Pipes of 75 mm are to be used to syphon water from a main canal to branch canal, the difference of water level between the two canals being 15 m. The length from the main canal to the summit of the pipe line is 20 m. The total length of the pipe line being 50 m. a) Determine the number of pipes required to discharge at least 50 l/sec of water to the branch canal. b) Find also the maximum height of the summit above the water level of the main canal in order the pressure at the summit may not fall below 25 kPa (absolute). Take f = 0.03 and ignore minor loss. ## Short answer Friction loss alone sets the flow; each 75 mm pipe delivers 16.95 L/s, so 3 pipes are needed. The maximum permissible summit height is 1.03 m above the main canal. ## Answer **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With minor losses ignored, the 15 m head equals: A. f L/D x V^2/2g only, giving V^2/2g = 0.75 m B. (1 + f L/D) V^2/2g C. 0.5 V^2/2g D. f L1/D x V^2/2g Answer: A. The whole fall is friction over 50 m: 15 = 20 V^2/2g. (Key point: Velocity with no minor losses) 2. A pipe carries 16.95 l/s. How many are needed for at least 50 l/s? A. 2 B. 4 C. 3 D. 1 Answer: C. 50/16.95 = 2.95, so 3 pipes are needed. (Key point: Number of pipes) 3. Which energy equation gives the maximum summit height z? A. Main surface to outlet B. p_a/gamma = z + p_s/gamma + (V^2/2g)(1 + f L1/D) C. z = 15 - 2.548 D. p_a/gamma = z + p_s/gamma Answer: B. Between the main canal surface and the summit, include velocity head and friction over 20 m, giving z = 1.03 m. (Key point: Summit height) 4. Why is 25 kPa absolute used? A. It is atmospheric pressure B. It is the exit pressure C. It is the average pressure D. It is the minimum allowable absolute pressure at the summit to avoid breaking the water column Answer: D. Pressure is lowest at the summit; the limit prevents cavitation. 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