--- title: "Siphon with 90 mm pipe, number and length" subject: "Hydraulics" chapter: 2 key: "semester_4/hydraulics:siphon-90mm-pipe-total-length" source: https://www.bctengineer.com/q/semester_4/hydraulics/siphon-90mm-pipe-total-length site: BCT Engineering --- # Siphon with 90 mm pipe, number and length Hydraulics · Chapter 2: Simple pipe flow problems and solution - Asked in: 2082 Kartik, 4+4 marks - Video: [Siphon Numerical to find discharge and pressure at summit (HYDRAULIC 4th sem)](https://www.youtube.com/watch?v=yPkTb2B2cwc) — CIVIL ENGINEERING - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/siphon-90mm-pipe-total-length - 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 wrought iron pipe of 90 mm diameter. The length of pipe up to summit is 25 m, height of summit above main canal is 2 m and the water pressure at the summit is equal to 20 kN/m² absolute. Water surface elevation in the branch canal is 10 m below that of main canal and assume friction factor as 0.03 and consider minor losses. (i) If total quantity of water required to be conveyed is 0.06 m³/s, how many pipelines are needed? (ii) What is the total length of the pipe needed? ## Short answer Using the summit pressure limit, each 90 mm pipe carries 0.02253 m³/s; three pipelines are required. The total pipe length needed is 42.4 m. ## Answer **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In (i), what fixes the velocity? A. The 10 m level difference B. The minimum summit pressure of 20 kN/m2 absolute, applied through the energy equation to the summit C. The friction factor alone D. The pipe diameter alone Answer: B. The summit pressure limit gives 6.287 = 9.833 V^2/2g, so V = 3.542 m/s. (Key point: Controlling condition for velocity) 2. How many pipelines are needed and why? A. 2, because 0.06/0.0225 rounds down B. 4, because of minor losses C. 3, since 0.06/0.02253 = 2.66 is rounded up D. 1, because the head is large Answer: C. One pipe carries 22.53 l/s, so 2.66 pipes are required; round up to 3. (Key point: Number of pipelines) 3. How is the total pipe length found? A. L = 25 m plus 10 m B. Total length equals twice the summit length C. L from the summit height divided by f D. Using 10 = (0.5 + f L/D + 1.0) V^2/2g at the same velocity, giving L = 42.4 m Answer: D. f L/D = 10/0.6394 - 1.5 = 14.14, so L = 14.14 x 0.09/0.03 = 42.4 m. 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