--- title: "Semi-modular pipe outlet design, flexibility 0.9" subject: "Irrigation and Drainage Engineering" chapter: 7 key: "semester_6/irrigation_and_drainage_engineering:semimodular-pipe-outlet-flexibility-09" source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/semimodular-pipe-outlet-flexibility-09 site: BCT Engineering --- # Semi-modular pipe outlet design, flexibility 0.9 Irrigation and Drainage Engineering · Chapter 7: Regulating Structures - Asked in: 2075 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/semimodular-pipe-outlet-flexibility-09 - Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7 ## Question A distributary channel having bed width 5.0 m, full supply depth of 1.20 m carries 3.0 cumec of water. A semi modular pipe outlet in this channel has a command area of 15 ha growing rice with a kor depth 20 cm and kor period of 3 weeks. Determine the size of the outlet and set it for sub proportionality with a flexibility of 0.9. Assume length of pipe as 3.0 m and friction factor as 0.03. ## Short answer Required discharge q = (15 ha × 0.20 m)/(3 weeks) = 0.01653 m³/s. For flexibility 0.9, set H = 0.40 m (centre 0.80 m above bed). Pipe diameter from head loss equation: adopt 110 mm, length 3.0 m, giving about 17.5 l/s. ## Answer A semi-modular pipe outlet discharges freely, so its discharge depends on the head $H$ between the distributary water level and the centre of the pipe: $q = K H^{m}$ with $m = 1/2$. The required discharge fixes the pipe size, and the flexibility fixes its setting (level). #### 1. Discharge required Kor depth $\Delta = 0.20$ m over the period $T = 3$ weeks $= 21\times86400 = 1\,814\,400$ s; area $A = 15$ ha $= 150\,000$ m²: $$ q = \frac{A\,\Delta}{T} = \frac{150000\times0.20}{1814400} = 0.01653\ \text{m}^3/\text{s} = 16.5\ \text{l/s} $$ #### 2. Setting for flexibility $F = 0.9$ For the outlet $q = KH^{m}$ and for the distributary $Q = kD^{n}$: $$ F = \frac{dq/q}{dQ/Q} = \frac{m}{n}\,\frac{D}{H} $$ For a pipe outlet $m = 1/2$, and for the distributary channel (wide, $B/D = 5/1.2 \approx 4$) $n = 5/3$: $$ \frac{H}{D} = \frac{m}{n\,F} = \frac{0.5}{(5/3)\times0.9} = 0.333 \Rightarrow H = 0.333\times1.2 = 0.40\ \text{m} $$ Centre of the pipe below FSL $= 0.40$ m, i.e. $1.2 - 0.40 = 0.80$ m above the distributary bed. ($F < 1$ gives a **sub-proportional** outlet; for proportionality the setting would be $H/D = 0.3$.) #### 3. Size of the pipe Head loss through a pipe of length $L = 3$ m with entry loss $0.5\dfrac{v^2}{2g}$, friction $f = 0.03$ and exit (velocity head) loss: $$ H = \left(1 + 0.5 + \frac{fL}{d}\right)\frac{v^2}{2g}, \qquad q = \frac{\pi}{4}d^2 v $$ Solving by trial with $H = 0.40$ m and $q = 0.01653$ m³/s gives $d = 0.107$ m. Adopt a **110 mm** diameter pipe. Check: $\left(1.5 + \dfrac{0.03\times3}{0.11}\right) = 2.32$, $v = \sqrt{\dfrac{2\times9.81\times0.40}{2.32}} = 1.84$ m/s, $q = \dfrac{\pi}{4}(0.11)^2\times1.84 = 17.5$ l/s, which is slightly above the 16.5 l/s required. **Answer:** pipe diameter 110 mm (theoretical 107 mm), length 3.0 m, with its centre set 0.40 m below FSL (0.80 m above the distributary bed), giving $F = 0.9$ and a discharge of about 17.5 l/s. ```text FSL ~~~~~~~~~~~~~~~~~~~~~~~~ H = 0.40 m water course D = 1.2 m --pipe 110 mm, L = 3 m--> free outfall bed ___________ 0.80 m above bed ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the outlet discharge computed? A. q = A Δ/T = 150000 x 0.20 / (21 x 86400) = 16.5 l/s B. q = A/T = 150000/21 days C. q = Δ/T x 3 m D. q = 3.0 m³/s x 15/100 Answer: A. Kor depth over the area divided by the kor period in seconds gives 0.01653 m³/s. (Key point: Required discharge) 2. For q = K sqrt(D - s), what head H over the pipe centre gives F = 0.9 with D = 1.2 m? A. H = D = 1.2 m B. H = 2FD = 2.16 m C. H = D/(2F) = 0.667 m D. H = D F = 1.08 m Answer: C. F = D/(2H), so H = D/(2F) = 1.2/1.8 = 0.667 m. (Key point: Head for given flexibility) 3. Where is the pipe centre set above the distributary bed? A. 0.667 m above the bed B. At the bed level C. 1.2 m above the bed D. s = D - H = 0.53 m above the bed, i.e. 0.67 m below FSL Answer: D. The pipe centre is H below FSL, so s = 1.2 - 0.667 = 0.533 m above the bed. (Key point: Setting of the pipe) 4. What does a flexibility of 0.9 indicate? A. Hyper-proportional outlet B. Sub-proportional outlet C. Proportional outlet D. Non-modular outlet Answer: B. F < 1 means the outlet discharge changes by a smaller fraction than the distributary discharge. 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