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Semi-modular pipe outlet design, flexibility 0.9

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.

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.

A semi-modular pipe outlet discharges freely, so its discharge depends on the head HH between the distributary water level and the centre of the pipe: q=KHmq = K H^{m} with m=1/2m = 1/2. The required discharge fixes the pipe size, and the flexibility fixes its setting (level).

1. Discharge required

Kor depth Δ=0.20\Delta = 0.20 m over the period T=3T = 3 weeks =21×86400=1 814 400= 21\times86400 = 1\,814\,400 s; area A=15A = 15 ha =150 000= 150\,000 m²:

q=A ΔT=150000×0.201814400=0.01653 m3/s=16.5 l/sq = \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=0.9F = 0.9

For the outlet q=KHmq = KH^{m} and for the distributary Q=kDnQ = kD^{n}:

F=dq/qdQ/Q=mn DHF = \frac{dq/q}{dQ/Q} = \frac{m}{n}\,\frac{D}{H}

For a pipe outlet m=1/2m = 1/2, and for the distributary channel (wide, B/D=5/1.2≈4B/D = 5/1.2 \approx 4) n=5/3n = 5/3:

HD=mn F=0.5(5/3)×0.9=0.333⇒H=0.333×1.2=0.40 m\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= 0.40 m, i.e. 1.2−0.40=0.801.2 - 0.40 = 0.80 m above the distributary bed. (F<1F < 1 gives a sub-proportional outlet; for proportionality the setting would be H/D=0.3H/D = 0.3.)

3. Size of the pipe

Head loss through a pipe of length L=3L = 3 m with entry loss 0.5v22g0.5\dfrac{v^2}{2g}, friction f=0.03f = 0.03 and exit (velocity head) loss:

H=(1+0.5+fLd)v22g,q=π4d2vH = \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=0.40H = 0.40 m and q=0.01653q = 0.01653 m³/s gives d=0.107d = 0.107 m. Adopt a 110 mm diameter pipe. Check: (1.5+0.03×30.11)=2.32\left(1.5 + \dfrac{0.03\times3}{0.11}\right) = 2.32, v=2×9.81×0.402.32=1.84v = \sqrt{\dfrac{2\times9.81\times0.40}{2.32}} = 1.84 m/s, q=π4(0.11)2×1.84=17.5q = \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=0.9F = 0.9 and a discharge of about 17.5 l/s.

 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

✓ Answer checked by Claude Opus 5.5 (high effort)

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