Skip to main content

Adjustable orifice semi-module design

Short answer

For q=0.22 m³/s and head 0.7 m, adopt orifice width 0.50 m, height Y=0.17 m. Sill level = 100.53 m, roof block at 100.70 m (FSL 101.40, bed 100.00).

Design an adjustable orifice semi-module for the following data: Discharge of the outlet = 0.22 m³/s; Working head = 0.7 m; F.S.L. of distributary = 101.40; Bed level of distributary = 100.00.

The Crump adjustable orifice semi-module (AOSM) has a rectangular orifice whose height YY can be changed by raising or lowering a roof block, while a flume d/s makes the outlet semi-modular.

Data

q=0.22q = 0.22 m³/s, working head H=0.7H = 0.7 m, FSL of distributary =101.40= 101.40 m, bed =100.00= 100.00 m (depth D=1.40D = 1.40 m). Discharge coefficient of the orifice Cd=0.7C_d = 0.7 (assumed, as is usual for the AOSM).

1. Size of the orifice

q=Cd B Y 2gHq = C_d\,B\,Y\,\sqrt{2gH}

Adopt orifice width B=0.50B = 0.50 m. Then

Y=qCd B2gH=0.220.7×0.5×2×9.81×0.7=0.220.7×0.5×3.706=0.170 m\begin{aligned} Y &= \frac{q}{C_d\,B\sqrt{2gH}} = \frac{0.22}{0.7\times0.5\times\sqrt{2\times9.81\times0.7}} \\ &= \frac{0.22}{0.7\times0.5\times3.706} = 0.170\ \text{m} \end{aligned}

Adopt Y=0.17Y = 0.17 m. Check: q=0.7×0.5×0.17×3.706=0.221q = 0.7\times0.5\times0.17\times3.706 = 0.221 m³/s.

2. Levels

  • Underside of the roof block =FSL−H=101.40−0.70=100.70= FSL - H = 101.40 - 0.70 = 100.70 m.
  • Sill of the orifice =100.70−0.17=100.53= 100.70 - 0.17 = 100.53 m, which is 0.53 m above the distributary bed (this lets the outlet draw bed silt in proportion).
  • The roof block is adjustable by ±\pm some centimetres, which changes YY and the discharge.

3. Other dimensions

  • Bell-mouthed entry and a parallel throat about 1.5Y1.5Y long (0.26 m).
  • Downstream expanding flume (1 in 5 to 1 in 6 flare) discharging into the water course; a hydraulic jump forms in the flume, so the discharge is independent of the water course level.
  • The jet should not be submerged: the minimum modular head is Hmin≈0.1H_{min} \approx 0.1 to 0.150.15 m below the roof block, so the outlet is modular for water course levels below the sill.

Answer: orifice B×Y=0.50×0.17B \times Y = 0.50\times0.17 m, sill at 100.53 m, roof block at 100.70 m.

 FSL 101.40 ~~~~~~~~~~~~~~~|
 roof block 100.70 ______|_______
                        <- Y ->  \____flume___ water course
 sill 100.53 ____________________\_________
 bed 100.00 _________________________________

All Irrigation and Drainage Engineering chapter 7 questions →