---
title: "Adjustable orifice semi-module design"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:adjustable-orifice-semimodule-022"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/adjustable-orifice-semimodule-022
site: BCT Engineering
---

# Adjustable orifice semi-module design

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2079 Chaitra, 4 marks
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/adjustable-orifice-semimodule-022
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

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.

## 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).

## Answer

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

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

#### 1. Size of the orifice
$$
q = C_d\,B\,Y\,\sqrt{2gH}
$$
Adopt orifice width $B = 0.50$ m. Then
$$
\begin{aligned} Y &= \frac{q}{C_d\,B\sqrt{2gH}} = \frac{0.22}{0.7\times0.5\times\sqrt{2\times9.81\times0.7}} \\ &= \frac{0.22}{0.7\times0.5\times3.706} = 0.170\ \text{m} \end{aligned}
$$
Adopt $Y = 0.17$ m. Check: $q = 0.7\times0.5\times0.17\times3.706 = 0.221$ m³/s.

#### 2. Levels
- Underside of the roof block $= FSL - H = 101.40 - 0.70 = 100.70$ m.
- Sill of the orifice $= 100.70 - 0.17 = 100.53$ 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 $Y$ and the discharge.

#### 3. Other dimensions
- Bell-mouthed entry and a parallel throat about $1.5Y$ 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 $H_{min} \approx 0.1$ to $0.15$ m below the roof block, so the outlet is modular for water course levels below the sill.

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

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does the Crump adjustable orifice semi-module vary discharge?
   A. By changing the pipe length
   B. By raising or lowering a roof block, which changes the orifice height Y
   C. By changing the water course level
   D. By changing the flume width only
   Answer: B. The roof block adjusts the orifice height; the downstream flume makes the outlet semi-modular. (Key point: Principle of AOSM)

2. Which equation gives the orifice height Y?
   A. Y = q Cd B sqrt(2gH)
   B. Y = H/B
   C. Y = q/(Cd B sqrt(2gH)) = 0.22/(0.7 x 0.5 x 3.706) = 0.17 m
   D. Y = q/(B H)
   Answer: C. From q = Cd B Y sqrt(2gH), Y = 0.17 m for B = 0.5 m, H = 0.7 m. (Key point: Sizing equation)

3. What is the level of the underside of the roof block?
   A. FSL + H = 102.10 m
   B. FSL - H = 101.40 - 0.70 = 100.70 m
   C. Bed + Y = 100.17 m
   D. Bed level 100.00 m
   Answer: B. The working head is measured below FSL, so the roof block underside is at 100.70 m. (Key point: Roof block level)

4. Why is the AOSM discharge independent of the water course level?
   A. The orifice is closed
   B. The pipe is long
   C. The sill is at the bed
   D. A hydraulic jump forms in the expanding flume downstream, so the jet is not submerged
   Answer: D. The flume d/s of the orifice forms a jump, so tail water changes do not reach the orifice (modular if head exceeds the minimum). (Key point: Why discharge is independent of water course)

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