---
title: "Outlet requirements; flexibility vs sensitivity"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:outlet-requirements-flexibility-sensitivity"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/outlet-requirements-flexibility-sensitivity
site: BCT Engineering
---

# Outlet requirements; flexibility vs sensitivity

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2075 Bhadra, 2+3 marks
- Video: [FLEXIBILITY , SENSITIVITY, PROPORTIONALITY,SETTING OF OUTLETS](https://www.youtube.com/watch?v=_3T_nqc4LIY) — Civil Engineer FBH
- 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/outlet-requirements-flexibility-sensitivity
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Write down essential requirement of outlet. Derive the relationship between Flexibility and Sensitivity.

## Short answer

Outlet must be simple, cheap, robust, tamper-proof, low head, fair silt share. Flexibility F = (fractional change in outlet discharge)/(fractional change in distributary discharge), Sensitivity S = (fractional change in q)/(fractional change in depth D). With Q proportional to D^n, dQ/Q = n dD/D, so F = S/n; S = nF. For wide channel n≈5/3.

## Answer

#### Essential requirements of an outlet (2 marks)
1. Simple and cheap construction, strong, with no moving parts and little maintenance.
2. Proportionality: the discharge should change in proportion to the distributary discharge (flexibility near unity), so the supply is equitable.
3. Low working head, to maintain command.
4. Tamper-proof, with the discharge not affected by the farmer's action in the water course.
5. It should draw its fair share of silt from the distributary.
6. Its discharge should be easy to measure or check.

#### Relationship between flexibility and sensitivity (3 marks)
Let the distributary discharge be $Q$ at depth $D$ (gauge reading $G$, so $dG = dD$), and the outlet discharge be $q$.

- **Flexibility:** $F = \dfrac{dq/q}{dQ/Q}$ (rate of change of outlet discharge to rate of change of distributary discharge).
- **Sensitivity:** $S = \dfrac{dq/q}{dG/D} = \dfrac{dq/q}{dD/D}$ (rate of change of outlet discharge to the change of distributary water level, referred to the normal depth $D$).

The distributary discharge is a function of its depth, $Q = k\,D^{n}$ ($n \approx 5/3$ for a wide channel). Then

$$
\frac{dQ}{Q} = n\,\frac{dD}{D}
$$

Substituting in $F$:

$$
F = \frac{dq/q}{n\,dD/D} = \frac{1}{n}\,\frac{dq/q}{dD/D} = \frac{S}{n}
$$

So

$$
S = n\,F
$$

Hence flexibility and sensitivity are directly proportional: an outlet with greater sensitivity has greater flexibility. For a wide channel ($n = 5/3$), $S = 1.67F$; a proportional outlet ($F = 1$) has $S = n$, and a rigid module has $F = S = 0$.

## Check yourself

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

1. How is flexibility F defined?
   A. F = (dq/q)/dD
   B. F = (dq/q)/(dQ/Q), the ratio of fractional change in outlet discharge to that in distributary discharge
   C. F = dQ/dq
   D. F = q/Q
   Answer: B. Flexibility compares the relative change of outlet discharge with the relative change of distributary discharge. (Key point: Definition of flexibility)

2. What does sensitivity S = (dq/q)/dD measure?
   A. Fractional change of outlet discharge per unit change in distributary depth
   B. Change of distributary discharge per unit depth
   C. Ratio of q to Q
   D. Silt drawn per unit discharge
   Answer: A. S is the relative change of outlet discharge per unit change in distributary water depth. (Key point: Definition of sensitivity)

3. Which relation links Q to D in the derivation?
   A. Q is independent of D
   B. Q ∝ D^n so dQ/Q = n dD/D
   C. Q ∝ 1/D
   D. Q = q D
   Answer: B. Writing Q ∝ D^n gives dQ/Q = n dD/D, which is substituted in F. (Key point: Key step in derivation)

4. What is the final relation between F and S?
   A. F = n/(S D)
   B. F = S/D
   C. F = S D/n, so for fixed D flexibility is directly proportional to sensitivity
   D. F = S n D
   Answer: C. Substituting gives F = S D/n, so a more sensitive outlet is more flexible. (Key point: Result)

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