Outlet requirements; flexibility vs sensitivity

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 $F = S = 0$.