Design cross and head regulator, Q 125

A cross regulator (in the parent canal, downstream of the off-take) holds the water level up for the distributary. The distributary head regulator controls the supply into the distributary. Both are designed as gated structures with Khosla's floor.

Data

Parent canal: $Q = 125$ m³/s, FSL u/s 120.00 m and d/s 119.70 m, depth 3 m, bed width 50 m (u/s) and 45 m (d/s), so bed levels are 117.00 m (u/s) and 116.70 m (d/s). Distributary: $Q = 20$ m³/s, FSL 119.00 m, bed width 20 m, depth 1.8 m, so bed 117.20 m. $f = 0.8$, safe exit gradient 1/6.

A. Cross regulator (6 marks)

1. Discharge through it: $Q_{cr} = 125 - 20 = 105$ m³/s. The permitted afflux (head loss) is $\Delta h = 120.00 - 119.70 = 0.30$ m.

2. Crest and waterway: the crest (sill) is kept at the d/s bed level, 116.70 m. With the gates fully raised the opening acts as an orifice of depth $D = 3.0$ m, $C_d = 0.8$:

$$\begin{aligned} B &= \frac{Q_{cr}}{C_d\,D\sqrt{2g\,\Delta h}} = \frac{105}{0.8\times3.0\times\sqrt{2\times9.81\times0.30}} = 18.03\ \text{m} \end{aligned}$$

Provide 4 bays of 5 m = 20 m clear with 3 piers of 1.5 m; overall width 24.5 m. With 20 m the actual afflux is 0.244 m $< 0.3$ m, safe.

3. Floor: discharge intensity $q = 105/24.5 = 4.29$ m²/s, Lacey scour $R = 1.35(q^2/f)^{1/3} = 3.84$ m. D/s cut-off bottom $= 119.70 - 1.5R = 113.94$ m, so $d = 116.70 - 113.94$, adopt $d = 2.8$ m (u/s cut-off 2.0 m). Maximum static head $H_s = 120.00 - 116.70 = 3.30$ m (d/s dry):

$$\frac{1}{\pi\sqrt{\lambda}} = \frac{d}{6H_s} = \frac{2.8}{6\times3.30} \Rightarrow \lambda = 5.07,\ \alpha = 9.08,\ b = \alpha d = 25.4\ \text{m}$$

Provide floor length 26 m with 1.2 m thickness tapering to 0.8 m, u/s and d/s cut-offs, d/s inverted filter and stone apron.

B. Distributary head regulator (6 marks)

1. Discharge $Q = 20$ m³/s; head loss from the parent FSL 120.00 m to the distributary FSL 119.00 m is $\Delta h = 1.00$ m.

2. Waterway: sill at distributary bed level 117.20 m, opening depth $D = 1.8$ m, $C_d = 0.8$:

$$B = \frac{20}{0.8\times1.8\times\sqrt{2\times9.81\times1.0}} = 3.14\ \text{m}$$

Provide 2 bays of 2.0 m (4.0 m clear, pier 1.2 m; overall 5.2 m). With the gates fully open the loss is only 0.61 m, so the gates throttle the flow to give the required discharge.

3. Floor: $q = 20/5.2 = 3.85$ m²/s, $R = 3.57$ m. D/s cut-off to $119.00 - 1.5R = 113.65$ m, so $d = 3.6$ m below the floor (117.20 m). $H_s = 120.00 - 117.20 = 2.80$ m. For $G_E = 1/6$: $\lambda = 2.21$, $\alpha = 3.26$, floor length $b = 11.7$ m; provide 12 m. Floor thickness 1.0 m.

Answer: cross regulator: crest 116.70 m, clear waterway 20 m (overall 24.5 m), floor 26 m, cut-off depth 2.8 m. Head regulator: sill 117.20 m, clear waterway 4.0 m, floor 12 m, d/s cut-off 3.6 m.

 Plan: parent canal ====>====[HR]====> distributary
                       |  [CR] across parent canal
                       v       (u/s of CR: FSL 120.0)
 Section: gates | crest 116.70 | floor b | d/s cut-off d