---
title: "Design cross and head regulator, Q 125"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:cross-head-regulator-design-q125"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/cross-head-regulator-design-q125
site: BCT Engineering
---

# Design cross and head regulator, Q 125

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2078 Baisakh, 6+6 marks
- Video: [DESIGN DISTRIBUTARY HEAD RAGULATORY | IRRIGATION NUMERICAL | MOST IMPORTANT](https://www.youtube.com/watch?v=g-fEq5UxpKc) — Sudip Dhakal
- 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/cross-head-regulator-design-q125
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Design a cross regulator and head regulator for a channel which takes off from the parent channel with following data: (i) Discharge of parent channel = 125 cumecs (ii) Discharge of distributary channel = 20 cumecs (iii) FSL of parent channel U/S = 120 m, D/S = 119.7 m (iv) Bed width of Parent channel U/S = 50 m, D/S = 45 m (v) Depth of water in the parent channel U/S = 3 m, D/S = 3 m (vi) FSL of distributary = 119 m (vii) Bed width of distributary = 20 m (viii) Full supply depth of distributary = 1.8 m (ix) Silt factor = 0.8 (x) Assume safe exit gradient = 1/6.

## Short 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.

## Answer

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.

```text
 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
```

## Check yourself

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

1. What discharge must the cross regulator pass in this design?
   A. 125 m³/s, the full parent discharge
   B. 105 m³/s, parent discharge minus the distributary discharge
   C. 20 m³/s, the distributary discharge
   D. 145 m³/s, parent plus distributary discharge
   Answer: B. The distributary draws 20 m³/s at the head regulator, so only 125 - 20 = 105 m³/s continues down the parent canal through the cross regulator. (Key point: Discharge through cross regulator)

2. Why is a cross regulator provided in the parent canal downstream of the off-take?
   A. To hold the water level up so the distributary can draw its supply
   B. To measure the discharge in the parent canal
   C. To lower the parent canal bed level
   D. To pass floods over the canal
   Answer: A. The cross regulator creates the afflux (here 0.30 m) that keeps the u/s level high enough for the distributary head regulator to command its supply. (Key point: Role of the cross regulator)

3. How is the clear waterway B of the cross regulator found?
   A. From Lacey's regime width formula
   B. From B = Q/(Cd D sqrt(2g Δh)) treating the opening as an orifice under the permitted afflux
   C. From Khosla's exit gradient
   D. By taking B equal to the parent bed width
   Answer: B. With gates raised the opening behaves as an orifice of depth D, so B = Q_cr/(Cd D sqrt(2gΔh)), giving about 18 m, and 20 m clear is provided. (Key point: Waterway from orifice flow)

4. How is the length of the impervious floor of the regulators obtained?
   A. From Bligh's creep only, ignoring the exit gradient
   B. As 1.5 times the scour depth R
   C. From the safe exit gradient (1/6) using Khosla's relation with the static head Hs and cut-off depth d to get alpha and b = alpha d
   D. As equal to the overall waterway width
   Answer: C. Setting 1/(π sqrt(λ)) = d/(6 Hs) gives λ, then α, and the floor length b = α d, which was 25.4 m (26 m provided) for the cross regulator. (Key point: Floor length from exit gradient)

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