---
title: "Canal head regulator design, Q 200"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:canal-head-regulator-design-q200"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/canal-head-regulator-design-q200
site: BCT Engineering
---

# Canal head regulator design, Q 200

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2076 Baisakh, 12 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/canal-head-regulator-design-q200
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Design a canal Head Regulator for the data given below. Crest level of under sluice = 300 m; Pond level = 304 m; Silt factor (f) = 1; U/S HFL = 307 m; Full Supply discharge = 200 m³/s; Full Supply level = 303 m; Bed level of canal = 299.50 m. A silt excluder is provided in the under sluice; Take GE = 1/6.

## Short answer

Crest level 301.20 m; waterway 25 m clear (5 bays of 5 m), overall 31.0 m; cut-offs u/s 2.5 m, d/s 3.6 m; impervious floor length 38 m.

## Answer

#### Data and assumptions
- Discharge $Q = 200$ m³/s; pond level $= 304$ m; full supply level of canal $= 303$ m; canal bed $= 299.5$ m; u/s HFL $= 307$ m; Lacey's $f = 1$; safe exit gradient $G_E = 1/6$.
- Crest of the head regulator is kept 1.2 m above the crest of the under-sluice (300.00 m): $300.00 + 1.2 = 301.20$ m, so bed load is excluded.
- The regulator is designed as a broad-crested weir for the free-flow condition, $Q = 1.705\,L\,H^{3/2}$.

#### 1. Crest level
$$
\text{Crest level} = 301.20\ \text{m}
$$
Head over the crest at pond level: $H = 304 - 301.20 = 2.80$ m. Depth of canal water over the crest: $FSL - \text{crest} = 1.80$ m, which is less than $\tfrac{2}{3}H$, so the weir is not drowned and the free-flow formula applies.

#### 2. Waterway
$$
L = \frac{Q}{1.705\,H^{3/2}} = \frac{200}{1.705\times2.80^{3/2}} = 25.04\ \text{m}
$$
Provide 5 bays of 5.0 m clear width (25 m) with 4 piers 1.5 m thick; overall width $= 31.0$ m. Check: $Q = 1.705\times25\times2.80^{3/2} = 199.7$ m³/s, which is accepted as $\approx 200$ m³/s.

#### 3. Impervious floor (Khosla's exit gradient)
Discharge intensity $q = Q/\text{width} = 200/31.0 = 6.45$ m²/s. Lacey's scour depth:
$$
R = 1.35\left(\frac{q^2}{f}\right)^{1/3} = 1.35\left(\frac{6.45^2}{1}\right)^{1/3} = 4.68\ \text{m}
$$
The d/s cut-off is taken down to $1.5R$ below the d/s water level: bottom level $= 303 - 1.5\times4.68 = 295.98$ m, so the d/s cut-off depth below the floor (canal bed) is $d = 299.5 - 295.98$, adopt **$d = 3.6$ m**. U/s cut-off $d_1 = 2.5$ m.
Maximum static head (canal dry, water at pond level): $H_s = 304 - 299.5 = 4.5$ m. The gates are closed in floods; the HFL (307 m) fixes the top of the piers and gates.
Exit gradient:
$$
G_E = \frac{H_s}{d}\,\frac{1}{\pi\sqrt{\lambda}} = \frac{1}{6} \Rightarrow \frac{1}{\pi\sqrt{\lambda}} = \frac{3.6}{6\times4.5} = 0.1333
$$
$$
\lambda = 5.70 \Rightarrow \alpha = \frac{b}{d} = 10.35 \Rightarrow b = 10.35\times3.6 = 37.3\ \text{m}
$$
(Using $\lambda = \frac{1+\sqrt{1+\alpha^2}}{2}$.) Check: $G_E = 0.1667 = 1/6.00$. Provide a floor of **38 m**: about 1/3 on the u/s side and 2/3 downstream, thickness 1.5 m under the gates tapering to 0.8 m.

#### 4. Other features
- Top of piers and gate: HFL + 1.0 m free-board. Vertical lift steel gates with a hoist deck.
- D/s of the floor: inverted filter and loose stone apron (length about $2D$), and wing walls splayed 45°.
- The silt excluder in the under-sluice keeps bed load away, so the regulator crest only needs to be 1.2 m above the under-sluice crest.

#### Summary
| Item | Value |
|---|---|
| Crest level | 301.20 m |
| Waterway | 25 m clear (5 bays of 5 m), overall 31.0 m |
| Cut-offs u/s / d/s | 2.5 m / 3.6 m |
| Floor length | 38 m |

Conceptual sketch (longitudinal section):
```text
 HFL / pond ~~~~~~~~~~|gates|~~~~~~ canal FSL ~~~~~
 river        ________|piers|________
 bed ________|  crest  (1.2 m above   |_____ canal bed
     |d1 u/s floor     U/S sluice crest) d/s floor | d2
       (impervious)                     inverted filter
       <------------ b (Khosla, GE = 1/6) --------->
```

## Check yourself

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

1. Why is the head regulator crest set at 301.20 m?
   A. It is placed 1.2 m above the under-sluice crest (300 m) so bed load is excluded
   B. It equals the canal bed level 299.5 m plus 1.7 m
   C. It is set at pond level minus 2.8 m by Lacey's formula
   D. It equals the full supply level minus 1.8 m for the gates
   Answer: A. With a silt excluder, the regulator crest is kept 1.2 m above the under-sluice crest (300 + 1.2 = 301.20 m) so heavy bed load passes the under-sluice. (Key point: Crest level of head regulator)

2. Why can the free-flow formula Q = 1.705 L H^(3/2) be used?
   A. Because the flow is always submerged
   B. Because the canal depth over the crest (1.8 m) is less than 2/3 of H (2.8 m), so the weir is not drowned
   C. Because H is less than 1 m
   D. Because the gates are closed
   Answer: B. The d/s water depth over the crest is 1.80 m, below 2/3 of 2.80 m, so the broad-crested weir discharges freely. (Key point: Free flow check)

3. What clear waterway is required, and how is it found?
   A. About 6.45 m from q = Q/width
   B. About 38 m from Khosla's theory
   C. About 25 m from L = Q/(1.705 H^(3/2)) with H = 2.80 m
   D. About 3.6 m from the cut-off depth
   Answer: C. L = 200/(1.705 x 2.8^1.5) = 25.04 m, so 5 bays of 5 m (25 m) are provided. (Key point: Waterway)

4. How is the d/s cut-off depth d = 3.6 m obtained?
   A. Bottom level = d/s water level - 1.5R with Lacey's R, measured below the canal bed
   B. Equal to the maximum static head 4.5 m
   C. Equal to the crest height above the bed
   D. Taken as half of the floor length
   Answer: A. R = 1.35(q^2/f)^(1/3) = 4.68 m; bottom level 303 - 1.5(4.68) = 295.98 m, which is 3.5 m, adopted as 3.6 m below the bed 299.5 m. (Key point: Cut-off depth)

5. With Hs = 4.5 m and d = 3.6 m, how is the floor length found for GE = 1/6?
   A. b = 6 x Hs
   B. b = Hs x d
   C. b = 1/GE
   D. 1/(π sqrt(λ)) = d/(6Hs) = 0.1333, so λ = 5.70, α = 10.35 and b = α d = 37.3 m (38 m provided)
   Answer: D. Khosla's exit gradient relation gives α from λ, and b = α d, i.e. about 38 m with the regulator closed and the canal dry. (Key point: Floor length)

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