---
title: "Canal head regulator design, Q 150"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:canal-head-regulator-design-q150"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/canal-head-regulator-design-q150
site: BCT Engineering
---

# Canal head regulator design, Q 150

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2074 Bhadra, 8 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-q150
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

A canal carrying 150 m³/s is to take off from the headwork. The HFL and average bed level of river is 257 m and 250 m respectively. The canal bed level, full supply level and pond level are 249.5 m, 253.0 m and 254.0 m respectively and Lacey's silt factor is equal to unity. Fix the crest level and water way of canal head regulator and also determine the length of impervious floor if safe exit gradient GE = 1/6. Draw the conceptual sketch of canal head regulator.

## Short answer

Crest level 251.20 m; waterway 20 m clear (4 bays of 5 m), overall 24.5 m; d/s cut-off 3.3 m; impervious floor length 42 m.

## Answer

#### Data and assumptions
- Discharge $Q = 150$ m³/s; pond level $= 254$ m; full supply level of canal $= 253$ m; canal bed $= 249.5$ m; u/s HFL $= 257$ m; Lacey's $f = 1$; safe exit gradient $G_E = 1/6$.
- The river bed is at 250.00 m; the crest is kept 1.2 m above the river bed (under-sluice crest level): $250.00 + 1.2 = 251.20$ m.
- The regulator is designed as a broad-crested weir for the free-flow condition, $Q = 1.705\,L\,H^{3/2}$.

#### 1. Crest level
$$
\text{Crest level} = 251.20\ \text{m}
$$
Head over the crest at pond level: $H = 254 - 251.20 = 2.80$ m. Depth of canal water over the crest: $FSL - \text{crest} = 1.80$ m, which is less than $\tfrac{2}{3}H = 1.87$ m, so the weir is not drowned and the free-flow formula applies.

#### 2. Waterway
$$
L = \frac{Q}{1.705\,H^{3/2}} = \frac{150}{1.705\times2.80^{3/2}} = 18.78\ \text{m}
$$
Provide 4 bays of 5.0 m clear width (20 m) with 3 piers 1.5 m thick; overall width $= 24.5$ m. Check: with the gates fully open, $Q = 1.705\times20\times2.80^{3/2} = 159.8$ m³/s $> 150$ m³/s, so the waterway is adequate and the gates are partly lowered to pass the design discharge.

#### 3. Impervious floor (Khosla's exit gradient)
Discharge intensity $q = Q/\text{width} = 150/24.5 = 6.12$ m²/s. Lacey's scour depth:
$$
R = 1.35\left(\frac{q^2}{f}\right)^{1/3} = 1.35\left(\frac{6.12^2}{1}\right)^{1/3} = 4.52\ \text{m}
$$
The d/s cut-off is taken down to $1.5R$ below the d/s water level: bottom level $= 253 - 1.5\times4.52 = 246.22$ m, so the d/s cut-off depth below the floor (canal bed) is $d = 249.5 - 246.22 = 3.28$ m; adopt **$d = 3.3$ m**. U/s cut-off $d_1 = 2.5$ m.
Maximum static head (canal dry, water at pond level): $H_s = 254 - 249.5 = 4.5$ m. The gates are closed during floods; the HFL (257 m) fixes the top of the piers and gates.
Exit gradient:
$$
G_E = \frac{H_s}{d}\,\frac{1}{\pi\sqrt{\lambda}} = \frac{1}{6} \Rightarrow \frac{1}{\pi\sqrt{\lambda}} = \frac{3.3}{6\times4.5} = 0.1222
$$
$$
\lambda = 6.78 \Rightarrow \alpha = \frac{b}{d} = 12.53 \Rightarrow b = 12.53\times3.3 = 41.3\ \text{m}
$$
(Using $\lambda = \frac{1+\sqrt{1+\alpha^2}}{2}$.) Check: $G_E = 0.1667 = 1/6.00$. Provide a floor of **42 m**: about 1/3 on the u/s side and 2/3 downstream, with thickness 1.5 m under the gates tapering to 0.8 m.

#### 4. Other features
- Top of piers and gates: 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 at 45°.

#### Summary
| Item | Value |
|---|---|
| Crest level | 251.20 m |
| Waterway | 20 m clear (4 bays of 5 m), overall 24.5 m |
| Cut-offs u/s / d/s | 2.5 m / 3.3 m |
| Floor length | 42 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       river bed)   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. What is the crest level of this head regulator and why?
   A. 249.50 m, same as the canal bed
   B. 254.00 m, same as the pond level
   C. 253.00 m, same as the full supply level
   D. 251.20 m, 1.2 m above the river bed/under-sluice crest (250 m)
   Answer: D. The crest is kept 1.2 m above the river bed level, 250 + 1.2 = 251.20 m, to keep out bed load. (Key point: Crest level)

2. Why is the head regulator treated as a free-flow broad-crested weir?
   A. Canal depth over the crest (1.8 m) is less than 2/3 of the head H = 2.8 m, so it is not drowned
   B. The HFL is lower than the crest
   C. The gates are fully closed
   D. The silt factor equals one
   Answer: A. FSL - crest = 253 - 251.2 = 1.8 m < 2/3 x 2.8 m = 1.87 m, so free flow Q = 1.705 L H^1.5 applies. (Key point: Free flow condition)

3. What is the computed waterway and what head H is used?
   A. L = 150/(1.705 x 4.5^1.5) = 9.2 m
   B. L = 150/(1.705 x 2.8^1.5) = 18.8 m, giving 4 bays of 5 m (20 m)
   C. L = 150/(1.705 x 1.8^1.5) = 36 m
   D. L = 150/6.12 = 24.5 m
   Answer: B. H is pond level minus crest = 254 - 251.2 = 2.8 m, so L = 18.78 m and 20 m clear is provided. (Key point: Waterway)

4. Which head is used as Hs in Khosla's exit gradient calculation?
   A. HFL minus canal bed = 7.5 m
   B. Pond level minus canal bed = 4.5 m, canal dry, water at pond level
   C. FSL minus canal bed = 3.5 m
   D. Pond level minus crest = 2.8 m
   Answer: B. The critical condition is canal dry with pond at 254 m, so Hs = 254 - 249.5 = 4.5 m; floods are handled by closing the gates. (Key point: Maximum static head for floor)

5. With d = 3.3 m, Hs = 4.5 m and GE = 1/6, what floor length results?
   A. About 3.3 m
   B. About 4.5 m
   C. About 20 m
   D. About 41-42 m, from 1/(π sqrt(λ)) = 3.3/(6 x 4.5), α = 12.53
   Answer: D. 1/(π sqrt λ) = 0.1222 gives λ = 6.78, α = 12.53 and b = 12.53 x 3.3 = 41.3 m, so 42 m is provided. (Key point: Floor length)

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