---
title: "Canal in alluvium, Q 64, d 0.5 mm"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:alluvial-canal-q64-d05mm"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/alluvial-canal-q64-d05mm
site: BCT Engineering
---

# Canal in alluvium, Q 64, d 0.5 mm

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2068 Baisakh (old course), 1+6+1 marks
- Video: [Lacey's sum | Type 1 when Silt factor is given & Side slope is 1/2 : 1 | Lacey's  Design of canal](https://www.youtube.com/watch?v=U4SwFYekkLg) — Professor Nidhi Trivedi
- 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/alluvial-canal-q64-d05mm
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

An irrigation canal passing through the alluvium soil (d_mean = 0.50 mm) carries a discharge of 64 m³/s. What major principle you apply while designing such a canal? Find out the principal dimension and bed slope of the canal and draw a sketch of the designed section.

## Short answer

Regime principle (Lacey) applied. f=1.245; dimensions: bed width 33.6 m, depth 1.96 m, side slope 0.5H:1V, bed slope 1 in 4639, velocity 0.94 m/s.

## Answer

#### Principle

An alluvial canal is designed on the **regime principle** (Lacey's theory): the channel carries silt-laden water through alluvium of its own kind, and its section, slope and velocity adjust until it neither silts nor scours. The silt grade is represented by the silt factor $f = 1.76\sqrt{d_{mm}}$. No side slope is given, so 0.5H:1V is assumed.

#### Silt factor

$$
f = 1.76\sqrt{d_{mm}} = 1.76\sqrt{0.5} = 1.245
$$

#### Lacey's regime values

Given: $Q = 64\ \text{m}^3/\text{s}$, $f = 1.245$, side slope 0.5H:1V.

$$
\begin{aligned}
V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{64\times 1.245^2}{140}\right)^{1/6} = 0.944\ \text{m/s} \\
A &= \frac{Q}{V} = \frac{64}{0.944} = 67.79\ \text{m}^2 \\
R &= \frac{5V^2}{2f} = \frac{5\times 0.944^2}{2\times 1.245} = 1.790\ \text{m} \\
P &= 4.75\sqrt{Q} = 4.75\sqrt{64} = 38.00\ \text{m} \\
S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.245^{5/3}}{3340\times 64^{1/6}} = 0.000216 = \frac{1}{4639}
\end{aligned}
$$

#### Trapezoidal section

For side slope $z = 0.5$: $A = (B + zD)D$ and $P = B + 2D\sqrt{1+z^2} = B + 2.2361D$.

Eliminating $B$:

$$
\begin{aligned}
1.7361\,D^2 - 38.00\,D + 67.79 &= 0 \\
D &= \frac{38.00 - \sqrt{38.00^2 - 4\times 1.7361\times 67.79}}{2\times 1.7361} = 1.959\ \text{m} \\
B &= P - 2.2361D = 38.00 - 2.2361\times 1.959 = 33.62\ \text{m}
\end{aligned}
$$

(The other root of the quadratic gives a very deep, narrow section, so the smaller root is taken.)

Check: $A = (33.62 + 0.5\times 1.959)\times 1.959 = 67.79\ \text{m}^2$ against required 67.79 $\text{m}^2$.

**Answer:** Bed width $B \approx 33.6$ m, depth of water $D \approx 1.96$ m, side slope 0.5H:1V, bed slope $S = 1$ in 4639 (0.000216), velocity $V = 0.94$ m/s, hydraulic mean radius $R = 1.79$ m.

#### Sketch of designed section

```text
      FSL
  ____|~~~~~~~~~~~~~~~~~~~~~~~|____
     /<--------- T --------->\
    /  D = 1.96 m            \  1V : 0.5H
   /___________________________\
        B = 33.6 m
```
Top width $T = B + 2zD = 35.58$ m (freeboard to be added above FSL).

## Check yourself

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

1. Which principle is used for canals in alluvium?
   A. Tractive force only
   B. Rigid-boundary Manning design
   C. Kennedy's trial and error only
   D. Regime principle (Lacey's theory) with silt factor
   Answer: D. The canal neither silts nor scours. (Key point: Principle)

2. For Q = 64 m^3/s, what wetted perimeter does Lacey's P = 4.75 sqrt(Q) give?
   A. 19.0 m
   B. 38.00 m
   C. 57.0 m
   D. 48.0 m
   Answer: B. P = 4.75 x sqrt(64) = 38.00 m, independent of silt factor. (Key point: Wetted perimeter)

3. Using S = f^(5/3)/(3340 Q^(1/6)) with f = 1.245, Q = 64, the bed slope is about
   A. 1 in 463
   B. 1 in 46390
   C. 1 in 4639
   D. 1 in 2319
   Answer: C. Slope is computed directly from Lacey's slope equation; a finer silt (smaller f) gives a flatter slope. (Key point: Regime slope)

4. After finding A and P, how is the depth D of the 0.5H:1V trapezoid obtained, and which root is taken (result D = 1.959 m, B = 33.62 m)?
   A. Assume D and iterate until V = V0
   B. Take D = R directly and find B from A
   C. Take the larger root, giving a deep narrow section
   D. Eliminate B using A = (B + 0.5D)D and P = B + 2.236D, solve the quadratic and take the smaller root
   Answer: D. The two equations in B and D give a quadratic; the smaller root is the sensible section, the larger gives a very deep narrow channel. (Key point: Trapezoidal section from A and P)

5. Top width of the designed section is T = B + 2zD, about
   A. 35.58 m
   B. 33.6 m
   C. 37.5 m
   D. 38.0 m
   Answer: A. 33.62 + 2 x 0.5 x 1.959 = 35.58 m. (Key point: Top width)

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