---
title: "Lacey design: Q 15, d 0.33 mm"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:lacey-q15-d033mm"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lacey-q15-d033mm
site: BCT Engineering
---

# Lacey design: Q 15, d 0.33 mm

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2076 Baisakh, 8 marks
- Video: [Design of Unlined(Earthen) Canal by Lacey's Theory | Irrigation Engineering | Detailed Explanation](https://www.youtube.com/watch?v=oX1fFUDsjfk) — Shivam Mehta
- 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/lacey-q15-d033mm
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Design an irrigation canal using Lacey's theory, when Q = 15 cumec, mean diameter of silt particle is 0.33 mm; side slope = ½:1 (H:V).

## Short answer

Silt factor f = 1.76√0.33 = 1.011. Lacey's regime values: V = 0.69 m/s, A = 21.7 m², R = 1.18 m, P = 18.4 m, slope S = 0.000194 (1/5150). Trapezoidal section (0.5:1) gives D = 1.35 m, B = 15.4 m.

## Answer

#### Silt factor

$$
f = 1.76\sqrt{d_{mm}} = 1.76\sqrt{0.33} = 1.011
$$

#### Lacey's regime values

Given: $Q = 15\ \text{m}^3/\text{s}$, $f = 1.011$, side slope 0.5H:1V.

$$
\begin{aligned}
V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{15\times 1.011^2}{140}\right)^{1/6} = 0.692\ \text{m/s} \\
A &= \frac{Q}{V} = \frac{15}{0.692} = 21.69\ \text{m}^2 \\
R &= \frac{5V^2}{2f} = \frac{5\times 0.692^2}{2\times 1.011} = 1.183\ \text{m} \\
P &= 4.75\sqrt{Q} = 4.75\sqrt{15} = 18.40\ \text{m} \\
S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.011^{5/3}}{3340\times 15^{1/6}} = 0.000194 = \frac{1}{5150}
\end{aligned}
$$

#### Trapezoidal section

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

Eliminating $B$:

$$
\begin{aligned}
1.7361\,D^2 - 18.40\,D + 21.69 &= 0 \\
D &= \frac{18.40 - \sqrt{18.40^2 - 4\times 1.7361\times 21.69}}{2\times 1.7361} = 1.351\ \text{m} \\
B &= P - 2.2361D = 18.40 - 2.2361\times 1.351 = 15.38\ \text{m}
\end{aligned}
$$

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

Check: $A = (15.38 + 0.5\times 1.351)\times 1.351 = 21.69\ \text{m}^2$ against required 21.69 $\text{m}^2$.

**Answer:** Bed width $B \approx 15.4$ m, depth of water $D \approx 1.35$ m, side slope 0.5H:1V, bed slope $S = 1$ in 5150 (0.000194), velocity $V = 0.69$ m/s, hydraulic mean radius $R = 1.18$ m.

## Check yourself

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

1. For d = 0.33 mm, f = 1.76 sqrt(d) is about
   A. 1.01
   B. 0.58
   C. 1.76
   D. 3.2
   Answer: A. sqrt(0.33) = 0.574, so f = 1.011. (Key point: Silt factor)

2. In the Lacey design procedure, which quantity is found first from Q and f?
   A. Slope S from Manning's formula
   B. Velocity V = (Qf^2/140)^(1/6)
   C. Depth D by trial and error
   D. Bed width B from Kutter's formula
   Answer: B. V is computed directly from Q and f, then A = Q/V, R = 5V^2/2f, P = 4.75 sqrt(Q) and S follow. (Key point: Order of Lacey design)

3. For Q = 15 m^3/s, what wetted perimeter does Lacey's P = 4.75 sqrt(Q) give?
   A. 9.2 m
   B. 27.6 m
   C. 18.40 m
   D. 28.4 m
   Answer: C. P = 4.75 x sqrt(15) = 18.40 m, independent of silt factor. (Key point: Wetted perimeter)

4. Using S = f^(5/3)/(3340 Q^(1/6)) with f = 1.011, Q = 15, the bed slope is about
   A. 1 in 515
   B. 1 in 51500
   C. 1 in 2575
   D. 1 in 5150
   Answer: D. Slope is computed directly from Lacey's slope equation; a finer silt (smaller f) gives a flatter slope. (Key point: Regime slope)

5. After finding A and P, how is the depth D of the 0.5H:1V trapezoid obtained, and which root is taken (result D = 1.351 m, B = 15.38 m)?
   A. Eliminate B using A = (B + 0.5D)D and P = B + 2.236D, solve the quadratic and take the smaller root
   B. Assume D and iterate until V = V0
   C. Take D = R directly and find B from A
   D. Take the larger root, giving a deep narrow section
   Answer: A. 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)

## Related in Irrigation and Drainage Engineering

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