---
title: "Lacey design: Q 18, f 1.5"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:lacey-q18-f15"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lacey-q18-f15
site: BCT Engineering
---

# Lacey design: Q 18, f 1.5

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2064 Kartik (old course), 4 marks
- Video: [Lacey's Theory | Canal Design | Irrigation Engineering](https://www.youtube.com/watch?v=6XZePHSkjoE) — APSEd
- 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-q18-f15
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Design a canal to carry a discharge of 18 m³/s, using Lacey's theory. Take silt factor = 1.5 and side slope = 0.5:1.

## Short answer

Lacey's regime equations give V = 0.81 m/s, A = 22.1 m², R = 1.10 m, P = 20.2 m, slope S = 0.000364 (1/2751). For 0.5:1 side slopes, depth D = 1.23 m and bed width B = 17.4 m.

## Answer

#### Lacey's regime values

Given: $Q = 18\ \text{m}^3/\text{s}$, $f = 1.500$, side slope 0.5H:1V.

$$
\begin{aligned}
V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{18\times 1.500^2}{140}\right)^{1/6} = 0.813\ \text{m/s} \\
A &= \frac{Q}{V} = \frac{18}{0.813} = 22.13\ \text{m}^2 \\
R &= \frac{5V^2}{2f} = \frac{5\times 0.813^2}{2\times 1.500} = 1.102\ \text{m} \\
P &= 4.75\sqrt{Q} = 4.75\sqrt{18} = 20.15\ \text{m} \\
S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.500^{5/3}}{3340\times 18^{1/6}} = 0.000364 = \frac{1}{2751}
\end{aligned}
$$

#### Trapezoidal section

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

Eliminating $B$:

$$
\begin{aligned}
1.7361\,D^2 - 20.15\,D + 22.13 &= 0 \\
D &= \frac{20.15 - \sqrt{20.15^2 - 4\times 1.7361\times 22.13}}{2\times 1.7361} = 1.228\ \text{m} \\
B &= P - 2.2361D = 20.15 - 2.2361\times 1.228 = 17.41\ \text{m}
\end{aligned}
$$

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

Check: $A = (17.41 + 0.5\times 1.228)\times 1.228 = 22.13\ \text{m}^2$ against required 22.13 $\text{m}^2$.

**Answer:** Bed width $B \approx 17.4$ m, depth of water $D \approx 1.23$ m, side slope 0.5H:1V, bed slope $S = 1$ in 2751 (0.000364), velocity $V = 0.81$ m/s, hydraulic mean radius $R = 1.10$ m.

## Check yourself

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

1. 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)

2. For Q = 18 m^3/s, what wetted perimeter does Lacey's P = 4.75 sqrt(Q) give?
   A. 10.07 m
   B. 30.22 m
   C. 20.15 m
   D. 30.15 m
   Answer: C. P = 4.75 x sqrt(18) = 20.15 m, independent of silt factor. (Key point: Wetted perimeter)

3. Using S = f^(5/3)/(3340 Q^(1/6)) with f = 1.5, Q = 18, the bed slope is about
   A. 1 in 275
   B. 1 in 27510
   C. 1 in 1375
   D. 1 in 2751
   Answer: D. 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 = 1.228 m, B = 17.41 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

- [Siphon design: canal 18, drainage 60](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Asiphon-q18-drain60-lacey) — Irrigation and Drainage Engineering, chapter 8
- [Lacey: max discharge at slope 1/4000](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alacey-max-discharge-slope-14000) — Irrigation and Drainage Engineering, chapter 4
- [Lacey design: Q 20, f 1.5](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alacey-q20-f15) — Irrigation and Drainage Engineering, chapter 4, asked 2×
- [Lacey design: Q 15, f 1.1](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alacey-q15-f11-slope) — Irrigation and Drainage Engineering, chapter 4
- [Lacey design: Q 15, d 0.33 mm](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alacey-q15-d033mm) — Irrigation and Drainage Engineering, chapter 4
- [Lacey: max discharge at slope 1/6000](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alacey-max-discharge-slope-16000) — Irrigation and Drainage Engineering, chapter 4

## Related in other subjects

- [Carry over factor and distribution factor](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Adefine-carry-over-distribution-factor) — Theory of Structures II, chapter 3
- [Regime channel design by Lacey's method](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aregime-channel-lacey-75-m3s) — Hydraulics, chapter 10
- [Design of side intake with trashrack](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aintake-side-design-3315masl) — Hydropower Engineering, chapter 5
- [Best hydraulic rectangular canal design](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Abest-rectangular-canal-design) — Hydro Power, chapter 1