---
title: "Lacey design: Q 20, f 1.5"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:lacey-q20-f15"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lacey-q20-f15
site: BCT Engineering
---

# Lacey design: Q 20, f 1.5

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2075 Baisakh, 4 marks; 2063 Asoj (old course), 4 marks (asked 2 times)
- Video: [Lecey's Theory & Solved Example | Design of Canal | lec-20 | Irrigation Engineering | wre-2](https://www.youtube.com/watch?v=xYsntpZMz2M) — GATE HUB
- 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-q20-f15
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Design a canal using Lacey's theory carrying a discharge of 20 cumec, silt factor = 1.5 and side slope is 0.5:1 (H:V).

## Short answer

For Q=20 m³/s, f=1.5: V=0.83 m/s, A=24.16 m², R=1.14 m, P=21.24 m, S=1/2800. For trapezoid with side slope 0.5H:1V, solving yields bed width 18.4 m and depth 1.27 m.

## Answer

#### Lacey's regime values

Given: $Q = 20\ \text{m}^3/\text{s}$, $f = 1.500$, side slope 0.5H:1V.

$$
\begin{aligned}
V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{20\times 1.500^2}{140}\right)^{1/6} = 0.828\ \text{m/s} \\
A &= \frac{Q}{V} = \frac{20}{0.828} = 24.16\ \text{m}^2 \\
R &= \frac{5V^2}{2f} = \frac{5\times 0.828^2}{2\times 1.500} = 1.142\ \text{m} \\
P &= 4.75\sqrt{Q} = 4.75\sqrt{20} = 21.24\ \text{m} \\
S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.500^{5/3}}{3340\times 20^{1/6}} = 0.000357 = \frac{1}{2800}
\end{aligned}
$$

#### Trapezoidal section

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

Eliminating $B$:

$$
\begin{aligned}
1.7361\,D^2 - 21.24\,D + 24.16 &= 0 \\
D &= \frac{21.24 - \sqrt{21.24^2 - 4\times 1.7361\times 24.16}}{2\times 1.7361} = 1.269\ \text{m} \\
B &= P - 2.2361D = 21.24 - 2.2361\times 1.269 = 18.40\ \text{m}
\end{aligned}
$$

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

Check: $A = (18.40 + 0.5\times 1.269)\times 1.269 = 24.16\ \text{m}^2$ against required 24.16 $\text{m}^2$.

**Answer:** Bed width $B \approx 18.4$ m, depth of water $D \approx 1.27$ m, side slope 0.5H:1V, bed slope $S = 1$ in 2800 (0.000357), velocity $V = 0.83$ m/s, hydraulic mean radius $R = 1.14$ 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. Velocity V = (Qf^2/140)^(1/6)
   B. Slope S from Manning's formula
   C. Depth D by trial and error
   D. Bed width B from Kutter's formula
   Answer: A. 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 = 20 m^3/s, what wetted perimeter does Lacey's P = 4.75 sqrt(Q) give?
   A. 10.62 m
   B. 21.24 m
   C. 31.86 m
   D. 31.24 m
   Answer: B. P = 4.75 x sqrt(20) = 21.24 m, independent of silt factor. (Key point: Wetted perimeter)

3. Using S = f^(5/3)/(3340 Q^(1/6)) with f = 1.5, Q = 20, the bed slope is about
   A. 1 in 280
   B. 1 in 28000
   C. 1 in 2800
   D. 1 in 1400
   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 = 1.269 m, B = 18.40 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)

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