---
title: "Lacey design: Q 40, f 0.9"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:lacey-q40-f09"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lacey-q40-f09
site: BCT Engineering
---

# Lacey design: Q 40, f 0.9

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2062 Kartik (old course), 10 marks
- 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-q40-f09
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Design a regime channel by Lacey's theory for 40 m³/s discharge and silt factor 0.9.

## Short answer

Assuming 0.5:1 side slopes, Lacey's equations yield V = 0.78 m/s, A = 51.1 m², R = 1.71 m, P = 30.0 m, slope S = 0.000136 (1/7362). Section dimensions: depth D = 1.91 m, bed width B = 25.8 m.

## Answer

No side slope is given, so 0.5H:1V is assumed.

#### Lacey's regime values

Given: $Q = 40\ \text{m}^3/\text{s}$, $f = 0.900$, side slope 0.5H:1V (assumed).

$$
\begin{aligned}
V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{40\times 0.900^2}{140}\right)^{1/6} = 0.784\ \text{m/s} \\
A &= \frac{Q}{V} = \frac{40}{0.784} = 51.05\ \text{m}^2 \\
R &= \frac{5V^2}{2f} = \frac{5\times 0.784^2}{2\times 0.900} = 1.705\ \text{m} \\
P &= 4.75\sqrt{Q} = 4.75\sqrt{40} = 30.04\ \text{m} \\
S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{0.900^{5/3}}{3340\times 40^{1/6}} = 0.000136 = \frac{1}{7362}
\end{aligned}
$$

#### Trapezoidal section

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

Eliminating $B$:

$$
\begin{aligned}
1.7361\,D^2 - 30.04\,D + 51.05 &= 0 \\
D &= \frac{30.04 - \sqrt{30.04^2 - 4\times 1.7361\times 51.05}}{2\times 1.7361} = 1.910\ \text{m} \\
B &= P - 2.2361D = 30.04 - 2.2361\times 1.910 = 25.77\ \text{m}
\end{aligned}
$$

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

Check: $A = (25.77 + 0.5\times 1.910)\times 1.910 = 51.05\ \text{m}^2$ against the required 51.05 $\text{m}^2$.

**Answer:** Bed width $B \approx 25.8$ m, depth of water $D \approx 1.91$ m, side slope 0.5H:1V, bed slope $S = 1$ in 7362 (0.000136), velocity $V = 0.78$ m/s, hydraulic mean radius $R = 1.71$ 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 = 40 m^3/s, what wetted perimeter does Lacey's P = 4.75 sqrt(Q) give?
   A. 15.02 m
   B. 45.06 m
   C. 30.04 m
   D. 40.04 m
   Answer: C. P = 4.75 x sqrt(40) = 30.04 m, independent of silt factor. (Key point: Wetted perimeter)

3. Using S = f^(5/3)/(3340 Q^(1/6)) with f = 0.9, Q = 40, the bed slope is about
   A. 1 in 736
   B. 1 in 73620
   C. 1 in 3681
   D. 1 in 7362
   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 = 1.910 m, B = 25.77 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)

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