---
title: "Lacey: max discharge at slope 1/4000"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:lacey-max-discharge-slope-14000"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lacey-max-discharge-slope-14000
site: BCT Engineering
---

# Lacey: max discharge at slope 1/4000

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2073 Bhadra, 6 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-max-discharge-slope-14000
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

The slope of a channel in alluvium is 1/4000, Lacey's silt factor is 0.9 and side slopes are 0.5:1 (H:V). Find the channel section and maximum discharge which can be allowed to flow in it.

## Short answer

Maximum regime discharge 1.03 m³/s. Channel section: bed width 3.4 m, depth 0.66 m, side slope 0.5H:1V, bed slope 1/4000, velocity 0.43 m/s.

## Answer

Given: $S = 1/4000$, $f = 0.9$, side slope 0.5H:1V. The maximum discharge is the one for which the channel is in regime at this slope.

#### Discharge from Lacey's slope equation

$$
S = \frac{f^{5/3}}{3340\,Q^{1/6}} \Rightarrow Q = \left(\frac{f^{5/3}}{3340\,S}\right)^6 = \left(\frac{0.9^{5/3}}{3340/4000}\right)^6 = 1.029\ \text{m}^3/\text{s}
$$

**Maximum discharge $Q \approx 1.03$ m³/s** (used for the section below).

Section for this discharge:

#### Lacey's regime values

Given: $Q = 1.03\ \text{m}^3/\text{s}$, $f = 0.900$, side slope 0.5H:1V.

$$
\begin{aligned}
V &= \left(\frac{Q f^2}{140}\right)^{1/6} = \left(\frac{1.03\times 0.900^2}{140}\right)^{1/6} = 0.426\ \text{m/s} \\
A &= \frac{Q}{V} = \frac{1.03}{0.426} = 2.42\ \text{m}^2 \\
R &= \frac{5V^2}{2f} = \frac{5\times 0.426^2}{2\times 0.900} = 0.504\ \text{m} \\
P &= 4.75\sqrt{Q} = 4.75\sqrt{1.03} = 4.82\ \text{m} \\
S &= \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{0.900^{5/3}}{3340\times 1.03^{1/6}} = 0.000250 = \frac{1}{4001}
\end{aligned}
$$

#### Trapezoidal section

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

Eliminating $B$:

$$
\begin{aligned}
1.7361\,D^2 - 4.82\,D + 2.42 &= 0 \\
D &= \frac{4.82 - \sqrt{4.82^2 - 4\times 1.7361\times 2.42}}{2\times 1.7361} = 0.657\ \text{m} \\
B &= P - 2.2361D = 4.82 - 2.2361\times 0.657 = 3.35\ \text{m}
\end{aligned}
$$

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

Check: $A = (3.35 + 0.5\times 0.657)\times 0.657 = 2.42\ \text{m}^2$ against the required 2.42 $\text{m}^2$.

**Answer:** Bed width $B \approx 3.4$ m, depth of water $D \approx 0.66$ m, side slope 0.5H:1V, bed slope $S = 1$ in 4000 (0.000250), velocity $V = 0.43$ m/s, hydraulic mean radius $R = 0.50$ m; maximum discharge $\approx 1.03$ m³/s.

## Check yourself

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

1. How is the maximum discharge found for S = 1/4000, f = 0.9?
   A. From Manning with assumed depth
   B. From Lacey's slope equation, Q = (f^(5/3)/(3340 S))^6
   C. From P = 4.75 sqrt(Q)
   D. From Kennedy's V0
   Answer: B. Rearranging S = f^(5/3)/(3340 Q^(1/6)) gives Q. (Key point: Method)

2. The discharge obtained is about
   A. 10.3 m^3/s
   B. 103 m^3/s
   C. 1.03 m^3/s
   D. 0.10 m^3/s
   Answer: C. (0.9^(5/3)/0.835)^6 = 1.03. (Key point: Result)

3. Then P = 4.75 sqrt(Q) is about
   A. 9.5 m
   B. 2.4 m
   C. 48 m
   D. 4.82 m
   Answer: D. 4.75 x sqrt(1.03) = 4.82 m. (Key point: Perimeter)

4. The resulting section is
   A. B about 3.4 m, D about 0.66 m
   B. B about 18 m, D about 1.3 m
   C. B about 0.7 m, D about 3.4 m
   D. B about 4.8 m, D about 2.4 m
   Answer: A. The smaller root of the quadratic gives D = 0.657 m and B = 3.35 m. (Key point: Section)

## Related in Irrigation and Drainage Engineering

- [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
- [Duty and head discharge, 4000 ha channel](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Aduty-head-discharge-4000ha-70pct) — Irrigation and Drainage Engineering, chapter 2
- [Kennedy and Lacey silt theories compared](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Akennedy-lacey-theories-compare) — Irrigation and Drainage Engineering, chapter 4, asked 4×
- [Concrete-lined canal: Q 70, slope 1 in 4000](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alined-canal-slope-14000-q70) — Irrigation and Drainage Engineering, chapter 4
- [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 30, f 1.1](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alacey-q30-f11) — Irrigation and Drainage Engineering, chapter 4

## Related in other subjects

- [Depth for maximum discharge, triangular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Atriangular-channel-depth-for-max-discharge) — Hydraulics, chapter 6
- [Drainage density, form factor, channel slope](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Adrainage-density-basin-140km2) — Engineering Hydrology, chapter 4, asked 2×
- [94% depth for maximum discharge in circular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acircular-channel-94-percent-depth-max-discharge) — Hydraulics, chapter 6
- [Efficient trapezoid with unequal side slopes](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aefficient-trapezoidal-two-side-slopes-critical-depth) — Hydraulics, chapter 6