---
title: "Regime channel design by Lacey's method"
subject: "Hydraulics"
chapter: 10
key: "semester_4/hydraulics:regime-channel-lacey-75-m3s"
source: https://www.bctengineer.com/q/semester_4/hydraulics/regime-channel-lacey-75-m3s
site: BCT Engineering
---

# Regime channel design by Lacey's method

Hydraulics · Chapter 10: Flow in mobile boundary channel

- Asked in: 2072 Asoj, 3 marks
- Video: [Irrigation - Lacey's theory // Design steps with numerical (In Hindi)](https://www.youtube.com/watch?v=hE8KOWBc6MA) — Yogendra Ram Meena
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/regime-channel-lacey-75-m3s
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-10

## Question

Design a regime channel for a discharge of 75 m³/s and soil particle size of 0.65 mm using Lacey's method. Assume suitable side slope of channel.

## Short answer

Lacey’s regime design: velocity 1.01 m/s, bed width 36.7 m, depth 1.96 m, side slope0.5H:1V, bed slope about1 in3830. Silt factor f=1.419; perimeter and area solved simultaneously.

## Answer

**Given:** $Q = 75\ \text{m}^3/\text{s}$, $d = 0.65$ mm. Assume side slope 0.5H : 1V (Lacey's usual value).

**Step 1: Silt factor**

$$
f = 1.76\sqrt{d_{mm}} = 1.76\sqrt{0.65} = 1.419
$$

**Step 2: Velocity**

$$
V = \left(\frac{Qf^2}{140}\right)^{1/6} = \left(\frac{75 \times 2.014}{140}\right)^{1/6} = 1.013\ \text{m/s}
$$

**Step 3: Area, perimeter, hydraulic radius, slope**

$$
A = \frac{Q}{V} = 74.06\ \text{m}^2,\qquad P = 4.75\sqrt{Q} = 41.14\ \text{m}
$$

$$
R = \frac{5V^2}{2f} = \frac{5(1.0256)}{2(1.419)} = 1.81\ \text{m}\quad (A/P = 1.80\ \text{m, consistent})
$$

$$
S = \frac{f^{5/3}}{3340\,Q^{1/6}} = \frac{1.419^{5/3}}{3340(75)^{1/6}} = 2.61 \times 10^{-4}\ (\approx 1\text{ in }3830)
$$

**Step 4: Bed width and depth.** For side slope 0.5H : 1V:

$$
A = (b + 0.5y)y = 74.06,\qquad P = b + 2y\sqrt{1 + 0.25} = b + 2.236y = 41.14
$$

Substituting $b = 41.14 - 2.236y$:

$$
(41.14 - 1.736y)y = 74.06 \;\Rightarrow\; 1.736y^2 - 41.14y + 74.06 = 0 \;\Rightarrow\; y = 1.96\ \text{m}
$$

$$
b = 41.14 - 2.236(1.963) = 36.75\ \text{m}
$$

**Answer: $V = 1.01$ m/s, $b \approx 36.7$ m, $y \approx 1.96$ m, side slope 0.5H : 1V, bed slope $\approx 1$ in 3830** (add freeboard).

## Check yourself

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

1. What is the silt factor for d = 0.65 mm?
   A. 1.42
   B. 0.65
   C. 1.76
   D. 2.01
   Answer: A. f = 1.76 sqrt(0.65) = 1.419. (Key point: Silt factor)

2. Which formulas give V and P in Lacey's method?
   A. V = Q/A and P = Q
   B. V = (Q f^2 / 140)^(1/6) and P = 4.75 sqrt(Q)
   C. V = 140 f and P = 2 sqrt(Q)
   D. V = f^2 and P = Q^2
   Answer: B. These give V = 1.013 m/s and P = 41.14 m. (Key point: Velocity and perimeter)

3. How are b and y found?
   A. Use R = 5V^2/2f only
   B. Use Shields diagram
   C. Solve A = (b + 0.5y)y and P = b + 2.236y using the Lacey A and P
   D. Set b = y
   Answer: C. Substituting gives y = 1.96 m and b = 36.75 m. (Key point: Finding b and y)

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## Related in other subjects

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