---
title: "Lacey's relation between R, Q and f"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:lacey-r-q-f-relation"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lacey-r-q-f-relation
site: BCT Engineering
---

# Lacey's relation between R, Q and f

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2070 Chaitra (old course), 6 marks; 2064 Jestha (old course), 8 marks (asked 2 times)
- 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-r-q-f-relation
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Using Lacey's basic equation, establish a relationship between R, Q and f.

## Short answer

From Lacey's V=(Qf²/140)^(1/6) and R=5V²/2f, equating and solving gives R≈0.48(Q/f)^(1/3). The regime hydraulic radius increases with the cube root of discharge and decreases with the cube root of the silt factor.

## Answer

Lacey's regime relations for a channel are

$$
V = \left(\frac{Qf^2}{140}\right)^{1/6}, \qquad R = \frac{5V^2}{2f}
$$

where $V$ = mean velocity (m/s), $Q$ = discharge (m³/s), $f$ = silt factor and $R$ = hydraulic mean radius (m).

#### Derivation

From the second equation, $V^2 = \dfrac{2fR}{5}$. Squaring the first equation gives

$$
V^2 = \left(\frac{Qf^2}{140}\right)^{1/3}
$$

Equating:

$$
\begin{aligned}
\frac{2fR}{5} &= \left(\frac{Qf^2}{140}\right)^{1/3} \\
R &= \frac{5}{2f}\left(\frac{Qf^2}{140}\right)^{1/3} = \frac{5}{2}\cdot\frac{Q^{1/3}f^{2/3}}{140^{1/3}\,f} \\
&= \frac{2.5}{5.19}\left(\frac{Q}{f}\right)^{1/3}
\end{aligned}
$$

$$
R \approx 0.48\left(\frac{Q}{f}\right)^{1/3}
$$

(Many textbooks write the constant as 0.47 because the constants 140 and 5/2 in the original regime relations are rounded.)

**Result:** the regime hydraulic mean radius increases with the cube root of the discharge and decreases with the cube root of the silt factor. For example, $Q = 20$ m³/s and $f = 1.5$ give $R = 0.48\,(13.33)^{1/3} = 1.14$ m, which agrees with $R = 5V^2/2f$ computed directly.

## Check yourself

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

1. From V = (Qf^2/140)^(1/6), what is V^2?
   A. (Qf^2/140)^(1/3)
   B. (Qf^2/140)^(1/12)
   C. (Qf^2/140)^2
   D. (Qf^2/140)^(1/6) x 2
   Answer: A. Squaring a sixth-root relation gives a cube-root relation. (Key point: Squaring step)

2. Which equation is equated with V^2 = (Qf^2/140)^(1/3)?
   A. V^2 = 5fR/2
   B. V^2 = 2fR/5 from R = 5V^2/2f
   C. V^2 = R/f
   D. V^2 = 5R/(2f)
   Answer: B. Rearranging R = 5V^2/(2f) gives V^2 = 2fR/5. (Key point: Equating)

3. What is the final relation?
   A. R = 0.48 (Q f)^(1/3)
   B. R = 0.48 (Q/f)^(1/6)
   C. R = 0.48 (Q/f)^(1/3)
   D. R = 4.75 sqrt(Q)
   Answer: C. f^(2/3)/f = f^(-1/3), so R varies as (Q/f)^(1/3). (Key point: Final form)

4. How does regime R change as silt factor f increases (coarser silt)?
   A. It increases with f
   B. It is independent of f
   C. It increases with f^2
   D. It decreases with the cube root of f
   Answer: D. R is proportional to (Q/f)^(1/3), so larger f lowers R. (Key point: Dependence)

## Related in Irrigation and Drainage Engineering

- [Lacey's P = 4.75 sqrt(Q) derivation](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alacey-wetted-perimeter-discharge) — Irrigation and Drainage Engineering, chapter 4, asked 7×
- [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

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