---
title: "Prove R = 1.35 (q^2/f)^(1/3)"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:lacey-scour-depth-formula"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lacey-scour-depth-formula
site: BCT Engineering
---

# Prove R = 1.35 (q^2/f)^(1/3)

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2072 Magh, 5 marks; 2062 Baisakh (old course), 6 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-scour-depth-formula
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Using Lacey's regime equations prove that $R = 1.35\left(\frac{q^2}{f}\right)^{1/3}$, where R = scour depth (hydraulic mean radius), q = specific (per unit width) discharge, f = silt factor.

## Short answer

For a wide channel q=VR. Using R=5V²/2f gives V=√(2fR/5), so q=√(2f/5)R^(3/2). Solving yields R=(5q²/2f)^(1/3)≈1.35(q²/f)^(1/3). This formula gives normal scour depth from unit discharge and silt factor.

## Answer

For a very wide channel the wetted perimeter is almost equal to the width. Since $A = PR$, the discharge per unit width is $q = Q/P = VR$ (the regime flow per metre width).

Lacey's regime equation connecting $R$, $V$ and $f$:

$$
R = \frac{5V^2}{2f} \quad\Rightarrow\quad V = \sqrt{\frac{2fR}{5}}
$$

#### Proof

$$
\begin{aligned}
q &= V R = \sqrt{\frac{2fR}{5}}\;R = \sqrt{\frac{2f}{5}}\;R^{3/2} \\
R^{3/2} &= q\sqrt{\frac{5}{2f}} \\
R^{3} &= \frac{5}{2}\cdot\frac{q^2}{f} \\
R &= \left(\frac{5}{2}\right)^{1/3}\left(\frac{q^2}{f}\right)^{1/3} = 1.357\left(\frac{q^2}{f}\right)^{1/3}
\end{aligned}
$$

$$
R \approx 1.35\left(\frac{q^2}{f}\right)^{1/3}
$$

where $R$ is the normal scour depth (m), $q$ the discharge per metre width (m³/s/m) and $f$ the silt factor. This is Lacey's scour depth formula. It is used to fix the depth of cutoff walls, guide bunds and aprons, with $R$ multiplied by a factor for the maximum scour (about 1.27 for a straight reach, up to 2-2.75 at the nose of guide bunds).

## Check yourself

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

1. What is assumed for a very wide channel?
   A. P is about the width so q = V R
   B. A = P/R so q = V/R
   C. The section is circular
   D. Side slopes control the flow
   Answer: A. For a wide channel A is about P R, so the discharge per unit width is q = V R. (Key point: Wide-channel assumption)

2. Which expression for V is used?
   A. V = sqrt(5fR/2)
   B. V = sqrt(2fR/5)
   C. V = 5R/(2f)
   D. V = (2f/5R)^(1/2)
   Answer: B. R = 5V^2/(2f) rearranged gives V^2 = 2fR/5. (Key point: Velocity from regime)

3. Substituting V in q = V R gives q proportional to R to the power
   A. 2
   B. 1/2
   C. 3/2
   D. 3
   Answer: C. q = sqrt(2f/5) x R^(1/2) x R = sqrt(2f/5) R^(3/2). (Key point: Power of R)

4. What constant results from the proof?
   A. (2/5)^(1/3) = 0.74
   B. (5/2)^(1/2) = 1.58
   C. 4.75
   D. (5/2)^(1/3) = 1.357, taken as 1.35
   Answer: D. Squaring and cubing give R^3 = (5/2) q^2/f. (Key point: Constant)

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