---
title: "Lacey's P = 4.75 sqrt(Q) derivation"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:lacey-wetted-perimeter-discharge"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lacey-wetted-perimeter-discharge
site: BCT Engineering
---

# Lacey's P = 4.75 sqrt(Q) derivation

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2079 Jestha, 4 marks; 2075 Baisakh, 4 marks; 2063 Asoj (old course), 4 marks; 2080 Chaitra, 3 marks; 2071 Magh, 3+2 marks; 2079 Chaitra, 4 marks; 2068 Chaitra (old course), 4 marks (asked 7 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-wetted-perimeter-discharge
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Using Lacey's theory (starting from Lacey's basic regime equations), derive the relationship of wetted perimeter with discharge, i.e. $P = 4.75\sqrt{Q}$.

## Short answer

From Lacey's regime equations V = (Qf²/140)^(1/6) and R = 5V²/(2f), use continuity Q = P*R*V. Substitute R and V³ = f√(Q/140) to get Q = (5P√Q)/(2√140). Solve for P = (2√140/5)√Q ≈ 4.75√Q. The silt factor cancels, making wetted perimeter depend only on discharge.

## Answer

**Lacey's regime equations (basic equations)** for a channel in regime:

$$
V = \left(\frac{Q f^2}{140}\right)^{1/6}\qquad (1)
$$

$$
R = \frac{5V^2}{2f}\qquad (2)
$$

where $V$ = mean velocity (m/s), $Q$ = discharge (m$^3$/s), $R$ = hydraulic mean radius (m), $f$ = silt factor ($f = 1.76\sqrt{d_{mm}}$), and the wetted perimeter $P$ in m.

#### Derivation of $P = 4.75\sqrt{Q}$
**Step 1.** Raise (1) to the 6th power, then take the square root:

$$
V^6 = \frac{Q f^2}{140}\;\Rightarrow\; V^3 = f\sqrt{\frac{Q}{140}}\qquad (3)
$$

**Step 2.** Continuity: $Q = A V$ and $A = P R$, so

$$
Q = P\,R\,V
$$

**Step 3.** Substitute $R$ from (2):

$$
Q = P\left(\frac{5V^2}{2f}\right)V = \frac{5}{2f}\,P\,V^3
$$

**Step 4.** Substitute $V^3$ from (3):

$$
Q = \frac{5}{2f}\,P\,f\sqrt{\frac{Q}{140}} = \frac{5\,P\sqrt{Q}}{2\sqrt{140}}
$$

**Step 5.** Solve for $P$ (note the silt factor $f$ cancels):

$$
P = \frac{2\sqrt{140}}{5}\sqrt{Q} = 4.733\sqrt{Q}
$$

Lacey rounded the coefficient to 4.75, so

$$
\boxed{P = 4.75\sqrt{Q}}
$$

**Remarks:** the wetted perimeter depends only on the discharge, not on the silt factor. Other Lacey relations follow: $A = Q/V$, $S = \dfrac{f^{5/3}}{3340\,Q^{1/6}}$.

## Check yourself

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

1. Which two Lacey equations are used in the derivation of P = 4.75 sqrt(Q)?
   A. V = (Qf^2/140)^(1/6) and R = 5V^2/(2f)
   B. V = 0.55 m D^0.64 and Kutter's formula
   C. V = (1/n)R^(2/3)S^(1/2) and S = f^(5/3)/3340Q^(1/6)
   D. P = 4.75 sqrt(Q) and A = Q/V only
   Answer: A. The derivation combines the two basic regime equations with continuity Q = P R V. (Key point: Starting equations)

2. Which continuity relation is used with the regime equations?
   A. Q = P V / R
   B. Q = A V with A = P R, so Q = P R V
   C. Q = A V with A = P/R
   D. Q = P R^2 V
   Answer: B. A = P x R by definition of hydraulic mean radius, so Q = P R V. (Key point: Continuity)

3. After substituting R, Q becomes (5/2f) P V^3. What is used for V^3?
   A. V^3 = Qf/140
   B. V^3 = 2fR/5
   C. V^3 = f sqrt(Q/140) from equation (1)
   D. V^3 = f^2 Q/140
   Answer: C. Raising (1) to the 6th power and square-rooting gives V^3 = f sqrt(Q/140). (Key point: Eliminating V^3)

4. What happens to the silt factor f in the final result?
   A. P is proportional to f
   B. P is inversely proportional to f
   C. P is proportional to f^2
   D. It cancels, so P depends only on Q
   Answer: D. f appears in both the (5/2f) term and V^3, so it cancels leaving P = (2 sqrt(140)/5) sqrt(Q) = 4.733 sqrt(Q), rounded to 4.75. (Key point: Silt factor)

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