---
title: "Relation between Chezy, Manning and Darcy coefficients"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:chezy-manning-darcy-relationship"
source: https://www.bctengineer.com/q/semester_4/hydraulics/chezy-manning-darcy-relationship
site: BCT Engineering
---

# Relation between Chezy, Manning and Darcy coefficients

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2069 Bhadra, 4 marks
- Video: [Condition for Uniform flow | Relation between Chezy's, Darcy & Manning's coefficient | Prashant YT |](https://www.youtube.com/watch?v=U531mwrVDj0) — Prashant YT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/chezy-manning-darcy-relationship
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Develop the relationship between Chezy's coefficient, Manning's coefficient and Darcy's coefficient.

## Short answer

From boundary shear τ₀ = γRS = (f/8)ρV²: Chezy’s C = √(8g/f). Manning’s C = R^(1/6)/n. Hence C = √(8g/f) = R^(1/6)/n, giving f = 8gn²/R^(1/3) and n = R^(1/6)√(f/(8g)).

## Answer

#### Uniform flow: force balance
For uniform flow in a channel of hydraulic radius $R$ and slope $S$, the boundary shear stress is
$$
\tau_0=\gamma RS=\rho gRS
$$

#### Darcy-Weisbach
In terms of the Darcy friction factor $f$ and mean velocity $V$:
$$
\tau_0=\frac{f}{8}\rho V^2
$$
Equating the two:
$$
\rho gRS=\frac f8\rho V^2\;\Rightarrow\;V=\sqrt{\frac{8g}{f}}\sqrt{RS}
$$

#### Chezy
Chezy's equation is $V=C\sqrt{RS}$. Comparing:
$$
\boxed{C=\sqrt{\frac{8g}{f}}}\qquad\left(f=\frac{8g}{C^2}\right)
$$

#### Manning
Manning's equation is $V=\dfrac1nR^{2/3}S^{1/2}=\dfrac{R^{1/6}}{n}\sqrt{RS}$. Comparing with Chezy:
$$
\boxed{C=\frac{R^{1/6}}{n}}
$$

#### Combined relation
$$
C=\sqrt{\frac{8g}{f}}=\frac{R^{1/6}}{n}\;\Rightarrow\;\boxed{f=\frac{8gn^2}{R^{1/3}}\quad\text{or}\quad n=R^{1/6}\sqrt{\frac{f}{8g}}}
$$
| Coefficient | Symbol | Relation |
|---|---|---|
| Chezy | $C$ | $C=\sqrt{8g/f}=R^{1/6}/n$ |
| Manning | $n$ | $n=R^{1/6}/C=R^{1/6}\sqrt{f/8g}$ |
| Darcy-Weisbach | $f$ | $f=8g/C^2=8gn^2/R^{1/3}$ |

For a pipe flowing full, $R=D/4$, so the same relation gives $h_f=\dfrac{fL}{D}\dfrac{V^2}{2g}$.

## Check yourself

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

1. For uniform flow a force balance gives the mean boundary shear stress as:
   A. tau0 = gamma A S
   B. tau0 = gamma R S
   C. tau0 = rho V^2
   D. tau0 = gamma y
   Answer: B. Gravity component gamma A L S equals tau0 P L, so tau0 = gamma R S. (Key point: Boundary shear stress)

2. Equating gamma R S with (f/8) rho V^2 and comparing with V = C sqrt(RS) gives:
   A. C = 8g f
   B. C = f/8g
   C. C = sqrt(f/8g)
   D. C = sqrt(8g/f)
   Answer: D. V = sqrt(8g/f) sqrt(RS), so C = sqrt(8g/f). (Key point: Chezy and Darcy)

3. Writing Manning's V = (1/n) R^(2/3) S^(1/2) as (R^(1/6)/n) sqrt(RS) shows:
   A. C = n R^(1/6)
   B. C = R^(1/6)/n
   C. C = R^(2/3)/n
   D. C = n/R^(1/6)
   Answer: B. Comparing with V = C sqrt(RS) gives C = R^(1/6)/n. (Key point: Chezy and Manning)

4. Combining the two expressions for C gives the Darcy factor as:
   A. f = 8 g n^2 / R^(1/3)
   B. f = 8 g n / R^(1/6)
   C. f = n^2 / (8 g R^(1/3))
   D. f = 8 g R^(1/3) / n^2
   Answer: A. Setting 8g/f = R^(1/3)/n^2 gives f = 8 g n^2/R^(1/3). (Key point: Combined relation)

## Related in Hydraulics

- [Darcy-Chezy-Manning link and composite roughness](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Adarcy-chezy-manning-shear-composite-roughness) — Hydraulics, chapter 6
- [Chezy's and Manning's coefficient for Seti River](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aseti-river-chezy-manning-coefficient) — Hydraulics, chapter 6
- [Half-hexagon channel dimensions and Manning n](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahalf-hexagon-channel-chezy-manning-n) — Hydraulics, chapter 6