---
title: "Darcy-Chezy-Manning link and composite roughness"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:darcy-chezy-manning-shear-composite-roughness"
source: https://www.bctengineer.com/q/semester_4/hydraulics/darcy-chezy-manning-shear-composite-roughness
site: BCT Engineering
---

# Darcy-Chezy-Manning link and composite roughness

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2068 Magh, 4+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/darcy-chezy-manning-shear-composite-roughness
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Establish the relationship between Darcy, Chezy and Manning's equations based on the shear stress distribution on the channel boundary for uniform flow. Explain the ways of estimating Manning's coefficient for composite boundary.

## Short answer

From τ₀ = γRS = (f/8)ρV², C = √(8g/f) = R^(1/6)/n. For composite boundaries, equivalent n can be estimated by Horton-Einstein (n_e = [ΣP_i n_i^(3/2)/P]^(2/3)), Pavlovskii (n_e = [ΣP_i n_i²/P]^(1/2)), Lotter’s method, or Cowan’s method.

## Answer

#### Part 1: Relationship between Darcy, Chezy and Manning
Consider uniform flow in a channel length $L$ with area $A$ and wetted perimeter $P$, bed slope $S_0$. Force balance along the flow (gravity component = boundary shear):
$$
\gamma AL\,S_0=\tau_0PL\;\Rightarrow\;\tau_0=\gamma RS_0
$$
- **Darcy-Weisbach:** shear stress in terms of friction factor $f$: $\tau_0=\dfrac f8\rho V^2$.
- Equating: $\rho gRS_0=\dfrac f8\rho V^2\;\Rightarrow\;V=\sqrt{\dfrac{8g}{f}}\sqrt{RS_0}$.
- **Chezy:** $V=C\sqrt{RS_0}$, therefore $C=\sqrt{\dfrac{8g}{f}}$.
- **Manning:** $V=\dfrac1nR^{2/3}S_0^{1/2}=\dfrac{R^{1/6}}{n}\sqrt{RS_0}$, therefore $C=\dfrac{R^{1/6}}{n}$.

Combining:
$$
\sqrt{\frac{8g}{f}}=C=\frac{R^{1/6}}{n}\;\Rightarrow\;f=\frac{8gn^2}{R^{1/3}},\quad n=R^{1/6}\sqrt{\frac{f}{8g}}
$$

#### Part 2: Manning's $n$ for a composite (mixed) boundary
When parts of the wetted perimeter have different roughness ($P_1,n_1$; $P_2,n_2$; ... $P_N,n_N$), the section is given an equivalent $n_e$, by assuming a rule about how the parts behave:

1. **Horton-Einstein (equal velocity):** every part has the same mean velocity as the whole section.
$$
n_e=\left[\frac{\sum P_in_i^{3/2}}{P}\right]^{2/3}
$$
2. **Pavlovskii, Muhlhofer and Einstein-Banks (equal total resistance):** the total resisting force is the sum of the resistance on the parts.
$$
n_e=\left[\frac{\sum P_in_i^2}{P}\right]^{1/2}
$$
3. **Lotter (equal discharge sum):** total discharge is the sum of the part discharges, each part with the same slope and with its own area and roughness.
$$
n_e=\frac{PR^{5/3}}{\sum\dfrac{P_iR_i^{5/3}}{n_i}}
$$
4. **Cowan's method:** when no data is available, $n=(n_0+n_1+n_2+n_3+n_4)m_5$ from the basic roughness, irregularity, variation of section, obstruction and vegetation, with a meandering factor.
5. **Compound channels:** the section is divided into main channel and flood plains by vertical lines; discharge is the sum of subsection discharges, and then the effective $n$ is found from $n_e=AR^{2/3}S^{1/2}/Q$.

Example: bed $n=0.015$, $P=3.6$ m; walls $n=0.03$, $P=2.4$ m. Horton-Einstein gives $n_e=0.0216$; Pavlovskii gives $0.0222$.

## Check yourself

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

1. Which balance gives tau0 = gamma R S0 in uniform flow?
   A. Pressure force = shear force
   B. Gravity component along the flow = boundary shear force, gamma A L S0 = tau0 P L
   C. Inertia = viscous force
   D. Weight = buoyancy
   Answer: B. In uniform flow there is no acceleration, so gravity along the slope balances boundary shear. (Key point: Shear stress balance)

2. The Horton-Einstein composite roughness formula assumes:
   A. Equal total resistance in each part
   B. Equal velocity in all parts of the section, n_e = [sum(P_i n_i^1.5)/P]^(2/3)
   C. Total discharge is the sum of part discharges
   D. Equal area in all parts
   Answer: B. Horton and Einstein take each part to have the same mean velocity as the whole. (Key point: Horton-Einstein assumption)

3. Which formula comes from adding resistances (Pavlovskii, Muhlhofer, Einstein-Banks)?
   A. n_e = [sum(P_i n_i^1.5)/P]^(2/3)
   B. n_e = sum(P_i n_i)/P
   C. n_e = [sum(P_i n_i^2)/P]^(1/2)
   D. n_e = P R^(5/3)/sum(P_i R_i^(5/3)/n_i)
   Answer: C. Summing the resistance forces gives n_e = sqrt(sum(P_i n_i^2)/P). (Key point: Pavlovskii method)

4. For a compound channel (main channel and flood plains), the effective n is found by:
   A. Averaging the n values
   B. Using Cowan's method only
   C. Dividing by vertical lines, summing subsection discharges, then n_e = A R^(2/3) S^(1/2)/Q
   D. Using the largest n
   Answer: C. The total discharge from the subsections is used to back out an effective n. (Key point: Compound channels)

## Related in Hydraulics

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- [Relation between Chezy, Manning and Darcy coefficients](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Achezy-manning-darcy-relationship) — Hydraulics, chapter 6
- [Boundary shear distribution in alluvial channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ashear-stress-distribution-alluvial-channel) — Hydraulics, chapter 10
- [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
- [Conditions for uniform flow in open channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aconditions-of-uniform-flow) — Hydraulics, chapter 6, asked 2×

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