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Relation between Chezy, Manning and Darcy coefficients

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)).

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

Uniform flow: force balance

For uniform flow in a channel of hydraulic radius RR and slope SS, the boundary shear stress is

τ0=γRS=ρgRS\tau_0=\gamma RS=\rho gRS

Darcy-Weisbach

In terms of the Darcy friction factor ff and mean velocity VV:

τ0=f8ρV2\tau_0=\frac{f}{8}\rho V^2

Equating the two:

ρgRS=f8ρV2  ⇒  V=8gfRS\rho gRS=\frac f8\rho V^2\;\Rightarrow\;V=\sqrt{\frac{8g}{f}}\sqrt{RS}

Chezy

Chezy's equation is V=CRSV=C\sqrt{RS}. Comparing:

C=8gf(f=8gC2)\boxed{C=\sqrt{\frac{8g}{f}}}\qquad\left(f=\frac{8g}{C^2}\right)

Manning

Manning's equation is V=1nR2/3S1/2=R1/6nRSV=\dfrac1nR^{2/3}S^{1/2}=\dfrac{R^{1/6}}{n}\sqrt{RS}. Comparing with Chezy:

C=R1/6n\boxed{C=\frac{R^{1/6}}{n}}

Combined relation

C=8gf=R1/6n  ⇒  f=8gn2R1/3orn=R1/6f8gC=\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}}}
CoefficientSymbolRelation
ChezyCCC=8g/f=R1/6/nC=\sqrt{8g/f}=R^{1/6}/n
Manningnnn=R1/6/C=R1/6f/8gn=R^{1/6}/C=R^{1/6}\sqrt{f/8g}
Darcy-Weisbachfff=8g/C2=8gn2/R1/3f=8g/C^2=8gn^2/R^{1/3}

For a pipe flowing full, R=D/4R=D/4, so the same relation gives hf=fLDV22gh_f=\dfrac{fL}{D}\dfrac{V^2}{2g}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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