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Darcy-Chezy-Manning link and composite roughness

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.

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.

Part 1: Relationship between Darcy, Chezy and Manning

Consider uniform flow in a channel length LL with area AA and wetted perimeter PP, bed slope S0S_0. Force balance along the flow (gravity component = boundary shear):

γAL S0=τ0PL  ⇒  τ0=γRS0\gamma AL\,S_0=\tau_0PL\;\Rightarrow\;\tau_0=\gamma RS_0
  • Darcy-Weisbach: shear stress in terms of friction factor ff: τ0=f8ρV2\tau_0=\dfrac f8\rho V^2.
  • Equating: ρgRS0=f8ρV2  ⇒  V=8gfRS0\rho gRS_0=\dfrac f8\rho V^2\;\Rightarrow\;V=\sqrt{\dfrac{8g}{f}}\sqrt{RS_0}.
  • Chezy: V=CRS0V=C\sqrt{RS_0}, therefore C=8gfC=\sqrt{\dfrac{8g}{f}}.
  • Manning: V=1nR2/3S01/2=R1/6nRS0V=\dfrac1nR^{2/3}S_0^{1/2}=\dfrac{R^{1/6}}{n}\sqrt{RS_0}, therefore C=R1/6nC=\dfrac{R^{1/6}}{n}.

Combining:

8gf=C=R1/6n  ⇒  f=8gn2R1/3,n=R1/6f8g\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 nn for a composite (mixed) boundary

When parts of the wetted perimeter have different roughness (P1,n1P_1,n_1; P2,n2P_2,n_2; ... PN,nNP_N,n_N), the section is given an equivalent nen_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.
ne=[∑Pini3/2P]2/3n_e=\left[\frac{\sum P_in_i^{3/2}}{P}\right]^{2/3}
  1. Pavlovskii, Muhlhofer and Einstein-Banks (equal total resistance): the total resisting force is the sum of the resistance on the parts.
ne=[∑Pini2P]1/2n_e=\left[\frac{\sum P_in_i^2}{P}\right]^{1/2}
  1. 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.
ne=PR5/3∑PiRi5/3nin_e=\frac{PR^{5/3}}{\sum\dfrac{P_iR_i^{5/3}}{n_i}}
  1. Cowan's method: when no data is available, n=(n0+n1+n2+n3+n4)m5n=(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.
  2. 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 nn is found from ne=AR2/3S1/2/Qn_e=AR^{2/3}S^{1/2}/Q.

Example: bed n=0.015n=0.015, P=3.6P=3.6 m; walls n=0.03n=0.03, P=2.4P=2.4 m. Horton-Einstein gives ne=0.0216n_e=0.0216; Pavlovskii gives 0.02220.0222.

✓ Answer checked by Claude Opus 5.5 (high effort)

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