Equivalent roughness of compound channel

Horton-Einstein method: each part of the section has the same mean velocity as the whole section, and the equivalent roughness is

$$n_e=\left[\frac{\sum P_in_i^{3/2}}{P}\right]^{2/3}$$

where $P_i$ is the wetted perimeter of the part having roughness $n_i$, and $P=\sum P_i$ is the total wetted perimeter.

Assumptions from the figure

Main channel: bed width 20 m, depth 4 m, side slopes 1V:4H. Flood plain on each side: horizontal part 5 m wide at the level of the top of the main channel, with an outer bank 1V:2H. Total depth $=5$ m, so the water depth on the flood plain is $5-4=1$ m. The vertical interface lines between the main channel and flood plain are not counted as wetted perimeter.

 flood plain   ~~~~~~~~~~~~~~~~~~~~~  flood plain
  \ 5 m    (1 m deep)      5 m /
   \_______      ________/
           \    /  4 m deep
            \__/  20 m

Wetted perimeters

• Main channel (n = 0.021): slope length each side $=4\sqrt{1+4^2}=4\times4.1231=16.492$ m

$$P_m=20+2\times16.492=52.985\ \text{m}$$

• Each flood plain (n = 0.039): horizontal part $5$ m + outer slope length $1\times\sqrt{1+2^2}=2.236$ m

$$P_f=5+2.236=7.236\ \text{m};\quad\text{both}=2P_f=14.472\ \text{m}$$

• Total: $P=52.985+14.472=67.457$ m.

Equivalent roughness

$$n_e=\left[\frac{52.985\,(0.021)^{1.5}+14.472\,(0.039)^{1.5}}{67.457}\right]^{2/3}$$ $$(0.021)^{1.5}=0.003043,\qquad (0.039)^{1.5}=0.007702$$ $$n_e=\left[\frac{0.2727}{67.457}\right]^{2/3}=(0.004043)^{2/3}=0.0254$$

Answer: equivalent Manning's roughness $n_e\approx0.0254$ (between 0.021 and 0.039, weighted towards the larger perimeter of the main channel).