Boundary shear stress and compound channel flow

The figure with dimensions and Manning's coefficients was not available. The method is given in full and applied to an assumed section with stated values, so that the numbers can be replaced by those of the figure.

Part 1: Boundary shear stress equation

Consider a uniform flow in a length $L$ of a prismatic channel of area $A$, wetted perimeter $P$ and bed slope $S_0$ (angle $\theta$).

  ----------------------------  WSL
  \  W sin(th) ->  /         
   \ ___ control volume / tau0 on boundary

Forces along the flow direction (pressure forces at the two ends are equal since depth is constant):

• Gravity component: $W\sin\theta=\gamma AL\sin\theta\approx\gamma AL\,S_0$
• Boundary resistance: $\tau_0\,P\,L$

For uniform flow (no acceleration):

$$\gamma AL\,S_0=\tau_0PL\;\Rightarrow\;\boxed{\tau_0=\gamma RS_0}$$

where $R=A/P$. This is the average shear stress on the boundary.

Part 2: Compound channel (method)

1. Divide the section by vertical lines into subsections (main channel and flood plains); the dividing lines are not counted in wetted perimeters.
2. For each subsection $i$: $A_i$, $P_i$, $R_i=A_i/P_i$ and $Q_i=\dfrac{1}{n_i}A_iR_i^{2/3}S_0^{1/2}$.
3. Total discharge $Q=\sum Q_i$.
4. Effective (equivalent) Manning coefficient from the whole section, treated as a single section:

$$n_e=\frac{A\,R^{2/3}S_0^{1/2}}{Q},\quad A=\sum A_i,\quad P=\sum P_i,\quad R=\frac AP$$

Illustration with assumed data

Main channel: bed 6 m wide, 2 m deep, $n=0.025$. Flood plains: 8 m wide each side, water depth 1 m over the plain, $n=0.040$, bank vertical. $S_0=0.002$. (Assumed.)

Answer (for the assumed section): $Q=44.67$ m³/s and $n_e=0.0295$. For the actual figure, repeat steps 1-4 with the given dimensions and $n$ values.