Normal depth in compound channel

Section (from the figure)

Central channel: bottom width 3 m, side slopes 1:1, depth 1 m (1 m horizontal each side), so top width at berm level $=3+2=5$ m. Two berms (flood plains) 2 m wide each. Let $Y$ = depth of water above the berm level, so total depth $=1+Y$. The outer edges of berms are assumed vertical.

 berm      Y (above berm)       berm
 2 m  ~~~~~~~~~~~~~~~~~~~~~~~~  2 m
 ____|                      |____
      \                    /   1 m
       \__________________/
              3 m

The section is divided by vertical lines at the berm edges; those lines are not counted as wetted perimeter. $S_0=0.00017$, $n=0.018$, $Q=8.97$ m³/s.

Subsection properties

Central channel:

$$A_c=\frac{(3+5)}{2}(1)+5Y=4+5Y,\qquad P_c=3+2\sqrt2=5.828\ \text{m}$$

Each berm (width 2 m, depth $Y$, vertical outer wall):

$$A_b=2Y,\qquad P_b=2+Y$$

Discharge

$$Q=\frac{\sqrt{S_0}}{n}\left[A_cR_c^{2/3}+2A_bR_b^{2/3}\right],\qquad \frac{\sqrt{0.00017}}{0.018}=0.72436$$

Trial values of $Y$:

Solving $Q(Y)=8.97$ gives $Y=0.821$ m.

Check at $Y=0.821$ m

Answer: depth of flow above the berm level $Y\approx0.82$ m (total depth from the bed $\approx1.82$ m).