Momentum coefficient for compound channel

Definition

The momentum correction factor for a section of total area $A$, discharge $Q$ and mean velocity $V=Q/A$ is

$$\beta=\frac{\int v^2\,dA}{V^2A}$$

For a compound channel divided into subsections, each with mean velocity $V_i$, area $A_i$ (velocity taken uniform in each subsection):

$$\beta=\frac{\sum V_i^2A_i}{V^2A}$$

Proof

1. Total discharge $Q=\sum Q_i$, $A=\sum A_i$ and $V=\dfrac{Q}{A}$.
2. For each subsection, conveyance $K_i$ is defined by $Q_i=K_i\sqrt{S_f}$. All subsections have the same energy slope $S_f$ (same water surface and energy line). Therefore

$$Q=\sqrt{S_f}\sum K_i,\qquad V_i=\frac{Q_i}{A_i}=\frac{K_i\sqrt{S_f}}{A_i}$$

3. Substituting,

$$\sum V_i^2A_i=\sum\frac{K_i^2S_f}{A_i^2}A_i=S_f\sum\frac{K_i^2}{A_i}$$ $$V^2A=\frac{Q^2}{A}=\frac{S_f\left(\sum K_i\right)^2}{\sum A_i}$$

4. Dividing,

$$\beta=\frac{S_f\sum\dfrac{K_i^2}{A_i}}{\dfrac{S_f\left(\sum K_i\right)^2}{\sum A_i}}$$ $$\boxed{\beta=\frac{\sum\left(\dfrac{K_i^2}{A_i}\right)\left(\sum A_i\right)}{\left(\sum K_i\right)^2}}$$

Hence proved. For a single uniform section, $\beta=1$; for compound sections $\beta>1$ because the velocities in the main channel and flood plains differ greatly.