Momentum Equation.

The momentum equation states that the rate of change in momentum is equal to the external forces acting on the system. From Appendix A, for a single channel:

1)	$\begin{array}{l}\displaystyle \displaystyle \frac{\partial Q}{\partial t} + \frac{\partial (VQ)}{\partial x} + gA \left( \frac{\partial z_s}{\partial x} +S_f \right) =0\end{array}$

The above equation can be written for the channel and for the floodplain:

2)	$\begin{array}{l}\displaystyle \displaystyle \frac{\partial Q_c}{\partial t} + \frac{\partial (V_c Q_c)}{\partial x_c} + gA_c \left( \frac{\partial z_s}{\partial x_c} +S_{fc} \right) =M_f\end{array}$

3)	$\begin{array}{l}\displaystyle \displaystyle \frac{\partial Q_f}{\partial t} + \frac{\partial (V_f Q_f)}{\partial x_f} + gA_f \left( \frac{\partial z_s}{\partial x_f} +S_{ff} \right) =M_c\end{array}$

where $\begin{array}{l}M_c\end{array}$ and $\begin{array}{l}M_f\end{array}$ are the momentum fluxes per unit distance exchanged between the channel and floodplain, respectively. Note that in (2) and (3) the water surface elevation is not subscripted. An assumption in these equations is that the water surface is horizontal at any cross section perpendicular to the flow. Therefore, the water surface elevation is the same for the channel and the floodplain at a given cross section

Using (.Implicit Finite Difference Scheme v6.0:4) through (.Implicit Finite Difference Scheme v6.0:6), the above equations are approximated using finite differences:

4)	$\begin{array}{l}\displaystyle \displaystyle \frac{\Delta Q_c}{\Delta t} + \frac{\Delta (V_c Q_c)}{\Delta x_c} + g \overline{A}_c \left( \frac{\Delta z_s}{\Delta x_c} +\overline{S}_{fc} \right) = M_f\end{array}$

5)	$\begin{array}{l}\displaystyle \displaystyle \frac{\Delta Q_f}{\Delta t} + \frac{\Delta (V_f Q_f)}{\Delta x_f} + g \overline{A}_f \left( \frac{\Delta z_s}{\Delta x_f} +\overline{S}_{ff} \right) = M_c\end{array}$

Note

Note that $\begin{array}{l}\Delta x_c M_c = -\Delta x_f M_f\end{array}$ (due to the horizontal water surface assumption).

Adding and rearranging the above equations yields:

6)	$\begin{array}{l}\displaystyle \displaystyle \frac{\Delta \left( Q_c \Delta x_c + Q_f \Delta x_f \right)}{\Delta t} + \Delta (V_c Q_c) + \Delta (V_f Q_f) + g(A_c + A_f) \Delta z_s + g \overline{A}_c \overline{S}_{fc} \Delta x_c + g \overline{A}_f \overline{S}_{ff} \Delta x_f =0\end{array}$

The final two terms define the friction force from the banks acting on the fluid. An equivalent force can be defined as:

7)	$\begin{array}{l}\displaystyle \displaystyle g \overline{A} \overline{S}_f \Delta x_e = g \overline{A}_c \overline{S}_{fc} \Delta x_c + g\overline{A}_f \overline{S}_{ff} \Delta x_f\end{array}$

Now, the convective terms can be rewritten by defining a velocity distribution factor:

8)	$\begin{array}{l}\displaystyle \displaystyle \beta = \frac{ V_c^2 A_c + V_f^2 A_f }{V^2 A} = \frac{ V_c Q_c + V_f Q_f }{QV}\end{array}$

then:

9)	$\begin{array}{l}\displaystyle \Delta (\beta VQ)=\Delta (V_c Q_c) + \Delta (V_f Q_f)\end{array}$

The final form of the momentum equation is:

10)	$\begin{array}{l}\displaystyle \displaystyle \frac{\Delta (Q_c \Delta x_c + Q_f \Delta x_f)}{\Delta t} + \Delta (\beta VQ) + g\overline{A} \Delta z_s + g \overline{A} \overline{S}_f \Delta x_e = 0\end{array}$

A more familiar form is obtained by dividing through by $\begin{array}{l}\Delta x_e\end{array}$ :

11)	$\begin{array}{l}\displaystyle \displaystyle \frac{\Delta (Q_c \Delta x_c + Q_f \Delta x_f)}{\Delta t \Delta x_e} + \frac{\Delta (\beta VQ)}{\Delta x_e} + g\overline{A} \left( \frac{\Delta z_s}{\Delta x_e} +\overline{S}_f \right) =0\end{array}$