Saint-Venant velocity and depth at (i, j+1)

For a wide rectangular channel the two governing equations in terms of velocity $V$ and depth $y$ are

$$\frac{\partial y}{\partial t}+V\frac{\partial y}{\partial x}+y\frac{\partial V}{\partial x}=0,\qquad \frac{\partial V}{\partial t}+V\frac{\partial V}{\partial x}+g\frac{\partial y}{\partial x}=g(S_0-S_f),\quad S_f=\frac{n^2V^2}{y^{4/3}}$$

Scheme (explicit, Lax diffusive form)

The old value at $i$ is replaced by the average of its neighbours, and central differences are used in space:

$$y_i^{j+1}=\frac{y_{i+1}^j+y_{i-1}^j}{2}-\frac{\Delta t}{2\Delta x}\left[V_i^j\left(y_{i+1}^j-y_{i-1}^j\right)+y_i^j\left(V_{i+1}^j-V_{i-1}^j\right)\right]$$ $$V_i^{j+1}=\frac{V_{i+1}^j+V_{i-1}^j}{2}-\frac{\Delta t}{2\Delta x}\left[V_i^j\left(V_{i+1}^j-V_{i-1}^j\right)+g\left(y_{i+1}^j-y_{i-1}^j\right)\right]+g\Delta t\,(S_0-S_{f,i}^j)$$

Data

$\Delta t=4\times60=240$ s, $\Delta x=1000$ m, so $\dfrac{\Delta t}{2\Delta x}=0.120$. $S_0=0.01$, $n=0.032$.

Step 1: Friction slope at $i$

$$S_f=\frac{n^2V_i^2}{y_i^{4/3}}=\frac{0.032^2(1.8)^2}{2.0^{4/3}}=0.001317$$

Step 2: Depth

$$y_i^{j+1}=\frac{2.4+1.6}{2}-0.120\left[1.8(2.4-1.6)+2.0(1.5-2.2)\right]=2.0-0.120\,(0.04)=1.9952\ \text{m}$$

Step 3: Velocity

• Average: $(1.5+2.2)/2=1.85$
• Convective: $-0.120\,[1.8(1.5-2.2)]=0.1512$
• Pressure: $-0.120\,[9.81(2.4-1.6)]=-0.9418$
• Gravity and friction: $9.81(240)(0.01-0.001317)=20.4441$

$$V_i^{j+1}=1.85+(0.1512)+(-0.9418)+(20.4441)=21.504\ \text{m/s}$$

Answer: $y_i^{j+1}=1.9952$ m and $V_i^{j+1}=21.50$ m/s.

Remark: the very large velocity arises because the gravity term ($S_0=1\%$) is far from balanced by friction and the pressure gradient for these data, and $\Delta t=240$ s exceeds the Courant limit $\Delta x/(V+\sqrt{gy})\approx161$ s. The values are the one-step result of the scheme; a real simulation would use a smaller $\Delta t$.