Linear kinematic wave, 30 m channel

The linear kinematic wave equation $\dfrac{\partial Q}{\partial t}+c\dfrac{\partial Q}{\partial x}=0$ is solved with the second-order four-point scheme using the three known values $Q_i^n,\ Q_i^{n+1},\ Q_{i+1}^n$.

Step 1: Celerity (wide channel, $P\approx B$)

Manning: $Q=\dfrac{1}{n}S_0^{1/2}B\,y^{5/3}$ (with $A=By$, $R\approx y$). A representative discharge is taken as the mean of the three known values: $Q=(20+28+18)/3=22$ m$^3$/s.

$$y=\left(\frac{nQ}{S_0^{1/2}B}\right)^{3/5}=0.3200\ \text{m},\qquad V=\frac{Q}{By}=2.2918\ \text{m/s}$$ $$c=\frac{dQ}{dA}=\frac53V=3.8197\ \text{m/s},\qquad C=\frac{c\Delta t}{\Delta x}=\frac{3.8197\times600}{1200}=1.9099$$

Step 2: Scheme

$$\frac{Q_i^{n+1}+Q_{i+1}^{n+1}-Q_i^n-Q_{i+1}^n}{2\Delta t}+c\,\frac{Q_{i+1}^{n+1}+Q_{i+1}^n-Q_i^{n+1}-Q_i^n}{2\Delta x}=0$$ $$Q_{i+1}^{n+1}=Q_i^n+\frac{1-C}{1+C}\left(Q_{i+1}^n-Q_i^{n+1}\right)$$

Step 3: Substitution

$$Q_{i+1}^{n+1}=20+\frac{1-1.9099}{1+1.9099}(18-28)=20+(-0.3127)(-10)=23.13\ \text{m}^3/\text{s}$$

Answer: $Q_{i+1}^{n+1}\approx23.13$ m$^3$/s. (Because $C>1$ the four-point scheme is still stable, as it is unconditionally stable; the value depends on the celerity assumed, here from the mean of the known flows.)