First-order kinematic wave, flood peak at community

The first-order accurate (implicit) scheme of the linear kinematic wave equation $\partial Q/\partial t+c\,\partial Q/\partial x=0$ is

$$\frac{Q_{i+1}^{n+1}-Q_{i+1}^n}{\Delta t}+c\,\frac{Q_{i+1}^{n+1}-Q_i^{n+1}}{\Delta x}=0 \ \Rightarrow\ Q_{i+1}^{n+1}=\frac{Q_{i+1}^n+C\,Q_i^{n+1}}{1+C},\qquad C=\frac{c\Delta t}{\Delta x}$$

Interpretation of the data

• The gauging station is point $i$; the community ($\Delta x=7.2$ km downstream) is $i+1$.
• $Q_i^{n+1}=150$ m$^3$/s (peak at the station at 12:00 noon).
• $Q_{i+1}^n=100$ m$^3$/s is the flow already at the community at the earlier level (taken as the 9:00 A.M. value, with $\Delta t=1$ h per step).
• The velocity given as 1.2 (m/s) is used as the wave celerity $c$; the channel width is then not needed, because the equation is written in terms of $Q$.

Calculation

$$C=\frac{c\,\Delta t}{\Delta x}=\frac{1.2\times3600}{7200}=0.6$$ $$Q_{i+1}^{n+1}=\frac{100+0.6\times150}{1+0.6}=\frac{190}{1.6}=118.75\ \text{m}^3/\text{s}$$

Answer: the discharge at the community at 12:00 noon is about 118.75 m$^3$/s (the peak is delayed and attenuated numerically). The data are incomplete in the question (width missing, unit of velocity printed as m/s$^2$), so this reading is assumed.