Maximum time step, kinematic wave routing

For an explicit kinematic wave scheme the time step is limited by the Courant condition:

$$C=\frac{c\,\Delta t}{\Delta x}\le1\ \Rightarrow\ \Delta t_{max}=\frac{\Delta x}{c}$$

where $c=dQ/dA$ is the kinematic wave celerity.

Step 1: Normal depth

With $R\approx y$ and $A=By$, Manning's equation gives

$$Q=\frac1nBy\,y^{2/3}S_0^{1/2}\ \Rightarrow\ y=\left(\frac{nQ}{BS_0^{1/2}}\right)^{3/5}=\left(\frac{0.03\times100}{40\times\sqrt{0.02}}\right)^{0.6}=0.6835\ \text{m}$$

Step 2: Velocity and celerity

$$V=\frac{Q}{By}=\frac{100}{40\times0.6835}=3.6577\ \text{m/s}$$

Since $Q=\alpha A^{5/3}$, $c=\dfrac{dQ}{dA}=\dfrac53V=\dfrac53(3.6577)=6.0962\ \text{m/s}$.

Step 3: Maximum time step

$$\Delta t_{max}=\frac{\Delta x}{c}=\frac{1500}{6.0962}=246.1\ \text{s}\approx4.1\ \text{min}$$

Answer: the time step should not exceed about 246 s (about 4.1 minutes); a practical choice is slightly less, e.g. 4 minutes (240 s).