Non-linear kinematic wave discharge at 1 km

Use the first-order implicit scheme

$$\frac{\Delta t}{\Delta x}Q_{i+1}^{n+1}+\alpha\left(Q_{i+1}^{n+1}\right)^{\beta}=\frac{\Delta t}{\Delta x}Q_i^{n+1}+\alpha\left(Q_{i+1}^n\right)^\beta$$

with $q=0$. Here $i$ is location X and $i+1$ is the point 1 km downstream; $n$ is 13:45 and $n+1$ is 14:00.

Data

$\Delta x=1000$ m, $\Delta t=15\times60=900$ s, $\dfrac{\Delta t}{\Delta x}=0.90$, $Q_i^{n+1}=14$, $Q_i^n=12$, $Q_{i+1}^n=11$ m$^3$/s.

Step 1: Kinematic coefficients ($P\approx B=20$ m)

$$\alpha=\left(\frac{nP^{2/3}}{S_0^{1/2}}\right)^{0.6}=\left(\frac{0.03\times20^{2/3}}{\sqrt{0.001}}\right)^{0.6}=3.2113,\qquad \beta=0.6$$

Step 2: Equation to solve

Right side: $0.90(14)+3.2113(11)^{0.6}=26.1369$

$$f(x)=0.90\,x+3.2113\,x^{0.6}-26.1369=0,\qquad f'(x)=0.90+1.9268\,x^{-0.4}$$

Step 3: Newton iteration from $x_0=Q_{i+1}^n=11$

Answer: the discharge 1 km downstream of X at 14:00 hrs is $Q_{i+1}^{n+1}\approx12.67$ m$^3$/s.