Non-linear kinematic wave, trapezoidal river

The non-linear kinematic wave model is used with $A=\alpha Q^\beta$ and the second-order four-point scheme (centred in space and time), solved by Newton's method.

Step 1: Channel properties at the mean flow

Mean discharge $=(420+580)/2=500$ m$^3$/s. Trapezoid: bottom width $b=350$ m, side slope $z=6$, $S_0=0.015$, $n=0.0359$.

$$Q=\frac{S_0^{1/2}}{n}\frac{A^{5/3}}{P^{2/3}},\quad A=(b+zy)y,\quad P=b+2y\sqrt{1+z^2}$$

Solving for $Q=500$: $y=0.592$ m, $A=209.30$ m$^2$, $P=357.20$ m.

Step 2: Kinematic coefficients

$$\alpha=\left(\frac{nP^{2/3}}{S_0^{1/2}}\right)^{0.6}=5.0280,\qquad \beta=0.6$$

Then $A_i^n=\alpha(420)^{0.6}=188.51$ m$^2$ and $A_i^{n+1}=\alpha(580)^{0.6}=228.80$ m$^2$. (Wave celerity $c\approx3.98$ m/s, so $C=c\Delta t/\Delta x\approx4.6>1$: an explicit scheme would be unstable, the four-point scheme is not.)

Step 3: Finite difference equation

Known: $Q_i^n=420$ (initial flow), $Q_{i+1}^n=420$, $Q_i^{n+1}=580$ (upstream flood). $\Delta x=4650$ m, $\Delta t=1.5\times3600=5400$ s, $\Delta t/\Delta x=1.1613$, no lateral inflow.

$$\frac{\Delta t}{\Delta x}x+\alpha x^{0.6}=K,\qquad K=\alpha\left[420^{0.6}+420^{0.6}-580^{0.6}\right]-\frac{\Delta t}{\Delta x}\left(420-580-420\right)=821.779$$

with $x=Q_{i+1}^{n+1}$.

Step 4: Newton iteration

$f(x)=1.1613x+5.0280x^{0.6}-821.779$, $f'(x)=1.1613+3.0168x^{-0.4}$.

Answer: $Q_{i+1}^{n+1}=522.6$ m$^3$/s at 4.65 km downstream after 1.5 h (it rises from 420 towards the upstream value of 580 m$^3$/s).