Linear kinematic wave, 200 ft channel

The linear kinematic wave model writes the kinematic wave equation as $\dfrac{\partial Q}{\partial x}+\alpha\beta Q^{\beta-1}\dfrac{\partial Q}{\partial t}=q$, where $A=\alpha Q^{\beta}$ comes from Manning's equation. It is solved with backward differences at the unknown point $(i+1,\ n+1)$, and the coefficient $Q^{\beta-1}$ is evaluated at the average $\bar Q=(Q_{i+1}^n+Q_i^{n+1})/2$, so the equation is linear in the unknown.

Step 1: Initial uniform flow (US units, Manning constant 1.49)

$$Q=\frac{1.49}{n}AR^{2/3}S_0^{1/2},\qquad A=200y,\ P=200+2y,\ R=A/P$$

Trial for $Q=2000$ cfs gives $y=1.680$ ft, so $A=336.09$ ft$^2$, $P=203.36$ ft, $R=1.653$ ft and $Q=2000$ cfs (check).

Step 2: Kinematic coefficients

$$\alpha=\left(\frac{nP^{2/3}}{1.49\sqrt{S_0}}\right)^{0.6}=\left(\frac{0.035\times203.36^{2/3}}{1.49\times\sqrt{0.01}}\right)^{0.6}=3.514,\qquad \beta=0.6$$

(Taking $P\approx B=200$ ft gives $\alpha=3.49$ and the same answer to the nearest cfs.)

Step 3: Finite difference equation

$$\frac{Q_{i+1}^{n+1}-Q_i^{n+1}}{\Delta x}+\alpha\beta\,\bar Q^{\beta-1}\,\frac{Q_{i+1}^{n+1}-Q_{i+1}^n}{\Delta t}=\bar q$$

Solving for the unknown:

$$Q_{i+1}^{n+1}=\frac{\dfrac{\Delta t}{\Delta x}Q_i^{n+1}+\alpha\beta\,Q_{i+1}^n\,\bar Q^{\beta-1}+\Delta t\,\bar q}{\dfrac{\Delta t}{\Delta x}+\alpha\beta\,\bar Q^{\beta-1}}$$

Here $\bar q=0$.

Step 4: Substitution

$Q_{i+1}^n=2000$ cfs (uniform flow at $t=0$), $Q_i^{n+1}=2250$ cfs (upstream, $t=3$ min); the point at 3000 ft is $i+1$. $\Delta t/\Delta x=180/3000=0.06$.

$$\bar Q=\frac{2000+2250}{2}=2125\ \text{cfs},\qquad \alpha\beta\,\bar Q^{-0.4}=3.514(0.6)(2125)^{-0.4}=0.09841$$ $$Q_{i+1}^{n+1}=\frac{0.06(2250)+0.09841(2000)}{0.06+0.09841}=\frac{135.00+196.82}{0.15841}=2094.7\ \text{cfs}$$

Answer: the discharge at 3000 ft downstream after 3 min is $Q\approx2095$ cfs. It lies between the initial 2000 cfs and the upstream 2250 cfs, because the flood wave is starting to arrive.

Note: if the second-order four-point linear scheme with a constant celerity $c=\tfrac53V=9.918$ ft/s ($V=5.951$ ft/s, $C=c\Delta t/\Delta x=0.5951$) is used instead, $Q_{i+1}^{n+1}=Q_i^n+\dfrac{1-C}{1+C}\left(Q_{i+1}^n-Q_i^{n+1}\right)=1936.5$ cfs. The dip below 2000 cfs is a numerical oscillation (dispersion) of that scheme, not a physical effect.