Example: Route a flood wave with the following characteristics: peak flood Qp= 1000 m3/s; baseflow Qb= 0 m3/s;
So = 0.000868; Ap = 400 m2; Tp = 100 m; β = 1.6; Δx = 14.4 km; Δt = 1 hour.
Solution: For lack of other data, base calculation on peak discharge.
The mean velocity V = Qp/Ap = 1000/400 = 2.5 m/s.
The wave celerity is: c = β Vp = 1.6 × 2.5 = 4 m/s.
The flow per unit width qo
(based on peak discharge)
= Qp/Tp = 1000/100 = 10 m2/s.
The Courant number C = c Δt/Δx = 4 m/s × 3600 s / 14400 m = 1.
The cell Reynolds number D = qo/(SocΔx) = 10 / (0.000868 × 4. × 14400) = 0.2.
The routing coefficients are: C0 = 0.091; C1 = 0.818; and C2 = 0.091.
The routing calculations are shown below.
Time | Inflows | C0I2 | C1I1 | C2O1 | O |
0 | 0 | - | - | - | 0 |
1 | 200 | 18.2 | 0.0 | 0.0 | 18.20 |
2 | 400 | 36.4 | 163.6 | 1.66 | 201.66 |
3 | 600 | 54.6 | 327.2 | 18.35 | 400.15 |
4 | 800 | 72.8 | 490.8 | 36.41 | 600.01 |
5 | 1000 | 91.0 | 654.4 | 54.60 | 800.00 |
6 | 800 | 72.8 | 818.0 | 72.80 | 963.60 |
7 | 600 | 54.6 | 654.4 | 87.69 | 796.69 |
8 | 400 | 36.4 | 490.8 | 72.50 | 599.70 |
9 | 200 | 18.2 | 327.2 | 54.57 | 399.97 |
10 | 0 | 0.0 | 163.6 | 36.4 | 200.00 |
11 | 0 | 0.0 | 0.0 | 18.2 | 18.2 |
12 | 0 | 0.0 | 0.0 | 1.66 | 1.66 |
13 | 0 | 0.0 | 0.0 | 0.16 | 0.16 |
Note that the peak outflow is 963.6 and it occurs at time 6 hr. The wave has diffused from a peak of 1000 to 963.6, and has translated from t = 5 hr to t = 6 hr.