∂Q/∂x + ∂A/∂t = 0
∂Q/∂x + B ∂y/∂t = 0
r = ∂y/∂t
∂Q/∂x + B r = 0
Qd - Qu = - B (Δx) r
Qu = Qd + B (Δx) r
Qu = 550 + 240 (20,000) (3 mm/hr / 3600 s/hr) (1 m / 1000 mm)
Qu = 554 m3
ANSWER.
∂Q/∂x + ∂A/∂t = 0
∂Q/∂x + B ∂y/∂t = 0
r = ∂y/∂t
∂Q/∂x + B r = 0
ra = 0.5 (ru + rd) = 4.5
Qd - Qu = - B (Δx) ra
Qu = Qd + B (Δx) ra
Qu = 550 + 240 (20,000) (4.5 mm/hr / 3600 s/hr) (1 m / 1000 mm)
Qu = 556 m3
ANSWER.
F2 = q2/(gy3)
y = [q2/(gF2 ]1/3 = 3.3 m.
F = v/(gy)1/2 = 0.16
v = 0.16 (gy)1/2
cr= (gy)1/2 = (9.81 × 3.3) 1/2 = 5.69 m/s
v = 0.16 × 5.69 = 0.91 m/s
cd = v + cr = 0.91 + 5.69 = 6.6 m/s ANSWER.
cu = v - cr = 0.91 - 5.69 = -4.78 m/s ANSWER.
Peak flood Qp= 600 m3/s;
baseflow Qb= 100 m3/s;
So = 0.0007;
Ap = 350 m2;
Tp = 85 m;
β = 1.65;
Δx = 9.8 km = 9,800 m;
Δt = 1 hour = 3,600 s.
The mean velocity Vp= Qp/Ap = 600/350 =
1.714 m/s.
The wave celerity is: The flow per unit width qo
= Qp/Tp = 600/85 = 7.059 m2/s. The Courant number C = c Δt/Δx = 2.828 m/s × 3600 s / 9,800 m =
1.039
The cell Reynolds number D = qo/(Soc Δx) = (7.059) / (0.0007 × 2.828 × 9,800) = 0.364
The routing coefficients are: C0 = 0.168; C1 = 0.697; and C2 = 0.135. The routing calculations are shown below.
The peak outflow is 581.351 and it occurs at time 11 hr.
The wave has diffused from a peak of 600 to 581, and has translated from t = 10 hr to t = 11 hr.
ANSWER.
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