CIV E 530 - OPEN-CHANNEL HYDRAULICS
SPRING 2005 - HOMEWORK 11 - SOLUTION

PROBLEM 1

∂Q/∂x + ∂A/∂t = 0

∂Q/∂x + B ∂y/∂t = 0

r = ∂y/∂t

∂Q/∂x + B r = 0

Q_d - Q_u = - B (Δx) r

Q_u = Q_d + B (Δx) r

Q_u = 650 + 280 (25,000) (4 mm/hr / 3600 s/hr) (1 m / 1000 mm)

Q_u = 657.78 m³ ANSWER.

PROBLEM 2

∂Q/∂x + ∂A/∂t = 0

∂Q/∂x + B ∂y/∂t = 0

r = ∂y/∂t

∂Q/∂x + B r = 0

r_a = 0.5 (r_u + r_d) = 6.0

Q_d - Q_u = - B (Δx) r_a

Q_u = Q_d + B (Δx) r_a

Q_u = 650 + 280 (25,000) (6.0 mm/hr / 3600 s/hr) (1 m / 1000 mm)

Q_u = 661.67 m³ ANSWER.

The second estimate is more accurate, because it is a central finite difference.

PROBLEM 3

F² = q²/(gy³)

y = [q²/(gF² ]^1/3 = 3.44 m.

F = v/(gy)^1/2 = 0.2

v = 0.2 (gy)^1/2

c_r= (gy)^1/2 = (9.81 × 3.44) ^1/2 = 5.81 m/s

v = 0.2 × 5.81 = 1.16 m/s

c_d = v + c_r = 1.16 + 5.81 = 6.97 m/s ANSWER.

c_u = v - c_r = 1.16 - 5.81 = -4.65 m/s ANSWER.

PROBLEM 4

Peak flood Q_p= 600 m³/s;

baseflow Q_b= 100 m³/s;

S_o = 0.0007;

A_p = 350 m²;

T_p = 85 m;

β = 1.65;

Δx = 9.8 km = 9,800 m;

Δt = 1 hour = 3,600 s.

The mean velocity V_p= Q_p/A_p = 600/350 = 1.714 m/s.

The wave celerity is: c = β V_p = 1.65 × 1.714 = 2.828 m/s.

The flow per unit width q_o = Q_p/T_p = 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 = q_o/(S_oc 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.

Time Inflows C₀I₂ C₁I₁ C₂O₁ O₂

0 100 - - - 100.000

1 130 21.795 69.721 13.514 105.030

2 150 25.148 90.637 14.193 129.979

3 180 30.178 104.581 17.565 152.324

4 220 36.884 125.497 20.585 182.966

5 250 41.914 153.386 24.726 220.025

6 300 50.296 174.302 29.734 254.332

7 360 60.356 209.162 34.370 303.888

8 450 75.445 250.995 41.067 367.506

9 520 87.180 313.744 49.664 450.588

10 600 100.593 362.548 60.891 524.032

11 550 92.210 418.325 70.816 581.351

12 490 82.151 383.464 78.562 544.177

13 370 62.032 341.632 73.539 477.203

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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