CIV E 445 - APPLIED HYDROLOGY SPRING 2003 SOLUTIONS TO HOMEWORK 11 , CHAPTER 9
Problem 11-1 Using Eqs. 9-4 to 9-6, the Muskingum routing coefficients are the following: C0 = 0.231, C1 = 0.538, C2 = 0.231
The peak outflow is 150.353 m3/s, and it occurs at 6 h. ANSWER. Problem 11-2 Storage is calculated with Eq. 9-7. Weighted flow is: [X • I + (1 - X) • O]. The results of the computation are shown in the following table.
The storage (Col. 4) vs. weighed flow relations (Col. 5 to 7) are plotted as shown in Fig. 9-4 of the text. The chosen value of X is that corresponding to the narrowest loop in the storage-weighted flow relation. In this case: X = 0.3. The value of the storage constant K is the slope of the storage-weighted flow relation for X = 0.3: K = 1.25 h. ANSWER. Problem 11-3 A rough estimate of the discharge at point 5 km upstream can be obtained by discretizing the equation of continuity, Eq. 9-9, using an off-centered numerical scheme: (Q2 - Q1) / Δx + (A2 - A1) / Δt = 0 in which Q = discharge, A = flow area, Δt = time interval, and Δx = space interval (reach length). Since the channel width is constant, the change is flow area per time interval is: ΔA = B Δy, in which B = channel width, and Δy = change in stage per time interval. Then : (Q2 - Q1) / Δx + B (Δy / Δt) = 0 Solving for the upstream discharge (Q1): Q1 = Q2 + B Δx (Δy / Δt) Q1 = 2000 + [600 m • 5 km • 1000 m/km • ( 9 cm • 0.01 m/cm) / (3600 s)] = 2075 m3/s. ANSWER. Problem 11-4 The results of flood routing with the Muskingum-Cunge method, using Program EH900 are summarized below.
THANK YOU FOR RUNNING PROGRAM EH900.
COURANT NUMBER C= 1.02
The Peak outflow is 1,093.39 m3/s, and it occurs in 8 h. ANSWER. THANK YOU FOR RUNNING EH900. PLEASE CALL AGAIN.
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