CIV E 445 - APPLIED HYDROLOGY
SPRING 2014
HOMEWORK No. 11 SOLUTION

  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.  The same answer is obtained with ONLINE ROUTING 04, shown below. ANSWER.


  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 below. 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.

  3. A rough estimate of the discharge at point 20 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 = 1175 + [300 m • 20 km • 1000 m/km • ( 3 mm/h • 0.001 m/mm) / (3600 s/h)] = 1175 + 5 = 1180 m3/s.  ANSWER.

  4. The results are summarized in the following table.

    The peak outflow is 1,093.39 m3/s, and it occurs at 8 h. The same answer is obtained with ONLINEROUTING05, shown below   ANSWER.