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.  The same answer is obtained with ONLINE ROUTING 04. 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 = 1250 + [600 m • 10 km • 1000 m/km • ( 3 mm • 0.001 m/mm) / (3600 s)] = 1255 m³/s.  ANSWER.

Problem 11-4

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 ONLINE ROUTING 05.   ANSWER.