Question 6.3: Figure 6-13 shows a channel network. The channel data are li......
Figure 6-13 shows a channel network. The channel data are listed in Table 6-3. All channels have subcritical flow and trapezoidal cross sections having side slopes of 1.5 horizontal to 1 vertical. The flow depth at the downstream end (Node F) is 5 m for a flow of 250 m³ /s. The form loss coefficient, k, for all junctions is 0.0 and the velocity-head coefficient, α, for all channels is 1. Compute the flows and depths in different channels of the system.
| Table 6-3 Channel Data | |||||
| Channel | Length (m) | Width (m) | Reach (m) | n | So |
| 1 | 200.0 | 30.0 | 50.0 | 0.013 | 0.0005 |
| 2 | 200.0 | 40.0 | 50.0 | 0.013 | 0.0005 |
| 3 | 200.0 | 20.0 | 50.0 | 0.012 | 0.0005 |
| 4 | 100.0 | 20.0 | 25.0 | 0.014 | 0.0005 |
| 5 | 100.0 | 20.0 | 25.0 | 0.013 | 0.0005 |
| 6 | 100.0 | 25.0 | 25.0 | 0.013 | 0.0005 |
| 7 | 100.0 | 30.0 | 25.0 | 0.014 | 0.0005 |
| 8 | 300.0 | 50.0 | 75.0 | 0.014 | 0.0005 |
The channels and nodes are numbered as shown in Fig. 6-13. The iterative procedure is started by assuming all flow depths equal to 5.0 m. The discharges are given random values without satisfying continuity at the branching junctions. The assumed flow directions in different channels are as shown by the arrows of Fig. 6-13. A tolerance of 0.0005 for both Δ_{y_{i,j}} and ΔQ_{i,j} is specified for the convergence of the solution procedure. The iterations converge after only six iterations. The computed discharges and flow depths are listed in Table 6-4.
\quad\quadThese values were verified by computing the flow profiles by using the fourth-order Runge Kutta method. Instead of using graphical or other manual methods to determine the discharge in various channels, the values calculated by the procedure of this section were used directly. The water levels computed by the Runge-Kutta method were identical (within the specified tolerance) to those listed in Table 6-4. The computational time for each individual channel was approximately the same for both methods. It is not possible to estimate the number of trial-and-error iterations required to determine the discharge in the Runge Kutta method. Assuming m such iterations, the computational time for the Runge-Kutta method is at least m times that required by the procedure presented in this section.
| Table 6-4 Network of Example 6-3 | ||||||||
| Section | Channel 1 Q = 95.748 m² /s | Channel 2 Q = 154.252 m² /s | Channel 3 Q = 55.003 m² /s | Channel 4 Q = 40.655 m² /s | ||||
| Depth (m) | Distance (m) | Depth (m) | Distance (m) | Depth (m) | Distance (m) | Depth (m) | Distance (m) | |
| 1 | 4.754 | 0.0 | 4.754 | 0.0 | 4.853 | 0.0 | 4.853 | 0.0 |
| 2 | 4.779 | 50.0 | 4.779 | 50.0 | 4.877 | 50.0 | 4.865 | 25.0 |
| 3 | 4.803 | 100.0 | 4.803 | 100.0 | 4.902 | 100.0 | 4.677 | 50.0 |
| 4 | 4.828 | 150.0 | 4.828 | 150.0 | 4.927 | 150.0 | 4.8 | 75.0 |
| 5 | 4.853 | 200.0 | 4.852 | 200.0 | 4.951 | 200.0 | 4.902 | 100.0 |
| Section | Channel 5 Q = 52.669 m² /s | Channel 6 Q = 12.014 m² /s | Channel 7 Q = 107.762 m² /s | Channel 8 Q = 142.238 m² /s | ||||
| Depth (m) | Distance (m) | Depth (m) | Distance (m) | Depth (m) | Distance (m) | Depth (m) | Distance (m) | |
| 1 | 4.902 | 0.0 | 4.852 | 0.0 | 4.051 | 0.0 | 4.352 | 0.0 |
| 2 | 4.914 | 25.0 | 4.864 | 25.0 | 4.986 | 25.0 | 4.860 | 75.0 |
| 3 | 4.927 | 50.0 | 4.877 | 50.0 | 4.979 | 50.0 | 4.926 | 150.0 |
| 4 | 4.939 | 75.0 | 4.889 | 75.0 | 4.988 | 75.0 | 4.963 | 225.0 |
| 5 | 4.952 | 100.0 | 4.902 | 100.0 | 5.000 | 100.0 | 5.000 | 300.0 |