Solve example 5.2.2 for discharges of 0, 25, 75, 125, and 200 ft3/s.
EXAMPLE 5.2.2
For a rectangular channel of 20 ft width, construct a family of specific energy curves for Q = 0 50, 100, and 300 cfs. Draw the locus of the critical depth points on these curves. For each flow rate, what is the minimum specific energy found from these curves?
The specific energy is computed using equation (5.2.1):
Computing critical depths for the flow rates using equation (5.2.14) with q = Q/B yields
Computed specific energies are listed in Table 5.2.1.
The specific energy curves are shown in Figure 5.2.3. The minimum specific energies are:
Table 5.2.1 Computed Specific Energy Values for Example 5.2.2
Depth, y (ft) | Specific energy, E (ft-lb/lb) | |||
Q = 0 | Q = 50 | Q = 100 | Q = 300 | |
0.5 | 0.50 | 0.89 | 2.05 | 14.86 |
0.6 | 0.60 | 0.87 | 1.68 | 10.57 |
0.8 | 0.80 | 0.95 | 1.41 | 6.41 |
1.0 | 1.00 | 1.10 | 1.39 | 4.59 |
1.2 | 1.20 | 1.27 | 1.47 | 3.69 |
1.4 | 1.40 | 1.45 | 1.60 | 3.23 |
1.6 | 1.60 | 1.64 | 1.75 | 3.00 |
1.8 | 1.80 | 1.83 | 1.92 | 2.91 |
2.0 | 2.00 | 2.02 | 2.10 | 2.90 |
2.2 | 2.20 | 2.22 | 2.28 | 2.94 |
2.4 | 2.40 | 2.42 | 2.47 | 3.02 |
2.6 | 2.60 | 2.61 | 2.66 | 3.13 |
2.8 | 2.80 | 2.81 | 2.85 | 3.26 |
3.0 | 3.00 | 3.01 | 3.04 | 3.40 |
3.5 | 3.50 | 3.51 | 3.53 | 3.79 |
4.0 | 4.00 | 4.50 | 4.2 | 4.22 |
4.5 | 4.50 | 5.00 | 4.52 | 4.68 |
5.0 | 5.00 | 5.00 | 5.02 | 5.14 |
Figure 5.2.3 Specific energy curves for example 5.2.2.
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