CE 4603: Water Resources Engineering — David Wagner 2007/03/26 13:18
Homework # 7 Due 3/26 (or 3/28)
Solve problems 5.74 and 5.75 of Water Resources Engineering by David Chin, 2nd edition, pages 597-598.
5.74
A stormwater detention basin has the following storage characteristics:
[Chin#460-462:Example 5.37] Δt=30 min.
| Stage | Storage | | 2S/Δt+O |
|---|---|---|---|
| 8.0 | 0 | 0 | 0 |
| 8.5 | 1041 | 1.16 | 2.32 |
| 9.0 | 2288 | 3.29 | 5.83 |
| 9.5 | 3761 | 6.04 | 10.22 |
| 10.0 | 5478 | 9.31 | 15.39 |
| 10.5 | 7460 | 13.00 | 21.29 |
| 11.0 | 9726 | 17.10 | 27.90 |
The discharge weir from the detention basin has a crest elevation of 8.0 m, and the weir discharge Q (m³/s) is given by
where h is the height of the water surface above the crest of the weir in meters. The catchment runoff hydrgraph is given by:
| Time | Runoff | 2S/deltat-O | 2S/deltat+O | O |
|---|---|---|---|---|
| (min) | (m³/s) | (m³/s) | (m³/s) | (m³/s) |
| 0 | 0 | 0 | 0.00 | 0.00 |
| 30 | 3.6 | -0.28 | 3.60 | 1.94 |
| 60 | 8.4 | -2.26 | 11.72 | 6.99 |
| 90 | 5.1 | -2.13 | 11.24 | 6.69 |
| 120 | 4.2 | -1.09 | 7.17 | 4.13 |
| 150 | 3.6 | -0.97 | 6.71 | 3.84 |
| 180 | 3.3 | -0.77 | 5.93 | 3.35 |
| 210 | 2.7 | -0.62 | 5.23 | 2.92 |
| 240 | 2.3 | -0.44 | 4.38 | 2.41 |
| 270 | 1.8 | -0.29 | 3.66 | 1.97 |
| 300 | 1.5 | -0.15 | 3.01 | 1.58 |
| 330 | 0.84 | -0.01 | 2.19 | 1.10 |
| 360 | 0.51 | 0 | 1.34 | 0.67 |
| 390 | 0 | 0 | 0.51 | 0.25 |
| 420 | 0 | 0 | 0.00 | 0.00 |
If the prestorm stage in the detention basin is 8.0 m, estimate the discharge hydrograph from the detention basin.
5.75
The flow hydrograph at a channel section is given by:
| Time | I | O |
|---|---|---|
| (min) | (m³/s) | (m³/s) |
| 0 | 0 | 0 |
| 30 | 12.5 | 1.42375 |
| 60 | 22.1 | 10.92960188 |
| 90 | 15.4 | 18.65038925 |
| 120 | 13.6 | 15.97669861 |
| 150 | 12.4 | 14.03491602 |
| 180 | 11.7 | 12.7134673 |
| 210 | 10.8 | 11.84122889 |
| 240 | 9.9 | 10.94790555 |
| 270 | 8.4 | 9.981171284 |
| 300 | 8.1 | 8.746101694 |
| 330 | 7.5 | 8.187047457 |
| 360 | 4.2 | 7.289364913 |
| 390 | 0 | 4.464612262 |
| 420 | 0 | 1.073739249 |
| 450 | 0 | 0.258234289 |
| 480 | 0 | 0.062105347 |
| 510 | 0 | 0.014936336 |
| 540 | 0 | 0.003592189 |
Use the Muskingum method to estimate the hydrograph 1000 m downstream from the channel section. Assume that X= 0.3 and K = 35 min.
[Chin#464-465:Example 5.38]
= 21
- 2KX ≤ Δt ≤ 2K(1-X) → 21 ≤ Δt ≤ 49 → Δt=30 min = 0.5 hr
= 0.1139
= 0.6456
= 0.2405- Assume O0=0

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