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 O=3.29(Stage-8)^{3/2} 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

Q = 3.29 h^{3/2}

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]

  1. 2KX = 2*35*0.3 = 21
  2. 2K(1-X) = 2*35*(1-0.3) = 49
  3. 2KX ≤ Δt ≤ 2K(1-X) → 21 ≤ Δt ≤ 49 → Δt=30 min = 0.5 hr
  4. C_1={delta t-2KX}/{2K(1-X)+30} = {30-21}/{49+30} = 9/79 = 0.1139
  5. C_1={delta t+2KX}/{2K(1-X)+30} = {30+21}/{49+30} = 51/79 = 0.6456
  6. C_1={2K(1-X)-delta t}/{2K(1-X)+30} = {49-30}/{49+30} = 19/79 = 0.2405
  7. Assume O0=0

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