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Tugas BA Pak Ivan

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Tugas bangunan air
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  9.5   A reservoir has a capacity of 5 x 10 6 m 3  and a drainage area of 190 km 2 . The average annual runoff from the watershed is 390 mm, which brings in a sediment quantity of 600 m 3 /km 2 . Determine the time required to reduce the reservoir capacity to 1 x 10 6  m 3 . Use the mean curve. Solution : 1.   Annual inflow ( I ) = 5 acre-ft 2.   Unfilled capacity =   x 4053.56 = 810.71 acre-ft 3.   Annual sediment production = 50 acre-ft Reservoir capacity (acre-ft) Capacity Lost (acre-ft) Capacity Inflow (C/I) Trap Efficiency (%) Average Trap Efficiency (%) Annually Trapped Sediment (acre-ft) Time to Fill (years) 4053.56 2.02678 92 810.71 90 45 18.0157778 3242.85 1.621425 88 810.71 86.5 43.25 18.7447399 2432.14 1.21607 85 810.71 82.5 41.25 19.6535758 1621.43 0.810715 80 810.71 74 37 21.9110811 810.72 0.40536 68 TOTAL 78.3251745 9.6   An earth dam has a crown (top) width of 20 ft. The height is 70 ft with a freeboard of 10 ft. The slpoes are 1 ( vertical ) : 3 ( horizontal ). It rests on an impervious foundation having a drainage blanket extending back 50 ft from the toe. The permeability of embankment material is 3.2 x 10 -5  ft/sec. Determine the seepage per foot length of the dam. Solution : H = 60 ft CB = 60(3) = 180 ft AB = 0.3 x 180 = 54 ft CA = 114 ft  d = base length  –   blanket length  –   CA = 440 - 70  –   114 = 356 ft   √                          9.7   An earth dam on an impervious foundation has a crown width of 25 ft, a total height of 80 ft, and a freeboard of 10 ft. It has an upstream slope of 20 o  and a downstream slope of 30 o . A horizontal blanket extends back 80 ft from the back toe. The dam has a permeability of 4 x 10 -4  cm/s. Determine the seapage throught the dam. Solution : Revised slope angle                   √         ()    √     √                              9.8   In the dam of Problem 9.7, if the material has a permeability of 4 x 10 -4  cm/s horizontally and 1 x 10 -4  cm/s vertically, determine the seapage throught the dam. Solution : Revised slope angle                    √         ()    √     √                              9.9   An earth dam of crown width 30 ft and height 85 ft rests on an impervious foundation. It has side slopes of 1 ( vertical ) : 4 ( horizontal ) and a permeability of 2 x 10 -4  cm/s. A center core is provide symmetrically about the vertical axis of the dam. The core has a top width of 20 ft, side slopes of 1 (vertical) to 1 (horizontal), and a permeability of 2 x 10 -5  cm/s. The water depth in the reservoir is 75 ft. Determine the quantity of seapage throught the dam. Solution : H = 85 ft CB = 85(4) = 340 ft AB = 0.3 x 340 = 102 ft CA = 187 ft d = base length  –   blanket length  –   CA = 340 - 85  –   187 = 68 ft   √                   H = 65 ft CB = 65(1) = 65 ft AB = 0.3 x 65 = 19.5 ft  CA = 84.5 ft d = base length  –   blanket length  –   CA = 170 - 65  –   84.5 = 20.5 ft   √                   

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