Detention Pond CalculationPhaseII

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<p>DRY SEDIMENTATION PONDA. Determine overland flow time of concentration 5.5 ha CHECK Catchment Area =CHECK Overland CHECK Overland CHECK Manning COMPUTE Overland COMPUTE Adopted</p> <p>slope = length = "n" = time, to =</p> <p>0.8 % 190 m 0.02 9.63 min 10 min</p> <p>time of concentration, tc =</p> <p>B Sizing sendimentation Table 39.4 type = Table 39.5 for a 3 month ARI surface area = Volume =</p> <p>CHECK Soils</p> <p>C</p> <p>CHECK Required CHECK Total COMPUTE Site COMPUTE Site</p> <p>333 m/ha 400 m/ha 1,831.50 m 2,200.00 m</p> <p>surface area required = total vomue required =</p> <p>i) Settling zoneCOMPUTE Settling</p> <p>zone depth, y1 = settling zone =</p> <p>0.6 m 1,100.00 m 17 m 107.84 m 1,833.33 m</p> <p>COMPUTE Required CHECK Assume</p> <p>average width, W1 = settling zone a. length, L1 =</p> <p>COMPUTE Required COMPUTE Average</p> <p>surface area =</p> <p>Check settling zone dimensionsCOMPUTE L1 / COMPUTE L1 /</p> <p>y1 = W1 =</p> <p>179.74 6.34</p> <p>&lt; 200 &gt;2</p> <p>OK OK</p> <p>Page 1 of 40</p> <p>C Sediment Storage ZoneCOMPUTE Required CHECK For COMPUTE .. COMPUTE ..</p> <p>sediment storage volume =</p> <p>1100 m 2 15.8 m 106.64 m</p> <p>a side slope Z = 2(H):1(V)</p> <p>W2 = W1 - 2 d1/2 x Z L2 = L1 - 2 d1/2 x Z</p> <p>Required depth for the sediment storage V2 = Z(y2) - Z (y2) (W2 + L2) + y2 ( W2 x L2 ) Trial and error to find y2 V2 1100 CHECKCHECK CHECK</p> <p>y2</p> <p>V2 0.6 923.68 0.7 1060.85 0.73 1101.08</p> <p>&gt; 0.3</p> <p>OK</p> <p>D Overall Basin Dimensions At top water levelCOMPUTE WTWL COMPUTE LTWL</p> <p>= W1 + 2 x Z x y1/2</p> <p>18.2 m 106.6 m</p> <p>say say</p> <p>18 m 107 m</p> <p>= L1 - 2 x Z x y1/2</p> <p>BaseCOMPUTE WB COMPUTE LB</p> <p>= W1 - 2 x Z x (y1/2 + y2)</p> <p>12.9 m 103.7 m</p> <p>say say</p> <p>13 m 104 m</p> <p>= L1 - 2 x Z x (y1/2 + y2)</p> <p>Depth:COMPUTE Settling</p> <p>Zone, y1 = storage zone, y2 =</p> <p>0.6 0.73</p> <p>m m</p> <p>COMPUTE Sediment</p> <p>Page 2 of 40</p> <p>E Sizing of outlet pipeCOMPUTE Total</p> <p>depth, y surface area, Aav =</p> <p>1.33 m 1,638.43 m 24 hr after filling</p> <p>Cd = 0.6</p> <p>COMPUTE Average</p> <p>Draining time, Orifice area, ATOTAL x y) /(tCd2g) orifice size of</p> <p>COMPUTE (2Aav CHECK Using COMPUTE Area</p> <p>0.0165 Equation 19.5, Volume 7 50 mm 0.00196 m</p> <p>of each orifice is, Ao =</p> <p>Total number of orifice required, COMPUTE ATOTAL / Ao =</p> <p>8</p> <p>nos</p> <p>At height of increments of 300mm, starting at the bottom of the pipes, put 2 rows orifices evenly around the pipe</p> <p>4</p> <p>nos</p> <p>of</p> <p>50 mm</p> <p>F Sizing of emergency spillway The emergency spillway must be designed for a 10 year ARI flood (see MSMA, Volume 15, Section 39.7.6c) The sill level must be set a minimum 300mm above the basin top water level. Parameter tc = 10 min For Penang, 10 year ARI Area, A = 5.5 ha A 3.73 B 1.44 Fd from Table 13.3 C -0.4 Interpolate for 140mm D 0.02 Fd = 1.048 From Equation 13,2 t=30 min, I30 = 136.648 mm/hr P30 = 68.32 mm</p> <p>Page 3 of 40</p> <p>t=60 min, I60 = P60 =</p> <p>92.833 mm/hr 92.83 mm</p> <p>Rainfall Intensity, I10 = [P30 - Fd(P60 - P30)] / (10/60) 255.83 mm/hr From Chart 14.3, Runoff Coefficient, C = Equation 14.7, Q10 = 0.84 3.28 m/s</p> <p>IF Equation 20.2 for Orifice flow An orifice discharge coefficient of = Outlet riser, perforated MS pipe = Ho = Qriser = Cd Ao 2 g Ho = Therefore Qspillway = Q10 - Qriser = From Equation 20.9, Qspillway = Csp B Hp^1.5 Trial Dimension, B 5</p> <p>0.64 0.9 m 0.3 m 0.99 m/s 2.29 m/s</p> <p>Hp 0.5</p> <p>D.Chart 20.2 Csp 1.65</p> <p>Qspillway = Therefore, the total basin depth including the spillway is =</p> <p>2.92</p> <p>&gt;</p> <p>2.29 OK</p> <p>2.13 m</p> <p>Page 4 of 40</p> <p>Page 5 of 40</p> <p>Page 6 of 40</p> <p>Volume 15, n top water</p> <p>for 140mm</p> <p>Page 7 of 40</p> <p>Page 8 of 40</p> <p>58925981.xls</p> <p>WET SEDIMENTATION PONDACHECK Catchment</p> <p>Area =</p> <p>5.2 ha 40 mm</p> <p>75th % 5-day storm event</p> <p>B Sizing sendimentation Table 39.4 type = Table 39.6 for a 3 month ARI Zone Volume Volume = Zone Volume =</p> <p>CHECK Soils</p> <p>F</p> <p>clay</p> <p>CHECK Settling CHECK Total</p> <p>200 m/ha 300 m/ha 1,040.00 m 1560 m</p> <p>COMPUTE Settling COMPUTE Total</p> <p>Volume</p> <p>i) Settling zoneCHECK Settling</p> <p>zone depth, y1 = settling zone =</p> <p>0.6 m 1560 m 30 m 87 m 2,600.00 m</p> <p>COMPUTE Required CHECK Try COMPUTE The</p> <p>settling zone average width, W1 = settling zone average length, L1 = surface area =</p> <p>COMPUTE Average</p> <p>Check settling zone dimensionsCOMPUTE L1 / COMPUTE L1 /</p> <p>y1 = W1 =</p> <p>144.44 2.89</p> <p>&lt; 200 &gt;2</p> <p>OK OK</p> <p>Page 9</p> <p>58925981.xls</p> <p>C Sediment Storage ZoneCOMPUTE Required CHECK For</p> <p>sediment storage volume =</p> <p>520 m</p> <p>a side slope Z = 2(H):1(V) 2 Dimensions at the top of the sediment storage zone are, 28.8 m COMPUTE .. W2 = W1 - 2 d1/2 x ZCOMPUTE ..</p> <p>L2 = L1 - 2 d1/2 x Z</p> <p>85.47 m</p> <p>Required depth for the sediment storage V2 = Z(y2) - Z (y2) (W2 + L2) + y2 ( W2 x L2 ) Trial and error to find y2 V2 520 CHECKCHECK CHECK</p> <p>y2 0.3 0.4 0.35 Taken, y2 =</p> <p>V2 717.97 948.27 833.68 0.35</p> <p>&gt; 0.3 m</p> <p>NOT OK OK OK</p> <p>D Overall Basin Dimensions At top water levelCOMPUTE WTWL COMPUTE LTWL</p> <p>= W1 + 2 x Z x y1/2</p> <p>31.2 m 87.9 m</p> <p>say say</p> <p>31 m 88 m</p> <p>= L1 + 2 x Z x y1/2</p> <p>BaseCOMPUTE WB COMPUTE LB</p> <p>= W1 - 2 x Z x (y1/2 + y2)</p> <p>27.4 m 84.1 m</p> <p>say say</p> <p>27 m 84 m</p> <p>= L1 - 2 x Z x (y1/2 + y2)</p> <p>Depth: COMPUTE Settling Zone, y1 =COMPUTE Sediment</p> <p>0.6 0.35</p> <p>m m</p> <p>storage zone, y2 =</p> <p>Page 10</p> <p>58925981.xls</p> <p>CHECK</p> <p>E Determine overland flow time of concentration Catchment Area, A = n=</p> <p>5.2 ha 0.02 0.03 % 870 m</p> <p>CHECK</p> <p>Overland slope, S = Overland length, L = = (107*n*L^1/3) / s^1/2 Assume velocity, v = Drain length = td = velocity/drain length =</p> <p>CHECK COMPUTE tc</p> <p>117.95 min 1 m/s 1275 m 0 min 118 min</p> <p>COMPUTE Adopted</p> <p>time of concentration, tc =</p> <p>F Sizing of emergency spillway The emergency spillway must be designed for a 10 year ARI flood (see MSMA, Volume 15, Section 39.7.6c) The sill level is set at the basin top water level. Therefore, Qspillway = Q10 Parameter tc = Area, A = 3.73 Fd = 1.44 -0.4 0.02 136.648 mm/hr 68.32 mm 92.833 mm/hr 92.83 mm</p> <p>CHECK CHECK CHECK CHECK COMPUTE</p> <p>For Penang, 10 year ARI A B C D From Equation 13,2 t=30 min, I30 = P30 = t=60 min, I60 = P60 =</p> <p>118 5.2 0.88</p> <p>min ha</p> <p>CHECK COMPUTE</p> <p>Page 11</p> <p>58925981.xls Rainfall Intensity, I12 = [P30 - Fd(P60 - P30)] / (13/60) 23.73 mm/hr From Chart 14.3, Runoff Coefficient, C = Equation 14.7, Q10 = Trial Dimension, 0.86 0.29 B 3 Qspillway = 0.96 Category 4 m/s Hp 0.35 &gt; D.Chart 20.2 Csp 1.55 0.29 OK</p> <p>COMPUTE</p> <p>Therefore, the total basin depth including the spillway is = COMPUTE</p> <p>1.6</p> <p>m</p> <p>Page 12</p> <p>58925981.xls</p> <p>Page 13</p> <p>58925981.xls</p> <p>Page 14</p> <p>58925981.xls</p> <p>Page 15</p> <p>58925981.xls</p> <p>Page 16</p> <p>WET SEDIMENTATION PONDACHECK Catchment</p> <p>Area =</p> <p>23.74 ha 40 mm</p> <p>75th % 5-day storm event</p> <p>B Sizing sendimentation Table 39.4 type =</p> <p>CHECK Soils</p> <p>F</p> <p>clay</p> <p>Table 39.6 for a 3 month ARI CHECK Settling Zone VolumeCHECK Total</p> <p>200 m/ha 300 m/ha 4,748.58 m 7122.87 m</p> <p>Volume = Zone Volume =</p> <p>COMPUTE Settling COMPUTE Total</p> <p>Volume</p> <p>i) Settling zoneCHECK Settling</p> <p>zone depth, y1 = settling zone =</p> <p>1.65 m 7122.87 m 30 m 144 m 4,316.89 m</p> <p>COMPUTE Required CHECK Try COMPUTE The</p> <p>settling zone average width, W1 = settling zone average length, L1 = surface area =</p> <p>COMPUTE Average</p> <p>Check settling zone dimensionsCOMPUTE L1 / COMPUTE L1 /</p> <p>y1 = W1 =</p> <p>87.21 4.80</p> <p>&lt; 200 &gt;2</p> <p>OK OK</p> <p>C Sediment Storage ZoneCOMPUTE Required CHECK For</p> <p>sediment storage volume =</p> <p>2374.29 m</p> <p>a side slope Z = 2(H):1(V) 2 Dimensions at the top of the sediment storage zone are, 26.7 m COMPUTE .. W2 = W1 - 2 d1/2 x ZCOMPUTE ..</p> <p>L2 = L1 - 2 d1/2 x Z</p> <p>140.60 m</p> <p>Required depth for the sediment storage V2 = Z(y2) - Z (y2) (W2 + L2) + y2 ( W2 x L2 ) Trial and error to find y2 V2 2374.29 CHECKCHECK CHECK</p> <p>y2</p> <p>V2 1.2 4029.81 0.6 2132.76 0.7 2465.17 0.7</p> <p>&gt; 0.3 m</p> <p>OK NOT OK OK</p> <p>Taken, y2 = D Overall Basin Dimensions At top water levelCOMPUTE WTWL COMPUTE LTWL</p> <p>= W1 + 2 x Z x y1/2</p> <p>33.3 m 147.2 m</p> <p>say say</p> <p>33 m 147 m</p> <p>= L1 + 2 x Z x y1/2</p> <p>BaseCOMPUTE WB COMPUTE LB</p> <p>= W1 - 2 x Z x (y1/2 + y2)</p> <p>23.9 m 137.8 m</p> <p>say say</p> <p>24 m 138 m</p> <p>= L1 - 2 x Z x (y1/2 + y2)</p> <p>Depth:COMPUTE Settling</p> <p>Zone, y1 =</p> <p>1.65</p> <p>m</p> <p>COMPUTE Sediment</p> <p>0.7 storage zone, y2 = m Sedimentation storage required is (top)33mx(top)291m. Sedimentation storage provided is (top)40mx(top)415m.Provided storage is more than required storage</p> <p>E Determine overland flow time of concentration Catchment Area, A = CHECK n=CHECK CHECK COMPUTE tc</p> <p>52 ha 0.02 0.05 % 870 m 91.36 min 2 m/s 1275 m 0 min</p> <p>Overland slope, S = Overland length, L = = (107*n*L^1/3) / s^1/2 Assume velocity, v = Drain length = td = velocity/drain length =</p> <p>COMPUTE Adopted</p> <p>time of concentration, tc =</p> <p>91</p> <p>min</p> <p>F Sizing of emergency spillway The emergency spillway must be designed for a 10 year ARI flood (see MSMA, Volume 15, Section 39.7.6c) The sill level is set at the basin top water level. Therefore, Qspillway = Q10 Parameter tc = Area, A = 3.73 Fd = 1.44 -0.4 0.02 69.646 mm/hr</p> <p>CHECK CHECK CHECK CHECK COMPUTE</p> <p>For Penang, 10 year ARI A B C D From Equation 13,2 t=118 min, I118 =</p> <p>91 52 0.88</p> <p>min ha</p> <p>From Chart 14.3, Runoff Coefficient, C = Equation 14.7, Q10 = Trial Dimension,</p> <p>0.52 5.23 B 3</p> <p>Category 4 m/s Hp 1.2 &gt; D.Chart 20.2 Csp 1.55 5.23 OK</p> <p>COMPUTE</p> <p>Qspillway =</p> <p>6.11</p> <p>COMPUTE</p> <p>Therefore, the total basin depth including the spillway is =</p> <p>3.85</p> <p>m</p> <p>Size of the spillway is 3m x 1.2m</p> <p>WET SEDIMENTATION PONDACHECK Catchment</p> <p>Area =</p> <p>23.74 ha 40 mm</p> <p>75th % 5-day storm event</p> <p>B Sizing sendimentation Table 39.4 type =</p> <p>CHECK Soils</p> <p>F</p> <p>clay</p> <p>Table 39.6 for a 3 month ARI CHECK Settling Zone VolumeCHECK Total</p> <p>200 m/ha 300 m/ha 4,748.58 m 7122.87 m</p> <p>Volume = Zone Volume =</p> <p>COMPUTE Settling COMPUTE Total</p> <p>Volume</p> <p>i) Settling zoneCHECK Settling</p> <p>zone depth, y1 = settling zone =</p> <p>1.65 m 7122.87 m 30 m 144 m 4,316.89 m</p> <p>COMPUTE Required CHECK Try COMPUTE The</p> <p>settling zone average width, W1 = settling zone average length, L1 = surface area =</p> <p>COMPUTE Average</p> <p>Check settling zone dimensionsCOMPUTE L1 / COMPUTE L1 /</p> <p>y1 = W1 =</p> <p>87.21 4.80</p> <p>&lt; 200 &gt;2</p> <p>OK OK</p> <p>C Sediment Storage ZoneCOMPUTE Required CHECK For</p> <p>sediment storage volume =</p> <p>2374.29 m</p> <p>a side slope Z = 2(H):1(V) 2 Dimensions at the top of the sediment storage zone are, 26.7 m COMPUTE .. W2 = W1 - 2 d1/2 x ZCOMPUTE ..</p> <p>L2 = L1 - 2 d1/2 x Z</p> <p>140.60 m</p> <p>Required depth for the sediment storage V2 = Z(y2) - Z (y2) (W2 + L2) + y2 ( W2 x L2 ) Trial and error to find y2 V2 2374.29 CHECKCHECK CHECK</p> <p>y2</p> <p>V2 1.2 4029.81 0.6 2132.76 0.7 2465.17 0.7</p> <p>&gt; 0.3 m</p> <p>OK NOT OK OK</p> <p>Taken, y2 = D Overall Basin Dimensions At top water levelCOMPUTE WTWL COMPUTE LTWL</p> <p>= W1 + 2 x Z x y1/2</p> <p>33.3 m 147.2 m</p> <p>say say</p> <p>33 m 147 m</p> <p>= L1 + 2 x Z x y1/2</p> <p>BaseCOMPUTE WB COMPUTE LB</p> <p>= W1 - 2 x Z x (y1/2 + y2)</p> <p>23.9 m 137.8 m</p> <p>say say</p> <p>24 m 138 m</p> <p>= L1 - 2 x Z x (y1/2 + y2)</p> <p>Depth:COMPUTE Settling</p> <p>Zone, y1 =</p> <p>1.65</p> <p>m</p> <p>COMPUTE Sediment</p> <p>0.7 storage zone, y2 = m Sedimentation storage required is (top)33mx(top)291m. Sedimentation storage provided is (top)40mx(top)415m.Provided storage is more than required storage</p> <p>E Determine overland flow time of concentration Catchment Area, A = 23.74 ha CHECK n=CHECK CHECK COMPUTE tc</p> <p>0.02 0.05 % 870 m 91.36 min 2 m/s 1275 m 0 min 91 min</p> <p>Overland slope, S = Overland length, L = = (107*n*L^1/3) / s^1/2 Assume velocity, v = Drain length = td = velocity/drain length =</p> <p>COMPUTE Adopted</p> <p>time of concentration, tc =</p> <p>F Sizing of emergency spillway The emergency spillway must be designed for a 10 year ARI flood (see MSMA, Volume 15, Section 39.7.6c) The sill level is set at the basin top water level. Therefore, Qspillway = Q10 Parameter tc = 91 Area, A = 23.74 3.73 Fd = 0.88 1.44 -0.4 0.02 69.646 mm/hr</p> <p>CHECK CHECK CHECK CHECK COMPUTE</p> <p>For Penang, 10 year ARI A B C D From Equation 13,2 t=91 min, I91 =</p> <p>min ha</p> <p>From Chart 14.3, Runoff Coefficient, C = Equation 14.7, Q10 = Trial Dimension,</p> <p>0.52 2.39 B 3</p> <p>Category 4 m/s Hp 1.2 &gt; D.Chart 20.2 Csp 1.55 2.39 OK</p> <p>COMPUTE</p> <p>Qspillway =</p> <p>6.11</p> <p>COMPUTE</p> <p>Therefore, the total basin depth including the spillway is =</p> <p>3.85</p> <p>m</p> <p>Size of the spillway is 3m x 1.2m</p> <p>WET SEDIMENTATION PONDACHECK Catchment</p> <p>Area =</p> <p>11.87 ha 40 mm</p> <p>75th % 5-day storm event</p> <p>B Sizing sendimentation Table 39.4 type =</p> <p>CHECK Soils</p> <p>F</p> <p>clay</p> <p>Table 39.6 for a 3 month ARI CHECK Settling Zone VolumeCHECK Total</p> <p>200 m/ha 300 m/ha 2,374.70 m 3562.05 m</p> <p>Volume = Zone Volume =</p> <p>COMPUTE Settling COMPUTE Total</p> <p>Volume</p> <p>i) Settling zoneCHECK Settling</p> <p>zone depth, y1 = settling zone =</p> <p>1.65 m 3562.05 m 30 m 72 m 2,158.82 m</p> <p>COMPUTE Required CHECK Try COMPUTE The</p> <p>settling zone average width, W1 = settling zone average length, L1 = surface area =</p> <p>COMPUTE Average</p> <p>Check settling zone dimensionsCOMPUTE L1 / COMPUTE L1 /</p> <p>y1 = W1 =</p> <p>43.61 2.40</p> <p>&lt; 200 &gt;2</p> <p>OK OK</p> <p>C Sediment Storage ZoneCOMPUTE Required CHECK For</p> <p>sediment storage volume =</p> <p>1187.35 m</p> <p>a side slope Z = 2(H):1(V) 2 Dimensions at the top of the sediment storage zone are, 26.7 m COMPUTE .. W2 = W1 - 2 d1/2 x ZCOMPUTE ..</p> <p>L2 = L1 - 2 d1/2 x Z</p> <p>68.66 m</p> <p>Required depth for the sediment storage V2 = Z(y2) - Z (y2) (W2 + L2) + y2 ( W2 x L2 ) Trial and error to find y2 V2 1187.35 CHECKCHECK CHECK</p> <p>y2</p> <p>V2 1.2 1932.16 0.6 1032.15 0.7 1191.19 0.7</p> <p>&gt; 0.3 m</p> <p>OK NOT OK OK</p> <p>Taken, y2 = D Overall Basin Dimensions At top water levelCOMPUTE WTWL COMPUTE LTWL</p> <p>= W1 + 2 x Z x y1/2</p> <p>33.3 m 75.3 m</p> <p>say say</p> <p>33 m 75 m</p> <p>= L1 + 2 x Z x y1/2</p> <p>BaseCOMPUTE WB COMPUTE LB</p> <p>= W1 - 2 x Z x (y1/2 + y2)</p> <p>23.9 m 65.9 m</p> <p>say say</p> <p>24 m 66 m</p> <p>= L1 - 2 x Z x (y1/2 + y2)</p> <p>Depth:COMPUTE Settling</p> <p>Zone, y1 =</p> <p>1.65</p> <p>m</p> <p>COMPUTE Sediment</p> <p>0.7 storage zone, y2 = m Sedimentation storage required is (top)33mx(top)291m. Sedimentation storage provided is (top)40mx(top)415m.Provided storage is more than required storage</p> <p>E Determine overland flow time of concentration Catchment Area, A = 11.87 ha CHECK n=CHECK CHECK COMPUTE tc</p> <p>0.02 0.05 % 870 m 91.36 min 2 m/s 1275 m 0 min 91 min</p> <p>Overland slope, S = Overland length, L = = (107*n*L^1/3) / s^1/2 Assume velocity, v = Drain length = td = velocity/drain length =</p> <p>COMPUTE Adopted</p> <p>time of concentration, tc =</p> <p>F Sizing of emergency spillway The emergency spillway must be designed for a 10 year ARI flood (see MSMA, Volume 15, Section 39.7.6c) The sill level is set at the basin top water level. Therefore, Qspillway = Q10 Parameter tc = 91 Area, A = 11.87 3.73 Fd = 0.88 1.44 -0.4 0.02 69.646 mm/hr</p> <p>CHECK CHECK CHECK CHECK COMPUTE</p> <p>For Penang, 10 year ARI A B C D From Equation 13,2 t=118 min, I118 =</p> <p>min ha</p> <p>From Chart 14.3, Runoff Coefficient, C = Equation 14.7, Q10 = Trial Dimension,</p> <p>0.52 1.19 B 3</p> <p>Category 4 m/s Hp 1.2 &gt; D.Chart 20.2 Csp 1.55 1.19 OK</p> <p>COMPUTE</p> <p>Qspillway =</p> <p>6.11</p> <p>COMPUTE</p> <p>Therefore, the total basin depth including the spillway is =</p> <p>3.85</p> <p>m</p> <p>Size of the spillway is 3m x 1.2m</p>