Post on 04-Jul-2015
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DRY SEDIMENTATION POND
A. Determine overland flow time of concentrationCHECK Catchment Area = 5.5 ha
CHECK Overland slope = 0.8 %
CHECK Overland length = 190 m
CHECK Manning "n" = 0.015
COMPUTE Overland time, to = 9.63 min
COMPUTE Adopted time of concentration, tc = 10 min
B Sizing sendimentation
Table 39.4CHECK Soils type = C
Table 39.5 for a 3 month ARICHECK Required surface area = 333
CHECK Total Volume = 400
COMPUTE Site surface area required = 1,831.50
COMPUTE Site total vomue required = 2,200.00
i) Settling zone
COMPUTE 0.6 m
COMPUTE Required settling zone = 1,100.00
CHECK 17 m
COMPUTE 107.84 m
COMPUTE Average surface area = 1,833.33
Check settling zone dimensions
COMPUTE 179.74 < 200 OK
COMPUTE 6.34 > 2 OK
m²/ha
m³/ha
m²
m³
Settling zone depth, y1 =
m³
Assume average width, W1 =
Required settling zone a. length, L1 =
m²
L1 / y1 =
L1 / W1 =
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C Sediment Storage Zone
COMPUTE Required sediment storage volume = 1100
CHECK For a side slope Z = 2(H):1(V) 2
COMPUTE 15.8 m
COMPUTE 106.64 m
Required depth for the sediment storage
CHECK 0.6 923.682CHECK 1100 0.7 1060.85CHECK 0.73 1101.08 > 0.3 OK
D Overall Basin Dimensions
At top water level
COMPUTE 18.2 m say 18 m
COMPUTE 106.6 m say 107 m
Base
COMPUTE 12.9 m say 13 m
COMPUTE 103.7 m say 104 m
Depth:
COMPUTE 0.6 m
COMPUTE 0.73 m
m³
.·. W2 = W1 - 2 d1/2 x Z
.·. L2 = L1 - 2 d1/2 x Z
V2 = Z²(y2)³ - Z (y2)² (W2 + L2) + y2 ( W2 x L2 )
Trial and error to find y2 y2 V2
V2
WTWL = W1 + 2 x Z x y1/2
LTWL = L1 - 2 x Z x y1/2
WB = W1 - 2 x Z x (y1/2 + y2)
LB = L1 - 2 x Z x (y1/2 + y2)
Settling Zone, y1 =
Sediment storage zone, y2 =
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E Sizing of outlet pipe
COMPUTE Total depth, y 1.33 m Cd = 0.6
COMPUTE Average surface area, Aav = 1,638.43
Draining time, 24 hr after filling
COMPUTE 0.0165 Equation 19.5, Volume 7
CHECK Using orifice size of 50 mm
COMPUTE Area of each orifice is, Ao = 0.00196
Total number of orifice required,
COMPUTE 8 nos
At height of increments of 300mm,starting at the bottom of the pipes, put 2 ro 4 nos of 50 mmorifices evenly around the pipe
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.
Parametertc = 10 min
For Penang, 10 year ARI Area, A = 5.5 haA 3.7277B 1.4393 Fd from Table 13.3C -0.4023 Interpolate for 140mmD 0.0241 Fd = ###
From Equation 13,2136.648 mm/hr
68.32 mm
m²
Orifice area, ATOTAL
(2Aav x √y) /(tCd√2g)
m²
ATOTAL / Ao =
t=30 min, I30 =
P30 =
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92.833 mm/hr92.83 mm
255.83 mm/hr
From Chart 14.3, Runoff Coefficient, C = 0.843.28
IF Equation 20.2 for Orifice flowAn orifice discharge coefficient of = 0.64
Outlet riser, perforated MS pipe = 0.9 m0.3 m
0.99
2.29
D.Chart 20.2Trial Dimension, B
5 0.5 1.65
2.92 > 2.29 OK
Therefore, the total basin depth including the spillway is = 2.13 m
t=60 min, I60 =
P60 =
Rainfall Intensity, I10 = [P30 - Fd(P60 - P30)] / (10/60)
Equation 14.7, Q10 = m³/s
Ho =
Qriser = Cd Ao √ 2 g Ho = m³/s
Therefore Qspillway = Q10 - Qriser = m³/s
From Equation 20.9, Qspillway = Csp B Hp^1.5
Hp Csp
Qspillway =
Page 5 of 25
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
Interpolate for 140mm
document.xls
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WET SEDIMENTATION POND
ACHECK Catchment Area = 5.2 ha
75th % 5-day storm event 40 mm
B Sizing sendimentation
Table 39.4CHECK Soils type = F clay
Table 39.6 for a 3 month ARICHECK Settling Zone Volume 200
CHECK Total Volume = 300
COMPUTE Settling Zone Volume = 1,040.00
COMPUTE Total Volume 1560
i) Settling zone
CHECK 0.6 m
COMPUTE Required settling zone = 1560
CHECK 30 m
COMPUTE 87 m
COMPUTE Average surface area = 2,600.00 m²
Check settling zone dimensions
COMPUTE 144.44 < 200 OK
COMPUTE 2.89 > 2 OK
m²/ha
m³/ha
m²
m³
Settling zone depth, y1 =
m³
Try settling zone average width, W1 =
The settling zone average length, L1 =
L1 / y1 =
L1 / W1 =
document.xls
Page 7
C Sediment Storage Zone
COMPUTE Required sediment storage volume = 520
CHECK For a side slope Z = 2(H):1(V) 2Dimensions at the top of the sediment storage zone are,
COMPUTE 28.8 m
COMPUTE 85.47 m
Required depth for the sediment storage
CHECK 0.3 717.972 NOT OKCHECK 520 0.4 948.267 OKCHECK 0.35 833.68 > 0.3 OK
0.35 mD Overall Basin Dimensions
At top water level
COMPUTE 31.2 m say 31 m
COMPUTE 87.9 m say 88 m
Base
COMPUTE 27.4 m say 27 m
COMPUTE 84.1 m say 84 m
Depth:
COMPUTE 0.6 m
COMPUTE 0.35 m
m³
.·. W2 = W1 - 2 d1/2 x Z
.·. L2 = L1 - 2 d1/2 x Z
V2 = Z³(y2)³ - Z (y2)² (W2 + L2) + y2 ( W2 x L2 )
Trial and error to find y2 y2 V2
V2
Taken, y2 =
WTWL = W1 + 2 x Z x y1/2
LTWL = L1 + 2 x Z x y1/2
WB = W1 - 2 x Z x (y1/2 + y2)
LB = L1 - 2 x Z x (y1/2 + y2)
Settling Zone, y1 =
Sediment storage zone, y2 =
document.xls
Page 8
E Determine overland flow time of concentrationCHECK Catchment Area, A = 5.2 ha
n = 0.02
CHECK Overland slope, S = 0.03 %
CHECK Overland length, L = 870 m
COMPUTE tc = (107*n*L^1/3) / s^1/2 117.95 min
Assume velocity, v = 1 m/s
Drain length = 1275 m
td = velocity/drain length = 1.3072E-05 min
COMPUTE Adopted time of concentration, tc = 118 min
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.
Parametertc = 118 min
For Penang, 10 year ARI Area, A = 5.2 haCHECK A 3.7277 Fd = 0.884CHECK B 1.4393CHECK C -0.4023CHECK D 0.0241
COMPUTE From Equation 13,2136.648 mm/hr
68.32 mm
CHECK 92.833 mm/hrCOMPUTE 92.83 mm
Therefore, Qspillway = Q10
t=30 min, I30 =
P30 =
t=60 min, I60 =
P60 =
Rainfall Intensity, I12 = [P30 - Fd(P60 - P30)] / (13/60)
document.xls
Page 9
23.73 mm/hr
From Chart 14.3, Runoff Coefficient, C = 0.86 Category 4
0.29D.Chart 20.2
Trial Dimension, B3 0.35 1.55
COMPUTE 0.96 > 0.29 OK
Therefore, the total basin depth including COMPUTE the spillway is = 1.6 m
Equation 14.7, Q10 = m³/s
Hp Csp
Qspillway =
document.xls
Page 10
WET SEDIMENTATION POND
ACHECK Catchment Area = 23.7429 ha
75th % 5-day storm event 40 mm
B Sizing sendimentation
Table 39.4CHECK Soils type = F clay
Table 39.6 for a 3 month ARICHECK Settling Zone Volume 200 m²/ha
CHECK Total Volume = 300 m³/ha
COMPUTE Settling Zone Volume = 4,748.58 m²
COMPUTE Total Volume 7122.87 m³
i) Settling zone
CHECK 1.65 m
COMPUTE Required settling zone = 7122.87 m³
CHECK 30 m
COMPUTE 144 m
COMPUTE Average surface area = 4,316.89 m²
Check settling zone dimensions
COMPUTE 87.21 < 200 OK
COMPUTE 4.80 > 2 OK
Settling zone depth, y1 =
Try settling zone average width, W1 =
The settling zone average length, L1 =
L1 / y1 =
L1 / W1 =
C Sediment Storage Zone
COMPUTE Required sediment storage volume = 2374.29 m³
CHECK For a side slope Z = 2(H):1(V) 2Dimensions at the top of the sediment storage zone are,
COMPUTE 26.7 m
COMPUTE 140.60 m
Required depth for the sediment storage
CHECK 1.2 4029.81 OKCHECK 2374.29 0.6 2132.76 NOT OKCHECK 0.7 2465.17 > 0.3 OK
0.7 mD Overall Basin Dimensions
At top water level
COMPUTE 33.3 m say 33 m
COMPUTE 147.2 m say 147 m
Base
COMPUTE 23.9 m say 24 m
COMPUTE 137.8 m say 138 m
Depth:
COMPUTE 1.65 m
COMPUTE 0.7 m
Sedimentation storage required is (top)33mx(top)291m.
Sedimentation storage provided is (top)40mx(top)415m.
Provided storage is more than required storage
.·. W2 = W1 - 2 d1/2 x Z
.·. L2 = L1 - 2 d1/2 x Z
V2 = Z³(y2)³ - Z (y2)² (W2 + L2) + y2 ( W2 x L2 )
Trial and error to find y2 y2 V2
V2
Taken, y2 =
WTWL = W1 + 2 x Z x y1/2
LTWL = L1 + 2 x Z x y1/2
WB = W1 - 2 x Z x (y1/2 + y2)
LB = L1 - 2 x Z x (y1/2 + y2)
Settling Zone, y1 =
Sediment storage zone, y2 =
E Determine overland flow time of concentrationCHECK Catchment Area, A = 52 ha
n = 0.02
CHECK Overland slope, S = 0.05 %
CHECK Overland length, L = 870 m
COMPUTE tc = (107*n*L^1/3) / s^1/2 91.36 min
Assume velocity, v = 2 m/s
Drain length = 1275 m
td = velocity/drain length = 2.6144E-05 min
COMPUTE Adopted time of concentration, tc = 91 min
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.
Parametertc = 91 min
For Penang, 10 year ARI Area, A = 52 haCHECK A 3.7277 Fd = 0.884CHECK B 1.4393CHECK C -0.4023CHECK D 0.0241
COMPUTE From Equation 13,269.646 mm/hr
From Chart 14.3, Runoff Coefficient, C = 0.52 Category 4
5.23 m³/sD.Chart 20.2
Trial Dimension, B3 1.2 1.55
COMPUTE 6.11 > 5.23 OK
Therefore, Qspillway = Q10
t=118 min, I118 =
Equation 14.7, Q10 =
Hp Csp
Qspillway =
Therefore, the total basin depth including COMPUTE the spillway is = 3.85 m
Size of the spillway is 3m x 1.2m
WET SEDIMENTATION POND
ACHECK Catchment Area = 23.7429 ha
75th % 5-day storm event 40 mm
B Sizing sendimentation
Table 39.4CHECK Soils type = F clay
Table 39.6 for a 3 month ARICHECK Settling Zone Volume 200 m²/ha
CHECK Total Volume = 300 m³/ha
COMPUTE Settling Zone Volume = 4,748.58 m²
COMPUTE Total Volume 7122.87 m³
i) Settling zone
CHECK 1.65 m
COMPUTE Required settling zone = 7122.87 m³
CHECK 30 m
COMPUTE 144 m
COMPUTE Average surface area = 4,316.89 m²
Check settling zone dimensions
COMPUTE 87.21 < 200 OK
COMPUTE 4.80 > 2 OK
Settling zone depth, y1 =
Try settling zone average width, W1 =
The settling zone average length, L1 =
L1 / y1 =
L1 / W1 =
C Sediment Storage Zone
COMPUTE Required sediment storage volume = 2374.29 m³
CHECK For a side slope Z = 2(H):1(V) 2Dimensions at the top of the sediment storage zone are,
COMPUTE 26.7 m
COMPUTE 140.60 m
Required depth for the sediment storage
CHECK 1.2 4029.81 OKCHECK 2374.29 0.6 2132.76 NOT OKCHECK 0.7 2465.17 > 0.3 OK
0.7 mD Overall Basin Dimensions
At top water level
COMPUTE 33.3 m say 33
COMPUTE 147.2 m say 147
Base
COMPUTE 23.9 m say 24
COMPUTE 137.8 m say 138
Depth:
COMPUTE 1.65 m
COMPUTE 0.7 m
Sedimentation storage required is (top)33mx(top)291m.
Sedimentation storage provided is (top)40mx(top)415m.
Provided storage is more than required storage
E Determine overland flow time of concentrationCHECK Catchment Area, A = 23.7429 ha
.·. W2 = W1 - 2 d1/2 x Z
.·. L2 = L1 - 2 d1/2 x Z
V2 = Z³(y2)³ - Z (y2)² (W2 + L2) + y2 ( W2 x L2 )
Trial and error to find y2 y2 V2
V2
Taken, y2 =
WTWL = W1 + 2 x Z x y1/2
LTWL = L1 + 2 x Z x y1/2
WB = W1 - 2 x Z x (y1/2 + y2)
LB = L1 - 2 x Z x (y1/2 + y2)
Settling Zone, y1 =
Sediment storage zone, y2 =
n = 0.02
CHECK Overland slope, S = 0.05 %
CHECK Overland length, L = 870 m
COMPUTE tc = (107*n*L^1/3) / s^1/2 91.36 min
Assume velocity, v = 2 m/s
Drain length = 1275 m
td = velocity/drain length = 2.6144E-05 min
COMPUTE Adopted time of concentration, tc = 91 min
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.
Parametertc = 91 min
For Penang, 10 year ARI Area, A = 23.7429 haCHECK A 3.7277 Fd = 0.884CHECK B 1.4393CHECK C -0.4023CHECK D 0.0241
COMPUTE From Equation 13,269.646 mm/hr
From Chart 14.3, Runoff Coefficient, C = 0.52 Category 4
2.39 m³/sD.Chart 20.2
Trial Dimension, B3 1.2 1.55
COMPUTE 6.11 > 2.39 OK
Therefore, the total basin depth including COMPUTE the spillway is = 3.85 m
Size of the spillway is 3m x 1.2m
Therefore, Qspillway = Q10
t=91 min, I91 =
Equation 14.7, Q10 =
Hp Csp
Qspillway =
NOT OK
m
m
m
m
The emergency spillway must be designed for a 10 year ARI flood (see MSMA, Volume 15,
WET SEDIMENTATION POND
ACHECK Catchment Area = 11.8735 ha
75th % 5-day storm event 40 mm
B Sizing sendimentation
Table 39.4CHECK Soils type = F clay
Table 39.6 for a 3 month ARICHECK Settling Zone Volume 200 m²/ha
CHECK Total Volume = 300 m³/ha
COMPUTE Settling Zone Volume = 2,374.70 m²
COMPUTE Total Volume 3562.05 m³
i) Settling zone
CHECK 1.65 m
COMPUTE Required settling zone = 3562.05 m³
CHECK 30 m
COMPUTE 72 m
COMPUTE Average surface area = 2,158.82 m²
Check settling zone dimensions
COMPUTE 43.61 < 200 OK
COMPUTE 2.40 > 2 OK
Settling zone depth, y1 =
Try settling zone average width, W1 =
The settling zone average length, L1 =
L1 / y1 =
L1 / W1 =
C Sediment Storage Zone
COMPUTE Required sediment storage volume = 1187.35 m³
CHECK For a side slope Z = 2(H):1(V) 2Dimensions at the top of the sediment storage zone are,
COMPUTE 26.7 m
COMPUTE 68.66 m
Required depth for the sediment storage
CHECK 1.2 1932.16 OKCHECK 1187.35 0.6 1032.15 NOT OKCHECK 0.7 1191.19 > 0.3 OK
0.7 mD Overall Basin Dimensions
At top water level
COMPUTE 33.3 m say 33
COMPUTE 75.3 m say 75
Base
COMPUTE 23.9 m say 24
COMPUTE 65.9 m say 66
Depth:
COMPUTE 1.65 m
COMPUTE 0.7 m
Sedimentation storage required is (top)33mx(top)291m.
Sedimentation storage provided is (top)40mx(top)415m.
Provided storage is more than required storage
E Determine overland flow time of concentrationCHECK Catchment Area, A = 11.8735 ha
.·. W2 = W1 - 2 d1/2 x Z
.·. L2 = L1 - 2 d1/2 x Z
V2 = Z³(y2)³ - Z (y2)² (W2 + L2) + y2 ( W2 x L2 )
Trial and error to find y2 y2 V2
V2
Taken, y2 =
WTWL = W1 + 2 x Z x y1/2
LTWL = L1 + 2 x Z x y1/2
WB = W1 - 2 x Z x (y1/2 + y2)
LB = L1 - 2 x Z x (y1/2 + y2)
Settling Zone, y1 =
Sediment storage zone, y2 =
n = 0.02
CHECK Overland slope, S = 0.05 %
CHECK Overland length, L = 870 m
COMPUTE tc = (107*n*L^1/3) / s^1/2 91.36 min
Assume velocity, v = 2 m/s
Drain length = 1275 m
td = velocity/drain length = 2.6144E-05 min
COMPUTE Adopted time of concentration, tc = 91 min
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.
Parametertc = 91 min
For Penang, 10 year ARI Area, A = 11.8735 haCHECK A 3.7277 Fd = 0.884CHECK B 1.4393CHECK C -0.4023CHECK D 0.0241
COMPUTE From Equation 13,269.646 mm/hr
From Chart 14.3, Runoff Coefficient, C = 0.52 Category 4
1.19 m³/sD.Chart 20.2
Trial Dimension, B3 1.2 1.55
COMPUTE 6.11 > 1.19 OK
Therefore, the total basin depth including COMPUTE the spillway is = 3.85 m
Size of the spillway is 3m x 1.2m
Therefore, Qspillway = Q10
t=118 min, I118 =
Equation 14.7, Q10 =
Hp Csp
Qspillway =
NOT OK
m
m
m
m
The emergency spillway must be designed for a 10 year ARI flood (see MSMA, Volume 15,