DRY SEDIMENTATION POND A. Determine overland flow time of concentration 5.5 ha CHECK Catchment Area = CHECK Overland
slope =
0.8 %
CHECK Overland
length =
19 0 m
"n" =
0.02
CHECK Manning
COMPUTE Overland COMPUTE Adopted
time, to =
9.63 9.63 min min
time of concentration, tc =
10 min min
B Sizing sendimentatio sendimentation n Table 39.4 CHECK Soils type =
C
Table 39.5 for a 3 month ARI CHECK Required surface area = CHECK
333 m²/ha m³/ha
Total Volume =
COMPUTE Site
surface area required =
1,831.50 m
²
COMPUTE Site
total vomue required =
2,200.00 m
³
400
i) Settling Settling zone zone COMPUTE Settling
zone depth, y1 =
COMPUTE Required
CHECK Assume
0.6 0.6 m
settling zone =
1,100.00 m
average width, W 1 =
COMPUTE Required COMPUTE Average
17 m
settling zone a. length, L 1 =
surface area =
³
107.84 107.84 m 1,833.33 m
²
Check settling zone dimensions COMPUTE L1 /
y1 =
COMPUTE L1 /
W1 =
179.74 6.34
Page 1 of 40
< 2 00
OK
>2
OK
C Sediment Sediment Storage Zone Zone COMPUTE Required
CHECK For
sediment storage volume =
1100 m
a side slope Z = 2(H):1(V)
COMPUTE .
·. W2 = W1 - 2 d1/2 x Z
COMPUTE .
·. L2 = L1 - 2 d1/2 x Z
³
2 15.8 15.8 m 106.64 106.64 m
Required depth for the sediment storage V2 = Z²(y2)³ - Z (y2)² (W2 + L2) + y2 ( W2 x L2 ) Trial and error to find y 2 V2 CHECK 1100 CHECK
y2
CHECK
V2 0.6 923.68 0.7 1060.85 0.73 1101.08
> 0.3
OK
D Overall Overall Basin Dimensions Dimensions At top water level COMPUTE W TWL COMPUTE LTWL
= W1 + 2 x Z x y 1/2
= L1 - 2 x Z x y 1/2
18.2 18.2 m
say
18 m
106.6 106.6 m
say
107 107 m
12.9 12.9 m
say
13 m
103.7 103.7 m
say
104 104 m
Base COMPUTE WB COMPUTE LB
= W1 - 2 x Z x (y 1/2 + y2)
= L1 - 2 x Z x (y1/2 + y2)
Depth: COMPUTE Settling Zone, y 1 = COMPUTE Sediment
storage zone, y 2 =
Page 2 of 40
0.6
m
0.73
m
E Sizing of of outlet pipe COMPUTE Total
depth, y
COMPUTE Average
surface area, Aav =
1,638.43 m
Draining time,
COMPUTE Area
²
24 hr after filling
Orifice area, A TOTAL /(tCd √2 √2g) COMPUTE (2Aav x √y) /(tC CHECK Using
Cd = 0.6
1.33 1.33 m
0.0165 Equation 19.5, Volume 7
orifice size of
50 mm
of each orifice is, Ao =
0.00196 m
Total number of orifice required, COMPUTE A TOTAL / Ao =
At height of increments of 300mm, starting at the bottom of the pipes, put 2 ro orifices evenly around the pipe
²
8
nos
4
nos
of
50 mm
F Sizing of emergency emergency spillway 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 68.32 mm
Page 3 of 40
t=60 min, I60 = P60 =
92.833 mm/hr 92.83 92.83 mm
Rainfall Intensity, I 10 = [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 m³/s
IF Equation 20.2 for Orifice Orifice flow An orifice discharge coefficient of = Outlet riser, perforated MS pipe = Ho =
0.64 0.9 m 0.3 0.3 m
Qriser = Cd Ao √ 2 g Ho =
0.99 m³/s
Therefore Q spillway = Q10 - Qriser =
2.29 m³/s
From Equation 20.9, Q spillway = Csp B Hp^1.5
Trial Dimension,
Hp 0.5
5
Qspillway = Therefore, the total basin depth including the spillway is =
Page 4 of 40
2.92
2.13 2.13 m
>
D.Chart 20.2 B Csp 1 .65
2.29 OK
Page 5 of 40
Page 6 of 40
,
Page 7 of 40
Page 8 of 40
58925981.xls
WET SEDIMENTATION POND A CHECK Catchment
Area =
5.2 ha
75th % 5-day storm event
40 mm
B Sizing sendimentation Table 39.4 CHECK Soils type =
F
Table 39.6 for a 3 month ARI CHECK Settling Zone Volume CHECK
clay
200 m²/ha m³/ha 300
Total Volume =
COMPUTE Settling COMPUTE Total
Zone Volume =
1,040.00 m
²
Volume
1560 m
³
i) Settling zone CHECK Settling
zone depth, y1 =
COMPUTE Required
CHECK Try COMPUTE The
0.6 m
settling zone =
1560 m
³
settling zone average width, W 1 =
30 m
settling zone average length, L 1 =
87 m
COMPUTE Average
surface area =
2,600.00 m²
Check settling zone dimensions COMPUTE L1 /
y1 =
COMPUTE L1 /
W1 =
144.44 2.89
Page 9
< 200
OK
>2
OK
58925981.xls
C Sediment Storage Zone COMPUTE Required
sediment storage volume =
520 m
³
CHECK For
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 Z COMPUTE .
·. L2 = L1 - 2 d1/2 x Z
85.47 m
Required depth for the sediment storage V2 = Z³(y2)³ - Z (y2)² (W2 + L2) + y2 ( W2 x L2 ) Trial and error to find y 2 V2 CHECK 520 CHECK
y2
V2 717.97 948.27 833.68
0.3 0.4 0.35
CHECK
0.35
Taken, y 2 =
> 0.3
NOT OK OK OK
m
D Overall Basin Dimensions At top water level COMPUTE W TWL COMPUTE LTWL
= W1 + 2 x Z x y 1/2
= L1 + 2 x Z x y 1/2
31.2 m
say
31 m
87.9 m
say
88 m
27.4 m
say
27 m
84.1 m
say
84 m
Base COMPUTE WB COMPUTE LB
= W1 - 2 x Z x (y 1/2 + y2)
= L1 - 2 x Z x (y1/2 + y2)
Depth: COMPUTE Settling Zone, y1 = COMPUTE Sediment
storage zone, y 2 =
0.6
m
0.35
m
Page 10
58925981.xls
E Determine overland flow time of concentration Catchment Area, A = CHECK n=
5.2 ha 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 = Drain length =
1 m/s 1275 m
td = velocity/drain length = COMPUTE Adopted
time of concentration, tc =
0 min 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. Therefore, Q spillway = Q10
CHECK CHECK CHECK CHECK COMPUTE
CHECK COMPUTE
Parameter tc = Area, A = 3.73 Fd = 1.44 -0.4 0.02
For Penang, 10 year ARI A B C D From Equation 13,2 t=30 min, I30 = P30 =
136.648 mm/hr 68.32 mm
t=60 min, I60 = P60 =
92.833 mm/hr 92.83 mm
Page 11
118 5.2 0.88
min ha
58925981.xls
Rainfall Intensity, I 12 = [P30 - Fd(P60 - P30)] / (13/60) 23.73 mm/hr From Chart 14.3, Runoff Coefficient, C =
0.86
Category 4
Equation 14.7, Q10 =
0.29
m /s
³
D.Chart 20.2 Hp B Csp 0.35 1.55
Trial Dimension, 3
COMPUTE
Qspillway =
>
0.96
Therefore, the total basin depth including the spillway is = COMPUTE
Page 12
1.6
m
0.29
OK
58925981.xls
Page 13
58925981.xls
Page 14
58925981.xls
Page 15
58925981.xls
Page 16
WET SEDIMENTATION POND A CHECK Catchment
Area =
23.74 ha
75th % 5-day storm event
40 mm
B Sizing sendimentation Table 39.4 CHECK Soils type = Table 39.6 for a 3 month ARI CHECK Settling Zone Volume CHECK
F
clay
200 m²/ha
Total Volume =
COMPUTE Settling COMPUTE Total
Zone Volume =
Volume
300 m³/ha 4,748.58 m² 7122.87 m³
i) Settling zone CHECK Settling
zone depth, y1 =
COMPUTE Required
Try CHECK COMPUTE The
settling zone =
1.65 m 7122.87 m³
settling zone average width, W 1 =
30 m
settling zone average length, L 1 =
144 m
COMPUTE Average
surface area =
4,316.89 m²
Check settling zone dimensions COMPUTE L1 /
y1 =
COMPUTE L1 /
W1 =
87.21 4.80
< 200
OK
>2
OK
C Sediment Storage Zone COMPUTE Required
sediment storage volume =
2374.29 m³
CHECK For
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 Z COMPUTE .·.
L2 = L1 - 2 d1/2 x Z
140.60 m
Required depth for the sediment storage V2 = Z³(y2)³ - Z (y2)² (W2 + L2) + y2 ( W2 x L2 ) Trial and error to find y 2 V2 CHECK 2374.29 CHECK
y2
V2 1.2 4029.81 0.6 2132.76 0.7 2465.17
CHECK
Taken, y 2 =
0.7
> 0.3
OK NOT OK OK
m
D Overall Basin Dimensions At top water level COMPUTE W TWL COMPUTE LTWL
= W1 + 2 x Z x y 1/2
= L1 + 2 x Z x y 1/2
33.3 m
say
33 m
147.2 m
say
147 m
23.9 m
say
24 m
137.8 m
say
138 m
Base COMPUTE WB COMPUTE LB
= W1 - 2 x Z x (y 1/2 + y2)
= L1 - 2 x Z x (y1/2 + y2)
Depth: COMPUTE Settling
Zone, y1 =
COMPUTE Sediment
storage zone, y 2 =
1.65
m
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 concentration Catchment Area, A = CHECK
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 = COMPUTE Adopted
time of concentration, tc =
0 min 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. Therefore, Q spillway = Q10
CHECK CHECK CHECK CHECK
COMPUTE
For Penang, 10 year ARI A B C D From Equation 13,2 t=118 min, I118 =
Parameter tc = Area, A = 3.73 Fd = 1.44 -0.4 0.02
0.52
Category 4
Equation 14.7, Q10 =
5.23
m³/s
Trial Dimension, 3
COMPUTE
Qspillway =
min ha
69.646 mm/hr
From Chart 14.3, Runoff Coefficient, C =
91 52 0.88
6.11
D.Chart 20.2 Hp B Csp 1.2 1.55 >
5.23
OK
Therefore, the total basin depth including the spillway is = COMPUTE
Size of the spillway is 3m x 1.2m
3.85
m
WET SEDIMENTATION POND A CHECK Catchment
Area =
23.74 ha
75th % 5-day storm event
40 mm
B Sizing sendimentation Table 39.4 CHECK Soils type = Table 39.6 for a 3 month ARI CHECK Settling Zone Volume CHECK
F
clay
200 m²/ha
Total Volume =
COMPUTE Settling COMPUTE Total
Zone Volume =
Volume
300 m³/ha 4,748.58 m² 7122.87 m³
i) Settling zone CHECK Settling
zone depth, y1 =
COMPUTE Required
Try CHECK COMPUTE The
settling zone =
1.65 m 7122.87 m³
settling zone average width, W 1 =
30 m
settling zone average length, L 1 =
144 m
COMPUTE Average
surface area =
4,316.89 m²
Check settling zone dimensions COMPUTE L1 /
y1 =
COMPUTE L1 /
W1 =
87.21 4.80
< 200
OK
>2
OK
C Sediment Storage Zone COMPUTE Required
sediment storage volume =
2374.29 m³
CHECK For
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 Z COMPUTE .·.
L2 = L1 - 2 d1/2 x Z
140.60 m
Required depth for the sediment storage V2 = Z³(y2)³ - Z (y2)² (W2 + L2) + y2 ( W2 x L2 ) Trial and error to find y 2 V2 CHECK 2374.29 CHECK
y2
V2 1.2 4029.81 0.6 2132.76 0.7 2465.17
CHECK
Taken, y 2 =
0.7
> 0.3
OK NOT OK OK
m
D Overall Basin Dimensions At top water level COMPUTE W TWL COMPUTE LTWL
= W1 + 2 x Z x y 1/2
= L1 + 2 x Z x y 1/2
33.3 m
say
33 m
147.2 m
say
147 m
23.9 m
say
24 m
137.8 m
say
138 m
Base COMPUTE WB COMPUTE LB
= W1 - 2 x Z x (y 1/2 + y2)
= L1 - 2 x Z x (y1/2 + y2)
Depth: COMPUTE Settling
Zone, y1 =
COMPUTE Sediment
storage zone, y 2 =
1.65
m
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 concentration Catchment Area, A = 23.74 ha CHECK 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 = COMPUTE Adopted
time of concentration, tc =
0 min 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. Therefore, Q spillway = Q10
CHECK CHECK CHECK CHECK
COMPUTE
For Penang, 10 year ARI A B C D From Equation 13,2 t=91 min, I91 =
Parameter tc = 91 Area, A = 23.74 3.73 Fd = 0.88 1.44 -0.4 0.02 69.646 mm/hr
From Chart 14.3, Runoff Coefficient, C =
0.52
Category 4
Equation 14.7, Q10 =
2.39
m³/s
Trial Dimension, 3
COMPUTE
Qspillway =
min ha
6.11
D.Chart 20.2 Hp B Csp 1.2 1.55 >
2.39
OK
Therefore, the total basin depth including the spillway is = COMPUTE
Size of the spillway is 3m x 1.2m
3.85
m
WET SEDIMENTATION POND A CHECK Catchment
Area =
11.87 ha
75th % 5-day storm event
40 mm
B Sizing sendimentation Table 39.4 CHECK Soils type = Table 39.6 for a 3 month ARI CHECK Settling Zone Volume CHECK
F
clay
200 m²/ha
Total Volume =
COMPUTE Settling COMPUTE Total
Zone Volume =
Volume
300 m³/ha 2,374.70 m² 3562.05 m³
i) Settling zone CHECK Settling
zone depth, y1 =
COMPUTE Required
Try CHECK COMPUTE The
settling zone =
1.65 m 3562.05 m³
settling zone average width, W 1 =
30 m
settling zone average length, L 1 =
72 m
COMPUTE Average
surface area =
2,158.82 m²
Check settling zone dimensions COMPUTE L1 /
y1 =
COMPUTE L1 /
W1 =
43.61 2.40
< 200
OK
>2
OK
C Sediment Storage Zone COMPUTE Required
sediment storage volume =
1187.35 m³
CHECK For
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 Z COMPUTE .·.
L2 = L1 - 2 d1/2 x Z
68.66 m
Required depth for the sediment storage V2 = Z³(y2)³ - Z (y2)² (W2 + L2) + y2 ( W2 x L2 ) Trial and error to find y 2 V2 CHECK 1187.35 CHECK
y2
V2 1.2 1932.16 0.6 1032.15 0.7 1191.19
CHECK
Taken, y 2 =
0.7
> 0.3
OK NOT OK OK
m
D Overall Basin Dimensions At top water level COMPUTE W TWL COMPUTE LTWL
= W1 + 2 x Z x y 1/2
= L1 + 2 x Z x y 1/2
33.3 m
say
33 m
75.3 m
say
75 m
23.9 m
say
24 m
65.9 m
say
66 m
Base COMPUTE WB COMPUTE LB
= W1 - 2 x Z x (y 1/2 + y2)
= L1 - 2 x Z x (y1/2 + y2)
Depth: COMPUTE Settling
Zone, y1 =
COMPUTE Sediment
storage zone, y 2 =
1.65
m
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 concentration Catchment Area, A = 11.87 ha CHECK 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 = COMPUTE Adopted
time of concentration, tc =
0 min 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. Therefore, Q spillway = Q10
CHECK CHECK CHECK CHECK
COMPUTE
For Penang, 10 year ARI A B C D From Equation 13,2 t=118 min, I118 =
Parameter tc = 91 Area, A = 11.87 3.73 Fd = 0.88 1.44 -0.4 0.02 69.646 mm/hr
From Chart 14.3, Runoff Coefficient, C =
0.52
Category 4
Equation 14.7, Q10 =
1.19
m³/s
Trial Dimension, 3
COMPUTE
Qspillway =
min ha
6.11
D.Chart 20.2 Hp B Csp 1.2 1.55 >
1.19
OK
Therefore, the total basin depth including the spillway is = COMPUTE
Size of the spillway is 3m x 1.2m
3.85
m
