# detention pond calculationphaseii

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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

slope = length = "n" = time, to =

0.8 % 190 m 0.02 9.63 min 10 min

time of concentration, tc =

B Sizing sendimentation Table 39.4 type = Table 39.5 for a 3 month ARI surface area = Volume =

CHECK Soils

C

CHECK Required CHECK Total COMPUTE Site COMPUTE Site

333 m/ha 400 m/ha 1,831.50 m 2,200.00 m

surface area required = total vomue required =

i) Settling zoneCOMPUTE Settling

zone depth, y1 = settling zone =

0.6 m 1,100.00 m 17 m 107.84 m 1,833.33 m

COMPUTE Required CHECK Assume

average width, W1 = settling zone a. length, L1 =

COMPUTE Required COMPUTE Average

surface area =

Check settling zone dimensionsCOMPUTE L1 / COMPUTE L1 /

y1 = W1 =

179.74 6.34

< 200 >2

OK OK

Page 1 of 40

C Sediment Storage ZoneCOMPUTE Required CHECK For COMPUTE .. COMPUTE ..

sediment storage volume =

1100 m 2 15.8 m 106.64 m

a side slope Z = 2(H):1(V)

W2 = W1 - 2 d1/2 x Z L2 = L1 - 2 d1/2 x Z

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

y2

V2 0.6 923.68 0.7 1060.85 0.73 1101.08

> 0.3

OK

D Overall Basin Dimensions At top water levelCOMPUTE WTWL COMPUTE LTWL

= W1 + 2 x Z x y1/2

18.2 m 106.6 m

say say

18 m 107 m

= L1 - 2 x Z x y1/2

BaseCOMPUTE WB COMPUTE LB

= W1 - 2 x Z x (y1/2 + y2)

12.9 m 103.7 m

say say

13 m 104 m

= L1 - 2 x Z x (y1/2 + y2)

Depth:COMPUTE Settling

Zone, y1 = storage zone, y2 =

0.6 0.73

m m

COMPUTE Sediment

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E Sizing of outlet pipeCOMPUTE Total

depth, y surface area, Aav =

1.33 m 1,638.43 m 24 hr after filling

Cd = 0.6

COMPUTE Average

Draining time, Orifice area, ATOTAL x y) /(tCd2g) orifice size of

COMPUTE (2Aav CHECK Using COMPUTE Area

0.0165 Equation 19.5, Volume 7 50 mm 0.00196 m

of each orifice is, Ao =

Total number of orifice required, COMPUTE ATOTAL / Ao =

8

nos

At height of increments of 300mm, starting at the bottom of the pipes, put 2 rows orifices evenly around the pipe

4

nos

of

50 mm

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

Page 3 of 40

t=60 min, I60 = P60 =

92.833 mm/hr 92.83 mm

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

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

0.64 0.9 m 0.3 m 0.99 m/s 2.29 m/s

Hp 0.5

D.Chart 20.2 Csp 1.65

Qspillway = Therefore, the total basin depth including the spillway is =

2.92

>

2.29 OK

2.13 m

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Volume 15, n top water

for 140mm

Page 7 of 40

Page 8 of 40

58925981.xls

WET SEDIMENTATION PONDACHECK Catchment

Area =

5.2 ha 40 mm

75th % 5-day storm event

B Sizing sendimentation Table 39.4 type = Table 39.6 for a 3 month ARI Zone Volume Volume = Zone Volume =

CHECK Soils

F

clay

CHECK Settling CHECK Total

200 m/ha 300 m/ha 1,040.00 m 1560 m

COMPUTE Settling COMPUTE Total

Volume

i) Settling zoneCHECK Settling

zone depth, y1 = settling zone =

0.6 m 1560 m 30 m 87 m 2,600.00 m

COMPUTE Required CHECK Try COMPUTE The

settling zone average width, W1 = settling zone average length, L1 = surface area =

COMPUTE Average

Check settling zone dimensionsCOMPUTE L1 / COMPUTE L1 /

y1 = W1 =

144.44 2.89

< 200 >2

OK OK

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58925981.xls

C Sediment Storage ZoneCOMPUTE Required CHECK For

sediment storage volume =

520 m

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 ..

L2 = L1 - 2 d1/2 x Z

85.47 m

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

y2 0.3 0.4 0.35 Taken, y2 =

V2 717.97 948.27 833.68 0.35

> 0.3 m

NOT OK OK OK

D Overall Basin Dimensions At top water levelCOMPUTE WTWL COMPUTE LTWL

= W1 + 2 x Z x y1/2

31.2 m 87.9 m

say say

31 m 88 m

= L1 + 2 x Z x y1/2

BaseCOMPUTE WB COMPUTE LB

= W1 - 2 x Z x (y1/2 + y2)

27.4 m 84.1 m

say say

27 m 84 m

= L1 - 2 x Z x (y1/2 + y2)

Depth: COMPUTE Settling Zone, y1 =COMPUTE Sediment

0.6 0.35

m m

storage zone, y2 =

Page 10

58925981.xls

CHECK

E Determine overland flow time of concentration Catchment Area, A = n=

5.2 ha 0.02 0.03 % 870 m

CHECK

Overland slope, S = Overland length, L = = (107*n*L^1/3) / s^1/2 Assume velocity, v = Drain length = td = velocity/drain length =

CHECK COMPUTE tc

117.95 min 1 m/s 1275 m 0 min 118 min

COMPUTE Adopted

time of concentration, tc =

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

CHECK CHECK CHECK CHECK COMPUTE

For Penang, 10 year ARI A B C D From Equation 13,2 t=30 min, I30 = P30 = t=60 min, I60 = P60 =

118 5.2 0.88

min ha

CHECK COMPUTE

Page 11

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 > D.Chart 20.2 Csp 1.55 0.29 OK

COMPUTE

Therefore, the total basin depth including the spillway is = COMPUTE

1.6

m

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58925981.xls

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58925981.xls

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WET SEDIMENTATION PONDACHECK Catchment

Area =

23.74 ha 40 mm

75th % 5-day storm event

B Sizing sendimentation Table 39.4 type =

CHECK Soils

F

clay

Table 39.6 for a 3 month ARI CHECK Settling Zone VolumeCHECK Total

200 m/ha 300 m/ha 4,748.58 m 7122.87 m

Volume = Zone Volume =

COMPUTE Settling COMPUTE Total

Volume

i) Settling zoneCHECK Settling

zone depth, y1 = settling zone =

1.65 m 7122.87 m 30 m 144 m 4,316.89 m

COMPUTE Required CHECK Try COMPUTE The

settling zone average width, W1 = settling zone average length, L1 = surface area =

COMPUTE Average

Check settling zone dimensionsCOMPUTE L1 / COMPUTE L1 /

y1 = W1 =

87.21 4.80

< 200 >2

OK OK

C Sediment Storage ZoneCOMPUTE Required CHECK For

sediment storage volume =

2374.29 m

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 ..

L2 = L1 - 2 d1/2 x Z

140.60 m

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

y2

V2 1.2 4029.81 0.6 2132.76 0.7 2465.17 0.7

> 0.3 m

OK NOT OK OK

Taken, y2 = D Overall Basin Dimensions At top water levelCOMPUTE WTWL COMPUTE LTWL

= W1 + 2 x Z x y1/2

33.3 m 147.2 m

say say

33 m 147 m

= L1 + 2 x Z x y1/2

BaseCOMPUTE WB COMPUTE LB

= W1 - 2 x Z x (y1/2 + y2)

23.9 m 137.8 m

say say

24 m 138 m

= L1 - 2 x Z x (y1/2 + y2)

Depth:COMPUTE Settling

Zone, y1 =

1.65

m

COMPUTE Sediment

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

E Determine overland flow time of concentration Catchment Area, A = CHECK n=CHECK CHECK COMPUTE tc

52 ha 0.02 0.05 % 870 m 91.36 min 2 m/s 1275 m 0 min

Overland slope, S = Overland length, L = = (107*n*L^1/3) / s^1/2 Assume velocity, v = Drain length = td = velocity/drain length =

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. Therefore, Qs

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