CHAPTER 3 SILTRAP AND TEMPORARY DRAINAGE
3.0
INTRODUCTION
The developer Kementerian Kewangan Malaysia is proposing to develop 3.2 hectar land on lot. The site location is attached on layout plan. The project implimentation will begin with site clearing and earthworks. The construction work will inevitably change the drainage pattern over the site dictated by the restructuring of earthworks to suits land use planning proposal. To cater for this development needs and erosion control control measures, the main and temporary drainage system will be establised concurrently including channelisation, silt trap and turfing before other infrastructures move in. 3.1
DESIGN RE REFERENCE
The design reference are base on the Planning and Design Procedure no 1, Standard and Procedu Procedures res for Peninsular Peninsular Malaysi Malaysia, a, Volumes Volumes 1 - 20
(Manual (Manual Saliran Saliran Mesra Mesra Alam Alam
Malaysia) Department of Irrigation and Drainage Malaysia, 2000. 3.2
DRAIN RAINAG AGE E OVE OVERA RALL LL DES DESIGN IGN SYS SYST TEM
The overall concept in the temporary drain and silt trap design is as follows:
i)
Prov Provid idin ing g site site per perim imet eter er tem tempo pora rary ry ear earth th dra drain in to to curb curb sil siltt wate waterr from from construction site and channel into the silt trap in order to prevent it from directly flowing into the nearby swapm or river.
ii) ii)
Silt Silt trap trap shal shalll be cons constr truc ucte ted d and and main maintai taine ned d to to effe effecti ctive vely ly tra trap p the the silt silt from the muddy water before the water flow into the drain.
iii) iii)
Fina Finall dis disch charg arges es all all the the filte filtere red d wate waterr (by (by silt silt trap trap)) from from this this development shall be into existing drain.
3.3 3.3
TEMP TE MPOR ORAR ARY Y EAR EARTH TH DRAI DRAIN N AND AND SILT SILT TRAP TRAP DESI DESIGN GN
3.3.7
Sediment Basin Design Report Area
:
125
acres
50.4
hectares
L = 850 m S = 0.5
STEP 1 : Determine overland flow time of concentration (minutes)
From equation 14.1,
t0
=
107 n L 1/3 / S1/2
=
107 (0.0275) (850) 1/3 / 0.51/2
=
39.42 minutes
Adopted Time of concentration = 40 minutes
STEP 2 : Sizing of Sediment Basin
From table 39.4, the predominant soil type is categorised as type C
From table 39.5, for a 3 month ARI, (Interpolation) Required Surface Area
= 172 m2/ha
Required Total Volume
= 206.67 m 3/ha
The Surface area required for this site
= Required Surface Area x Area = 172 x 50.4 = 8668.8 m 2
(Note : this is the average surface area for the settling zone volume, i.e. at middepth)
The total basin volume required for the site
= Required Total Volume x area = 206.67 x 50.4
= 10416.17 m 3 a) Settling zone
From table 39.5 The settling zone depth,
y1
= 0.6 m = 5208.09 m 3
The required settling zone, V1 Try a settling zone average width,
W1 = 73m
Required settling zone average length, L1 = V1/W1y1 = 5208.09 / (73 x 0.6) = 118.91 m, say 119m
Average surface area = 119 x 73 = 8687 m 2 > 8668.8 m 2
OK!
Check settling zone dimension (page 39-16) : (L1 / y1) ratio = 119 / 0.6 = 198.33
< 200 OK!
(L1 / W1) ratio = 119 / 73 = 1.63 < 2 (baffles should be provided to prevent short Circuiting)
b) Sediment storage zone
The required sediment storage zone volume,
V2 = 5208.09 m3
For a side slope, Z = 2.0 ( H ) : 1 ( V ), the dimension at the top of the sediment storage zone are, W2 = W1-2(y1/2)Z
L2 = L1-2(y1/2)Z
=
73 – 2 (0.6/2)2
=
71.8
=
119 – 2(0.6/2)2
=
117.8 m
m
The required depth for the sediment storage zone, which must be at least 0.3 m, can be calculated from the following relationship,
V2 =
Z2y23 - Zy22 (W2 + L2) + y2(W2L2)
Which gives, = 4y 23 – 379.2y 22 + 8458.04y 2
5208.09
Use trial and error to find y2 For y2 = 0.5 m , V2 = 4134.72 m 3 For y2 = 0.6 m ,V2 = 4939.18 m 3 For y2 = 0.65 m ,V2 = 5338.61 m 3 For y2 = 0.64 m ,V2 = 5258.87 m 3 *
y2
>
0.3 OK!
V2
>
5208.09 OK!
c) Overall Basin Dimensions At top water level: WTWL
= W1 + 2Z(y1/2) = 74.2
, say
= 74
LTWL
= L1 + 2Z(y1/2) = 120.2 , say = 120
m m
Base : WB
= W1 - 2Z[(y1/2) + y2]
= 69.24, say =69 m
LB
= L1 - 2Z[(y1/2) + y2]
=115.24, say =115 m
Depth : Settling zone, y1
= 0.6 m
Sediment storage, y2 = 0.64 m Side Slope, Z =2 (H): 1 (V)
STEP 3 : Sizing of Outlet Pipe
Outlet riser
= 900 mm diameter perforated MS pipe
The pipe is to be provided with sufficient small orifice openings to ensure that the basin will completely drain within 24 hours after filling. Average surface area, Aav = (WTWL x LTWL + WB x LB)/2 = 8407.5 m 2 Orifice area, total =
2Aav(y)1/2 / [t Cd(2g)1/2]
= 0.082 m2
Using an orifice size of 50 mm, the area of each orifice is A0 = 1.96 x 10 -3 m2 Total no of orifices required = 0.082 / 1.96 x 10 -3 = 42
At height of 200 mm, starting at the bottom of the pipe, put 2 rows of 7 x 50 mm orifices evenly spaced around the pipe. STEP 4 : Sizing of Emergency Spillway
The emergency spillway must be design for a 5 year ARI flood. The silt level must be set a minimum 300 mm above the basin top water level. The following assumptions are made in the calculations. * assume riser pipe flow is orifice flow through the top of pipe only. * riser pipe is 300 mm, the height between the top of the pipe and the spillway crest level. Qspillway
=
Q10 - Qriser
From equation 14.7, Q10 = C.10I10A / 360
The 10 year ARI rainfall intensity for Kuala Lumpur is derived from equation 13.3 and table 13.3 for a 40 minutes duration. 10I60 = 316.24 mm/hr, 10I30 = 291.02mm/hr and Fd = 0.376. Therefore,
40
I10
= (145.51 - 0.376(316.24 – 145.51)) / (45/60) = 108.42 mm/hr
From equation 14.7, with C = 0.915 ( Refer Design Chart 14.3, Category 1) Q10
= C.10I10A / 360 = 0.915(108.42) 50.4 / 360 = 13.89 m 3 / s
From equation 20.2 for orifice flow and assuming an orifice discharge coefficient of 0.6 , C0 A0 (2gH0)1/2 = 0.6 x ((π x 0.9 2)/4) x (2 x 9.81x0.3)1/2
Qriser =
= 1.05 m3/s
Therefore, allowing for the riser pipe flow required spillway capacity is Qspillway
=
Q10 - Qriser = 13.89 – 1.05 = 12.84 m 3/s
From equation 20.9, Q spillway = CspBH p1.5
Trial Dimensions : (from design chart 20.2) B
=
9
H p
=
0.6
Csp
=
1.70
Qspillway
= 7.111
Therefore, the total basin depth including the spillway is = y1 + y2 + HO + HP = 0.6 + 0.64 + 0.3 + 0.6 = 2.14 m
