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 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 REFERENCE

The design reference are base on the Planning and Design Procedure no 1, Standard

and Procedures for Peninsular Malaysia, Volumes 1 - 20 (Manual Saliran Mesra Alam

Malaysia) Department of Irrigation and Drainage Malaysia, 2000.

3.2 DRAINAGE OVERALL DESIGN SYSTEM

The overall concept in the temporary drain and silt trap design is as follows:

i) Providing site perimeter temporary earth drain to curb silt water from

construction site and channel into the silt trap in order to prevent it from

directly flowing into the nearby swapm or river.

ii) Silt trap shall be constructed and maintained to effectively trap the silt

from the muddy water before the water flow into the drain.

iii) Final discharges all the filtered water (by silt trap) from this

development shall be into existing drain.

3.3 TEMPORARY EARTH DRAIN AND SILT TRAP DESIGN

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 L1/3 / 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 m3/ha

The Surface area required for this site = Required Surface Area x Area

= 172 x 50.4

= 8668.8 m2

(Note : this is the average surface area for the settling zone volume, i.e. at mid-

depth)

The total basin volume required for the site = Required Total Volume x area

= 206.67 x 50.4

= 10416.17 m3

a) Settling zone

From table 39.5

The settling zone depth, y1 = 0.6 m

The required settling zone, V1 = 5208.09 m3

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 m2 > 8668.8 m2 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 = 73 – 2 (0.6/2)2

= 71.8 m

L2 = L1-2(y1/2)Z = 119 – 2(0.6/2)2

= 117.8 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 - Zy2

2 (W2 + L2) + y2(W2L2)

Which gives,

5208.09 = 4y23 – 379.2y2

2 + 8458.04y2

Use trial and error to find y2

For y2 = 0.5 m , V2 = 4134.72 m3

For y2 = 0.6 m ,V2 = 4939.18 m3

For y2 = 0.65 m ,V2 = 5338.61 m3

For y2 = 0.64 m ,V2 = 5258.87 m3

* 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 m

LTWL = L1 + 2Z(y1/2) = 120.2 , say = 120 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 m2

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,

40I10 = (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 m3 / s

From equation 20.2 for orifice flow and assuming an orifice discharge coefficient of

0.6 ,

Qriser = C0 A0 (2gH0)1/2 = 0.6 x ((π x 0.92)/4) x (2 x 9.81x0.3)1/2

= 1.05 m3/s

Therefore, allowing for the riser pipe flow required spillway capacity is

Qspillway = Q10 - Qriser = 13.89 – 1.05

= 12.84 m3/s

From equation 20.9, Qspillway = CspBHp1.5

Trial Dimensions : (from design chart 20.2)

B = 9

Hp = 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