BUILDING

Area = 221.57m2,Perimeter = 64 mFrom table 13.A1, minor system design ARI = 5 years (Terengganu, Kuala Dungun )

Table 13.A1 Coefficients for the IDF Equations for the Different Major Cities and Towns inMalaysia (30 t 1000 min)

Determine tc :

Overland flow =32 mCatchment area average slope = 0.5%From design chart 14.1 for bare soil surface, to = 13.5 ~ 8 minDrain length = 64mAssume, v = 1.0 m/s

td = L/V = 64/1.0 = 64 s = 1.07 min

Total tc = to + td

= 8 + 1.07 =9.07~ 10 min

Determine I and C :

t = 30

ln (5I30) = a + b (ln t) + c (ln t2) + d (ln t3)= 5.5077- 0.0310(ln 30) – 0.0899(ln 30)2 +0.0050(ln 30)3

= 4.559 mm/hr(5I30) =95.49 / 2 mm/hr P30 =47.75mm

t = 60

ln (5I60) = a + b (ln t) + c (ln t2) + d (ln t3)= 5.5077 – 0.0310 (ln 60) -0.0899(ln 60)2 +0.0050 (ln 60)3

= 4.217 (5I60) =4.217 mm/hrP60=67.83mm

Rainfall depth

Using Equation 13.3

Pd= P30 – FD ( P60 - P30 )= 47.75 mm –1.28(67.83mm - 47.75 mm) =22.05 mm

where,

Pd = design rainfall depth

P30, P60 = 30 and 60 minutes duration rainfall depths respectively

FD = adjustment factor for storm duration from Table 13.3

To determine FD :

1) Determination of tc byrefer to Design Chart 14.1

Length of overland flow (m) = 32 m

Average surface slopes = 0.5%

tc = 10 min

2) Value of FD for Equation 13.3, from Table 13.3 – East Coast

FD = 1.28

Intensity

Equation 13.4, I = Pd / d

=22.05mm/(10/60)

=132.3mm/hr

Determination of Qy

Qy = C y I EA

360

where,

Runoff coefficient, C = 0.83

Refer Design Chart 14.3, Category 3

Intensity, I =132.3 mm/hr ~ 133 mm/hr

Drainage area, A = 221.57m2= 0.0548ha

Peak Flow for 5 year ARI :

Qy = C y I EA

360

= (0.83 x 133 x 0.0548)

360

= 0.016 m

3

/s

Proposed Building Perimeter Drain Section

0.45 m

0.45 m

Check :

Area of build up drain section, A = 0.2025 m2

Wetted perimeter of drain section, P = 1.35m

R = A/P = 0.2025/1.35= 0.15

S = 1 : 300 = 0.003

Manning coefficient, n = 0.04

Therefore, Qcapacity= 1.49AR2/3 S1/2 / n

= 1.49 (0.2025)(0.15) 2/3 (0.003) 1/2 / 0.04

= 0.116m3/s

Qdischarge<Qcapacity

0.016 m3/s <0.116 m3/s OK!

BOUNDRY

Area = 32000m2, Perimeter = 800mFrom table 13.A1, minor system design ARI = 5 years (Johor Bahru, Johor )

Determine tc :

Overland flow =400 mCatchment area average slope = 0.5%From design chart 14.1 for bare soil surface, to = 8 minDrain length = 800m

Assume, v = 1.0 m/std = L/V = 800m /1.0 = 800s = 13.3 min ~ 14min

Total tc = to + td

= 8 + 14 =22 min

Determine I and C :

t = 30

ln (5I30) = a + b (ln t) + c (ln t2) + d (ln t3)= 5.5077- 0.0310(ln 30) - 0.0899(ln 30)2 +0.0050(ln 30)3

= 4.559 mm/hr(5I30) = 95.49 / 2 mm/hr P30 = 47.74 mm

t = 60

ln (5I60) = a + b (ln t) + c (ln t2) + d (ln t3)= 5.5077 – 0.0310(ln 60) – 0.0899(ln 60)2 +0.0050(ln 60)3

= 4.217 (5I60) = 67.83mm/hrP60=67.83mm

Rainfall depth

Using Equation 13.3

Pd = P30 – FD ( P60 - P30 )

= 47.74 mm - 0.376(67.83mm-47.74 mm)

=40.19mm

where,

Pd = design rainfall depth

P30, P60 = 30 and 60 minutes duration rainfall depths respectively

FD = adjustment factor for storm duration from Table 13.3

To determine FD :

Determination of tc byrefer to Design Chart 14.1

Length of overland flow (m) = 400 m

Average surface slopes = 0.5%

tc = 22 min

Value of FD for Equation 13.3, from Table 13.3 – East Coast

FD = 0.376

Intensity

Equation 13.4, I22 = Pd / d

= 40.19/(22/60)

= 109.61 mm/hr

Determination of Qy

Qy = C y I EA

360

where,

Runoff coefficient, C = 0.8

Refer Design Chart 14.3, Category 3

Intensity, I = 40.19mm/hr ~ 41mm/hr

Drainage area, A = 32000m2 = 7.907 ha

Peak Flow for 2 year ARI :

Qy = C y I EA

360

= (0.80 x 41 x 7.90)

360

= 0.72 m

3

/s

Proposed Remaining Road Drain Section

Type of channel = bare soil surfaceManning coefficient, n = 0.04Slope gradient, So = 1:300 = 0.003Proposed U drain = 1000 x 1000 mm

1m

1m

where,

Cross sectional area channel, A = 1 m × 1 m = 1 m2

Hydraulic Radius, R = A/P = (1/3) = 0.33 m

Slope of drain, S = 1:300

Manning’s roughness coefficient, n = 0.04

Flow :

Therefore, Qcapacity = 1.49AR2/3 S1/2 / n

= 1.49 (1.0)(0.33) 2/3 (0.003) 1/2 / 0.04

= 0.974m3/s

Qdischarge<Qcapacity

0.72 m3/s <0.974 m3/s OK!

As a conclusion, this drain is adequate to cater for the overall requirement

Q = 1.49 A R2/3S1/2 / n

Proposed Size of Drain Sump

Drain sump basically acting as a connection for all the drainage. Sump is defined as a drain consisting of an outer tube which is vented to the outside with a smaller tube within it that is attached to a suction pump. Both tube have multiple perforations that allows fluid and air to be carried away through the suction tube.

Refer to road and drainage plan for the proposed location for all the drain sump.

Maximum flow from the drainage, Q = 0.639m3/s

The velocity of water flow in drain, v = Q/A = 0.639 / 0.25

= 2.56 m/s

Calculation size of sump

The width of drainage size = 0.50 m

The length => 0.25 m2 / 0.5 m = 0.50 m

The minimum size of sump:

Length = 0.50 m

Width = 0.50 m

Depth = 0.50 m

Thus, the minimum size of sump is 500mm×500mm

Therefore, the propose size of sump is 1100mm×1100mm according to standard size.

Refers to road and drainage plan for the detailing sump.