BUILDING Area = 221.57m2,Perimeter = 64 m From 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 in Malaysia (30 t 1000 min) Determine tc : Overland flow =32 m Catchment area average slope = 0.5% From design chart 14.1 for bare soil surface, to = 13.5 ~ 8 min Drain length = 64m Assume, 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 5 ( I30) =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 5 ( I60) =4.217 mm/hr P60=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
tc = 10 min
= 0.5%
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 = CyIEA 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 = CyIEA 360 =
(0.83 x 133 x 0.0548) 360 3
=
0.016 m /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
<0.116 m3/s
OK!
BOUNDRY Area = 32000m2, Perimeter = 800m From table 13.A1, minor system design ARI = 5 years (Johor Bahru, Johor )
Determine tc : Overland flow =400 m Catchment area average slope = 0.5% From design chart 14.1 for bare soil surface, to = 8 min Drain length = 800m Assume, v = 1.0 m/s td = 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 5 ( I30) = 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 5 ( I60) = 67.83mm/hr P60=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
tc = 22 min
= 0.5%
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 = CyIEA 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 = CyIEA 360 =
(0.80 x 41 x 7.90) 360 3
=
0.72 m /s
Proposed Remaining Road Drain Section Type of channel Manning coefficient, n Slope gradient, So Proposed U drain
= bare soil surface = 0.04 = 1:300 = 0.003 = 1000 x 1000 mm
1m
1m
= 1.49 A R2/3S1/2 / n
Q
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
<0.974 m3/s
OK!
As a conclusion, this drain is adequate to cater for the overall requirement
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.
