Drainage System Calculation MASMA

September 3, 2017 | Author: Asyraf Malik | Category: Surface Runoff, Drainage Basin, Civil Engineering, Liquids, Hydrology
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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
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