1. The document provides calculations to determine the pre-development and post-development discharge rates (Qp and Qa) for a site using the rational method. 2. It is determined that Qa is greater than Qp, requiring hydraulic structures. The permissible site discharge is calculated to be 0.316 m3/s. 3. Based on calculations, the required site storage is determined to be 42.577 m3. Dimensions for an above-ground storage tank are provided to accommodate this storage.

1. 1/7
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DETAIL CALCULATION
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Solution:
Qp for pre development
a) Determine the time concentration for natural catchment, tc
tc = Fc × L / (A1/10 × S1/5)
= 92.5 × 0.10000 / (0.41/10 × 1341/5)
= 3.81 min
Hence, tc = 5 min
b) Determine the average rainfall intensity, RIt
ln RIt = a + b(ln t) + c(ln t)² + d(ln t)³
ln 50I5 = 5.1922 + 0.3652(ln 5) + (-0.1224)(ln 5)² + (0.0027)(ln 5)³
= 5.474
tc < 30 min, Pd = P30 - FD(P60 - P30)
ln 50I30 = 5.1922 + 0.3652(ln 30) + (-0.1224)(ln 30)² + (0.0027)(ln 30)³
= 5.125
50I30 = 168.11 mm/hr
ln 50I60 = 5.1922 + 0.3652(ln 60) + (-0.1224)(ln 60)² + (0.0027)(ln 60)³
= 4.821
50I60 = 124.08 mm/hr
PA = (IA × tA) / 60
P30 = (I30 × t30) / 60
= (168.11 × 30) / 60
= 84.05 mm
P60 = (I60 × t60) / 60
= (124.08 × 60) / 60
= 124.08 mm
FD = 1.39
Pd = P30 - FD(P60 - P30)
= 84.05 - 1.39(124.08 - 84.05)

2. 2/7
= 28.42 mm
I = Pd / d
= 28.42 / (5/60)
= 341.07 mm/hr
c) Determine runoff coefficient, C
C = Refer to design chart 14.4,
= 0.90 (Steep Rocky Slopes; Clay Soil, Open Crop, Close Crop or
Forest - Type A)
C = Refer to design chart 14.4,
= 0.81 (Sandy Soil, Forest - Type F)
d) Determine discharge, Qp (Pre Development)
Qp = C × yIt × A / 360
= [(0.000 × 0.90) + (0.400 × 0.81)] × 341.075 / 360
= 0.305 m³/s
Qa for post development
a) Determine the overland sheet flow travel time, to
to = (107 × n × L1/3) / S0.2
= (107 × 0.011 × 5.001/3) / 0.110.2
= 3.13 min
b) Determine the drain time, td
td = L/V
= 100.00 / (1 × 60)
= 1.67 min
c) Determine time of concentration
tc = to + td
= 4.80 min
d) Determine the average rainfall intensity, RIt
tc < 30 min, Pd = P30 - FD(P60 - P30)
ln 50I30 = 5.1922 + 0.3652(ln 30) + (-0.1224)(ln 30)² + (0.0027)(ln 30)³
= 5.125
50I30 = 168.11 mm/hr
ln 50I60 = 5.1922 + 0.3652(ln 60) + (-0.1224)(ln 60)² + (0.0027)(ln 60)³
= 4.821

3. 3/7
50I60 = 124.08 mm/hr
PA = (IA × tA) / 60
P30 = (I30 × t30) / 60
= (168.11 × 30) / 60
= 84.05 mm
P60 = (I60 × t60) / 60
= (124.08 × 60) / 60
= 124.08 mm
Take tc as nearest = 5 min, to table 13.3, ²P24h, FD = 1.39
Pd = P30 - FD(P60 - P30)
= 84.05 - 1.39(124.08 - 84.05)
= 28.42 mm
I = Pd / d
= 28.42 / (5/60)
= 341.07 mm/hr
e) Determine runoff coefficient, C
C = Refer to design chart 14.3,
= 0.90 (Impervious Roofs, Concrete; City Areas Fulls and Solidly
Built Up - Type 1)
C = Refer to design chart 14.3,
= 0.89 (Semi Detached Houses on Bare Earth - Type 3)
f) Determine discharge, Qa (Post Development)
Qa = C × yIt × A / 360
= [(0.110 × 0.90) + (0.290 × 0.89)] × 341.075 / 360
= 0.339 m³/s
Hence, Qpost > Qpre, hydraulic structure need to be proposed.
Development status I (mm/hr)
Impervious area Pervious area Total
C × A
Q (m³/s) Q (l/s)
C A (ha) C A (ha)
Pre-development, Qp 341.07 0.90 0.000 0.81 0.400 0.32 0.305 305.34
Post-development, Qa 341.07 0.90 0.110 0.89 0.290 0.36 0.339 339.46
From the study of overland flow time including travel time on roof and gutter,
minimum time of concentration adopted:
tcs = 3 minutes

4. 4/7
tc = 5 minutes
Permissible Site Discharge (PSD)
For above-ground storage,
a = (4Qa/tc) (0.333tcQp/Qa + 0.75tc + 0.25tcs)
= (4 × 0.339/5) (0.333 × 5 × 0.305 / 0.339 + 0.75 × 5 + 0.25 × 3)
= 1.629
b = 4QaQp
= 4 × 0.339 × 0.305
= 0.415
PSD = [a - √(a² - 4b)] / 2
= [1.629 - √(1.629² - 4 × 0.415)] / 2
= 0.316 m³/s
Site Storage Required (SSR)
For above-ground storage,
c = 0.875 PSD (1 - 0.459(PSD / Qd))
d = 0.214 (PSD² / Qd)
Required SSR,
SSR = 0.06td (Qd - c - d)
td (min)
I
(mm/hr)
Impervious area Pervious area
∑CA Qd (l/s) PSD (l/s) c d SSR (m³)
C A (ha) C A (ha)
5 341.07 0.90 0.11 0.89 0.29 0.3583 339.46 315.77 158.3268 62.860 35.48
10 256.99 0.90 0.11 0.89 0.29 0.3559 254.03 315.77 118.6517 84.001 30.82
15 217.75 0.90 0.11 0.88 0.29 0.3547 214.55 315.77 89.6486 99.454 22.91
20 194.53 0.90 0.11 0.88 0.29 0.3533 190.89 315.77 66.5104 111.783 15.12
25 178.68 0.90 0.11 0.87 0.29 0.3505 173.96 315.77 46.0906 122.664 7.80
30 168.11 0.90 0.11 0.86 0.29 0.3485 162.76 315.77 30.2495 131.104 2.53
35 158.34 0.90 0.11 0.85 0.29 0.3466 152.46 315.77 13.6285 139.961 -2.37
40 149.80 0.90 0.11 0.85 0.29 0.3448 143.49 315.77 -2.7921 148.710 -5.83
45 142.26 0.90 0.11 0.84 0.29 0.3431 135.60 315.77 -19.0380 157.367 -7.38
50 135.55 0.90 0.11 0.84 0.29 0.3415 128.59 315.77 -35.1386 165.945 -6.66
55 129.52 0.90 0.11 0.83 0.29 0.3399 122.29 315.77 -51.1740 174.490 -3.39
60 124.08 0.90 0.11 0.82 0.29 0.3382 116.57 315.77 -67.2314 183.046 2.73
Max capacity SSR = 35.48
Note: * indicates the default value of I determined by program has been changed by
user.
From the table, a maximum of SSR for 35.48 m³ occurs at a time of 5 minutes.

5. 5/7
An additional 20% is added to the volume to account for inaccuracies in construction
and future loss of storage due to the build up of the lawn surface.
Therefore, required SSR = 42.577 m³
Primary Outlet
Primary outlet discharge should not exceed the lesser of pre-development Qp and PSD
flow.
Qp = 305.34 l/s
PSD = 315.77 l/s
Provide 2 nos. 200 mm diameter orifice:
Ao = No. of Orifice π × (D/2)2
= 2 × π × (0.200/2)2
= 0.062832 m2
Ho = Depth - Diameter/2 - H
Ho = 0.57 - 0.200/2 - 0.000
Ho = 0.467 m
Q = CdAo(2gHo)0.5
=0.62 × 0.062832(2 × 9.81 × 0.466666666666668)0.5
=0.11788 m3/s
=117.88 1/s
Secondary Outlet
For ARI = 100 years and tc = 5 minutes.
ln 100I30 = 0.0000 + 0.0000(ln 30) + (0.0000)(ln 30)² + (0.0000)(ln 30)³
= 0.000
100I30 = 1.00 mm/hr
ln 100I60 = 0.0000 + 0.0000(ln 60) + (0.0000)(ln 60)² + (0.0000)(ln 60)³
= 0.000
100I60 = 1.00 mm/hr
PA = (IA × tA) / 60
P30 = (I30 × t30) / 60
= (1.00 × 30) / 60
= 0.50 mm
P60 = (I60 × t60) / 60

6. 6/7
= (1.00 × 60) / 60
= 1.00 mm
From table 13.3, 2P24h, FD = 1.39
Pd = P30 - FD(P60 - P30)
= 0.50 - 1.39(1.00 - 0.50)
= -0.20 mm
I = Pd / d
= -0.20 / (5/60)
= -2.34 mm/hr
Using the Rational Method, the pre and post development flows are derived using the
equation:
Q = CIA
Development status I (mm/hr)
Impervious area Pervious area Sum
C × A
Q (l/s)
C A (m²) C A (m²)
Post-development -2.34 0.000 1100.00 0.000 2900.00 0.0 0.00
Qoverflow = Q100 - Qprimary outlet
= 0.00 - 117.88
= -117.88 l/s
Discharge over board crested weir
Qsecondary outlet capacity = CBCWBH1.5
= 1.7 × 1.00 × 0.5001.5
= 0.60104 m3/s
= 601.04l/s
Discharge over board crested weir (601.04 l/s) > Qoverflow(-117.88 l/s), OK
Conclusion
The dimension of the proposed above-ground storage tank for the proposed
development:
V = (dstart + dend)/2 × Width × Length
= (0.60 + 0.57)/2 × 0.60 × 140.00
= 49.00
Item Width (m) Length (m) Depth (m) Vol. (m³)
OSD 0.6 140.0 0.6 49.0

7. 7/7
APPENDIX