Question 1. Given:. Total amount of water = 17 litres.

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Question 1
Given:
Total amount of water = 17 litres
Area = 1.27 m2
Time = 2 hrs 30 min
i) Needed: Average rainfall in mm/hr
Average rainfall= 17∗10−3
1.27∗10−4∗2.5 =53.54 mm/ hr (Bonnin, 2010)
ii) Rainfall intensity for 5 minute duration
Given:
Total amount of water = 1.8 litres
Rainfallintensity = 1.8∗10−3
1.27∗10−4∗5
60
=170.08 mm/ hr (Newby, 2015)
iii) Required: Annual Exceedance Probability (AEP)
Given:
Average Recurrence Interval ARI=50 years
Annual Exceedance Probability AEP= 1
ARI ∗100 % (Bonnin, 2011)
AEP= 1
50∗100 %=2%
iv) Required: Rainfall Intensity
Given:
ARI=20 years
5-minute duration
ln ( I ) =4.978+0.7533 ln 5−0.2796 ( ln 5 ) 2+0.0166 ( ln 5 ) 3
I =253.50 mm/hr
Question 2
Given:
Stormwater generated = 300 ML
Suspended solids = 12 mg/l
Amount of solids discharged = 2.16 tonnes
Needed: Reduction capacity

Amount of suspended solids in stormwater ¿ 12∗300∗106 =3.6∗109 mg
¿ 3.6 tonnes
Reduction capacity ¿ 2.16
3.6 ∗100 %=60 % (Moore, 2016)
Question 3
Q= AV
V = 1
n R
2
3 S
1
2
R= A
P = 10
155∗2 =0.032 m
s=0.021
n=0.013
V = 1
0.013∗0.032
2
3 ∗0.021
1
2 =1.124 m/s
Thus,
Q=10∗1.124=11.24 m3 /s
Working for dimensions;
A R
2
3 =10∗0.032
2
3 =1.008
1.008= √3∗( y8
4 )1
3
y=0.97 m
A=2 y2 =2∗0.972=1.885 m2
Breadth B=2 y=2∗0.97=1.94 m
Q= AV
V = Q
A = 11.24
1.885 =5.963 m/s
Drawing representation
V = 5.963 m/s H = 0.97 m
B = 1.94 m

Question 4
Given:
Initial investment ¿ $ 120000
Annual Benefits = $ 13200
T = 7 years
Salvage value = $ 34000
Interest rate r =7.5 %
Needed: Net present worth NPV
NPV =∑ cashflow
( 1+i )t −initial investment +salvage value (Siddique, 2014)
NPV =∑
t=1
7 13200
( 1+ 0.075 ) t −120000+34000
NPV =−$ 16084.86
Question 5
Given:
Capacity = 230 KW
Capital cost = $5500/KW
Interest rate = 3.8%
T = 25 years
Capacity factor = 0.72
Required: Annualised capital cost and levelised cost.
Annualised capital cost At = Asset prices∗interest rate
1− ( 1+interest rate )−t (Zakeri, 2015)
¿ ( 230∗5500 )∗0.038
1− ( 1+0.038 )−25
¿ $ 79271.11

Levelised cost=
I +∑
t =1
25 At
( 1+i ) t
∑
t =1
25 M
( 1+i ) t
(Larsson, 2014)
¿
( 230∗5500 )+∑
t =1
25 79271.11
( 1+0.038 )t
∑
t =1
25 230∗0.72
( 1+0.038 )t
¿ $ 984.74
References
Bonnin, G. M. (2010). Trends in heavy rainfalls in the observed record in selected areas of the
US. In World Environmental and Water Resources Congress 2010: Challenges of Change (pp.
2432-2440).
Bonnin, G. M., Maitaria, K., & Yekta, M. (2011). Trends in Rainfall Exceedances in the
Observed Record in Selected Areas of the United States 1. JAWRA Journal of the American
Water Resources Association, 47(6), 1173-1182.
Larsson, S., Fantazzini, D., Davidsson, S., Kullander, S., & Höök, M. (2014). Reviewing
electricity production cost assessments. Renewable and Sustainable Energy Reviews, 30, 170-
183.
Moore, T. L., Gulliver, J. S., Stack, L., & Simpson, M. H. (2016). Stormwater management and
climate change: vulnerability and capacity for adaptation in urban and suburban contexts.
Climatic change, 138(3-4), 491-504.
Newby, M., Franks, S. W., & White, C. J. (2015). Estimating urban flood risk–uncertainty in
design criteria. Proceedings of the International Association of Hydrological Sciences, 370, 3-7.
Siddique, A. R. M., Khondokar, A. A., Patoary, M. N. H., Kaiser, M. S., & Imam, A. (2014,
February). Financial feasibility analysis of a micro-controller based solar powered rickshaw. In
2013 International Conference on Electrical Information and Communication Technology
(EICT) (pp. 1-5). IEEE.
Zakeri, B., & Syri, S. (2015). Electrical energy storage systems: A comparative life cycle cost
analysis. Renewable and sustainable energy reviews, 42, 569-596.