A small village in Africa with wants to implement a 4-hour evening lighting sche
ID: 2248716 • Letter: A
Question
A small village in Africa with wants to implement a 4-hour evening lighting scheme using 50 x 15W LED lights supplied by solar energy. The clear sky solar irradiation at their location is 15 kWh/day/m2 . Size a system of solar panels and deep-cycle batteries to meet this demand: (a) assuming clear sky every day; (b) taking into account the possibility of 2 cloudy days in a row. Assume the solar panels are 15% efficient the batteries have a round trip (i.e. charge and discharge) efficiency of 75% and a loss rate of 1%/day.
Explanation / Answer
Given clear sky solar irradiation = 15 kWh/day/m^2
Energy required per day E = 50*15*4 = 3 kWh / day
Efficiency of Solar Panels = 15% and efficiency of battery = 75% and Loss rate = 1% /day
Battery capacity required EB = (50*15*4)/(0.75*0.99) = 4040 Wh / day
Considering 12 V battery , Ah = 4040 / 12 = 337 Ah
To 337 Ah battery energy need to be generated by solar panel without considering cloudy days is = 4040/(0.15) = 26.93 kWh / day .
Area of solar panels required = 26.93 / 15 = 1.795 m2
Considering two days of cloudy capacity battery required is = 3*337 = 1011 Ah and EB = 12,131 Wh
Solar panel need to generate = 12,131 / 0.15 = 80.88 kWh
Area of Solar panels required = 80.88 / 15 = 5.382 m2
a)
Without considering cloudy days
Size solar panels required = 1.795 m2
Size of battery = 331 Ah @ 12 V
b)
Cosidering two days cloudy
Size of solar panels required = 5.382 m2
Size of battery required = 1011 Ah @ 12 V
To decrease the size of solar panels required we can charge the battery for 5 days to charge fully when day is clear
= (26930 + 53860/3) / 15000 = 2.99 m2 . Consider that the extra battery size need to supply energy on cloudy days will charge for three days, by this assumption the size of solar panels required can be decreased and thus initial investment will be decreased.
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