calculate the battery and energy and solar panel power requied for the DC loads for 500 Wh power per day. battery DOD is 80% battery losses are 15%, controller losses are 5%, inverter losses are 15%, solar radiation is a 5.25 kWh/m2/day, solar panel losses are 25%.
Answers
Solar panel requirement = 150 W
Explanation:
DC Load = 500 Wh per day
Charge controller losses at load side = 5%
So 5% of 500 Wh = 25 Wh.
Battery output = 500 + 25 = 525 Wh
Battery capacity = Battery output / DOD = 525 / 50% = 525 / 0.5 = 1050 Wh
Battery losses = 15% of 525 Wh = 78.75 Wh
Input for Battery = 525 + 78.75 = 603.75 Wh
Charge controller losses at panel side = 5% of 603.75 Wh = 30.19 Wh
Output from solar panel = 603.75 + 30.19 = 633.94 Wh
Losses at panel = 25 % of 633.94 Wh = 158.48 Wh
Input at panel = 633.94 + 158.48 = 792.42 Wh
Solar Radiation = 5.25 kWh / m2 / day
Solar panel requirement = Input at panel / sunshine hours = 792.42 / 5.25 = 150.93 W ~ 150 W
Answer:
DC Load = 500 Wh per day
Charge controller losses at load side = 5%
So 5% of 500 Wh = 25 Wh.
Battery output = 500 + 25 = 525 Wh
Battery capacity = Battery output / DOD
= 525 / 50%
= 525 / 0.5
= 1050 Wh
Battery losses = 15% of 525 Wh
= 78.75 Wh
Input for Battery = 525 + 78.75
= 603.75 Wh
Charge controller losses at panel side = 5% of 603.75 Wh = 30.19 Wh
Output from solar panel = 603.75 + 30.19
= 633.94 Wh
Losses at panel = 25 % of 633.94 Wh
= 158.48 Wh
Input at panel = 633.94 + 158.48
= 792.42 Wh
Solar Radiation = 5.25 kWh / m2 / day
Solar panel requirement = Input at panel / sunshine hours = 792.42 / 5.25 = 150.93 W ~ 150 W
Explanation:
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ÌÑŚŤÂĞŘÂM Srinivas_Rao_2008