Question

Calculate the battery energy and solar panel power required for the AC loads of 500 Wh per day. Battery DoD is 80%, Battery losses are 15%, Controller losses are 5%, inverter losses are 15%, Solar radiation is 5.25 kWh/m2/day, Solar panel losses are 25%.

Answer 1

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