Question

Calculate the battery energy and solar panel power required for the DC 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:

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 +25525 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

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