KISSAN EDUCATION www.kissanedu.com You are going to install a solar power system in our home for a total load of 800W where the required backup time of the battery is 3 hours. Optimize it. If you are living in Kolkata, West Bangla. Find out Charging Current for Battery . Take Solar Inverter, efficiency = 95% ( Luminious Inverter) Battery, efficiency = 85 % DoD = 80 % Solution : Since , Total Wattage = 800 W According to saftey Factor = 800 * 1.25 = 1000 Watt Step 1 : Size of inverter = 1 KW Step 2 : Size of Battery: Backup = 3 hour Back up = Total energy / Total wattage Total input energy in battery = 3 * 800 = 2400 wh Output of Battery = 2400 * efficiency = 2400 * 0.80 = 1920 wh if Voltage = 24v, Ah = 1920 / 24 = 80 Ah Charging current of battery = 10 % of Ah = 16 Amp(ideal value) = 20 amp(real value) 24V, 80V, Step 3 : Solar irradiance in kolkata = 4.12 kwh panel size after losses = 2400 / 0.70 = 3428 wh size of pan
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