Assignment 2: Design a Off Grid Solar System For Your Home

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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 panel = 3428/4.12 = 832 watt

Let us take Size of Panel 270 watt

Capacity = 1000 Va

Battery System = 24 V

Battery Support = 80 Ah - 200 Ah

Panel Support = 1000 Wp

Solar Panel Current = 40 amp

Solar Voc = 30 V - 85 V

Total panel = 832 / 200 = 4

series connection = 24.6 * 2 = 49.2 v

Parallel = 8.13 * 2 = 16.26 amp

power = 49.2 * 16.26 = 800watt

Battery costing = 5649

panel costing = 38800

Inverter = 15000

mis = 10000

Total costing = Rs 70000

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