Physics, asked by harshkumarbariya1073, 9 months ago

Calculate the battery energy and solar panel power required for the D. c loads of 500 Wh per day. Depth of discharge of battery 15% Controller 5%. Inverter 15% solar radiation is 2.5 Kwh m square solar panel losses are 25% A) 718 Wh, 180 W. B) 656 Wh, 150 w. C). 1050 Wh, 150 w. D) 1148 Wh, 180 w​

Answers

Answered by 232ishanee
12

Lets assume that you want to install 10 solar panels rated at 100 Watts each and having a conversion efficiency of 18%. The total power output of the solar system can be calculated as:

Total Power Output=Total Area x Solar Irradiance x Conversion Efficiency

We know the required Total Output Power is 1000 Watts (10 panels x 100 Watts), the Solar Irradiance for a surface perpendicular to the Sun’s rays at sea level on a clear day is about 1000 Watt/m2 and the Conversion Efficiency is 18%. Plugging these number in the above equation we get:

1000 Watts = Total Area x 1000 Watts/m2 x 0.18

or Total Area = 5.56 m2

I you are going to install all the panels in one line you would need a space of approximately 1 m x 5.56 m (each panel having a size of 1 m x 0.556 m) on your rooftop. There you go. You have a rough estimate of the space required by the solar panels of your system.

In addition

1. Do remember that solar panels are usually installed at an angle to the earth surface and this may change the results somewhat.

2. Imagine a solar panel has a conversion efficiency of 100% i.e. it converts all the solar energy into electrical energy then all you would need is a 1 m2 solar panel to produce 1000 Watts of electrical energy.

Based on the above in my opinion the best formula is the first one

Area =(kw)/((kw/m^2)) but you have to consider the conversion efficiency in this formula

:

Answered by abhifun101
20

Answer:

150W

Explanation:

DC Load = 500 Wh per day Charge controller losses at load side = 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

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