How many meter of solar panel i need?

Question

my monthly house used is 800kwh, i what to know how many solar panel i need? and how much per meter?

Answer 1

Let's assume that your house is located in the Sahara Desert on the Equator, where the annual average length of daylight is 12 hours. We also assume that there is no cloud cover year round, and ignore sandstorms.

The energy density impinging on the surface of the Earth from sunlight normal to the Earth at local noon is about 1254 W/m^2.  The average daily energy density can be shown to be (2/π)(1254 W/m^2).  Estimating that atmospheric attenuation follows the same pattern, we have

(1254 W/m^2)(2/π)^2 per day. Unless your array tracks the Sun throughout the year, the annual variation reduces the average energy to (1254 W/m^2)(2sin(23.45°))(2/π)^2.  The maximum efficiency of solar cells I have encountered is 45%, reducing the output to (0.45)(1254 W/m^2)(2sin(23.45°))(2/π)^2.  Yet one more factor needs to be included that is often totally ignored -- maintenance.  Solar cells are delicate, and myst be protected from falling objects, windblown debris, and hail by glass or transparent plastic.  Disregarding the losses due to aging, you will experience about a 90% loss in efficiency due to dust accumulation, leaving (0.90)(0.45)(1254 W/m^2)(2sin(23.45°))(2/π)^2 ≈ 9.8696 W/m^2

(850 KWH/mo)(12 mo/yr) / [(9.8696 W/m^2)(12 hr/da)(365.26 da/yr)(1 KW/1,000 W)] ≈ (10,200 KWH/yr / (718.0475 KWH/m^2•yr) ≈ 14.205 m^2 ≈ 14.2 m^2

Your local weather bureau can probably give you the percentage of "clear days".  Divide the above by this percentage.  The need for an inverter to get usable AC voltages requires doubling the solar panel energy, as inverters are about 50% efficient.  You will also need a fairly large battery bank.  The above figures should be close to adequate for any latitude if the solar cell array is mounted normal to the equatorial plane at your  latitude.