The answer depends on if you intend to be connected to the grid (power company) or not. If you are off-grid, you need to either size it for worst case of winter sun hours, or plan on a generator to help make up the difference of available sun in the winter.
On grid is much easier, because if your system is too small, or you have a stretch of bad weather, you can just buy any extra you need from the electric company.
We’ve got some calculators for both scenarios at http://www.altestore.com/howto/Calculato…
If we use 5 sun hours for an example, 200kwh / month = 6.666 kwh / day. 6666 / 5 = 1333 W of PV panels. However, you also need to figure in system inefficiencies, and add about 30% to the equation, for 1700W. You can look at a package right around that size that includes the panels, inverter, mounting hardware, safety equipment and everything at http://www.altestore.com/store/Kits-and-…
Off grid adds complexity and expense in adding batteries to store the power. If you are looking at off-grid, there is a package that includes additional equipment needed, including batteries at http://www.altestore.com/store/Kits-and-…
Congratulations on your conservative usage.
When you use only 7 kWh of electricity a day, your bills are likely to be low, and it’s not certain that solar would pay for itself.
But to answer, try taking a look here http://rredc.nrel.gov/solar/old_data/nsr… . Enter Average, Annual, Flat Plate Tilted South at Latitude, then click View the Map. This will show you how many peak sun hours you get per day.
Example: Where I live, there are 5 peak sun hours per day, or 5 x 360 = 1800 peak sun hours per year. If my usage was like yours, it would be 200 x 12 = 2400 kWh / year. 2400/1800 = 1.3 kW of panels.
figure 50w/m^2 for 8 hr of sun/day = 400 w-hr/m^2-day
200 kwh/mo = 6700 w-hr/day
6700 /400 = 17 m^2
Check out what the solar panels you expect to use will provide and what you will do if you need power when the sun does not shine.
The answer depends on if you intend to be connected to the grid (power company) or not. If you are off-grid, you need to either size it for worst case of winter sun hours, or plan on a generator to help make up the difference of available sun in the winter.
On grid is much easier, because if your system is too small, or you have a stretch of bad weather, you can just buy any extra you need from the electric company.
We’ve got some calculators for both scenarios at http://www.altestore.com/howto/Calculato…
If we use 5 sun hours for an example, 200kwh / month = 6.666 kwh / day. 6666 / 5 = 1333 W of PV panels. However, you also need to figure in system inefficiencies, and add about 30% to the equation, for 1700W. You can look at a package right around that size that includes the panels, inverter, mounting hardware, safety equipment and everything at http://www.altestore.com/store/Kits-and-…
Off grid adds complexity and expense in adding batteries to store the power. If you are looking at off-grid, there is a package that includes additional equipment needed, including batteries at http://www.altestore.com/store/Kits-and-…
Congratulations on your conservative usage.
When you use only 7 kWh of electricity a day, your bills are likely to be low, and it’s not certain that solar would pay for itself.
But to answer, try taking a look here http://rredc.nrel.gov/solar/old_data/nsr… . Enter Average, Annual, Flat Plate Tilted South at Latitude, then click View the Map. This will show you how many peak sun hours you get per day.
Example: Where I live, there are 5 peak sun hours per day, or 5 x 360 = 1800 peak sun hours per year. If my usage was like yours, it would be 200 x 12 = 2400 kWh / year. 2400/1800 = 1.3 kW of panels.
figure 50w/m^2 for 8 hr of sun/day = 400 w-hr/m^2-day
200 kwh/mo = 6700 w-hr/day
6700 /400 = 17 m^2
Check out what the solar panels you expect to use will provide and what you will do if you need power when the sun does not shine.