How to estimate Solar energy on certain direction pointed surface?
If I mount a flat surface which points to certain constant direction. At certain place of the world. I now need to estimate the maximum amount of Solar radiation energy which this surface can get during certain time.
Is there a PC program or an algorithm that can make this estimate for me?
The actual problem.
I need to know how much maximum energy can I get during a year, if I put solar cells or solar collectors on my house tilted roof?
Is there anyone who can somehow help me?
Thank you in advance.
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It depends where on the planet you are talking about. The amount of solar energy you will be able to obtain will be very different if you are installing a system in Seattle/Portland in the North West of the US, vs in the deserts of California, Arizona, and New Mexico. Once you determine the location then you will be able to obtain the data on the maximum amount of energy obtainable, type of system needed etc etc. When and if you get serious about installing a system contact the guys at Sun Electronics as they did my Uncle’s house in the Florida Keys, In Laws place outside of Chicago and Brothers boat in the Caribbean and everybody has been very impressed with their finished products.
References :
http://www.sunelec.com
Comment by solar answers — @
Try looking here: http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/atlas/
That’s the National Renewable Energy Labs site, and will give you insolation (amount of sun) for your area. If you have an ideal location and roof, you would choose, Average, Annual, Flat plate tilted south at latitude. Then a map will come up. This map takes into account weather conditions, too, because it is derived from actual measured data. Thus, a dry, cloudless location such as Phoenix shows more sun than a place with more clouds like Honolulu.
Let’s say the map says you get 5 hours equivalent sun per day. The word "equivalent" is important, because it factors in that morning and evening sun are weaker than noonday. The number you get means the equivalent of direct, noonday sun for 5 hours, even though it may be 12 hours from sunrise to sunset. And it’s an annual average, so you will get more in the summer, less in the winter. Ok, so you take that figure, say 5 hours, and multiply it my the size of your array in watts, and multiply it by 365, the number of days in a year.
Example:
5 hours/day X 3 kW array X 365 days/year = 5475 kWh/year.
If you want a more complex calculation, you can try http://www.pvwatts.org/ , and take into account exact tilt and orientation of your roof.
References :
Comment by roderick_young — @