If I get 2 KiloWattHours (KWH) per hour then in North Little Rock (where I live) I get 11,700 of sun a year so I would get 23,400 KWH of energy minus obstructions. The price for electricity is around 10 cents per KWH(A little over 11 right now, but the average is SUPPOSED to be 9.7 so I'll go with the easy number a little over it.) This means I have to divide by 10 if I want to get dollars. So it's obviously $2,340 a year where I am minus cost of obstructions.(If it doesn't move then time of shade, If I don't clean it then dirtiness, I can't stop the clouds, and any form of partial eclipse... I'm sure there are more and will keep trying to figure them in.)
If we could find a way to buy in bulk it's a lot cheaper I'm sure. I'm going to call these guys
http://sunelec.com/index.ph
p?main_page=index&cPath
=32 tomorrow to see what I can talk out.
If we use the prices of $0.61 per watt, (I heard $0.85 was supposed to be good) then we could get an idea of what it should begin to cost when it all comes together.
It takes 1000 watts to make a kilowatt. So it takes 1000 watts activated for an hour to make a kilowatt hour or KWH. To get 2 KWH per hour you would need 2000 watts of solar paneling. You can find the cells cost by multiplying by it's price per cell. Since 2000 x .61 is 1220 you know it's precisely $1220.00 to buy the solar cells (Not full panels yet) to get 2KWH per hour.
So add that with other costs and obstructions subtract from $2,340 to find out precisely how much ground you'll cover in the first year. (Of course I know some of these things like the weather can't be predetermined, but we can resort to the almanac and meteorology to ball park it.)
So far I've got $2,340 - 1220.00 + $165.00 for inverter and ? for all other costs and obstructions, lets try to figure out and or delete as many of these as possible
by
EuphoriagenicCause on
07/10/2009 05:06:17 AM EST
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