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Chris J Dixon Chris J Dixon is offline
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Default Solar power calculations, please help!

Chris Hogg wrote:

On Wed, 15 Jul 2015 14:38:46 -0700 (PDT), David Paste
wrote:

Thanks all for the replies.

The amount oelectricity available seems dismally small.


Hence the very and depressingly large areas of countryside taken up by
solar farms for only modest return.

If you believe Harry's figures, he has a 4kW array, and estimates you
need ~5m^2 of actual panels per 1kW, i.e. from which you deduce he has
about 20m^2 of panels. He says he gets about 4000kWh/yr, i.e.
4000000/365/24 W =~460 W from 20m^2 of panels, or ~ 23 W/m^2, a lot
better than Bavaria, and even better than California!


Maybe you would prefer my numbers.

I have 14 panels each 260 Wp, and about 1.8 m^2s, giving 25.2 m^2
overall. As I said earlier, averaged over 4.6 years, the output
is 400 W, so that gives around 15.9 W/m^2.

Chris
--
Chris J Dixon Nottingham UK


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