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Insolation question - Page 2

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Posted by Ron Rosenfeld on February 13, 2009, 2:35 pm

If you are converting from sq m to kWh/day, then just multiply by
efficiency of the panels (varies by make/model) and insolation:

kWh =  Area * Eff * Insolation

Assuming panels with 14% efficiency:

kWh/day    =  17 sq M *
       14 %  *
       4 kWh/sq M /day

That would be the expected DC output from the panels, averaged over the
same period of time as the insolation figures.

Posted by stevey on February 14, 2009, 6:34 pm

Ron's answer would be 9.52 KWh.  If your heater works with DC power,
as with
your LED nightlight. You should not tilt your panel at Lat +15
degrees. You'll
lose much energy that way.  A tilt of Lat - 15 should give you better
energy production.

Posted by lemur on February 14, 2009, 7:44 pm
 Thanks Ron...thats what i was looking for

Steve...not trying to make electricity....im making heat and +15 deg
works MUCH better in winter...which is when we need it.


Posted by Ron Rosenfeld on February 15, 2009, 3:01 am

One further comment.  When you wrote

I did make the assumption that the 4 hrs daily insolation was at +15.  If
that's not the case, the result will be different.

Posted by Morris Dovey on February 15, 2009, 3:17 am
 lemur wrote:

If you're making heat where the ground is snow covered, then you may get
better results with vertical panels.

Morris Dovey
DeSoto Solar
DeSoto, Iowa USA

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