"Real" Sun Hours [RE-wrenches]

[Bill Brooks]

Windy,

My associate Tim Townsend knows more about solar data than anybody I know
and he tells me that 96% of the insolation available can actually be used by
a well-designed PV system. That means that if the Red Books says you get an
average of 5.3 kWh/m2/day, you can use 96% of that or 5.1 kWh/m2/day.

It has little to do with thermal or electric. The solar resource is the
solar resource. Systems of various types and styles use that resource
effectively or not.

A slight correction on my previous post. It was pointed out to me by Mark
Mrohs that the Red Books does not discuss "peak sun hours". However, to
convert a daily number like that mentioned above--5.3 kWh/m2/day--one simply
divides by 1 kW/m2 (peak irradiance) to get the equivalent number of hours
the sun would have shown if it only shined at peak intensity (5.3 hours).

Converting that to estimate energy output is extremely tricky and the source
of many huge mistakes. A system capable of 1 kW AC output (1450 WattDCSTC
array) Will produce something near 5.1 kWh per day with a solar resource of
5.3 sun hours. However, this is only true if there is no shading, the
orientation matches the resource data, and the inverter operates at an
average efficiency comparable to the efficiency when the measurement
occurred (not likely).

I like methods that identify were the power goes. Take a look at the
approach in my guide sometime.

Bill.

-----Original Message-----
From: Windy Dankoff, Dankoff Solar [mailto:windy at dankoffsolar.com]
Sent: Wednesday, September 26, 2001 8:23 AM
To: RE-wrenches at topica.com
Subject: RE: "Real" Sun Hours [RE-wrenches]


>
>Measurements are made with pyranometers and the insolation recorded is
>listed in units of peak sun hours. This is an average daily energy
delivered
>by the sun ...
>Bill.


I've heard that most solar data (pyranometer, says Bill) relates
directly to THERMAL energy and it differs from data that is properly
gathered for PV yield. I hope somebody can give us the low-down
finding data appropriate to PV. I need a simplified method for
estimating PV yield around the world, based on the NASA site.

Regarding using a C-40 AH readout for solar data, it measures charge
current, so I presume it would be reduced by the unit's charge
controlling function when batteries are high.

Windy

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