Re: Solar flux calculation

On Sat, 2 Aug 2008 16:00:10 -0700 (PDT), mundt <(E-Mail Removed)> wrote:

>I wish to calculate the solar flux falling on my panels (Ip) for
>modeling purposes.
>
>I have:
>1. the insolation from a local meterology station (at local solar
>noon) = Is
>2. the solar declination (at solar noon) = Td
>3. my latitude = Tl
>4. my panel tilt (azimuth) =Tp
>5. my panels face due south
>
>To correct the insolation for the declination of the sun, I believe
>that I need to divide the measured insolation (Is) by the cosine of
>the actual sun angle (Td-Tl). Thus the solar flux on a surface NORMAL
>to the sun is Is/cos(Td-Tl).
>
>To correct for the tilt angle of my panels, I believe I need to
>multiply the NORMAL flux by the cosine of the angle between the
>normal surface and the actual panel surface, i.e. cos(Td-Tl-Tp).
>
>The flux my panels then see is thus: Ip = Is *cos(Td-Tl-Tp)/cos(Td-
>Tl).
>
>Is this a correct formulation?
>
>Thank you in advance for your comments.

Not so simple.

There are two major measurements made by many observing stations,
global irradiation on a horizontal surface (using a pyranometer) and direct
normal irradiation (pyrheliometer or pyranometer with a shadow band).

Sunshine on a tilted surface has three components, direct normal (the
component that can be concentrated with a magnifying glass), diffuse
(the blue sky and/or reflections from clouds), and reflected from the
ground (albedo).

There is some good information at
http://www.nrel.gov/rredc/models_tools.html
and the related site.

Your calculations apply to the direct beam component on a clear day.
You need to make sure you understand the origin of the data you are using
before applying calculations.

There are computer programs that do these calculations, but
the old links I have no longer function. There are solar radiation
handbooks with full details.


Bill Kaszeta
Photovoltaic Resources Int'l
Tempe Arizona USA
(E-Mail Removed)