PV Watt potential (as a percentage) vs. the suns angle from solar noon

Than you. But the link is not working for me (Firefox browser.)

It does not work for me either. I am also using Firefox web browser.

??? What is this ?

Maybe its an estimate of the output of a panel in outer space. It certainly doesn't describe my panels. All mine put out more than zero
anytime its not pretty dark. With a couple panels back to back, the one facing AWAY from the sun does 5 to 10% of the other.

Bruce Roe

What I mean is, what is this describing ? I'm all for information, but this needs some definition before going further, and I'd guess a lot of work after that to be useful.

After reading it appears he is saying the angle of the sun striking a solar panel is a function of Cosine Law which is just plain false. If that were true a 2-axis tracker would produce 100% power from dawn to dusk every day of the year.

Cosine of 90 degrees = 0 or 0% power.
Cosine of 0 degrees = 1 or 100% power.

Sorry Silver but that is Junk Science.

Here we go.

If anyone cares, and perhaps to clear things up some some:

1.) The Cosine of the angle of beam radiation striking a flat surface ~ = (Cos(Zenith angle)*Cos(surface tilt angle)) + (Sin(Zenith angle)* Sin(surface tilt angle)*Cos(solar azimuth angle - surface azimuth angle)) (see Duffie & Beckman/others for how the angles are defined). The angle of incidence can be described in terms of several angles and trigonometric functions of those angles.

If what the OP wrote is meant to represent some f("cosine law"), whatever that means, it is probably correct to say that what the OP wrote is incorrect.

However, the solar position relative to a surface of arbitrary orientation on the earth is a function of several trigonometric relationships, three of which are cosine functions.

If the surface in question is stationary, the received solar radiation over the course of a day will be a function of many things, one of which is the Cos of the angle of incidence of beam radiation. Diffuse insolation is a loose f(incidence angle). Other influencing variables are cloud cover, atmospheric quality and others. Seasonal variations in solar position are accounted for in the zenith angle and the azimuth angle which are f(declination angle and hour angle functions). Also see NREL or a nautical almanac for more info.

2.) As for a dual axis (or gimbaled) tracking system, a dual axis tracker can be made to track the sun with a fairly low day long incidence angle and will thus intercept pretty close to 100% of the incident beam insolation, and pretty close to 100 % of the diffuse insolation. However, since the intensity of both of those varies, the solar device's output will vary.

3.) It looks to me anyway, that indeed, at least without more information, Silver is Money's post qualifies as junk science, if for no other reasons than poor definition and lack of detail.

Today I finally got clear skies all day and played around with my oscilloscope and the LAN web pages of my charge controllers and Outback Radian. After some chainsaw work, I've eliminated most of the shading even when the sun is as low on the horizon as it is here (around 48 degrees N) now (three weeks from Winter solstice).

Well, I was looking at my 24 panels that are tilted at a 45 degree elevation angle to make the most of our meager winter sunlight and seeing them all nice and bright, and was wondering why I was only getting a lousy 4.5A from panels that are rated at Isc=9A. The cosine of the incidence angle of the sunlight hitting the panels was around 0.9, which would only account for a 1A reduction in current.

After doing a google search on "solar insolation elevation angle" and a subsequent visit to the "Now I Feel Stupid" department, I came across this web page on the impact of air mass: http://www.ftexploring.com/solar-ene...nsolation2.htm

The web page provides a table that shows my solar insolation even on an optimally-oriented panel to be 34% of what it would be with the sun straight up. That would result in just 3A. However, I'm at about 2000 feet above sea level, so the effect is less. Clearly, that's what's cutting my PV current in half.

I feel satisfied to have figured out the mystery, but a little annoyed at yet another strike against trying to generate usable PV power in the wintertime at a northern latitude. A sunny winter day around here isn't much better than a light cloudy sky scattering the photons everywhere, when you consider the shading problems you get from from all the tall pines around here otherwise.

The 4.5 A/9.0 Amp Isc on an instantaneous measurement, would indicate that, if all's well with the array, the P.O.A. irradiance is something less than ~~ 500 W/m^2 or thereabouts. The ratio of string or individual panel instantaneous current to rated current is a pretty good 1st approx. of P.O.A irradiance after reflection and absorption losses from the glazing, or P.O.A /1000 W/m^2 STC irradiance.

If you want good information that will allow you to understand array irradiance and how it varies with conditions, do this: Consult a copy of Duffie & Beckman. More information than can be explained in this forum format is there for the understanding. Doing so will allow you, for example, to understand why the statement you reference of air mass lowering the sun's intensity by 20 % is, at best, incomplete, and otherwise confusing and misleading. That would be at a condition called airmass 2 = (1/cos (solar elevation angle)). A clear (and clean) sky with the sun directly overhead (@ solar elevation angle == 90 deg. and airmass == 1.0) at sea level will attenuate about 28% or 29% of the solar irradiance. I've measured clear day direct irradiance on 12/21 near sea level of to close 900 W/m^2. Various models (i.e. the Bird clear sky broadband model, see NREL) seem to be about the same or a bit more.