Solar Standard Test Conditions Question

[BFL13]

Need help understanding this or else it is just wrong, am very confused, any help appreciated---

"By convention, solar cell efficiencies are measured under standard test conditions (STC) unless stated otherwise. STC specifies a temperature of 25 °C and an irradiance of 1000 W/m2 with an air mass 1.5 (AM1.5) spectrum. These conditions correspond to a clear day with sunlight incident upon a sun-facing 37°-tilted surface with the sun at an angle of 41.81° above the horizon.[2][3] This represents solar noon near the spring and autumn equinoxes in the continental United States with surface of the cell aimed directly at the sun."

http://en.wikipedia.org/wiki/Solar_cell_efficiency

I am confused about how the STC 37/42 angles can be with the panel aimed directly. IMO if the sun's altitude is 42 degrees then the panel tilt to be direct would be 48 degrees, not 37.

Meanwhile they then say that that STC set-up is equivalent to what the panel will do in the USA. I am confused about that too.

You can get your latitude at noon from the sun's altitude of course. You actually use the Zenith Distance which is 90-Alt. So you get

Lat = ZD - Declination and at the Equinoxes, Dec is zero, so in the above quote, where Alt is 42, then ZD and Lat is 90-42 = 48

They say that puts it in the continental US, which it does (barely)

But now they say the panel is tilted 37 degrees but it is aimed directly at the sun at noon. How can that be? Panel tilt at noon direct sun would be the same as ZD and at the time of the two Equinoxes, panel tilt would also be the same as your Latitude.

With latitude at 48 and tilt at 37, how can that be "direct" and so how can that be the same as STC?

I know to get my Isc from a panel I have to aim it right at the sun with the sun high enough to give enough light. I also know from doing it, that at 49.3N Lat, with Dec at say 20N, then my panel tilt at noon should be ZD = Lat -Dec or 49.3 - 20 = 29.3 degrees. On 21 June, with Dec maxed at 23.3ish, then my panel would be tilted 49.3-23.3 = 26 degrees. On 21 Mar or 21 Sep, the panel would be tilted the same as the latitude or 49.3 degrees to be direct aimed at noon.

So what are they trying to say in that Wiki article? I need more coffee or else it is wrong or what?

Edit--Hmmm they didn't say that the equivalent USA example was when the sun was at 42 altitude or that the panel was at 37. Hmmmm

Thanks

[BFL13]

Thanks for the clarification. Right. Latitude would be 48 not 42.

I should clarify that the "vertical down to the sun" is the zenith distance and from the sun down to the horizon is the sun's altitude, so 90-Alt=ZD

With Dec zero (twice a year at the Equinox) noon latitude = ZD

What you can't have anytime, anywhere, is ZD 48, tilt direct at sun 37, and altitude 42. Direct tilt always = ZD

The convention is to measure "tilt" as the angle at the low end of the panel from the high end of the panel to the ground (with the ground flat to the horizon)

BTW it may be no coincidence that Lat 48N is also where the nautical mile of 6,080 feet is taken from at 1 minute of latitude, since this is about half the way from the pole to the equator--the Earth is not an exact sphere, being fatter in the north half.

Maybe they took solar STC at 48N for the same reason, to get an average of some sort.

[12thgenusa]

Altitude has nothing to do with it. What is an altitude of 42 anyway?

I see my mistake though. I meant equinox instead of solstice. And an air mass of 1.5 equates to a sun angle of 41.81° above the horizon which is the angle of the sun at its highest point at the equinoxes at latitude 48.19°.

Sorry. Much ado about nothing.

[BFL13]

12thgenusa wrote:
No. AM of 1.5 is at a latitude of 41.81.

The calculation was for ZD at 48 so altitude was 42 whatever the latitude, date, or time of day.

For noon latitude to be 42N, and ZD 48 (to get that 1.5):

Lat = ZD -Dec so 42= 48-Dec so Dec must be 6N so sometime (three weeks?) after the 21 March Equinox.

(Dec runs from zero to 23 in three months so about 8 degrees a month change of tilt needed to be aimed at noon panel facing South)

[12thgenusa]

No. AM of 1.5 is at a latitude of 41.81 when the sun is at the highest point.

[BFL13]

Thanks. (I think you meant altitude, not latitude at the end there) Don't get solstice part. The 37 degrees is a puzzle.

[12thgenusa]

Not sure about the 37 degrees, but note 2 in the Wiki article references an ASTM that talks about that. The 41.81° I believe comes from the calculation of Air Mass of 1.5. Air mass of 1 indicates the sun is directly overhead. The calculation for AM is 1/cos of angle of sun from vertical. So 1/cos 48.19° = 1.5. 90-48.19 = 41.81° latitude at solar noon on solstice.