Solar energy calculation

System

I’m trying to determine how much energy is lost if the PV panel is mounted horizontal (no tilt). I now see there is actual data that makes this decision easy.
http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/23183.txt

However, at the time, I didn’t know the data existed and I made my own calculation/algorithm. Comparing the results, it looks like the nrel.gov data is wrong. LOL. Anybody know why the discrepancy?
Edit, my error in reading the table of data that doesn't lineup.

My algorithm:

The amount of energy the panel collects is dependent on the sine of the solar altitude.

E(theta) = P * t * sin(theta)

Where
theta is the sun’s altitude.
P is the power capability of the PV panel, or I(battery) * V(battery) when PV is 90degrees to sun.
t is time.

For example, at sunrise and sunset, the altitude is zero degrees, and the E = sin(0) = 0.
At noon, the energy is the greatest.

Now it’s just a matter of calculating the average value of the sine wave from sunrise to sunset. We would integrate sin(theta) from zero to PI and divide by PI to get an equivalent average energy quantity.

(Correction: addition in bold)
Av E = 2/PI * sin(max altitude) * P * t = 0.64 * sin(max altitude) * P * t

Total Energy = 0.64 * sin(max altitude) * P * daylight hours




I used this program to determine max altitude and number of daylight hours:

http://www.susdesign.com/sunposition/

It’s limited to a resolution of 15 minutes. I think it’s close enough.

Data for Phoenix (33.43 N latitude) for June 15 and Nov 15. The 15th is used as an average for the entire month. This should correlate to nrel.gov data.

June 15: Sunrise = 5:00, Sunset = 19:00, Hours = 14, sin(79.81) = 0.98
Nov. 15: Sunrise = 6:45, Sunset = 17:15, Hours = 10.5, sin(38.45) = 0.62

Total Energy(June) = 0.64 * 0.98 * 14 * P = 8.8 * P
Total Energy(Nov) = 0.64 * 0.62 * 10.5 * P = 4.2 * P

My calculation shows you get twice the energy in June as you do in November.

That’s not the case for nrel.gov. The 0 tilt maximum data is:
June = 8.7 8.1
Nov = 4.0 5.1

After correcting my reading error, the nrel.gov data coinsides with my calculation. We can have faith in goverment run projects again. ;-)

This data shows only a 59% difference between June and Nov. I calculated over 100%. Is it a matter of cloud cover? Is June a bad solar month for Phoenix? May also shows larger energy potential than June (8.7 vs 8.1). Yet May has a lower sun altitude and less daylight hours.

Anyways, I believe my simple algorithm can be used anywhere on earth with excellent results for a cloudless day.

Cal

BB.

Re: Solar energy calculation

I am not sure that you find the "home page" for the data... But there is quite a bit of background on the issues of collecting (or at least attempting to collect) 30 years of solar day on a shoe string budget and with changing technologies.

http://rredc.nrel.gov/solar/pubs/redbook/HTML/techbkgd.html

A comparison of the old and new data bases provides the incentive for developing new solar radiation resource products such as this data manual. On an annual basis, 40% of the NSRDB and SOLMET/ERSATZ stations are in disagreement for global horizontal radiation by more than 5%, with some stations showing disagreement of up to 18% (Marion and Myers 1992). For direct beam radiation, 60% of the NSRDB and SOLMET/ERSATZ stations are in disagreement by more than 5%; some stations show disagreement of up to 33%. Disagreement between the two data bases is even greater when compared on a monthly basis. Most of the disagreement is attributed to differences in the instruments' calibration procedures and models.

Also, there are PDF station summaries that may be a bit easier to understand, for example Phoenix is in the Arizona file:

http://rredc.nrel.gov/solar/pubs/redbook/ Home Page of Solar Data Set
http://rredc.nrel.gov/solar/pubs/redbook/PDFs/AZ.PDF Page 2

They show scatter plots of the data set overlaid on a 1 year plot... Summer, apparently pretty cloudless has the smallest scatter, other months are much higher (and even higher solar irradiation than June/July).

You would also have to account for the amount of air in the path of light, amount of humidity, as well as the other "unpredictables" to get closer.

It is also possible that the text data and the PDF data differ... The PDF data is not the same as what you have here for a fixed plate collector at zero degrees (June=8.4, November=3.6)... May also account for some of the confusion.

-Bill

crewzer

Re: Solar energy calculation

The local environment (altitude, weather patterns, humidity, etc.) has a great effect on the local insolation. For example, El Paso, TX, and Savannah, GA are at about the same latitude (~32 N), but their insolation numbers are significantly different:

http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/23044.txt
http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/03822.txt

With respect to higher average daily insolation in Phoenix in May than in June, I suspect that's due to weather patterns and/or pollution (guesses on my part... I have no data to cite).

HTH,
Jim / crewzer

Roderick

Re: Solar energy calculation

The original reasoning of power proportional to sin(theta) sounds like its on the right track. However, as others have noted, the actual empirical data is worth far more.

I find that the biggest determinant of how much power I will get in a given day is the cloud cover. Second after that is the time of year (well accounted for by the theoretical model), and third, the ambient temperature.

That nrel map is probably the best start available, but you would have to adjust it for local geography. For example, if I was at the bottom of a foggy/smoggy valley, I would adjust he average insolation downward.

If you have the luxury of having a neighbor nearby with photovoltaic, you could ask them if they would share their records. I'm fortunate to have a number of other installations nearby (the closest, just 5 houses away). I got a spreadsheet of monthly production from one person, and our array seems to be tracking in proportion to that quite well.

System2

Re: Solar energy calculation/ my error

I corrected the data from nrel.gov. It was my reading error of their table. The data doesn't line up with the heading. My first post is corrected.

The energy ratio June/Nov is:
nrel.gov: 8.7/4.0 = 2.17
my calculation: 8.8/4.2 = 2.10

I would say that's extremely close. Hell, could have saved the government a lot of money with one quick calculation. LOL

Feel free to use Calbiker's energy algorithm.

Cal

System2

Re: Solar energy calculation

The local environment (altitude, weather patterns, humidity, etc.) has a great effect on the local insolation. For example, El Paso, TX, and Savannah, GA are at about the same latitude (~32 N), but their insolation numbers are significantly different:

There are significant differences in the Savannah solar output. El Paso and Phoenix (as well as my calculation) have very similar data.

For June 0 degrees tilt, maximum:
Phoenix: 8.7
El Paso: 8.7
Savannah: 7.0
My calculation = 8.8

I would wager the Savannah data is not accurate. It needs calibration!

Cal

crewzer

Re: Solar energy calculation

Cal,

I'd wager that you'd lose your bet! I've no doubt that Savannah's average insolation is affected by low elevation (thicker atmosphere) and local weather patterns (SW winds = Gulf of Mexico warm water influence = clouds, rain and HIGH humidity -- YUCK!).

For example, average June precipitation in El Paso is 0.87 inches. In Savannah, it's 5.49 inches. (Source: www.weather.com)

Regards,
Jim / crewzer

System2

Re: Solar energy calculation

Jim-

There is a reason I took the MAXIMUM data for my comparison. It is not an average but hand picked data for days having clear sky (or should be). With that in mind, I don't think I would lose the bet. Unless there's actually never a cloudless day in Savannah.

As I stated in my original post, my calculation is for a cloudless day. I would expect the MAXIMUM data to represent that. I'm back to a not so warm fuzzy feeling for gov work. ;-)

Cal

crewzer

Re: Solar energy calculation

It is not an average

Ah, but it is an average. The monthly "maximum" insolation value in represents the highest average daily insolation for that month over a 30 year period from 1961 to 1990. The detailed data for Savannah for each month over those 30 years is available here:

http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/mon2/03822.txt

It's not the maximum insolation theoretically possible under ideal conditions. If it was, I would think that there'd be a single number for each month for all locations at a specific latitude, and perhaps then we'd be in agreement.

Regards,
Jim / crewzer

System2

Re: Solar energy calculation

Glad we got it cleared up that Savannah is NOT the place I would want to live. ;-)

In regards to our discussion, Savannah would be the wrong data to validate my max daily solar energy calculation. Phoenix or El Paso would do a better job of that. Incidentally, it would be easy to add an average cloud factor (for given regions) to my calculation.

It's not the maximum insolation theoretically possible under ideal conditions.

I'm certainly not looking for that. Solar enthusiasts want to know what's the best they can achieve on any given day of the month. The MAXIMUM data falls woefully short in that regard.

Cal

crewzer

Re: Solar energy calculation

8-)

System2

Re: Solar energy calculation

The humidity difference between southern GA and similar latitudes in TX/AZ is huge! Those of us that live in the southeastern US sometimes feel that we are living in a tropical rain forest during the warmer months ;-) Even though the days are shorter in cooler weather, we usually have less clouds to help make up for it.

Bob