Solar Energy - Insolation Value

From the Wikipedia article:

The Sun's rays are attenuated as they pass through the atmosphere, thus reducing the insolation at the Earth's surface to approximately 1,000 watts per square meter for a surface perpendicular to the Sun's rays at sea level on a clear day.

The actual figure varies with the Sun angle at different times of year, according to the distance the sunlight travels through the air, and depending on the extent of atmospheric haze and cloud cover. Ignoring clouds, the average insolation for the Earth is approximately 250 watts per square meter (6 (kW·h/m2)/day), taking into account the lower radiation intensity in early morning and evening, and its near-absence at night.

So the 1,000 watts per sq meter value is only for the surface of the earth where the Sun is directly overhead, and only at local noon. Averaged over the entire hemispheric surface of the Earth facing the Sun, where most of the Sun's energy strikes the Earth at an oblique angle, the value is 250 Watts per sq meter.

I read that but the solar panel retailers argue that if you tilt a PV panel so the suns rays strike the panel perpendicular then you get the full rated output. I question this but so far have not found anywher to back up my position in a verifiaable way. People think I am just anti solar if I make that argument.

From the official goverment website that tabulates solar insulation values it gets equally confusing. Although they do list reduced values for more northerly locations they say the insolation hours is for when the panel delivers its rated output assuming it has the correct orientation. So for a northern location they may list 1.9 hours in January. If you read the fine print on their website you could draw the conclusion that for 1.9 hours a solar panel might in fact deliver the rated output.

I question this but cannot find anywhere to support this supposition.

There is also the issue of 'air mass'. The longer the path length of the sunlight through the atmosphere, the more attenuation the light suffers. For an object far above the atmosphere at a person's zenith, the air mass is 1. The farther the object is from your zenith, the greater the air mass roughly according to the secant of the angle from the zenith. If the angle is 60 degrees the cosine of 60 is 0.5 and the secant is 1/cos(angle) = 1/0.5 = 2. So there is roughly twice as much air between you and the object if it is 60 degrees from your zenith (30 degrees above the horizon).

Tilting the panel so that the Sunlight strikes it perpendicularly helps, but all other things being equal, a solar panel in Florida will always outperform a solar panel in North Dakota because of the lower air mass.

Here is more info from Wikipedia under the subject Air Mass (Astronomy). It mentions sources that will help determine the amount of atmospheric attenuation.

Atmospheric attenuation of solar radiation is not the same for all wavelengths; consequently, passage through the atmosphere not only reduces intensity but also alters the spectral irradiance. Photovoltaic modules are commonly rated using spectral irradiance for an air mass of 1.5 (AM1.5); tables of these standard spectra are given in ASTM G 173-03. The extraterrestrial spectral irradiance (i.e., that for AM0) is given in ASTM E 490-00a.

BTW, an air mass of 1.5 corresponds to a zenith angle of 48 degrees.

Thanks for the useful information. Much better than what I have seen so far.

It helps support my contention that solar is not especially useful in the far north during winter. The official insolation tables gives us 1.9 hours in January and I have measured the sun altitude above the horizon as being just under 30 degrees at noon.

That makes it very hard to develop useful power during winter months. Naturally my quotes come out far more expensive that what a dealer almost 1000 kilometers south of here is quoting.

Hi elnav

Just as a side note, you have to consider that at high noon the solar panels will loose some of their efficiency due to them heating up. I forgot the exact amount but it was considerable.

The more astute participants in this thread possibly know that figure off the top of their heads. So, the more sun hitting the panels doesn't equal higher out put. If I remember correctly anything above 35 deg.C will reduce the output.

I'm to lazy to find the link now but others will have it on the screen or recite these figures. I don't deal with this every day so sorry I can't be of more help at this time. Besides this would hardly be a problem in the areas you have quoted.

Just thought I'd mention it, Ky.

KY

That depends on the ambient temperature. At -30F I am not going to worry about derating for temp. About two year or so ago I attempted to start a serious thread on the subject of running coolant pipes on the back side of solar panels.

Things wer going great until some smart ass decided to hijack the thread with silly remarks making fun of the concept.

Six months or a year later I find some Aussie company was makeing them and advertising on Comspec data sheets.

Sorry to hear. I have something on offer myself but that will take a while still. It's been on the cards and in hotter climates it will work.I should be able to publish soon.

Keep at it, Ky.

Probably me - It has been tried and creates more problems than solved though I will look up that company - tried but Comspec didn't get me there do you have a web address?

Just because some company sells something certainly does not make it viable.

The only place I have seen cooling work is with concentrators where it is required to preserve the cells.

Something like 0.35 to 0.5% loss of production per degree C over 25° C. Some panels are at the 0.35% number but most are closer to the 0.5% number.

Panels are rated at 1000 watts/m2 insolation and at 25° C

Hours are meaningless and misleading - you need to consider kW/m2/day

All the time people get hung up on some hour figure for some reason.

The reason people get hung up on hours is because it is a critical piece of information. If left out, you get a theoretical calculation designed to mislead, much like rating a wind generator's usefulness by looking at it's maximum momentary production capacity in its maximum wind speed environment. This is a meaningless number unless you know the hourly windspeed numbers, and predict the generation environment based on the many points of information that you have surveyed over many months or years of time. This data is rarely available for the exact location a mill might go.

The beauty of full sun hour numbers is that they actually work. Where I live, I go to one of many sources, like PV watts, and see that at my latitude, longitude and elevation that I receive the equivalent of 4.63 full sun hours per day, at perfect 189 degree (magnetic) orientation, at 39 degrees of tilt, with no shade at any time. That only means that the 1 square meter surface referenced above will receive on average, 4630 watts of insolation, every day of the year. (of course, it will likely never receive that in a given day) This is a number that will let you produce estimates of potential energy for the site, with about 60 different requirements for adjustment. Each of those adjustments, based on equipment, temperature, site conditions, orientation, voltage loss, etc, will reduce the final energy projection.

Lest you are not convinced of this methodology, and the correctness of the average sun hour figure, we have observed that on grid tied systems, even relatively small arrays, the accuracy is astonishing. The key is a very careful and honest de rating process can produce estimates that match within a 1% window of actual energy production on an annual basis, measured with a revenue grade meter.

PTC (not manufacturer supplied STC) numbers are the industry standard now, so use them as equipment rating. Most inverter efficiency numbers should be derated by at least 5%, as they list maximum efficiency. Incidentally, if you follow the full analysis route, you will note that any gain that micro inverters provide for mismatch and shadowing issues is more than overshadowed by the increased voltage loss in transmission. Wire sizing is critical. Shade impact is critical. Good Luck!

How in the world do you manage to get some wind turbine data mixed up in solar - apples and oranges or rather apples and road apples. Keep your luck - better to do the homework upfront without and WAGs.

The insolation figure takes available sun into account - hours are of interest only to say when you will be generating. PV Watts gives you hours? Here is a data run from PV Watts - where are the hours?

"Station Identification"
"City:","San"
"State:","California"
"Lat (deg N):", 32.73
"Long (deg W):", 117.17
"Elev (m): ", 9
"PV System Specifications"
"DC Rating:"," 1.0 kW"
"DC to AC Derate Factor:"," 0.770"
"AC Rating:"," 0.8 kW"
"Array Type: Fixed Tilt"
"Array Tilt:"," 32.7"
"Array Azimuth:","180.0"

"Energy Specifications"
"Cost of Electricity:","12.5 cents/kWh"

"Results"
"Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value ($)"
1, 4.83, 109, 13.62
2, 5.35, 108, 13.50
3, 5.91, 132, 16.50
4, 6.52, 140, 17.50
5, 6.05, 134, 16.75
6, 5.98, 127, 15.88
7, 6.26, 135, 16.88
8, 6.62, 142, 17.75
9, 6.02, 126, 15.75
10, 5.88, 130, 16.25
11, 5.16, 112, 14.00
12, 4.68, 104, 13.00

"Year", 5.77, 1498, 187.25

Your blurb is meaningless with many wrong points.

down at the bottom, average of all the months

"Year" 5.77

That is average sun hours

second number 1498 is annual total sun hours

third number 187.25, or $ value of the electricity that the 1KW system you plugged in, with those derates, in that orientation, would produce.

It's not the best source, but it's pretty good. NREL is better.

Care to elaborate on the errors?

Look at the headings - that is kWh/m2/day on an annual average.

If you don't know the difference between sun hours and kW hours I think you need to quit smoking whatever it you are on.

You make no sense whatsoever.

watts/m2 or derivatives of that are referred to as insolation

yes, that is a sun hour.

A kilowatt of energy directed at one square meter for one hour. It may be a little confusing because a sun hour does not imply that it only takes one hour to produce. If the sun insolation were at 500 watts per square meter of intensity, it would take 2 hours of exposure to make one sun hour. Think of it this way, in the PV watts example you just posted, on average, for the amount of time the sun shines in CA, at that location you selected, from dawn to dusk, that plane you defined will collect 5.77 sun hours. Now you know the sun doesn't pop up directly over head with no clouds. On a clear sunny day, the insolation occurs in a big bell curve. If it rains, it's a smaller curve. Sun hours let you develop energy models, and if done correctly, are very accurate. I'm not going to insult you back. It took me a while to get it.

Insult me - I could care less. You are somewhat challenged it seems. Your definition is pure blather.

There are standard terms in the industry that you really should learn. Maybe games work in house but in dealing with the world they don't.

A sun hour - first time I have ever heard such tripe.

I know what a daily insolation curve looks like - I have a pyranometer readout available on my desk.

Sun hours

you have a pyrometer. That pyrometer will not help you design and specify a PV system. It will only give you a spot reading.

A sun hour represents a window of time, at a given location, required to gather 1 KWH. The PV watts number is telling you that, on average, for every 24 hours of exposure at that location, the sun is providing, 5770 watt hours of insolation on your selected plane.

It can even take less than one hour to collect a sun hour! (We have an array built next to a lake that occasionally reflects additional insolation onto the array) It could take all day! You need to know the energy potential of a spot on the earth, and details about the plane your collectors reside on, and details about the mechanical and electrical components, to make a prediction of performance. Sun hours, available for every spot that we work (USA), tell us the potential.

Are you annoyed that this does not make sense to you? Because this is how we do it in the US.

Hey my friend - forget your superior crap - I am an American citizen from the US - born and grew up there. The past 30 years I have spent most of my time working in many countries around the world - running projects that you may some day dream of.

I guess you must be a salesman - or maybe a tech for some small solar company? The knowledge would be a good fit.

That is not how most in the US do it - I know for a fact. I probably spend more hours per day on solar topics than you do - if you are working full time.

The rest of the US and world use insolation as shown on PV Watts or any other program I have seen. Sun hours are a term you fellows have dreamed up - guess it impresses heck out of your customers.

We also use insolation as shown in PV Watts. You only demonstrated that you don't know how to use PV watts, or the meaning of sun hours, which are distinctly different than KWH's. It's not likely that we are friends, as I have never met you. I do dream of working on projects, like the ones we are building. They are generally small, from 25-to 50KW, mostly grid tied, with many solar thermal plants as well. I am happy to say that our 350 projects have extremely high customer satisfaction rates, and much of our current work is federal and municipal projects that we developed, engineered and installed. I founded the company, although now some of our employees own some of it. So why the bitter replies? I just saw that you were posting from Turkey, no disrespect at all was meant.

GA.

"Tilting the panel so that the Sunlight strikes it perpendicularly helps, but all other things being equal, a solar panel in Florida will always outperform a solar panel in North Dakota because of the lower air mass."

not true. A mono crystalline solar panel at ND will easily outperform that panel in Florida, if they are both in full sun, oriented at their longitude, if it is clear. If it is 90 degrees in FL, and 10 degrees in ND. Your math is correct, but you didn't solve the equation, simply listed the theory. Very common these days, especially in politics. If you are talking about annual production, your prediction may be correct, but it is not because of "lower air mass" It is because of fewer sun hours in ND.

Your interpretation is not correct. The wattage rating on a PV panel is derived from standard test conditions, giving the DC output of the panel when sun light of the average spectrum, at the intensity of 1000 watts per square meter, at 75 degrees F, strikes the surface. (This is a 1 in a thousand scenario)

In reality, the intensity of sun is between 0 and 1600 watts per square meter at various times, conditions and locations. The output is related to the conditions, though not linear, in either PV panel brand and type or in ratio of intensity to output, so it is difficult to analyze spot experimentation for broad application or comparison.

I know. Just telling you what the local solar panel dealers are saying. I don't really believe them. Hence my questions hrre.

great elnav. Your original post says that you are comparing what you heard on TV with what you read on the web. I guess this is additional info about local dealers.

I'll try to help, by asking, Are you trying to determine what the solar PV system requirements would be to produce x amount of power at your location, and should you believe someone or try to figure it out yourself?

The Value you refer to is an average over the whole hemisphere surface of the Earth. It is NOT a measure of any place on the surface. Each place will have a different value. It is from the 1 watt per square metre at the poles to the 1,500 watt per square metre at the equator. See http://www.oksolar.com/abctech/world_solar_radiation.pdf

This map gives you some indication of the differing values around the World. If you know a good PV installer they should have a Pyronometer and will have measurements for the real W/M2 value in your area.

Someone mentioned that even when you have the real value you need to know the Air mass as well. Then if you use the old Crystalline PV you have to derate the output when above 25C by -0.05% per degree Centigrade. Thin Film PV does not have these losses and will average 30% more energy per year for the same Wp value. Note: The 25C temperature at STC is surface temperature. Surface temperature is air temperature plus 25C. So if Air temperature is 20C then surface temperature is 45C so derate by 20C at -0.05% per degree in a 20C climate.

Using Thin Film PV in Hot climates is cheaper than cooling down Crystalline PV with cooling systems.

Funny thing - the manufacturers of thin film list a temperature derating for above 25°C

About the same as for silicon

The irradiation value according IPCC is 1360W/m2 or therebout outside the atmosphere. On Earth's surface it is 350W/m2, or therebouts, influenced by many factors, mainly by moisture.

Silicon, being a wide bandgap material, can convert only a certain frequency light perfectly to DC power. Lower frequency heats, only. Higher frequency gets converted at the set rate, the rest is heat. There are multi layer cells taking advantage of the various material's conversions, raising the total conversion efficiency.

Ballpark:

polycrystalline = cheapest = 10%+

monocrystalline = moderate = 20%+

multilayer = expensive = 30%-

for comparison, solar Stirling world record is 31,25%

Solar insolation above the atmosphere = 1357 watts/m2 as I remember the number

On earths surface it varies as noted from nothing to maybe 1000 watts/m2 under perfect conditions when the panel has the correct inclination for the latitude.

Panels are tested (PTC conditions) at a 1000 watts/m2 insolation value - this is one reason panels rarely put out their name plate value.

STC conditions are more correct - that is 800 watts/m2

There are cells Boeing makes that are 43% efficient I believe - I am not interested in paying for them. Cell efficiency only means how much area is required to produce a kW of power and nothing more. If you have ground mount or a very large roof it is silly to pay a premium for efficiency.

The insolation outside of Earth's atmosphere is 1360W/m2 give or take a few for Solar variability. That value is not open to debate or opinions. Mentioning 1600 and some such is arrant nonsense, in particular for surface applications.

On the surface it is a fraction. Air is a filter and an absorbent. At the equator the rays cross less at noon, more evening and morning, hence some additional drop in output. Closer to the poles, the rays have to cross more air masses at low slant angles. Hence further drop in output. That again is not debatable, your wishful thinking nonwithstanding.

Add to it the lower number of hours available sunlight, and the power production closer to the poles is miserable. It only takes highschool knowledge, or plain common sense to follow it.

But, I never stood between a man and his phantasies. Go ahead, and knock yourself out. Invent, what you want, build what you want. Iust spare us the nonsense.

Do you REALLY want to know?!? Get a small cell, maybe the size of a page, and measure. Weep, and report back to us.

The nonsense is that while it is clear you know enough to read Wikipedia, ( I stand corrected, 1430 wsm has been measured) the OP is trying to harvest that insolation at the surface of the PV panel. (That would be here on Earth) Your analysis of the "air" is very nice. It's very informative to know there is some "dropout", due to the time of day and where you live. If it's ok with you, I'd like to borrow a little of that expertise for my next "Phantasy" proposal. I just hope the owners have a little highschool knowledge.

Many years ago, seeking similar information, I phoned the local (Vancouver) airport met office and received the 20-year record for solar radiation in, I seem to remember, Kj/sq. meter/month.

Maybe that would serve your purpose?

Not much help if the data is for Vancouver airport. The places I am interested in are not near airports unfortunately. I have copies of government maps giving general data but none are specific and detailed enough for some indiividual location.

My original question had to do with why a national Geographic article provided data that differeed so much from what I was seeing from other sources. Atmospheric absorption is probably the best explanation. I was aware of the industry standard testing program that seeeks to provide a level playing field and a fair comparison of all manufacturers.