Silly WATTS Question
SolarSailor
Solar Expert Posts: 49 ✭
Our local weather stations' Solar Radiation readings suggest, for example, that on a given day, we averaged 100 W/M^2 of Solar Radiation. (this suggests January, but 100 was a good round number).
These readings come from SR detector devices planted on a outpost weather station. If the particular instrument is crucial, I can get it.
Anyway, question is whether that 100 W/M^2 is what I can expect to be able to collect, or is that the number I use to apply my particular solar panels' efficiency factor to?
Many solar panels are in the 10-20% efficiency rating, so does that mean I'd could achieve only 10-20 W/M^2 with them under these conditions, or does the 100 W/M^2 apply?
These readings come from SR detector devices planted on a outpost weather station. If the particular instrument is crucial, I can get it.
Anyway, question is whether that 100 W/M^2 is what I can expect to be able to collect, or is that the number I use to apply my particular solar panels' efficiency factor to?
Many solar panels are in the 10-20% efficiency rating, so does that mean I'd could achieve only 10-20 W/M^2 with them under these conditions, or does the 100 W/M^2 apply?
Comments
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Re: Silly WATTS Question
The table of insolation usually specify as equivalent average hours of full sun. This may manifest itself as six to seven hours of partial sun. In winter, a given northern latitude may only average 2-3 hrs. of summation of full 1000 w/m^2 sun.
Average of 100 w/m^2 seems a bit low even for northern U.S., southern Canada latitudes. The tables do not take into account a unique situation like a valley location with a mountain to the south that blocks most of day's sun.
You may check your location from the tables at:
http://rredc.nrel.gov/solar/pubs/redbook/ -
Re: Silly WATTS Question
The readings I'm trying to use are Solar Radiation values provided by a realtime monitoring station. See attachment.
But, back to the question, and per the attached graph, would my 1-SqM collector collect 1031 Watts at 13:55? -
Re: Silly WATTS Question
Believe that in summer time it should be closer to 1000 watts/m2. Right - your graph shows that.
Here in İzmir, today, we saw right at 7000 watts total per m2 for the day according to my Davis weather station. İn March the average was 4230 watts/m2 per day - summer is on the way! -
Re: Silly WATTS Question
The Average is based upon a 24-hr collection period, 5-Minute averages of 1-sec readings. (3,600 readings per hour, 300 readings per 5-minute average, 12 5-min averages per hour)
About half that time, it's totally dark.
The 259.26 W/M^2 on the attachment in my last post would be approx double that rate averaged over only sunlight hours.
But, when figuring total collection, 259.26 x 24 is a lot easier than trying to determine sunup/sundown times for each chart.
....which is beside the point. -
Re: Silly WATTS Question
İ figure (others here are better at this) That about 85% of the available sun is in the usable range. Forget early and late hours. This way İ learn if İ am thinking correctly
So say 260*24= for a total of (useful) 5300 watts/m2/day -
Re: Silly WATTS Question
Trivia:
There's an old rule in photography about colour temperature, and how it is 'off' (colours won't reproduce properly on film) two hours after sunrise and two hours before sunset. This is strangely coincidental with insolation intensity on PV's and how "8 hours of sunshine" ends up being 4 hours of usable solar each day.
In the end, I wouldn't get obsessed with how much radiance is there to capture. Quite a number of factors get in the way of converting that to panel output, and you've really no recourse if your data says they should be putting out "X Watts" and all you get is "y/X Watts". -
Re: Silly WATTS Question
How much is available vs what the expectation might be was the original question.
Though, was hoping for something a little more precise than a couple hours pre/post or 85% rule of thumb.
Actually, it was unclear to me if the 'available' readings were anything close to what the actual capture might be at any moment. That is, do panels reduce avail by its' own efficiency factor (solar panels in the 10-20% efficiency range). So, there would be a large difference.
If a panel is only 20% efficient, does that mean it could capture only 1/5th of available? I think that is more what was originally asked. -
Re: Silly WATTS Question
Any two days in a row with the same irradiance can have quite different results in actual panel output. Changes in ambient temperature will have a big effect, as will humidity.
In short, trying to predict results from W/M^2 is picking at nits in my opinion. Might be interesting from a technical point of view, but is it really a practical thing to do? -
Re: Silly WATTS Question
That's kind of the question. What can one expect to achieve?
As for relevance, was hoping it a means to pre-plan a site.
Real, long-term availability produces real averages.
Probably the same means used to create the redbook charts.
Wasn't trying to build expectation from one reading.
The particular question was more related to a general relation of panel to availability. If 100W/M^2 is avail, will a 1M^2 panel produce something near that,
or does it's own efficiency depleat that by that factor?
Could I expect something close to 100W, or is it going to be more like 20W?
That's a big difference.
Maybe I'm not stating it correctly. -
Re: Silly WATTS QuestionSolarSailor wrote: »The particular question was more related to a general relation of panel to availability. If 100W/M^2 is avail, will a 1M^2 panel produce something near that, or does it's own efficiency deplete that by that factor?
Could I expect something close to 100W, or is it going to be more like 20W?
That's a big difference.
First, as far as I know (unless your are using different units)--"Full Sun" is called 1,000 Watts per square meter--not 100 W/M^2...
You can can look at this generic data sheet which has a few extra graphs that not many other data sheets do not usually have:
data sheet for a crystalline silicon panel
So, a 1 meter square panel in full sun will generate somewhere between ~8% and 18% of that 1,000 watts/sq.mtr. depending on lots of things... For example amorphous panels (the "cheaper" kind) are closer to 8% and the more expensive crystalline panels (poly/mono) are closer to 12-18% of that value (80 watts to 180 watt output).
And then you have the temperature issues (hot panels produce less power). And controller types (PWM and MPPT have different power conversion efficiencies and operational parameters).
And for off-grid systems, the state of the charge of the battery bank also will limit the overall power collected (you don't dump 100% of panel power into already fully charged batteries).
I can give you some examples of snips of power generated from my 3.5kWatts of solar panels and a Grid Tied inverter:Date..........kWHours per day
04/18/08.........17.307
04/19/08.........18.237
04/20/08.........19.534
04/21/08.........18.675
04/22/08.........12.866
04/23/08.........12.915
04/24/08.........19.380
04/25/08.........19.018
04/26/08.........18.750
04/27/08.........19.015
04/28/08.........17.451
04/29/08.........18.271
04/30/08.........19.155
05/01/08.........19.400
05/02/08.........15.595
05/03/08.........18.405
05/04/08.........15.325
05/05/08.........19.357
05/06/08.........19.022
05/07/08.........18.917
05/08/08.........19.144
05/09/08.........18.856
05/10/08.........18.534
05/11/08.........17.042
05/12/08.........18.908
12/23/07.........11.489
12/24/07.........11.709
12/25/07.........12.447
12/26/07.........12.491
12/27/07.........5.509
12/28/07.........1.312
12/29/07.........1.688
12/30/07.........9.544
12/31/07.........12.617
01/01/08.........11.383
01/02/08.........10.916
01/03/08.........1.260
01/04/08.........0.297
01/05/08.........1.990
01/06/08.........5.061
01/07/08.........9.662
01/08/08.........2.477
01/09/08.........2.201
01/10/08.........1.354
01/11/08.........3.444
01/12/08.........10.623
01/13/08.........12.674
-BillNear San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset -
Re: Silly WATTS Question
That does answer the question. Thanks.
The 100W/M^2, stated in the original post, was a 24-hr day average Watts/Hour (NOT Watt-Hour) availability for a Jan date. So, 2,400W total day available, as an hourly average.
What that really means is yet to be determined.
If realtime weather monitoring can't be used for something, why do it? -
Re: Silly WATTS Question
To be clear on units... Watt is already a rate (like miles per hour). Watts/Hour is meaningless...
Watts = rate (like miles per hour or miles per gallon)
Watt*Hours = actual work being done (like miles, gallons, etc.)
It is confusing at first--And I have made my share of Watts/Watt*Hour unit mistakes in my life too.
Typically, we talk about "Hours of Sun" or--Hours of "full sun" which is 1,000 Watts per square meter...
So, when we say 2 hours of sun in winter and 5-7 hours of sun in summer--we really mean 2-7 hours of "noon time sun" at 1,000 Watts per sq.mtr.
Averaging the "dark time" into average watts per sq. meter is sort of not helpful for designing a solar system... You need to know the" peak power" (to size equipment) and you need to know "how long of sun" to size the system to the load.
Most of the capture is between 9am and 3pm (6 hours)... My GT inverter will run upwards of 12-13 hours a day---but much of that is beginning and ending of the day. That is lots of 0-100 watt run time into the grid--not very much compared to its 3,000 watt peak rating (less than 3% of peak power). That residual power is almost within the margin of error of 1-2% accuracy for a utility billing meter.
In the end, why are you doing the weather monitoring? I have a 3.5 kW "weather monitor" already on my roof... And I am watching for large failures (open string, failed inverter)--Those are 50% to 100% energy loss signals.
I am sure that I could never measure/prove a 10% energy loss in my system... The various sources for error and variability is just too much to expect more accurate numbers.
If I was looking to monitor my system more accurately--I would install an extra solar panel (ideally right in the middle of my array) and monitor its performance and temperature and compare the rest of the system to it.
And even then, I have choice to make in monitoring... I can monitor Amp*Hours pretty easily and accurately (just short the pilot panel with a current shunt)--And that will give me accurate solar irradiation numbers.
However, a significant amount of power variation (5-25%) is based on the panel temperature and Vmp output voltage. So, I would have to monitor the panel temperature and scale Vmp into the P=V*I*Time equation, or get an Enphase like GT inverter and use it to provide a "temperature compensated" MPPT measurement value to compare with the rest of my system... Not worth it to me.
A large system, industrial sized installation--Those type of monitoring options make sense.
-BillNear San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset -
Re: Silly WATTS Question
I am reminded of the Dylan lyric "You don't need a weather man to tell you which way the wind blows!"
It seems pretty simple in the real world. Define the problem (loads) create a solution that works in the real world. Real world estimates from solar calculators are fairly good places to find the ball park. Once in the ball park, look at your local conditions. For example, my situation is my panels get sun very early, from sunrise, but they lose it early because of my local anomaly. I also have considerable reflection off the water in he summer, and off of snow in the winter. So PV watts would give me a general idea of what to expect, it wouldn't cover those anomalies.
So if PV watts suggest that you can harvest say 1kwh/day with X sized system, but you know that you have a bit better site than average then you could downsize a bit. Or vise-versa.
It is not a strictly calcuable equation that works in every instance. As Bill suggests, there are daily and (locally) regional variations to everything. Simple rule of thumb,, build a little bigger than you think you need, with perhaps a bit of head room to grow. If your calcs turn out too pessimistic then you are nicely surprised. If you are a bit short, add on.
The other simple rule, is that people more often than not, OVER estimate the amount of Pv they can harvest on average. More days have some adverse effect than not, and a little here and a little there and you end up short.
Tony -
Re: Silly WATTS Question
Since the sun shines the way the sun shines and you're going to get what you're going to get ...
You might want to compare the physical size of the panels to their rated output. This would give you an idea of how efficient one panel is over another. You won't see too much difference, because they're all made of silicon in pretty much the same fashion. You're looking at 16-20% efficiency.
For all our numbers and data and careful calculations, the real world still manages to be a crap shoot.
-
Re: Silly WATTS Question
While I can see, and agree with, the comments on the 'slop' associated with actual vs reality on collection, there are reasons to at least start from what is known. First, and foremost, one would never really know how well the panels are performing if there weren't a basis for evaluation.
The monitoring is close enough to be considered 'local', though not exactly 'on-site', so on-site conditions could affect results. Those can be factored in. For example, shade for two hours during a particular time. Of course, if it from 8-10am, it's a different deal than from 1-2pm.
Anyway, the original question was what to expect of panels in relation to available sun, and that was answered.
btw, I agree also that watts is a rate. So is MPH. But, the difference between MPH and MPH/hour avg is 110 miles at 55MPH over two hours. Same relation exists with watts. So, Watts/Hour is relevant, and even can be translated directly to Watt-Hours on occaision.
But, then, I'm kind of new to this. So, I'm also trying to respect the protocols. -
Re: Silly WATTS Question
This is actually quite a good discussion.
The whole Watt/hours thing is a matter of bad English, I'm afraid. A Watt is a rate, yes. But when people write "Watts per hour" half the time they mean Watt hours. Half the time ... who knows what they mean. Semantics. Grammar. Think "it's" is the possessive form. Misspelling. Discarding "affect" and "effect" for the incorrect "impact" (as noun, verb, adverb, possibly personal pronoun the way things are going). Not to sound like a teacher, but quite often misunderstandings and mistakes are made as a result of such miscommunication. Got to get the nomenclature right!
But to get back to the interesting bit. As an experiment I "ran the numbers" on my array. It is approximately 3 square meters, so that would mean a possible output of 3000 Watts based on irradiance. But panels are only about 18% efficient, right? And 18% of 3000 is ... 540. Guess what? I can in fact count on 540 Watts from the panels on a sunny day. In fact it's usually about 560, which would be 18.6% efficiency. Not bad, eh?
Keep in mind I've got some things going for me: higher than average elevation, clearer skies (low pollution), and generally moderate to cool to downright cold temps.
This brings us to Two Days In November. Roughly the same irradiance, but on the first day the temp is about 10C (50F). It's clear, and the temperature drops overnight. The next day it's -10C (14F) as a result. Guess what happens to the panel output? And yes we really do get those wild temp swings up here. Also our days are 16 hours in July and 6 in December, with the sun angle shifting drastically.
So yes you can measure the area of a panel or panels and compare its rating to the maximum possible output. But without all the site specific derating factors you still have just an arbitrary number. And the amount of variation due to local conditions will make such a number rather useless in my opinion. -
Re: Silly WATTS Question
Actually, it is:SolarSailor wrote: »btw, I agree also that watts is a rate. So is MPH. But, the difference between MPH and MPH/hour avg is 110 miles at 55MPH over two hours. Same relation exists with watts. So, Watts/Hour is relevant, and even can be translated directly to Watt-Hours on occaision.- 55 MPH * 2 hours = 110 miles
- D = starting point + V*T + 1/2 * A * T^2
So, Watts/Hour is a mixed unit (seconds and hours) and almost never makes sense. The proper units would be Watts/second or Joules/sec^2 (would be a rate of change or 2nd derivative of work).
Watt*Hours is a more useful number because it is more human scale... A typical home may use between 100 and 10,000 Watt*Hours...
If we used the real SI units, we would be using Watt*Seconds and the 300 kWhr per month power bill would be x3,600 seconds/hour or 1,080,000 WS per month--kind of an absurd number.But, then, I'm kind of new to this. So, I'm also trying to respect the protocols.
No protocols to worry about here--We try to be helpful and "talk the same language" to ensure that we are talking about the same thing.
Every industry has its technical terms and in-exact usage can cause lots of confusion... That is why we are always asking questions and trying to clarify--So as to reduce errors and frustrations in our communications.
-BillNear San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
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