Energy captured from incoming solar radiation on PV cells 1

[Tsiolkovsky]

Hi all.

On the following link:

http://lightbucket.wordpress.com/2008/02/24/insolation-and-a-solar-panels-true-power-output/

scrolling down a little bit from the top, gives a table of average daily 24hr insolation. Due to it being a 24hr average, to calculate the energy /m2 over a total day we multiply those values by 24hrs.

So if Athens gets 68 W/m2 daily average insolation for the month of December, we simply multiply that by 24 hrs to get joules/m2 of solar radiation. What I want to know however is:

Can I simply take that energy value and multiply it by the PV total efficiency (say 10%) and that is the energy/metre squared that the PV cells would capture for the month of December?

Theory makes sense but im still skeptical. Something is giving me the feeling that the 10% efficiency is not a constant but a variable. I somehow have a feeling that PV cells efficiency increase under more incoming radiation than low radiation. My ultimate question. If a manufacturer quotes X% efficiency, can that value be used for all conditions or is it a function of other variables. If so, can anybody forward me to online literature of explanations or graphs showing this variable nature/relationship?

Thanks

 

[GregLocock]

PV cells operate at their highest efficincy when cold. One hour after dawn we would see about 40% of peak power at mid day. If there is a fall off at low insolation it is masked to some extent or dominated by the temperature effect.

Typically on a clear day in Oz a panel with a nominal peak power of 1 kW will generate about 7 kWh over the day.

Cheers

Greg Locock


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[Tsiolkovsky]

Do you have any of these efficiency relationships?

Or is it a safe calculation to simply multiply the 24hr isolation by 24 to get the total energy per area that the PV cell can convert (before efficiency losses of coarse)?

 

[GregLocock]

I imagine the data sheet for the cell will give the temperature sensitivity. failing that google is your friend.

The 7 kWh/day/kWpeak number is handy, use that to calibrate your measurements, I'm not going to sift through it to understand your approach.



Cheers

Greg Locock


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[vpl]

Without knowing much about the subject, I would say that simply dividing by 24 is not very accurate. This is because the relationship does not appear to be linear.

Patricia Lougheed

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[racookpe1978]

Typically - and the above "7 KWatt/day for a 1 KWatt peak output" shows just that - you get only 6 hours of effective power output from any solar PV cell over any 24 hour period.

Max will be at local solar noon - when the sun is as high above the horizon as it can possibly get for that day. Every other minute of the day will generate less power, but the power out will vary as the cosine wave of the sun's elevation above the horizon. So, 8:00 to 9:00 will generate wvery little. 9:00 to 10:00 local solar time will generate more (about 75% to 80% of max), 10:00 to 11:00 will generate 85-90% of max, 11:00 through 1:00 will be near 90 to 95 of max. 1:00 to 2:00 will be the same as 10:00 to 11:00, etc. (All of this will be less if you have buildings or trees or natural features to the south. All will vary by time-of-year as the sun's path varies monthly.

From 3:00 PM to 9:00 AM you will generate essentially nothing. So why are you looking at a 24 hour number?

To get your 24-hour value, you'd need to multiply your max daily number (1 KWatt ?) by 4: You need 24 hours of power but only receive that power over a 6 hour period. (Actually for a northern hemisphere winter, you get less than 6 hours per day. Assume between 4.5 to 5 hours in December with even less as you go further north, but between 7 hours to 7.5 hours (maybe) in the summer.)

To deliver 1 KWatt of energy every hour, you need to build your system (wires, fuses, clamps, converters, controllers, etc.) to be able to receive, transmit and store 5x to 6x the average delivered power.

This is because, to store that 1 KWatt of power, you need to compensate for a 70 - 80% conversion loss each way to and from the storage battery. You need to compensate for a 90-95% conversion efficiency from the PV panel low-voltage DC to a useable AC voltage. And all this assumes perfectly clear ("solar max") days with no clouds, no haze, no humidity, no dirty panels, etc.

 

[GregLocock]

"So if Athens gets 68 W/m2 daily average insolation for the month of December, we simply multiply that by 24 hrs to get joules/m2 of solar radiation. What I want to know however is:Can I simply take that energy value and multiply it by the PV total efficiency (say 10%) and that is the energy/metre squared that the PV cells would capture for the month of December?"

Yes, insolation per day = 68*24*3600 J, and at 10% efficincy you'd expect to generate 68*24*0.1*.001 kWh for one sq m of panel, per day.



Cheers

Greg Locock


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[Tsiolkovsky]

Thanks Greg that's exactly what I wanted to know.

racookpe1978, I multiply by 24 because those values posted in the link are 24hr averaged values. That doesn't mean they averaged it over 6 hours peak. they didn't average it around noon they didnt average it from 6am to 6pm. they took the whole 24hrs. Thats why youll notice that the values are low compared to values we normally find on the internet because other sources only average over peak periods. I think the table I linked is therefore most convenient.