Solar Panel Output verse rating

[StuM]

Originally Posted by rom

Right let's be clear about *STC* rating and allow me to correct you : 44-25 = 19*-0.5 = -9.5% makes 90.5W not 95W.
And let's also be clear that this is just an example ! Sunpower specs says 0.29% per °C, which in your example would make a 94.5W ouput.
And let's also be clear that STC also talks about brigthness and AM1.5, not only cell temp, and that AM1.5 especially can make a huge difference : please see post #72 by Barnakiel : http://www.all-science-fair-projects...91/0080-b3.jpg

Is that clear enough ? Is that useful for sailorboy1 ? I doubt it

AM1.5 is a just a way of saying 1.5 times the depth of the atmosphere. What do you mean by it "make a huge difference"? Difference to what?

[rom]

Originally Posted by StuM

AM1.5 is a just a way of saying 1.5 times the depth of the atmosphere. What do you mean by it "make a huge difference"? Difference to what?

All right !! I got got confused and thought AM1.5 implied that panels were rated with a 48° angle to the sun ! how stupid is that !
Thanks Stu ! I think I get it now, and so the OP should receive about 86% of about 1kW/m2 now, which if we let other parameters aside makes a max output of 250W. Correct ?

[StuM]

Originally Posted by rom

All right !! I got got confused and thought AM1.5 implied that panels were rated with a 48° angle to the sun ! how stupid is that !
Thanks Stu ! I think I get it now, and so the OP should receive about 86% of about 1kW/m2 now, which if we let other parameters aside makes a max output of 250W. Correct ?

Don't know where you get 86% from.

The calculation is quite simple. The ratio to sun overhead is simply Sin(Sun Altitude). For 48° that is 74.3%.
(In my earlier diagram, 721/970).

Reducing that by say 5% to compensate for panel temperature makes it about 70% as the potential raw output. Knock a bit more off because the radiation has to get through 50% more atmosphere to reach the panels at that altitude. Take into account all the other inefficiencies in wiring losses, efficiency of solar controller etc, and that would put it closer to 60% overall.

60% of his nominal 290W panels is 174 Watts. The OP said he is getting 170-180 Watts.

As I said in post #12 - he's getting about what is expected for his latitude at this time of year.

[StuM]

Originally Posted by sailorboy1

Still seeing 210-220W from my 290W panel under good conditions

When you were getting the previous 170-180W in Post#1, the Sun's latitude was about 9°S, it's now about 3°S, so your previous noon altitude of 48° is now 54°. The higher it gets, the higher your output will be.

Give it another 3 months and tell us what you see.

[newhaul]

Originally Posted by StuM

When you were getting the previous 170-180W in Post#1, the Sun's latitude was about 9°S, it's now about 3°S, so your previous noon altitude of 48° is now 54°. The higher it gets, the higher your output will be.

Give it another 3 months and tell us what you see.

Heck stu in 3 months I will be near my 100 watt max for my system at 48°north lat. That is if I ever see the sun

[noelex 77]

Originally Posted by StuM

The calculation is quite simple. The ratio to sun overhead is simply Sin(Sun Altitude). For 48° that is 74.3%.
(In my earlier diagram, 721/970).

Hi Stu I enjoy your input on electrical matters, but I don't think the calculation is as simple as you suggest. Sunlight is scattered in the atmosphere so the radiation is not a point source. You are assuming all the power is generated from what is sometimes calłed beam solar radiation. We also need to allow for diffuse solar radiation which has a significant effect on solar production. Diffuse solar radiation is the radiation that does not arrive on a direct path from the sun, but has been scattered by molecules and particles in the atmosphere. There are also other complicating factors.

In practice, the irradiance drops less rapidly when you move away from the equator, than your simple geometry method suggests.

I don't know where Rom obtained the number of 970 W/m2, but the fact that he quotes a date, time and location, suggests this number already takes latitude into account. I dont know if this is correct but it makes sense as the number would be too low otherwise. If so, applying a second correction for the angle of the sun to arrive at your figure of 721 W/m2 is not correct, even if your simple geometry method was accurate, which I don't think it is.

[rom]

Originally Posted by StuM

Don't know where you get 86% from.

The calculation is quite simple. The ratio to sun overhead is simply Sin(Sun Altitude). For 48° that is 74.3%.
(In my earlier diagram, 721/970).

Reducing that by say 5% to compensate for panel temperature makes it about 70% as the potential raw output. Knock a bit more off because the radiation has to get through 50% more atmosphere to reach the panels at that altitude. Take into account all the other inefficiencies in wiring losses, efficiency of solar controller etc, and that would put it closer to 60% overall.

60% of his nominal 290W panels is 174 Watts. The OP said he is getting 170-180 Watts.

As I said in post #12 - he's getting about what is expected for his latitude at this time of year.

hey Stu, thanks again for the explanation and nice graphic however you lost me here ...

From the excellent link you posted (Calculation of Solar Insolation | PVEducation) I gather that *today*, at Palm beach, noon local time, the OP is getting 1023 W/m2.

Now to translate that to an horizontal panel I need to know the angle "z" relative to the normal to the Earth's surface, which is the sum of local latitude and declination (?) of the sun, which is about 3° + 26°, that is 29°. The direct radiation on 1sqm horizontal is then given by cos(z) x 1023 = 0.87 x 1023 = 895 W/m2.

Hence the 86% or 87% I was talking about. Isn't that so ?

Anyway, totally agree with noelex though that using this alone to predict the output of our solar panels is an over simplification, as demonstrated also by barnakiel in a previous post.

[StuM]

Originally Posted by noelex 77

Hi Stu I enjoy your input on electrical matters, but I don't think the calculation is as simple as you suggest. Sunlight is scattered in the atmosphere so the radiation is not a point source. You are assuming all the power is generated from what is sometimes calłed beam solar radiation. We also need to allow for diffuse solar radiation which has a significant effect on solar production. Diffuse solar radiation is the radiation that does not arrive on a direct path from the sun, but has been scattered by molecules and particles in the atmosphere. There are also other complicating factors.
.

You are correct. It is not as simple in reality, but it is near enough to a first approximation. (The relative proportion of direct to diffuse solar radiation becomes very apparent when you look at what happens when just the direct radiation on a panel is blocked by a boom etc, but there is still full diffuse illumination)

It's actually very complex - if you want to get into all the intricate calculations you can find relevant formulae in Sections 1.2 and 1.3 here:

New Daylight Algorithms (Task 21 - Subtask C2)

[rom]

Originally Posted by noelex 77

I don't know where Rom obtained the number of 970 W/m2, but the fact that he quotes a date, time and location, suggests this number already takes latitude into account. I dont know if this is correct but it makes sense as the number would be too low otherwise. If so, applying a second correction for the angle of the sun to arrive at your figure of 721 W/m2 is not correct, even if your simple geometry method was accurate, which I don't think it is.

Hey Noelex, would you be so kind to explain that part ? If we simplify and stick to direct radiation, what would be your calculation ?

[StuM]

Originally Posted by rom

hey Stu, thanks again for the explanation and nice graphic however you lost me here ...

From the excellent link you posted (Calculation of Solar Insolation | PVEducation) I gather that *today*, at Palm beach, noon local time, the OP is getting 1023 W/m2.

Now to translate that to an horizontal panel I need to know the angle "z" relative to the normal to the Earth's surface, which is the sum of local latitude and declination (?) of the sun, which is about 3° + 26°, that is 29°. The direct radiation on 1sqm horizontal is then given by cos(z) x 1023 = 0.87 x 1023 = 895 W/m2.

Hence the 86% or 87% I was talking about. Isn't that so ?

Anyway, totally agree with noelex though that using this alone to predict the output of our solar panels is an over simplification, as demonstrated also by barnakiel in a previous post.

Agree for today. That's why in a recent post the OP is now getting 220W - 230W ( or about 78%) rather than the 170-180W (60%) he was getting.

At the time of Post #1, the sun was 16 ° South, making your calculation 16+26 = 42. COS(42) = 0.743 or 74%

[noelex 77]

Originally Posted by rom

Hey Noelex, would you be so kind to explain that part ? If we simplify and stick to direct radiation, what would be your calculation ?

I would interpret the data you presented as indicating that on the earth's surface at that latitude and time of day the solar radiation, ignoring clouds, would be 970 W/m2. Therefore, this is what would also be falling on a flat solar panel at that latitude. I suspect you are essentially applying a latitude correction to data that already has this factor built in, but it would help to have a link to see exactly what is said. Certainly around 970W/m2 as the peak falling on the earths surface is the sort of number I would expect at that latitude.

I plugged the numbers for Palm Beach in from Stu's link (thanks Stu). I doubt I have the absolute peak time, but you can see the results for a flat panel in Palm Beach at the time and date indicated is:

945 W/m2

This is a long way above the 721 W/m2 that is being suggested.

Therefore, at the OP's location the maximum solar output at the peak time of year from a 290w flat panel should be:

290x945/1000= 274w

This ignores temperature effects, wires losses etc, but at the best time of year the OP should be getting a peak of a lot more than 170-180w. However, this is at the best time of year, not now.

This is important because the absolute peak output from the panels that is recorded I think gives a good indication of the health of the system.

[rom]

Originally Posted by noelex 77

I would interpret the data you presented as indicating that on the earth's surface at that latitude and time of day the solar radiation, ignoring clouds, would be 970 W/m2. Therefore, this is what would also be falling on a flat solar panel at that latitude. I suspect you are essentially applying a latitude correction to data that already has this factor built in, but it would help to have a link to see exactly what is said. Certainly around 970W/m2 as the peak falling on the earths surface is the sort of number I would expect at that latitude.

Now that is interesting and probably where we have a problem. I do not understand why you choose the 20th of june in your example, please look at the difference between these 2 calculations. Both for today, at OP's position, one with horizontal panel and one with a panel titled towards the sun. So I might have missed something but to me the latitude correction is not built in.

[noelex 77]

Originally Posted by rom

Now that is interesting and probably where we have a problem. I do not understand why you choose the 20th of june in your example, please look at the difference between these 2 calculations. Both for today, at OP's position, one with horizontal panel and one with a panel titled towards the sun. So I might have missed something but to me the latitude correction is not built in.

20th of June is the solstice so this is about when the peak output will occur.

Both of your tables are corrected for the latitude of Palm Beach, but one is for a tilted panel and one for a flat panel.

[SV THIRD DAY]

There are very few real life cruiser who's pre-cruise solar "predictions" match his cruise real life reality. What's even more interesting is just how many cruisers don't even have the measuring capabilities to know how many AH per day their solar system is producing so it's all just a WAG or at best a SWAG.

I didn't sharing my AH total averages and solar Wattage to imply something is wrong with my system (there isn't, it's working perfectly) and solicit advice and help...no...I shared it because it real damn life data from a sailboat with things called masts, rigging, shading, floating on the water with the boat and solar array not caring about your angle of panels to the sun. It's hilarious to listen to all the solar experts chime away about the angles this, angles that and give calculations as if the wind and tide care.

This friends is why you can't have too much solar.
If you have more room, put on another panel. Once you have no more room...you are done and have to live with what you get. For all the huff and puff it really IS as simple as that. Who would only put on 2 panels (or whatever size) when they could put on 3? Solar is so cheap these days why in the world do you not max out what can reasonably fit on your boat...it's crazy talk and crazy planning.

Now go back to debating calculations and I'll boil it down to reality:
If you have more room on your Boat for it...then add more Solar leave the calculations and BS for those that never leave the dock.

[john61ct]

There are other contexts where more panels are always possible, limitations other than space then have to kick in, not least budget.