Solar Panel Power Output 1

[cllrdoor]

I'm thinking of mounting a solar panel to the top of an ATV to charge an external battery for whatever use. I am trying to model such a scenario by knowing the time of day and direction my solar panel is facing. As I assume the direction of the solar panel will make a difference on the output. With the data, I have been able to calculate the angle of incidence. I found information online describing how to calculate the global aperture irradiance which is a function of the diffuse/direct irradiation, angle of incidence and tilt angle of the aperture (solar panel). I was able to find an hourly average for the diffuse/direct irradiation in W/m^2 from an online database for my location to calculate global irradiation at different times of day. If I have done this correctly this would be the solar energy available over a specified one hour period.

If I have calculated an average of 500W/m^2 over one hour. How would I determine my max solar panel power output, assuming the max peak is approximately 135W or would it be best to determine the size in m^2 to get W. Also, I have to throw in efficiency somewhere in there which is usually between 10-15%. The other thing is, is the thought process I described above correct to some extent? I would appreciate any help. Thanks.

 

[BigInch]

Be careful. The "hourly average" in many data bases is the average output per hour for a specific number of sunlit hours/day that is available over a given month. The data base should also state that number of "average" hours available for each month of the year. Thus, for March, the average useable hours per day might be 8, mabye 9 am to 5 pm, or whatever, thus that is 8 hours for each day in March x the w/m2. That is much easier to calculate on that basis, rather than finding the number of hours for each day of the year and calculating the average radiation available for each of those hours. It is not clear what type of data s contained in your data base. If you get 100 w/m2 at 9 am and 5 pm, but 800 w/m2 between 11:30 pm to 3 pm, the average might be 500 for the useable daylight period of the day, but the actual production each hour would vary considerably.

Continuing with the same example, if at 11:30 you get 800 W/m2, multiply by the area of your cells x the efficiency of the cells x 800 W/m2. That's all.

If you have a 135 W rated panel, that is determined by assuming it is at a 90 deg angle to the incoming sunlight at 1000 W/m2 x efficiency of say 15% x area of the panel. Also note that the efficiency figure is given assuming something like 20 deg C, 68 F, I forget the actual rating temperature, but that efficiency figure also varies with temperature. So obviously the efficiency can drop a lot, if at 3pm the temperature of the panel in full sun has reached 160 F!!! I think that's why southern Germany has relatively good production figures, good sunlight when it's not cloudy, wich may not be much time, but in the mountains the cool temperatures make up for lack of sun during cloudy times. Best actual solar cell production figures are often obtained on a sunny, but very cold winter day without adding up to a great amount because of the relatively few hours of usable daylight on a given winter day.

Temperature is a very important parameter that must be considered whenever your temperatures vary away from 60-80 degrees. Especially important when the cells are on a rooftop in summer as I presume will be the case for when you will be out the most in your RV.

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[BigInch]

I forgot. A given panel, say the one with the 135 W rating, already accounts for the area of the given panel, which might be .9 m2, so area exposed to sunlight does not enter into the calculations when evaluating the output for a given specific panel, ie the 135 W output was calculated by 1000 W/m2 x 0.15 eff x the panel's area of 0.9 m2. The 0.9 m2 was already included when designing the panel to deliver that its rated output.

And, if I'm remembering correctly, a standard radiation value of 1000 W/m2 is used for calculating that rating, so if you have only 500 W/m2 available, that will reduce the actual output to 500/1000, or 50% of 135 W = 67.5 W



"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

 

[IRstuff]

As with most things we buy, performance quotes are written for the best case, so the 135W is indeed the maximum you can expect at high noon in the desert. The maximum calculated energy delivery for the MIL-HDBK-310 hottest day is about 5.5kWh/m^2, given the maximum irradiance of 1120 W/m^2.

That means that your 135W panel would output a peak of about 67W for the irradiance you've calculated, and a maximum of about 335Wh for that day.

TTFN
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