Solar Irradiance -> Watts

quique
quique Solar Expert Posts: 259 ✭✭
in Solar Beginners Corner #1
If my location's solar irradiance average is 4.93 (kWh/m2/day)...

monthly and daily variations notwithstanding, I can get 4.93kWh/m2 each day.

how do i convert this to energy produced by a panel?

If I have a 300W panel, (lets ignore efficiency losses), this means the panel produces 300W of power at STC which includes 1000w/m2.

I mean, if I have 5 hours of sunlight, ok, Ill get 5 x 300W = 1500Wh in a day ,but how does 4.93 vs 4.00 vs 6.00 affect my 300W production of electricity?
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  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    #2
    Re: Solar Irradiance -> Watts

    You can not ignore the efficiency losses because of the great big one: panels themselves are less than 20% efficient. Most poly panels are about 16% and most monos about 18%.

    You want real data: PV Watts provides; http://rredc.nrel.gov/solar/calculators/pvwatts/version1/
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  • Photowhit
    Photowhit Solar Expert Posts: 6,006 ✭✭✭✭✭
    #3
    Re: Solar Irradiance -> Watts

    First are you grid tied? or off grid?

    Grid tied I think you can expect close to 75% of the panel rating x the irradiance as an average for the year.

    Off grid other factors come into play and 52% max depending on how much loss you need in your system for battery health.
    Home system 4000 watt (Evergreen) array standing, with 2 Midnite Classic Lites,  Midnite E-panel, Magnum MS4024, Prosine 1800(now backup) and Exeltech 1100(former backup...lol), 660 ah 24v Forklift battery(now 10 years old). Off grid for 20 years (if I include 8 months on a bicycle).
    - Assorted other systems, pieces and to many panels in the closet to not do more projects.
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  • NorthGuy
    NorthGuy Solar Expert Posts: 1,913 ✭✭
    #4
    Re: Solar Irradiance -> Watts

    (300 W) / (1000 W/m2) x (4.93 kWh/m2/day) = 1.48 Wh/day
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  • quique
    quique Solar Expert Posts: 259 ✭✭
    #5
    Re: Solar Irradiance -> Watts

    Ok so with pvWatts, 1000kW provides whatever the data here says it does.

    For Central America, La Mesa/Pedro Sula:

    So 4kW worth of solar panels (about 13 of these 300w panels), with the 0.77 derate factor, with tilt equal to its latitude, 15.45, produces:

    4,962 kWh per year.

    I guess what I wanted to know is how the solar radiation of say, January (4.61) results in 386 kWh for that month.
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