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So I know that the Sun's luminosity is $3.839 \cdot 10^{26}$.

What I want is the luminosity per square meter at the Earth's surface. So this is what I have got so far:

I know the distance to the Earth from the Sun is $149.60 \cdot 10^9$ m. So imagining the Sun creates a sphere with a radius of $149.60 \cdot 10^9$ m, the surface area of that sphere would be $2.812 \cdot 10^{23} \mathrm{m}^2$. If I divide the luminosity of the Sun by that surface area, I get 1356 W/m$^2$.

Assuming this is right (if not, please correct me), is this actually the amount of Watts per square meter that would reach a solar panel on Earth under ideal conditions? I didn't account for the atmosphere of the Earth, and the amount of light it blocks - as far as I understand.

So what calculations am I missing here, and how would I go about finding what I want (that is, how many Watts per square meter hit a solar panel on Earth due to the Sun under ideal conditions)?

Thanks!

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I was once assigned this problem for homework and every single person in the class got a different answer; the teacher and the book had different answers still. –  AdamRedwine Oct 12 '12 at 12:03
    
Your answer is quite correct as an order-of-magnitude estimation; after that you do need to worry very much about local conditions. The quantity you're calculating is called the solar constant. –  Emilio Pisanty Oct 12 '12 at 17:16

2 Answers 2

up vote 3 down vote accepted

What will be the proportion of this correct number hitting a 1 meter solar panel will depend on the atmosphere and there is a geographic dependence: clouds (albedo is on average 30% but higher the more northern the country), amount of daylight available, inclination of sun and locality ... .

In Greece the maximum in summer at noon is about 800 out of those watts.

Since the interest is for energy concerns the tables found on the internet are averaged over all these factors over a year, but one can get the relative rates all over from these maps.

Edit Here is a plot of the live measurement of solar radiation in Thessaloniki, Greece. You can see that the maximum is almost at 800watts/m^2 (power). To get the daily energy one has to integrate over the plot.

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Great! I found one for my region, but it is listed in kWh/m^2, not kW (as I had calculated above). So how do I compare the two? Is what I calculated in my question (1356 W/m^2) equivalent to 1356 J/m^2 considering one Joule is a Watt-second and in an ideal situation, we are getting that fraction of the Sun's luminosity per square meter every second? –  capcom Sep 13 '12 at 11:55
    
Essentially, how can I relate the number I retrieved from the map in kWh to 1356 W/m^2? –  capcom Sep 13 '12 at 12:06
    
To go from watts( power) to joules ( energy) you integrate over time the curve of energy coming from the sun which has a sort of Gaussian shape ( really sort of)over the time the sun shines and is zero during the rest 24 hours, and also takes into account albedo etc. There are mathematical models that do this and that is how the map is made, checked against some data at some points. –  anna v Sep 13 '12 at 14:01
    
    
also at this for a simplified formula for the whole earth home.iprimus.com.au/nielsens/solrad.html –  anna v Sep 13 '12 at 14:07

That's exactly right (in the ideal case). Of course, this energy is spread out all over the EM spectrum (with some blocked by the atmosphere), and photovoltaic cells can't make much use of photons that are too low-energy (they don't excite electrons into the conduction band), and they can't fully utilize too-high-energy photons either. There are other inefficiencies too.

I once checked a new solar panel installation at my school, and saw that the ~$1\ \mathrm{m}^2$ panels were rated for (though not necessarily producing) 2 to 3 hundred watts, so there's a good check of your numbers with an actual physical installation.

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