Calc Power Output of 1 Square Meter Solar Panel on Earth

[guss]

Let's say we have a 1 square meter solar panel on Earth's surface. The sun is shining brightly and it's the middle of the summer. Let's say this solar panel is 100% efficient.

What is the total power output by this solar panel? How would you go about calculating it? Would it matter if it was summer or winter (in the middle of the day)?

We obviously only have to worry about the visible spectrum because that's all that reaches Earth from the Sun.

 

[diazona]

Actually radiation of all wavelengths reaches the Earth from the Sun. A lot of frequencies are blocked by the atmosphere, but still, what reaches the surface is more than just visible light.

According to http://www.grida.no/publications/other/ipcc_tar/?src=/climate/ipcc_tar/wg1/041.htm#121 by the UN Environmental Program, if you average the solar radiation intensity received on the Earth's surface over the entire surface over a year, it comes out to 168 W/m^2. Realistically, I've heard that solar panels run around 5% efficient (constantly increasing, though) so you'd be looking at about 8 watts.

Of course, that's just an average which includes nighttime, cloudy days, etc. when the solar panel would be receiving little or no energy. For a blog post I wrote a while ago I had occasion to calculate that the maximum intensity of solar radiation at the surface is around 1 kW/m^2, basically a factor of about 6 larger. To get that value you can integrate the ASTM reference spectrum with respect to wavelength. That suggests the maximum power output of a 5% efficient solar panel would be around 50 W.

The actual output would vary between 0 and the maximum value, mostly depending on angle of inclination (higher, which means more radiation, at noon and during summer), and atmospheric composition (e.g. cloud cover).

 

[guss]

Actually radiation of all wavelengths reaches the Earth from the Sun. A lot of frequencies are blocked by the atmosphere, but still, what reaches the surface is more than just visible light.

According to http://www.grida.no/publications/other/ipcc_tar/?src=/climate/ipcc_tar/wg1/041.htm#121 by the UN Environmental Program, if you average the solar radiation intensity received on the Earth's surface over the entire surface over a year, it comes out to 168 W/m^2. Realistically, I've heard that solar panels run around 5% efficient (constantly increasing, though) so you'd be looking at about 8 watts.

Of course, that's just an average which includes nighttime, cloudy days, etc. when the solar panel would be receiving little or no energy. For a blog post I wrote a while ago I had occasion to calculate that the maximum intensity of solar radiation at the surface is around 1 kW/m^2, basically a factor of about 6 larger. To get that value you can integrate the ASTM reference spectrum with respect to wavelength. That suggests the maximum power output of a 5% efficient solar panel would be around 50 W.

The actual output would vary between 0 and the maximum value, mostly depending on angle of inclination (higher, which means more radiation, at noon and during summer), and atmospheric composition (e.g. cloud cover).

Thanks, great info. One additional question, do you think you could combine traditional solar panels with the Seebeck Effect to get a higher efficiency? Whenever I touch solar panels, I notice that they are very hot.

 

[Pythagorean]

If you're curious about the solar radiation in a particular geographic location, find the "solar map" for it.

And yes, it varies throughout the day and year and depending on atmosphere (we get an inversion layer in the winter and forest fire smoke in the summer here)

 

[sophiecentaur]

Thanks, great info. One additional question, do you think you could combine traditional solar panels with the Seebeck Effect to get a higher efficiency? Whenever I touch solar panels, I notice that they are very hot.

The seebeck effect isn't very efficient, I believe - thermodynamical efficiency depends on temperature difference, which is very little (although, subjectively, it may seem a lot).
Using the heating effect to produce hot water might be an option but the heat output would almost certainly be a lot less than for a conventional thermal panel and they are only just worth installing. The additional cost would probably militate against it. If you have the area, it would probably be better to instal a separate thermal array.

 

[OmCheeto]

The seebeck effect isn't very efficient, I believe - thermodynamical efficiency depends on temperature difference, which is very little (although, subjectively, it may seem a lot).
Using the heating effect to produce hot water might be an option but the heat output would almost certainly be a lot less than for a conventional thermal panel and they are only just worth installing. The additional cost would probably militate against it. If you have the area, it would probably be better to instal a separate thermal array.

I am tending to agree. At $50/watt, this is 10 to 20 times less cost effective than a solar panel.

The TG12-8 sells for ~$50.
http://www.marlow.com/power-generators/tg12-8-01l.html

And from the way I interpolate the power to thermal graph;
http://www.marlow.com/media/marlow/product/downloads/tg12-8-01l/TG12-8.pdf
you would get around 1 watt max from each unit, when attached to a solar panel.
(50'C Tc vs 100'C Th)

Though I did just buy a pair of TEC1-12706 Thermoelectric Peltier Coolers (12 Volt 92 Watt) about 3 minutes ago. They are only ~$10 each with shipping and handling.

I consider first hand experimental knowledge priceless. Unless of course it comes second hand from someone you trust. That's even better than priceless. (They had to buy all that junk, spend countless hours testing it, interpret the data, and publish their findings. Geesh!)

 

[Antiphon]

The correct answer to the original question is 1000 Watts/square meter.

 

[sophiecentaur]

The correct answer to the original question is 1000 Watts/square meter.

The OP is discussing OUTPUT / INPUT of a 100% efficient solar panel - not the Solar Constant. We don't get 1kW down here, do we?

Also, that graph is interesting. I suspect that the 'really hot' PV panel may be well below 100C (would it fizz if you spat on it?) 60C is reckoned to be scalding temperature..

 

[guss]

There has to be some simple way doing something with that heat to get power out of it. You could build some type of steam engine, I suppose, but that wouldn't really be realistic. I was thinking you could run some metal through the glass in the solar panel horizontally and vertically at 1 foot intervals or so, kind of making a screen (except the "holes" would be large). I'd say the glass there can easily reach 50 celsius/150 Fahrenheit on a moderately sunny day.

 

[sophiecentaur]

Any system you could propose would be a Heat Engine* and would have a maximum efficiency based on the temperature difference between a Hot source and Cold sink. You'd be talking the ratio of 300K and about 330K. Not much use to anyone, I'm afraid. Hot water would be the only useful output. Nice to have but, as I said before, you'd be better off with a dedicated thermal panel.
But, with the available Feed In Tariff, PV gets my vote. 43p a unit !

*This is really fundamental thermodynamics and you can't get round it.

 

[OmCheeto]

The OP is discussing OUTPUT / INPUT of a 100% efficient solar panel - not the Solar Constant. We don't get 1kW down here, do we?

Also, that graph is interesting. I suspect that the 'really hot' PV panel may be well below 100C (would it fizz if you spat on it?) 60C is reckoned to be scalding temperature..

I interpolated Artman's solar array to be close to 100'C last July.

Ah ha! You could almost interpolate the panel temperatures from http://www.schottsolar.com/fileadmin/media/us/data_sheets/SCHOTT%20POLY%20220-235%20Data%20Sheet%20US%200510.pdf" :

30%/0.45% + 25'C = 92'C

Yup. Probably could have fried eggs on them.

 

[Antiphon]

The OP is discussing OUTPUT / INPUT of a 100% efficient solar panel - not the Solar Constant. We don't get 1kW down here, do we?

Yes we do. The output of a 100% efficient solar panel at noon on a sunny day is 1000 Watts.

 

[pallidin]

Yes we do. The output of a 100% efficient solar panel at noon on a sunny day is 1000 Watts.

That sounds about right. The conversion of commercial solar cells per 1-square meter is 0-300watts potential depending on conditions(if I recall correctly) versus the 1000 watt maximum with 100% efficient conversion under super ideal conditions(sunny sky, equator, high noon, solar panel perpendicular, and an as-yet 100% efficient solar panel))

 

[sophiecentaur]

Yes we do. The output of a 100% efficient solar panel at noon on a sunny day is 1000 Watts.

Woops! you're right.

 

[QuantumPion]

How about a solar stirling engine, like http://www.stirlingengines.org.uk/sun/sola3.html" !

25 kw max output, not too shabby. Although the neighbors might complain about the 11 m dish in your back yard...

 

[sophiecentaur]

How about a solar stirling engine, like http://www.stirlingengines.org.uk/sun/sola3.html" !

25 kw max output, not too shabby. Although the neighbors might complain about the 11 m dish in your back yard...

Impressive - but you need to remember (re the OP) that it only works as well as it does because the sunlight is focused onto a small area - thus achieving a high temperature and a consequent high efficiency. On a 'hottish' PV array the temperature situation is very different.

 

[thopsy]

The direct normal irradiance (DNI) on a sunny day is around 1000 W/m^2, though values from 600-900 are more common due atmospheric attenuation. Generally a DNI of 1000 W/m^2 is used as a standard value and is termed "1 sun". The total irradiance can only be intercepted for a 2-axis tracking panel, i.e. the panel is continuously tracked to face the direction of the sun. Otherwise the sunrays will approach at non-normal incidence and there will be some cosine loss.

Yes it does matter if it is summer or winter. There are 3 main contributing factors to the seasonal dependence: 1. greater air mass (distance passed through the atmosphere) in winter, 2. change in spectral distribution as a result in change of atmospheric attenuation, 3. different incident angles on the panel (the 3rd effect is eliminated for a 2-axis tracking panel).

The power output of a solar cell is calculated from:

P = efficiency*I*cos(theta);

where I is the direct normal irradiance (W/m^2), and theta is the incidence angle, i.e. the angle that the sunrays make with the normal vector of the panel.

Typical values for efficiency are around 10% for crystalline-Silicon panels, but newer triple-junction cells have efficiencies surpassing 40% (under concentrated irradiation).

If you wanted to use the waste heat to produce additional electricity, the most probable solution would probably be either a Stirling engine or an organic Rankine engine. Keep in mind that the maximum heat-to-electric efficiency (Carnot efficiency) of such a system is: 1 - TL/TH, where TH is around the cell temperature, and TL is around the ambient temperature both in Kelvin. At TH = 100 C = 373 K, and TL = 25 C = 298 K, this gives a maximum heat engine efficiency of 20%. A realistic value would be around 10%. A thermoelectric (Seebeck) device would also be limited to the Carnot efficiency, but the actual attainable efficiencies would be much lower, I think around a few percent.

Hope this helps.

 

[guss]

Yes, great post, thanks.