# The Truth About Solar Energy?

I have been trying to find some values for how much watts per square meter there is of energy on the ground due to the sun, but everywhere there are different values ranging from 100-1000 Watts so I have no clue of what to believe... And also are these values per second/minute/day???

So however due to all the uncertainties I really appreciate some help to find the answer for this one.

Another thing I also wish to know is how large area of solar cells (let's say they can absorb all the energy that hits them) is required to power the whole north America for a year, also this question has been a struggle to find.

I'm also aware of energy loss due to the atmosphere, clouds and such.

Robin.

## Answers and Replies

tony873004
Gold Member
The Sun's luminosity is 3.8E26 Watts. An imaginary sphere with Earth's distance to the Sun as its radius has a surface area of 4pi r^2 = (4 pi 150000000000 m)^2 = 2.8E23 m^2. So at Earth's distance, these 3.8E26 Watts are spread out over an area of 2.8E23 m^2. So each of these square meters receives 3.8E26/2.8E23 = 1360 W/m^2.

It doesn't make sense to ask if this is per second/minute/day because a Watt already has a per time built into it. 1 Watt = 1 Joule/second. So the correct way to say it would be that each square meter of Earth's surface receives 1360 Joules per second.

Earth's atmosphere will diminish this somewhat, even on the sunniest of days. I've heard it lowers it to about 1000 W/m^2. That's probably where the high value in the figures you quoted came from.

This also assumes that the Sun's rays are striking Earth's surface directly (meaning the Sun is directly overhead). The Sun usually isn't directly overhead. Outside the tropics, it can't be. So you need to add a sin(altitude) to your wattage, where altitude is the degrees above the horizon where the Sun is. So here in San Francisco, the Sun is currently about 30 degrees above the horizon. (1000 W/m^2)*sin(30)=500 W/m^2.

If there's cloud cover, the number obviously goes way down. And at night it is 0.

So the true value can be anywhere in between 0 and 1000 W/m^2. The value of 100 W/m^2 you mention might be an average. The average will tend to be much lower than the maximum possible value of 1000 since the Sun is usually never very close to overhead, it's night half the time, and its cloudy 1/3 - 1/2 of the time.

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Pengwuino
Gold Member
And at night it is 0.

Unless you add in the Moon reflecting the Sun's light :D muahaha!

@OP: When people talk about solar energy, they're going to talk about the numbers when the Sun is directly perpendicular to the PV panel as that's where you're really concerned about power generation. Another reason this is more important is because some systems have trackers that will actually rotate the panels so that they're facing the Sun for longer durations compared to a stationary panel. In that situation, you don't have to worry about the Sun making an angle with your panels and reducing the available energy. As the Sun moves towards the horizon, though, you lose energy due to the atmosphere.

tony873004
Gold Member
...Another thing I also wish to know is how large area of solar cells (let's say they can absorb all the energy that hits them) is required to power the whole north America for a year, also this question has been a struggle to find...

According to Wikipedia, the USA used 26.6E3 TWh in 2008. Per capita that works out to 89000 kWh / year. That seems a bit high to me for just electricity consumption. I'm guessing they're counting all energy consumption, including burning natural gas and gasoline.

For electricity only, let's do a 'back-of-the-envelope' calculation. At any given time, I'm probably consuming about 500 Watts in my home. So I personally use about 12 kWh per day, or 4400 kWh per year. Multiply this by 300 million Americans (assuming I'm typical), and this is 1.3E12 kWh. Each square meter of ground can provide 1 kW for 4400 hours (hours of daylight in a year) = 4400 kWh/m^2

So the number of square meters needed is 1.3e12kWh / 4400 kWh/m^2 = 3E8 m^2. Taking the square root of this and multiplying by 1000 to get kilometers gives us a grid 17 km x 17 km.

That's a best-case scenario. It assumes solar panels are 100% efficient (they're not. 15% is more like it). It also assumes the Sun is always directly overhead, and it is never cloudy. Tracking the solar panels won't do any good unless you increase your array size, since any time you tilt one, you cast a shadow on the one next to it.

I also used the population of the USA, and not North America like you asked. I'll let you adjust for all these factors.

Just curious, is that about what you expected?

Cool! Thanks for the replies. And I don't really know at all what I suspected but I heard from TV recently that a solar cell of just 40 square meters(if the solar cells had an efficiency of 100%) was enough to power while Sweden (I live in Sweden), but I had no clue if that was a reasonable amount of area that was required, it felt a bit small. Thanks though for doing the math I really appreciate it.

By the way is there anything by the laws of physics that makes it impossible to have solar cells with an efficiency of 100%?

tony873004
Gold Member
If you can see it, then it's reflecting some light. That's light that's not being converted. If sunlight warms it up, then that's also a loss.

Regardless, even if 100% were possible, 40 square meters would provide 40 kilowatts. That's enough to power about 25 hairdryers.

Drakkith
Staff Emeritus
By the way is there anything by the laws of physics that makes it impossible to have solar cells with an efficiency of 100%?

Nope. There will always be losses.

Damn... I hope though we will get the efficiency of solar cells above at least 50% (why I said 50% I don't know, but it sounds good! soon the coming years. I think the energy "problem" in the world today is an immense question.

Nope. There will always be losses.

Always, is really no way to avoid losses? :/ I'm completely down with it but I'm just curious, I can understand it's HARD (or maybe even impossible) to avoid making solar cells heat up due to sun light.

Pengwuino
Gold Member
Cool! Thanks for the replies. And I don't really know at all what I suspected but I heard from TV recently that a solar cell of just 40 square meters(if the solar cells had an efficiency of 100%) was enough to power while Sweden (I live in Sweden), but I had no clue if that was a reasonable amount of area that was required, it felt a bit small. Thanks though for doing the math I really appreciate it.

By the way is there anything by the laws of physics that makes it impossible to have solar cells with an efficiency of 100%?

Yes, if it sounds ridiculous and it's science based, the TV is probably lying. Now, 40 square KILOmeters... now you might have something. 40 square kilometers would give you production on the order of a single nuclear power plant with a few reactors. So you're really talking about numbers in the 10's of square kilometers, not 10s of square meters.

Damn... I hope though we will get the efficiency of solar cells above at least 50% (why I said 50% I don't know, but it sounds good! soon the coming years. I think the energy "problem" in the world today is an immense question.

Highly unlikely. I hear most people seem to agree that 25% is probably the highest we'll ever get in terms of economically viable solar panels.

Always, is really no way to avoid losses? :/ I'm completely down with it but I'm just curious, I can understand it's HARD (or maybe even impossible) to avoid making solar cells heat up due to sun light.

You'll lose energy to the cells heating up, reflection of the sunlight (a big problem), the fact that solar cells are unable to capture the whole spectrum that is emitted from the Sun, and a couple others. Not that that's a horrible thing. A 10% improvement in solar cell technology (say from 20% to 22%) is a big thing and couple it with 10% improvements in electricity transmission, energy efficient appliances, improvements in other areas, and it all adds up. Some are easier to do than others, but there's certainly no great need to see vastly increased solar technology efficiencies.

Wow there were much more factors than I thought. Incredible! And yeah you are right Pengwuino, they might have said 40 km^2.

I must say I'm amazed by this conversation, thanks once again for the replies everyone.

Another question, do you think nuclear energy will be our main energy source for at least another 100 years?
Or maybe even fossil fuels are giving more energy than our nuclear reactors today but I think you know what I'm trying to ask, thanks.

Pengwuino
Gold Member
Another question, do you think nuclear energy will be our main energy source for at least another 100 years?
Or maybe even fossil fuels are giving more energy than our nuclear reactors today but I think you know what I'm trying to ask, thanks.

Well, this is a country-dependent question. In the US, nuclear power is not the main source of power whereas it is in countries like Sweden and France. I think in the US we'll slowly move away from Coal and increase our nuclear/solar/wind use until it's a nice mix of various sources and no one dominating source. That's just my opinion though.

Theoretical efficiency of photovoltaics varies quite a bit. Experimental efficiencies have been pushed to more than 40%. In thermodynamic terms the maximum efficiency - the Carnot limit - is (5780-300)/5780 = 94.8% because the Sun is at 5780 K and Earth is typically at about 300 K. Thus we can double or triple the top efficiency of commercial PV cells and still have room for improvement. A new cell design claims a theoretical maximum of 66% efficiency, though the experimental version only gets 8% presently.

Of course since we're at the bottom of the atmosphere we only get ~73% of the Solar constant reaching us. Plus the diurnal cycle reduces the total to just 40% of that figure. Thus solar collectors in space receive about 3.5 times more than collectors on the ground. If one factors in storage efficiency of ~85% for ground power, then space systems can supply 4 times what is made on the ground if averaged over the day. And that's for near-equatorial sites - further North or South of the tropics means a lower average supply per unit area per day. Some countries get 1/6-1/12 of what a surface in space would receive.

Of course space power has the difficulty of getting power to the ground - satellite to grid efficiency is estimated at 50%. Thus a space power collector can deliver twice what a ground facility can deliver - and it needs no storage. It's always on.

So that's one possible solar option.

Cool I had no idea about that but it's sounds amzing!

Imagine if we had the technology to have huge solar cells in orbit that could transfer all the energy that is recieved down to Earth, of course in this idea I imagine solar cells with very huge efficiency compared to solar cells today.
But even to transfer the energy requires energy, I guess?

Ryan_m_b
Staff Emeritus
Cool I had no idea about that but it's sounds amzing!

Imagine if we had the technology to have huge solar cells in orbit that could transfer all the energy that is recieved down to Earth, of course in this idea I imagine solar cells with very huge efficiency compared to solar cells today.
But even to transfer the energy requires energy, I guess?
See this: http://en.wikipedia.org/wiki/Space-based_solar_power though to be honest I've never been convinced that the advantages of putting solar panels in space outweigh the costs. A country might as well use that money to build far more solar panels and accept the loss in efficiency per unit.

Interestingly over the past couple of years in the UK there's been a bit of a gold rush in solar power, the number of homes with solar panels on their roofs have shot up. Indeed just in my local area about 5 houses have got it in the past few months.

See this: http://en.wikipedia.org/wiki/Space-based_solar_power though to be honest I've never been convinced that the advantages of putting solar panels in space outweigh the costs. A country might as well use that money to build far more solar panels and accept the loss in efficiency per unit.

Interestingly over the past couple of years in the UK there's been a bit of a gold rush in solar power, the number of homes with solar panels on their roofs have shot up. Indeed just in my local area about 5 houses have got it in the past few months.

Thanks for the wiki link, gonna save it and read the rest for a better time!

However, I quote now from wiki:

Space-based solar power systems convert sunlight to microwaves outside the atmosphere, avoiding these losses, and the downtime

Microwaves, they pass through the atmosphere without interacting with anything on the way down? => No loss of energy/"signal loss"?

Edit:

And sorry for asking so much questions but it's interesting! :)

Drakkith
Staff Emeritus
Microwaves, they pass through the atmosphere without interacting with anything on the way down? => No loss of energy/"signal loss"?

Edit:

And sorry for asking so much questions but it's interesting! :)

There is a little bit of interaction, but I believe the atmosphere is mostly transparent to them, depending on the wavelength, so most of it would make it through. Still, there will be losses in both converting the power to microwaves and converting the microwaves back to electricity.

Pengwuino
Gold Member
See this: http://en.wikipedia.org/wiki/Space-based_solar_power though to be honest I've never been convinced that the advantages of putting solar panels in space outweigh the costs. A country might as well use that money to build far more solar panels and accept the loss in efficiency per unit.

Interestingly over the past couple of years in the UK there's been a bit of a gold rush in solar power, the number of homes with solar panels on their roofs have shot up. Indeed just in my local area about 5 houses have got it in the past few months.

Without giving it too much thought, I don't understand their claim that the panels could operate 24/7. The Earth would have to get in the way at least some of the time. I'll assume it's not that hard to constantly rotate a satellite given the fact that it's generating power continually, though.

tony873004
Gold Member
Without giving it too much thought, I don't understand their claim that the panels could operate 24/7.
It's easy, you just
...add in the Moon reflecting the Sun's light :D muahaha!

You could also put them in a "walking orbit" that caused them to orbit over Earth's terminator, where the Sun is always visible. Or you could put them in a geosychronous orbit with a small bit of inclination. That way the Sun would always pass above or below the Earth except twice a year when the nodes crossed. Might not exactly be 24/7/365. Call it 24/7/358 (my guess), where the 24 drops to 23.5 on the 7 exception days. It's kinda like lunar eclipses. The Earth only occasionally shadows the Moon.

Cool I had no idea about that but it's sounds amzing!

Imagine if we had the technology to have huge solar cells in orbit that could transfer all the energy that is recieved down to Earth, of course in this idea I imagine solar cells with very huge efficiency compared to solar cells today.
But even to transfer the energy requires energy, I guess?

As you've now read on Wikipedia about half the energy generated by the satellite is lost in the process of getting it to the ground as microwaves. Yet even that factor can be reduced and the efficiency can be pushed higher, at least in theory. Generating and transmitting microwaves is very efficient.

As for the sunlight levels, in geosynchronous orbit the Earth's shadow only shades a solar power satellite for about 1 hour, twice per year. Thus it's almost continuous sunlight.

Without giving it too much thought, I don't understand their claim that the panels could operate 24/7. The Earth would have to get in the way at least some of the time. I'll assume it's not that hard to constantly rotate a satellite given the fact that it's generating power continually, though.

If geosynchronous orbit perfectly shared Earth's axial inclination, then most of the time the power-sats will be in sunlight. Twice per year the Earth's cone of shadow crosses the orbital plain and there's shadowing for about 1 hour. Thus about 8764 hours of the 8766 hours in a Julian Year produce energy from power-sats. With power-sats spread around geosynchronous orbit evenly and it means for that brief period twice a year about 1/24 of the total power supplied drops out.

As for rotating a satellite, with the right configuration you only have to set it spinning once and it will keep rotating the same way, forever. No frictional braking. However there are slight perturbations to account for from solar radiation pressure, which roughly cancels over the course of a year, so slight station-keeping impulses are needed to keep perfectly steady rotation. A proper gravitationally anchored design will keep its transmitter pointed at the ground and its panels forever pointed at the Sun.