Posted on 10/30/2009 7:11:10 AM PDT by Reaganesque
Dye-sensitized cells get a double boost from nanowires and optical fiber.
Dye-sensitized solar cells are flexible and cheap to make, but they tend to be inefficient at converting light into electricity. One way to boost the performance of any solar cell is to increase the surface area available to incoming light. So a group of researchers at Georgia Tech has made dye-sensitized solar cells with a much higher effective surface area by wrapping the cells around optical fibers. These fiber solar cells are six times more efficient than a zinc oxide solar cell with the same surface area, and if they can be built using cheap polymer fibers, they shouldn't be significantly more expensive to make.
The advantage of a fiber-optic solar-cell system over a planar one is that light bounces around inside an optical fiber as it travels along its length, providing more opportunities to interact with the solar cell on its inner surface and producing more current. "For a given real estate, the total area of the cell is higher, and increased surface area means improved light harvesting and more energy," says Max Shtein, an assistant professor of materials science and engineering at the University of Michigan who was not involved with the research.
Fiber-optic solar cells could also be used in ways that aren't possible currently. Zhong Lin Wang, professor of materials science and engineering at Georgia Tech, says fiber solar cells would take up less roof area than planar cells because long lengths of the fibers could be nestled into the walls of a house like electrical wiring.
Dye-sensitized solar cells use dye molecules to absorb light and generate electrons. The Georgia Tech group first removes the cladding from optical fibers and then grows zinc-oxide nanowires along their surface, like bristles on a pipe cleaner. Next, the fibers are treated with dye molecules, which the zinc-oxide structures absorb. The advantage of coating nanowires, rather than a smooth surface, with the dye is that the wires collectively have a very large surface area. The more dye molecules there are over a given area of such a cell, the more light it can absorb, says Wang. The dye-coated fibers are then surrounded by an electrolyte and a metal film that carries electrons off the device. The work is described online in the journal Angewandte Chemie International Edition.
"The question is, can you absorb all the light using a small amount of materials?" says Yi Cui, assistant professor of materials science at Stanford University. Building a nanostructured cell on an optical fiber provides a way to do this by increasing both the surface area covered by the dye and the effective path length of the light, he says. The longer a photon travels through a solar cell, the more opportunities it has to interact and generate an electron.
One potential stumbling block for fiber cells is getting enough light inside them in the first place. Wang's devices only collect light at their tips, so to get enough light into such a solar cell without having to track the sun, smaller fibers might be bundled together. Cui says the tips of the fibers could be made of materials that are very effective at directing light into the fiber. Another way to overcome this problem is to build fiber cells that can absorb light along their entire length, not just at the tips--which Michigan's Shtein is working on. This is tricky, because it means the cells' coatings need to be both electrically conductive and transparent, an unusual combination.
However, Shtein says that fibers that absorb light from the sides offer "an interesting architecture for light capture, because you can distribute the fibers in space in a way that helps you capture more photons more effectively than you can in a planar device." The shallower the angle at which light hits a planar cell, the more light reflects off its surface. But the light reflecting off the curved surface of a fiber at a shallow angle will hit an adjacent fiber. These cells could be designed so that it's not necessary to install them with sun-tracking systems, and they would work on cloudy days when the light is diffuse, Shtein says.
Wang says the next step is to try different materials. So far, he has built the cells on quartz optical fibers, which are relatively expensive. Next he plans to try making the cells using cheaper polymer fibers.
Now this IS cool.
Arrange cheap, but low-efficiency, solar cells around an optical fiber.
Result would be high efficiency on the surface exposed to sunlight.
Now this IS cool.
Arrange cheap, but low-efficiency, solar cells around an optical fiber.
Result would be high efficiency on the surface exposed to sunlight.
When designing optical fibers for telecommunications engineers try to avoid light "bouncing around" inside the fiber because it causes loss in amplitude of the signal beig transmitted down the fiber. Leave it to my homies at my Alma Mater to come up with a way to turn loss into gain.....again.
You are still going to need significant square footage, to get power. The 24/7, day-night mean for the CONUS is 7 Watts per square foot (averages Phoenix and Chicago, winter-summer, etc) so if you want a kilowatt you’ll need 150 square feet, more or less, even with 80% efficiency.
Note also that the comparison is between zinc oxide planar cells and zinc nanowire cells. What’s the comparison between the nanowire cells and Cu-Si cells or crystalline Si cells?
How much would a million-square-foot array of the new technology cost?
As an investor I read these periodic announcements with caution. Go back ten years and look up how many of these incredible breakthroughs actually worked, in practical terms.
And if you go back to 1957, we should have moon colonies, flying cars, robot lawn mowers (we have robot carpet vacs now), solar power, nuclear power too cheap to meter, and so forth.
“Result would be high efficiency on the surface exposed to sunlight.”
That would be the end face of the fiber. For a 1mm dia fiber, the area is small.
Fiber manufacturers have to keep in mind the use of CWDM and DWDM(multiple light wavelengths over a single strand of fiber) when constructing fiber cables. They can’t have lightwaves “bouncing around” inside fiber and have it work.
Fiber manufacturers have to keep in mind the use of CWDM and DWDM(multiple light wavelengths over a single strand of fiber) when constructing fiber cables. They can’t have lightwaves “bouncing around” inside fiber and have it work.
I think you’d need a high loss cable, or a leaky cable, so that the light hits the converter in a short distance. If you pick up your 80% of the incoming in a kilometer of coated cable you’ll go broke building a 100 KW array.
Like the guitar strings on a Hard Rock Casino sign, they are made to have strong side emissions to be visible. Um, the strings are fibers, yano.
There always is, though. There is no such thing as a perfect fiber, or connector or splice. All induce some loss into the system. This loss is measurable and determines the distance between repeaters, among other things.
With this solar power application, though, it might be desirable to scatter the light inside the fiber.....
The physics is obvious, but the practical problem is going to be making the electrical connection(s).
I’ve always heard there was about 1000 watts/square meter of potential solar power at noon on the equater and between 600-800 in the U.S.
This link tends to agree.
http://hypertextbook.com/facts/1998/ManicaPiputbundit.shtml
600 W/M2 is 60W ft2, then average that for a 24 hour cycle, and also from North to South in the CONUS, over all the seasons and account for “average” daytime cloud cover.
The nunber I used, 7 W/ft2, is from an engineering textbook.
Solar proponents like to use Phoenix AZ at noon in July for a benchmark but in Troy NY at 10AM in November the number is a bit different.
Perhaps at noon but you need the average value combined with energy storage to compare to power sources.
Unless you we only going to use the power at noon at the equator.
I think the 7w/ft2 is off by a factor of 5 for areas where large solar power facilities would actually be installed.
His response was helpful, the number was for the average CONUS.
Just thinking out loud here and I apologize as this is not my field of engineering (Civil Engineers build targets), but I would think that you would need to lay the fiber bundle in a small, highly reflective tray similar to a oversize rain gutter so that you can focus the light on to the fiber array.
But I am sure they already thought of this...
Thanks for the map. It looks like the southwest would average 30 watts/ft2.
Portland could be problematic.
I don’t think a fiber accepts much light in from the sides. Fibers are designed to take light from the ends and shoot it down the length.
Going in from the side with a trough reflector, you would not need the FO, but you could do it with a wire or pipe (for cooling).
My math says the scale tops out at 25W/sq ft.
But we are in the same ballpark.
=6800/8/10/3
watts/hours/m2 to ft2 hours of darkness
28.8
Yup ballparks are all close.
There’s always one joker with a calculator. ; )
We could have argued all day and you’ve gone and ruined it.
I didn’t integrate the power curve, though, I just assumed the sun snapped on for 8 hours at one angle. A poor model, overly simplistic and not representative.
So we can continue!
I still have a chance to be RIGHT, (a little tear welling up in the corner of they eye), hear that thackney, I’m coming for ya.
Yes ... but I believe you'd be exposing the ends of the fibers to the sun ... so to some extent that extra square footage would be made up by making deeper solar arrays.
Making it up in volume, as it were.
I see what you did there.
You did not need to integrate for the power curve. The number is already a 24 hour average for the entire year. I was curious why you devided by 8 and then by 3, to get the hourly rate instead of just deviding by 24.
The World Meteorological Organization defines “sunshine” as direct irradiance from the sun measured on the ground of at about 120 watts per square meter. In the case of scattered clouds (cumulus, stratocumulus), the steepness of the transition is high and the irradiance measured from the cloudy sky with a pyrheliometer is generally lower than 80 Watts per square meter.
That means that under the best of circumstances (no clouds and using the entire light spectrum with no loss - an impossibility) a square-meter collector will only power two 60-watt bulbs during noon time.
You can’t make a dollar out of fifteen cents (without government assistance).
I’m not sure why, but you are off by a factor of 10. Please see the link on my first post above.
Ah, I see the problem, you looked up the definition of sunlight which they define as being brighter than the background sky and give a number rather than the energy of direct sunlight which we are discussing.
I think the “average power” number in the chart is for daylight only. It’s too high to be a 24 hr average.
As for the syntax, I try to show my work.
A good use of flat-panel collectors, I agree.
You could even put the long trough type in parking lots.
You could probably use surrounding buildings as reflectors, too, as long as I get the sunglasses concession for NYC!
The article was suggesting that efficiency alone could get significant power, which we know is not the case. The fiber type would need significant depth (and weight) to get an increase in efficiency. If it takes a half-meter to get 15% more out of the panel, then the one meter square panel must be at least a half-meter deep, packed full of glass and Zn nanofibers.
I haven’t looked back at the illustration but I think the nomenclature was kilowatt hours. Which would be an integration of power over 24 hours. Simply deviding by 24 gives you the average kilowatts over 24 hours. Dividing by 10 gives you approximate kilowatts per square foot over 24 hours. Multiple by 1000 should give you watts per square foot averaged over 24 hours.
KWH is an energy term, like Joules, that does not include a time span. I have a laser that puts out 0.9 Joules in 10 nanoseconds, which is 90 megawatts, but only 0.000000025
KWh.
The chart shows how many KWH you get in “a day”, which would be for as long as the sun is shining. It’s the integral of the sunlight curve. The number they show is consistent with a standard crystalline solar array. If they made it the TWA (Time Weighted Average) then they’d divide by the number of hours.
I think we just drove right past each other. The number is kwh for a 24 hour period taking into account factors such as latitude and weather. Details dividing by 24 hours removes the time factor leaving average power measured in kilowatts.
In the southwest, you can expect about 7 kilowatt hours per day per square meter solar input.
Posting from phone is giving me a headache.
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