Footage is no problem if the price is right. It all boils down to price. There are so many existing places you can put solar: house roofs, garages, factories, even parking lots:
Cool cars + solar power = dual use at its best.
Posted on 10/30/2009 7:11:10 AM PDT by Reaganesque
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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