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Solar cells project

  1. Feb 22, 2014 #1
    How can I calculate the total diffused and incident light of the sun?Edit
    I am doing this project with solar panels. I thought, watching as well several youtube videos, that I could improve the panels with mirrors, and looking at some plants, I realized that some of their leaves bended towards the wall and the floor. The wall was white and that made me realize that they were receiving the light that bounced off the wall. I did some plant measurements and found out that the radius from the center of the plant to the tip of the leaves decreased with height, and that the angles of the leaves increased with height, so I came out with an arrangement sort of this way :

    (what I was saying about the radius and the angle depending on height, I'm seeing height as increased if it goes down.)

    (attached files)




    The ones that are at the bottom, although they don't have the incident angle of the sun, they have partial reflected rays from the mirror and I thought they would be good as well because they could receive the scattered light for the clouds, so my model would work good in partially cloud days. However, I know that my plant measurements are not going to reveal me the most efficient way for me to settle these solar cells, and there comes the tricky part. I know that the way the rays enter depends on several factors, latitude ( my model would work best in higher latitudes since the reflected rays would have more angle), the tilt of the earth's axis, although I am only focusing for now with the latitude. They say that you should face your panels south ( if you are in northern hemisphere that is), and tilt them depending on the latitude. I need to know how to arrange my panels, so that they have just the right angle and just the right radius, for them to receive the maximum amount of power.

    In order to do such calculations, I know that the pink ray has higher energy because it travels less distance, and so the power would depend on how far the rays are. I know that the angle the ray has would be given by the latitude. I would need to calculate the area that the sun rays hit the earth, with that and the power given by the reflected rays, looking at their behavior, I guess I could then maximize the power received in the arrangement of the solar panels by putting them at certain angles, and at certain distances, but i'm not sure how to get there.

    The other thing, which I see even trickier, is to calculate how would the panels absorb the scattered rays of the clouds, as this is a process more random that depends on the clouds, which are different every day, but well, one issue at a time I guess.

    Thank you for your help
     

    Attached Files:

  2. jcsd
  3. Feb 23, 2014 #2

    etudiant

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    Gold Member

    The plant leaf has better ability to use sunlight than standard solar cells, so diffuse light which is ineffective for solar cells is still quite valuable for plants.
    Your desire to make your solar panels more efficient is good, so it is well worth while to set the cells on the right tilt to maximize their sunlight capture. Because sunlight is a pretty diffuse resource, mirrors and similar structures are marginal economically, they cost a lot to add for the added watts they contribute. Tracking and focusing mechanisms are even more iffy economically. In short, simple is best, just maximize the cell area getting direct sunlight between 9am and 3pm for the most return on your investment.
     
  4. Apr 29, 2014 #3
    I'd argue in favor of inputting a very very lightweight servo with light-level detection so that the time of day could orient the solar cell toward the sunlight's through this device. it might not account for seasonal differences but this could be very useful, especially if utilized for a focusing mechanism in the event one is utilized.
    I wonder if polarity could be advantageous also?
     
  5. Apr 29, 2014 #4

    Bobbywhy

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    Gold Member

    1. I would start this project by selecting the most efficient solar cells possible. In the early days a typical solar cell operated at only 3 or 5 percent efficiency. Today’s modern cells now provide around 30% conversion efficiency.

    2. Then mount your fixed panels at the best angle for your latitude. This gives approximately “one sun” input energy.
    http://homeguides.sfgate.com/figure-correct-angle-solar-panels-79489.html
    http://www.solarpoweristhefuture.com/how-to-figure-correct-angle-for-solar-panels.shtml
    http://www.mysolarpannels.com/optimum-angle-for-solar-panels/

    3. Then consider concentrating the sunlight for more than “one sun”. The panels remain fixed, but solar energy input is increased. Remember, when the sun’s energy is concentrated onto the solar cells they become hot. This heat degrades their performance and excess heat needs to be removed. Also, the cells may be damaged by too much heat. Several methods are available:
    http://www.ioserver.com/fresnelx/
    http://www.acrylite-polymers.com/pr...out/news/Pages/news-details.aspx?newsid=23897
    http://www.solargenix.com/hot_water_products.cfm [Broken]

    4. Tracking systems are complicated and expensive, but gather the most energy. The entire panels are driven during the daylight hours so as to point directly at the sun. You may use Google to search for these systems.

    Cheers, Bobbywhy

    Edit: Welcome to Physics Forums!
     
    Last edited by a moderator: May 6, 2017
  6. May 4, 2014 #5
    Thanks you all!! I did some measurements with the mirrors, and no, they don't add efficiency. So far, I've seen tracking devices are the best, but a way for them to not using energy is necessary so that they can be implemented. What do you mean by polarity ? (1ledzepplin1) I've seen some that work using just some water bottles like the sun saluter from a Princeton girl, you should check that one, once again thanks !!
     
  7. May 4, 2014 #6
    For the motor I was thinking that the additional input would balance out to equal a net gain of energy. It could be calculated If you simply measured the energy yielded in direct perfect sunlight versus the efficiency losses over the day. Then compare the difference between a normal run, with the cell in a single place and orientation, versus a a calculation of max voltage for however many additional hours the motor would yield minus the energy costs of the motor. The setup for this would be very easy too. You could use something like a daylight sensor for a simple electronic device like an outdoor lawn light that comes on at night and then tweak it to provide more sensitive adjustments or better yet just use a lawn sprinkler timer to operate the motor in increments where it could adjust something like every hour to give a generalized tracking of the sun. Could pay off if you did It right I think :)

    But as far as polarization, I'm really no expert on light or solar cell design but I can't help but wonder if there would be some technical advantage that could be employed with the polarity of the incident light. I have an idea, though it's simply conjecture really, that light polarity would influence its infrared presence. I have a polarized lens and I always notice a color difference looking through them. Otherwise another efficiency gain could be to utilize color to mitigate or distribute heat. Idk just possiby relevant conceptsfor ya!
     
  8. May 5, 2014 #7
    They are good ideas, using polarization seems a promising idea for further research, thank you :)
     
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