1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about solar energy: angle of incidence

  1. Feb 15, 2015 #1
    My understanding is that the seasonal variation in temperature, amongst many other things, is a reflection of the Earth's tilt.

    Simply put, at the equator, sun rays fall "straight down" (100% vertical), and all of their energy hits the surface of the Earth. At different latitudes, these rays begin to have a horizontal component, so only the vertical portion (cosine of angle of incidence, I believe) hits the surface.

    I assume this generalization also applies to solar panels: sun rays falling on the panel completely vertically will have more energy than those falling at an agle.

    What would happen if you tilted your solar panel to match the sun ray's angle of incidence? Would it be just as effective in the middle of winter in Wisconsin (assuming no snow was on it) as in the summer in Ecuador?

    Perhaps solar panels just don't work in the cold, even if high energy sun rays are falling upon it.
     
  2. jcsd
  3. Feb 15, 2015 #2

    russ_watters

    User Avatar

    Staff: Mentor

    Welcome to PF!

    Minor correction first:
    Since the earth's tilt varies, it only falls straight down on the equator (and every other latitude between the tropics) twice a year.
    Correct.
    Yes. And this is done.
    http://en.wikipedia.org/wiki/Solar_tracker#Dual_axis_trackers

    But single axis trackers are much more common than two-axis trackers, since, obviously the sun moves all the way across the sky every day.

    There are two issues this can't get around completely though:
    1. There are fewer hours of sunlight in January in Wisconsin than in Equador in January.
    2. Eventually, the solar panels will shade each other if they aren't far enough apart. So that either reduces the amount of light they collect or increases the area they have to be buildt on.
     
  4. Feb 15, 2015 #3

    davenn

    User Avatar
    Science Advisor
    Gold Member

    Russ beat me to the minor correction :)


    and
    3. even with tilting the panels, the sun's energy at higher latitudes is still dispersed over a wider area than at the equator
    decreasing the efficiency

    sunlight_path.jpg

    cheers
    Dave
     
  5. Feb 15, 2015 #4

    russ_watters

    User Avatar

    Staff: Mentor

    No, that's just #2 repeated in a different - and slightly wrong implying - way. To be explicit: a 1 kW solar panel on the equator and perpendicular to the sun and a 1 kW solar panel in Wisconsin that is also perpendicular to the sun will both produce 1 kW.
     
  6. Feb 16, 2015 #5

    davenn

    User Avatar
    Science Advisor
    Gold Member

    have you some links to that please ? :)
     
  7. Feb 16, 2015 #6

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    This argument applies only if the light hits level ground. Not when it hits panels which are perpendicular to the light. What does affect light intensity though, is the different distance through the atmosphere.
     
    Last edited: Feb 16, 2015
  8. Feb 16, 2015 #7

    OmCheeto

    User Avatar
    Gold Member
    2016 Award

    Solar panels work better when cold. [ref]
    Also, if they are tilted towards the sun, in the northern latitudes, and there is clean snow on the ground, their output will improve again. [ref]
    Snow has a reflectance between 90 & 100% in the wavelengths of 400 - 700 nm. http://www2.hawaii.edu/~jmaurer/albedo/ [Broken]

    According to the following website:

    So I think solar panels would be well suited for Wisconsin. They get lots of snow don't they? I haven't been there since 1978, and only stayed a couple of months.
     
    Last edited by a moderator: May 7, 2017
  9. Feb 16, 2015 #8

    russ_watters

    User Avatar

    Staff: Mentor

    Your photo shows it just fine. The two beams are the same width and contain the same amount of light, with the difference being that the light for the higher latitdue is spread-out over land that isn't perpendicular to the sun. So all you have to do is turn your detector to be perpendicular and it will catch the entire beam.
     
  10. Feb 16, 2015 #9

    davenn

    User Avatar
    Science Advisor
    Gold Member

    yes agreed, and I was assuming ( maybe incorrectly) that that is an important consideration between panels at equatorial regions and at much higher latitudes ?
     
  11. Feb 17, 2015 #10
    Thank you for the very informative responses. Very helpful!

    To follow up on russ and davenn's sidepoint: is the light more "dilute" if it covers more area? Davenn's initial picture showed two illuminated areas, one at an equatorial region, one at a northern (tilted region). Also, each area is demarcated by two yellow lines. If the two yellow lines (at each illuminated area) are evenly spaced, would the light be more "dilute" at the illuminated area where more surface area is covered?
     
  12. Feb 18, 2015 #11

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    If distance through atmosphere is what you meant in post #3, then it was incomprehensible. Instead you were talking about a "wider area", even when the panels are perpendicular to the light rays. That doesn't make any sense.
     
  13. Feb 18, 2015 #12

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    There is only more surface area if the panels are not perpendicular to the light rays. If they are perpendicular, then distance through atmosphere is the only factor that differs between the latitudes.
     
  14. Feb 18, 2015 #13
    AT: I see that now, and agree.

    Is the decrease in energy as the rays go through more atmosphere negligible, or a pertinent factor?
     
  15. Feb 18, 2015 #14

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    I have no data on this. You can test it by orienting a solar panel perpendicularly to the light at different times of day. I assume that there will be quite a difference between midday sun and just before sunset.
     
  16. Feb 18, 2015 #15

    anorlunda

    Staff: Mentor

    A rule of thumb that I keep in my head is that a panel at 45 degrees latitude needs 50% more panel area for the same output as a panel at 25 degrees latitude.

    An Internet search should locate some formulas and graphs to give more accurate estimates.

    I would be surprised to find many solar panel installations above the arctic circle. Perhaps others can tell me I'm wrong about that.

    I do remember that at 65 degrees latitude at noon on midsummer, the Sun was not powerful enough to make my photochromstic glasses turn dark.
     
  17. Feb 18, 2015 #16

    OmCheeto

    User Avatar
    Gold Member
    2016 Award

    Interesting. Where and when did you come up with that rule of thumb?
    (I would do the maths, but, well, see below. Research takes like an hour or two sometimes.)

    And after much searching, I finally found some:

    The relevant equations were listed under "Air Mass" and "Air Mass and Direct Solar Radiation"

    I ran the numbers through my spreadsheet, and came to the following conclusions:

    If the sun is directly overhead, a solar panel will put out 100% of its rated power.
    At the horizon, it is attenuated by a factor of ≈38.
    Attenuation is 50% with the sun at ≈11.5° above the horizon.

    Normally I wouldn't go to the following trouble, but Menomonee Falls Wisconsin is relatively close to my latitude, so it's relevant to me.

    pf.2015.02.18.1052.solar.panel.output.vs.sun.elevation.seasonal.png
    Winter and Summer plots are for the solstices.​

    The sun's elevation vs hour of the day, was derived from a NOAA website, which is where I came up with Menomonee Falls: Solar Position Calculator
    (Never heard of the place. I'm guessing they chose it for the "say this three times fast" factor)
     
    Last edited by a moderator: May 7, 2017
  18. Feb 18, 2015 #17
    Thank you for all that valuable information, OmCheeto!

    So anything below 45 degrees gives limited returns.

    In addition, at location on higher latitudes, sun elevation may even reach >45 degrees, but the sun will likely not spend much time there before diving down to low-yield elevations.
     
  19. Feb 18, 2015 #18

    OmCheeto

    User Avatar
    Gold Member
    2016 Award

    Everything below 90° is by definition "limited".
    I'm actually surprised at how little atmospheric caused attenuation there is over such large angles.
    On the winter solstice in Menomonee Falls, the sun peaks at 23.45° above the horizon, yet the panels still put out 74% of full power.

    Code (Text):
    sun’s
    elevation    
    above         % of full rated
    horizon °     output power
    0             2.15%
    5             25.2%
    10            45.5%
    15            59.0%
    20            68.5%
    25            75.5%
    30            80.8%
    35            85.0%
    40            88.3%
    45            91.0%
    50            93.2%
    55            95.0%
    60            96.4%
    65            97.6%
    70            98.5%
    75            99.2%
    80            99.6%
    85            99.9%
    90           100.0%
    If you want to figure out what happens at higher latitudes, be my guest.
    Barrow Alaska might be a fun one to plot on the Summer solstice, as they get 24 hours of sun.
    My dad lived there for a brief period.

    Code (Text):
    Time          °       Output
    Midnight:     6.11       30%
    6 AM:        15.48       60%
    Noon:        40.49       89%
    6 PM:        29.21       80%

    Of course, they get no sun on the winter solstice. Might be why dad didn't stay long.
     
  20. Feb 21, 2015 #19
    does it take into account the extra power needed to drive the mounting system?
     
  21. Feb 21, 2015 #20

    OmCheeto

    User Avatar
    Gold Member
    2016 Award

    I would imagine that a well designed mounting system would consume a negligible amount of power.
    I think the real world concerns would be the cost of a dual tracking system, and the shielding factor, that Russ mentioned above.

    Though, the alignment system doesn't have to be very accurate, as the panels have to be out of alignment with the sun by more than 15° to reduce output by more than 5%.
    Code (Text):

    misalignment   percent
    angle °        of rated
                   output
    30               86.6%
    25               90.6%
    20               94.0%
    15               96.6%
    10               98.5%
     5               99.6%
     0              100.0%
     
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question about solar energy: angle of incidence
Loading...