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Solar Constant?

  1. Dec 3, 2008 #1
    Hey guys, so I missed like the last week of classes due to me being sick all the time but I got a homework problem that im pretty sure I would have learned if I were in class last week but here it is:

    The area of the continental United States is approximately 3.5 x 10^6 mi^2, and its population is apporximately 305 million people. (a). Estimate the total solar power falling on the continental U.S. The "solar constant????" there (averaged over a 24 hour period) is 180 W/m2. (b) What area would have to be devoted to a solar collectors with a 10 percent conversion efficiency in orderr to get all of people's powers need from solar power? Each person requires, on average, 1300 W of power. (c). What percent of the area of the U.S. would be covered with solar collectors?

    Again, I have no idea where to begin, I spent about an hour or 2 looking online for helpful hints or clues but nothing really came up. Anything that would help would be greatly appreciated.
     
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  3. Dec 3, 2008 #2

    Dick

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    You don't need any special knowledge to do this. Just do it. The problem is telling you you can get 180W (watts) for every square meter you use to collect solar power. If one person needs 1300W, how many square meters do you need for each person? Etc, etc.
     
  4. Dec 3, 2008 #3
    Okay so im guessing you do 1300W/180w to reach the amount of sq meters needed for each person which is 7.22. Im not sure what to do for part A still, but for part b would you just do .10(3500000) then divide by 7.22? Could you please explain it a little more
     
  5. Dec 3, 2008 #4

    mgb_phys

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    For part a - you are given the total area of the US and the power / area. All you need to do is convert the units.
     
  6. Dec 3, 2008 #5

    Dick

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    For (a) you know you get 180W per square meter. To know total power over the US, wouldn't you just find the area of the US in square meters and multiply by 180W?? Can you convert mi^2 to m^2? For part (b) you'd better think about it some more. 0.10 efficiency means of the 180W hitting each square meter, you only get 18W out (180W*0.10).
     
  7. Dec 3, 2008 #6
    Alright I coverted 3500000mi^2 into 9.07m^2 and then did (9.07m^2)(180w) and reached 1632.6w/m^2?

    For part b do you do .10(180w) then multiply that by 1300W and then take 305 million people/(18)(1300)?

    Im still not understanding part b and c
     
  8. Dec 3, 2008 #7

    Dick

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    9.07m^2 is a pretty small area for a country of 305 million people, isn't it? I don't think you are thinking about what these numbers mean. Can you explain how you managed that conversion? And the 180 is watts per square meter, not just watts. And how did you get m^2*w=w/m^2??
     
  9. Dec 3, 2008 #8
    idk im getting really frusturated with this, ive been at this for hours and have gotten no where, IM LOSING MY MIND I HATE THIS!
     
    Last edited: Dec 3, 2008
  10. Dec 3, 2008 #9

    mgb_phys

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    This isn't even physics it's basic arithmatic - it's really nothing to do with solar panels etc - it could just as easily be a question about how many parking spaces do you need.

    The area is 3.5 x 10^6 square miles, a mile is roughly 1600m so a square mile is 1600*1600 m^2,
    so you can easily work out the number of square m.
    To check you answer you can do the conversion in google, enter "3.5 x 10^6 square miles in square m"

    a, Once you have the number of sq m, you know that 180W falls on each of them every day.

    b, If the efficeny is 10% then you only get 18W for each sq m instead of 180W. Each person needs 1300W so how many sq m do you need?

    c, If you multiply the answer in b by the number of people you get the total area needed. In part a you worked out the total area of the USA - so just divide them to get the fraction that needs to be covered.
     
  11. Dec 3, 2008 #10
    let me check this

    i got 9.07x10^12 m^2 for the area in meters squared

    for part a, I multiply 9.07x10^12(180w)
    and get 1.63e15 rounded?

    for part b do I do 1300w/18w?
     
  12. Dec 3, 2008 #11

    mgb_phys

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    Yes, you can check what to do by the units.
    a, You have solar constant = w/m^2 and area= m^2 and you want power =W
    So it must be W = W/m^2 * m^2 = solar constant * area

    b, You want area, m^2 = W / (W / m^2) = W so divide total power by power/solar constant.
     
  13. Dec 3, 2008 #12

    turin

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    1300 W per person! That's ridiculous! No wonder there is an energy crisis. I use, on average, 300 W from the grid.
     
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