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Solar energy cell question

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Current solar-panel technology has the efficiency of converting 12% of the solar energy to electricity. Given that on a sunny day in Arizona the average solar radiation energy flux is 1.0 kW/m^2 , and that the area of Arizona is 3.0 × 10^5 km^2 ,


    1. What percentage of that area needs to be covered with solar panels in order to supply all of the electrical requirements of the United States, estimated at 5 × 10^20 J/yr? (Assume that each day has 12 hours of cloudless daylight, and round your answer to the nearest percent):



    2. At the Niagara Falls water is falling an average of 52 m at a rate of r =dm/dt = 1.8 × 10^6 kg/s. If 50% of that potential energy could be converted into electricity, what is the solar-panel area equivalent to the Niagara Falls in km^2 ? (round to one decimal place):

    2. Relevant equations



    3. The attempt at a solution


    What I did:

    We get 1000 W/m^2 per second of input. As we run it only half the time, and the efficiency is 0.12, we get a net power of 0.12*500 W/m^2 = 60W/m^2.

    A year has 365*86400 seconds, which are roughly 300*10^5 = 3*10^7 seconds. So the power needed by the US is
    5*10^20 J/yr = 5*10^20 / (3*10^7) J/s = 1.66*10^13 W.

    So
    166*10^11 W / 60 W/m^2 =approx 3*10^11 m^2 = 3*10^5 km^2

    Is it right? How do I convert to % after this?


    2. Per second, m*g*h = 1.8*10^6 * 9.81 * 52 * 0.5 Joule =approx 450 MJ (megajoule),

    Area=
    450 MW / 60W/m² = 450/60 km² = 7.5km²
     
  2. jcsd
  3. Oct 11, 2009 #2

    Pengwuino

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    Looks about right. Simplify divide the area needed by the area of arizona and multiply by 100% for part a).
     
  4. Oct 11, 2009 #3
    So its 100 %?
    But if I don't approximate the seconds, i will get a different answer. Which answer should I take?
     
  5. Oct 11, 2009 #4

    Pengwuino

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    No it's [tex]\frac{Area of solar panels}{Area of Arizona} * 100\%[/tex].

    I don't follow what you mean by approximating the seconds.
     
  6. Oct 11, 2009 #5
    In my statement:

    A year has 365*86400 seconds, which are roughly 300*10^5 = 3*10^7 seconds.

    I have approximated to 3 * 10^7
    The real value is 31536000 = 3.15 * 10^7

    So power needed by US will be : 5*10^20 / (3.15 *10^7) J/s = 1.59 * 10^13

    So area =
    159*10^11 W / 60 W/m^2 = 2.64 * 10^11

    So % will be in this case: 2.64 * 10^11 / (3.0 × 10^5 km^2) *100 = 0.88* 100 = 88%

    Or if i don't approximate, it will be 100%. So which one should I take?
     
  7. Oct 11, 2009 #6

    Pengwuino

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    Ohh I see what you're saying. Don't approximate the seconds since conversions are exact. 365 days = 3.1536*107 s. Do your calculations with as much accuracy as possible and at the end, use 2 significant figures.
     
  8. Oct 11, 2009 #7
    So is it 88%?
     
  9. Oct 11, 2009 #8

    Pengwuino

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