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Please Help with my question on insolation and power

  1. Jul 22, 2008 #1
    The average insolation in London in July is approx. 200Wm-2. Calculate:

    The maximum power that could be produced by a solar panel of area 2m2?


    Im realy confused, I dont realy know how to start it, im guessing i might need power=work done/time but I dont know how to work out those numbers.

    Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 22, 2008 #2

    G01

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    You need another formula for power. You can find that formula by considering the units in this problem.

    HINT: insolation=intensity (It has units of intensity.)

    Notice the units of insolation- W/m^2

    Now, what is the unit for power?

    Using this information about the units can you remember a formula for Power involving intensity?
     
  4. Jul 22, 2008 #3
    I think power is measured in Watts so does that mean to work out the power I do:
    W/m^2 x m^2 ??
    which is 200 x 2^2 which is 800w?

    Thats probably completely wrong cause Im so confused

    Thanks for your help...
     
  5. Jul 22, 2008 #4

    G01

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    Your formula is right, but you shouldn't square the 2. The units of area are m^2, the area itself isn't squared in the formula:

    [tex]I=\frac{P}{A}[/tex]

    Therefore, P=IA.

    Your intensity is 200[itex]\frac{W}{m^2}[/itex], and your area is 2[itex]m^2[/itex].

    Your looking for the maximum possible power a solar cell will produce. This will be equal to the total amount of power impinging on it(if only we had solar cells this good!)

    So your answer should be P=IA= 400W, not 800W. Does this make sense?
     
  6. Jul 22, 2008 #5
    so its 200x2 = 400 W?

    Thank you so much for your help, I realy appreciate it
     
  7. Jul 22, 2008 #6

    G01

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    Yes. Do you understand why it is 400 and not 800? Feel free to ask about anything that may be confusing you.
     
    Last edited: Jul 22, 2008
  8. Jul 22, 2008 #7
    Yeah, before I had posted this question I had actually written 400 down as the answer but talked myself out of it because I started to confuse myself. I understand that m^2 is the measurement so i shoudlent square the number 2. Thanks again for your help :)
     
  9. Jul 22, 2008 #8

    G01

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    Anytime.:smile:
     
  10. Jul 22, 2008 #9
    I do actually have one more question, I might be being stupid, but further down there is another question which is:

    The power available if the solar panels are approx 12% efficient.

    Im ok with this Q I think, I did (12/100)x400=48W

    But the next part of the Question:

    Calculate the number of kWh produced in one day.

    I think I need to convert the watts to kilowatts by dividing by 1000. But the thing that is confusing me is whether in the original part of the question which states that the average insolation in July is 200w/m^2 I dont know whether this means the amount that is produced in July overall? Or whether that is per hour/per minute/per day?
     
  11. Jul 22, 2008 #10

    G01

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    48W is correct.

    For the next part, kWh is a unit of energy, so they are asking for the energy produced for one "average" day in July. So, use the "average" insolation they give you. This make sense? (And yes, you convert the power from W to kW by dividing by 1000.)
     
  12. Jul 22, 2008 #11
    Thanks alot again. Im working this out as 48/1000 = 0.048
    Hours in a day =24.
    0.048/24 = 0.002 kWh

    I think this is right but I will probably talk myself out of it again

    Thank u :)
     
  13. Jul 22, 2008 #12

    G01

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    No, but close. Notice your looking for energy, which is equal to power times time, not divided by time.

    So you should multiply by 24, not divide by it.

    You can also tell from the units. The units of your answer are kWh, not kW/h.
     
  14. Jul 23, 2008 #13
    Thanks :)
     
  15. Jul 23, 2008 #14

    G01

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    No problem.:smile:
     
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