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Homework Help: Lux and power

  1. Dec 19, 2009 #1
    Is there are way to find out power from lux entering a solar cell?



    I currently have the area of a solar cell 127.2 cm square, and the lux reading 8434 lux


    I have heard of the equation watts per metre square. However that's not the equation for lux right?
     
  2. jcsd
  3. Dec 19, 2009 #2
    Are you trying to compute the efficiency of the solar cell? In short, the answer is no--generally there is not a one to one corrspondence between lux and watts/m-m. This is because lux (based on the lumen, candela) takes into the account the spectral sensitivity of the retina which is most responsive to green light with a wavelength of 555nm. If your source was at this frequency, the conversion is simple: watts/m-m=lux/683

    In theory if you knew the exact composition of the source, you could de-emphasize the emphasis assigned to different wavelengths in coming up with lumens. This would allow for a calculation of watts/m. The http://www.electro-optical.com/eoi_page.asp?h=What Is Visible Radiation?" can be found here.
     
    Last edited by a moderator: Apr 24, 2017
  4. Dec 19, 2009 #3
    Okay here's the problem, my light source is a 100W bulb.
    There's no distance between the solar cell and the bulb (meaning that they are next to each other)
    How could you calculate the wavelength of an incandescent light source?

    I've also heard of this term called efficacy, which is measured in lumen per watt. So could I use this term? Because I have the lumen and the efficacy of a 100W bulb.
     
  5. Dec 19, 2009 #4
    yes, efficacy is the key. It relates how the watts of light power relate to the apparent brighness perceived by the eye. In the above example, using a monochromatic 555 nm source it is 100%.In other words the 683 conversion factor will be made worse by the efficacy--say it is 10% then watts=lux / 6830.

    There is still the issue of geometry in this problem. I guess we could assume some kind of parabolic mirror is used so that the entire light emitted is captured by the cell. Otherwise I'm a little fuzzy on how to use the 100 W number provided. Besides which there are going to be a range of wavelengths, not just one, given off by your incandescent source.
     
  6. Dec 19, 2009 #5
    I am doing this solar cell efficiency for my IB Physics extended essay.
    I can calculate the power out through voltage and current.
    However I have problems in calculating power in from a 100W light bulb that is placed directly next to my solar cell.
    The efficacy number for 100W bulb was from wikipedia, however I'm questioning this source. Should I use it?
     
  7. Dec 19, 2009 #6
    yes it is ok to use the source. You could also make sume assumptions about the distance away, r and know that the area/(4pi r^2) = amount of incident energy on cell/100W.
     
  8. Dec 20, 2009 #7
    where did u get that area/(4pi r^2) equation from?
    I'm finding out how much power goes into the solar cell, so I don't really need to find out incident energy on 100W bulb right?
     
  9. Dec 20, 2009 #8
    Well if the light is coming from a point source in all directions equally--at any distance r there is a ball of light with surface area 4 pi r ^2. The sum of all of the light is 100 watts but only a portion of the sphere strikes the solar cell. Then from a purely proportionate view

    100w/4 pi r^2 = x watts/ area of panel

    It may be helpfulto think of the solar panel as a curved tile that forms a portion of the imaginary spheres shell.
     
  10. Dec 20, 2009 #9
    Thanks for the help, it's really helped me in writing my extended essay!
     
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