What total solar cell area would you need to provide energy needs? Mastering Physics problem?

Solar cells convert the energy of incoming light to electric energy; a good quality cell operates at an efficiency of 15%. Each person in the United States uses energy (for lighting, heating, transportation, etc.) at an average rate of 11 kilowatts . Although sunlight varies with season and time of day, solar energy falls on the United States at an average intensity of 200 W/m^2.

Assuming you live in an average location, what total solar-cell area would you need to provide all of your energy needs with energy from the sun?

The correct rate is 11kWhr per day. 11kw would be a horrendous amount of power.(maybe Al Gore uses it at this rate)

I will calc it at your rate for problem sake. (this is not real life by the way, there are lots more variables, length of day, etc)

200W/m^2 * 15% = 30w/meter

11kw / 30w/meter = 366m^2

On a side note I think mastering physics is a crappy program, made for lazy physics teacher that dont ever care about application of knowledge.

Share and Enjoy:
  • Digg
  • del.icio.us
  • Facebook
  • NewsVine
  • Reddit
  • StumbleUpon
  • Google Bookmarks
  • Yahoo! Buzz
  • Twitter
  • Technorati
  • Live
  • LinkedIn
  • MySpace

One Response to “What total solar cell area would you need to provide energy needs? Mastering Physics problem?”

  • ericnutsch:

    The correct rate is 11kWhr per day. 11kw would be a horrendous amount of power.(maybe Al Gore uses it at this rate)

    I will calc it at your rate for problem sake. (this is not real life by the way, there are lots more variables, length of day, etc)

    200W/m^2 * 15% = 30w/meter

    11kw / 30w/meter = 366m^2

    On a side note I think mastering physics is a crappy program, made for lazy physics teacher that dont ever care about application of knowledge.
    References :
    If you want to know the real calculation visit: http://www.aurorapower.net/alternative-energy/solar-electric.aspx

Leave a Reply

Security Code: