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March 25, 2017

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When the sun is in the zenith and the sky is perfectly clear, the solar power we receive on the surface of the earth is roughly 1 kW per square meter (on a horizontal surface that is normal to the direction of the sun).

Calculate the average solar power per square meter (in Watts) on a horizontal surface for a day when the sun goes through the zenith at noon.

The sun goes through the zenith exactly twice a year on latitudes that are close to the equator. Take the angle to the sun into account.

  • Physics pls anyone can help me? its urgent - ,

    The incident power is given by

    power=Powermax*sinTheta

    where theta is the angle measured from horizontal East to the sun. At noon, Theta=90 degrees This takes 12 hours, so theta=time*PI/12, where time is in hours, ie, noon is time 6

    avg=total energy /time

    lets measure energy in kw-hr
    total energy= INTEGRal power*dtime from time=0 to 12

    total energy= INT 1kw*sin(PI*t/12 )dt
    = -1kw* (cosPIt/12)*12/PI from zero to 12
    = -12/PI * (cos (PI) -cos0)
    = -12/PI ( -1 -1)=24/PI kw-hr

    average power= total energy/(12)
    = 2/PI kw-hrs

    check this.

  • Physics pls anyone can help me? its urgent - ,

    It should be in watts. so 2/PI kw-hr should be 2*1000/(PI*3600)watts or ? please

  • Physics pls anyone can help me? its urgent - ,

    something got mangled in the units

    kw-hr is energy
    kw is power

    Better check the units in the math above.

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