Physics pls anyone can help me? its urgent

posted by .

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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. astrophysics

    1) Every second the Sun outputs 3.90* 10^26 Joules of energy. Power is energy output per unit time and is given in units of Watts (W) where 1 W = 1 Joule per second. The Sun's power output is therefore 3.90*10^26 Watts. This is called …
  2. astrophysics

    1) Every second the Sun outputs 3.90* 10^26 Joules of energy. Power is energy output per unit time and is given in units of Watts (W) where 1 W = 1 Joule per second. The Sun's power output is therefore 3.90*10^26 Watts. This is called …
  3. physics help

    At a certain location, the solar power per unit area reaching the Earth's surface is 180 W/m2 averaged over a 24-hour day. Suppose you live in a solar powered house whose average power requirement is 3.2 kW. At what rate must solar …
  4. physics

    At a certain location, the solar power per unit area reaching the Earth's surface is 180 W/m2 averaged over a 24-hour day. Suppose you live in a solar powered house whose average power requirement is 3.2 kW. At what rate must solar …
  5. chem

    1. The total power used by humans worldwide is approximately 15 TW (terawatts). Sunlight striking the Earth provides 1.336 kW per square meter (assuming no clouds). The surface area of Earth is approximately 197,000,000 square miles. …
  6. Physics

    The sun radiates like a perfect blackbody with an emissivity of exactly 1. (a) Calculate the surface temperature(K) of the sun, given it is a sphere with a 7.00 multiplied by 108 m radius that radiates 3.80 multiplied by 1026 W into …
  7. physics

    The sun radiates like a perfect blackbody with an emissivity of exactly 1. (a) Calculate the surface temperature(K) of the sun, given it is a sphere with a 7.00 multiplied by 108 m radius that radiates 3.80 multiplied by 1026 W into …
  8. Physics pls anyone can help me? its urgent

    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 …
  9. physics

    he sun radiates like a perfect blackbody with an emissivity of exactly 1. (a) Calculate the surface temperature of the sun, given it is a sphere with a 7.00 ✕ 108 m radius that radiates 3.80 ✕ 1026 W into 3 K space. K …
  10. physics

    The power output in the sun is about 4.00 x 10^26W. (a). Calculate the power per unit area (intensity), in kilowatts per square meter, reaching Earth's upper atmosphere from the sun. The radius of the Earth's orbit is 1.5 x 10^11m.

More Similar Questions