Posted by **saru** on Friday, October 18, 2013 at 1:25am.

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 -
**bobpursley**, Friday, October 18, 2013 at 3:56am
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 -
**saru**, Friday, October 18, 2013 at 4:15am
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 -
**Steve**, Friday, October 18, 2013 at 4:28am
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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