Physics pls anyone can help me? its urgent
posted by saru on .
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.

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 kwhr
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 kwhr
average power= total energy/(12)
= 2/PI kwhrs
check this. 
It should be in watts. so 2/PI kwhr should be 2*1000/(PI*3600)watts or ? please

something got mangled in the units
kwhr is energy
kw is power
Better check the units in the math above.