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 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.

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

something got mangled in the units

kw-hr is energy
kw is power

Better check the units in the math above.

To calculate the average solar power per square meter on a horizontal surface for a day when the sun goes through the zenith at noon, we need to take into account the angle to the sun at different times of the day.

First, we need to determine the length of the day when the sun goes through the zenith at noon. At the equator, the length of the day is approximately 12 hours. Since the sun goes through the zenith twice a year, we can assume that each time it goes through the zenith, it occurs for half of the day. Therefore, for each occurrence, the sun goes through the zenith for approximately 6 hours.

Next, we need to consider the angle of the sun's rays with respect to the horizontal surface. When the sun is in the zenith, the angle of the sun's rays is 90 degrees, meaning it is perpendicular to the surface. As the sun moves away from the zenith, the angle of the sun's rays decreases, resulting in less power received per square meter.

To calculate the average solar power per square meter for the entire day, we can estimate the solar power received at different angles by assuming a linear relationship between the angle and the power received. We can use the proportion: average solar power = (power at zenith * time at zenith + power at 0 degrees * time at 0 degrees) / total time.

Given that we receive 1 kW per square meter on a horizontal surface at the zenith, and assuming that at 0 degrees (horizon), we receive almost no power (0 W), we can use these values to calculate the average solar power.

Using our previous estimation of 6 hours at the zenith and 6 hours at 0 degrees, we can calculate the average solar power per square meter for the entire day:

Average solar power = (1 kW * 6 hours + 0 kW * 6 hours) / 12 hours
= (6 kWh + 0 kWh) / 12 hours
= 0.5 kW
= 500 W

Therefore, the average solar power per square meter on a horizontal surface for a day when the sun goes through the zenith at noon is approximately 500 Watts.