Question

4. Solar power (15 points). This problem leads you through some basic aspects of solar power. a) (5pt) Imagine a giant sphere with a radius of 1AU surrounding the Sun. What is the surface area of this sphere, in units of m2? b) (5pt) Calculate the solar power passing through each square meter of this imaginary sphere in units of watts/m2 c) (5pt) If we wanted to power all the electric consumption of an average household in Alberta (The average household in Alberta uses about 600 kWh per month) with solar power, what is the (flat) surface area that we would need to cover with solar panels, assuming constant power consumption, perfect efficiency, and the Sun shining for 24 hours a day directly over the panels?

Answer 1

1 AU = 150 times 10^9 m = Radius Area of sphere = 4 pi R^2 = 4 pi times (150 times 10^9)^2 = 2.827 times 10^23 m^2 Ans power output of sun = 3.846 times 10^26 Watt solar intensity = Power/Area = 3.846 times 10^26/2.827 times 10^23 W/m^2 = 1360.2 W/m^2 Ans 600 kwh in one month (requirement) Find requirement per hour. = > 600/30 times 24 = 5/6 KW = 833.3 Watt Area = 833.3 Watt/1360.2 Watt/m^2 = 0.612 m^2 Ans