Question

Jane wants to be able to power her home with 100% solar power at least part of the time using rooftop-mounted solar panels. Her roof can host 64 m2 of 23% efficient solar panels, and with her installation the peak (mid-day) insolation is about 690 W/m2 in July. a) If her typical daytime power draw - including a running air conditioner - is 5 kW, can her home be fully powered by these panels? What if she chooses to run a 6 kW clothes dryer at the same time as everything else? [2 Marks] b) There may be times when the solar panels will produce more power than is needed in the home. What are Jane’s options for handling the excess power produced? [1 Mark]

Answer 1

Answer a) mid day solar insolation in July = 690 W/m2

Efficiency of panels = 23% ; Area of solar panels = 64 m2 thus solar power produced = 0.23*64*690 W = 10156.8 W = 10.1568 KW > 5KW thus the home could be fully powered by solar power. However when 6 KW clothes dryer then power required = 5+6= 11 KW > 10.1568 KW thus additional source required if she uses air conditioner and clothes dryer at same time.

Answer b) To handle excess power she could use storage batteries or could have an arrangement that excess power is sent to grid of electricity department and she be compensated accordingly.