Averaged over the entire year, the average solar radiation incident on a horizontal surface in Vancouver is approximately 11 MJ/m2 per day. Due to the average position of the sun in the sky in Vancouver, a surface facing to the south and tilted at 60 degrees from the horizontal will receive even more incident solar radiation: an average of approximately 13.5 MJ/m2 per day.
Average power consumption for a typical household in British Columbia in 2001 was approximately 10,300 kWh per year. Assuming that a solar cell has a 10% efficiency, determine the average roof area that a home would need to cover with solar cells to produce as much power as is consumed. Assume that the solar cells are facing to the south and tilted at 60 degrees from the horizontal.
Express your answer to 3 significant figure
There are approximately 1.6 million homes in British Columbia. Assume that your number from Part A could be used as an estimate for all the homes in British Columbia. What is the total area that would need to be covered in solar cells in order to produce the amount of power consumed by 1.6 million homes? Numbers:
Express your answer to 3 significant figures
Originally posted 2009-08-21 20:45:45.
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Every metre square of 10% efficient solar panels would harvest an average of 1.35MJ/day. Multiply by 365.25 to get energy per year (accounting for leap years) = 492.75MJ per year.
Convert the 10,300kWh to MJ: 1 kWh is a kW running for an hour = 1000W x 3600s = 3.6MJ, so 10,300kWh = 10,300 x 3.6 = 37080MJ.
37080MJ / 492.75 MJ/year = 75.25114m^2 per house.
Multiply by 1.6million to get total: 120401826 m^2
To 3sf = 120 million m^2.
(note that 1 million m^2 is 1km^2, because that’s 1000m x 1000m, so the answer is 120 square kilometres).
Out of interest: 10,300kWh is very heavy power consumption. We live in the UK and use less than 3,000kWh. Do you know if BC has a gas grid!? Ofc, the numbers could have just been made up…