[Peppersrule]

Hello

I was working on an assignment question and I was wondering if somebody can check my work. The question is:

The best solar panels currently available are about 15% efficient in converting sunlight to electricity. A typical home will use about 40 kWh of electricity per day. Assuming 8 hrs of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home's electricity. (Assume also the sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area)


So far I have:

Energy the home requires per day: (40 kWh * 3600 kJ)/1kWh = 144000 kJ per day

Considering the panel's 15% efficiency, the amount of energy needed for conversion is: 960,000 kJ
So, energy to be converted per hour is: 960,000 kJ / 24 hrs = 40,000 kJ/h

I am unable to proceed from here as I'm unsure of what to do. Your assistance will be much appreciated.

Thanks


 

[Borek]

So, energy to be converted per hour is: 960,000 kJ / 24 hrs = 40,000 kJ/h

No - you will be not converting energy for 24 hours a day, but just for 8 hours a day (which means you have to store it somehow, but it is not part of the problem).

1 m² collects 1 kW - how much energy can it collect per hour? (You will have to assume 100% efficiency here, as you already took care of 15% earlier).