How Much Solar Panel Area is Needed for Household Electricity Needs?

[physicsss]

Another "Easy" Question

The household experiences an average sunlight intensity of 700 W/m2 for 6 hours each day. If the efficiency of converting solar energy to electricity is 12%, what area of photovoltaic panels would be necessary to supply the household’s electricity? A typical household uses 5000 kWh/year of electricity.

P = ηIA
P = power
η= efficiency
I = intensity of sunlight (in W/m2)
A = area of solar collector

How do you account for the fact that there's only light intensity 6 hours each day?

 

[Diane_]

Which question did you want answered?

 

[physicsss]

How do you account for the fact that there's only light 6 hours each day?

 

[Diane_]

I can think of a number of possibilities. Perhaps it's at a latitude and time of year where that's all the sunlight it receives.

 

[Skippy]

I think they mean mathematically...? You'll need to convert seconds to years. When you do that instead of multiplying by 24 (hours in a day) use 6.

 

[physicsss]

tell me if the following is correct:
5000kWh/yr x 3.60E6 J/1 kWh=1.8E10 J/yr

700W/m^2 x (1 J/s)/(1 W) x 3600 s/h x 6h=1.512E7 J/m^2...

pluggin in formula:
1.8E10=0.12 x 1.512E7 J/m^2 x A
solve for A

 

[Skippy]

Unit wise in the last line you have J/yr = J/h.

 

[physicsss]

how do I do it then?

 

[Skippy]

Ok, sorry I said that wrong. For the solar panel you solved how much Power you get from it in one day, but you want to compare it to how much power you use in one year. So either convert days to years or years to days.

In the calculation you did 3600 s/h x 6 h. That should be 3600 s/h x 6 h/d. Since we're calling one day 6 hours (as that's all the light you get in a day).

 

[physicsss]

5000kWh/yr x 3.60E6 J/1 kWh=1.8E10 J/yr

700W/m^2 x (1 J/s)/(1 W) x 3600 s/h x 6h/d x 365 d/yr = number

pluggin in formula:
1.8E10=0.12 x number J/m^2 x A
solve for A

Is it correct now? :)

 

[Skippy]

Looks good to me ;)