Intensity and power question help

[Zeynaz]

Homework Statement

The solay Intensity in the Netherlands is 1.0e3 W/m^2 at a maximum sunshine and a perpendicular incidence.
On average, about 10% of this peak power can be utilized.
The efficiency of the solar panel is 13%.
An average household in the Netherlands consumes approximately 3.5 kWh of the electrical energy in one year.
Calculate how many m^2 solar panel is required to produce the energy of one household.

Relevant Equations

E=Pt
n= E-useful/E-in (or power)

First, I calculated the power generated by the solar panel in one day. So since the intensity is 1.0e3 watts per 1 m2.
10% of this peak power is utilized so

(1000)(0.10)=P-utilized=100W (per 1 m2)
the solar panel has the efficiency of 13% so of this power only 13% is useful.
P-useful= (100)(0.13)= 13W (or J per second per m2)
Now, since the household uses 3.5e3 kwh in one year. P-useful multiplied by 365 days gives me the Power generated in one year.
(13)(365)= 4745 W or J per second for 1 m2

So i divided the household power by the value i find so i could find how many m2 i needed.
3.5e3 kwh (or kilo Joules)
3.5e6/ 4745 = 737 m^2

However this answer is wrong. I should be getting 31 m^2. Could you help me where i went wrong? or how to find this answer?
thanks!

 

[kuruman]

You have correctly calculated that a household needs 13 W per m². If the house uses 3.5 kWh of energy in one year, what is its consumption of energy in one second? Hint: How many Joules in one kWh or in one 1000 (J/s)*(1 hour)?

 

[Zeynaz]

to be able to get W from Wh (or Joules), i should divide 3500 kWh by 3600 right?
so i get 972 W or (joules per second is used by the house)
since the solar panel produces 13 W per m2 i divide the watts needed by watts produced
971w/13w = 74 m2
the correct answer is 31 m2

 

[kuruman]

to be able to get W from Wh (or Joules), i should divide 3500 kWh by 3600 right?

Not right. You can see what's going on better if you use units after your numbers. The 3600 (I assume) is seconds/hour. Now 3.5 kWh = 3500 Wh = 3500 (Joules/s)*(1 hour). If you divide that by 3600 (seconds/hour) what do you get?

On edit: The yearly energy consumption of 3.5 kWh that you posted should be 3500 kWh.

 

[Zeynaz]

if i divide it like that then i would get Joules/ s^2 * 1hour^2 which not what i want.
so the h in there equals to the number of hours in 1 year? because if i multiply wh with 3600s/1h i would get joules. So to be able to get W only, i would have to divide 3500 kwh by the number of hours in 1 year??

 

[kuruman]

if i divide it like that then i would get Joules/ s^2 * 1hour^2 which not what i want.
so the h in there equals to the number of hours in 1 year? because if i multiply wh with 3600s/1h i would get joules. So to be able to get W only, i would have to divide 3500 kwh by the number of hours in 1 year??

Yes.

 

[Zeynaz]

yes, because when i do that and divide the value i got by 13 i get 31m2. which is the correct answer. Thank you!