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Solar energy cell question 
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#1
Oct1109, 08:36 PM

P: 138

1. The problem statement, all variables and given/known data
Current solarpanel technology has the efficiency of converting 12% of the solar energy to electricity. Given that on a sunny day in Arizona the average solar radiation energy flux is 1.0 kW/m^2 , and that the area of Arizona is 3.0 × 10^5 km^2 , 1. What percentage of that area needs to be covered with solar panels in order to supply all of the electrical requirements of the United States, estimated at 5 × 10^20 J/yr? (Assume that each day has 12 hours of cloudless daylight, and round your answer to the nearest percent): 2. At the Niagara Falls water is falling an average of 52 m at a rate of r =dm/dt = 1.8 × 10^6 kg/s. If 50% of that potential energy could be converted into electricity, what is the solarpanel area equivalent to the Niagara Falls in km^2 ? (round to one decimal place): 2. Relevant equations 3. The attempt at a solution What I did: We get 1000 W/m^2 per second of input. As we run it only half the time, and the efficiency is 0.12, we get a net power of 0.12*500 W/m^2 = 60W/m^2. A year has 365*86400 seconds, which are roughly 300*10^5 = 3*10^7 seconds. So the power needed by the US is 5*10^20 J/yr = 5*10^20 / (3*10^7) J/s = 1.66*10^13 W. So 166*10^11 W / 60 W/m^2 =approx 3*10^11 m^2 = 3*10^5 km^2 Is it right? How do I convert to % after this? 2. Per second, m*g*h = 1.8*10^6 * 9.81 * 52 * 0.5 Joule =approx 450 MJ (megajoule), Area= 450 MW / 60W/m² = 450/60 km² = 7.5km² 


#2
Oct1109, 08:47 PM

PF Gold
P: 7,120

Looks about right. Simplify divide the area needed by the area of arizona and multiply by 100% for part a).



#3
Oct1109, 09:02 PM

P: 138

So its 100 %?
But if I don't approximate the seconds, i will get a different answer. Which answer should I take? 


#4
Oct1109, 09:10 PM

PF Gold
P: 7,120

Solar energy cell question
No it's [tex]\frac{Area of solar panels}{Area of Arizona} * 100\%[/tex].
I don't follow what you mean by approximating the seconds. 


#5
Oct1109, 09:26 PM

P: 138

In my statement:
A year has 365*86400 seconds, which are roughly 300*10^5 = 3*10^7 seconds. I have approximated to 3 * 10^7 The real value is 31536000 = 3.15 * 10^7 So power needed by US will be : 5*10^20 / (3.15 *10^7) J/s = 1.59 * 10^13 So area = 159*10^11 W / 60 W/m^2 = 2.64 * 10^11 So % will be in this case: 2.64 * 10^11 / (3.0 × 10^5 km^2) *100 = 0.88* 100 = 88% Or if i don't approximate, it will be 100%. So which one should I take? 


#6
Oct1109, 09:37 PM

PF Gold
P: 7,120

Ohh I see what you're saying. Don't approximate the seconds since conversions are exact. 365 days = 3.1536*107 s. Do your calculations with as much accuracy as possible and at the end, use 2 significant figures.



#7
Oct1109, 09:39 PM

P: 138

So is it 88%?



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