Power Problem Math

1. May 27, 2005

verd

So this isn't a typical power problem... But I'm stuck on a simple mathematical section of the problem. I've gotten most of it right, but am just getting stuck on the last section-- which is worded a bit strangely-- atleast for me.

So here it is:

The total consumption of electrical energy in the United States is about 1.0*10^{19} joules per year.
a.) What is the average rate of electrical energy consumption in watts?
b.) If the population of the United States is 260 million, what is the average rate of electrical energy consumption per person?
c.) The sun transfers energy to the earth by radiation at a rate of approximately 1.0 kW per square meter of surface. If this energy could be collected and converted to electrical energy with 40% efficiency, how great an area (in square kilometers) would be required to collect the electrical energy used by the United States?

Okay, so... I got part a and b correct, with the following answers:
a.) 3.17×1011 W
b.) 1.22 kW/person
And I know these are correct because they were automatically graded-- Now... I'm having difficulty with part c of the problem-- It's just difficiult to understand, for the most part. I'm not quite sure at all what to do.

...I have to use one of the two values, either or, that I got in the first two sections, to yield a result for part c. ...So I take the 1.22 KW/person, multiply that again by 260,000,000, and get 3.172x10^{8}. I need to find the total area that the in km^{2}, that this would cover, given the 40% efficiency rate-- ...So, I multiply 3.172x10^{8} by 2, and then add a half-- then multiply by 1000. And I get the wrong answer. Which I had a feeling I was going to get. Hah.

I'm just severely confused here. Any suggestions?

2. May 27, 2005

FredGarvin

Your problem lies in the efficiency. It is stating that even though 1.0 kW/m^2 is radiated to the Earth, only 40% of that number is useful in this case. That means that .4 kW/m^2 is what you should use as the radiation rate from the sun. So what area do you have to multiply that radiation rate by to get the power requirement of 3.17x10^11 W? You will then have to convert that answer from m^2 to km^2...

3. May 27, 2005

verd

Wow... great. I get it. Yeah-- now that I'm looking at it it, it does seem a bit more obvious, with the percentage and all. Turns out that I was getting the right answer after all-- just wasn't yet converted into km^2

Thank you.