A calculate the power per square meter reaching earths upper...

(a) Calculate the power per square meter reaching Earth's upper atmosphere from the Sun. (Take the power output of the Sun to be 4.00×10^26 W.) (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of 1.30 kW/m^2 reaches Earth's surface. Calculate the area in km^2 of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States' energy needs (1.05×10^20 J)? Australia's energy needs (5.4×10^18 J)? China's energy needs (6.3×10^19 J)? (These energy consumption values are from 2006.)

Transcript

00:02 And this problem, we're looking at the sun earth system.

00:05 We're told that the power output of the sun is 4 times 10 to the 26 watts.

00:10 And if we look in the appendix of our textbook, we see that the average distance the earth is away from the sun is 1 .496 times 10 to the 11 meters.

00:21 The first part, we want to calculate the power per square meter reaching earth's atmosphere.

00:26 This is also known as the intensity.

00:29 So i'll note this with i.

00:31 Is the power reaching the atmosphere divided by the area on which it falls, the area of earth facing the sun.

00:44 So not all of this output power from the sun is going to reach the earth.

00:48 It's only going to be the fractional amount of earth's area to this entire sphere at this distance.

00:56 So to help visualize, i'll kind of draw what that sphere looks like.

00:59 You have this massive sphere out here.

01:07 The power that reaches the atmosphere is only going to be the fractional ratio of earth's disk as seen from the sun over its entire projection.

01:20 So we've got power reaching the atmosphere will be the fractional amount of earth's disk pi r squared, so that being the radius of earth over the whole area of this sphere, which is 4 pi d squared.

01:42 And this would be multiplied by p out.

01:46 Now do we need to look up what the radius of the earth is? well, we're looking at intensity here.

01:53 So with this area, that's going to drop out.

02:00 We can see this.

02:04 Sorry about that.

02:08 This, if we plug it in, we have this fractional amount of power.

02:19 And then the area is the area on the area.

02:21 Which that power output is landing on and that is the same disk of earth.

02:30 So that's pi r squared.

02:33 We don't have to worry about the radius of the earth.

02:37 So this intensity is just the power output of the sun divided by 4 pi d squared.

02:46 Here we've got 4 times 10 to the 26 watts.

02:54 4 pi d is 1.

02:57 1 .496 times 10 to the 11 meters.

03:04 And this to three significant figures is 1 ,420 watts per meters squared.

03:13 For the second part, we want to calculate the area of solar collection needed in kilometers squared to achieve 750 megawatts of power output.

03:23 Assuming two different things.

03:25 One is that the efficiency of the solar panel conversion is 2%, and that the maximum intensity, uh, intensity reaching the earth's surface is 1 .3 kilowatts for meter squared.

03:37 So we see that's a little bit less than what we calculated in part a, and that's because some of this energy is being absorbed by the atmosphere.

03:48 So in order to get this power output of 750 megawatts from absorption from solar energy, we need this p, i'll call this p achieve, which is our, 750 megawatt and our efficiency is given in percentage here so this to be our efficiency of this maximum power per square meter landing on earth surface which is said to be 1 .3 kilowatts per meter squared multiplied by the area that we need to achieve this power our output.

04:44 So if we solve for our area, which is what we're interested in here, we see that that is 100 times this p.

04:56 I actually just call this pa.

04:59 So i don't have to write the entire thing.

05:02 And that is divided by our efficiency multiplied by this i max.

05:09 So here we've got 750 megawatts.

05:19 We've got an efficiency of 2%.

05:21 And the max...