# Homework Help: Units question

1. Feb 2, 2008

### frasifrasi

The average American uses electrical energy at the rate of about 1.5 kilowatts . Solar energy reaches Earth's surface at an average rate of about 300 watts on every square meter.What fraction of the United States' land area would have to be covered with solar cells to provide all of our electrical energy? Assume the cells are 20% efficient at converting sunlight to electricity.

this is from the first chapter on units. Don't I need the us's size and population for this? Can anyone give me hints on how to get the answer?

2. Feb 2, 2008

### G01

You know how much electrical power is needed to power the US- 1.5kW per person.

So, we need enough solar cells to collect enough energy from the sun to produce 1.5kW *(US Population).

HINT:

Let's assume we have one square meter of land covered in solar cells. How much power does the light hitting those cells carry? How much of that power is actually converted to electrical power by the solar cells?

Last edited: Feb 2, 2008
3. Feb 2, 2008

### frasifrasi

I convert 1.5 to 1500 W. So, 20% of 300 is 60 W/SQ m. SO, the light converted is 60W.
what is the next step?

4. Feb 2, 2008

### Antenna Guy

I think you do - at least an average population density.

Regards,

Bill

5. Feb 2, 2008

### TVP45

You'r quite right. So, how can you find the population and land area?

6. Feb 2, 2008

### Dick

I think you'll have to look them up if they don't give them to you. You can't compute them.

7. Feb 2, 2008

### TVP45

So, OP, where would you look for this?

8. Feb 2, 2008

### frasifrasi

I looked it up but the calculations got very large.

Basically, I multiplied the population by 1500W to get the total energy needed. Than got 20% of 300W = 60W. Now, I divided the total energy by 60 to get the rerritory needed. then, I divided the total landmass by this number, did I do the right calculations?

9. Feb 2, 2008

### Dick

Sorry, mistaken identity. I thought you were the OP and the mistake seems to be mutual. Sorry again.

10. Feb 2, 2008

### Dick

Those are the right calculations.

11. Feb 2, 2008

### frasifrasi

so, just for confirmation, dividing [big number of people X 1500w]/60W/m will give you the number of meters needed to be covered with solar panels right? I am very trepid when it comes to units.

PS. I aced the calc II final last semester (If any of you remember I was asking about 10 questions a day and getting berated, but it was worth it--I beat ~678 students)--thank you everyone.

12. Feb 2, 2008

### Antenna Guy

Since this is a units class, it might be worth while to point out that using population density (e.g. $\frac{people}{kilometer^2}$ should give the same answer.

$$\frac{person}{area\_available} * \frac{energy}{person} * \frac{area\_used}{energy} = \frac{area\_used}{area\_available}$$

Regards,

Bill

[P.S. "person"=1 "people"=?]

Last edited: Feb 2, 2008
13. Feb 2, 2008

### frasifrasi

I just wanted a direct confirmation of my last question, can anyone please verify it?

As for your suggestion, what is area available (the total landmass?) and which value is "energy"? what about area used?

14. Feb 2, 2008

### Antenna Guy

area_available=total_area/(# of people) <-each person gets the same area

I think Dick already answered your question, but you may want to re-write the process more clearly. I can't tell from what you wrote if you used the same units of area for both landmass and solar panels.

Regards,

Bill

15. Feb 2, 2008

### frasifrasi

I used m^2 for total area and m for solar panels, which cancelled to give me m. is this correct?

16. Feb 2, 2008

### Antenna Guy

no - both should be units of area (m^2)

note: "square meter"=m^2

your answer should have no units - they should all cancel in the case of a %

Regards,

Bill

17. Feb 2, 2008

### frasifrasi

how do I make m --> m^2 ?

regards.

just for the heck of it, would you mind working it out with the set up you proposed?

Population: 301,139,947
Area (Land): 9,161,923 SQ KM

This was an online homework and I already sent out the wrong submission (the right answer was . 06%), I just want to learn how to do this effectively once and for all.

Thank you.

Last edited: Feb 2, 2008
18. Feb 2, 2008

### Antenna Guy

The note I made in my last post should show you how to convert the m you used into the m^2 it should have been. Glance over the problem again to see why.

Since "the right answer" you show appears to be close enough, I'll take your word for having submitted this already. Here it goes:

$$\frac{301139947 person}{9161923000000m^2}*\frac{1000 W}{person}*\frac{m^2}{0.2*300W}=0.00055=0.055\%$$

Do you follow what I did?

Regards,

Bill

19. Feb 2, 2008

### frasifrasi

Yes, I was able ti follow, but did you not use the fact the 1.5 kw is the average usage--I do not see 1500W anywhere in the solution. anyway, thank you for taking the time to help me with this, it is just that this one question was bothering me. Good night.

20. Feb 3, 2008

### Antenna Guy

Good catch - I didn't transcribe the problem accurately, and by some coincidence calculated the "correct" answer you posted. The correct answer using your population/landmass should be 1.5 times as large (.055% *1.5=.082%).

The population density I get from your numbers checks out with a quick google of my own, so I don't see where a "correct" answer of .06% could come from.

Regards,

Bill

21. Feb 3, 2008

### frasifrasi

the next lowest answer was .6%, so i think that would be acceptable.

I am just curious, what is that 1000W/person term you have, I don't know what it means--is that what should have been 1500W?

Thanks and that is my final question!! : )

Last edited: Feb 3, 2008
22. Feb 3, 2008

### Antenna Guy

??

That should be 1500W/person as you pointed out. It is the average usage. For example:

$$\frac{1500W}{person}=\frac{451709920500W}{301139947 person}$$

Total energy over number of people equals average energy.

Regards,

Bill