How much energy per person?

sonhouse
Posers and Puzzles 07 Mar '07 16:12
1. sonhouse
Fast and Curious
07 Mar '07 16:12
It's way in the future. I am reading a story called "Pandora's Star"
and in it there is this piece about the earth going green energy-wise by having a belt of solar cells all the way round the moon and shipping the energy to earth. So if the belt of solar cells is 800 Km wide, and we assume half the available energy actually makes it to earth and the population in that era is 10 billion, how much energy per person would be available? Since the moon turns round the sun once in 28 days, it means only a certain percentage of the cells are illuminated, so what would be that percentage, where 100% would be as if all the cells in the equatorial belt of cells were illuminated at once. Then figure out how much gets to earth.
2. 07 Mar '07 18:54
-the watts per area supplied by sunlight on the moon
-the efficiency of the solar cells in converting light to energy
-the diameter of the moon

I don't think the length of time the moon takes to complete an orbit matters.

Let's say
D = the diameter of the moon in meters
W = energy in watts per square meter of sunlight on the surface of the moon
E = efficiency of solar cells (0 = no energy is collected, 1 = perfect efficiency at converting sunlight to energy)

The total area of solar cells that will be illuminated is D * 800000 m^2. The total energy collected is then E * W * D * 800000 watts. Only half of this energy makes it to earth: divide by 2 to get E * W * D * 400000 watts. Divide by 10,000,000,000 people to get

0.00004 * E * W * D watts per person. We still need to find E, W, and D.

The moon's diameter is about 3476000 meters.

According to Wikipedia the efficiency of today's standard solar cells is about 15%.

Also according to Wikipedia, solar energy at the equator of the Earth is about 1000 W/m^2. On the moon there is no atmosphere, so the energy density will be higher, let's say 2000 W/m^2.

0.00004 * .15 * 2000 * 3476000 = about 42 kilowatts per person.

I think that must be pretty low: think about all the energy used by industry and cars and such and consider that that is just 420 100-watt lightbulbs per person...
3. sonhouse
Fast and Curious
07 Mar '07 20:052 edits
Originally posted by GregM
-the watts per area supplied by sunlight on the moon
-the efficiency of the solar cells in converting light to energy
-the diameter of the moon

I don't think the length of time the moon takes to complete an orbit matters.

Let's say
D = the diameter of the moon in meters
W = energy in watts per stry and cars and such and consider that that is just 420 100-watt lightbulbs per person...
I mentioned the total efficiency, 50%, thats energy to the home, so it covers all the inefficiencies in the system. You can google for info about the incident radiation reaching the moon.
Did you correct for the fact that at any given point on the moon, only half is illuminated and of that half, only so much of it will be fully illuminated? So near the horizon, the power drops to zero and goes up to max dead on to the sun. So what is that function, how much do you lose compared to the same surface area fully facing the sun?
The time it takes for the moon to effectively rotate around the sun at 28 days just means that some cells will be leaving sunlight and going into darkness and on the other horizon, new ones coming into view. So the system as describes will give power 24/7. BTW, 2000 W/M^2 is too high a number. Look up the solar intensity at the TOP of Earth's atmosphere, which is the same average as what you get on the moon's surface.
4. 08 Mar '07 00:12
I mentioned the total efficiency, 50%, thats energy to the home, so it covers all the inefficiencies in the system.

OK, I interpreted 50% as the efficiency of the system that transmits the energy from the Moon to the Earth. Correct my answer up by a factor of 7 if 50% is the efficiency of the whole system -- obviously we have some amazing solar panels. We're now at about 250 KW per person.

Did you correct for the fact that at any given point on the moon, only half is illuminated and of that half, only so much of it will be fully illuminated? So near the horizon, the power drops to zero and goes up to max dead on to the sun. So what is that function, how much do you lose compared to the same surface area fully facing the sun?

I considered this. Imagine you are standing on the sun, looking at the moon. Then the moon looks like a flat circle with an 800 km wide band running across its middle. The amount of solar energy absorbed by the band is proportional to its area. The length of the band is the diameter of the Moon. Since 800 km is small compared to the diameter of the moon, the band is like a rectangle with slightly curved sides. So the area of the band, as seen from the Sun, is about 800 km * (diameter of Moon).

The time it takes for the moon to effectively rotate around the sun at 28 days just means that some cells will be leaving sunlight and going into darkness and on the other horizon, new ones coming into view. So the system as describes will give power 24/7.

The rotation doesn't matter because the same total area of solar cells is exposed to the sun at any given time.

BTW, 2000 W/M^2 is too high a number. Look up the solar intensity at the TOP of Earth's atmosphere, which is the same average as what you get on the moon's surface.

According to http://en.wikipedia.org/wiki/Solar_radiation the correct value is 1366 W/m^2 -- about 30% less than my value.

The new average energy per person is about 175 kW. It would be nice to be able to compare this to a real value...
5. AThousandYoung
08 Mar '07 04:341 edit
In developed countries, people use about 3 - 8 "tonnes of oil equivalent (TOE)" of energy per year.

http://www.nationmaster.com/graph/ene_usa_per_per-energy-usage-per-person

How many Joules in a TOE?

EDIT - This site says the US uses about 9 x 10^19 J per year.

http://hypertextbook.com/facts/1998/TommyZhou.shtml

We have about 300 million (3x10^8) people in the US.

http://www.census.gov/main/www/popclock.html

Thus each person uses 3 x 10^11 J per year right?
6. AThousandYoung
08 Mar '07 05:271 edit
A year gas 31,536,000 seconds. 3x10^11 J divided by this number gives 9.512 kW.

42 kW is plenty.
7. sonhouse
Fast and Curious
08 Mar '07 14:27
Originally posted by AThousandYoung
A year gas 31,536,000 seconds. 3x10^11 J divided by this number gives 9.512 kW.

42 kW is plenty.
That figure of 9.5 Kw includes all industrial, transportation, home heating and cooking and other incidentals?
8. sonhouse
Fast and Curious
08 Mar '07 14:30
Originally posted by GregM
[b]I mentioned the total efficiency, 50%, thats energy to the home, so it covers all the inefficiencies in the system.

OK, I interpreted 50% as the efficiency of the system that transmits the energy from the Moon to the Earth. Correct my answer up by a factor of 7 if 50% is the efficiency of the whole system -- obviously we have some amazing solar pane ...[text shortened]... y per person is about 175 kW. It would be nice to be able to compare this to a real value...[/b]
So at 175Kw/person, the number I also got, even dividing the photocell width by ten, to 80 Km, would give 17.5 Kw, about twice the energy consumed now, which is listed at about 10Kw/person (24/7)
9. 08 Mar '07 14:55
Originally posted by sonhouse
It's way in the future. I am reading a story called "Pandora's Star"
and in it there is this piece about the earth going green energy-wise by having a belt of solar cells all the way round the moon and shipping the energy to earth. So if the belt of solar cells is 800 Km wide, and we assume half the available energy actually makes it to earth and the popul ...[text shortened]... e equatorial belt of cells were illuminated at once. Then figure out how much gets to earth.
Just out of curiosity, how did the author propose transferring that energy to Earth? It seems that to do this exercise properly you would have to take into consideration the energy loss during transfer.
10. sonhouse
Fast and Curious
08 Mar '07 15:27
Originally posted by jebry
Just out of curiosity, how did the author propose transferring that energy to Earth? It seems that to do this exercise properly you would have to take into consideration the energy loss during transfer.
In the case of my problem, I lumped all the inefficiencies together, at 50%, so that allows for losses along the way, including solar cell efficiency, which I assume in a couple hundred years would be more like 80% or so, right now, there are lab level cells getting 70%. If you use microwave beams you will lose about half getting power from the moon. In the story, the first manned mars mission is interrupted by a team who invents a kind of projectable wormhole so one of them steps out on the surface of mars right next to the just landed mars craft, and a form of interstellar transport is invented. So in this story, they use a mini wormhole to send the power through, with superconductors, the loss is neglegible. So with 80% photocells on the moon, you would have at least 75% of that energy available on earth, something I felt unlikely to ever be invented so I just made a guestemate and put it at 50% because the time frame is several hundred years in the future and in my scenerio, they use more realistic
technology consistant with that era.
11. 08 Mar '07 15:35
Originally posted by sonhouse
In the case of my problem, I lumped all the inefficiencies together, at 50%, so that allows for losses along the way, including solar cell efficiency, which I assume in a couple hundred years would be more like 80% or so, right now, there are lab level cells getting 70%. If you use microwave beams you will lose about half getting power from the moon. In the ...[text shortened]... in the future and in my scenerio, they use more realistic
technology consistant with that era.
I'll accept that since you aren't actually building the thing! ðŸ™‚

Otherwise we would have a lot to discuss! ðŸ˜€
12. sonhouse
Fast and Curious
08 Mar '07 16:12
Originally posted by jebry
I'll accept that since you aren't actually building the thing! ðŸ™‚

Otherwise we would have a lot to discuss! ðŸ˜€
The interesting thing is, it could be built now. Or within this century anyway. With automated cell manufacturing on the moon using lunar materials and solar powered, they could crank out millions of cell a year and pave the equator just like in the book, but using microwaves to get the power to earth. There are greenies who would fight the microwave thing because several stations would have to be built due to the moon rotating around the earth once a month. Greenies don't like the idea of gigawatts or terawatts of microwave energy in a beam even if it's over a kilometer across. There are plans afoot to have satellites in geosyncronous orbit around the earth to do much the same thing but with a lot less energy per satellite, using microwave beams to get the energy back to earth. I think the average power level in the beam was supposed to be something like what you get on your head from using a cell phone, which won't fry you anytime soon or birds or aircraft.
13. AThousandYoung
09 Mar '07 07:26
Originally posted by sonhouse
That figure of 9.5 Kw includes all industrial, transportation, home heating and cooking and other incidentals?
Yep. It's the US energy usage divided by the population.
14. AThousandYoung