Is Solar Power Enough?
Posted by softestpawn on June 19, 2008
(Last edited 3rd July 2008; still looking for a factor of 5 error!)
First I look at the supply and how much electrical energy we can expect to extract for each square meter of ground, assuming ground solar panel collectors in fixed geographic locations.
Total radiation arriving outside the atmosphere (ie, at all wavelengths) is 1368Watts/m2 (ref: NASA).
Assume (?) that only radiation reaching the ground can be harvested by wind and solar panels; as I understand it winds etc are caused by ground heat conducting into air rather than direct solar heating.
Radiation arriving at ground (refs: squ1.org, Wikipedia, Encyclopedia of Earth) appears to be less than half of this as some is reflected by cloud or absorbed by the atmosphere but it’s not entirely clear. So:
Approx 500W/m2 (after weather) on the flat plane facing the sun
x Total area facing sun = pi * (radius = ~6000km)2 = ~100 million sq km = 1014 m2
-> 5×1016 Watts total solar energy arriving on earth at ground.
Incindence & Night
As the above radiation is spread over the curved surface of the ground, half (ish) of which is in the dark, the ‘average’ radiation received over a day on a square meter of surface is the total surface radiation divided by the total surface. So:
Above total solar energy = 5×1016 Watts
Earth’s surface area (radius = ~6000km) = 500,000,000 km2 = 5×108 km2 = ~5×1014 m2
-> about 100W/surface m2, averaged over a day. (ie about 1/5 of the flat plane incidence above)
Inefficiency in gathering and transporting the energy: solar photoelectric cells are currently between 12-40% (wikipedia), losses over the national grid are a few percent. Assume a total of 50% overall (including transport) for future direct-collection technologies (ie not wind or wave).
-> 50W/m2 collected averaged over a day.
-> 2.5×1016 Watts total global harvestable solar energy
Sanity Math Check
If we wanted all the energy harvestable, demand would be 2.5×1016 watts, @ 50W/m2 surface use = 5×1014 m2 = 5×108 km2 = 500,000,000km2 = total surface area. Good.
Here I look at various energy demand estimates and calculate how much surface area is required for each one:
= 200 GJ per person per year
= 200 x109 J / 3 x107 seconds/year (Joules = watts x time)
= ~7000 Watts per rich person
@50W/m2 -> 140 m2 per rich person.
From a BBC article demand by 2030 (ie not ‘final demand’) will be 15,000 million tons of oil equivelant per year global total
-> 1.5 x1010 toe @ (40GJ/toe = 4 x1010 J/toe)
-> 6 x1020 J/year / 3 x107 seconds/year
-> 2 x 1013 watts total global demand in 2030
@50W/m2 -> 4 x1011 m2 = 4 x105 km2 = 400,000 km2
Naive Predicted Demand
Naively assuming 10 billion people world population using only the current ‘rich country’ scaling (ie ignoring air conditioning, etc in emerging warmer countries):
x 7000 Watts = 7 x 1013 watts total global demand
x 140 m2 = 1.4×1012 m2 = 1.4 x106 km2 = 1,400,000 km2
= 1014 watts total global demand.
@ 50W/m2 -> 2 x1012 m2 = 2 x106 km2 = 2,000,000 km2
Worldwide USA demand
USA energy use is currently around 12kW/person and covers a range of climates. For simplicity, assuming this rises to 14kW within a few decades and that we would like the world to have the same benefit, this doubles the Naive Predicted Demand to require 2,800,000 km2 to capture sufficient solar power.
So if we look at the proportion of the world’s surface area required by the above demand:
Total earth’s surface = ~5×1014 m2 = 5×108 km2 -> using 0.3% to 0.4% of the earth’s surface
About 30% is land -> 1.5×108 km2 -> 0.5% to 1.3% of the earth’s land.
The Sahara desert is about 9 million km2, and has better than average harvest rate per square meter. ie, we if we covered around a third of the sahara desert with solar panels we’d expect to generate enough electricity for the forseeable future. (All the same, that is a lot of area to cover in manufactured equipment)
5×1016 Watts total arrives, predicted demand is 1014 watts, so we would want to use 0.2% of all incoming radiation. As the harvesting is only 50% efficient we would need to intercept 0.4% of it. Allowing for angles of incidence/night time we would need to cover 5 times that of the surface, ie about 2% of the surface.
But 2% of the surface is 1013 m2 , and at 50W/m2 = 5x 10 ^14 Watts. So something here is wrong!
UK Coverage Required
Ground radiation is about 100W/m2 during the day time, so over 24 hours is about 50W/m2, with about 50% efficiency this is a harvestable 25W/m2. Population 60 million (6×107), using 7000W/person -> 4×1011W -> ~ 2×1010m2 = 20,000 km2
Surface area is about 250,000 km2 so we’d need to use about 8% of the land (assuming we didn’t use the surrounding sea surface)
Wherever my factor of five has gone, it’s certainly “feasible”. Requiring a few percent of the surface it is likely – ignoring manufacturing the gathering equipment – that just the presence of it may have a widespread environmental impact.