# SoftestPawn’s Weblog

## Is Solar Power Enough?

Posted by softestpawn on June 19, 2008

(Last edited 3rd July 2008; still looking for a factor of 5 error!)

From this badscience thread and particularly this post, a quick set of rough calculations to see if solar power (wind, solar panels, wave, biofuels, etc) is sufficient to supply our energy needs.

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.

## Solar Supply

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)

#### Efficiency

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.
and
-> 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.

## Demand

Here I look at various energy demand estimates and calculate how much surface area is required for each one:

### Current Demand

Current use in rich countries is ~4-5 Tonnes of oil equivelent per person per year (BERR and Economist’s “Pocket World in Figures”). 1 TOE = 42 GJ.

= 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.

### Transition Demand

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):

1010 people
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

### Predicted Demand

= 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.

### Global Impact

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)

### Sanity Check

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)

### Conclusion

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.