## 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/m^{2} (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/m

^{2}(after weather) on the flat plane facing the sun

x Total area facing sun = pi * (radius = ~6000km)^{2}= ~100 million sq km = 10^{14}m^{2}

->5×10.^{16}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 = 5×10

^{16}Watts

Earth’s surface area (radius = ~6000km) = 500,000,000 km^{2}= 5×10^{8}km^{2}= ~5×10^{14}m^{2}

-> about 100W/surface m^{2}, 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/m

^{2 }collected averaged over a day.

and

-> 2.5×10^{16}Watts total global harvestable solar energy

#### Sanity Math Check

If we wanted all the energy harvestable, demand would be 2.5×10^{16} watts, @ 50W/m^{2 }surface use = 5×10^{14} m^{2 }= 5×10^{8} km^{2 }= 500,000,000km^{2 }= 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 x10^{9}J / 3 x10^{7}seconds/year (Joules = watts x time)

= ~7000 Watts per rich person

@50W/m^{2 }->140 m.^{2}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 x10

^{10}toe @ (40GJ/toe = 4 x10^{10}J/toe)

-> 6 x10^{20}J/year / 3 x10^{7}seconds/year

->2 x 10^{13}watts total global demand in 2030

@50W/m^{2}-> 4 x10^{11}m^{2 }= 4 x10^{5}km^{2 }=400,000 km^{2 }

### 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):

10

^{10}people

x 7000 Watts =7 x 10^{13}watts total global demand

x 140 m^{2 }= 1.4×10^{12}m^{2 }= 1.4 x10^{6}km^{2 }=1,400,000 km^{2 }

### Predicted Demand

This article, from UNDP data, predicts 102 TW demand by 2050:

=

10.^{14}watts total global demand

@ 50W/m^{2}-> 2 x10^{12}m^{2 }= 2 x10^{6}km^{2 }=2,000,000 km^{2}

### 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 km ^{2}** 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 = ~5×10^{14} m^{2} = 5×10^{8} km^{2} -> using 0.3% to 0.4% of the earth’s surface

About 30% is land -> 1.5×10^{8} km^{2} -> 0.5% to 1.3% of the earth’s land.

The Sahara desert is about 9 million km^{2}, 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

5×10^{16} Watts total arrives, predicted demand is 10^{14} 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 10^{13} m^{2} , and at 50W/m^{2} = 5x 10 ^14 Watts. So something here is wrong!

### UK Coverage Required

Ground radiation is about 100W/m^{2} during the day time, so over 24 hours is about 50W/m^{2}, with about 50% efficiency this is a harvestable 25W/m^{2}. Population 60 million (6×10^{7}), using 7000W/person -> 4×10^{11}W -> ~ 2×10^{10}m^{2} = 20,000 km^{2}

Surface area is about 250,000 km^{2} 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.

(A much better explained article from withouthotair)

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