Is Solar Power Enough?

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.

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

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

= 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 2,800,000 km²

Global Impact

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.

Sahara desert is about 9 million km², and has better than average harvest rate per square meter.

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)