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Covering up large areas of a desert with solar panels may be more harmful than you realize. I've heard & use the term solar strip-mining. Solar power maxes ou

Covering up large areas of a desert with solar panels may be more harmful than you realize. I've heard & use the term solar strip-mining. Solar power maxes out at no better than 1kW/m^2 (sun at zenith, clear day, practically 100% efficient conversion), we're talking an absolute minimum of 1 sq km to match a baseline 1GW power plant ... and likely 100 times larger for real-world scenarios (10% efficiency, night/clouds, non-zenith air densities, dust, etc). So yeah, the net effect is "strip-mining" a vast area by removing all (or a seriously impactful percentage) natural power input and thus killing everything off. Rooftop solar should indeed be the focus. There's even solar panels intended as road/pavement surfaces, super-durable interlinked hexes grabbing power where we've already wrecked the natural surface.

honkhonkpants 10y ago
> 1 sq km to match a baseline 1GW power plant Does that sound like a lot to you? A single airport runway is much larger than 1 sq km.
- ctdonath 10y ago
  As I noted: real-world conditions demand 100x as much space.
- jcranmer 10y ago
  Uh, no. Chicago Midway happens to fit very nicely inside 1 mi² = 2.56 km². Its main runways (5 of them) with feeder ramps collectively take up at best ½ the space, so a single runway itself is not 1 km² but more like .2-.3 km².
  - honkhonkpants 10y ago
    Midway is an airport for ants. 10/28 at SFO is 2 sq km, and there are two of them, and 1/19 are 1.5 sq km, and there are two of them. Then you also have OAK and SJC. Point: a square kilometer to generate 1GW peak of electricity would be an obviously excellent trade off, well within the cost parameters that humans are willing to spend on far less useful endeavors like runways and parking lots.
    - jcranmer 10y ago
      Wikipedia says that 10/28 is 3.618 km. For the runway itself to be 2 km², it would have to be 0.552 km wide. A quick glance at an aerial map indicates that a width of 0.552 km gives approximately the width of both runways, relevant taxiways, and the surrounding ground. Wikipedia says that the runways are in fact .061 km wide, for a total per-runway size of .221 km². I picked Midway because it's a very conveniently sized airport, being almost exactly 1 mi² in a square-shaped with obvious meter marks (due to surrounding grid system).
      - honkhonkpants 10y ago
        Ok you got me. Bad at math.
    - ctdonath 10y ago
      Except that 1GW/km^2 is the absolute upper limit. As dreadmorbius computed, reality gives ~1.5% of that.
jsight 10y ago
Why do you assume 10% efficiency? That seems awfully low? I agree with regards to rooftop being an important part, though. It makes sense to use land that is already occupied to a large degree.
- jcranmer 10y ago
  The efficiency of modern commercial photovoltaic solar cells is about 10-20%.
- dredmorbius 10y ago
  If you start looking at effective achieved energy delivery from solar, 10% is exceptional. Consider that plants are lucky to return 1-10%. Algae, under exceptional conditions, may also reach 10%, but that takes some doing. Insolation: 1 kW/m^2 Panel efficiency: 10-20%, nominally 17%. Maximum for a single-layer cell is 37% for quantum physics reasons, with an "infinite" layer panel, 87%. Capacity factor is the amount of time a panel actually recieves sunlightlight. Typically 20-30%, with 30% as a peak. Spacing factor. PV doesn't work if shaded -- the entire circuit cuts out. So you have to angle and space your panels. On net, this reduces availble area by about 50%. Less near the equator, more at high latitudes. Degredation. For numerous reasons, solar panels degrade with time, mostly due to physical wearing (broken cells, wiring, hazing of panels, stone and hail damage, etc.) When nominal output falls to about 85% new, panels are deemed due for replacement. This is about 20 years based on multiple studies. So you're replacing 5% of your stock every year, and losing a few percent of output. Inverter losses. Converting panel (DC) to transmission (AC) power costs about 10%. Transmission losses. Another 6% or so for high-voltage transmission. Multiply all that out, and your 1 kW/m^2 * 0.17 * 0.50 * 0.9 * 0.9 * 0.25 leaves you with 17 watts of 24/7/365 capacity per square meter. Or about 1.7% net efficiency. This doesn't account for storage which can cut into that some more.

honkhonkpants 10y ago

> 1 sq km to match a baseline 1GW power plant Does that sound like a lot to you? A single airport runway is much larger than 1 sq km.

ctdonath 10y ago
As I noted: real-world conditions demand 100x as much space.
jcranmer 10y ago
Uh, no. Chicago Midway happens to fit very nicely inside 1 mi² = 2.56 km². Its main runways (5 of them) with feeder ramps collectively take up at best ½ the space, so a single runway itself is not 1 km² but more like .2-.3 km².
- honkhonkpants 10y ago
  Midway is an airport for ants. 10/28 at SFO is 2 sq km, and there are two of them, and 1/19 are 1.5 sq km, and there are two of them. Then you also have OAK and SJC. Point: a square kilometer to generate 1GW peak of electricity would be an obviously excellent trade off, well within the cost parameters that humans are willing to spend on far less useful endeavors like runways and parking lots.
  - jcranmer 10y ago
    Wikipedia says that 10/28 is 3.618 km. For the runway itself to be 2 km², it would have to be 0.552 km wide. A quick glance at an aerial map indicates that a width of 0.552 km gives approximately the width of both runways, relevant taxiways, and the surrounding ground. Wikipedia says that the runways are in fact .061 km wide, for a total per-runway size of .221 km². I picked Midway because it's a very conveniently sized airport, being almost exactly 1 mi² in a square-shaped with obvious meter marks (due to surrounding grid system).
    - honkhonkpants 10y ago
      Ok you got me. Bad at math.
  - ctdonath 10y ago
    Except that 1GW/km^2 is the absolute upper limit. As dreadmorbius computed, reality gives ~1.5% of that.

jsight 10y ago

Why do you assume 10% efficiency? That seems awfully low? I agree with regards to rooftop being an important part, though. It makes sense to use land that is already occupied to a large degree.

jcranmer 10y ago
The efficiency of modern commercial photovoltaic solar cells is about 10-20%.
dredmorbius 10y ago
If you start looking at effective achieved energy delivery from solar, 10% is exceptional. Consider that plants are lucky to return 1-10%. Algae, under exceptional conditions, may also reach 10%, but that takes some doing. Insolation: 1 kW/m^2 Panel efficiency: 10-20%, nominally 17%. Maximum for a single-layer cell is 37% for quantum physics reasons, with an "infinite" layer panel, 87%. Capacity factor is the amount of time a panel actually recieves sunlightlight. Typically 20-30%, with 30% as a peak. Spacing factor. PV doesn't work if shaded -- the entire circuit cuts out. So you have to angle and space your panels. On net, this reduces availble area by about 50%. Less near the equator, more at high latitudes. Degredation. For numerous reasons, solar panels degrade with time, mostly due to physical wearing (broken cells, wiring, hazing of panels, stone and hail damage, etc.) When nominal output falls to about 85% new, panels are deemed due for replacement. This is about 20 years based on multiple studies. So you're replacing 5% of your stock every year, and losing a few percent of output. Inverter losses. Converting panel (DC) to transmission (AC) power costs about 10%. Transmission losses. Another 6% or so for high-voltage transmission. Multiply all that out, and your 1 kW/m^2 * 0.17 * 0.50 * 0.9 * 0.9 * 0.25 leaves you with 17 watts of 24/7/365 capacity per square meter. Or about 1.7% net efficiency. This doesn't account for storage which can cut into that some more.