Earth Solar Efficiency

Photovoltaic panels

Because of the broad spectrum of solar light, from ultraviolet to the far infrared, no photovoltaic system will ever approach 100% efficiency. The best space-grade triple-junction gallium arsenide cells in production (Spectrolab XTJ) approach 30% efficiency and cost many thousands of dollars per square meter.

Terrestrial arrays, such as the SolarWorld SW 285-300 produce a maximum of 300 W brand new, and 240 W (at 25 years) from 1000 W/m² "standard" sunlight. The panel is 1.61 m², so the brand new efficiency with maximum illumination is 18.6% . The efficiency drops to 17.4% with 800 W/m² illumination, and presumably lower still when the sun is at low elevation (which is attenuated further by absorption, Rayleigh scattering, and more clouds when observed sideways) - lets optimistically assume 18%.

Interconnection bus

Now lets look at the rest of the system. Assume 33 panels in series, making a facility bus voltage of 1000 volts - and a 3.8 gigawatt peak summer power facility, connected to Chicago 2000 km away (a proxy for the national grid with cities at various distances) by an HVDC power line like the 2016-upgraded Pacific DC intertie, running 560 kV bipolar at 3400 A. The two ACSR conductors each have cross sections of 1,170 mm², for a resistance of 26 milli-ohms per kilometer, 51 ohms end to end. A voltage drop of 87 volts per kilometer, or 175 kV end to end - a 30% loss.

Assume the panels are latitude-oriented for 33°N, and spaced to not shade each other in winter, when the sun is at 34 degrees above the horizon (56° from zenith). If the panels are 0.91 meters wide, they stick up 0.496 meters, and the north-south panel spacing is 1.5 meters. Each panel's "footprint" is 2.4 m².

In the summer, the fixed panels are 23 degrees south of optimum, the same for winter - they are running at 92% illumination and 89% of 300 rated Watts using 18% instead of 18.6%, and produce an estimated 266 watts each. To produce 3.8 GW peak, we will need 14 million of them (now retailing for perhaps $400, assume $200 in huge quantities with mounting hardware. Note that Solarworld's entire 2014 panel output was "530 MW", which would be only 1.8 million panels the way they count them. The total array footprint is 34 km² - if it is circular, with the HVDC converter in the center for minimum system bus length, the radius of the array will be 3.3 km.

Presume a two stage conversion, with the local 1 kV bus stepped up to 200 kV for power transmission across the array, perhaps ten thousand such converters, 380 kW each, each fed by a hexagonal (approximately circular) area 66 meters in diameter, with 1,400 panels in it, about 43 strings of 33 panels in series. This produces a peak summer current per string of about 8.8 amps, which we can move with ordinary #12 aluminum wire, insulated for the higher voltage, and with special aluminum-capable connectors. Assume 100 meters of wire in each string, about 0.8 ohms and 7.2 watts loss per string, 310 W total, about 0.1% loss for the wiring. Assume the inverter has 0.7% loss.

The first inverter out power, 200 kV at 1.9 A, is added in parallel to large buses connecting the whole array. assume 1000 such busses feeding the center, each connecting in parallel to 100 inverters spaced 66 meters apart, a total current feeding the center of 3800 amps. If we use cables with 500 mm² cross section, the resistance per leg is 60 mΩ per kilometer, and the length is 6.6 km, for an end-to-end resistance of 0.4 ohms. Almost all of the resistive loss occurs near the center; using calculus, the total power loss per bus is 2 * 0.4Ω * ( 1900A )² / 3 or 1 MW loss for both legs of each bus, 1 GW loss total. By making the ends skinnier and the core end fatter, we might cut that in half. By increasing the number of 560 KV inverters, and paralleling thise we can probably cut the loss in half again. So, assume 7% 200kV wiring loss, and an additional 0.7% 560 KV intertie-driving inverter loss.

Pumped Hydro: That's all very nice - but we are not done yet. During worst case times when the sun isn't shining, we will be drawing down pumped hydro reservoirs that were filled beforehand; all of our power from late afternoon and early morning power will be coming from pumped hydro. Typical PH efficiencies are 80% - lets assume half our power comes that way, for an efficiency loss of 10%.

Further, let's include the degradation of the panels, to 80% at the 25 year end-of-life. For a mature system, with efficiencies averaging 90% of new, another 10% loss.

All that aluminum may need replacing every 100 years. How much aluminum?

A total of 11,420 cubic meters of aluminum, at 2700 kg/m³ that is 31,000 metric tons. Towers estimated to weigh as much as the wire.

Panels are 5 kg of aluminum (estimated out of 18 kg including glass, wiring, and solar cells). Frames and ground anchors are an estimated 5 kg. Times 14 million panels, that is 140,000 metric tons.

This does not include power towers, converter stations, etc.; as a wild guess, assume 200,000 metric tons. At 50MJ/kg manufacturing energy, that is 1E16 J. Divided by 100 years, that is only 3 MW, about 0.1 %

There are probably other things reducing efficiency - so I will round that down to 7%.