If solar panels covered 1/4 of Utah, those alone would be able to power the entire United States. US consumption is very high, weighing in at 12,000 kWh per year per capita.
So even without reducing consumption at all, we have the technology right now to produce ample clean energy. This is the message. There is hope. #solarpunk #directaction #doyourpart
Sources:
https://ecotality.com/how-many-solar-panels-to-power-the-us/
13750000 acres = 21,484.4 sq mi
https://www.justintools.com/unit-conversion/area.php?k1=acres&k2=square-miles
https://en.m.wikipedia.org/wiki/Utah 82,144 sq mi
A real-world case for PV is https://en.wikipedia.org/wiki/Ouarzazate_Solar_Power_Station in Morocco which occupies 2500 hectares with nameplate power of 582 MW, which gives an actual nameplate capability of 23.28 W/m^2 which is ~400x less than the theoretical irradiation.
Another important factor is that the plant uses vast amounts of water - 1.7 million m3 per year or 4.6 liters per kWh - for cleaning the mirrors, which is 23x more than a coal plant.
Jacobson is the opium of renewables :) Have a look at this table - do you see anything wrong about it from engineering point of view?
source: http://web.stanford.edu/group/efmh/jacobson/Articles/I/143WWSCountries.pdf
No, just look at percentages in column E and absolute numbers in column F.
These numbers are completely unrealistic - it's like saying that we now have built a 828 m tall building (Burj Khalifa) so what's the problem to just add 1999 km and reach low Earth orbit.
The model also assumes "perfectly interconnected grid", so basically ignores the largest engineering challenge when building a grid based on RE, as it assumes electricity from Mexico will reach Alaska instantly and magically
@kravietz No. What is it you see? That all the numbers in column E are based on the numbers from columns C and D?