Thank you for diving into the details with me, and continuing to ask probing questions!
The water brought-in by the Sahara doesn’t depend upon the area of the source; it’s the humidity times the m3 per second arriving. Humidity is low on arrival, reaching only 50% right now in Tunisia, their winter drizzles! The wind speed is roughly 2m/sec coming in from the sea, which is only 172,800m/day of drift. Yet! That sea-breeze is a wall of air a half kilometer high—that is why it can hold quite a bit.
If we need +10% of a 500m tall drift, that’s 50m; if we can use solar concentrators to accelerate convection, we can get away with less. And, we’re allowed to do an initial row that follows the shoreline closely, while a second row is a quarter kilometer inland, running parallel to the shore, where mixing of air lets you add another round of evaporate. So, we could have four rows across the northern edge of the Sahara, each row as thick as it needs to be to hit high humidity, and 10m tall, to send +10% moisture over the entire 9 million km2 of the Sahara.
How much water would we be pumping? The Sahara carries 172,800m/day flow per m2 intake surface x 500m tall x 4,000km coastline at 10g h2o per m3 = 3.5 billion tons per day, a thousand or so dead seas. (About 1.25 Trillion tons a year, enough to cover the 9 Million km2 with 139mm of rain, on average, if it had fallen instead of being sopped-up by adiabatic heat.)
We need 10% of that, or a hundred and eighty dead seas. It seems monstrous, but much of the coastline there is low for miles, so pumping 1 ton to the top of 10m at even just 20% efficiency costs 500kJ. If you want to pump that in a day, using solar, you’ll need 1/4th of a square foot of solar. That 1 ton, if we cross the threshold and it becomes surplus rain, will water 3 square meters their annual budget… and the solar is paying for that amount of irrigation every day; 1,000 m2 of rains from a dinner plate of solar, each year. It’s that energy efficiency, combined with dead simple capital expenditures, which would make something so insane potentially feasible. I’d pick California to try, first!
500kJ per ton, for 350Mil tons per day—that’s 175TJ per day, or 2 GW. That’s a nuclear power plant. To pump enough water, continuously, to irrigate 9 million km2, potentially feeding a billion people, once we dig swales! (Check out Africa’s better-than-trees plan: “Demi-Lune” swales that catch sparse, seasonal rain, to seep into the ground, with minimal tools and labor!)
Thank you for diving into the details with me, and continuing to ask probing questions!
The water brought-in by the Sahara doesn’t depend upon the area of the source; it’s the humidity times the m3 per second arriving. Humidity is low on arrival, reaching only 50% right now in Tunisia, their winter drizzles! The wind speed is roughly 2m/sec coming in from the sea, which is only 172,800m/day of drift. Yet! That sea-breeze is a wall of air a half kilometer high—that is why it can hold quite a bit.
If we need +10% of a 500m tall drift, that’s 50m; if we can use solar concentrators to accelerate convection, we can get away with less. And, we’re allowed to do an initial row that follows the shoreline closely, while a second row is a quarter kilometer inland, running parallel to the shore, where mixing of air lets you add another round of evaporate. So, we could have four rows across the northern edge of the Sahara, each row as thick as it needs to be to hit high humidity, and 10m tall, to send +10% moisture over the entire 9 million km2 of the Sahara.
How much water would we be pumping? The Sahara carries 172,800m/day flow per m2 intake surface x 500m tall x 4,000km coastline at 10g h2o per m3 = 3.5 billion tons per day, a thousand or so dead seas. (About 1.25 Trillion tons a year, enough to cover the 9 Million km2 with 139mm of rain, on average, if it had fallen instead of being sopped-up by adiabatic heat.)
We need 10% of that, or a hundred and eighty dead seas. It seems monstrous, but much of the coastline there is low for miles, so pumping 1 ton to the top of 10m at even just 20% efficiency costs 500kJ. If you want to pump that in a day, using solar, you’ll need 1/4th of a square foot of solar. That 1 ton, if we cross the threshold and it becomes surplus rain, will water 3 square meters their annual budget… and the solar is paying for that amount of irrigation every day; 1,000 m2 of rains from a dinner plate of solar, each year. It’s that energy efficiency, combined with dead simple capital expenditures, which would make something so insane potentially feasible. I’d pick California to try, first!
500kJ per ton, for 350Mil tons per day—that’s 175TJ per day, or 2 GW. That’s a nuclear power plant. To pump enough water, continuously, to irrigate 9 million km2, potentially feeding a billion people, once we dig swales! (Check out Africa’s better-than-trees plan: “Demi-Lune” swales that catch sparse, seasonal rain, to seep into the ground, with minimal tools and labor!)