Yeah, I know it is feasible for domestic use. I was trying to replace the Colorado river! My thought was what would it take for agriculture and everything including restoring flow into the Sea of Cortez. That would be a job for nuclear.
The Colorado River is what I was going to post about next.
The flow of the Colorado is highly variable through the seasons; it has been higher than 100,000 cubic feet per second, and lower than 2,500 c.ft./s. Therefore, we have to use its annual flow rate. This annual flow rate is not easy to define, however.
The 1920 Compact that divided the water among the Western states assumed a water flow of 17.5 million acre-feet. Later, it was thought that the above number was based on only 30 years of record, and those were wet years. The long-term average may be as low as 13.5 million acre-feet. The water allocation was based on 16.5 million acre-feet (15 to the USA, and 1.5 to Mexico), so I will use this number.
16.5 million acre-feet per year = 20.35 billion cubic meters per year
From the range of 3-5.5 kWh/cubic meter for desalination in my earlier post, let's use the average of 4.25 kWh for the energy cost of desalination.
The annual energy requirement is then 20.35e9 m3/year x 4.25 kWh/m3 = 86.5 billion kWh/year.
The output of the Palo Verde plant is quoted as 31,920 GWh/year. That's 31.92 billion kWh/year.
It then takes 2.7x the output of the Palo Verde plant to desalinate seawater to match the flow of the Colorado River. This seems low, so I checked and checked the numbers, but did not find an error.
Now, this has not included the cost of pumping the desalinated water for distribution. Let's say we want to pump the water from sea level up to Lake Mead for distribution.
The top of the Hoover Dam is at 376m above sea level. I don't know about the actual energy needed for pumping with losses from friction, but a simple fact from physics is that lifting 20.35 billion cubic meters of water weighting 20.35 trillion kilograms up 376 m takes a minimum of 7.5e16 Joules, or 20.8 billion kWh/year. Some of this energy will be recovered when the water goes back down through the Hoover Dam.
And so, the minimum energy required is (86.5+20.8)/31.92 = 3.4 x Palo Verde plant.