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Comment: Re:3T gallons at a Superfund site with earthquakes (Score 1) 110

I am not saying to use that site. It is still an active mine . I am just using it as an example of the size of the hole needed.

(you may be able to halve the number by using the amount dug out to form a upper reservoir too)

The main problem with this method is of course the cost.

Comment: Re:Ludinton pumped storage facility X 1,500,000 (Score 1) 110

(We actually only need 3.28767123 × 10^14 BTU for daily storage)

Capacity of Bingham Canyon Mine = 3,220 billion gallons. (or 1.2 *10^10 m^3) (120 times more water)

Height 0.97 km ( Or an average of 5*100m) compared with an average of (1 .21* 100m) for Ludington (~5 times as much power for each L)

Using the calculations on pumped storage from wikipedia if you pumped water out of Bingham and let it flow back in.

1.2 *10^10 * 5 * .272 Kwh * 70% in BTU = 3.8980306 × 10^13 BTU / day (so probably 10 or so not 5)

(size in m^3) * (number of 50m height increments) * .272Kwh * 70%

using the same calculations for Ludington
8.36645409 × 10^10 BTU (Average fall of 120m)

(based on the local one here , it is 10 hours max output not 13 hours)

(And in your case the lower reservoir would be the ocean)

Comment: Re:Well, you COULD flood most of the country (Score 1) 110

Rough estimate

120 quadrillion BTU /year http://www.eia.gov/totalenergy...

using https://en.wikipedia.org/wiki/... 0.272 kWh per 1m^3 per 100m height

square root of (120 quadrillion btu / 365 / 70% / 0.272 kWh * 1 m^3 /100m) = 71km.

(or in other words 71km *71km *100m at a height of 100m from the lower lake will stall enough power for the average day.

Or to put it in perspective about 5 holes the size of Bingham Canyon Mine

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