Originally Posted by popcorn_karate
or perhaps you missed that you would need a 180,000 gallon tank to run it for 1.5 hours each day of the month as you suggested?
Perhaps. From your BOE calculations, the 10k gallon tank delivers:
3.5 kW for 5 min - close enough for BOE for sure
3.5 kW x 5 min /60 min/hr = .2916667 kWh/day reflecting the energy recovered each time you run the upper tank through the generator.
.291666. kWh/day x 30 d/mo = 8.75 kWh/mo
for the 10k gallon tank you suggested. Perhaps you are referring to my suggestion that with a slightly greater head, and a larger water store one could easily envision a home system with a storage capacity in the neighborhood of 100 kWh/mo?
100/8.75 ~ 11.4
suggesting you need a bit more than an order of magnitude increase in head, gallons of storage, or some combination thereof. i.e gallons x N and head x P with N x P ~ 11.4 < the factor of 18 as you suggest. (i.e. 1.5 hours/day = 90 min at 5 cfs = 18 x 5 min) , presumably how you got to the 180k gallons figure). As I recall I suggested that a combination of a modest increase in head (20' vs. 15' gives a factor of 4/3, and a larger water store such as swimming pool size: 11.4 x 3/4 x 10k gallons ~ 85.5k gallons for example). Or bump the head up to 25 feet. So on.
Now my algebra and ability to click the right buttons on Microsoft's calculator are both always suspect so please feel free to expound in greater detail on what error I am making here. Lacking such, we could pretend to be physicists who, being within an octave let alone an order of magnitude, could consider ourselves to be in agreement that the technique seems applicable, but of questionable economic efficiency at the household scale outside of specific users with strong economic motivations and/or situations likely including topographic advantage in reducing implementation cost. On the other hand, a technique of such efficiency at larger scales that we are a rather sad and stupid bunch if we fail to use it collectively at the grid-level scale.