Photo Credit: Mitsui & Co., Ltd. |

The technology is, in principal, capable of turning an intermittent source into one suitable for base load power. Of course, you might recognize that hydropower from dams utilizes this principle with mother nature providing the pump via the hydrologic cycle.

In the thread, I mentioned that geography places limits on the wide-scale adoption of pumped hydro storage because of the necessity for two very large reservoirs with a large elevation difference in close proximity. This is something that we don't find naturally around every corner and that, to construct, would entail an extraordinary infrastructure expenditure. Of course, given that one of my Company's main Divisions is dedicated to quality assurance and quality control of infrastructure projects, I'm not at all opposed to such an undertaking!

But I want to "bring this storage method home," so to speak. Were I to install a solar array and utilize pumped hydro in my household, what would this entail? Let me assume that I'd like to have the ability to store sufficient energy to provide electricity to my home for three days. Such a need will most likely arise in winter when I won't need air conditioning - hot summer days usually provide plenty of sunshine and the three days' worth of winter storage should also suffice for summer nights when the sun is down.

As it happens, I've calculated my home's average continuous energy use before, and it's an embarrassingly large 2.1 kilowatts. Therefore, I need to provide 2.1*24*3 or 151.2 kilowatt hours of energy in the system's water tank (that my homeowners association will never let me build no matter how the numbers come out). Energy units are energy units, so I can convert these 151.2 kilowatt hours to 5.443*10^8 or 544.3 million joules. My system is likely to run, optimistically, at about 75% efficiency, so I'll need to store 7.258*10^8 joules. As a side note, this is the total thermal energy available from burning about six gallons of gasoline. To actually use this energy and assuming a Home Depot generator runs at 25% efficiency, I'd need to have 24 gallons available. Keep this number in mind!

Now there are two ways to increase the energy stored in a pumped hydro system: increase the quantity; and increase the height to which it's pumped. A cubic meter of water is 1000 kilograms and raising this 1000 kilograms one meter will add 1000*9.8*1 or 9800 joules. Let's say instead that I could raise it 10 meters. Note that this is 33 feet so that the average height of the water in the tank is above the tops of the roofs in my neighborhood. Now I've added 9.8*10^4 joules. This means I need to raise (7.258*10^8)/(9.8*10^4) or 7406 cubic meters of water. This is just shy of two million gallons (1.96 million) so I'll go with the two million gallon tank. This is the part where I suggest recalling the 24 gallons of gasoline.

Photo credit: San Patricio Municipal Water District |

There's much more on the practicalities (or lack thereof) for utilizing pumped hydro storage on a wide scale to store energy from intermittent sources at one of my favorites sites. It's called "Do the Math" and is published by Dr. Tom Murphy, a professor of physics at UC San Diego.

## 2 comments:

Typo there -- should be 9.8x 10^3 joules.

I don't think so. 9.8*10^3 joules for one meter(i.e., 9,800), 9.8*10^4 joules for 10 meters.

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