The calculations show that the storage requirement ranges from 141 to 386 billion KWH, depending on how much comes from solar and how much from wind.
There is no way to store that amount of energy. In fact, we'll have to devise a fictional example to illustrate the problem.
Imagine that a lake exists, named Upper Lake Fead, which is equal in size to Lake Mead. Lower Lake Fead is the same size and is located at the bottom of Foover Dam, which is identical to Hoover Dam. However, all the water in Upper Lake Fead can drain through the water turbines.
Lake Volume = 30,000,000 acre-feet
Average head at dam = 520 feet
If the efficiency were 100%, then
Energy = volume x pressure = volume x head x weight-density = 30,000,000 acre-feet x 43560 sq-ft/acre x 520 feet x 62.4 lb/cu-ft = 4.24 x 10^16 ft-lb = 16 billion KWH
We'll set the turbine efficiency at 85% and account for pump inefficiency by upsizing where necessary. Thus, Upper Lake Fead is good for 13.6 billion KWH.
So we have calculated that the US would need between 10 and 28 Foover Dams, each with Upper and Lower Lake Feads, depending on how much electricity is generated with solar energy. There are, in fact, no Foover Dams and no locations for building any.
Another scheme that sometimes is mentioned is storing compressed air in caves. There is a facility in Huntorf, Germany that we can use for an example.[5] It compresses air to 1000 pounds per square inch pressure.
The data show that it stores 3 x 290 = 870 MWH of energy and the cave volume is 310,000 cubic meters.
To store one billion KWH, the cave's volume would have to be 1,000,000 / 870 x 310,000 cubic meters = 356 million cubic meters. Suppose a cave averaged 20 meters wide and 20 meters high (65 feet x 65 feet); then the length required would be 356 million / 400 = 890,000 meters = 890 km = 552 miles. To save 141 to 386 billion KWH would require a total cave-length of 78,000 to 213,000 miles. Granted that most big caves have never been surveyed, it's still safe to say that there aren't tens of thousands of miles of caves in the US.
As our reference battery, we shall use the vanadium-redox type of flow battery, which has an energy density of up to 40 watt-hour/Kg. This is the best battery currently available for bulk storage.
Let's say that batteries are limited to 10 meters diameter and 5 meters height. That's roughly the volume of a house. The volume of each is 393 cubic meters. Sulphuric acid, the main constituent, has almost twice the density of water, so the weight of each tank would be 786,000 Kg and its capacity is 31,440 KWH. To store 141 to 386 billion KWH would requre 4,485,000 to 12,277,000 batteries, or one battery for every 25 to 69 persons. Another tank the same size is needed to hold the spent acid.
A company called Ausra is proposing to build concentrated-solar power plants in the deserts of the American Southwest. They intend to get around the problem of night-time electrical demand by storing hot water.
It would be helpful if Ausra would provide some technical information. In its absence, we'll have to make some assumptions. To avoid accusations of unfairness, we'll make all the assumptions favorable to the solar plant.
We don't know, for example, what the storage temperature will be. Higher temperatures allow more energy storage but require thicker container walls because of the higher pressure. We'll assume 600ºF because density starts dropping off quickly at higher temperatures, and pressure rises quickly. Even at that temperature the collector efficiency will be very low and the pressure would be about 1550 pounds per square inch. At that temperature, a pound of water holds 618 BTU, compared to 70 BTU at 100º.[6]
A modest thermal plant would be rated at 1000 MW of electricity. Imagine we wished to average that power over 16 hours, for a total of 16000 MWHe. Assuming a generous 45% efficiency, the energy required would be 16000 x 3,413,000 / 0.45 = 121 billion BTU of heat. Each pound of water can give up 618 - 70 = 548 BTU, so that means the storage has to be 221 million pounds. Density at 600º is about 42 lbs / cubic foot, so the volume required would be 5,262,000 cubic feet. Suppose the tanks were 20 feet in diameter, so their area would be 3.14 x 20 ft x 20 ft / 4 = 314 sq feet. To hold the required amount of water would require a total tank length of 16,750 feet, or 3.2 miles. By the way, if we say steel is good for 50,000 psi stress, the tank thickness would have to be 1550 / 50000 x 20 ft / 2 = 0.31 feet or 3.7 inches.[6]
So we need 3.2 miles of tank for 16 hours of storage. But what if the sky is cloudy for one day? Then we need 40 hours of storage, or 8 miles of tank. And that's for just one smallish thermal plant. In fact, there could be weeks of cloudy days, or days with only a few hours of sunshine, even in Arizona.
[1] http://www.eia.doe.gov/cneaf/electricity/epa/epat1p1.html
[2] http://www.nypa.gov/facilities/niagara.htm
[3] http://www.great-lakes.net/lakes/ref/eriefact.html
[4] http://www5.ncdc.noaa.gov/documentlibrary/pdf/wind1996.pdf
[5] http://www1.ceit.es/asignaturas/tecener1/Lesson6.pdf
[6] Baumeister and Marks, Standard Handbook for Mechanical Engineers